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		<title><![CDATA[Tetration Forum - All Forums]]></title>
		<link>http://math.eretrandre.org/tetrationforum/</link>
		<description><![CDATA[Tetration Forum - http://math.eretrandre.org/tetrationforum]]></description>
		<pubDate>Sun, 05 Feb 2012 22:16:34 +0000</pubDate>
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		<item>
			<title><![CDATA[Introduction]]></title>
			<link>http://math.eretrandre.org/tetrationforum/showthread.php?tid=722</link>
			<pubDate>Sat, 24 Dec 2011 02:55:09 +0000</pubDate>
			<guid isPermaLink="false">http://math.eretrandre.org/tetrationforum/showthread.php?tid=722</guid>
			<description><![CDATA[Hey everyone,<br />
<br />
My name's Lee, and I'm currently a senior in high school. I just finished Calculus III, and I'm planning to double major in mathematics and computer science in college next year.<br />
<br />
Needless to say, I have an undying love for math, and tetration really interests me. Obviously, much of this stuff is going to be way over my head until I learn a lot more, but I'm looking forward to learning and trying to figure it out!<br />
<br />
Hopefully I won't make too much of a fool of myself. See you guys around. <img src="images/smilies/blush.gif" style="vertical-align: middle;" border="0" alt="Blush" title="Blush" />]]></description>
			<content:encoded><![CDATA[Hey everyone,<br />
<br />
My name's Lee, and I'm currently a senior in high school. I just finished Calculus III, and I'm planning to double major in mathematics and computer science in college next year.<br />
<br />
Needless to say, I have an undying love for math, and tetration really interests me. Obviously, much of this stuff is going to be way over my head until I learn a lot more, but I'm looking forward to learning and trying to figure it out!<br />
<br />
Hopefully I won't make too much of a fool of myself. See you guys around. <img src="images/smilies/blush.gif" style="vertical-align: middle;" border="0" alt="Blush" title="Blush" />]]></content:encoded>
		</item>
		<item>
			<title><![CDATA[imho a core issue]]></title>
			<link>http://math.eretrandre.org/tetrationforum/showthread.php?tid=721</link>
			<pubDate>Fri, 16 Dec 2011 19:51:07 +0000</pubDate>
			<guid isPermaLink="false">http://math.eretrandre.org/tetrationforum/showthread.php?tid=721</guid>
			<description><![CDATA[one of the most important things imho is the sequence of derivatives of half-iterates of exp(x).<br />
<br />
i have been thinking about it for a long time and its about time i ask about this.<br />
<br />
there are some partial results but in general the question is quite open and tetration seems to be almost " immune  " to standard calculus " tricks ".<br />
<br />
in my imagination i always conjecture<br />
<br />
tommysexp(tommyslog(x)+1/2) (around x=0) = a0^2 + a1^2 x + a2^2 x^2 + ...<br />
<br />
where the squares indicate POSITIVE and the radius is assumed to be at least 1/3.<br />
<br />
although the POSITIVE part seems unlikely , i am considering it.<br />
<br />
now keep in mind , tetration is tricky and so is calculus.<br />
<br />
for instance if someone writes a program it might make roundoff errors.<br />
<br />
if someone uses a slighty different method to compute the derivates or to define the half-iterate that might give different results.<br />
<br />
and one of the simple reasons is that asymptotics might have very different derivatives.<br />
<br />
or might not be differentiable.<br />
<br />
or might have a different radius.<br />
<br />
or even more weird , might have fractals nth derivatives that finally cause the iterations to converge to the same function , be differentiable everywhere BUT the derivate of the function is not equal the the limit of derivatives.<br />
<br />
for instance lim n-&gt;00  sin(n^2 * x)/n^2 gives f(x) = 0 but alot of nonzero derivatives.<br />
<br />
i believe such things might occur in the research of tetration as well though i cannot prove it.<br />
<br />
i remember having puzzling results looking like paradoxes and illusionary fractals when doing research.<br />
<br />
for instance ln^[2](2sinh^[a](exp^[2](x))) seems very similar to my 2sinh method and numerically close , but seems to give very different results in many ways ...<br />
<br />
im very skeptical of computer computations and computer graphics for the above mentioned reasons and others.<br />
<br />
however i dont find much theory , references or talk about this.<br />
<br />
is it obviously wrong to assume positivity ?<br />
<br />
is it numerically easier to compute than slog or sexp ?<br />
<br />
can we conclude something about the signs of half-iterates from the super or invsuper signs without simply composing those 2 ?<br />
<br />
i tried many calculus tricks such as cauchy intermediate and also matrix methods and i dont know what else i can do ...<br />
<br />
if someone would be willing to give the derivatives of a method mentioned here on the forum , in particular my own , that would be appreciated.<br />
<br />
but as mentioned above , plz use the exact same method because it might deviate.<br />
<br />
for instance sheldon has many algoritms including one too compute " a tommysexp " but it is slightly different from the way i compute it ...<br />
<br />
i dont trust my own results at the moment ...<br />
<br />
i wonder when it is allowed to use differences instead of derivatives for the series expansion since the half-iterate of exp(x) grows slower than exp(x) ...<br />
<br />
in particular i experimented with newtons forward difference formula , q-difference , q-derivatives and other " q-stuff "<br />
<br />
but although seemingly promising at first , finally without results.<br />
<br />
one of the main issues - or so it appeared - is well demonstrated by the following :<br />
<br />
ln( a^2 + b^2 * exp(x) )<br />
<br />
for various a and b we might get quite different taylor series with even negative terms ( unexpected ? ) at different ( unexpected ? ) n'th derivatives.<br />
<br />
i am aware of formula's for the n'th derivative of exp(f(x)) or ln(f(x)) and similar but against all odds it didnt seem to help me much ???<br />
<br />
it seems this example is similar to the problems i came across.<br />
<br />
wonder if im alone with that and what others did.<br />
<br />
it would be nice to see my coefficients of my own method <img src="images/smilies/smile.gif" style="vertical-align: middle;" border="0" alt="Smile" title="Smile" /><br />
<br />
i know the coefficients of (non-linear)  2*sinh^[sin^2(z)](x) change sign oo often and so do the signs of any slog.<br />
<br />
thanks in advance<br />
<br />
tommy1729]]></description>
			<content:encoded><![CDATA[one of the most important things imho is the sequence of derivatives of half-iterates of exp(x).<br />
<br />
i have been thinking about it for a long time and its about time i ask about this.<br />
<br />
there are some partial results but in general the question is quite open and tetration seems to be almost " immune  " to standard calculus " tricks ".<br />
<br />
in my imagination i always conjecture<br />
<br />
tommysexp(tommyslog(x)+1/2) (around x=0) = a0^2 + a1^2 x + a2^2 x^2 + ...<br />
<br />
where the squares indicate POSITIVE and the radius is assumed to be at least 1/3.<br />
<br />
although the POSITIVE part seems unlikely , i am considering it.<br />
<br />
now keep in mind , tetration is tricky and so is calculus.<br />
<br />
for instance if someone writes a program it might make roundoff errors.<br />
<br />
if someone uses a slighty different method to compute the derivates or to define the half-iterate that might give different results.<br />
<br />
and one of the simple reasons is that asymptotics might have very different derivatives.<br />
<br />
or might not be differentiable.<br />
<br />
or might have a different radius.<br />
<br />
or even more weird , might have fractals nth derivatives that finally cause the iterations to converge to the same function , be differentiable everywhere BUT the derivate of the function is not equal the the limit of derivatives.<br />
<br />
for instance lim n-&gt;00  sin(n^2 * x)/n^2 gives f(x) = 0 but alot of nonzero derivatives.<br />
<br />
i believe such things might occur in the research of tetration as well though i cannot prove it.<br />
<br />
i remember having puzzling results looking like paradoxes and illusionary fractals when doing research.<br />
<br />
for instance ln^[2](2sinh^[a](exp^[2](x))) seems very similar to my 2sinh method and numerically close , but seems to give very different results in many ways ...<br />
<br />
im very skeptical of computer computations and computer graphics for the above mentioned reasons and others.<br />
<br />
however i dont find much theory , references or talk about this.<br />
<br />
is it obviously wrong to assume positivity ?<br />
<br />
is it numerically easier to compute than slog or sexp ?<br />
<br />
can we conclude something about the signs of half-iterates from the super or invsuper signs without simply composing those 2 ?<br />
<br />
i tried many calculus tricks such as cauchy intermediate and also matrix methods and i dont know what else i can do ...<br />
<br />
if someone would be willing to give the derivatives of a method mentioned here on the forum , in particular my own , that would be appreciated.<br />
<br />
but as mentioned above , plz use the exact same method because it might deviate.<br />
<br />
for instance sheldon has many algoritms including one too compute " a tommysexp " but it is slightly different from the way i compute it ...<br />
<br />
i dont trust my own results at the moment ...<br />
<br />
i wonder when it is allowed to use differences instead of derivatives for the series expansion since the half-iterate of exp(x) grows slower than exp(x) ...<br />
<br />
in particular i experimented with newtons forward difference formula , q-difference , q-derivatives and other " q-stuff "<br />
<br />
but although seemingly promising at first , finally without results.<br />
<br />
one of the main issues - or so it appeared - is well demonstrated by the following :<br />
<br />
ln( a^2 + b^2 * exp(x) )<br />
<br />
for various a and b we might get quite different taylor series with even negative terms ( unexpected ? ) at different ( unexpected ? ) n'th derivatives.<br />
<br />
i am aware of formula's for the n'th derivative of exp(f(x)) or ln(f(x)) and similar but against all odds it didnt seem to help me much ???<br />
<br />
it seems this example is similar to the problems i came across.<br />
<br />
wonder if im alone with that and what others did.<br />
<br />
it would be nice to see my coefficients of my own method <img src="images/smilies/smile.gif" style="vertical-align: middle;" border="0" alt="Smile" title="Smile" /><br />
<br />
i know the coefficients of (non-linear)  2*sinh^[sin^2(z)](x) change sign oo often and so do the signs of any slog.<br />
<br />
thanks in advance<br />
<br />
tommy1729]]></content:encoded>
		</item>
		<item>
			<title><![CDATA[Integer tetration and convergence speed rules]]></title>
			<link>http://math.eretrandre.org/tetrationforum/showthread.php?tid=720</link>
			<pubDate>Sun, 11 Dec 2011 23:50:18 +0000</pubDate>
			<guid isPermaLink="false">http://math.eretrandre.org/tetrationforum/showthread.php?tid=720</guid>
			<description><![CDATA[Hi all,<br />
This is my first post on this wonderful place and I have to confess I'm feeling like a kid in the playground. <br />
<br />
I've just published a quite simple book (in Italian) about integer tetration. It's mainly focused on its ending digits (perfect p-adic convergence, regular displacements of the most important constrained digits, and so on), plus some related topics.<br />
It's possible to read the first (introductive) chapter online: <a href="http://www.uni-service.it/images/stories/product_uploaded_file/Coda%20Preview.pdf" target="_blank">http://www.uni-service.it/images/stories...review.pdf</a><br />
<br />
The required skill level is quite low (high school proficiency), but I hope it could be a nice amusement for recreational math lovers.<br />
<br />
All the best,<br />
Marco]]></description>
			<content:encoded><![CDATA[Hi all,<br />
This is my first post on this wonderful place and I have to confess I'm feeling like a kid in the playground. <br />
<br />
I've just published a quite simple book (in Italian) about integer tetration. It's mainly focused on its ending digits (perfect p-adic convergence, regular displacements of the most important constrained digits, and so on), plus some related topics.<br />
It's possible to read the first (introductive) chapter online: <a href="http://www.uni-service.it/images/stories/product_uploaded_file/Coda%20Preview.pdf" target="_blank">http://www.uni-service.it/images/stories...review.pdf</a><br />
<br />
The required skill level is quite low (high school proficiency), but I hope it could be a nice amusement for recreational math lovers.<br />
<br />
All the best,<br />
Marco]]></content:encoded>
		</item>
		<item>
			<title><![CDATA[fractional iteration by schröder and by binomial-formula]]></title>
			<link>http://math.eretrandre.org/tetrationforum/showthread.php?tid=719</link>
			<pubDate>Wed, 23 Nov 2011 15:45:00 +0000</pubDate>
			<guid isPermaLink="false">http://math.eretrandre.org/tetrationforum/showthread.php?tid=719</guid>
			<description><![CDATA[Just a small result from some playing around in a break; I'll look at the confirmation for a wider range of parameters later.<br />
<br />
This days I'm re-considering the differences of the fractional iterates of <img src="/cgi-bin/mimetex.cgi? b^x -1 " align="middle" /> using the schröder-function by their carleman-matrices. At the moment I use b=2, comparing the results when I use the power series centered around fixpoint t0=0 versus that around fixpoint t1=1. The range of 0&lt;x0&lt;1 can be analyzed well using both power series and we get the known effect, that the iterates disagree at fractional heights (but not at integer heights), and that the differences form a sinusoidal curve.   <br />
<br />
