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 Taylor polynomial. System of equations for the coefficients. marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 01/14/2016, 01:22 AM (This post was last modified: 01/14/2016, 01:24 AM by marraco.) Here I make an expansion of a row, in hope that it helps somebody to digest the equation. (01/13/2016, 04:32 AM)marraco Wrote: We are now very close to the solution. The only obstacle remaining is the product: $\vspace{15}{ \frac{1}{ \prod_{j=1}^{i} c_{n,j}!} }$ The product is what I called "the integer divisor" (05/03/2015, 04:35 AM)marraco Wrote: $\mathbf{ -The \,\ integer \,\ divisor \,\ is \,\ the \,\ product \,\ of \,\ the \,\ factorials \,\ of \,\ the \,\ exponents \,\ of \,\ a_i }$ $ +a.(lna.a_9 + lna^2.a_4 .a_5 + lna^2 .a_3 .a_6 + \frac{lna^3}{6} .a_3^3 + lna^2.a_2 .a_7 + lna^3.a_2 .a_3 .a_4 + \frac{lna^3}{2}.a_2^2 .a_5 + \frac{lna^4}{6} .a_2^3 .a_3 + lna^2.a_1 .a_8 + \frac{lna^3}{2} .a_1 .a_4^2 + \\ lna^3 .a_1 .a_3 .a_5 + lna^3 .a_1 .a_2 .a_6 + \frac{lna^4}{2} .a_1 .a_2 .a_3^2 + \frac{lna^4}{2} .a_1 .a_2^2 .a_4 + \frac{lna^5}{24}.a_1 .a_2^4 + \frac{lna^3}{2}.a_1^2 .a_7 + \frac{lna^4}{2}.a_1^2 .a_3 .a_4 + \frac{lna^4}{2} .a_1^2 .a_2 .a_5 + \frac{lna^5}{4}.a_1^2 .a_2^2 .a_3 + \frac{lna^4}{6} .a_1^3 .a_6 +\\ \frac{lna^5}{12} .a_1^3 .a_3^2 + \frac{lna^5}{6} .a_1^3 .a_2 .a_4 + \frac{lna^6}{36}.a_1^3 .a_2^3 + \frac{lna^5}{24}.a_1^4 .a_5 + \frac{lna^6}{24}.a_1^4 .a_2 .a_3 + \frac{lna^6}{120}.a_1^5 .a_4 + \frac{lna^7}{240} .a_1^5 .a_2^2 + \frac{lna^7}{720} .a_1^6 .a_3 + \frac{lna^8}{5040}.a_1^7 .a_2 + \frac{lna^9}{362880}.a_1^9 ).x^9 $ $\small { 9\,\,\right\,\, a_9^1\,\,\right\,\, 1! \,=\, 1 \\ 1,\, 8\,\,\right\,\,a_1^1\,.\,a_8^1\,\,\right\,\,1!.1! \,=\, 1 \\ 2,\, 7\,\,\right\,\, a_2^1\,.\,a_7^1\,\,\right\,\, 1!.1! \,=\, 1 \\ 3,\, 6\,\,\right\,\, a_3^1\,.\,a_6^1\,\,\right\,\, 1!.1! \,=\, 1 \\ 4,\, 5\,\,\right\,\, a_4^1\,.\,a_5^1\,\,\right\,\, 1!.1! \,=\, 1 \\ 1,\, 1,\, 7\,\,\right\,\,a_1^2\,.\,a_7^1\,\,\right\,\,2!.1! \,=\, 2 \\ 1,\, 2,\, 6\,\,\right\,\,a_1^1\,.\,a_2^1\,.\,a_6^1\,\,\right\,\,1!.1!.1! \,=\, 1 \\ 1,\, 3,\, 5\,\,\right\,\,a_1^1\,.\,a_3^1\,.\,a_5^1\,\,\right\,\,1!.1!.1! \,=\, 1 \\ 1,\, 4,\, 4\,\,\right\,\,a_1^1\,.\,a_4^2\,\,\right\,\,1!.2! \,=\, 2 \\ 2,\, 2,\, 5\,\,\right\,\, a_2^2\,.\,a_5^1\,\,\right\,\, 2!.1! \,=\, 2 \\ 2,\, 3,\, 4\,\,\right\,\, a_2^1\,.\,a_3^1\,.\,a_4^1\,\,\right\,\, 1!.1!.1! \,=\, 1 \\ 3,\, 3,\, 3\,\,\right\,\, a_3^3\,\,\right\,\, 3! \,=\, 6 \\ 1,\, 1,\, 1,\, 6\,\,\right\,\,a_1^3\,.