No problem tommy. The tex is failing because you are writing Exp instead of \exp, not sure why it's formatting like that.
I apologize I thought the result was obvious.
The function:
\[
h(z) = \sum_{n=0}^\infty \frac{a_n}{W'(n)(z-n)}\\
\]
Has a simple pole at \(z=n\). Similarly, \(W(z)\) has a simple zero at \(z=n\). Thereby:
\[
W(z+n)h(z+n) = a_n + Qz\\
\]
I mean this result from complex analysis.
\[
\frac{W(z+n) a_n}{W'(n)z} = a_n \,\,\text{as}\,\,z\to 0\\
\]
All the other coefficients of the sum defining \(h\) disappear because they are finite and \(W(n)\) is zero...
So,
\[
h(z+n)W(z+n) = 0 + 0 + 0....\frac{W(z+n) a_n}{W'(n)z}....+0+0\\\
\]
I'm sorry, I refuse to believe you've never seen this before Tommy! This is a common consequence of Weierstrass/Mittlag Leffler theory.
We are setting all the other terms of the sum to zero, and the only ones which evaluate to a value are those with simple poles....
Come On, you know this!!!!
I apologize I thought the result was obvious.
The function:
\[
h(z) = \sum_{n=0}^\infty \frac{a_n}{W'(n)(z-n)}\\
\]
Has a simple pole at \(z=n\). Similarly, \(W(z)\) has a simple zero at \(z=n\). Thereby:
\[
W(z+n)h(z+n) = a_n + Qz\\
\]
I mean this result from complex analysis.
\[
\frac{W(z+n) a_n}{W'(n)z} = a_n \,\,\text{as}\,\,z\to 0\\
\]
All the other coefficients of the sum defining \(h\) disappear because they are finite and \(W(n)\) is zero...
So,
\[
h(z+n)W(z+n) = 0 + 0 + 0....\frac{W(z+n) a_n}{W'(n)z}....+0+0\\\
\]
I'm sorry, I refuse to believe you've never seen this before Tommy! This is a common consequence of Weierstrass/Mittlag Leffler theory.
We are setting all the other terms of the sum to zero, and the only ones which evaluate to a value are those with simple poles....
Come On, you know this!!!!