If we compare the according fractional iterates which we get using the Newton-formula (the binomial composition of integer-height-iterates only) which does not use a fixpoint, then it seems, that the results approximate that of the powers series around zero. I think this means for more general bases (other than 2): around the attracting fixpoint, which is zero only if log(base)&lt;1 but is the negative real fixpoint if log(base)&gt;1.<br />
<br />
The point x0, at which the most symmetric sinusoidal curve occurs seems to be x0~0.382160520000... (not rational!) which has a special property for base b=2 which I'll discuss in a later post.<br />
<br />
Gottfried]]></description>
			<content:encoded><![CDATA[Just a small result from some playing around in a break; I'll look at the confirmation for a wider range of parameters later.<br />
<br />
This days I'm re-considering the differences of the fractional iterates of <img src="/cgi-bin/mimetex.cgi? b^x -1 " align="middle" /> using the schröder-function by their carleman-matrices. At the moment I use b=2, comparing the results when I use the power series centered around fixpoint t0=0 versus that around fixpoint t1=1. The range of 0&lt;x0&lt;1 can be analyzed well using both power series and we get the known effect, that the iterates disagree at fractional heights (but not at integer heights), and that the differences form a sinusoidal curve.   <br />
<br />
If we compare the according fractional iterates which we get using the Newton-formula (the binomial composition of integer-height-iterates only) which does not use a fixpoint, then it seems, that the results approximate that of the powers series around zero. I think this means for more general bases (other than 2): around the attracting fixpoint, which is zero only if log(base)&lt;1 but is the negative real fixpoint if log(base)&gt;1.<br />
<br />
The point x0, at which the most symmetric sinusoidal curve occurs seems to be x0~0.382160520000... (not rational!) which has a special property for base b=2 which I'll discuss in a later post.<br />
<br />
Gottfried]]></content:encoded>
		</item>
		<item>
			<title><![CDATA[using cosh(x) ?]]></title>
			<link>http://math.eretrandre.org/tetrationforum/showthread.php?tid=718</link>
			<pubDate>Sat, 19 Nov 2011 22:56:21 +0000</pubDate>
			<guid isPermaLink="false">http://math.eretrandre.org/tetrationforum/showthread.php?tid=718</guid>
			<description><![CDATA[no , im not kidding , apart from my sinh method im considering a cosh method.<br />
<br />
although it might be more complicated or speculative.<br />
<br />
it might be insightfull to plot things.<br />
<br />
consider<br />
<br />
exp(1/e)&lt;b&lt;sqrt(e)<br />
<br />
f(x) - f^[-1](x) = 2*cosh(ln(b)*x)<br />
<br />
for variable x and fixed b ( base ).<br />
<br />
i bet you can find more functional equations from this one.<br />
<br />
for instance to express f^[-1](x) in terms of f(x) and elementary functions.<br />
<br />
the solution is unique for R-&gt; R.<br />
<br />
replace x by f(x) and let g(x) be the super of f(x).<br />
<br />
then we get a difference equation for g(x).<br />
<br />
we solve that difference equation , now we go from g(x) to f(x) or we find the taylor for f(x) from the equation ( or both , in programming ).<br />
<br />
similar to the sinh method : since f is close to 2*cosh(ln(b)*x)) and 2*cosh(ln(b)*x) is close to b^x , we have a new method.]]></description>
			<content:encoded><![CDATA[no , im not kidding , apart from my sinh method im considering a cosh method.<br />
<br />
although it might be more complicated or speculative.<br />
<br />
it might be insightfull to plot things.<br />
<br />
consider<br />
<br />
exp(1/e)&lt;b&lt;sqrt(e)<br />
<br />
f(x) - f^[-1](x) = 2*cosh(ln(b)*x)<br />
<br />
for variable x and fixed b ( base ).<br />
<br />
i bet you can find more functional equations from this one.<br />
<br />
for instance to express f^[-1](x) in terms of f(x) and elementary functions.<br />
<br />
the solution is unique for R-&gt; R.<br />
<br />
replace x by f(x) and let g(x) be the super of f(x).<br />
<br />
then we get a difference equation for g(x).<br />
<br />
we solve that difference equation , now we go from g(x) to f(x) or we find the taylor for f(x) from the equation ( or both , in programming ).<br />
<br />
similar to the sinh method : since f is close to 2*cosh(ln(b)*x)) and 2*cosh(ln(b)*x) is close to b^x , we have a new method.]]></content:encoded>
		</item>
		<item>
			<title><![CDATA[Nonation Nopt structure Pictures 3 & 4]]></title>
			<link>http://math.eretrandre.org/tetrationforum/showthread.php?tid=717</link>
			<pubDate>Fri, 18 Nov 2011 00:23:09 +0000</pubDate>
			<guid isPermaLink="false">http://math.eretrandre.org/tetrationforum/showthread.php?tid=717</guid>
			<description><![CDATA[Nonation Nopt structure<br />
Pictures 3 &amp; 4<br />
Picture 3<br />
By careful inspection argument, one can order ellipsis lengths by increasing order of magnitude.  Although it is very difficult to quantify “how much” longer one ellipsis is compared to another, the magnitudes increase very quickly as is the nature of fast-growing functions.  The idea of visualising the ellipsis lengths in increasing magnitude order has been color-coded in Picture 3.<br />
Picture 4<br />
Finally, Picture 4 shows the induced Multi-layered Ellipsis Structure from a NOPT structure and is color-coded in the same way as in Picture 3.  The basic structural pattern of the well-known H-Fractal can be seen somewhat clearly.<br />
<br />

<br /><img src="images/attachtypes/image.gif" border="0" alt=".jpg" />&nbsp;&nbsp;<a href="attachment.php?aid=920" target="_blank">p3_seedncolor.jpg</a> (Size: 80.04 KB / Downloads: 45)
<br />

<br /><img src="images/attachtypes/image.gif" border="0" alt=".jpg" />&nbsp;&nbsp;<a href="attachment.php?aid=921" target="_blank">p4_Hfrac.jpg</a> (Size: 87.39 KB / Downloads: 45)
]]></description>
			<content:encoded><![CDATA[Nonation Nopt structure<br />
Pictures 3 &amp; 4<br />
Picture 3<br />
By careful inspection argument, one can order ellipsis lengths by increasing order of magnitude.  Although it is very difficult to quantify “how much” longer one ellipsis is compared to another, the magnitudes increase very quickly as is the nature of fast-growing functions.  The idea of visualising the ellipsis lengths in increasing magnitude order has been color-coded in Picture 3.<br />
Picture 4<br />
Finally, Picture 4 shows the induced Multi-layered Ellipsis Structure from a NOPT structure and is color-coded in the same way as in Picture 3.  The basic structural pattern of the well-known H-Fractal can be seen somewhat clearly.<br />
<br />

<br /><img src="images/attachtypes/image.gif" border="0" alt=".jpg" />&nbsp;&nbsp;<a href="attachment.php?aid=920" target="_blank">p3_seedncolor.jpg</a> (Size: 80.04 KB / Downloads: 45)
<br />

<br /><img src="images/attachtypes/image.gif" border="0" alt=".jpg" />&nbsp;&nbsp;<a href="attachment.php?aid=921" target="_blank">p4_Hfrac.jpg</a> (Size: 87.39 KB / Downloads: 45)
]]></content:encoded>
		</item>
		<item>
			<title><![CDATA[Nonation Nopt structure Pictures 2]]></title>
			<link>http://math.eretrandre.org/tetrationforum/showthread.php?tid=716</link>
			<pubDate>Fri, 18 Nov 2011 00:13:06 +0000</pubDate>
			<guid isPermaLink="false">http://math.eretrandre.org/tetrationforum/showthread.php?tid=716</guid>
			<description><![CDATA[Nonation Nopt structure Pictures 2<br />
with seedvalue of the form "n"<br />
<br />

<br /><img src="images/attachtypes/image.gif" border="0" alt=".jpg" />&nbsp;&nbsp;<a href="attachment.php?aid=918" target="_blank">p2_seedn.jpg</a> (Size: 73.57 KB / Downloads: 30)
<br />

<br /><img src="images/attachtypes/image.gif" border="0" alt=".jpg" />&nbsp;&nbsp;<a href="attachment.php?aid=919" target="_blank">p2_seedn2.jpg</a> (Size: 97.74 KB / Downloads: 31)
]]></description>
			<content:encoded><![CDATA[Nonation Nopt structure Pictures 2<br />
with seedvalue of the form "n"<br />
<br />

<br /><img src="images/attachtypes/image.gif" border="0" alt=".jpg" />&nbsp;&nbsp;<a href="attachment.php?aid=918" target="_blank">p2_seedn.jpg</a> (Size: 73.57 KB / Downloads: 30)
<br />

<br /><img src="images/attachtypes/image.gif" border="0" alt=".jpg" />&nbsp;&nbsp;<a href="attachment.php?aid=919" target="_blank">p2_seedn2.jpg</a> (Size: 97.74 KB / Downloads: 31)
]]></content:encoded>
		</item>
		<item>
			<title><![CDATA[Nonation Nopt structure Pictures 1]]></title>
			<link>http://math.eretrandre.org/tetrationforum/showthread.php?tid=715</link>
			<pubDate>Fri, 18 Nov 2011 00:07:51 +0000</pubDate>
			<guid isPermaLink="false">http://math.eretrandre.org/tetrationforum/showthread.php?tid=715</guid>
			<description><![CDATA[Nonation Nopt structure Pictures 1<br />
with seedvalue of the form "theta ) n"<br />
<br />

<br /><img src="images/attachtypes/image.gif" border="0" alt=".jpg" />&nbsp;&nbsp;<a href="attachment.php?aid=916" target="_blank">p1_thetan.jpg</a> (Size: 80.54 KB / Downloads: 41)
<br />

<br /><img src="images/attachtypes/image.gif" border="0" alt=".jpg" />&nbsp;&nbsp;<a href="attachment.php?aid=917" target="_blank">p1_thetan2.jpg</a> (Size: 92.24 KB / Downloads: 40)
]]></description>
			<content:encoded><![CDATA[Nonation Nopt structure Pictures 1<br />
with seedvalue of the form "theta ) n"<br />
<br />

<br /><img src="images/attachtypes/image.gif" border="0" alt=".jpg" />&nbsp;&nbsp;<a href="attachment.php?aid=916" target="_blank">p1_thetan.jpg</a> (Size: 80.54 KB / Downloads: 41)
<br />

<br /><img src="images/attachtypes/image.gif" border="0" alt=".jpg" />&nbsp;&nbsp;<a href="attachment.php?aid=917" target="_blank">p1_thetan2.jpg</a> (Size: 92.24 KB / Downloads: 40)
]]></content:encoded>
		</item>
		<item>
			<title><![CDATA[Nept and Nopt structures (Part 3).]]></title>
			<link>http://math.eretrandre.org/tetrationforum/showthread.php?tid=714</link>
			<pubDate>Thu, 17 Nov 2011 23:51:36 +0000</pubDate>
			<guid isPermaLink="false">http://math.eretrandre.org/tetrationforum/showthread.php?tid=714</guid>
			<description><![CDATA[Nept and Nopt structures (Part 3).<br />
HEFTY NOPT Structures<br />
H-(Ellipsis)-Fractal Type Nested Operational Power Towers.<br />
Transitional numbers between finite and infinite realm.<br />
Keywords: hyperoperations, seed values (starting and controlling), linear nopt structures, gi-sequence from Graham’s number, g-subscript towers.<br />
Summary: NOPT structures give a standard approach to looking at the bizarre world of numbers that straddle the finite and infinite divide. The induced Multi-layered Ellipsis Structure from a NOPT structure has similar component-connection structure to the well-known H-Fractal, given a corresponding degree of resolution.<br />
Moreover, The H-fractal structure is emergent from the NOPT structure and guides the multi-layered nestedly embedded computational pathways. In the representations I give (see the pictures below) the computation starts at the bottom right corner and moves towards the top left corner.<br />
Caveat: This mathematical material is very abstract and not practically computable on any computer, if accurate magnitude is a desired goal, because it’s in the area of hyperoperations, the boundary of practical maths.<br />
When the operation is not specified, NOPT structures are floaty and esoteric in nature, but nevertheless they still say something about the patterns arising from unlimited multilayered nested recursion, with SeedValue being the standard stage level phase transition marker.<br />
A comparison can be made with Hereditary Base n, where the base value is used at other exponent levels apart from the level of standard positional notation, resulting in a treelike number structure.  Perhaps, another comparison can be made with CantorNormalForm for infinite ordinals.<br />
In the pictures below:<br />
The NOPT structure we consider below and is shown in the 4 pictures below is the nonation NOPT structure, remembering that OrderType=9 is derived from considering the corresponding NEPT structure arising from the nonation hyperoperation.<br />
All of the “formal power towers”  in the NOPT structure are given the symbol “theta” (Note: by “formal” expression I mean an unevaluated symbolic expression that could be evaluated given more information). <br />
When the SeedValue is supplied, in theory, the power tower can be evaluated because the height information about the power tower is given, then each succeeding power tower in the linear Nopt structure can be evaluated with each power tower evaluation supplying the height of the next power tower to be evaluated.<br />
In the first picture the SeedValues have the form “theta ) n”<br />
In the second picture the SeedValues have the form “n”.<br />
Sometimes, we may want to ensure that the next linear component in the NOPT structure has nontrivial ellipsis value and so we use the first style of NOPT structure.<br />
However, I think that for a standard definition of NOPT structure it is better to use the second style of NOPT structure where SeedValue=n.<br />
By some kind of careful inspection argument, one can notice ellipsis values that are the same magnitude within a NOPT structure, for example where the ellipsis is equal to SeedValue (a controlling seed value, not a starting seed value).<br />
(Remember, starting seed values inform the height of the adjacent power tower, theta, in the linear NOPT component. Controlling seed values give the length of an ellipsis within a component of the NOPT structure.)<br />
Also, by careful inspection argument, one can order ellipsis lengths by increasing order of magnitude.  Although it is very difficult to quantify “how much” longer one ellipsis is compared to another, the magnitudes increase very quickly as is the nature of fast-growing functions.  The idea of visualising the ellipsis lengths in increasing magnitude order has been color-coded in Picture 3.<br />
Finally, Picture 4 shows the induced Multi-layered Ellipsis Structure from a NOPT structure and is color-coded in the same way as in Picture 3.  The basic structural pattern of the well-known H-Fractal can be seen somewhat clearly.]]></description>
			<content:encoded><![CDATA[Nept and Nopt structures (Part 3).<br />
HEFTY NOPT Structures<br />
H-(Ellipsis)-Fractal Type Nested Operational Power Towers.<br />
Transitional numbers between finite and infinite realm.<br />
Keywords: hyperoperations, seed values (starting and controlling), linear nopt structures, gi-sequence from Graham’s number, g-subscript towers.<br />
