closed form of Taylor series expansion

Question

Is there a closed form of this summation? $A = \sum_{j=0}^\infty \frac{1}{j!}\times \frac{1}{j!}x^j$

or can it be derived as multiple of two function? ( ex) $A = \cos x\times e^x$ )

Answer 1

Let $$y(x)=\sum\frac{x^j}{j!^2}.$$

We have $$y'(x)=\sum\frac{x^{j-1}}{j!(j-1)},$$

$$xy'(x)=\sum\frac{x^j}{j!(j-1)!},$$ and $$(xy')'(x)=\sum\frac{x^{j-1}}{(j-1)^2}=y(x).$$

This finally leads us to the differential equation

$$xy''+y'-y=0.$$

By a change of variable $t=2\sqrt x$, we can convert it to the modified Bessel type (of order $0$):

$$t^2y''+ty'-t^2y=0.$$

https://en.wikipedia.org/wiki/Bessel_function#Modified_Bessel_functions:_I%CE%B1,_K%CE%B1

Then from the initial condition $y(0)=1$, we can infer

$$y(x)=I_0(2\sqrt x).$$

https://www.wolframalpha.com/input/?i=plot+I_0(2sqrt+x)+from+0+to+10

Answer 2

Yes.

$$\sum_{k = 0}^{+\infty} \frac{x^k}{(k!)^2} = I_0\left(2 \sqrt{x}\right)$$

Where $I_0$ is the modified Bessel Function of the first kind.

http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html