• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 Matrix Operator Method Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 08/26/2007, 10:54 PM (This post was last modified: 08/27/2007, 07:54 AM by Gottfried.) bo198214 Wrote:As Andrew mentioned the Carlemann-Matrix is transposed to the Bell-Matrix (which is basicly the power derivation matrix but without the 0th power and without the 0-th coefficent), thatswhy I asked about the Carlemann-Matrix. Indeed for a parabolic fixed point at 0 your formula (in PDM form) $P_{f^{\circ t}}=\exp(t\cdot\log(P_f))$ gives the regular iteration. So this a wonderful idea to simply extend this also to series with no fixed point. Hendryk - first I want to say: I feel very comfortable to see this connection now. I couldn't have express this relation myself (although Andrew had already pointed this out to me). bo198214 Wrote:But what is still not clear to me (I certainly am not that acquainted with matrices) how fits the Eigenanalysis in here. Perhaps you can also answer a question that recently occured to me: In Maple there is a function "MatrixFunction" where you can apply an arbitrary real function to a matrix, by somehow manipulating the Eigenvalues. Maybe, the eigenanalysis is in relation to the fixed-point-approach, while the matrix-logarithm is an alternative, where eigenvalues are degenerate and/or not accessible (as, for instance with the C-matrix for "exp(x)-1" -iteration, see below). -------------- Something about eigensystems, I hope this fits your remark. There are two ways to introduce the eigensystem-idea. (a) One is the question: is there any vector, which, when multiplied with a matrix A, makes the result equal to itself? In matrix-notation, with a matrix A and a column-vector X: A * X = X This may be extended to allow an arbitrary scaling v (the associated eigenvalue) of X, so for instance A * X = v * X If X occurs in a context, where its entries are to be understood as euclidean coordinates, then this means, that X is a vector, whose direction is not changed by this specific matrix-multiplication, and thus expresses an interesting characteristic also of the matrix A. (b) Another way is the question: if I want to analyze functions on A, like for instance powers, then all entries change. But in this mathematical complex entity of a matrix: can we separate constant parts from variable parts and find a description for A and its powers, where the variable part is optimally reduced, for instance factored out? For instance, in the above example A * X = 1 * X it means, that also A*A*X = 1 * X or A^2 * X = 1 * X and A and its powers have in this sense a common characteristic. For a wide set of (square) matrices there is this amazing property, that A and its powers can be expressed by a multiplication of two constant parts (specific to A, however) and the variation by functions applied to A occurs only in a diagonal-matrix D so that A = Q * D * Q^-1 Q is here the square-matrix of eigenvectors, and D is a diagonal-matrix containing the eigenvalues. (Note, that not all matrices can be "diagonalized" this way, but Jordan proved, that -shortly sketched- all square matrices can be reduced to one at least near diagonal canonical form) The advantage of this matrix-diagonalization is then, that sums, products, powers, series of powers (like exponential-function, sinus, etc) can be expressed in terms of the function, when applied to the diagonal-matrix only. A simple example is: A^2 = A * A = (Q * D * Q^-1)*(Q * D * Q^-1) = Q * D*D * Q^-1 = Q*D^2*Q^-1 and all except the diagonal-matrix remains constant. Sums or series can be evaluated this nice way as long as all the matrices in the formula have the same eigenvectors Q, - which is given, if the series consists, for example, of powers of A only, or are the identity-matrix I. (Regarding your Maple-question: I don't know Maple, but possibly the function call is just a request for diagonalization, possibly caching the eigen-matrices and applying the function to the diagonal) ---------------------------- The eminent advantage of this is (regardless of applying the logic of the approaches (a) or (b)), that we can apply any such functions to a diagonal-matrix, by simply applying it to the entries of the diagonal, and can thus reduce the very complex task of applying a function to a matrix to the simple calculation of the same function applied to the scalar values of the diagonal-matrix D. This includes then fractional, real or complex powers, exponentials, logarithms to matrices (which were intractable otherwise), but also geometric series and the like. The explicite application of a function like multiplying a matrix A times itself (integral powers or even inversion as negative power), computing the exponential or the matrix-logarithm by the common series-expansion like A^2 = A*A or exp(A) = I + A + A*A/2! + A*A*A/3! + ... and A^2 = Q*D*D*Q^-1 or exp(A) = Q*( I + D + D^2/2! + D^3/3! + ...)