(Bessel Functions) I'm trying to show a general property of a certain Bessel Function and I'm running into some confusion. More info below.

Question

mendicant_bias · Answer

Original prompt: $$\text{Show that $ J_0^{'} (x)=-J_1(x). $ }$$

Workings so far: $$J_n(x)=\sum_{k=1}^{\infty}\frac{(-1)^k}{k! \Gamma(k+1+n)}\bigg(\frac{x}{2}\bigg)^{2k+n}$$\
$$J_0(x)=\sum_{k=1}^{\infty}\frac{(-1)^k}{k! \ \Gamma(k+1)}\bigg(\frac{x}{2}\bigg)^{2k}$$
$$J'_{0}(x)=\sum_{k=1}^{\infty}\frac{d}{dx}\bigg(\frac{(-1)^k}{k! \ \Gamma(k+1)}\bigg( \frac{x}{2} \bigg)^{2k}\bigg)$$
$$= \sum_{k=1}^{\infty}\frac{2k (-1)^k}{k! \ \Gamma(k+1)}\bigg(\frac{x}{2}\bigg)^{2k-1.}$$

mendicant_bias · Answer

$$-J_1(x)= -\sum_{k=1}^{\infty}\frac{(-1)^k}{k! \ \Gamma(k+1+1)}\bigg( \frac{x}{2}\bigg)^{2k+1}$$

mendicant_bias · Answer

How are these equal? I can't figure it out. The negative sign in the second one makes me think that it might involve shifting of exponents, but I'm not entirely sure.

mendicant_bias · Answer

@SithsAndGiggles

mendicant_bias · Answer

I'm also wondering if this is the kind of situation where I should choose common properties of the Bessel or Gamma functions to work this out, or if that isn't necessary. I'd first and foremost try to do this through plain algebra.

mendicant_bias · Answer

@wio

mendicant_bias · Answer

@dan815

dan815 · Answer

http://prntscr.com/6kmsgi

mendicant_bias · Answer

Why? Because of the factorial, or what?

dan815 · Answer

d/dx(x/2) = 1/2

mendicant_bias · Answer

So you still have the k from differentiation in the numerator, or what.

dan815 · Answer

yes

dan815 · Answer

simply it out, and then see what happens when you shift it by 2 indeces

mendicant_bias · Answer

Or do you have k-1 in the denominator due to k factorial.

dan815 · Answer

right

dan815 · Answer

and the missing (k+1+1) multiplication is compensated by the fact that (x/2) start at lower power

dan815 · Answer

write out the first few terms and ull see where the comesnsation is coming from

mendicant_bias · Answer

compensation?

mendicant_bias · Answer

(And I'll follow up on what you've said and write that stuff out here, just working on other stuff right now simultaneously)

anonymous · Answer

$$\begin{align*}
J_0(x)&=\sum_{k=\color{red}0}^{\infty}\frac{(-1)^k}{k! \Gamma(k+1)}\left(\frac{x}{2}\right)^{2k}\\

\frac{d}{dx}[J_0(x)]&=\frac{d}{dx}\left[\sum\cdots\right]\\

&=\sum_{k=0}^\infty\frac{(-1)^k}{k!\Gamma(k+1)}\frac{d}{dx}\left[\left(\frac{x}{2}\right)^{2k}\right]\\

&=\sum_{k=0}^\infty\frac{(-1)^k}{k!\Gamma(k+1)}2k\left(\frac{x}{2}\right)^{2k-1}\frac{d}{dx}\left[\frac{x}{2}\right]\\

&=\sum_{k=0}^\infty\frac{(-1)^k}{(k-1)!\Gamma(k+1)}\left(\frac{x}{2}\right)^{2k-1}\\

&=\sum_{p+1=0}^\infty\frac{(-1)^{p+1}}{((p+1)-1)!\Gamma((p+1)+1)}\left(\frac{x}{2}\right)^{2(p+1)-1}\\

&=-\sum_{p=-1}^\infty\frac{(-1)^{p}}{p!\Gamma(p+2)}\left(\frac{x}{2}\right)^{2p+1}\end{align*}$$
At this point, I believe we assume that $p!=0$ for negative $p$, which gives the desired result.

mendicant_bias · Answer

That's a really strange assumption, I've never heard of something like that before, huh.

anonymous · Answer

I've seen it used in probability and stats when computing binomial probabilities. I have to correct myself though, since this makes the first term undefined. Instead, we consider the limit
$$\lim_{p\to-1^+}\Gamma(p)=-\infty$$
which means the first term (when $p=-1$) is zero. At least that's how I would read it.

mendicant_bias · Answer

(1) I also still don't get how the index shift is okay/acceptable, the sums have the exact same general form, but k = 0 =/= p = -1. Even with the shift, wouldn't p somehow have to be equal to zero for the expression to hold or be okay? Or is it just since for negative integers, since the factorial is equal to zero....well then, the expression wouldn't be defined. 