Summary: NOPT structures give a standard approach to looking at the bizarre world of numbers that straddle the finite and infinite divide. The induced Multi-layered Ellipsis Structure from a NOPT structure has similar component-connection structure to the well-known H-Fractal, given a corresponding degree of resolution.<br />
Moreover, The H-fractal structure is emergent from the NOPT structure and guides the multi-layered nestedly embedded computational pathways. In the representations I give (see the pictures below) the computation starts at the bottom right corner and moves towards the top left corner.<br />
Caveat: This mathematical material is very abstract and not practically computable on any computer, if accurate magnitude is a desired goal, because it’s in the area of hyperoperations, the boundary of practical maths.<br />
When the operation is not specified, NOPT structures are floaty and esoteric in nature, but nevertheless they still say something about the patterns arising from unlimited multilayered nested recursion, with SeedValue being the standard stage level phase transition marker.<br />
A comparison can be made with Hereditary Base n, where the base value is used at other exponent levels apart from the level of standard positional notation, resulting in a treelike number structure.  Perhaps, another comparison can be made with CantorNormalForm for infinite ordinals.<br />
In the pictures below:<br />
The NOPT structure we consider below and is shown in the 4 pictures below is the nonation NOPT structure, remembering that OrderType=9 is derived from considering the corresponding NEPT structure arising from the nonation hyperoperation.<br />
All of the “formal power towers”  in the NOPT structure are given the symbol “theta” (Note: by “formal” expression I mean an unevaluated symbolic expression that could be evaluated given more information). <br />
When the SeedValue is supplied, in theory, the power tower can be evaluated because the height information about the power tower is given, then each succeeding power tower in the linear Nopt structure can be evaluated with each power tower evaluation supplying the height of the next power tower to be evaluated.<br />
In the first picture the SeedValues have the form “theta ) n”<br />
In the second picture the SeedValues have the form “n”.<br />
Sometimes, we may want to ensure that the next linear component in the NOPT structure has nontrivial ellipsis value and so we use the first style of NOPT structure.<br />
However, I think that for a standard definition of NOPT structure it is better to use the second style of NOPT structure where SeedValue=n.<br />
By some kind of careful inspection argument, one can notice ellipsis values that are the same magnitude within a NOPT structure, for example where the ellipsis is equal to SeedValue (a controlling seed value, not a starting seed value).<br />
(Remember, starting seed values inform the height of the adjacent power tower, theta, in the linear NOPT component. Controlling seed values give the length of an ellipsis within a component of the NOPT structure.)<br />
Also, by careful inspection argument, one can order ellipsis lengths by increasing order of magnitude.  Although it is very difficult to quantify “how much” longer one ellipsis is compared to another, the magnitudes increase very quickly as is the nature of fast-growing functions.  The idea of visualising the ellipsis lengths in increasing magnitude order has been color-coded in Picture 3.<br />
Finally, Picture 4 shows the induced Multi-layered Ellipsis Structure from a NOPT structure and is color-coded in the same way as in Picture 3.  The basic structural pattern of the well-known H-Fractal can be seen somewhat clearly.]]></content:encoded>
		</item>
		<item>
			<title><![CDATA[Nept and Nopt structures (Part 2).]]></title>
			<link>http://math.eretrandre.org/tetrationforum/showthread.php?tid=713</link>
			<pubDate>Thu, 17 Nov 2011 23:24:39 +0000</pubDate>
			<guid isPermaLink="false">http://math.eretrandre.org/tetrationforum/showthread.php?tid=713</guid>
			<description><![CDATA[Nept and Nopt structures (Part 2).<br />
It can be noticed that the induced SeedValues from a NOPT Structure have a similar role to the omega (w) limit which is used as a “default way” of describing the length of an infinite list. I think that omega is a shorthand for saying “we don’t know how long an unending list of items goes on for, but we’ll call it omega”.  This seems more sensible than saying “yes we do know, the list goes on for as many natural numbers there are”. This is because, while the process of generating naturals (into Standard Positional Notation) is well-defined and clear, the extent of the natural numbers depends on how creative we get by combining powerful operations with NOPT structures. In other words, the set of Natural Numbers is well-defined at the level of (+1)-iterative description, but not at the level of unbounded descriptive capabilities.<br />
As the examples shown above demonstrate, there are plenty of intermediary ellipsis stages that large numbers need to pass through in order to reach an elusive target magnitude goal.<br />
The omega limit is used for limit ordinals in theory of infinite ordinals, but it is independent of the values of the symbols in the list.<br />
So lim&lt;w^1, w^2, …&gt; = w^w, and is understood as an w-limit.<br />
And lim&lt;w, w^w, w^w^w, …&gt; = e0, and is understood as an w-limit.<br />
And lim&lt;e0, e0^e0, e0^e0^e0, …&gt; = e1, and is understood as an w-limit.<br />
And lim&lt;e0, e1, e2, …&gt; = e_w, and is understood as an w-limit.<br />
In the last example, it doesn’t matter that all the component symbols are much larger than omega - we have “large machinery at our disposal” (eg e0) – but we always use the “small infinite value” of omega to decide the list length of any of these infinite sequences whenever a limit to the sequence is desired. Omega is like a convenient yardstick that is used in evaluating limit points for horizonal phenomena.<br />
Whenever a list structure emerges and a limit is desired then the idea that an infinite list of symbols can be indexed or sequenced by the natural numbers is called upon.<br />
So in theory of infinite ordinals, omega is the convenient yardstick.<br />
In NOPT structures, the induced SeedValues have the “omega role” by deciding how long a stage should be lingered upon before the transitioning to other levels becomes necessary.]]></description>
			<content:encoded><![CDATA[Nept and Nopt structures (Part 2).<br />
It can be noticed that the induced SeedValues from a NOPT Structure have a similar role to the omega (w) limit which is used as a “default way” of describing the length of an infinite list. I think that omega is a shorthand for saying “we don’t know how long an unending list of items goes on for, but we’ll call it omega”.  This seems more sensible than saying “yes we do know, the list goes on for as many natural numbers there are”. This is because, while the process of generating naturals (into Standard Positional Notation) is well-defined and clear, the extent of the natural numbers depends on how creative we get by combining powerful operations with NOPT structures. In other words, the set of Natural Numbers is well-defined at the level of (+1)-iterative description, but not at the level of unbounded descriptive capabilities.<br />
As the examples shown above demonstrate, there are plenty of intermediary ellipsis stages that large numbers need to pass through in order to reach an elusive target magnitude goal.<br />
The omega limit is used for limit ordinals in theory of infinite ordinals, but it is independent of the values of the symbols in the list.<br />
So lim&lt;w^1, w^2, …&gt; = w^w, and is understood as an w-limit.<br />
And lim&lt;w, w^w, w^w^w, …&gt; = e0, and is understood as an w-limit.<br />
And lim&lt;e0, e0^e0, e0^e0^e0, …&gt; = e1, and is understood as an w-limit.<br />
And lim&lt;e0, e1, e2, …&gt; = e_w, and is understood as an w-limit.<br />
In the last example, it doesn’t matter that all the component symbols are much larger than omega - we have “large machinery at our disposal” (eg e0) – but we always use the “small infinite value” of omega to decide the list length of any of these infinite sequences whenever a limit to the sequence is desired. Omega is like a convenient yardstick that is used in evaluating limit points for horizonal phenomena.<br />
Whenever a list structure emerges and a limit is desired then the idea that an infinite list of symbols can be indexed or sequenced by the natural numbers is called upon.<br />
So in theory of infinite ordinals, omega is the convenient yardstick.<br />
In NOPT structures, the induced SeedValues have the “omega role” by deciding how long a stage should be lingered upon before the transitioning to other levels becomes necessary.]]></content:encoded>
		</item>
		<item>
			<title><![CDATA[Introduction to NEPT and NOPT structures]]></title>
			<link>http://math.eretrandre.org/tetrationforum/showthread.php?tid=712</link>
			<pubDate>Thu, 17 Nov 2011 23:19:50 +0000</pubDate>
			<guid isPermaLink="false">http://math.eretrandre.org/tetrationforum/showthread.php?tid=712</guid>
			<description><![CDATA[Introduction to NEPT and NOPT structures<br />
Username: MikeSmith<br />
Email: aliwww@mail.com<br />
<br />
Hi everybody at tetration forum. Thanks to Bo for all the administration help.<br />
I hope that there are some new ways of looking at familiar things, and some of the material below can be understood and is reasonably clear. <br />
Warning: The material is jam-packed with ideas, but quite difficult to check, so please take your time reading it, and if you’re reading this and new to hyperoperations you should try to work out 3^^3, 3^^^3, 3^^^^3 to get some intuition about the patterns I’ll be discussing below.  Also, if you understand Graham’s number construction that can help as well. <br />
Post or email any comments or queries as you wish.<br />
<br />
Nept and Nopt structures (Part 1).<br />
A NEPT structure is the representation of a large number as multi-layered Nested Exponentional Power Towers.<br />
A NOPT structure is a generalisation of multi-layered nested power towers. Usually we think of power towers where the operation is exponentiation, but the concept of nested power towers could be used with other operations from the hyperoperation hierarchy as well as any others where height of power tower is well-defined and each power tower in the expression produces a natural number and the operation is a strictly increasing function on the natural numbers.<br />
Other operations are possible for putting into nested power towers for example, such as (+, iterated addition) and (*, iterated multiplication) and (^^, tetration, iterated tetration)  and so on. Considering very large numbers, other applications of NOPT structures are possible by using powerful operations such as the gi-sequence from Graham’s number construction where g1=(3)^4(3) (3 hexated to 3) and g64=Graham’s number. Even more powerful operations can be considered such as g-subscript towers where each “subscript power tower” has a well-defined height (of nested g-subscript symbols, ending with g1).<br />
The canonical NOPT structure is the NEPT structure where the operation is exponentiation. The canonical relationship between NEPT and NOPT structures is that an OrderType of a NOPT structure can be well-defined by comparing with the NEPT structure of same order number. In other words, the OrderType of a NOPT structure borrows from the canonical NEPT structure related to some hyperoperation.<br />
So the relationship can be seen like this:<br />
Hyperoperation (n&gt;=4) &lt;-&gt; NEPT (n&gt;=4) &lt;-&gt; NOPT (n&gt;=4)<br />
For example, NOPT structures with OrderType 4 or 5 correspond to the appearance of tetration and pentation numbers when they are written into NEPT from (the form of Nested Exponential Power Towers).<br />
NOPT structures with OrderType 4 or 5 are (the only) Linear NOPT structures.<br />
Hexation NOPT structure has OrderType 6 and has 2-dimensional array structure.<br />
Hepation NOPT structure has OrderType 7 and has (2,1)-dimensional array structure.<br />
Octation NOPT structure has OrderType 8 and has (2,2)-dimensional array structure.<br />
Nonation NOPT structure has OrderType 9 and has (2,2,1)-dimensional array structure. By zigzagging from bottom right corner, in left-up-left-up-left-up-… fashion we can build up NOPT structures with higher dimensions. The computation pathway proceeds from bottom right corner in left-up-left-up-left-up-… fashion until the top left corner. <br />
Linear NOPT structures have Linear Ellipsis Structure.<br />
Non-linear NOPT structures (OrderType&gt;=6) have Multi-layered Ellipsis Structure.<br />
The Ellipsis corresponding to the final top-left corner component of the NOPT structure is where the final computation is achieved. Superficially, it looks like a Linear Nopt structure, but looking at it carefully, notice that the Ellipsis length (of linearly arranged nested power towers) is given by a small number in Tetration and Pentation NOPT structures, but in larger NOPT structures is given by a Multi-layered Ellipsis expression, that is the combined effort of the rest of the interlinked components of the NOPT structure.<br />
The induced Multi-layered Ellipsis Structure from a NOPT structure has a very similar structure to the well-known H-Fractal.<br />
<br />
A NOPT-6 structure (hexation) has a 2-dimensional array structure and the induced Multi-layered Ellipsis Structure corresponds with simple H-Fractal at the first level.<br />
A NOPT-7 structure (heptation) has a (2,1)-dimensional array structure and the induced Multi-layered Ellipsis Structure corresponds with the H-Fractal at the second level of resolution.<br />
A NOPT-8 structure (octation) has a (2,2)-dimensional array structure and the induced Multi-layered Ellipsis Structure corresponds with the H-Fractal at the third level of resolution.<br />
A NOPT-9 structure (nonation) has a (2,2,1)-dimensional array structure and the induced Multi-layered Ellipsis Structure corresponds with the H-Fractal at the fourth level of resolution.<br />
You can observe that in terms of written symbolic requirements, going from NOPT-(I) to NOPT-(I+1) structure requires twice as many symbol expressions, due to the symbol folding that proceeds zigzagging L-U-L-U-L-U-L-U from bottom right corner to top right corner.<br />
