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 Tetration series for integer exponent. Can you find the pattern? marraco Fellow Posts: 100 Threads: 12 Joined: Apr 2011 01/30/2016, 09:06 PM (This post was last modified: 02/12/2016, 11:06 PM by marraco.) $^0(1+x) \,=\,$${\color{Red} 1}$$+ 0+ 0+0...$ $^1(1+x) \,=\,$${\color{Red} 1}+ x$$+ 0+0+0...$ $^2(1+x) \,=\,$${\color{Red} 1}+ x+ x^2$$+\frac {1}{2}x^{3}+\frac {1}{3}x^{4}+\frac {1}{12}x^{5}+\frac {3}{40}x^{6}+\frac {-1}{120}x^{7}+\frac {59}{2520}x^{8}+\frac {-71}{5040}x^{9}+\frac {131}{10080}x^{10}+\frac {-53}{5040}x^{11}+O(x^{12})$ $^3(1+x) \,=\,$${\color{Red} 1}+x+x^2 +\frac {3}{2}x^{3}$$+\frac {4}{3}x^{4}+\frac {3}{2}x^{5}+\frac {53}{40}x^{6}+\frac {233}{180}x^{7}+\frac {5627}{5040}x^{8}+\frac {2501}{2520}x^{9}+\frac {8399}{10080}x^{10}+\frac {34871}{50400}x^{11}+O(x^{12})$ $^4(1+x) \,=\,$${\color{Red} 1}+x+x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}$$+3x^{5}+\frac {163}{40}x^{6}+\frac {1861}{360}x^{7}+\frac {33641}{5040}x^{8}+\frac {8363}{1008}x^{9}+\frac {22391}{2160}x^{10}+\frac {7589}{600}x^{11}+O(x^{12})$ $^5(1+x) \,=\,$${\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}$$+\frac {243}{40}x^{6}+\frac {3421}{360}x^{7}+\frac {71861}{5040}x^{8}+\frac {54371}{2520}x^{9}+\frac {69281}{2160}x^{10}+\frac {7200983}{151200}x^{11}+O(x^{12})$ $^6(1+x) \,=\,$${\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}+\frac {283}{40}x^{6}$$+\frac {4321}{360}x^{7}+\frac {102941}{5040}x^{8}+\frac {85871}{2520}x^{9}+\frac {61333}{1080}x^{10}+\frac {886763}{9450}x^{11}+O(x^{12})$ $^7(1+x) \,=\,$${\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}+\frac {283}{40}x^{6}+\frac {4681}{360}x^{7}$$+\frac {118061}{5040}x^{8}+\frac {106661}{2520}x^{9}+\frac {81583}{1080}x^{10}+\frac {10169449}{75600}x^{11}+O(x^{12})$ $^8(1+x) \,=\,$${\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}+\frac {283}{40}x^{6}+\frac {4681}{360}x^{7}+\frac {123101}{5040}x^{8}$$+\frac {115481}{2520}x^{9}+\frac {93013}{1080}x^{10}+\frac {12169699}{75600}x^{11}+O(x^{12})$ $^9(1+x) \,=\,$${\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}+\frac {283}{40}x^{6}+\frac {4681}{360}x^{7}+\frac {123101}{5040}x^{8}+\frac {118001}{2520}x^{9}$$+\frac {97333}{1080}x^{10}+\frac {13165099}{75600}x^{11}+O(x^{12})$ $^{10}(1+x) \,=\,$${\color{Red} 1}+x+ x^2+\frac {3}{2}x^{3}+\frac {7}{3}x^{4}+{4}x^{5}+\frac {283}{40}x^{6}+\frac {4681}{360}x^{7}+\frac {123101}{5040}x^{8}+\frac {118001}{2520}x^{9}+\frac {98413}{1080}x^{10}$$+\frac {13505299}{75600}x^{11}+O(x^{12})$ with each integer power of n, the tetration $\vspace{15}{^n(x+1)}$ converges, one extra coefficient (highlighted in red), to the Taylor series of $\vspace{15}{\frac {lambertW(-ln(x+1))}{ln(x+1)} }$, which gives the known asymptotic limit for $\vspace{15}{e^{-e}<(x+1) the converged coefficients are 1, followed by the sequence http://oeis.org/A033917 (divided by n!) Some formulas for some of those tetrations are $\vspace{24}{ ^1(x+1) \,=\, 1+ \sum_{n=1}^{\infty} \frac{x^n}{n!}\sum_{j=1}^{n} \begin{bmatrix} & n & \\ & j & \end{bmatrix} \sum_{k=0}^{n} {{n} \choose {k}}*(n-k)^k \\ }$ $\vspace{24}{ ^2(x+1) \,=\, 1+ \sum_{n=1}^{\infty} \frac{x^n}{n!}\sum_{j=1}^{n} \begin{bmatrix} & n & \\ & j & \end{bmatrix} \sum_{k=0}^{j} {{j} \choose {k}}*(j-k)^k \\ }$ $\vspace{24}{ ^\infty(x+1) \,=\,1+ \sum_{n=1}^{\infty} \frac{x^n}{n!}\sum_{j=1}^{n} \begin{bmatrix} & n & \\ & j & \end{bmatrix} \sum_{k=0}^{j-1} {{j-1} \choose {k}}*{j^k} \\ \\ }$ where $\vspace{24}{ \begin{bmatrix} & n & \\ & j & \end{bmatrix}}$ are the Stirling numbers of the first kind. From the definition of the inverse Stirling transform, it looks like each tetration can be written as (1 plus...) a polynomial whose coefficients are inverse Stirling transform of the coefficients $\vspace{15}{b_j}$ of some relatively simple function (coefficients not including the factorial denominator in the Taylor series) $\vspace{24}{ ^z(x+1) \,=\, ^0a + \sum_{n=1}^{\infty} \frac{x^n}{n!