Power Series Problem

  • Thread starter kingturtle
  • Start date
  • #1

Homework Statement



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

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:

[itex]
\begin{equation}
(xg(x)')' + g(x) =0 \nonumber
\end{equation}
[/itex]

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 ?
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
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Bessel function? Yikes :eek:

It's way easier than that. Hint: what is [tex]\frac{(-1)^0 x^0}{0!}[/tex]? How about [tex]\frac{(-1)^1 x^1}{1!}[/tex]? How about [tex]\frac{(-1)^2 x^2}{2!}[/tex]? Do you see the resemblance to the LHS?:wink:
 
  • #3
If you are trying to point to [itex]e^{-x}[/itex] by any chance, then let me bring to your attention the powers that are present over the factorial terms, which unlike [itex]e^{-x}[/itex] are not equal to one. Or else, I am too ignorant to "see" anything substantial for now. :confused:

Thanks for replying!
 
  • #4
gabbagabbahey
Homework Helper
Gold Member
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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 [tex]\frac{(-1)^n x^n}{(n!)^2}[/tex] just by inspection. And so your LHS can be written as [tex]\sum_{n=0}^{\infty}\frac{(-1)^n x^n}{(n!)^2}[/tex]

If you compare that to the zero order Bessel function : [tex]J_0(u)=\sum_{n=0}^{\infty}\frac{(-1)^n (u/2)^{2n}}{(n!)^2}[/tex] You should see that your LHS is [tex]J_0(2\sqrt{x})[/tex]
 
  • #5
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.
 
  • #6
gabbagabbahey
Homework Helper
Gold Member
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I'm not sure what you mean by "show this algebraically". It is easy to show that your LHS is equivalent to [itex]J_0(2\sqrt{x})[/itex] 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 [itex]e^{-x}[/itex]? 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.
 
  • #7
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 !
 
  • #8
1,631
4
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 ?

[tex]e^{-x}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^n[/tex]

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

[tex]\frac{e^{-x}}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(n!)^2}x^n[/tex]
 
  • #9
gabbagabbahey
Homework Helper
Gold Member
5,002
7
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 ?

[tex]e^{-x}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}x^n[/tex]

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

[tex]\frac{e^{-x}}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(n!)^2}x^n[/tex]

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!).
 

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