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 Taylor series of upx function Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 12/05/2008, 12:30 AM sheldonison Wrote:What does this equation mean? $\mathrm{upx}(z^*)= \mathrm{upx}(z)^*$The asterisc * means the complex conjugation. You can evaluate function at the complex–conjugated argument, and the result is the conjugated value of the function. In particular, for real values of the argument, the function is real. sheldonison Wrote:I was expecting more along the lines of upx(x+1) = exp ( upx(x) )?? These are not just lines, this is all the range of complex values of x, except part of the real axis $x\le -2$. sheldonison Wrote:The next line is also intriguing.... $\mathrm{upx'}(x)>0 \forall x>-2$ Does this mean that other holomorphic solutions exist, but they all must have a negative derivative at some point x>-2?No, but condition that the derivative is positive simplifies my proof of the uniqueness. Recently, Henryk Trappmann suggested the proof through the inverse function; it does not use the requirement that, at the real argument, the derivative is positive. I see no errors in his proof. So, we may remove this claim about the first derivative. sheldonison Wrote:I started working on the algebra to generate the general equations for all 5th and 6th order upx approximations that have smooth 1st, 2nd, and 3rd derivatives, but not necessarily smooth 4th and 5th derivatives. You would claim that all members of that set, if extended with higher order coefficients so that all derivatives would be smooth, would have a negative 1st derivative somewhere?Your question is not clear. Instead of "somewhere" you could say "at some real value of the argument, larger, than $-2$". Then the answer is "not". You may consider the function $ f(x)=\mathrm{upx}(x+s\cdot\sin(2 \pi x))$ at small real values of parameter $s$, id est, $-1. Smoothness of all the derivatives at the real axis is not sufficient for the uniqueness. We have to claim the holomorphism at least for the domain with real part of the argument larger than $-2$. I have pictures for the singular solutions too. Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 02/18/2009, 05:59 AM (This post was last modified: 02/18/2009, 06:06 AM by Kouznetsov.) Ansus Wrote:So your solution is universal?Yes. Ansus Wrote:- Is there compact analytic expression for Taylor serie coefficients? No. Until now, only the integral equation. Ansus Wrote:- Can you find analytic expressions for some particular values of tetration (with natural base) and its derivatives, for example, its derivative in 0 (= derivative in 1)?In some sense, the representation through the Cauchi integral equation is analytic. However, I would not say that it is very "compact" or explicit. Ansus Wrote:- Do you derive new properties of the function?I derive the asymptotic behavior. See http://en.citizendium.org/wiki/Tetration Are these properties new? Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 02/18/2009, 04:03 PM Ansus Wrote:Quote:No. Until now, only the integral equation. It it possible to compute Tetration in Mathematica using your method?I would not recommend Mathematica for the numerical evaluation of the Cauchi integrals; it is slow. For the base e, the Taylor expansions are already calculated; with these expanssions, the evaluation of tetration and its derivatives is fast. You may use the code from http://en.citizendium.org/wiki/Tetration...l.jpg/code Quote:Is it possible to find an expression of, say $(\text{sexp}_e x)'$ in 0? Why I ask it? Because there is a property: $\ln f'(x)=f(-1)+f(0)+...