MHB Derivative of bessel function informal proof

[skate_nerd]

Not exactly sure where this post belongs, but it is a problem from my P.D.E. class so I'll leave it here. Feel free to move it if you like...

I need to prove the differentiation theorem for the Bessel function, 1st kind. I've gotten considerably close, but the last bit is really making my brain itch for the last few days. I'm at a roadblock. It would be nice if somebody told me not just the answer, but the property of exponents/factorials that I am forgetting about that would lead me in the right direction.

So far I am pretty positive I did everything right, because I started with

$$\frac{d}{dz}J_m(z)=\frac{d}{dz}\sum_{k=0}^{\infty}\frac{(-1)^k}{k!(k+m)!}(\frac{z}{2})^{2k+m}$$
$$\frac{d}{dz}J_m(z)=\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^{k}(2k+m)}{k!(k+m)!}(\frac{z}{2})^{2k+m-1}$$
$$\frac{d}{dz}J_m(z)=\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^{k}(k+(k+m))}{k!(k+m)!}(\frac{z}{2})^{2k+m-1}$$
Then using the fact that
$$\frac{k}{k!}=\frac{1}{(k-1)!}$$
and
$$\frac{(k+m)}{(k+m)!}=\frac{1}{(k+m-1)!}$$
I got to
$$\frac{d}{dz}J_m(z)=\frac{1}{2}[J_{m-1}(z)+\sum_{k=0}^{\infty}\frac{(-1)^k}{(k-1)!(k+m)!}(\frac{z}{2})^{2k+m-1}$$
This is why I think I've done everything right, because I at least got the first term of the formula right. Since I am trying to show ultimately that
$$\frac{d}{dz}J_m(z)=\frac{1}{2}[J_{m-1}(z)-J_{m+1}(z)]$$
Then this means I need to show that
$$\frac{1}{(k-1)!(k+m)!}(\frac{z}{2})^{2k+m-1}=-\frac{1}{k!(k+m+1)!}(\frac{z}{2})^{2k+m+1}$$
but I just really do not see how that is possible.

Any hints would be appreciated! Thanks guys!

 

[I like Serena]

$$\frac{(k+m)}{(k+m)!}=\frac{1}{(k+m-1)!}$$

This is off by a minus sign.

Then this means I need to show that
$$\frac{1}{(k-1)!(k+m)!}(\frac{z}{2})^{2k+m-1}=-\frac{1}{k!(k+m+1)!}(\frac{z}{2})^{2k+m+1}$$
but I just really do not see how that is possible.

That won't be true.
Substitute for instance $k=2, m=1$ and you'll see there's a mismatch.

 

[skate_nerd]

When you say
$$\frac{(k+m)}{(k+m)!}=\frac{1}{(k+m-1)!}$$
is off by a minus sign, do you mean it is really supposed to be
$$\frac{(k+m)}{(k+m)!}=-\frac{1}{(k+m-1)!}$$
and if so, I don't see how that would be the case...

Also, I had an inkling that the last point I got to couldn't be right, but how do you explain how I got the first term of the formula? I had to have done that much right, wouldn't you say?

 

[I like Serena]

When you say
$$\frac{(k+m)}{(k+m)!}=\frac{1}{(k+m-1)!}$$
is off by a minus sign, do you mean it is really supposed to be
$$\frac{(k+m)}{(k+m)!}=-\frac{1}{(k+m-1)!}$$
and if so, I don't see how that would be the case...

Sorry.
My mistake. It is correct.

 

[I like Serena]

I believe the key is to match the powers of $\frac z 2$.

Substitute $k = l - 1$ in:
\begin{aligned}J_{m+1}
&= \sum_{k=0} \frac{(-1)^k}{k!(k+m+1)!} \left( \frac z 2 \right)^{2k+m+1} \\
&= \sum_{l=1} \frac{(-1)^{l-1}}{(l-1)!(l+m)!} \left( \frac z 2 \right)^{2l+m-1} \\
&= -\sum_{k=1} \frac{(-1)^k}{(k-1)!(k+m)!} \left( \frac z 2 \right)^{2k+m-1}
\end{aligned}

Now that we have matching powers:
\begin{aligned}
J_{m-1} - J_{m+1}
&= \sum_{k=0} \frac{(-1)^k}{k!(k+m-1)!} \left( \frac z 2 \right)^{2k+m-1}
+ \sum_{k=1} \frac{(-1)^k}{(k-1)!(l+m)!} \left( \frac z 2 \right)^{2k+m-1} \\
&= \frac 1 {(m-1)!} \left( \frac z 2 \right)^{m-1}
+ \sum_{k=1} (-1)^k \left( \frac{1}{k!(k+m-1)!} + \frac{1}{(k-1)!(k+m)!} \right)
\left( \frac z 2 \right)^{2k+m-1} \\
&= \frac 1 {(m-1)!} \left( \frac z 2 \right)^{m-1}
+ \sum_{k=1} \frac{(-1)^k (2k+m)}{k!(k+m)!} \left( \frac z 2 \right)^{2k+m-1} \\
&= \sum_{k=0} \frac{(-1)^k (2k+m)}{k!(k+m)!} \left( \frac z 2 \right)^{2k+m-1}
\end{aligned}

 

[skate_nerd]

How did you get that negative sign in your working out of \(J_{m+1}(z)\)?

 

[I like Serena]

How did you get that negative sign in your working out of \(J_{m+1}(z)\)?

$(-1)^{l-1} = -1 \cdot (-1)^l$

 

[skate_nerd]

Ah, my bad. I should have noticed that.

Well thanks for the idea though! Changing the index of the sum seems obvious now, since it doesn't change anything about the final result for the Bessel function. I just worked through it and got the desired result, actually substituting \(k=l+1\) instead. Finally, that problem is done (Whew)