*Q^-1 is then in one-to-one relation: doing it this or that way is completely equivalent and leads to the same solution. Only the numerical aspect is different in terms of numerical accuracy and may play a role this way. Also the two motivations (finding invariant (except scaling) vectors vs. finding the eigensystem-decomposition) principally lead to the same solutions, except for numerical advantages or disadvantages due to the order of computation. So if we find a tetration-solution via the explicit matrix-logarithm or via the eigensystem-decomposition is only a matter of convenience and of numerical stability. However, this is not exactly true. There are "borderline"-cases. If eigenvalues are of similar value, or even equal value, or if some are extremely high and others near at zero, or even worse: exactly zero, then the eigensystem-analysis is numerically instable or impossible, or at least indeterminate. So, if an eigenvalue is zero, then the inversion is impossible because inverting zero would result in infinity values. Or if some eigenvalues are equal, then their associated eigenvectors (the columns in Q) are indeterminate, at least in a rotational sense. Then general eigen-decomposition-routines mostly give up and don't provide a usable, uniquely defined result. This is for instance the case for the C-matrix for base e, which implements the (exp(x)-1) -iteration: all eigenvalues are equal and actually equal to 1. An eigensystem-solver refuses his job, and we are lost. But in this case we may apply the matrix-logarithm-procedure implemented by the naive series using A1 = (A-I) and then log(A) = A1 - A1*A1/2 + A1*A1*A1/3 - ... or some more enhanced versions, and find the matrix-logarithm this way. For infinite matrices we need approximations by finite dimensions, and it is not always obvious, which procedure gives the better approximation. Will the fractional power of matrix A by A^0.5 = Q * D^0.5 * Q^-1 or by A^0.5= exp(0.5 * log(A)) give the better/the more convenient approximation, if we have only dimension dim=32 accessible? Theoretically the results must be equal, but numerically one may be preferable. And if we have equal eigenvalues (which may happen for base e in the "exp(x)-1" -tetration or base e^(1/e) in the standard tetration-iteration) eigensystem-solvers should be principally helpless because of the internal indeterminacy of the associated eigenvectors in our matrices. ----------------------------------- Well, I made a long comment up to here. This is also due to Daniel's request in the other thread for hints, how his method is related to others, for which I may give a guess: Since the idea of a "fixpoint" is the engine of his derivations, I assume, that his way is (at least implicitely) equivalent to the eigensystem analysis, whose one question was: is there one vector X which remains unchanged by matrix-multiplication A * X = X * 1 which would then simply reflect the computation of and the dependency of the eigenvector associated to the eigenvalue 1. But this is only a guess, since I didn't understand Daniel's method until now. But - if this is the case, then the results should be identical to that of the matrix-exponential-method (besides optimality of approximation and "borderline"-cases), since they are mutually one-to-one exchangable by matrix-theory. ----------------------------------------------- One more point, to make this post even one paragraph longer and to give an example for a consideration of this eigenvalue-properties in our context. In my first contact with tetration, where I derived the method for computing numerical approximations for infinite series of powertowers of increasing height (as(s) means alternating sum of powertowers of like base s) as(s) = 1 - s + s^s - s^s^s + s^s^s^s - ... + ... + s^^k - s^^(k+1) +...