NOPT structures require a fixed SEEDVALUE, that has the roles of (1) either initiating a Linear NOPT Component OR (2) Controlling Ellipsis Length of a Component.]]></description>
			<content:encoded><![CDATA[Introduction to NEPT and NOPT structures<br />
Username: MikeSmith<br />
Email: aliwww@mail.com<br />
<br />
Hi everybody at tetration forum. Thanks to Bo for all the administration help.<br />
I hope that there are some new ways of looking at familiar things, and some of the material below can be understood and is reasonably clear. <br />
Warning: The material is jam-packed with ideas, but quite difficult to check, so please take your time reading it, and if you’re reading this and new to hyperoperations you should try to work out 3^^3, 3^^^3, 3^^^^3 to get some intuition about the patterns I’ll be discussing below.  Also, if you understand Graham’s number construction that can help as well. <br />
Post or email any comments or queries as you wish.<br />
<br />
Nept and Nopt structures (Part 1).<br />
A NEPT structure is the representation of a large number as multi-layered Nested Exponentional Power Towers.<br />
A NOPT structure is a generalisation of multi-layered nested power towers. Usually we think of power towers where the operation is exponentiation, but the concept of nested power towers could be used with other operations from the hyperoperation hierarchy as well as any others where height of power tower is well-defined and each power tower in the expression produces a natural number and the operation is a strictly increasing function on the natural numbers.<br />
Other operations are possible for putting into nested power towers for example, such as (+, iterated addition) and (*, iterated multiplication) and (^^, tetration, iterated tetration)  and so on. Considering very large numbers, other applications of NOPT structures are possible by using powerful operations such as the gi-sequence from Graham’s number construction where g1=(3)^4(3) (3 hexated to 3) and g64=Graham’s number. Even more powerful operations can be considered such as g-subscript towers where each “subscript power tower” has a well-defined height (of nested g-subscript symbols, ending with g1).<br />
The canonical NOPT structure is the NEPT structure where the operation is exponentiation. The canonical relationship between NEPT and NOPT structures is that an OrderType of a NOPT structure can be well-defined by comparing with the NEPT structure of same order number. In other words, the OrderType of a NOPT structure borrows from the canonical NEPT structure related to some hyperoperation.<br />
So the relationship can be seen like this:<br />
Hyperoperation (n&gt;=4) &lt;-&gt; NEPT (n&gt;=4) &lt;-&gt; NOPT (n&gt;=4)<br />
For example, NOPT structures with OrderType 4 or 5 correspond to the appearance of tetration and pentation numbers when they are written into NEPT from (the form of Nested Exponential Power Towers).<br />
NOPT structures with OrderType 4 or 5 are (the only) Linear NOPT structures.<br />
Hexation NOPT structure has OrderType 6 and has 2-dimensional array structure.<br />
Hepation NOPT structure has OrderType 7 and has (2,1)-dimensional array structure.<br />
Octation NOPT structure has OrderType 8 and has (2,2)-dimensional array structure.<br />
Nonation NOPT structure has OrderType 9 and has (2,2,1)-dimensional array structure. By zigzagging from bottom right corner, in left-up-left-up-left-up-… fashion we can build up NOPT structures with higher dimensions. The computation pathway proceeds from bottom right corner in left-up-left-up-left-up-… fashion until the top left corner. <br />
Linear NOPT structures have Linear Ellipsis Structure.<br />
Non-linear NOPT structures (OrderType&gt;=6) have Multi-layered Ellipsis Structure.<br />
The Ellipsis corresponding to the final top-left corner component of the NOPT structure is where the final computation is achieved. Superficially, it looks like a Linear Nopt structure, but looking at it carefully, notice that the Ellipsis length (of linearly arranged nested power towers) is given by a small number in Tetration and Pentation NOPT structures, but in larger NOPT structures is given by a Multi-layered Ellipsis expression, that is the combined effort of the rest of the interlinked components of the NOPT structure.<br />
The induced Multi-layered Ellipsis Structure from a NOPT structure has a very similar structure to the well-known H-Fractal.<br />
<br />
A NOPT-6 structure (hexation) has a 2-dimensional array structure and the induced Multi-layered Ellipsis Structure corresponds with simple H-Fractal at the first level.<br />
A NOPT-7 structure (heptation) has a (2,1)-dimensional array structure and the induced Multi-layered Ellipsis Structure corresponds with the H-Fractal at the second level of resolution.<br />
A NOPT-8 structure (octation) has a (2,2)-dimensional array structure and the induced Multi-layered Ellipsis Structure corresponds with the H-Fractal at the third level of resolution.<br />
A NOPT-9 structure (nonation) has a (2,2,1)-dimensional array structure and the induced Multi-layered Ellipsis Structure corresponds with the H-Fractal at the fourth level of resolution.<br />
You can observe that in terms of written symbolic requirements, going from NOPT-(I) to NOPT-(I+1) structure requires twice as many symbol expressions, due to the symbol folding that proceeds zigzagging L-U-L-U-L-U-L-U from bottom right corner to top right corner.<br />
NOPT structures require a fixed SEEDVALUE, that has the roles of (1) either initiating a Linear NOPT Component OR (2) Controlling Ellipsis Length of a Component.]]></content:encoded>
		</item>
		<item>
			<title><![CDATA[e^f(ln(ln(f(exp(x))))) = f(x)  f ''(x) =/= 0]]></title>
			<link>http://math.eretrandre.org/tetrationforum/showthread.php?tid=711</link>
			<pubDate>Wed, 16 Nov 2011 20:18:12 +0000</pubDate>
			<guid isPermaLink="false">http://math.eretrandre.org/tetrationforum/showthread.php?tid=711</guid>
			<description><![CDATA[hi all<br />
<br />
if i didnt ask or post this before i do now :<br />
<br />
e^f(ln(ln(f(exp(x))))) = f(x) and  f ''(x) =/= 0<br />
<br />
if it exists ...<br />
<br />
btw does anyone know where i can find an easy online ( free ) applet for working with carleman matrices ??<br />
<br />
maple doesnt seem to know it , right ?<br />
<br />
its quite timeconsuming to program it or fill large matrices ( ok im not so good at it yet <img src="images/smilies/tongue.gif" style="vertical-align: middle;" border="0" alt="Tongue" title="Tongue" /> )]]></description>
			<content:encoded><![CDATA[hi all<br />
<br />
if i didnt ask or post this before i do now :<br />
<br />
e^f(ln(ln(f(exp(x))))) = f(x) and  f ''(x) =/= 0<br />
<br />
if it exists ...<br />
<br />
btw does anyone know where i can find an easy online ( free ) applet for working with carleman matrices ??<br />
<br />
maple doesnt seem to know it , right ?<br />
<br />
its quite timeconsuming to program it or fill large matrices ( ok im not so good at it yet <img src="images/smilies/tongue.gif" style="vertical-align: middle;" border="0" alt="Tongue" title="Tongue" /> )]]></content:encoded>
		</item>
		<item>
			<title><![CDATA[rumor result]]></title>
			<link>http://math.eretrandre.org/tetrationforum/showthread.php?tid=710</link>
			<pubDate>Mon, 14 Nov 2011 20:35:18 +0000</pubDate>
			<guid isPermaLink="false">http://math.eretrandre.org/tetrationforum/showthread.php?tid=710</guid>
			<description><![CDATA[hi all<br />
<br />
i dont have time to explain or post alot.<br />
<br />
but i managed to get some partial results about tetration i believe.<br />
<br />
i seem to have gotten a similar equation as andrew's slog ( the infinite matrix ) for my tommysexp(tommyslog(x)+k) in base e.<br />
<br />
also this seems to relate and possibly answer many uniqueness and existance questions that seemed somewhat unrelated and unattackable before.<br />
<br />
im still working on it but it seems promising and i might even solve some of those hard matrix questions.<br />
<br />
in a sense i also seem to have found a trick to attach a fixpoint to an infinite matrix equation. ( yes this is vague and informal , i dont have time to explain fully )<br />
<br />
regards<br />
<br />
tommy1729]]></description>
			<content:encoded><![CDATA[hi all<br />
<br />
i dont have time to explain or post alot.<br />
<br />
but i managed to get some partial results about tetration i believe.<br />
<br />
i seem to have gotten a similar equation as andrew's slog ( the infinite matrix ) for my tommysexp(tommyslog(x)+k) in base e.<br />
<br />
also this seems to relate and possibly answer many uniqueness and existance questions that seemed somewhat unrelated and unattackable before.<br />
<br />
im still working on it but it seems promising and i might even solve some of those hard matrix questions.<br />
<br />
in a sense i also seem to have found a trick to attach a fixpoint to an infinite matrix equation. ( yes this is vague and informal , i dont have time to explain fully )<br />
<br />
regards<br />
<br />
tommy1729]]></content:encoded>
		</item>
		<item>
			<title><![CDATA[A support for Andy's (P.Walker's) slog-matrix-method]]></title>
			<link>http://math.eretrandre.org/tetrationforum/showthread.php?tid=709</link>
			<pubDate>Mon, 14 Nov 2011 03:01:24 +0000</pubDate>
			<guid isPermaLink="false">http://math.eretrandre.org/tetrationforum/showthread.php?tid=709</guid>
			<description><![CDATA[Hi -<br />
 <br />
 in a selfstudy of the possibility of defining a "bernoulli-polynomial"-like solution for the problem of summing of like powers of logarithms <img src="/cgi-bin/mimetex.cgi? s_p(a,b)=\log(1+a)^p + \log(2+a) ^p+ \ldots + \log(b)^p " align="middle" /> and its  generalizations to arbitrary lower and upper summation-bounds <span style="font-style: italic;">a</span> and <span style="font-style: italic;">b</span> I tried the method of indefinite summation. There I had to find an infinite-sized matrix-reciprocal (inverse) in the same spirit as the matrix-reciprocal which occurs in the slog-ansatz of Andy Robbins(also used earlier by P.Walker).       <br />
<br />
Interestingly the matrix-reciprocal, which can be defined in the same way, gives not only meaningfully approximated values. That would be nice enough, but we might not be able to check, whether the computed values are always true approximations to the expected values. Actually we get even more: we seem to get exactly the coefficients of the most meaningful closed-form-function for this sums-of-like-powers-problem, namely involving the lngamma-function.         <br />
<br />
This occurence of the lngamma here is interesting in twofold manner:      <br />
<br />
a) it supports Andy's/P.Walker's matrix-ansatz for the solution of the tetration/slog     <br />
<br />
b) it supports the meaningfulness of the choice of the L. Euler's gamma-definition for the interpolation of the factorial besides of the criterion of log convexity (maybe this has then a similar effect for the solution of tetration).          <br />
<br />
I began to write a small article about that for my "mathematical miniatures" website, but am a bit distracted currently by my teaching duties and my weak health, and do not know when I'll have time to polish it up fully for presentation. However I thought it might already be useful/interesting to be accessible here in the current state; I think it should be readable, be selfcontained enough and understandable so far. If not, I'd like to answer/elaborate on specific questions.      <br />
   <br />
I uploaded the *.pdf to this forum, see attachment <br />
<br />
<br />
Gottfried<br />
<br />
P.s.: this is very near to that first-time observation in the thread <a href="http://math.eretrandre.org/tetrationforum/showthread.php?tid=632" target="_blank">http://math.eretrandre.org/tetrationforu...hp?tid=632</a> where I used that slog-matrix-computation rather as a curiosity, where here I'm focusing specifically on it.         <br />
<br />

<br /><img src="images/attachtypes/pdf.gif" border="0" alt=".pdf" />&nbsp;&nbsp;<a href="attachment.php?aid=915" target="_blank">BernoulliForLogSums.pdf</a> (Size: 123.46 KB / Downloads: 16)
]]></description>
			<content:encoded><![CDATA[Hi -<br />
 <br />
 in a selfstudy of the possibility of defining a "bernoulli-polynomial"-like solution for the problem of summing of like powers of logarithms <img src="/cgi-bin/mimetex.cgi? s_p(a,b)=\log(1+a)^p + \log(2+a) ^p+ \ldots + \log(b)^p " align="middle" /> and its  generalizations to arbitrary lower and upper summation-bounds <span style="font-style: italic;">a</span> and <span style="font-style: italic;">b</span> I tried the method of indefinite summation. There I had to find an infinite-sized matrix-reciprocal (inverse) in the same spirit as the matrix-reciprocal which occurs in the slog-ansatz of Andy Robbins(also used earlier by P.Walker).       <br />
<br />
Interestingly the matrix-reciprocal, which can be defined in the same way, gives not only meaningfully approximated values. That would be nice enough, but we might not be able to check, whether the computed values are always true approximations to the expected values. Actually we get even more: we seem to get exactly the coefficients of the most meaningful closed-form-function for this sums-of-like-powers-problem, namely involving the lngamma-function.         <br />
<br />
This occurence of the lngamma here is interesting in twofold manner:      <br />
<br />
a) it supports Andy's/P.Walker's matrix-ansatz for the solution of the tetration/slog     <br />
<br />
b) it supports the meaningfulness of the choice of the L. Euler's gamma-definition for the interpolation of the factorial besides of the criterion of log convexity (maybe this has then a similar effect for the solution of tetration).          <br />
<br />
I began to write a small article about that for my "mathematical miniatures" website, but am a bit distracted currently by my teaching duties and my weak health, and do not know when I'll have time to polish it up fully for presentation. However I thought it might already be useful/interesting to be accessible here in the current state; I think it should be readable, be selfcontained enough and understandable so far. If not, I'd like to answer/elaborate on specific questions.      <br />