\,a_6^1\,\,\right\,\,3!.1! \,=\, 6 \\ 1,\, 1,\, 2,\, 5\,\,\right\,\,a_1^2\,.\,a_2^1\,.\,a_5^1\,\,\right\,\,2!.1!.1! \,=\, 2 \\ 1,\, 1,\, 3,\, 4\,\,\right\,\,a_1^2\,.\,a_3^1\,.\,a_4^1\,\,\right\,\,2!.1!.1! \,=\, 2 \\ 1,\, 2,\, 2,\, 4\,\,\right\,\,a_1^1\,.\,a_2^2\,.\,a_4^1\,\,\right\,\,1!.2!.1! \,=\, 2 \\ 1,\, 2,\, 3,\, 3\,\,\right\,\,a_1^1\,.\,a_2^1\,.\,a_3^2\,\,\right\,\,1!.1!.2! \,=\, 2 \\ 2,\, 2,\, 2,\, 3\,\,\right\,\, a_2^3\,.\,a_3^1\,\,\right\,\, 3!.1! \,=\, 6 \\ 1,\, 1,\,1,\, 1,\, 5\,\,\right\,\,a_1^4\,.\,a_5^1\,\,\right\,\,4!.1! \,=\, 24 \\ 1,\, 1,\, 1,\, 2,\, 4\,\,\right\,\,a_1^3\,.\,a_2^1\,.\,a_4^1\,\,\right\,\,3!.1!.1! \,=\, 6 \\ 1,\, 1,\, 1,\, 3,\, 3\,\,\right\,\,a_1^3\,.\,a_3^2\,\,\right\,\,3!.2! \,=\, 12 \\ 1,\, 1,\, 2,\, 2,\, 3\,\,\right\,\,a_1^2\,.\,a_2^2\,.\,a_3^1\,\,\right\,\,2!.2!.1! \,=\, 4 \\ 1,\, 2,\, 2,\, 2,\, 2\,\,\right\,\,a_1^1\,.\,a_2^4\,\,\right\,\,1!.4! \,=\, 24 \\ 1,\, 1,\, 1,\, 1,\, 1,\, 4\,\,\right\,\,a_1^5\,.\,a_4^1\,\,\right\,\,5!.1! \,=\, 120 \\ 1,\, 1,\, 1,\, 1,\, 2,\, 3\,\,\right\,\,a_1^4\,.\,a_2^1\,.\,a_3^1\,\,\right\,\,4!.1!.1! \,=\, 24 \\ 1,\, 1,\, 1,\, 2,\, 2,\, 2\,\,\right\,\,a_1^3\,.\,a_2^3\,\,\right\,\,3!.3! \,=\, 36 \\ 1,\, 1,\, 1,\, 1,\, 1,\, 1,\, 3\,\,\right\,\,a_1^6\,.\,a_3^1\,\,\right\,\,6!.1! \,=\, 720 \\ 1,\, 1,\, 1,\, 1,\, 1,\, 2,\, 2\,\,\right\,\,a_1^5\,.\,a_2^2\,\,\right\,\,5!.2! \,=\, 240 \\ 1,\, 1,\, 1,\, 1,\, 1,\, 1,\, 1,\, 2\,\,\right\,\,a_1^7\,.\,a_2^1\,\,\right\,\,7!.1! \,=\, 5040 \\ 1,\, 1,\, 1,\, 1,\, 1,\, 1,\, 1,\, 1,\, 1\,\,\right\,\,a_1^9\,\,\right\,\,9! \,=\, 362880 \\ }$^^ Here I expanded the row for i=9 of the equation: $\vspace{15}{ \left [ \sum_{n=1}^{P(9)} \frac{a.ln(a)^{\sum_{j=1}^{9}c_{n,j}}}{ \prod_{j=1}^{9} c_{n,j}!}\prod_{j=1}^{9}a_j^{c_{n,j}} \right ]}$ after the substitution $\vspace{25}{a_i=\frac {b_i} {ln(a)} }$: $ = ln(^2a) \, \left [ \sum_{n=1}^{P(9)} \prod_{j=1}^{9}{\frac{b_j^{c_{n,j}}} { c_{n,j}!} \right ] \,=\, ln(^2a) \, \left [\frac {b_9^1} {1!}+\frac {b_1^1} {1!}\frac {b_8^1} {1!}+\frac {b_2^1} {1!}\frac {b_7^1} {1!}+\frac {b_3^1} {1!}\frac {b_6^1} {1!}+\frac {b_4^1} {1!}\frac {b_5^1} {1!}+\frac {b_1^2} {2!}\frac {b_7^1} {1!}+\frac {b_1^1} {1!}\frac {b_2^1} {1!}\frac {b_6^1} {1!}+\frac {b_1^1} {1!}\frac {b_3^1} {1!}\frac {b_5^1} {1!}+\frac {b_1^1} {1!}\frac {b_4^2} {2!}+\frac {b_2^2} {2!}\frac {b_5^1} {1!}+\frac {b_2^1} {1!}\frac {b_3^1} {1!}\frac {b_4^1} {1!}+\frac {b_3^3} {3!}+\frac {b_1^3} {3!}\frac {b_6^1} {1!}+\frac {b_1^2} {2!}\frac {b_2^1} {1!