}\sum_{j=1}^{n} \begin{bmatrix} & n & \\ & j & \end{bmatrix} \,.\, b_j \\ \\ }$ Here $\vspace{15}{b_j}$ represents the last summation on the previous set of equations. The point of them being "relatively simple", compared to tetration, is that if we can find a general expression for $\vspace{15}{b_j}$, for any exponent n,it probably will be easier to generalize it to real exponents of tetration. the $\vspace{15}{b_j}$ can be obtained by the direct Stirling transform of the coefficients shown at the start of the post. For example, the coefficients $c_j$ of the Taylor series of $\vspace{15}{ ^3(x+1)}$ are 1 followed by [ 1, 1, 3/2, 4/3, 3/2, 53/40, 233/180, 5627/5040, 2501/2520, ...] (as shown at the start of this post). To get the $\vspace{15}{b_j}$, we need to remove the n! denominators of the Taylor series, and apply the Stirling transform. $c_j$ = [1, 2, 9, 32, 180, 954, 6524, 45016, 360144, 3023640, 27617832, 271481880, ...] $b_n=\sum_{j=1}^n \left\{\begin{matrix} n \\ j \end{matrix} \right\} c_j$ and we get $b_j$= [1, 3, 16, 101, 756, 6607, 65794, 733833, 9046648, 121961051, 1782690174, 28055070397, ...] Piece of code for calculation of $b_j$ Code:tet=x+O(x^21)+1 { for (iter=1,10,     tet=(x+O(x^21)+1)^tet;     c_j=vector( 12,n,polcoeff(tet,n)*n! );     b_j=vector( 12,n,sum(j=1,n,stirling(n,j,2)*c_j[j]));          print("^"iter+1"(x+1): b_j="b_j); ); /*for iter*/ } gives this output: ^2(x+1): b_j=[1, 3, 10, 41, 196, 1057, 6322, 41393, 293608, 2237921, 18210094, 157329097] ^3(x+1): b_j=[1, 3, 16, 101, 756, 6607, 65794, 733833, 9046648, 121961051, 1782690174, 28055070397] ^4(x+1): b_j=[1, 3, 16, 125, 1176, 12847, 160504, 2261289, 35464816, 612419291, 11539360944, 235469524237] ^5(x+1): b_j=[1, 3, 16, 125, 1296, 16087, 229384, 3687609, 66025360, 1303751051, 28151798544, 659841763957] ^6(x+1): b_j=[1, 3, 16, 125, 1296, 16807, 257104, 4480569, 87238720, 1874561291, 44057589984, 1124459440117] ^7(x+1): b_j=[1, 3, 16, 125, 1296, 16807, 262144, 4742649, 96915520, 2197675691, 54640864224, 1476693931957] ^8(x+1): b_j=[1, 3, 16, 125, 1296, 16807, 262144, 4782969, 99637120, 2323474091, 59755204224, 1676301882037] ^9(x+1): b_j=[1, 3, 16, 125, 1296, 16807, 262144, 4782969, 100000000, 2354318891, 61498237824, 1760945456437] ^10(x+1): b_j=[1, 3, 16, 125, 1296, 16807, 262144, 4782969, 100000000, 2357947691, 61877447424, 1786651875637] ^11(x+1): b_j=[1, 3, 16, 125, 1296, 16807, 262144, 4782969, 100000000, 2357947691, 61917364224, 1791681392437] According of OEIS, the $b_j$ sequence for $\vspace{15}{^z(x+1)}$ seems to be number of "forests" with n nodes and height at most z-1 (OEIS A210725) For n=3, $\vspace{25}{b_j=1+\sum_{k=0}^{j-1} {{j}\choose{k}} \sum_{i=1}^{j-k} {{j-k}\choose{i}}(i^{(j-k-i)}*k^i)}$ I made everything I could to simplify this last expression, to make it look as similar as possible to the others (for n=1, 2, and $\infty$). Can you make it simpler? Can you identify a pattern? Can you identify the exponent z in $b_j$ from $\vspace{24}{ ^z(x+1) \,=\, ^0a + \sum_{n=1}^{\infty} \frac{x^n}{n!}\sum_{j=1}^{n} \begin{bmatrix} & n & \\ & j & \end{bmatrix} \,.\, b_j \\ \\ }$ ? (I wrote here the formulas for n=1,2,3 and ∞) I have the result, but I do not yet know how to get it. « Next Oldest | Next Newest »

 Messages In This Thread Tetration series for integer exponent. Can you find the pattern? - by marraco - 01/30/2016, 09:06 PM RE: Tetration series for integer exponent. Can you find the pattern? - by Gottfried - 02/14/2016, 03:43 AM

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