- I came across this eigensystem-problem first time. 1) I described this as() as the result of V(1)~ * (Bs^0 - Bs^1 + Bs^2 - Bs^3 + ... - ...) = Y~ and y = Y[1] should be the result, if this can be summed at all. 2) The parenthese represents the alternating geometric series of Bs, and according to the remarks about the eigensystem above, this can be rewritten by the formula for geometric series: V(1)~ * (I + Bs)^-1 = V(y)~ This involves a matrix-inversion; and a judgement, whether this is principally possible (for the ideal infinite case, not only in finite approximations), involves (at least) consideration of the eigenvalues. 3) From numerical approximations with increasing dimensions I could observe: for s in the admissible range for infinite powertowers 1/e^e+eps < s < e^(1/e)-eps the eigenvalues of the matrix Bs approximated to the consecutive powers of log(h(s)), where h(s) is Ioannis Galidakis' h()-function, defined implicitely by t^(1/t) = s ====> h(s)= t Then let v = log(t). The set of eigenvalues converge (heuristically) to the set of (1,v,v^2,v^3,...) So I took(and take) this as a hypothesis for the Bs-matrices of those parameters s, with which I worked on. 4.a) for abs(v)<1 this means, we have in the limit for infinite dimension eigenvalues lim{k->inf} v^k = 0, so the used matrix Bs is *principally* not invertible due to the occurence of 1/0 in the infinite case. But what has to be inverted, when discussing this series, is not Bs but (I + Bs), so the limit for the smallest eigenvalue of this sum is 1, and (I + Bs) is invertible. Then in the inverse M=(I + Bs)^-1 we have eigenvalues between 1 and 1/2 and the matrix should behave very well (which was well backed by the numerical computations) 4.b) Even if v is negative but abs(v)<1 , so 1/e < t < 1 and also 1/e^e < s < 1 this should thus work.(in fact the terms were more difficult to manage, they seemed to diverge, or at least to increase and oscillate in their sign) 4.c) For s> e^(1/e) we have for finite dimensions only a subset of the eigenvalues stabilizing to fixed values > 0, and another subset of empirical eigenvalues, whose characteristic I was not able to derive; but they seemed to be in the range 0 < v_k without upper bound. But again here: the addition of the identity-matrix in the geometric-sum-formula shifts the smallest eigenvalue up to 1 as the new minimum, and the inversion provides then a comfortably behaving matrix M, which allows summation of the described form for s beyond the (in-)famous limit e^(1/e) This allowed then (in my opinion) the conjectures concerning approximation of series like 1 - 10 + 10^^2 - 10^^3 + 10^^4 - ... + ... = as(10) to values about as(10) ~ 0.164280... as a non-completely-unfounded estimate ;-) -------------- I keep the further details away here, because this is not the focus of this post. The last paragraphs should serve as a more intense introduction into one specific consideration, where the eigenvalues-concept should be unavoidable for argumentation besides its usability for the approximation and analytical description of contiunuous tetration. (If you don't know this article already but want to read into it, it is here: http://go.helms-net.de/math/binomial/10_...rtower.pdf It's not much sophisticated, to say the least; in fact some mathematicians found it horrible, but it was my first exploring, so...) To add one conclusive statement here: I expect, that all methods, which involve analysis of tetration *in terms of powerseries*, or more precisely of exponential-series, will arrive at the same values; they seem to be only different views into the same matter, and may have relative advantages under certain circumstances. For me the first place is eigenvalue-decomposition, then, where it is principally impossible or has indeterminacies, I use the matrix-logarithm method. The "fix-point"-discussion seems to involve/imply the eigensystem-approach, and the first rowvector of Q^-1, associated with the eigenvalue 1, seems just to reflect this fixpoint. For s = sqrt(2), t=2, it is just the vector [1,2,4,8,16,...] in arbitrary scaling which makes [1,2,4,8,16,...] * Bs = [1,2,4,8,16,... ] ( * 1 ) and "is an invariant vector under transformation by Bs" . So much as the current opinion of mine - Gottfried Gottfried Helms, Kassel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/27/2007, 08:29 AM Thanks for this repetition course, Gottfried, now I first time understand what you are talking about all the time I mean I still knew what an Eigenvector was from my base lessons, but it was never clear to me what you mean by Eigenanalysis etc, now this is clear. Did you anyway already realize that instead of $\exp(t\cdot\log(A))$ you can directly use the binomial series for computation? $A^t = \sum_{n=0}^\infty \left(t\\n\right) (A-I)^n$ Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 