+f(x-1) + \ln f'(-1)$ for tetration with natural base $\text{sexp}_e x$ and integer x (note also that $\ln f'(-1)=\ln f'(0)$). I cannot yet answer. I hope to answer this question later. Quote: did you discover some new functional/differential equations that tetration may satisfy, such as change-of-base formula, properties of derivatives/integral or relationship with other functions. Not yet. Use the code above, try to guess some properties and try to prove them. Quote:By the way, can you plot a graph of tetration $\text{sexp}_{\sqrt 2} x$? Yes. Red dotted curve at http://en.citizendium.org/wiki/Image:TetrationReal.jpg Quote:Is it symmetric against y=-x line? Wow! I did not check.. Perhaps, I should. Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 02/19/2009, 02:06 AM Ansus Wrote:Is it possible to find an expression of, say $(\text{sexp}_e x)'$ in 0? Why I ask it? Because there is a property: $\ln f'(x)=f(-1)+f(0)+...+f(x-1) + \ln f'(-1)$Dear Ansus. Do you mean that $f$=sexp ? I cannot understand the equation you wrote. The similar equation I can deduce is $\ln(f'(x))=f(-1)-f(0)+f(x-1) + \ln(f'(x-1))$, but I do not see, how does it help to get a simple expression for $f'(0)$. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/19/2009, 03:27 PM (This post was last modified: 03/03/2009, 12:46 PM by bo198214.) Ansus Wrote:Is it symmetric against y=-x line? No it is not. We already saw here that the 3 rules: $f(0)=1$ $f(x+1)=\exp_{\sqrt{2}}(f(x))$ $f(-f(x))=-x$ imply a non-smooth graph. Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 02/19/2009, 10:17 PM Ansus Wrote:Well I cannot deduce your expression.My expression is $\ln(f'(x))=f'(-1)-f'(0)+f(x-1) + \ln(f'(x-1))$ It can be deduced in the following way: $ \ln\left( f'(x) \right) = \ln\left( \frac{{\rm d}}{{\rm d} x} f(x) \right) = \ln\left( \frac{{\rm d}}{{\rm d} x} \exp(f(x-1)) \right) = \ln\left( \exp(f(x-1)) \cdot f'(x-1) \right) = \ln\Big( \exp(f(x-1)) \Big) + \ln\Big( f'(x-1) \Big) = f(x-1)+ \ln\Big( f'(x-1) \Big) = f(x-1)+ \ln\Big( f'(x-1) \Big) + f'(-1)-f'(0)$ , because $f'(-1)=f'(0)$. Ansus Wrote:I meant sexp with natural base. What I can derive is: $ \ln \frac{f'(x)}{f'(0)} = f(0)+...+f(x-1)$What is sense of ... in this expression? How does Ansus get this espressionan? Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 02/20/2009, 08:43 AM Ansus Wrote:... $ f_a(x+1)=a^{f_a(x)}$ differentiate $ f'_a(x+1)=a^{f_a(x)}\ln a f'_a(x)$ induction $ \frac{f'_a(x)}{f'_a(0)(\ln a)^x}=\prod_{k=0}^x f_a(k)$ ...The last statement looks doubtful. Is it for real values of $x>-2$? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/20/2009, 11:44 AM Ansus Wrote:You can compare your solution with a new analytic solution suggested here: http://math.eretrandre.org/tetrationforu...hp?tid=237 Dmitrii uses two distinct methods for $b>e^{1/e}$ and for $b. I think he uses regular iteration for $b so for $b=\sqrt{2}$ your graphs should be equal, considering what I replied to your newton method. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 02/24/2009, 04:22 AM Ansus Wrote:By the way, can you plot a graph of tetration $\text{sexp}_{\sqrt 2} x$? Is it symmetric against y=-x line? Just wanted to remind that we also discussed this here as well. Andrew Robbins Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 03/03/2009, 05:41 AM Ansus Wrote:Dmitriy, can you please give a mathematica notebook for your method and base Sqrt(2)? b=Sqrt[2] RGB[x_,y_,z_]=RGBColor[x,y,z] M=9 f=2-y+Sum[A[n]*y^n,{n,2,M}] g=Series[ReplaceAll[f,y\[Rule]y*Exp[q]],{y,0,M}] h=Simplify[Series[Exp[Log[b]*f],{y,0,M}]] u=g-h c1=Coefficient[u,y] s1=Solve[c1\[Equal]0,q] s11=Extract[s1,1] Q=ReplaceAll[q,s11] v[1]=Simplify[ReplaceAll[u,s11]] For[i=2,i

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