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I'm trying to see that $J_0(x)$ is indeed a solution for the Bessel equation $x^2y''+xy'+x^2y=0$, so: $$J_0(x)=\sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(k!)^22^{2k}}$$ Pluging it in the equation and and operating let's me with this instead of $0$: $$\sum_{k=0}^\infty \frac{4k^2(-1)^kx^{2k}}{(k!)^22^{2k}}+\sum_{k=0}^\infty \frac{(-1)^k x^{2k+2}}{(k!)^22^{2k}}$$

The differentiation was done (by computer) term by term, and that's the only think that comes to my mind that may be wrong, but power series can be differentiated term by term, right?

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2 Answers

up vote 3 down vote accepted

The first sum is really from $k=1$, because the term with $k=0$ is zero. It also simplifies: $$\sum_{k=1}^\infty \frac{4k^2(-1)^kx^{2k}}{(k!)^22^{2k}} = \sum_{k=1}^\infty \frac{(-1)^k x^{2k}}{((k-1)!)^2 2^{2(k-1)}} \tag1$$ Now change the index, writing $m$ instead of $k-1$: $$ \dots = \sum_{m=0}^\infty \frac{(-1)^{m+1} x^{2(m+1)}}{(m!)^2 2^{2m}} = -\sum_{m=0}^\infty \frac{(-1)^{m} x^{2(m+1)}}{(m!)^2 2^{2m}} \tag2$$ Of course, the index of summation is a dummy variable. We can just as well call it $k$ again. Then the sum in (2) is exactly your second sum, with the opposite sign*

except that your second sum has a typo in the exponent.

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I totally missed that, thank you very much. –  MyUserIsThis Feb 28 '13 at 20:51
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You can use the differentiation theorem for power series with the Bessel function as the radius of convergence is $\infty$

If you let $y = \displaystyle\sum_{k=0}^{\infty} \frac{(-1)^k}{(k!)^2} \left(\frac{x}{2}\right)^{2k}$

Then

$y' = \displaystyle\sum_{k=0}^{\infty} \frac{k(-1)^k}{(k!)^2} \left(\frac{x}{2}\right)^{2k-1}$

and

$y'' = \displaystyle\sum_{k=0}^{\infty} \frac{k(2k-1)(-1)^k}{2(k!)^2} \left(\frac{x}{2}\right)^{2k-2}$

Thus

$x^2y'' + xy' + x^2y = x^2\displaystyle\sum_{k=0}^{\infty} \frac{k(2k-1)(-1)^k}{2(k!)^2} \left(\frac{x}{2}\right)^{2k-2} + x\displaystyle\sum_{k=0}^{\infty} \frac{k(-1)^k}{(k!)^2} \left(\frac{x}{2}\right)^{2k-1} + x^2\displaystyle\sum_{k=0}^{\infty} \frac{(-1)^k}{(k!)^2} \left(\frac{x}{2}\right)^{2k} $

Note that

$\displaystyle\sum_{k=0}^{\infty} \frac{k(2k-1)(-1)^k}{2(k!)^2} \left(\frac{x}{2}\right)^{2k-2} = \displaystyle\sum_{k=0}^{\infty} \frac{k(k)(-1)^k}{(k!)^2} \left(\frac{x}{2}\right)^{2k-2} - \frac{k(-1)^k}{2(k!)^2} \left(\frac{x}{2}\right)^{2k-2} $

So

$x^2y'' + xy' + x^2y = x^2\displaystyle\sum_{k=0}^{\infty} \frac{k(k)(-1)^k}{(k!)^2} \left(\frac{x}{2}\right)^{2k-2} - x^2\frac{k(-1)^k}{2(k!)^2} \left(\frac{x}{2}\right)^{2k-2} + x\displaystyle\sum_{k=0}^{\infty} \frac{k(-1)^k}{(k!)^2} \left(\frac{x}{2}\right)^{2k-1} + x^2\displaystyle\sum_{k=0}^{\infty} \frac{(-1)^k}{(k!)^2} \left(\frac{x}{2}\right)^{2k} $

$ = x^2\displaystyle\sum_{k=0}^{\infty} \frac{k(k)(-1)^k}{(k!)^2} \left(\frac{x}{2}\right)^{2k-2} - x^2\frac{k(-1)^k}{2(k!)^2} \left(\frac{x}{2}\right)^{2k-2} + x\displaystyle\sum_{k=0}^{\infty} \frac{k(-1)^k}{(k!)^2} \left(\frac{x}{2}\right)^{2k-2}\left(\frac{x}{2}\right) + x^2\displaystyle\sum_{k=0}^{\infty} \frac{(-1)^k}{(k!)^2} \left(\frac{x}{2}\right)^{2k} $

Now you should be able to see the second and third term cancel. Thus

$x^2y'' + xy' + x^2y = x^2\displaystyle\sum_{k=0}^{\infty} \frac{k(k)(-1)^k}{(k!)^2} \left(\frac{x}{2}\right)^{2k-2} + x^2\displaystyle\sum_{k=0}^{\infty} \frac{(-1)^k}{(k!)^2} \left(\frac{x}{2}\right)^{2k}$

$= x^2\left[\displaystyle\sum_{k=0}^{\infty} \frac{k^2(-1)^k}{(k!)^2} \left(\frac{x}{2}\right)^{2k-2} + \displaystyle\sum_{k=0}^{\infty} \frac{(-1)^k}{(k!)^2} \left(\frac{x}{2}\right)^{2k}\right]$

$= x^2\left[\displaystyle\sum_{k=1}^{\infty} \frac{(-1)^k}{((k-1)!)^2} \left(\frac{x}{2}\right)^{2k-2} + \displaystyle\sum_{k=0}^{\infty} \frac{(-1)^k}{(k!)^2} \left(\frac{x}{2}\right)^{2k}\right]$

$= x^2\left[\displaystyle\sum_{m=0}^{\infty} \frac{(-1)^{m+1}}{((m)!)^2} \left(\frac{x}{2}\right)^{2m} - \displaystyle\sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{(k!)^2} \left(\frac{x}{2}\right)^{2k}\right]$

The two series are the same, thus cancel giving $0$

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The radius of convergence is $\infty$, not $2$. –  Antonio Vargas Feb 28 '13 at 20:55
    
@AntonioVargas Sorry, yes, it is $\infty$ I've corrected my post. –  Noble. Feb 28 '13 at 20:59
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