# Derivative of a term within a sum

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1. Feb 13, 2019

### CricK0es

1. The problem statement, all variables and given/known data

From the Rodrigues’ formulae, I want to derive nature of the spherical Bessel and Neumann functions at small values of p.

2. Relevant equations

I'm going to post an image of the Bessel function where we're using a Taylor expansion, which I'm happy with and is as far as I got...

https://imgur.com/a/LmDovch

3. The attempt at a solution

I'm recovering some of the terms, but the jump from line 1 to 2 and 2 to 3 have me confused because I'm not recovering a similar answer.

I need to attempt it for the Neumann function at small p also, but I'm sure that a similar issue will arise and I'd rather fix it here and use the Neumann as practice for my newfound knowledge

Last edited by a moderator: Feb 13, 2019
2. Feb 13, 2019

### kuruman

How about posting your work so far? You may have made a mistake that a pair of fresh eyes might be able to diagnose.

3. Feb 13, 2019

### SammyS

Staff Emeritus
I presume that the word "Derivative" in your title should be changed to some form of the word "derive" as in the context of a "derivation".

Boldface type should only be used for the three headings supplied by the Homework Template and in rare cases for items in your post which need emphasis.
(I realize that it is easy to include text in the supplied headings by mistake .​

Please use either the "Image" icon in the banner heading or the upload button found below - right of the Input Box. (See figure below.)

So it seems that you want to know why
$(2n)(2n-2)(2n-4) \dots (2n-2\lambda +2)$​
is equal to
$\dfrac {2^n (n!)} {(n-\lambda)!}$​
Right?

Actually, it looks to me as if that last expression should be $\dfrac {2^\lambda (n!)} {(n-\lambda)!}$ .

Try expanding $\dfrac { (n!)} {(n-\lambda)!}$ and see what you get.

How many terms are in $(2n) ,~ (2n-2),~(2n-4),~ \dots ,~ (2n-2\lambda +2) ~?$

(& fixed a typo)

Last edited: Feb 14, 2019 at 4:09 PM
4. Feb 15, 2019 at 8:44 AM

### CricK0es

I can't see where the (2n)(2n-2).... term is coming from, and then from the solutions, I don't see how we go from line 2 to 3.

So, this final question has not gone well... at all ha! So there's some very key thing in my understanding that I'm missing here, as my original attempt above indicates

5. Feb 15, 2019 at 9:59 AM

### kuruman

If series confuse you, I recommend that you write out a few terms and see how things go together.

The series is $S(\rho)=a_0+a_2\rho^2+a_4\rho^4+\cdots+a_{2n-2}\rho^{2n-2}+ a_{2n}\rho^{2n}+\cdots$
For $\lambda=1$,$$\frac{1}{\rho}\frac{\partial S(\rho)}{\partial \rho}=\frac{1}{\rho}(2a_2\rho^1+4a_4\rho^3+\cdots+(2n-2)a_{2n-2}\rho^{2n-2-1}+ 2na_{2n}\rho^{2n-1}+\cdots)\\=2a_2\rho^0+4a_4\rho^2+\cdots+(2n-2)a_{2n-2}\rho^{2n-4}+ 2na_{2n}\rho^{2n-2}+\cdots$$For $\lambda=2$, take the first derivative of the above and divide by another $\rho$ because you have already divided by one when you did $\lambda=1$. Thus,
$$\frac{1}{\rho^2}\frac{\partial^2 S(\rho)}{\partial \rho^2}=4\cdot 2a_4\rho^0+\cdots+(2n-2)(2n-4)a_{2n-2}\rho^{2n-6}+ 2n(2n-2)a_{2n}\rho^{2n-4}+\cdots$$See how it works? If not, do $\lambda=3$ in the same manner.

Last edited: Feb 15, 2019 at 10:16 AM