   <br />
I uploaded the *.pdf to this forum, see attachment <br />
<br />
<br />
Gottfried<br />
<br />
P.s.: this is very near to that first-time observation in the thread <a href="http://math.eretrandre.org/tetrationforum/showthread.php?tid=632" target="_blank">http://math.eretrandre.org/tetrationforu...hp?tid=632</a> where I used that slog-matrix-computation rather as a curiosity, where here I'm focusing specifically on it.         <br />
<br />

<br /><img src="images/attachtypes/pdf.gif" border="0" alt=".pdf" />&nbsp;&nbsp;<a href="attachment.php?aid=915" target="_blank">BernoulliForLogSums.pdf</a> (Size: 123.46 KB / Downloads: 16)
]]></content:encoded>
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			<title><![CDATA[generalizing the problem of fractional analytic Ackermann functions]]></title>
			<link>http://math.eretrandre.org/tetrationforum/showthread.php?tid=708</link>
			<pubDate>Sun, 13 Nov 2011 22:33:56 +0000</pubDate>
			<guid isPermaLink="false">http://math.eretrandre.org/tetrationforum/showthread.php?tid=708</guid>
			<description><![CDATA[Well there's an obvious connection between the Ackermann function and superfunctions, but now I've found a way to generalize this to what I am labeling "meta-superfunctions".  These have to do with iteration but in a very different complicated light.<br />
<br />
consider the definition:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?f^{\diamond n}(g^{\diamond n}(x)) = g^{\diamond n}(f^{\diamond n}(x)) = x" align="middle" /><br />
<br />
<br />
<img src="/cgi-bin/mimetex.cgi?f^{\diamond n}(x) = f^{\diamond n+1}(g^{\diamond n+1}(x) + 1)" align="middle" /><br />
<br />
or so that<br />
<br />
<img src="/cgi-bin/mimetex.cgi?f^{\diamond n + 1}(x)" align="middle" /> is the super function of <img src="/cgi-bin/mimetex.cgi?f^{\diamond n}(x)" align="middle" /><br />
<br />
and<br />
<img src="/cgi-bin/mimetex.cgi?g^{\diamond n + 1}(x)" align="middle" /> is the abel function of <img src="/cgi-bin/mimetex.cgi?f^{\diamond n}(x)" align="middle" /><br />
<br />
this gives the relation:  <br />
<br />
<img src="/cgi-bin/mimetex.cgi?f^{\diamond n + 1}(x) = (f^{\diamond n} \circ f^{\diamond n} \circ ...\text{x amount of times}... \circ f^{\diamond n}) ( C ) = (f^{\diamond n})^{\circ x} ( C )" align="middle" /><br />
<br />
for some arbitrary constant C, that gives no singularities and such.<br />
<br />
we'd center the equation at zero, so that:<br />
<img src="/cgi-bin/mimetex.cgi?f^{\diamond 0}(x) = f(x)" align="middle" /><br />
<img src="/cgi-bin/mimetex.cgi?g^{\diamond 0}(x) = f^{\circ -1}(x)" align="middle" /><br />
<br />
the question I ask is, how do we find fractional values of n?  Or put more linguistically, what function is in between a function and its super function?<br />
<br />
The connection to the Ackermann function is simple:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?\vartheta(a,b,\sigma) = a\,\,\bigtriangleup_\sigma\,\,b" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(\vartheta(a,b,\sigma)) = b" align="middle" /><br />
<br />
so that we get<br />
<br />
<img src="/cgi-bin/mimetex.cgi?\vartheta(a,L_a^{\sigma + 1}(b) + 1, \sigma + 1) = \vartheta(a,b,\sigma)" align="middle" /><br />
<br />
which is a simple exercise in super functions<br />
<br />
now to split confusion we'll rewrite this in "meta superfunction" notation, fixing a to some arbitrary constant<br />
<br />
<img src="/cgi-bin/mimetex.cgi?f^{\diamond \sigma}(b) = \vartheta(a,b,\sigma)" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?g^{\diamond \sigma}(b) = L_a^{\sigma}(b)" align="middle" /><br />
<br />
we get the simple result from out previous formula above<br />
<br />
<img src="/cgi-bin/mimetex.cgi?f^{\diamond \sigma}(b) = f^{\diamond \sigma + 1}(g^{\diamond \sigma + 1}(b) + 1)" align="middle" /><br />
<br />
and this defines the Ackermann function in terms of "meta superfunctions"<br />
<br />
<br />
Now, the difficulty with defining fractional values of these "meta superfunctions" is that we do not have a recurrence relation for fractional values...  The difference between that and regular superfunctions is that we do, namely<br />
<br />
<img src="/cgi-bin/mimetex.cgi?f^{\circ \frac{1}{2}}(f^{\circ \frac{1}{2}}(x)) = f(x)" align="middle" /><br />
<br />
there is no such property or relation for "meta superfunctions"<br />
<br />
the only property we know it must satisfy is:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?f^{\diamond \frac{1}{2}}(b) = f^{\diamond \frac{3}{2}}(g^{\diamond \frac{3}{2}}(b) + 1)" align="middle" /><br />
<br />
but this is no good because it doesn't give an expression in terms of an even integer superfunction.  So my question is, are meta-superfunctions by nature <span style="font-style: italic;">discrete</span>?  and inconclusively continuous?  Since there is no external property they need satisfy other than maybe<br />
<br />
<img src="/cgi-bin/mimetex.cgi?R(x) = f^{\diamond x}(b)" align="middle" /> <br />
<br />
with R(x) being analytic across x.<br />
<br />
<br />
So, I believe, before the Ackermann function can be continued analytically we must define a recurrence relation across meta superfunctions that applies to fractional values.<br />
<br />
My rather meagre attempt at this recurrence relation comes from defining a new operator.  I call it "meta composition".  It's rather vague and does not have a concrete definition, but it's the best I can come up with.<br />
<br />
<img src="/cgi-bin/mimetex.cgi?(f^{\diamond m} \diamond f^{\diamond n})(x) = f^{\diamond m + n}(x)" align="middle" /><br />
<br />
this would mean that the identity of meta composition is the function itself<br />
<br />
<img src="/cgi-bin/mimetex.cgi?(f \diamond f^{\diamond n})(x) = f^{\diamond n}(x)" align="middle" /><br />
<br />
which implies idempotency and also that <img src="/cgi-bin/mimetex.cgi?\diamond" align="middle" /> is defined relatively based on each function.  Actually, <img src="/cgi-bin/mimetex.cgi?\diamond" align="middle" /> is very much like the composition operator <img src="/cgi-bin/mimetex.cgi?\circ" align="middle" /> except it is reformulated with <img src="/cgi-bin/mimetex.cgi?f" align="middle" /> as the identity function rather than the normal identity function <img src="/cgi-bin/mimetex.cgi?f(x) = x" align="middle" /><br />
<br />
Actually thinking about it, I'm sure I've read somewhere of composition operators redefined to suit different identity functions.  If this was the case this could perhaps be an approach to a solution for analytic fractional operators!<br />
<br />
So I'm wondering, is this just absurd?  It's really the best I can come up with...<br />
<br />
any other possible "meta superfunction" recurrence relations would be welcomed with a warm heart!]]></description>
			<content:encoded><![CDATA[Well there's an obvious connection between the Ackermann function and superfunctions, but now I've found a way to generalize this to what I am labeling "meta-superfunctions".  These have to do with iteration but in a very different complicated light.<br />
<br />
consider the definition:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?f^{\diamond n}(g^{\diamond n}(x)) = g^{\diamond n}(f^{\diamond n}(x)) = x" align="middle" /><br />
<br />
<br />
<img src="/cgi-bin/mimetex.cgi?f^{\diamond n}(x) = f^{\diamond n+1}(g^{\diamond n+1}(x) + 1)" align="middle" /><br />
<br />
or so that<br />
<br />
<img src="/cgi-bin/mimetex.cgi?f^{\diamond n + 1}(x)" align="middle" /> is the super function of <img src="/cgi-bin/mimetex.cgi?f^{\diamond n}(x)" align="middle" /><br />
<br />
and<br />
<img src="/cgi-bin/mimetex.cgi?g^{\diamond n + 1}(x)" align="middle" /> is the abel function of <img src="/cgi-bin/mimetex.cgi?f^{\diamond n}(x)" align="middle" /><br />
<br />
this gives the relation:  <br />
<br />
<img src="/cgi-bin/mimetex.cgi?f^{\diamond n + 1}(x) = (f^{\diamond n} \circ f^{\diamond n} \circ ...\text{x amount of times}... \circ f^{\diamond n}) ( C ) = (f^{\diamond n})^{\circ x} ( C )" align="middle" /><br />
<br />
for some arbitrary constant C, that gives no singularities and such.<br />
<br />
we'd center the equation at zero, so that:<br />
<img src="/cgi-bin/mimetex.cgi?f^{\diamond 0}(x) = f(x)" align="middle" /><br />
<img src="/cgi-bin/mimetex.cgi?g^{\diamond 0}(x) = f^{\circ -1}(x)" align="middle" /><br />
<br />
the question I ask is, how do we find fractional values of n?  Or put more linguistically, what function is in between a function and its super function?<br />
<br />
The connection to the Ackermann function is simple:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?\vartheta(a,b,\sigma) = a\,\,\bigtriangleup_\sigma\,\,b" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(\vartheta(a,b,\sigma)) = b" align="middle" /><br />
<br />
so that we get<br />
<br />
<img src="/cgi-bin/mimetex.cgi?\vartheta(a,L_a^{\sigma + 1}(b) + 1, \sigma + 1) = \vartheta(a,b,\sigma)" align="middle" /><br />
<br />
which is a simple exercise in super functions<br />
<br />
now to split confusion we'll rewrite this in "meta superfunction" notation, fixing a to some arbitrary constant<br />
<br />
<img src="/cgi-bin/mimetex.cgi?f^{\diamond \sigma}(b) = \vartheta(a,b,\sigma)" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?g^{\diamond \sigma}(b) = L_a^{\sigma}(b)" align="middle" /><br />
<br />
we get the simple result from out previous formula above<br />
<br />
<img src="/cgi-bin/mimetex.cgi?f^{\diamond \sigma}(b) = f^{\diamond \sigma + 1}(g^{\diamond \sigma + 1}(b) + 1)" align="middle" /><br />
<br />
and this defines the Ackermann function in terms of "meta superfunctions"<br />
<br />
<br />
Now, the difficulty with defining fractional values of these "meta superfunctions" is that we do not have a recurrence relation for fractional values...  The difference between that and regular superfunctions is that we do, namely<br />
<br />
<img src="/cgi-bin/mimetex.cgi?f^{\circ \frac{1}{2}}(f^{\circ \frac{1}{2}}(x)) = f(x)" align="middle" /><br />
<br />
there is no such property or relation for "meta superfunctions"<br />
<br />
the only property we know it must satisfy is:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?f^{\diamond \frac{1}{2}}(b) = f^{\diamond \frac{3}{2}}(g^{\diamond \frac{3}{2}}(b) + 1)" align="middle" /><br />
<br />
but this is no good because it doesn't give an expression in terms of an even integer superfunction.  So my question is, are meta-superfunctions by nature <span style="font-style: italic;">discrete</span>?  and inconclusively continuous?  Since there is no external property they need satisfy other than maybe<br />
<br />
<img src="/cgi-bin/mimetex.cgi?R(x) = f^{\diamond x}(b)" align="middle" /> <br />
<br />
with R(x) being analytic across x.<br />
<br />
<br />
So, I believe, before the Ackermann function can be continued analytically we must define a recurrence relation across meta superfunctions that applies to fractional values.<br />
<br />
My rather meagre attempt at this recurrence relation comes from defining a new operator.  I call it "meta composition".  It's rather vague and does not have a concrete definition, but it's the best I can come up with.<br />
<br />
<img src="/cgi-bin/mimetex.cgi?(f^{\diamond m} \diamond f^{\diamond n})(x) = f^{\diamond m + n}(x)" align="middle" /><br />
<br />
this would mean that the identity of meta composition is the function itself<br />
<br />
<img src="/cgi-bin/mimetex.cgi?(f \diamond f^{\diamond n})(x) = f^{\diamond n}(x)" align="middle" /><br />
<br />
which implies idempotency and also that <img src="/cgi-bin/mimetex.cgi?\diamond" align="middle" /> is defined relatively based on each function.  Actually, <img src="/cgi-bin/mimetex.cgi?\diamond" align="middle" /> is very much like the composition operator <img src="/cgi-bin/mimetex.cgi?\circ" align="middle" /> except it is reformulated with <img src="/cgi-bin/mimetex.cgi?f" align="middle" /> as the identity function rather than the normal identity function <img src="/cgi-bin/mimetex.cgi?f(x) = x" align="middle" /><br />
<br />
Actually thinking about it, I'm sure I've read somewhere of composition operators redefined to suit different identity functions.  If this was the case this could perhaps be an approach to a solution for analytic fractional operators!<br />
<br />
So I'm wondering, is this just absurd?  It's really the best I can come up with...<br />
<br />
any other possible "meta superfunction" recurrence relations would be welcomed with a warm heart!]]></content:encoded>
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			<title><![CDATA[Derivative of E tetra x]]></title>
			<link>http://math.eretrandre.org/tetrationforum/showthread.php?tid=707</link>
			<pubDate>Sat, 12 Nov 2011 06:08:48 +0000</pubDate>
			<guid isPermaLink="false">http://math.eretrandre.org/tetrationforum/showthread.php?tid=707</guid>
			<description><![CDATA[Hello!  First off, I have to thank everybody who contributed to the thread at <a href="http://math.eretrandre.org/tetrationforum/showthread.php?tid=47" target="_blank">http://math.eretrandre.org/tetrationforu...php?tid=47</a>, as that was what started me on this line of reasoning.  In the attached pdf (I don't know how to use TeX yet, so I did it in Mathematica), I derived what I thought was a logical derivative of e[4]x, but upon actually running the numbers, it turned out not to work.  I can't figure out why, but hopefully somebody here can.  Thanks in advance for reading through this post (and hopefully the pdf as well).  :-)<br />
<br /><img src="images/attachtypes/pdf.gif" border="0" alt=".pdf" />&nbsp;&nbsp;<a href="attachment.php?aid=914" target="_blank">Derivative of E tetra x.pdf</a> (Size: 66.27 KB / Downloads: 35)
]]></description>
			<content:encoded><![CDATA[Hello!  First off, I have to thank everybody who contributed to the thread at <a href="http://math.eretrandre.org/tetrationforum/showthread.php?tid=47" target="_blank">http://math.eretrandre.org/tetrationforu...php?tid=47</a>, as that was what started me on this line of reasoning.  In the attached pdf (I don't know how to use TeX yet, so I did it in Mathematica), I derived what I thought was a logical derivative of e[4]x, but upon actually running the numbers, it turned out not to work.  I can't figure out why, but hopefully somebody here can.  Thanks in advance for reading through this post (and hopefully the pdf as well).  :-)<br />
