# Power Series Problem

## Homework Statement

$\begin{equation} 1 - x + \frac{x^2}{(2!)^2} - \frac{x^3}{(3!)^2} + \frac{x^4}{(4!)^2} +.... = 0 \nonumber \end{equation}$

## Homework Equations

To find out the power series in the LHS of the given equation.

## The Attempt at a Solution

I have tried to solve it by constructing a differential equation for the LHS expression (=g(x) say) as:

$\begin{equation} (xg(x)')' + g(x) =0 \nonumber \end{equation}$

which gives the solution for g(x) as Bessel function of first kind and zero order.
But, I am still not fully convinced regarding the idea of "recognising" the power series in the LHS. Is there any other algebraic approach towards this problem ?

gabbagabbahey
Homework Helper
Gold Member
Bessel function? Yikes It's way easier than that. Hint: what is $$\frac{(-1)^0 x^0}{0!}$$? How about $$\frac{(-1)^1 x^1}{1!}$$? How about $$\frac{(-1)^2 x^2}{2!}$$? Do you see the resemblance to the LHS? If you are trying to point to $e^{-x}$ by any chance, then let me bring to your attention the powers that are present over the factorial terms, which unlike $e^{-x}$ are not equal to one. Or else, I am too ignorant to "see" anything substantial for now. gabbagabbahey
Homework Helper
Gold Member
Oops, I didn't see the powers in the factorial before.

Anyways, it should be clear that the (n-1)th term on the LHS is $$\frac{(-1)^n x^n}{(n!)^2}$$ just by inspection. And so your LHS can be written as $$\sum_{n=0}^{\infty}\frac{(-1)^n x^n}{(n!)^2}$$

If you compare that to the zero order Bessel function : $$J_0(u)=\sum_{n=0}^{\infty}\frac{(-1)^n (u/2)^{2n}}{(n!)^2}$$ You should see that your LHS is $$J_0(2\sqrt{x})$$

Yes, that's right. But is there any way to show this algebraically, and not by jumping directly to the form of the Bessel function per se.

gabbagabbahey
Homework Helper
Gold Member
I'm not sure what you mean by "show this algebraically". It is easy to show that your LHS is equivalent to $J_0(2\sqrt{x})$ by simply writing out the terms in the Bessel function's power series. I don't see any non-algebraic steps in my method?

Do you mean that you want to explicitly sum the series on the LHS the way you would for the series of $e^{-x}$? If so, I don't see how.

I think the only way is to simply recognize what the series is by inspection, or set up a differential equation for the LHS.

Yes exactly. All I want to know is if it is remotely possible to deliver the same solution by explicitly summing up the series, or splitting the series into seemingly identifiable products. Anyways, I think the approach that we have with us should do it for now. Thank you so much !

Pardone my ignorance, for i am not that much experienced with power series, so i might just be throwing usless stuff here.

However, would the following be correct ?

$$e^{-x}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^n$$

So,from here would the following hold(i'm not sure though):

$$\frac{e^{-x}}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(n!)^2}x^n$$

gabbagabbahey
Homework Helper
Gold Member
Pardone my ignorance, for i am not that much experienced with power series, so i might just be throwing usless stuff here.

However, would the following be correct ?

$$e^{-x}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^n$$

So,from here would the following hold(i'm not sure though):

$$\frac{e^{-x}}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(n!)^2}x^n$$

No. On the RHS of your expression, n is a dummy variable; an index that is being summed over. On the LHS, n is not a dummy variable (in fact, you haven't defined what n is!).