I found this sum while trying solving the Hermite differential equation by series.
$$x+x*(4-x)+x*(4-x)*(8-x)+..... $$
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3Well, it never converges (except when $x$ is a multiple of $4$.) – Thomas Andrews May 03 '17 at 01:50
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I think this works:
$x*(1+\sum_{n=1}^\infty\prod_{k=1}^n(4k-x))$
Or maybe (apparently on a technicality)
$-\sum_{n=0}^\infty\prod_{k=0}^n(4k-x)$
John Lou
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That's missing the very first term $x,$. You probably want the sum to start at $n=0$ with the convention that an empty product is considered to equal $1,$ (see e.g. here why). – dxiv May 03 '17 at 01:55
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First one is missing a parenthesis. The second one does technically work. – dxiv May 03 '17 at 02:08
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so $\prod_{k=0}^0 f(x) = 1$ always? Is that why? If not, I get that at n = 0, the entire thing gives $x$. – John Lou May 03 '17 at 02:13
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No, but $,\prod_{k=\color{red}{1}}^0 a_k=1$ which is the multiplicative identity, like $,\sum_{k=1}^0 a_k=0$ which is the additive identity. – dxiv May 03 '17 at 02:39
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Then I'm not sure why the answer doesn't work. Could you write out the first couple terms? – John Lou May 03 '17 at 02:44
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It does work, read my comment again. Reason why I said that it
*technically* worksis that I find the alternative a bit easier to follow, where you factor out the $x,$, instead. – dxiv May 03 '17 at 02:48
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This works: $$x\sum_{n=0}^{\infty}4^n\frac{\Gamma(1-\frac{x}{4}+n)}{\Gamma(1-\frac{x}{4})}$$
The fraction is effectively doing $(1 - \frac{x}{4})(2 - \frac{x}{4})\dots (n - \frac{x}{4})$, which when combined with the $4^n$ gives $4(1 - \frac{x}{4})\cdot4(2 - \frac{x}{4})\cdot\dots \cdot4(n - \frac{x}{4})$ or $(4 - x)(8 - x)\dots (4n - x)$.
Only $x$ is missing for every term, but that can be put outside the sum. In fact, the $\Gamma(1-\frac{x}{4})$ can be put outside the sum, giving:
$$\frac{x}{\Gamma(1-\frac{x}{4})}\sum_{n=0}^{\infty}4^n\Gamma(1-\frac{x}{4}+n)$$
orlp
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