Power Series Problem: Bessel Function Solution

[kingturtle]

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]

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?

 

[kingturtle]

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.

Thanks for replying!

 

[gabbagabbahey]

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 $$J_0(2\sqrt{x})$$

 

[kingturtle]

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]

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.

 

[kingturtle]

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 !

 

[sutupidmath]

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]

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!).