}\frac {b_5^1} {1!}+\frac {b_1^2} {2!}\frac {b_3^1} {1!}\frac {b_4^1} {1!}+\frac {b_1^1} {1!}\frac {b_2^2} {2!}\frac {b_4^1} {1!}+\frac {b_1^1} {1!}\frac {b_2^1} {1!}\frac {b_3^2} {2!}+\frac {b_2^3} {3!}\frac {b_3^1} {1!}+\frac {b_1^4} {4!}\frac {b_5^1} {1!}+\frac {b_1^3} {3!}\frac {b_2^1} {1!}\frac {b_4^1} {1!}+\frac {b_1^3} {3!}\frac {b_3^2} {2!}+\frac {b_1^2} {2!}\frac {b_2^2} {2!}\frac {b_3^1} {1!}+\frac {b_1^1} {1!}\frac {b_2^4} {4!}+\frac {b_1^5} {5!}\frac {b_4^1} {1!}+\frac {b_1^4} {4!}\frac {b_2^1} {1!}\frac {b_3^1} {1!}+\frac {b_1^3} {3!}\frac {b_2^3} {3!}+\frac {b_1^6} {6!}\frac {b_3^1} {1!}+\frac {b_1^5} {5!}\frac {b_2^2} {2!}+\frac {b_1^7} {7!}\frac {b_2^1} {1!}+\frac {b_1^9} {9!} \right ]$ The problematic terms come from the factors $\vspace{25}{\frac {b_i^q}{q!}$. The q! divisors may emerge not from the term raised to q. q! could emerge from the absence of the other terms: $\vspace{25}{b_i^0}$. For example, the term $\vspace{25}{\frac {b_1^2} {2!} .\frac {b_2^2} {2!} .\frac {b_3^1} {1!} }$ is actually $\vspace{25}{\frac {b_1^2} {2!} .\frac {b_2^2} {2!} .\frac {b_3^1} {1!} \,.\, \frac {b_4^0} {0!}.\frac {b_5^0} {0!}.\frac {b_6^0} {0!}.\frac {b_7^0} {0!}.\frac {b_8^0} {0!}.\frac {b_9^0} {0!}}$ I have the result, but I do not yet know how to get it. « Next Oldest | Next Newest »

 Messages In This Thread Taylor polynomial. System of equations for the coefficients. - by marraco - 04/30/2015, 03:24 AM RE: Taylor polinomial. System of equations for the coefficients. - by tommy1729 - 05/01/2015, 08:37 AM RE: Taylor polinomial. System of equations for the coefficients. - by marraco - 05/01/2015, 09:42 AM RE: Taylor polinomial. System of equations for the coefficients. - by tommy1729 - 05/01/2015, 09:43 PM RE: Taylor polinomial. System of equations for the coefficients. - by marraco - 05/03/2015, 04:46 AM RE: Taylor polinomial. System of equations for the coefficients. - by marraco - 05/03/2015, 12:07 PM RE: Taylor polinomial. System of equations for the coefficients. - by Gottfried - 05/05/2015, 07:40 AM RE: Taylor polinomial. System of equations for the coefficients. - by marraco - 05/06/2015, 02:42 PM RE: Taylor polinomial. System of equations for the coefficients. - by Gottfried - 05/06/2015, 04:17 PM RE: Taylor polynomial. System of equations for the coefficients. - by marraco - 05/07/2015, 09:45 AM RE: Taylor polynomial. System of equations for the coefficients. - by marraco - 01/14/2016, 12:47 AM

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