f( x/[(1 - x)^2] ) = A g(x) f(x) + B g(x) + C.

[tommy1729]

Ah, I see you're looking at Euler's Pentagonal Number Theorem/Modular functions!

I suggest picking up an analytic number theory book and seeing the sharp modern bounds and techniques we use now--to prove something Euler did in the 18th century!

Sounds interesting Tommy. Excited to see. You may be interested in Modular functions, if you're looking at this--which are constructed from Euler's results The function \(f\) has been studied before; and it does have a name:

\[
f(x) = \prod_{j=1}^\infty (1-x^{2j-1})\\
\]

is nothing new; and attempts to understand this, are similar to understanding the Pentagonal number theorem.

\[
E(x) = E(x^2) f(x) = \prod_{j=1}^\infty (1-x^j)\\
\]

And the coefficients of \(E\) are well studied.... Where:

\[
\frac{1}{E(x)} = \sum_{n=0}^\infty \#n x^n\\
\]

Where \(\# n\) is the partitions of \(n\). Upon which:

\[
f(x) = \sum_{n=0}^\infty f_n x^n\\
\]

Which is given as:

\[
E(x) = \sum_{n=0}^\infty p_n x^n\\
\]

And:

\[
f(x) = \frac{E(x)}{E(x^2)}\\
\]

Where now we are doing some kind of Cauchy convolution of \(p_n\) and \(\# n\) to get \(f_n\).


The values \(p_n\) are well studied; and can be expressed as:

\[
\sum_{k=-\infty}^\infty (-1)^k x^{k(3k-1)/2} =\sum_{n=0}^\infty p_n x^n\\
\]

So if \(c_n = 0\) if \(n \equiv 1\,\,\text{mod}\,\,2\) and \(c_n = \# \frac{n}{2}\) if \(n \equiv 0\,\,\text{mod}\,\,2\)--then your values \(f_n\) are written:

\[
f_n = \sum_{d=0}^n c_{n-d}p_d\\
\]

Because:

\[
\begin{align}
f(x) &= \frac{E(x)}{E(x^2)}\\
&= \left(\sum_{n=0}^\infty p_n x^n\right) \left(\sum_{n=0}^\infty \#n x^{2n}\right)\\
&= \left(\sum_{n=0}^\infty p_n x^n\right) \left(\sum_{n=0}^\infty c_n x^{n}\right)\\
&= \sum_{n=0}^\infty \left(\sum_{d=0}^n c_{n-d}p_d\right) x^n\\
&= \sum_{n=0}^\infty f_n x^n\\
\end{align}
\]

x/[(1 - x)^2] and g(x) are both famous functions.

x/[(1 - x)^2] is the köbe function.

James , do you believe in coincidence ?


regards

tommy1729