(12/11/2009, 11:16 AM)bo198214 Wrote: \( G_0(x)=g_0(x)=\phi(0) \)
\( G_n(x)=g_n(x)-g_{n-1}(x) \) for \( n>0 \),
\( \phi(x)=\sum_{n=0}^\infty G_n(x) \)
Well this G is of no use because
\( \phi(x)=\lim_{n\to\infty} g_n(x)=\sum_{\lambda_1=0}^{n^{2n}}\lim_{n\to\infty}\sum_{\lambda_2=0}^{n^{2n-2}}
\dots\sum_{\lambda_n=0}^{n^2} \frac{\phi^{(\lambda_1+\dots+\lambda_n)}(0)}{\lambda_1!\dots\lambda_n!} \left(\frac{x}{n}\right)^{\lambda_1+\dots+\lambda_n}
\)
If we sort the terms for \( m=\lambda_1+\dots+\lambda_n \) we can see that this is a sequence of polynomials
\( g_n(x)=\quad\sum_{m=0}^{n^2+\dots+n^{2n}}\quad c^{(n)}_m \frac{\phi^{(m)}(0)}{m!} x^m \), where the coefficients \( c^{(n)}_m \) are independent on \( \phi \) and on the region.
I guess these huge number of terms is the drawback of the formula and the reason why you nearly can find it anywhere in the net (if it was of big practical use one would get hundrets of hits for the formula).
Other coefficients
I guess there are also other coefficients possible, the only thing that is needed in the proof is a sequence of polynomials \( P_n(z)=\sum_{m=0}^{M(n)}c^{(n)}_m z^m \) that converge compactly to \( \frac{1}{1-z} \) on the region \( \mathbb{C}\setminus \{z: z\ge 1\} \). The basic idea is that
\( f(z)=\frac{1}{2\pi i} \int_{\gamma} \frac{f(\zeta)}{\zeta-z}d\zeta \), where you replace \( \frac{1}{\zeta-z}=\frac{1}{(1-\frac{z}{\zeta})\zeta} \) with the \( P_n(\frac{z}{\zeta})\frac{1}{\zeta}=\sum_{m=0}^{M(n)} c^{(n)}_m z^m \frac{1}{\zeta^{m+1}} \)
Then
\( f(z)=\lim_{n\to\infty} \sum_{m=0}^{M(n)} c^{(n)}_m z^m \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)}{\zeta^{m+1}}d \zeta = \lim_{n\to\infty} \sum_{m=0}^{M(n)} c^{(n)}_m \frac{\phi^{(m)}(0)}{m!} z^m \)
So give me any such polynomial sequence and I give you a Mittag-Leffler expansion.
Computing the original expansion
To get an impression I computed the original expansion for f(z)=log(z+1). And this is the result:
MLE \( g_2 \) has degree \( 2^2+2^4=20 \):
MLE \( g_3 \) has degree \( 3^2+3^4+3^6=819 \):
Normal powerseries of log(z+1) with as many terms as \( g_3 \):
As comparison:
\( g_2(x)=-\frac{969}{4194304} x^{20} + \frac{255}{524288} x^{19} - \frac{1343}{1572864} x^{18} + \frac{189}{131072} x^{17} - \frac{2517}{1048576} x^{16} + \frac{647}{163840} x^{15} - \frac{1471}{229376} x^{14} +\\ \frac{1093}{106496} x^{13} - \frac{397}{24576} x^{12} + \frac{281}{11264} x^{11} - \frac{193}{5120} x^{10} + \frac{1}{18} x^{9} - \frac{163}{2048} x^{8} + \frac{99}{896} x^{7} - \frac{19}{128} x^{6} + \frac{31}{160} x^{5} - \frac{1}{4} x^{4} + \frac{1}{3} x^{3} - \frac{1}{2} x^{2} + x
\)
one sees that up to \( x^4 \) the development matches the development of log(x+1).
Here the direct coefficients each sequence starting with index 0:
\( c^{(1)}=(1,1) \)
\( c^{(2)}=(1, 1, 1, 1, 1, \frac{31}{32}, \frac{57}{64}, \frac{99}{128}, \frac{163}{256}, \frac{1}{2}, \frac{193}{512}, \frac{281}{1024}, \frac{397}{2048}, \frac{1093}{8192}, \frac{1471}{16384}, \frac{1941}{32768}, \frac{2517}{65536}, \frac{3213}{131072}, \frac{4029}{262144}, \frac{4845}{524288}, \frac{4845}{1048576}) \)
\( c^{(3)}=(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, \frac{59048}{59049}, \frac{177124}{177147}, \frac{531152}{531441}, \frac{1591696}{1594323}, \frac{4763648}{4782969}, \frac{14226880}{14348907}, \frac{42360320}{43046721}, \frac{125616640}{129140163}, \frac{370626560}{387420489}, \frac{1086986240}{1162261467}, \frac{3166363648}{3486784401}, \frac{9155108864}{10460353203}, \frac{26261389312}{31381059609},\dots) \)
