# Homework Help: Jackson eqs. 3.25 and 3.26

1. Oct 18, 2009

### Bill Foster

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

How do you get from (3.25) to (3.26) in Jackson?

2. Relevant equations

Equation 3.25:

$$A_l=\left(2l+1\right)\int_0^1P_l\left(x\right)dx$$

Equation 3.26:

$$A_l=\left(-\frac{1}{2}\right)^{\frac{l-1}{2}}\frac{\left(2l+1\right)\left(l-2\right)!!}{2\left(\frac{l+1}{2}\right)!}$$

Rodriques:

$$P_l\left(x\right)=\frac{1}{2^l l!}\frac{d^l}{dx^l}\left(x^2-1\right)^l$$

3. The attempt at a solution

Put Rodrigues into (3.25):

$$A_l=\frac{\left(2l+1\right)}{2^l l!}\int_0^1\frac{d^l}{dx^l}\left(x^2-1\right)^ldx$$

Evaluate:

$$\frac{d^l}{dx^l}\left(x^2-1\right)^l$$

$$=\frac{d^l}{dx^l}\sum_k{\left(\frac{l!}{k!\left(l-k\right)!}\right)\left(x^2\right)^{l-k}\left(-1\right)^k}$$

We worked this out in class, and somehow that sum goes away. But I forgot how.

As I work out the derivatives, I end up with:

$$=\sum_k{\left(\frac{l!}{k!\left(l-k\right)!}\right)2\left(l-k\right)\left(2\left(l-k\right)-1\right)\left(2\left(l-k\right)-2\right)\left(2\left(l-k\right)-3\right)...\left(2\left(l-k\right)-\left(l-1\right)\right)x^{l-2k}\left(-1\right)^k$$

Any value of $$k>\frac{l}{2}$$ and the expression is zero. So the sum only goes from $$k=0$$ to $$\frac{l-1}{2}$$.

How do I get rid of that sum completely?

Last edited: Oct 18, 2009
2. Oct 18, 2009

### gabbagabbahey

Looks like Jackson contains a typo; this should should be:

$$A_l=(-1)^{l-1}\left(\frac{1}{2}\right)^{\frac{l-1}{2}}\frac{\left(2l+1\right)\left(l-2\right)!!}{2\left(\frac{l+1}{2}\right)!}$$

You've messed up the binomial expansion; you should have:

$$=\frac{d^l}{dx^l}\sum_{k=0}^{\infty}{\left(\frac{l!}{k!\left(l-k\right)!}\right)x^{2k}\left(-1\right)^{l-k}}$$

3. Oct 18, 2009

### Bill Foster

That still doesn't get rid of the sum. It only gets rid of the values $$k=0$$ to $$k=\frac{l-1}{2}$$.

4. Oct 19, 2009

### gabbagabbahey

Okay, so that leaves you with

$$A_l=\frac{2l+1}{2^ll!}\sum_{k=\frac{l-1}{2}}^\infty\frac{l!}{k!(l-k)!}}(-1)^{l-k}\int_0^1\frac{d^l}{dx^l}(x^{2k})dx$$

Now, I think what you want to do is make the substitution $j=k-\frac{l-1}{2}$...

Last edited: Oct 20, 2009
5. Oct 20, 2009

### Bill Foster

$$\frac{d^l}{dx^l}\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)x^{2k}\left(-1\right)^{l-k}}$$

Differentiate once (1st time):

$$\frac{d^{l-1}}{dx^{l-1}}\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2kx^{2k-1}\left(-1\right)^{l-k}}$$

Differentiate again (2nd time):

$$\frac{d^{l-2}}{dx^{l-2}}\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)x^{2k-2}\left(-1\right)^{l-k}}$$

Differentiate again (third time):

$$\frac{d^{l-3}}{dx^{l-3}}\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)x^{2k-3}\left(-1\right)^{l-k}}$$

...
...

Differentiate lth time:

$$\frac{d^{l-l}}{dx^{l-l}}\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}$$

That leaves me with:

$$\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}$$

Now, if $$k=0$$, the whole expression is zero. So we can rewrite it as:

$$\sum_{k=1}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}$$

But if $$k=\frac{1}{2}$$, then the expression is also zero. But k can only be an integer. However, due to the term $$2k-2$$, if $$k=1$$ the expression is zero. So rewrite as:

$$\sum_{k=2}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}$$

Now let's just cut to the chase:

$$(2k-(l-1)$$

If $$k=\frac{l-1}{2}$$ or less, the expression is zero.