08/27/2007, 11:04 AM (This post was last modified: 08/27/2007, 09:16 PM by Gottfried.) bo198214 Wrote:Thanks for this repetition course, Gottfried, now I first time understand what you are talking about all the time I mean I still knew what an Eigenvector was from my base lessons, but it was never clear to me what you mean by Eigenanalysis etc, now this is clear.Well, thanks, so it was useful, although my terminology is often a bit handwaved... :-). Good! bo198214 Wrote:Did you anyway already realize that instead of $\exp(t\cdot\log(A))$ you can directly use the binomial series for computation? $A^t = \sum_{n=0}^\infty \left(t\\n\right) (A-I)^n$ No, but it looks very good. I'll give it a deeper look, thanks for the hint! [update] Well, in a second thought: in the form of $(A+I)^t = \sum_{n=0}^\infty \left(t\\n\right) A^n$ actually I used it extensely for the scalar and also for the vectorial case. Applied to a vector of type V(x) = [1,x,x^2,...] it is P * V(x) = V(x+1) where P is the pascal-matrix containing the binomial-coefficients. I dealt a lot with this, and analoguously I derived the fractional and complex powers of P for the general complex solution P^s * V(x) = V(x + s) via matrix-logarithm of P (which is an extremely basic object, btw! see this on a t-shirt :     and in wikipedia at "pascal-triangle" ) Then, with iterated application of P as an operator, just equivalently as described here related to tetration, one can find the interesting Faulhaber/Bernoulli-matrix and zeta/eta-values at negative exponents. Summing of like powers which may be seen as a preliminary training :-) Gottfried Gottfried Helms, Kassel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/27/2007, 11:35 AM Gottfried Wrote:bo198214 Wrote:Did you anyway already realize that instead of $\exp(t\cdot\log(A))$ you can directly use the binomial series for computation? $A^t = \sum_{n=0}^\infty \left(t\\n\right) (A-I)^n$No, but it looks very good. I'll give it a deeper look, thanks for the hint!But it seems as if A must have a diagonal of 1. So probably you have to convert A into a Jordan normal form first as you must do with the logarithm (or how do you compute the matrix logarithm?) Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 08/27/2007, 11:58 AM (This post was last modified: 08/27/2007, 09:19 PM by Gottfried.) bo198214 Wrote:Gottfried Wrote:bo198214 Wrote:Did you anyway already realize that instead of $\exp(t\cdot\log(A))$ you can directly use the binomial series for computation? $A^t = \sum_{n=0}^\infty \left(t\\n\right) (A-I)^n$No, but it looks very good. I'll give it a deeper look, thanks for the hint!But it seems as if A must have a diagonal of 1. So probably you have to convert A into a Jordan normal form first as you must do with the logarithm (or how do you compute the matrix logarithm?)If in a triangular matrix A the diagonal are all ones, the eigenvalues of (A-I) are zero. So I have to compute the matrix-logarithm by the ordinary series expression With A1 = A-I then log(A) = A1 - A1*A1/2 + A1*A1*A1/3 .... which is a nice exercise... since it comes out, that A1 is nilpotent and we can compute an exact result using only as many terms as the dimension of A is. For the infinite dimensional case one can note, that the coefficients are constant when dim is increased step by step, only new coefficients are added below the previously last row. So even the infinite case is analytically accessible. Gottfried Gottfried Helms, Kassel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/27/2007, 12:13 PM Gottfried Wrote:If in a triangular matrix A the diagonal are all ones, the eigenvalues of (A-I) are zero. So I have to compute the matrix-loagarithm by the ordinary series expression With A1 = A-I then log(A) = A1 - A1*A1/2 + A1*A1*A1/3 .... which is a nice exercise... since it comes out, that A1 is nilpotent and we can compute an exact result using only as many terms as the dimension of A is. Yes, and the diagonal is 1 in the parabolic case. Because the matrix $A-I$ is nilpotent also the binomial series $\sum_{k=0}^\infty \left(t\\k\right) (A-I)^k$ is finite and this is the way Jabotinsky derived the double binomial formula. Quote:For the infinite dimensional case one can note, that the coefficients are constant when dim is increased step by step, only new coefficients are added below the previously last row. So even the infinite case is analytically accessible. Yes, but how do you compute the infinite case. You can not simply do $\sum_{n=0}^\infty (-1)^{n+1} \frac{(A-I)^n}{n}$ as this does not converge. So do you compute it as described here via the Jordan form? Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 08/27/2007, 12:43 PM bo198214 Wrote:Gottfried Wrote:bo198214 Wrote:Did you anyway already realize that instead of $\exp(t\cdot\log(A))$ you can directly use the binomial series for computation? $A^t = \sum_{n=0}^\infty \left(t\\n\right) (A-I)^n$No, but it looks very good. I'll give it a deeper look, thanks for the hint!But it seems as if A must have a diagonal of 1. So probably you have to convert A into a Jordan normal form first as you must do with the logarithm (or how do you compute the matrix logarithm?) Second try. You say $\hspace{24} A^t = \sum_{n=0}^\infty \left(t\\n\right) (A-I)^n$ Let $\hspace{24} A = Q * D * Q^{-1}$ if A is diagonalizable. Also, since $\hspace{24} I = Q * I * Q^{-1}$ it is also $\hspace{24} (A - I) = Q *(D - I) * Q^{-1}$ Then the above means, for example t = 3 $\hspace{24} A^3 = Q * ( 1*(D -I)^0 + 3*(D-I)^1 + 3*(D-I)^2 + 1*(D-I)^3 ) * Q^{-1}$ Let's look at the parenthese only as the diagonal matrix E. The entries e_k can directly be computed: by binomial-theorem $\hspace{24} e_0 = ((d_0 - 1) + 1)^3$ $\hspace{24} e_1 = ((d_1 - 1) + 1)^3$ $\hspace{24}$ $\hspace{24} ...$ $\hspace{24} e_0 = d_0^3$ $\hspace{24} e_1 = d_1^3$ $\hspace{24}$ $\hspace{24} ...$ and $\hspace{24} E = D^3$ So the rhs is $\hspace{24} Q * E * Q^{-1} = Q * D^3 * Q^{-1}$ and the lhs is $\hspace{24} A^3 = Q * D^3 * Q^{-1}$ and the both sides are equal. Here was no assumption needed concerning the actual eigenvalues of A, except the assumption, that it is diagonalizable at all. It is perhaps important to note, that symbolically we may write formulae with equal eigenvalues even if a numerical solver would not produce a solution because of the mentioned indeterminacy of the eigenvectors. But this does not affect the analytical description. Well, here I assumed A is diagonalizable (with no assumptions about the actual value of its eigenvalues). If it is not, then at least it is possible to derive the Jordan-canonical form, and in A = Q*J*Q^-1 we get J non-diagonal, but in its Jordan-canonical form. All what was said about the diagonal form is still valid in principle, only that powers and functions on J have not the simple form as in the true diagonal case. But still it is a simpler computation than with the original matrix, especially if infinite series are assumed. Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 08/27/2007, 01:19 PM (This post was last modified: 08/27/2007, 01:25 PM by Gottfried.) bo198214 Wrote:Yes, but how do you compute the infinite case. You can not simply do $\sum_{n=0}^\infty (-1)^{n+1} \frac{(A-I)^n}{n}$ as this does not converge. So do you compute it as described here via the Jordan form? Hmm, I observed that I'm not good in answering correctly in speedy discussions. But may be I've got it right in this case. If the sequence of corresponding entries of the powers of (A-I)^n/n do not converge well with finite dimension, but at least alternate in sign, I try to approximate the sums $s^{^{(\infty)}}_{i,k}$ of individual entries $b^{^{(n)}}_{i,k}$ of the n'th powers of the matrix B=(A-I) by Euler-summation via partial sums, which is sometimes an option. If the growthrate of the entries is below a certain constant limit $b^{^{(n+1)}}_{i,k} / b^{^{(n)}}_{i,k} =q_{i,k}$ constant, or $| q^{^{(n)}}_{i,k} |$ is a decreasing sequence (which is for instance the case in the scalar series for the logarithm) and no further exotic behaviour has to be expected, then Euler-summation of order q can give the final approximation to arbitrary precision for each individual entry for $s^{^{(\infty)}}_{i,k}$ of the intended result-matrix by finitely many partial sums (where "exotic" has to be precisely specified,well). Is it that what you meant? Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 08/27/2007, 02:29 PM (This post was last modified: 08/27/2007, 02:32 PM by Gottfried.) Gottfried Wrote:If the sequence of corresponding entries of the powers of (A-I)^n/n do not converge well with finite dimension, but at least alternate in sign, I try to approximate the sums $s^{^{(\infty)}}_{i,k}$ of individual entries $b^{^{(n)}}_{i,k}$ of the n'th powers of the matrix B=(A-I) by Euler-summation via partial sums, which is sometimes an option. Well, I should add a remark concerning the appropriateness of the Euler-summation here. Ideed I observed some cases, where Euler-summation did not satisfyingly well with the series, which occur in