<br /><img src="images/attachtypes/pdf.gif" border="0" alt=".pdf" />&nbsp;&nbsp;<a href="attachment.php?aid=914" target="_blank">Derivative of E tetra x.pdf</a> (Size: 66.27 KB / Downloads: 35)
]]></content:encoded>
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			<title><![CDATA[kouznetsov billiard ?]]></title>
			<link>http://math.eretrandre.org/tetrationforum/showthread.php?tid=706</link>
			<pubDate>Fri, 11 Nov 2011 22:41:52 +0000</pubDate>
			<guid isPermaLink="false">http://math.eretrandre.org/tetrationforum/showthread.php?tid=706</guid>
			<description><![CDATA[on the page <br />
<br />
<a href="http://en.wikipedia.org/wiki/Dynamical_billiards" target="_blank">http://en.wikipedia.org/wiki/Dynamical_billiards</a><br />
<br />
reference 6 , i noticed Kouznetsov D.<br />
<br />
i was wondering if it is the same Kouznetsov who posts here.<br />
<br />
it is the same guy who made these <br />
<br />
<a href="http://www.tandfonline.com/action/doSearch?action=runSearch&amp;type=advanced&amp;result=true&amp;prevSearch=%2Bauthorsfield%3A(Kouznetsov%2C+Dmitrii)" target="_blank">http://www.tandfonline.com/action/doSear...C+Dmitrii)</a><br />
<br />
so i suppose it is.<br />
<br />
nice.]]></description>
			<content:encoded><![CDATA[on the page <br />
<br />
<a href="http://en.wikipedia.org/wiki/Dynamical_billiards" target="_blank">http://en.wikipedia.org/wiki/Dynamical_billiards</a><br />
<br />
reference 6 , i noticed Kouznetsov D.<br />
<br />
i was wondering if it is the same Kouznetsov who posts here.<br />
<br />
it is the same guy who made these <br />
<br />
<a href="http://www.tandfonline.com/action/doSearch?action=runSearch&amp;type=advanced&amp;result=true&amp;prevSearch=%2Bauthorsfield%3A(Kouznetsov%2C+Dmitrii)" target="_blank">http://www.tandfonline.com/action/doSear...C+Dmitrii)</a><br />
<br />
so i suppose it is.<br />
<br />
nice.]]></content:encoded>
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			<title><![CDATA[Proof Ackermann function cannot have an analytic identity function]]></title>
			<link>http://math.eretrandre.org/tetrationforum/showthread.php?tid=705</link>
			<pubDate>Fri, 11 Nov 2011 01:26:35 +0000</pubDate>
			<guid isPermaLink="false">http://math.eretrandre.org/tetrationforum/showthread.php?tid=705</guid>
			<description><![CDATA[Well I've been having suspicions for quite a while that the Ackermann function cannot be analytic.  I was having trouble visualizing it and then one step to the answer came to me through the identity function.  It's actually incredibly simple the formulas involved.  It's a proof by contradiction.<br />
<br />
First, we start off by defining what we mean by the "Ackermann function", so as not to cause any confusion.<br />
<br />
<img src="/cgi-bin/mimetex.cgi?\vartheta(a,b,\sigma) = a\,\,\bigtriangleup_\sigma\,\,b" align="middle" /><br />
<br />
where:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_{\sigma - 1} (a\,\,\bigtriangleup_\sigma\,\,b) = a\,\,\bigtriangleup_\sigma\,\,b+1" align="middle" /><br />
<br />
and if <br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_{\sigma}\,\,S(\sigma) = a" align="middle" /><br />
<br />
or <br />
<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma)" align="middle" /> is the identity function<br />
<br />
we set:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_{1}\,\,b = a + b" align="middle" /><br />
<br />
so that<br />
<br />
<img src="/cgi-bin/mimetex.cgi?S(1) = 0" align="middle" /><br />
<br />
and we make the final two requirements:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?n \in \mathbb{Z}\,\,;n \ge 2" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?S(n) = 1" align="middle" /> so that we have <img src="/cgi-bin/mimetex.cgi?\bigtriangleup_2" align="middle" /> as multiplication and <img src="/cgi-bin/mimetex.cgi?\bigtriangleup_3" align="middle" /> as exponentiation so on and so forth.<br />
<br />
<br />
if <br />
<img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(a\,\,\bigtriangleup_\sigma\,\,b) = b" align="middle" /><br />
<br />
or <img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(b)" align="middle" /> is the "logarithmic" inverse operator of order <img src="/cgi-bin/mimetex.cgi?\sigma" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(a) = S(\sigma)" align="middle" />  must be analytic across <img src="/cgi-bin/mimetex.cgi?\sigma" align="middle" /><br />
<br />
therefore<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(b)" align="middle" /> is analytic across <img src="/cgi-bin/mimetex.cgi?\sigma" align="middle" /> at all values of b if and only if our identity function (<img src="/cgi-bin/mimetex.cgi?S(\sigma)" align="middle" />) is analytic.<br />
<br />
We will work to prove that <img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(b)" align="middle" /> cannot be analytic by <img src="/cgi-bin/mimetex.cgi?\sigma" align="middle" /> at all values of b.<br />
<br />
So we come to the weaker conclusion than the Ackermann function cannot be analytic across all values, to the "logarithmic" Ackermann function cannot be analytic across all values.<br />
<br />
or the series<br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{1}(b) = b - a" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{2}(b) = \frac{b}{a}" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{3}(b) = \log_a(b)" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{4}(b) = \text{slog}_a(b)" align="middle" /><br />
<br />
cannot be analytic if it holds the property:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(a\,\,\bigtriangleup_\sigma\,\,b) = b" align="middle" /><br />
<br />
Firstly, we make the argument that <img src="/cgi-bin/mimetex.cgi?S(\sigma)" align="middle" /> cannot be periodic if we want it to be analytic.<br />
<br />
Or simply put if:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma + 1) = S(\sigma)" align="middle" /><br />
<br />
this is instantly invalid for <img src="/cgi-bin/mimetex.cgi?S(1) = 0" align="middle" /> and <img src="/cgi-bin/mimetex.cgi?S(2) = 1" align="middle" /> and an analytic periodic function <span style="font-style: italic;">must</span> be periodic for all values, however, we do have it to be true for integer values of <img src="/cgi-bin/mimetex.cgi?\sigma" align="middle" /> greater than 1.<br />
<br />
Now we write, <br />
<br />
since <img src="/cgi-bin/mimetex.cgi?S(2) = S(3) = S(4) = ... = 1" align="middle" /> and since <img src="/cgi-bin/mimetex.cgi?S(\sigma) \neq 1" align="middle" />, <img src="/cgi-bin/mimetex.cgi?S(\sigma)" align="middle" /> must oscillate similarly to a periodic function only it is not periodic since if it was periodic <span style="font-style: italic;">and</span> analytic <img src="/cgi-bin/mimetex.cgi?S(1) = 1" align="middle" /> therefore we designate it cannot be analytic and periodic.<br />
<br />
So let us write what I call the "pseudo period" as<br />
<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma + \ell(\sigma)) = S(\sigma)" align="middle" /><br />
<br />
evidently <img src="/cgi-bin/mimetex.cgi?\ell(\sigma)" align="middle" /> is multivalued and not analytic and does not always return values.<br />
<br />
for example <img src="/cgi-bin/mimetex.cgi?\ell(0)" align="middle" /> does not necessarily exist, because it is fully possible that<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma) &gt; 0" align="middle" /> ; <img src="/cgi-bin/mimetex.cgi?\sigma &gt; 1" align="middle" /><br />
<br />
however, it is evident that <img src="/cgi-bin/mimetex.cgi?\ell(\sigma)" align="middle" /> must exist for <span style="font-style: italic;">some</span> values of <img src="/cgi-bin/mimetex.cgi?\sigma" align="middle" /> in order that <img src="/cgi-bin/mimetex.cgi?S(\sigma)" align="middle" /> be analytic.<br />
<br />
it is trivial that<br />
<br />
<img src="/cgi-bin/mimetex.cgi?n, k \in \mathbb{Z} ; n \ge 2 ; k \ge 0" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?\ell(n) = k" align="middle" /><br />
<br />
<br />
Now we must derive a few lemmas<br />
<br />
consider<br />
<br />
<img src="/cgi-bin/mimetex.cgi?(1 + S(\sigma))\,\, \bigtriangleup_{\sigma - 1}\,\, (1 + S(\sigma)) = (1 + S(\sigma))\,\,\bigtriangleup_{\sigma - 1}\,\,((1 + S(\sigma))\,\,\bigtriangleup_{\sigma}\,\,S(\sigma))" align="middle" /><br />
<br />
by the first axiom it's easy to deduce:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?\text{I:}\,\,(1 + S(\sigma))\,\, \bigtriangleup_{\sigma-1}\,\, (1 + S(\sigma)) = (1 + S(\sigma))\,\, \bigtriangleup_{\sigma}\,\, (1 + S(\sigma))" align="middle" /><br />
<br />
continuing this process again we get:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?(1 + S(\sigma))\,\, \bigtriangleup_{\sigma-1}\,\, (1 + S(\sigma)) = (1 + S(\sigma))\,\, \bigtriangleup_{\sigma}\,\, (1 + S(\sigma)) = (1 + S(\sigma))\,\, \bigtriangleup_{\sigma+1}\,\, (1 + S(\sigma+1))" align="middle" /><br />
<br />
this is how we get the famed identity <img src="/cgi-bin/mimetex.cgi?2\,\,\bigtriangleup_n\,\,2 = 4" align="middle" /><br />
<br />
<br />
Now we move on to the "logarithmic" inverse function and defining a new function I call Q.<br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(b) + Q^{\sigma}(t, a, b) = L_a^{\sigma}(a\,\,\bigtriangleup_{\sigma + t}\,\,b)" align="middle" /><br />
<br />
it's obvious that<br />
<br />
<img src="/cgi-bin/mimetex.cgi?Q^{\sigma}(-1, a, b) = 1" align="middle" /> which is a simple exercise in super operators<br />
<br />
and<br />
<br />
<img src="/cgi-bin/mimetex.cgi?Q^{\sigma}(t, a, b) = L_a^{\sigma}(a\,\,\bigtriangleup_{\sigma + t}\,\,b) - L_a^{\sigma}(b)" align="middle" /><br />
<br />
before continuing we need to acknowledge that<br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(a\,\,\bigtriangleup_{\sigma+1}\,\,b) = a\,\,\bigtriangleup_{\sigma+1} (b-1)" align="middle" /><br />
<br />
which is again a simple exercise.<br />
<br />
<br />
now we have to make some odd observations that may get confusing<br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_{1+S(\sigma)}^{\sigma + \ell(\sigma)}((1 + S(\sigma))\,\,\bigtriangleup_{\sigma + \ell(\sigma) + 1}\,\,(1+S(\sigma + \ell(\sigma) + 1)) = (1 + S(\sigma))\,\,\bigtriangleup_{\sigma + \ell(\sigma) + 1}\,\,S(\sigma + \ell(\sigma) + 1) = 1 + S(\sigma)" align="middle" /><br />
<br />
from the "logarithmic" inverse law<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(a\,\,\bigtriangleup_{\sigma+1}\,\,b) = a\,\,\bigtriangleup_{\sigma+1} (b-1)" align="middle" /><br />
<br />
this implies:<br />
<img src="/cgi-bin/mimetex.cgi?L_{1+S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma)) + Q^{\sigma + \ell(\sigma)}(1, 1 + S(\sigma), 1 + S(\sigma + \ell(\sigma) + 1)) = 1 + S(\sigma)" align="middle" /><br />
<br />
which in turn gives<br />
<br />
<img src="/cgi-bin/mimetex.cgi?Q^{\sigma + \ell(\sigma)}(1, 1 + S(\sigma), 1 + S(\sigma + \ell(\sigma) + 1)) = 1" align="middle" /><br />
<br />
since<br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_{1+S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma)) = S(\sigma + \ell(\sigma)) = S(\sigma)" align="middle" /><br />
<br />
by the identity<br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(a) = S(\sigma)" align="middle" /><br />
<br />
<br />
but we also know by Q's definition<br />
<img src="/cgi-bin/mimetex.cgi?Q^{\sigma}(t, a, b) = L_a^{\sigma}(a\,\,\bigtriangleup_{\sigma + t}\,\,b) - L_a^{\sigma}(b)" align="middle" /><br />
<br />
so that<br />
<br />
<img src="/cgi-bin/mimetex.cgi?Q^{\sigma + \ell(\sigma)}(1, 1 + S(\sigma), 1 + S(\sigma + \ell(\sigma) + 1)) = 1 = L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}((1 + S(\sigma))\,\,\bigtriangleup_{1 + \sigma + \ell(\sigma)} (1 + S(\sigma + \ell(\sigma) + 1))) - L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma + \ell(\sigma) + 1))" align="middle" /><br />
<br />
we know that from earlier<br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}((1 + S(\sigma))\,\,\bigtriangleup_{1 + \sigma + \ell(\sigma)} (1 + S(\sigma + \ell(\sigma) + 1))) = 1 + S(\sigma)" align="middle" /><br />
<br />
so that we get<br />
<br />
<img src="/cgi-bin/mimetex.cgi?1 = 1 + S(\sigma)  - L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma + \ell(\sigma) + 1))" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma) = L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma + \ell(\sigma) + 1))" align="middle" /><br />
<br />
but we also know, by the definition of the "logarithmic" inverse:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_\sigma\,\,L_a^{\sigma}(b) = b" align="middle" /><br />
<br />
so therefore<br />
<br />
<img src="/cgi-bin/mimetex.cgi?(1 + S(\sigma))\,\,\bigtriangleup_{\sigma + \ell(\sigma)}\,\,S(\sigma) = (1 + S(\sigma))\,\,\bigtriangleup_{\sigma + \ell(\sigma)}\,\,L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma + \ell(\sigma) + 1))" align="middle" /><br />
<br />
which since<br />
<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma) = S(\sigma + \ell(\sigma))" align="middle" /><br />
<br />
and<br />
<br />
<img src="/cgi-bin/mimetex.cgi?(1 + S(\sigma))\,\,\bigtriangleup_{\sigma + \ell}\,\,L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma + \ell(\sigma) + 1)) = 1 + S(\sigma + \ell(\sigma) + 1)" align="middle" /><br />
<br />
we get the beautiful result:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma) = S(\sigma + \ell(\sigma) + 1)" align="middle" /><br />
<br />
which is the core identity of our proof<br />
<br />
since <img src="/cgi-bin/mimetex.cgi?\ell(\sigma)" align="middle" /> is multivalued, we find that<br />
<br />
for some value <img src="/cgi-bin/mimetex.cgi?q &gt; 0" align="middle" /><br />
<br />
if<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma + q) = S(\sigma)" align="middle" /><br />
<br />
then<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma + q + 1) = S(\sigma)" align="middle" /><br />
<br />
which in turn applies<br />
<br />
<img src="/cgi-bin/mimetex.cgi?\ell(\sigma) = q + 1" align="middle" /><br />
<br />
which iterated gives <img src="/cgi-bin/mimetex.cgi?k \in \mathbb{Z} ; k \ge 0" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma + q + k) = S(\sigma + q)" align="middle" /><br />
<br />
