Evaluate $f(x)=\sum_0^\infty\frac{x^n}{n!!}$.

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A question from Introduction to Analysis by Arthur Mattuck:

Let $n!!=n(n-2)(n-4)\cdot…\cdot k$, where $k=1$ or $2$,depending on whether n is odd or even. (define $0!!=1$.)

Evaluate the sum $f(x)=\sum_0^\infty\frac{x^n}{n!!},$ using term-by-term differentiation and integration.

I think what the question asked is to give an explicit form for this sum.

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If you differentiate,

$$f'(x)=\sum_{k=1}^\infty\frac{nx^{n-1}}{n!!}=1+\sum_{k=2}^\infty\frac{x^{n-1}}{(n-2)!!}=1+xf(x).$$

This is a linear differential equation. The homogeneous part is separable and yields

$$f'(x)=xf(x)\to f(x)=Ke^{x^2/2}.$$

Then by variation of the constant,

$$K(x)=\int e^{-x^2/2}dx+C,$$

$$f(x)=e^{x^2/2}\int e^{-x^2/2}dx+Ce^{x^2/2}.$$

With the initial condition $f(0)=1$,

$$f(x)=e^{x^2/2}\int_{t=0}^x e^{-t^2/2}dt+e^{x^2/2}.$$