our cases. So to say, the "exotic" cases occured in fact. That was for instance at some difficult-cases of the (s^x-1)-iteration Thus I tried other variants similar to the Euler-summation and possibly found a tool, which is more appropriate. While Euler-summation employs binomial-coefficients to "limit" a sequence of oscillating divergent partial sums, I tried to apply the Stirling-numbers 2'nd kind themselves, since the whole computation involves powers of a matrix containing just that numbers. That seemed to be promising; I plotted a comparision of the two summation methods, where the theoretical limit was known. To come near that limit I needed Euler-summation of very high order; very high means here of about the same order as the dimension of my matrices are. But that means: I have only dim=32 or dim=64 terms and Euler-summation of order 27 or 60 to get the partial sums to converge to a limit - but what after that number of terms/size of dimension? I learned, not to trust Euler-summation if it does not converge in, say the last 8 partial sums with orders of 3...7 when only 32 terms of a series are available. So I didn't rely my computations on such results at nasty parameters s (near the critical bounds). If I preconditioned the occuring series with a transformation using Stirling-numbers of 2'nd kind, a much more clear convergence occured with an appended Euler-transform of low order like 3 or 4. An example graph, with which I studied a special case with Euler-orders of about to 10, can be seen here: Summation-Comparision. The situation is not so difficult for our usual cases, where the base-parameter s is in a "friendly" region, like 1+eps < s < e^(1/e)-eps and the like. So this is another path of exploration: to find optimal methods for approximation of the occuring series in our tetration-context; that's where I am involved currently at about 50% . Gottfried Gottfried Helms, Kassel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/27/2007, 02:36 PM Hm, then I somewhere must have an error in my computation. I wanted to compute the Matrix logarithm via the formula $\log(A)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} (A-I)^n$. For A being the power derivation matrix of $e^x$ at 0, i.e. truncated to 6x6 $ A=\left( \begin{matrix} 1&0&0&0&0&0\\ 1&1&1/2&1/6&1/24&{\frac {1}{120}}\\ 1&2&2&4/3&2/3&{\frac {4}{ 15}}\\ 1&3&9/2&9/2&{\frac {27}{8}}&{\frac {81}{40}} \\ 1&4&8&{\frac {32}{3}}&{\frac {32}{3}}&{\frac {128} {15}}\\ 1&5&{\frac {25}{2}}&{\frac {125}{6}}&{\frac { 625}{24}}&{\frac {625}{24}} \end{matrix} \right) $ (which is transposed to the matrix you usually uses). This Matrix is diagonalizable, but if I compute the logarithm via the above sum then the entries do not converge. For example: $\sum_{n=1}^{10} \frac{(-1)^{n+1}}{n} (A-I)^n\\= \left( \begin{matrix} 0.0& 0.0& 0.0& 0.0& 0.0& 0.0\\ - 44101769800.0&- 145224229100.0&- 279795780200.0& - 388116737400.0&- 423988471200.0&- 383041240500.0\\ - 594655919700.0&- 1958162913000.0&- 3772688135000.0&- 5233257693000.0 &- 5716942159000.0&- 5164821148000.0\\ - 3307574479000.0&- 10891625760000.0&- 20984315030000.0&- 29108244400000.0&- 31798577400000.0&- 28727589090000.0 \\ - 11906282180000.0&- 39206606270000.0&- 75537279400000.0&- 104781003800000.0&- 114465400800000.0&- 103410758300000.0\\ - 33057719360000.0&- 108856901800000.0&- 209728793100000.0&- 290923814800000.0&- 317812483800000.0&- 287119336600000.0 \end{matrix} \right)$ « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post My interpolation method [2020] tommy1729 1 128 02/20/2020, 08:40 PM Last Post: tommy1729 Kneser method question tommy1729 9 787 02/11/2020, 01:26 AM Last Post: sheldonison Half-iterates and periodic stuff , my mod method [2019] tommy1729 0 490 09/09/2019, 10:55 PM Last Post: tommy1729 A fundamental flaw of an operator who's super operator is addition JmsNxn 4 7,249 06/23/2019, 08:19 PM Last Post: Chenjesu Tetration and Sign operator tetration101 0 521 05/15/2019, 07:55 PM Last Post: tetration101 2 fixpoints , 1 period --> method of iteration series tommy1729 0 1,785 12/21/2016, 01:27 PM Last Post: tommy1729 Tommy's matrix method for superlogarithm. tommy1729 0 1,786 05/07/2016, 12:28 PM Last Post: tommy1729 [split] Understanding Kneser Riemann method andydude 7 8,403 01/13/2016, 10:58 PM Last Post: sheldonison [2015] New zeration and matrix log ? tommy1729 1 3,361 03/24/2015, 07:07 AM Last Post: marraco Kouznetsov-Tommy-Cauchy method tommy1729 0 2,136 02/18/2015, 07:05 PM Last Post: tommy1729

Users browsing this thread: 1 Guest(s)