or that <img src="/cgi-bin/mimetex.cgi?S(\sigma + q)" align="middle" /> is periodic.  However we already noted that if <img src="/cgi-bin/mimetex.cgi?S(\sigma)" align="middle" /> is periodic then <img src="/cgi-bin/mimetex.cgi?S(\sigma)" align="middle" /> cannot be analytic since by definition of an analytic function that is periodic it must be periodic over every value in the domain and <img src="/cgi-bin/mimetex.cgi?S(1) \neq S(2)" align="middle" /><br />
<br />
Therefore<br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(b)" align="middle" /> is not analytic by <img src="/cgi-bin/mimetex.cgi?\sigma" align="middle" /> for all values of b<br />
<br />
and <br />
<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma)" align="middle" /> is not analytic by <img src="/cgi-bin/mimetex.cgi?\sigma" align="middle" /><br />
<br />
Q.E.D.<br />
<br />
<br />
This puts me one step closer to proving non analycity across <img src="/cgi-bin/mimetex.cgi?\sigma" align="middle" /> for some value of a and b in the Ackermann function <img src="/cgi-bin/mimetex.cgi?\vartheta(a,b,\sigma)" align="middle" /><br />
<br />
Personally I think that non-analycity is intuitively implied since <img src="/cgi-bin/mimetex.cgi?\vartheta(a,b,\sigma)" align="middle" /> is not primitive recursive, but this is yet to be rigorously decided.  To disprove universal analycity, it suffices to show that for some fixed <img src="/cgi-bin/mimetex.cgi?u, v \in C" align="middle" />, <img src="/cgi-bin/mimetex.cgi?\vartheta(u,v,\sigma)" align="middle" /> is not analytic across <img src="/cgi-bin/mimetex.cgi?\sigma" align="middle" />.<br />
<br />
I understand the notation in this proof maybe a bit confusing, but when working with three variables I think that's inevitable.<br />
<br />
any questions, comments?]]></description>
			<content:encoded><![CDATA[Well I've been having suspicions for quite a while that the Ackermann function cannot be analytic.  I was having trouble visualizing it and then one step to the answer came to me through the identity function.  It's actually incredibly simple the formulas involved.  It's a proof by contradiction.<br />
<br />
First, we start off by defining what we mean by the "Ackermann function", so as not to cause any confusion.<br />
<br />
<img src="/cgi-bin/mimetex.cgi?\vartheta(a,b,\sigma) = a\,\,\bigtriangleup_\sigma\,\,b" align="middle" /><br />
<br />
where:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_{\sigma - 1} (a\,\,\bigtriangleup_\sigma\,\,b) = a\,\,\bigtriangleup_\sigma\,\,b+1" align="middle" /><br />
<br />
and if <br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_{\sigma}\,\,S(\sigma) = a" align="middle" /><br />
<br />
or <br />
<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma)" align="middle" /> is the identity function<br />
<br />
we set:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_{1}\,\,b = a + b" align="middle" /><br />
<br />
so that<br />
<br />
<img src="/cgi-bin/mimetex.cgi?S(1) = 0" align="middle" /><br />
<br />
and we make the final two requirements:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?n \in \mathbb{Z}\,\,;n \ge 2" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?S(n) = 1" align="middle" /> so that we have <img src="/cgi-bin/mimetex.cgi?\bigtriangleup_2" align="middle" /> as multiplication and <img src="/cgi-bin/mimetex.cgi?\bigtriangleup_3" align="middle" /> as exponentiation so on and so forth.<br />
<br />
<br />
if <br />
<img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(a\,\,\bigtriangleup_\sigma\,\,b) = b" align="middle" /><br />
<br />
or <img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(b)" align="middle" /> is the "logarithmic" inverse operator of order <img src="/cgi-bin/mimetex.cgi?\sigma" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(a) = S(\sigma)" align="middle" />  must be analytic across <img src="/cgi-bin/mimetex.cgi?\sigma" align="middle" /><br />
<br />
therefore<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(b)" align="middle" /> is analytic across <img src="/cgi-bin/mimetex.cgi?\sigma" align="middle" /> at all values of b if and only if our identity function (<img src="/cgi-bin/mimetex.cgi?S(\sigma)" align="middle" />) is analytic.<br />
<br />
We will work to prove that <img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(b)" align="middle" /> cannot be analytic by <img src="/cgi-bin/mimetex.cgi?\sigma" align="middle" /> at all values of b.<br />
<br />
So we come to the weaker conclusion than the Ackermann function cannot be analytic across all values, to the "logarithmic" Ackermann function cannot be analytic across all values.<br />
<br />
or the series<br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{1}(b) = b - a" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{2}(b) = \frac{b}{a}" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{3}(b) = \log_a(b)" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{4}(b) = \text{slog}_a(b)" align="middle" /><br />
<br />
cannot be analytic if it holds the property:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(a\,\,\bigtriangleup_\sigma\,\,b) = b" align="middle" /><br />
<br />
Firstly, we make the argument that <img src="/cgi-bin/mimetex.cgi?S(\sigma)" align="middle" /> cannot be periodic if we want it to be analytic.<br />
<br />
Or simply put if:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma + 1) = S(\sigma)" align="middle" /><br />
<br />
this is instantly invalid for <img src="/cgi-bin/mimetex.cgi?S(1) = 0" align="middle" /> and <img src="/cgi-bin/mimetex.cgi?S(2) = 1" align="middle" /> and an analytic periodic function <span style="font-style: italic;">must</span> be periodic for all values, however, we do have it to be true for integer values of <img src="/cgi-bin/mimetex.cgi?\sigma" align="middle" /> greater than 1.<br />
<br />
Now we write, <br />
<br />
since <img src="/cgi-bin/mimetex.cgi?S(2) = S(3) = S(4) = ... = 1" align="middle" /> and since <img src="/cgi-bin/mimetex.cgi?S(\sigma) \neq 1" align="middle" />, <img src="/cgi-bin/mimetex.cgi?S(\sigma)" align="middle" /> must oscillate similarly to a periodic function only it is not periodic since if it was periodic <span style="font-style: italic;">and</span> analytic <img src="/cgi-bin/mimetex.cgi?S(1) = 1" align="middle" /> therefore we designate it cannot be analytic and periodic.<br />
<br />
So let us write what I call the "pseudo period" as<br />
<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma + \ell(\sigma)) = S(\sigma)" align="middle" /><br />
<br />
evidently <img src="/cgi-bin/mimetex.cgi?\ell(\sigma)" align="middle" /> is multivalued and not analytic and does not always return values.<br />
<br />
for example <img src="/cgi-bin/mimetex.cgi?\ell(0)" align="middle" /> does not necessarily exist, because it is fully possible that<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma) &gt; 0" align="middle" /> ; <img src="/cgi-bin/mimetex.cgi?\sigma &gt; 1" align="middle" /><br />
<br />
however, it is evident that <img src="/cgi-bin/mimetex.cgi?\ell(\sigma)" align="middle" /> must exist for <span style="font-style: italic;">some</span> values of <img src="/cgi-bin/mimetex.cgi?\sigma" align="middle" /> in order that <img src="/cgi-bin/mimetex.cgi?S(\sigma)" align="middle" /> be analytic.<br />
<br />
it is trivial that<br />
<br />
<img src="/cgi-bin/mimetex.cgi?n, k \in \mathbb{Z} ; n \ge 2 ; k \ge 0" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?\ell(n) = k" align="middle" /><br />
<br />
<br />
Now we must derive a few lemmas<br />
<br />
consider<br />
<br />
<img src="/cgi-bin/mimetex.cgi?(1 + S(\sigma))\,\, \bigtriangleup_{\sigma - 1}\,\, (1 + S(\sigma)) = (1 + S(\sigma))\,\,\bigtriangleup_{\sigma - 1}\,\,((1 + S(\sigma))\,\,\bigtriangleup_{\sigma}\,\,S(\sigma))" align="middle" /><br />
<br />
by the first axiom it's easy to deduce:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?\text{I:}\,\,(1 + S(\sigma))\,\, \bigtriangleup_{\sigma-1}\,\, (1 + S(\sigma)) = (1 + S(\sigma))\,\, \bigtriangleup_{\sigma}\,\, (1 + S(\sigma))" align="middle" /><br />
<br />
continuing this process again we get:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?(1 + S(\sigma))\,\, \bigtriangleup_{\sigma-1}\,\, (1 + S(\sigma)) = (1 + S(\sigma))\,\, \bigtriangleup_{\sigma}\,\, (1 + S(\sigma)) = (1 + S(\sigma))\,\, \bigtriangleup_{\sigma+1}\,\, (1 + S(\sigma+1))" align="middle" /><br />
<br />
this is how we get the famed identity <img src="/cgi-bin/mimetex.cgi?2\,\,\bigtriangleup_n\,\,2 = 4" align="middle" /><br />
<br />
<br />
Now we move on to the "logarithmic" inverse function and defining a new function I call Q.<br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(b) + Q^{\sigma}(t, a, b) = L_a^{\sigma}(a\,\,\bigtriangleup_{\sigma + t}\,\,b)" align="middle" /><br />
<br />
it's obvious that<br />
<br />
<img src="/cgi-bin/mimetex.cgi?Q^{\sigma}(-1, a, b) = 1" align="middle" /> which is a simple exercise in super operators<br />
<br />
and<br />
<br />
<img src="/cgi-bin/mimetex.cgi?Q^{\sigma}(t, a, b) = L_a^{\sigma}(a\,\,\bigtriangleup_{\sigma + t}\,\,b) - L_a^{\sigma}(b)" align="middle" /><br />
<br />
before continuing we need to acknowledge that<br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(a\,\,\bigtriangleup_{\sigma+1}\,\,b) = a\,\,\bigtriangleup_{\sigma+1} (b-1)" align="middle" /><br />
<br />
which is again a simple exercise.<br />
<br />
<br />
now we have to make some odd observations that may get confusing<br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_{1+S(\sigma)}^{\sigma + \ell(\sigma)}((1 + S(\sigma))\,\,\bigtriangleup_{\sigma + \ell(\sigma) + 1}\,\,(1+S(\sigma + \ell(\sigma) + 1)) = (1 + S(\sigma))\,\,\bigtriangleup_{\sigma + \ell(\sigma) + 1}\,\,S(\sigma + \ell(\sigma) + 1) = 1 + S(\sigma)" align="middle" /><br />
<br />
from the "logarithmic" inverse law<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(a\,\,\bigtriangleup_{\sigma+1}\,\,b) = a\,\,\bigtriangleup_{\sigma+1} (b-1)" align="middle" /><br />
<br />
this implies:<br />
<img src="/cgi-bin/mimetex.cgi?L_{1+S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma)) + Q^{\sigma + \ell(\sigma)}(1, 1 + S(\sigma), 1 + S(\sigma + \ell(\sigma) + 1)) = 1 + S(\sigma)" align="middle" /><br />
<br />
which in turn gives<br />
<br />
<img src="/cgi-bin/mimetex.cgi?Q^{\sigma + \ell(\sigma)}(1, 1 + S(\sigma), 1 + S(\sigma + \ell(\sigma) + 1)) = 1" align="middle" /><br />
<br />
since<br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_{1+S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma)) = S(\sigma + \ell(\sigma)) = S(\sigma)" align="middle" /><br />
<br />
by the identity<br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(a) = S(\sigma)" align="middle" /><br />
<br />
<br />
but we also know by Q's definition<br />
<img src="/cgi-bin/mimetex.cgi?Q^{\sigma}(t, a, b) = L_a^{\sigma}(a\,\,\bigtriangleup_{\sigma + t}\,\,b) - L_a^{\sigma}(b)" align="middle" /><br />
<br />
so that<br />
<br />
<img src="/cgi-bin/mimetex.cgi?Q^{\sigma + \ell(\sigma)}(1, 1 + S(\sigma), 1 + S(\sigma + \ell(\sigma) + 1)) = 1 = L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}((1 + S(\sigma))\,\,\bigtriangleup_{1 + \sigma + \ell(\sigma)} (1 + S(\sigma + \ell(\sigma) + 1))) - L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma + \ell(\sigma) + 1))" align="middle" /><br />
<br />
we know that from earlier<br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}((1 + S(\sigma))\,\,\bigtriangleup_{1 + \sigma + \ell(\sigma)} (1 + S(\sigma + \ell(\sigma) + 1))) = 1 + S(\sigma)" align="middle" /><br />
<br />
so that we get<br />
<br />
<img src="/cgi-bin/mimetex.cgi?1 = 1 + S(\sigma)  - L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma + \ell(\sigma) + 1))" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma) = L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma + \ell(\sigma) + 1))" align="middle" /><br />
<br />
but we also know, by the definition of the "logarithmic" inverse:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_\sigma\,\,L_a^{\sigma}(b) = b" align="middle" /><br />
<br />
so therefore<br />
<br />
<img src="/cgi-bin/mimetex.cgi?(1 + S(\sigma))\,\,\bigtriangleup_{\sigma + \ell(\sigma)}\,\,S(\sigma) = (1 + S(\sigma))\,\,\bigtriangleup_{\sigma + \ell(\sigma)}\,\,L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma + \ell(\sigma) + 1))" align="middle" /><br />
<br />
which since<br />
<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma) = S(\sigma + \ell(\sigma))" align="middle" /><br />
<br />
and<br />
<br />
<img src="/cgi-bin/mimetex.cgi?(1 + S(\sigma))\,\,\bigtriangleup_{\sigma + \ell}\,\,L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma + \ell(\sigma) + 1)) = 1 + S(\sigma + \ell(\sigma) + 1)" align="middle" /><br />
<br />
we get the beautiful result:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma) = S(\sigma + \ell(\sigma) + 1)" align="middle" /><br />
<br />
which is the core identity of our proof<br />
<br />
since <img src="/cgi-bin/mimetex.cgi?\ell(\sigma)" align="middle" /> is multivalued, we find that<br />
<br />
for some value <img src="/cgi-bin/mimetex.cgi?q &gt; 0" align="middle" /><br />
<br />
if<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma + q) = S(\sigma)" align="middle" /><br />
<br />
then<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma + q + 1) = S(\sigma)" align="middle" /><br />
<br />
which in turn applies<br />
<br />
<img src="/cgi-bin/mimetex.cgi?\ell(\sigma) = q + 1" align="middle" /><br />
<br />
which iterated gives <img src="/cgi-bin/mimetex.cgi?k \in \mathbb{Z} ; k \ge 0" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma + q + k) = S(\sigma + q)" align="middle" /><br />
<br />
or that <img src="/cgi-bin/mimetex.cgi?S(\sigma + q)" align="middle" /> is periodic.  However we already noted that if <img src="/cgi-bin/mimetex.cgi?S(\sigma)" align="middle" /> is periodic then <img src="/cgi-bin/mimetex.cgi?S(\sigma)" align="middle" /> cannot be analytic since by definition of an analytic function that is periodic it must be periodic over every value in the domain and <img src="/cgi-bin/mimetex.cgi?S(1) \neq S(2)" align="middle" /><br />
<br />
Therefore<br />
<br />
<img src="/cgi-bin/mimetex.cgi?L_a^{\sigma}(b)" align="middle" /> is not analytic by <img src="/cgi-bin/mimetex.cgi?\sigma" align="middle" /> for all values of b<br />
<br />
and <br />
<br />
<img src="/cgi-bin/mimetex.cgi?S(\sigma)" align="middle" /> is not analytic by <img src="/cgi-bin/mimetex.cgi?\sigma" align="middle" /><br />
<br />
Q.E.D.<br />
