Bessel functions of the first kind

[John 123]

Homework Statement

Can anyone tell me if:
<br /> \frac{d}{dx}J_k(ax)=aJ&#039;_k(x)<br />
where a is a real positive constant and
<br /> J_k(x) <br />
is the Bessel function of the first kind.
Regards
John

Homework Equations

The Attempt at a Solution

 

[Char. Limit]

I believe that the answer to that is no. I believe that...

\frac{d}{dx}J_k(ax) = a J_k^{&#039;}(ax)

 

[John 123]

But doesn't
<br /> J_k(ax)=aJ_k(x)?<br />

 

[Char. Limit]

Sorry, but that is not the case.

 

[John 123]

Then that is my error.
To find the derivative of
<br /> J_k(ax)<br />
presumably one has to differentiate the series expansion?
John

 

[Dick]

Then that is my error.
To find the derivative of
<br /> J_k(ax)<br />
presumably one has to differentiate the series expansion?
John

You can express the derivative of a bessel function in terms of other bessel functions. See the "Selected identities" list at the end of http://en.wikipedia.org/wiki/Bessel_function

 

[John 123]

Hi Dick
The question I am asking derives from proving the integral property of Bessel functions of the first kind. This amounts to showing the orthogonal properties of Bessel functions.
Part way through the proof I need to show that:
<br /> u&#039;_1(x)=r_ix^(\frac{1}{2})J&#039;_k(r_ix)+\frac{1}{2}x^{\frac{-1}{2}}J_(r_ix)<br />
where
<br /> r_i<br />
is a distinct positive zero of
<br /> J_k(x)<br />

 

[John 123]

Hi Dick
I need to show that if:
<br /> u(x)=x^{\frac{1}{2}}J_k(r_ix)<br />
Then:
<br /> u&#039;=r_ix^{\frac{1}{2}}J&#039;_k(r_ix)+\frac{1}{2}x^{\frac{-1}{2}}J_k(r_ix)<br />
This, of course, uses product rule but the derivative of:
<br /> J_k(r_ix)<br />
I am unclear about?
Incidentally
<br /> r_i<br />
is the ith distinct positive zero of
<br /> J_k(x)<br />
This is part of proving the orthogonality of Bessel functions
<br /> J_k(r_ix)<br />
with respect to the weight function x.
John

 

[Dick]

It's just the chain rule. If h(x)=f(a*x) then h'(x)=f'(a*x)*a. It's a special case of the general chain rule, if h(x)=f(g(x)) then h'(x)=f'(g(x))*g'(x). It's just like saying the derivative of sin(2*x) is cos(2*x)*2.

 

[John 123]

Thanks Dick
John