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 Uniqueness of half-iterate of exp(x) ? hixidom Junior Fellow Posts: 20 Threads: 3 Joined: Feb 2014 05/08/2014, 07:39 PM (This post was last modified: 05/08/2014, 08:11 PM by hixidom.) Ahhh. Thank you. That makes a lot of sense. I assume b is a constant, then I solve for b as a function of x! I am pretty dense. Anyways, here's something else I found that may be of use, Let's say that $f(x)=e^{g(x)}$. Then by definition $f(f(x))=e^{g(e^{g(x)})}=e^x\\\Rightarrow g(e^{g(x)})=x\\\Rightarrow e^{g(x)}=g^{-1}(x)$ Now substitute g(x) for x, $e^{g(g(x))}=g^{-1}(g(x))=x\\\Rightarrow g(g(x))=ln(x)$ So, in conclusion, the half-iterate of e^x is the exponential of the half-iterate of ln(x). Not sure if that is useful but I think it's kinda neat. Hopefully I didn't make another mistake, but it's always possible. Xorter Fellow Posts: 93 Threads: 30 Joined: Aug 2016 01/07/2017, 11:00 PM (This post was last modified: 01/07/2017, 11:02 PM by Xorter. Edit Reason: wrong tex ) Hello! I found an interesting site about the Carleman matrix: https://en.m.wikipedia.org/wiki/Carleman_matrix $M[f]_i_,_j = 1/k! [D^k f(0)^j]$ $f(x) = \sum_{k=0}^{\infty} M[f]_1,k x^k$ And the most important of this case: M[fog] = M[f] M[g] So these matrices convert composition to matrix multiplication. Thus $M[f^o ^N] = M[f]^N$ Therefor $f^o ^0.5 (x) = \sum_{k=0}^{\infty} sqrt(M[f])_1_,_k x^k$ So we get the M[exp(x)], it was the easy part of the thing. We need to get the squered root of this matrix, and I could find a program for it: http://calculator.vhex.net/calculator/li...quare-root And I got another matrix, which satisfies that: $sqrt(M[exp(x)])_1_,_k = [0.606, 0.606, 0.303 ...]^T$ So the function is: $exp^o ^0.5 (x) ~= 0.606 + 0.606x + 0.303x^2 + 0.101x^3$ But it is not the half-iterate of exp(x), Could you help me why not, please? What was my mistake? Xorter Unizo Gottfried Ultimate Fellow Posts: 786 Threads: 121 Joined: Aug 2007 01/08/2017, 01:31 AM (This post was last modified: 01/15/2017, 10:14 PM by Gottfried. Edit Reason: added information ) (01/07/2017, 11:00 PM)Xorter Wrote: Hello! I found an interesting site about the Carleman matrix: https://en.m.wikipedia.org/wiki/Carleman_matrix (...) So these matrices convert composition to matrix multiplication. Thus $M[f^o ^N] = M[f]^N$ Therefor $f^o ^0.5 (x) = \sum_{k=0}^{\infty} sqrt(M[f])_1_,_k x^k$ So we get the M[exp(x)], it was the easy part of the thing. We need to get the squered root of this matrix, and I could find a program for it: http://calculator.vhex.net/calculator/li...quare-root And I got another matrix, which satisfies that: $sqrt(M[exp(x)])_1_,_k = [0.606, 0.606, 0.303 ...]^T$ So the function is: $\exp^o ^{0.5} (x) ~= 0.606 + 0.606x + 0.303x^2 + 0.101x^3$ But it is not the half-iterate of exp(x), Could you help me why not, please? What was my mistake? Well, using the truncated series of the exp(x)-function up to 16 terms (Carleman-matrix-size) I get, using my own routine for matrix-square-root in Pari/GP with arbitrary numerical precision (here 200 decimal digits for internal computation) , the following truncated series-approximation: $\exp^{[0.5]}(x) \approx 0.498568472273 + 0.876337510066*x + 0.247418943917*x^2 + 0.0248068936680*x^3 - 0.00112303037149*x^4 \\ + 0.000361451686885*x^5 + 0.0000337024986252*x^6 - 0.0000517784266699*x^7+ 0.0000259224188256*x^8 - 0.00000189354770473*x^9 \\ - 0.00000360748613972*x^{10} + 0.00000411482178000*x^{11} - 0.00000216221756598*x^{12} + 0.000000558540558241*x^{13} \\ - 0.0000000635173773180*x^{14} + 0.00000000192054352361*x^{15}+ O(x^{16})$  This gives, for $0 \le x \lt 1$ eight correct digits when applying this two times (and should approximate Sheldon's Kneser-implementation).                                    The reason, why your function is badly misshaped might be: matrix is too small (did you only take size 4x4?) and/or the matrix-squareroot-computation is not optimal. To crosscheck: one simple approach to the matrix-square-root is the "Newton-iteration".                        Let M be the original Carleman-matrix and N denote its approximated square-root                  initialize ...                  $N=Id \qquad \qquad \text{ /* Identity matrix of some finite dimension dim */}$ iterate  ...                                 $N = (M * N^{-1} + N)/2$. until convergence   . Unfortunately, the matrix N shall not be "Carleman" unless M were of infinite size; nitpicking this means, the function $\exp^{0.5}(x)$ with coefficients taken from the second row (or in my version:column) is not really well suited for iteration. (But this problem has not yet been discussed systematically here in the forum, to my best knowledge) Gottfried Gottfried Helms, Kassel Xorter Fellow Posts: 93 Threads: 30 Joined: Aug 2016 01/08/2017, 10:55 AM (01/08/2017, 01:31 AM)Gottfried Wrote: Well, using the truncated series of the exp(x)-function up to 16 terms (Carleman-matrix-size) I get, using my own routine for matrix-square-root in Pari/GP with arbitrary numerical precision (here 200 decimal digits for internal computation) , the following truncated series-approximation: $\exp^{[0.5]}(x) \approx 0.498568472273 + 0.876337510066*x + 0.247418943917*x^2 + 0.0248068936680*x^3 - 0.00112303037149*x^4 \\ + 0.000361451686885*x^5 + 0.0000337024986252*x^6 - 0.0000517784266699*x^7+ 0.0000259224188256*x^8 - 0.00000189354770473*x^9 \\ - 0.00000360748613972*x^{10} + 0.00000411482178000*x^{11} - 0.00000216221756598*x^{12} + 0.000000558540558241*x^{13} \\ - 0.0000000635173773180*x^{14} + 0.00000000192054352361*x^{15}+ O(x^{16})$  This gives, for $0 \le x \lt 1$ eight correct digits when applying this two times (and should approximate Sheldon's Kneser-implementation).                                    