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Is it possible to find a sum of a series:

$$f(x) = \sum_n \frac{x^n}{\sqrt{n!}}$$

at least as a combination of some special functions?

Ilya
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  • You might be out of luck when Wolfram|Alpha has no idea. There might be a chance, though. – M. Winter Nov 09 '17 at 13:26
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    There is no known closed form for that series. You can name it the Ilya function and publish a paper on its interesting properties (if there are any, which I doubt). – GEdgar Nov 09 '17 at 13:26
  • Did this arise from the quantum harmonic oscillator? – user438666 Nov 09 '17 at 13:58
  • Yes. I am trying to calculate the "coherence" (https://arxiv.org/abs/1311.0275), which is the sum of non-diagonals of the density matrix, of an actual coherent state. – Ilya Nov 09 '17 at 14:05
  • No way, but the first terms of asymptotic behaviour of $f(x)$ (for $x\to +\infty$) can be simply computed by squaring both sides. – Jack D'Aurizio Nov 09 '17 at 14:17

  • "first terms of asymptotic behaviour of f(x) (for x→+∞) can be simply computed by squaring both sides", @JackD'Aurizio, can you, please, show this in more detail?

     – Ilya Nov 10 '17 at 17:06
  • @Ilya: have a look at my answer here: https://math.stackexchange.com/a/2510830/44121 – Jack D'Aurizio Nov 10 '17 at 17:17

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As already pointed out in the comments the function $f(x)$ does not have a closed form, however I'll post here my way of finding its asymptotic behaviour at $x \to \infty$, in case it is useful to anyone.

$$f'(x) = \frac{1}{x} \sum_n \frac{x^n }{\sqrt{n!}}n.$$ Suppose that $f'(x) =\frac{1}{x}g(x) f(x)$. $$g(x)=\frac{\sum \frac{x^n }{\sqrt{n!}}n}{\sum \frac{x^n }{\sqrt{n!}}}$$ Using the fact that: $$ \lim_{x\to \infty} \frac{a_n x^n}{b_n x^n} =\lim_{n\to \infty} \frac{a_n}{b_n},$$ $$\lim_{x\to \infty} \frac{g(x)}{x^2}=\frac{\sum \frac{x^n }{\sqrt{n!}}n}{\sum \frac{x^{n+2} }{\sqrt{n!}}} = \lim_{n\to \infty} \frac{n}{\sqrt{n(n-1)}} =1$$ Therefore $g(x) = x^2 + o(x^2)$. $$\lim_{x\to \infty} g(x) - x^2 =\frac{\sum \frac{x^n }{\sqrt{n!}}n -\sum \frac{x^{n+2} }{\sqrt{n!}} }{\sum \frac{x^{n} }{\sqrt{n!}}} = \lim_{n\to \infty} n -\sqrt{n(n-1)} = \frac{1}{2}$$ Therefore $g(x) = x^2 + \frac{1}{2}+o(1)$. Now we have a simple differential equation: $$f' = \left(x + \frac{1}{2x}\right)f,$$ which gives the answer: $$f \sim \sqrt{x}e^{\frac{x^2}{2}}, \, x\to \infty$$ Next terms of the approximation of $g(x)$ will be $\sim c/x^k$, $k>1$ which gives multipliers of $\exp(c/x^{k+1})$ in $f$, which quickly goes to 1.

Ilya
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