03/13/2016, 02:58 AM
(01/13/2016, 04:32 AM)marraco Wrote: So, we want the vector \( \\[15pt]
{[a_i]} \), from the matrix equation:
\( \color{Blue}\\[15pt]
{\left [ {{i} \choose {r}} \right ]\cdot \left [a_i \right ] \)\( \,=\, \)\( {\color{Red}
\left [ \sum_{n=1}^{P(i)} \frac{a.ln(a)^{\sum_{j=1}^{i}c_{n,j}}}{ \prod_{j=1}^{i} c_{n,j}!}\prod_{j=1}^{i}a_j^{c_{n,j}} \right ]} \)
Thanks to Daniel advice, is easy to see that the red side can be derived as a direct application of Faà di Bruno's formula
\( {d^i \over dx^i} f(g(x))
=\sum \frac{i!}{m_1!\,m_2!\,\cdots\,m_i!} \cdot
f^{(m_1+\cdots+m_i)}(g(x)) \cdot
\prod_{j=1}^i\left(\frac{g^{(j)}(x)}{j!}\right)^{m_j} \)
in the blue red equation,
(04/30/2015, 03:24 AM)marraco Wrote: We want the coefficients aᵢ of this Taylor expansion:is easy to see on the red side, that
\( {^xa}=\sum_{n=0}^{\infty}{a_n .x^n} \)
They should match this equation:
\( {\color{Blue} {^{x+1}a} \)\( \,=\, \)\( {\color{Red} a^{^xa}} \)
\( {m_j=c_{n,j}} \)
\( {f(x)=a^x} \)
\( {g(x)={^xa}=\sum_{n=0}^{\infty}{a_n .x^n}} \)
\( f^{(m_1+\cdots+m_i)}\,_{(g(0))}=a.ln(a)^{\sum_{j=1}^{i}c_{n,j}} \)
\( {\frac{g^{(j)}(0)}{j!}=a_j} \)
I have the result, but I do not yet know how to get it.

