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 Taylor polynomial. System of equations for the coefficients. marraco Fellow Posts: 100 Threads: 12 Joined: Apr 2011 01/13/2016, 04:32 AM (This post was last modified: 01/14/2016, 12:41 AM by marraco.) So, we want the vector $\vspace{15}{[a_i]}$, from the matrix equation: $\vspace{15}{\left [ {{i} \choose {r}} \right ]\cdot \left [a_i \right ] = \left [ \sum_{n=1}^{P(i)} \frac{a.ln(a)^{\sum_{j=1}^{i}c_{n,j}}}{ \prod_{j=1}^{i} c_{n,j}!}\prod_{j=1}^{i}a_j^{c_{n,j}} \right ]}$ where "r" is the row index of the first matrix at left, and "i" his column index. Note that in the last equation, both r and i start counting from zero for the first row and column. ______________________________________ P(i) is the partition function The first few values of the partition function are (starting with p(0)=1): 1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 3718, 4565, 5604, … (sequence A000041 in OEIS; the link has valuable information about the partition function). ______________________________________ $\vspace{15}{c_{n,j}}$ is the number of repetitions of the integer j in the $\vspace{15}{n^{th}}$ partition of the number i ______________________________________ Solving the equation If we do the substitution $\vspace{25}{a_i=\frac {b_i} {ln(a)} }$, we simplify the first equation to: $\vspace{15}{\left [ {{i} \choose {r}} \right ]\cdot \left [b_i \right ] = ln(^2a) \, \left [ \sum_{n=1}^{P(i)} \frac{1}{ \prod_{j=1}^{i} c_{n,j}!}\prod_{j=1}^{i}b_j^{c_{n,j}} \right ]}$ ______________________________________ Special base. This equation suggest a special number, which is m=1.7632228343518967102252017769517070804... m is defined by $\vspace{15}{^2m=e}$ For the base a=m, the equation gets simplified to: $\vspace{35}{\left [ {{i} \choose {r}} \right ]\cdot \left [b_i \right ] = \left [ \sum_{n=1}^{P(i)} \frac{1}{ \prod_{j=1}^{i} c_{n,j}!}\prod_{j=1}^{i}b_j^{c_{n,j}} \right ]}$ But let's forget about m for now. ______________________________________ We are now very close to the solution. The only obstacle remaining is the product: $\vspace{15}{ \frac{1}{ \prod_{j=1}^{i} c_{n,j}!} }$ If we can do a substitution that get us rid of him, we have the solution: $\vspace{15}{\left [ {{i} \choose {r}} \right ]\cdot \left [b_i \right ] = ln(^2a) \, \left [ \sum_{n=1}^{P(i)}\prod_{j=1}^{i}b_j^{c_{n,j}} \right ]}$ At this point we only need to substitute $\vspace{25}{b_i=f^i}$, where f is arbitrary, to get: $\vspace{15}{\left [ {{i} \choose {r}} \right ]\cdot \left [f^i \right ] = ln(^2a) \, \left [ \sum_{n=1}^{P(i)} f^i \right ] \,=\, ln(^2a) . [P(i) . f^i]}$ ... and we get: $\vspace{15}{ \left [f^i \right ] = \left [ {{i} \choose {r}} \right ]^{-1} \cdot [ ln(^2a) . P(i) . f^i]}$ The choice of f, very probably, determines the value for °a, and the branch of tetration. (01/03/2016, 11:24 PM)marraco Wrote: 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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