<br />
<br />
This puts me one step closer to proving non analycity across <img src="/cgi-bin/mimetex.cgi?\sigma" align="middle" /> for some value of a and b in the Ackermann function <img src="/cgi-bin/mimetex.cgi?\vartheta(a,b,\sigma)" align="middle" /><br />
<br />
Personally I think that non-analycity is intuitively implied since <img src="/cgi-bin/mimetex.cgi?\vartheta(a,b,\sigma)" align="middle" /> is not primitive recursive, but this is yet to be rigorously decided.  To disprove universal analycity, it suffices to show that for some fixed <img src="/cgi-bin/mimetex.cgi?u, v \in C" align="middle" />, <img src="/cgi-bin/mimetex.cgi?\vartheta(u,v,\sigma)" align="middle" /> is not analytic across <img src="/cgi-bin/mimetex.cgi?\sigma" align="middle" />.<br />
<br />
I understand the notation in this proof maybe a bit confusing, but when working with three variables I think that's inevitable.<br />
<br />
any questions, comments?]]></content:encoded>
		</item>
		<item>
			<title><![CDATA[Fixpoints of the dxp() - has there been discussion about it?]]></title>
			<link>http://math.eretrandre.org/tetrationforum/showthread.php?tid=704</link>
			<pubDate>Thu, 10 Nov 2011 19:29:07 +0000</pubDate>
			<guid isPermaLink="false">http://math.eretrandre.org/tetrationforum/showthread.php?tid=704</guid>
			<description><![CDATA[Hi, <br />
<br />
 I was just reviewing some properties of the dxp-function, the map: x-&gt; b^x-1 .        <br />
<br />
For functions without constant terms it is widely common to accept the fractional iterate via the powerseries generated by fractional matrix-powers as "natural" embedding into the continuous flow -  and to ignore issues of other fixpoints (besides the fixpoint zero). However, I looked at the powerseries, for instance 2^x-1, developed at the other fixpoint - and not surprisingly, the fractional iterates differ in the range for x, where the range of convergence of the different power series for their schroeder-functions overlap, as well as we've discussed this with x-&gt; b^x  (where b=e or b=sqrt(2) etc) for instance in the "bummer" thread.      <br />
<br />
So at this moment I became curious about the common acceptance of the "natural" embedding of iterations of 2^x-1 or exp(x)-1 or in general of functions having f(0)=0 and f'(0)&lt;&gt;0 in a flow based on their powerseries around zero.       <br />
<br />
Has someone come across that discussion in the context of the dxp? A discussion about the "wobbling" - the sinusoidal different solutions per unit-height-iteration-interval when centered around the different fixpoints? And an explicite justification for the common choice?]]></description>
			<content:encoded><![CDATA[Hi, <br />
<br />
 I was just reviewing some properties of the dxp-function, the map: x-&gt; b^x-1 .        <br />
<br />
For functions without constant terms it is widely common to accept the fractional iterate via the powerseries generated by fractional matrix-powers as "natural" embedding into the continuous flow -  and to ignore issues of other fixpoints (besides the fixpoint zero). However, I looked at the powerseries, for instance 2^x-1, developed at the other fixpoint - and not surprisingly, the fractional iterates differ in the range for x, where the range of convergence of the different power series for their schroeder-functions overlap, as well as we've discussed this with x-&gt; b^x  (where b=e or b=sqrt(2) etc) for instance in the "bummer" thread.      <br />
<br />
So at this moment I became curious about the common acceptance of the "natural" embedding of iterations of 2^x-1 or exp(x)-1 or in general of functions having f(0)=0 and f'(0)&lt;&gt;0 in a flow based on their powerseries around zero.       <br />
<br />
Has someone come across that discussion in the context of the dxp? A discussion about the "wobbling" - the sinusoidal different solutions per unit-height-iteration-interval when centered around the different fixpoints? And an explicite justification for the common choice?]]></content:encoded>
		</item>
		<item>
			<title><![CDATA[extension of the Ackermann function to operators less than addition]]></title>
			<link>http://math.eretrandre.org/tetrationforum/showthread.php?tid=703</link>
			<pubDate>Sun, 06 Nov 2011 00:31:48 +0000</pubDate>
			<guid isPermaLink="false">http://math.eretrandre.org/tetrationforum/showthread.php?tid=703</guid>
			<description><![CDATA[Well the result is surprisingly simple, and is derived from a single defining axiom.  I feel if there are any qualms with the proof it comes from the axiom itself.<br />
<br />
we assume for now that all variables can be extended to the complex domain<br />
<br />
<img src="/cgi-bin/mimetex.cgi?a, b, \sigma \in \text{C}" align="middle" /><br />
<br />
and thus, the Ackermann series of operators are defined by the single property:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?\text{(I):}\,\,a\,\,\bigtriangleup_\sigma\,\,(a\,\,\bigtriangleup_{\&#8203;sigma + 1}\,\,b) = a\,\,\bigtriangleup_{\sigma + 1} (b+1)" align="middle" /><br />
<br />
starting the Ackermann function with addition at 1 gives us:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_1\,\,b = a + b" align="middle" /><br />
<br />
therefore, if we plug in from our defining axiom (I) and try to solve for when sigma is zero we get:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_0\,\,(a\,\,\bigtriangleup_1\,\,b) = a\,\,\bigtriangleup_1\,\, (b + 1)" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?\text{(II):} a\,\,\bigtriangleup_0\,\,(a + b) = a + b + 1" align="middle" /><br />
<br />
<br />
with this, we try to solve for the more general:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_0\,\,b = a\,\,\bigtriangleup_0\,\,(a + (b - a))" align="middle" /><br />
<br />
which by (II) gives<br />
<br />
<img src="/cgi-bin/mimetex.cgi?\text{(III):}\,\,a\,\,\bigtriangleup_0\,\,b = a + b - a + 1 = b + 1" align="middle" /><br />
<br />
this was derived from only a single defining axiom and no other assumptions were made.  But that's not it, we can continue this sequence and try to solve for when sigma is negative one.<br />
<br />
Again, we'll start with the defining axiom:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_{-1}\,\,(a\,\,\bigtriangleup_{0}\,\,b) = a\,\,\bigtriangleup_{0}\,\,(b+1)" align="middle" /><br />
<br />
of course, by (III) we have the simple formula:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_{-1}\,\,(b+1) = b+2" align="middle" /><br />
<br />
which of course, gives the fantastic equation:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_{-1}\,\,b = b + 1" align="middle" /><br />
<br />
now, by induction, we can make the complete argument:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?k \in \mathbb{Z}; k \le 0" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_k\,\,b = b + 1" align="middle" /><br />
<br />
Or basically, we have the result that every operator less than addition which is equal to an integer is the equivalent to successorship.<br />
<br />
Q.E.D.<br />
<br />
Now if we add the additional property that sigma plus one be an iteration count of sigma, we run into many problems.  And I will shortly argue here:<br />
<br />
The second axiom which need not be in play is the axiom of iteration given as<br />
<br />
<img src="/cgi-bin/mimetex.cgi?n \in \mathbb{Z}; n &gt; 0" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?\text{(IV):}\,\,a\,\,\bigtriangleup_{\sigma + 1}\,\,n = a_1\,\,\bigtriangleup_{\sigma}\,\,(a_2\,\,\bigtriangleup_{\sigma} \,\, (a_3 \,\, \bigtriangleup_{\sigma}\,\,...\,\,(a_{n-1}\,\, \bigtriangleup_\sigma \,\,a_n)" align="middle" /><br />
<br />
with this axiom we instantly have a contradiction with the first axiom when we extend operators less than addition:<br />
<br />
by (III)<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_{0}\,\,a = a + 1" align="middle" /><br />
<br />
by(IV)<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_{0}\,\,a = a + 2" align="middle" /><br />
<br />
therefore if we allow the axiom of iteration we cannot extend the Ackermann function to operators less than addition.<br />
<br />
Furthermore, if we allow the axiom of iteration we also have the cheap result:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?x \in \mathbb{R}; x \ge 2" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_x\,\,1 = a" align="middle" /><br />
<br />
this causes the function<br />
<br />
<img src="/cgi-bin/mimetex.cgi?f(\sigma) = a\,\,\bigtriangleup_\sigma\,\,1" align="middle" /> to no longer possibly be analytic.<br />
<br />
we can however accept a modified axiom of iteration which is written as:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?n \in \mathbb{Z}; n &gt; 1" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?\text{(V):}\,\,a\,\,\bigtriangleup_{\sigma + 1}\,\,n = a_1\,\,\bigtriangleup_{\sigma}\,\,(a_2\,\,\bigtriangleup_{\sigma} \,\, (a_3 \,\, \bigtriangleup_{\sigma}\,\,...\,\,(a_{n-1}\,\, \bigtriangleup_\sigma \,\,a_n)" align="middle" /><br />
<br />
this allows for different identities at non-integer values of sigma, however, this still disallows sigma to be extended to values less than addition.<br />
<br />
So, I leave it open ended, hoping someone else has some comment.  Do we use axioms (I) and (IV), or (I) and (V), or, as I prefer, (I) alone?]]></description>
			<content:encoded><![CDATA[Well the result is surprisingly simple, and is derived from a single defining axiom.  I feel if there are any qualms with the proof it comes from the axiom itself.<br />
<br />
we assume for now that all variables can be extended to the complex domain<br />
<br />
<img src="/cgi-bin/mimetex.cgi?a, b, \sigma \in \text{C}" align="middle" /><br />
<br />
and thus, the Ackermann series of operators are defined by the single property:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?\text{(I):}\,\,a\,\,\bigtriangleup_\sigma\,\,(a\,\,\bigtriangleup_{\&#8203;sigma + 1}\,\,b) = a\,\,\bigtriangleup_{\sigma + 1} (b+1)" align="middle" /><br />
<br />
starting the Ackermann function with addition at 1 gives us:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_1\,\,b = a + b" align="middle" /><br />
<br />
therefore, if we plug in from our defining axiom (I) and try to solve for when sigma is zero we get:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_0\,\,(a\,\,\bigtriangleup_1\,\,b) = a\,\,\bigtriangleup_1\,\, (b + 1)" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?\text{(II):} a\,\,\bigtriangleup_0\,\,(a + b) = a + b + 1" align="middle" /><br />
<br />
<br />
with this, we try to solve for the more general:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_0\,\,b = a\,\,\bigtriangleup_0\,\,(a + (b - a))" align="middle" /><br />
<br />
which by (II) gives<br />
<br />
<img src="/cgi-bin/mimetex.cgi?\text{(III):}\,\,a\,\,\bigtriangleup_0\,\,b = a + b - a + 1 = b + 1" align="middle" /><br />
<br />
this was derived from only a single defining axiom and no other assumptions were made.  But that's not it, we can continue this sequence and try to solve for when sigma is negative one.<br />
<br />
Again, we'll start with the defining axiom:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_{-1}\,\,(a\,\,\bigtriangleup_{0}\,\,b) = a\,\,\bigtriangleup_{0}\,\,(b+1)" align="middle" /><br />
<br />
of course, by (III) we have the simple formula:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_{-1}\,\,(b+1) = b+2" align="middle" /><br />
<br />
which of course, gives the fantastic equation:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_{-1}\,\,b = b + 1" align="middle" /><br />
<br />
now, by induction, we can make the complete argument:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?k \in \mathbb{Z}; k \le 0" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_k\,\,b = b + 1" align="middle" /><br />
<br />
Or basically, we have the result that every operator less than addition which is equal to an integer is the equivalent to successorship.<br />
<br />
Q.E.D.<br />
<br />
Now if we add the additional property that sigma plus one be an iteration count of sigma, we run into many problems.  And I will shortly argue here:<br />
<br />
The second axiom which need not be in play is the axiom of iteration given as<br />
<br />
<img src="/cgi-bin/mimetex.cgi?n \in \mathbb{Z}; n &gt; 0" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?\text{(IV):}\,\,a\,\,\bigtriangleup_{\sigma + 1}\,\,n = a_1\,\,\bigtriangleup_{\sigma}\,\,(a_2\,\,\bigtriangleup_{\sigma} \,\, (a_3 \,\, \bigtriangleup_{\sigma}\,\,...\,\,(a_{n-1}\,\, \bigtriangleup_\sigma \,\,a_n)" align="middle" /><br />
<br />
with this axiom we instantly have a contradiction with the first axiom when we extend operators less than addition:<br />
<br />
by (III)<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_{0}\,\,a = a + 1" align="middle" /><br />
<br />
by(IV)<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_{0}\,\,a = a + 2" align="middle" /><br />
<br />
therefore if we allow the axiom of iteration we cannot extend the Ackermann function to operators less than addition.<br />
<br />
Furthermore, if we allow the axiom of iteration we also have the cheap result:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?x \in \mathbb{R}; x \ge 2" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?a\,\,\bigtriangleup_x\,\,1 = a" align="middle" /><br />
<br />
this causes the function<br />
<br />
<img src="/cgi-bin/mimetex.cgi?f(\sigma) = a\,\,\bigtriangleup_\sigma\,\,1" align="middle" /> to no longer possibly be analytic.<br />
<br />
we can however accept a modified axiom of iteration which is written as:<br />
<br />
<img src="/cgi-bin/mimetex.cgi?n \in \mathbb{Z}; n &gt; 1" align="middle" /><br />
<br />
<img src="/cgi-bin/mimetex.cgi?\text{(V):}\,\,a\,\,\bigtriangleup_{\sigma + 1}\,\,n = a_1\,\,\bigtriangleup_{\sigma}\,\,(a_2\,\,\bigtriangleup_{\sigma} \,\, (a_3 \,\, \bigtriangleup_{\sigma}\,\,...\,\,(a_{n-1}\,\, \bigtriangleup_\sigma \,\,a_n)" align="middle" /><br />
<br />
this allows for different identities at non-integer values of sigma, however, this still disallows sigma to be extended to values less than addition.<br />
<br />
So, I leave it open ended, hoping someone else has some comment.  Do we use axioms (I) and (IV), or (I) and (V), or, as I prefer, (I) alone?]]></content:encoded>
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	</channel>
</rss>