The reason, why your function is badly misshaped might be: matrix is too small (did you only take size 4x4?) and/or the matrix-squareroot-computation is not optimal.                 To crosscheck: one simple approach to the matrix-square-root is the "Newton-iteration".                        Let M be the original Carleman-matrix and N denote its approximated square-root                  initialize ...                  $N=Id \qquad \qquad \text{ /* Identity matrix of some finite dimension dim */}$                   iterate  ...                                  $N = (M * N^{-1} + N)/2$                 until convergence   .                 Unfortunately, the matrix N shall not be "Carleman" unless M were of infinite size; nitpicking this means, the function $\exp^{0.5}(x)$ with coefficients taken from the second row (or in my version:column) is not really well suited for iteration. (But this problem has not yet been discussed systematically here in the forum, to my best knowledge) Gottfried Did I take size 4x4? No, of course not, It was 20x20 later 84x84. I made and recognised my mistake: I generate wrong Carleman matrix instead of M[exp(x)]_i,j = i^j/j!. Now I regenerate the matrix and I got approximately the same solution. It works, yuppie!  Thank you very much. Could you tell me what you wrote into pari to calculate it out, please? I am not so good at pari codes. Xorter Unizo Gottfried Ultimate Fellow Posts: 786 Threads: 121 Joined: Aug 2007 01/09/2017, 02:41 AM (This post was last modified: 01/09/2017, 03:10 AM by Gottfried.) Here is Pari/GP - code Code: default(realprecision,200)   \\ increase internal precision to 800 digits or higher when matrixsize more than, say, 64 ...  default(format,"g0.12")      \\ only 12 digits for display of float numbers  dim=16                       \\ increase later, when everything works, but stay less then, say, 128  M = matrix(dim,dim,r,c,(c-1)^(r-1)/(r-1)!)   \\ carlemanmatrix, transposed, series-coefficients along a column!  M=1.0*M           \\it is better to have float-values in M otherwise the number-of-digits in N explodes over iterations  N = matid(dim)  N = (M * N^-1 + N) / 2  N = (M * N^-1 + N) / 2  N = (M * N^-1 + N) / 2  /* ... do this a couple of times to get convergence; careful: not too often to avoid numerical errors/overflow*/ /* note, N's expected property of being Carleman-type shall be heavily distorted. */ M - N*N   \\ check for sanity, the difference should be near zero /* define the function;   */ exp05(x) = sum(k=1,dim, x^k * N[k,2])  \\ only for x in interval with good convergence (0<=x<1 ) /* try, 6 to eight digits might be correct when dim is at least 32 x32 /* x0 = 0 x05 = exp05(x0)   \\ this should be the half-iterate about   0.498692160537 x1 = exp05(x05)  \\ this should be the full iterate and equal exp(x0)=1 and is  about 1.00012482606 x1 - exp(x0)   \\ check error Example. With dim=8 I got after 8 iterations for N: Code:N=               1.00000000000        0.498692160537     0.248258284527    0.123313067961  0.0613783517169  0.0309705773951  0.0161518156415  0.00900178983873               0        0.876328584414     0.875668009082    0.651057846300   0.427494354197   0.262472285853   0.155983031832   0.0925625176728               0        0.246718723415      1.01708995680     1.34271949385    1.26030722116   0.991600974872   0.698202964055    0.456331269934               0       0.0248938874134     0.453724180460     1.35543926159    2.03890540259    2.20219490711    1.93455428276     1.45660923676               0    -0.000559114024252     0.101207623371    0.716292108758    1.94548221867    3.15424069454    3.69849457008     3.40094743897               0     0.000132927042876    0.0119615040464    0.219667721568    1.11333664588    2.97146364527    5.09318298067     6.20122030134               0    0.0000114543791108   0.00113904404837   0.0431682924811   0.396012087313    1.78179566161    5.01623097009     9.19538628629               0  -0.00000540376918712  0.000111767358395  0.00661697871499  0.0933915529553   0.665946396103    3.15521910471     11.3987254320 Of course, to make this more flexible for varying fractional powers of M you'll need diagonalization - but then the required "realprecision" becomes exorbitant for dim=32 and more. For reference, I call this method the "polynomial method" because by the matrix being of finite size this is a polynomial approximation and no attempt is done to produce N in a way, that it basically maintains the structure of a Carlemanmatrix when fractional powers are computed. If this is wanted, the conjugacy using the complex fixpoint is needed before the diagonalization and the generation of a power series with complex coefficients to have the famous Schröder-function by the eigenvectors-matrices. After that, Sheldon has the method to proceed backwards to a real-to-real solution after H. Kneser (which seems to be possibly the limit of the above construction when the matrix size goes to infinity). Gottfried Gottfried Helms, Kassel « Next Oldest | Next Newest »

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