Verifying Bessel's Differential Equation

[taiello805]

Hey everyone, I am trying to use the Bessel function:

f(x)= \sum_{k=0}^\infty \frac{(-1)^{k}x^{2k+1}}{2^{2k+1}(k!)(k+1)!}

To verify the Differential Equation:

x^2y''+xy'+(x^2-1)y=0

So I thought this was going to be really easy, but I am having some difficulties. First I calculated y,y',y'' to be:

y= \sum_{k=0}^\infty \frac{(-1)^{k}x^{2k+1}}{2^{2k+1}(k!)(k+1)!}

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

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

And then I just pluged those into the differential equation to yield:

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

Then I just put the x terms in the summation since they are constant to get:

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

And since the summations are all indexed the same and all have the same denominator, I combined them to one summation with one fraction:

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

But now that I have one huge fraction, after all the simplification I have done, I can't get the numerator to zero out since the DE is homogeneous.

Did I make an error in my calculation? Have I done everything correct, but there is even MORE simplification? And if so, can you help me? Or am I just approaching this problem completely wrong?

Thanks a TON!Tony

 

[ross_tang]

You should have expanded the term:

\left(x^2-1\right)\sum _{k=0}^{\infty } \frac{(-1)^kx^{2k+1}}{2^{2k+1}k!(k+1)!}

into 2 terms:

x^2\sum _{k=0}^{\infty } \frac{(-1)^kx^{2k+1}}{2^{2k+1}k!(k+1)!}-\sum _{k=0}^{\infty } \frac{(-1)^kx^{2k+1}}{2^{2k+1}k!(k+1)!}

You shift the index, and combine it again.

 

[taiello805]

I did that at first, but I thought that would'd help at all. Expanding that, putting the x's in the summation and then combining, I got:

\sum_{k=0}^\infty \frac{(-1)^k(2k+1)(2k)x^{2k+1}+(-1)^k(2k+1)x^{2k+1}+(-1)^kx^{2k+3}-(-1)^kx^{2k+1}}{2^{2k+1}(k!)(k+1)!}

But that still looks like it is going no where. The numerator doesn't look like it is zero, unless it is really cryptic and I am not catching it. Help?

 

[ross_tang]

You didn't do the shifting of the summation variables. You should do the following:

\sum _{k=0}^{\infty } \frac{(-1)^k(2k+1)(2k)x^{2k+1}+(-1)^k(2k+1)x^{2k+1}+(-1)^kx^{2k+3}-(-1)^kx^{2k+1}}{2^{2k+1}k!(k+1)!}

\sum _{k=0}^{\infty } \frac{(-1)^k(2k+1)(2k)x^{2k+1}+(-1)^k(2k+1)x^{2k+1}-(-1)^kx^{2k+1}}{2^{2k+1}k!(k+1)!}+\sum _{k=0}^{\infty } \frac{(-1)^kx^{2k+3}}{2^{2k+1}k!(k+1)!}

Do the shifting in this step:

\sum _{k=0}^{\infty } \frac{(-1)^k(2k+1)(2k)x^{2k+1}+(-1)^k(2k+1)x^{2k+1}-(-1)^kx^{2k+1}}{2^{2k+1}k!(k+1)!}+\sum _{k=1}^{\infty } \frac{(-1)^{k-1}x^{2k+1}}{2^{2k-1}(k-1)!k!}

Grouping it together again:

\sum _{k=1}^{\infty } \frac{(-1)^k(2k+1)(2k)x^{2k+1}+(-1)^k(2k+1)x^{2k+1}-(-1)^kx^{2k+1}+4k(k+1)(-1)^{k-1}x^{2k+1}}{2^{2k+1}k!(k+1)!}

You should be able to show that it is indeed zero.

 

[taiello805]

Ah, I see. That is indeed zero. Thanks a MILLION! However, I am a little fuzzy how you went from your 3rd to 4th step in your last post. I understand that you let j=k-1 to shift the end summation, but how were you able to recombine it back with the first summation when that one was indexed at k=0? I don't know if this is a stupid question, but that transition from step 3 to 4 just doesn't look natural to me.

 

[ross_tang]

Since I take out the k=0 term, and it is again zero! I just skip a bit.

 

[taiello805]

AHH! I see. Duh. Thanks soooooo much! :D