(2) Oh, and since we're only dealing with integer values of p/k, saying that the expression is equal to zero at p=-1 by limits is okay enough to say that the sums are effectively the same.

anonymous · Answer

To get a feel for why index shifts work, consider the power series for $e^x$:
$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\cdots$$
Take the derivative:
$$\frac{d}{dx}e^x=\frac{d}{dx}\sum_{n=0}^\infty\frac{x^n}{n!}=\sum_{n=0}^\infty \frac{nx^{n-1}}{n!}=0+1+x+\frac{1}{2}x^2+\cdots$$
In particular, when $n=0$, you have the first term being $0$, so we can ignore the $0$th term and the series above is equivalent to
$$\frac{d}{dx}e^x=\frac{d}{dx}\sum_{n=\color{red}1}^\infty\frac{x^n}{n!}=\sum_{n=\color{red}1}^\infty \frac{nx^{n-1}}{n!}$$
We can also cancel out a factor of $n$:
$$\frac{d}{dx}e^x=\sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!}$$
Now, if we let $k=n-1$, we have
$$\frac{d}{dx}e^x=\sum_{\color{red}{n-1=0}}^\infty \frac{x^{\color{red}k}}{\color{red}k!}=\sum_{\color{red}{k=0}}^\infty \frac{x^{\color{red}k}}{\color{red}k!}=e^x$$
You're saying that when $k=0$ it is not true that $p=-1$, but this is in fact true. It must be, since I'm setting $k=p+1$! In the same way, the derivative of $e^x$ is also $e^x$ because $k=0$ when $n=1$.

You're right that $p$ must start at $0$ in order for the series to exactly match up with the Bessel function definition, which is why it's necessary to show that the series indexed by $p=-1,0,1,2,\ldots$ must be identically zero for $p=-1$.

Assuming my limit-reasoning holds up (which I'm looking into currently), where I say that
$$p!=\Gamma(p+1)\to\color{red}+\infty\quad\text{as}\quad p\to-1^+$$
we are done, since this makes the first term ($p=-1$) approach $0$. (Notice the sign correction, though the result is the same.)

anonymous · Answer

Forget what I said about limits, I don't think we can use that sort of logic. Instead, look back at the derivative of $e^x$. We used the fact that for $n=0$, the derivative's series' first term is $0$. Similarly, for $k=0$ in the derivative of $J_0(x)$, the first term is $0$:
$$\begin{align*}
\frac{d}{dx}[J_0(x)]&=\frac{d}{dx}\left[\sum\cdots\right]\\

&=\sum_{k=0}^\infty\frac{(-1)^k}{k!\Gamma(k+1)}2k\left(\frac{x}{2}\right)^{2k-1}\frac{d}{dx}\left[\frac{x}{2}\right]\\

&=\sum_{k=\color{red}1}^\infty\frac{(-1)^k}{k!\Gamma(k+1)}2k\left(\frac{x}{2}\right)^{2k-1}\frac{d}{dx}\left[\frac{x}{2}\right]\\

&=\sum_{k=1}^\infty\frac{(-1)^k}{(k-1)!\Gamma(k+1)}\left(\frac{x}{2}\right)^{2k-1}\\

&=\sum_{k-1=0}^\infty\frac{(-1)^{k-1+1}}{(k-1)!\Gamma(k-1+2)}\left(\frac{x}{2}\right)^{2(k-1)+1}\\

&=\sum_{\color{red}p=0}^\infty\frac{(-1)^{\color{red}p+1}}{\color{red}p!\Gamma(\color{red}p+2)}\left(\frac{x}{2}\right)^{2\color{red}p+1}\\
&=-J_1(x)
\end{align*}$$

anonymous · Answer

We end up using the same index shift of $k-1=p\iff k=p+1$.

anonymous · Answer

We actually can use the limit reasoning, as it turns out. We just need to establish that $\dfrac{1}{\Gamma(x)}$ has zeroes for negative integers $x$.

mendicant_bias · Answer

That e^x derivative example I think is both good for what you directly/plainly stated, and the fact that d/dx (e^x) is always e^x. It's pretty much the one thing that's a crazy excellent example for that, because the power series obviously has to change, yet it has to be effectively the exact same series, term-by-term, as the original one.