So rewrite as:

$$\sum_{k=\frac{l-1}{2}+1}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}$$

Which can be written as:

$$\sum_{k=\frac{l+1}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}$$

That only works if l is odd. If l is even, then it would be written as:

$$\sum_{k=\frac{l}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}$$

So we have the following:

$$A_l=\frac{\left(2l+1\right)}{2^l l!}\int_0^1\frac{d^l}{dx^l}\left(x^2-1\right)^ldx$$
$$=\frac{\left(2l+1\right)}{2^l l!}\int_0^1\sum_{k=\frac{l}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))x^{2k-l}\left(-1\right)^{l-k}}dx$$
$$=\frac{\left(2l+1\right)}{2^l l!}\sum_{k=\frac{l}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))\left(-1\right)^{l-k}\int_0^1x^{2k-l}}dx$$ for even l
$$=\frac{\left(2l+1\right)}{2^l l!}\sum_{k=\frac{l+1}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))\left(-1\right)^{l-k}\int_0^1x^{2k-l}}dx$$ for odd l

Now: $$\int_0^1x^{2k-l}dx=\frac{1}{2k-l+1}x^{2k-l+1}$$ evaluated from 1 to 0:

That would equal:

$$\int_0^1x^{2k-l}dx=\frac{1}{2k-l+1}$$

So now we have:

$$A_l=\frac{\left(2l+1\right)}{2^l l!}\sum_{k=\frac{l}{2},\frac{l+1}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-1))\left(-1\right)^{l-k}\frac{1}{2k-l+1}$$

Which can be written as:

$$A_l=\frac{\left(2l+1\right)}{2^l l!}\sum_{k=\frac{l}{2},\frac{l+1}{2}}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-2))\left(-1\right)^{l-k}$$

Now I still have that sum.

Last edited by a moderator: Apr 24, 2017
6. Oct 20, 2009

### gabbagabbahey

You might want to re-read my last post; summing over $j$ instead of $k$ will probably make things easier for you...

7. Oct 21, 2009

### Bill Foster

I don't see how. It changes the limits on the sum:

$$A_l=\frac{\left(2l+1\right)}{2^l l!}\sum_{j=1}^{\frac{l+1}{2}}{\left(\frac{l !}{k!\left(l-k\right)!}\right)2k(2k-1)(2k-2)...(2k-(l-2))\left(-1\right)^{l-k}$$

And all the k's become more complicated:

$$A_l=\frac{\left(2l+1\right)}{2^l l!}\sum_{j=1}^{\frac{l+1}{2}}{\left(\frac{l !}{\left(j+\frac{l-1}{2}\right)!\left(l-\left(j+\frac{l-1}{2}\right)\right)!}\right)2\left(j+\frac{l-1}{2}\right)(2\left(j+\frac{l-1}{2}\right)-1)(2\left(j+\frac{l-1}{2}\right)-2)...(2\left(j+\frac{l-1}{2}\right)-(l-2))\left(-1\right)^{l-\left(j+\frac{l-1}{2}\right)}$$

8. Oct 21, 2009

### gabbagabbahey

Why are you summing from $j=1$ to $j=(l+1)/2$?

You started out with something like

$$A_l=\sum_{k=0}^{\infty}c_k\int_0^1\frac{d^l}{dx^l}x^{2k}dx=\sum_{k=0}^{\infty}c_k\left.\left[\frac{d^{l-1}}{dx^{l-1}}x^{2k}\right]\right|_0^1$$

The terms for which $2k<l-1$ are all zero, and $l$ is odd (Remember, Jackson showed the $A_l=0$ for even $l$, so this sum is just to determine the odd coefficients), So the first $\frac{l-1}{2}-1$ terms of the sum are zero and you are left with:

$$A_l=\sum_{k=\frac{l-1}{2}}^{\infty}c_k\left.\left[\frac{d^{l-1}}{dx^{l-1}}x^{2k}\right]\right|_0^1$$

So the substitution $j=k-\frac{l-1}{2}$should give you

$$A_l=\sum_{j=0}^{\infty}c_{j+\frac{l-1}{2}}\left.\left[\frac{d^{l-1}}{dx^{l-1}}x^{2j+l-1}\right]\right|_0^1=\sum_{j=0}^{\infty}c_{j+\frac{l-1}{2}}\left.\left[(2j+l-1)(2j+l-2)\ldots(2j+1)x^{2j}\right]\right|_0^1=\sum_{j=0}^{\infty}c_{j+\frac{l-1}{2}}\frac{(2j+l-1)!}{(2j)!}$$

Last edited: Oct 21, 2009
9. Oct 21, 2009

### Bill Foster

The sum doesn't go to infinity. Using k as the index variable, it goes from 0 to l:

$$\frac{d^l}{dx^l}\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)x^{2k}\left(-1\right)^{l-k}}$$

10. Oct 21, 2009

### gabbagabbahey

No, it goes to infinity. You need to review the binomial expansion.

$$(x^2-1)^l=\sum_{k=0}^{\infty}{\left(\frac{l !}{k!\left(l-k\right)!}\right)x^{2k}\left(-1\right)^{l-k}}\neq\sum_{k=0}^{l}{\left(\frac{l !}{k!\left(l-k\right)!}\right)x^{2k}\left(-1\right)^{l-k}}$$

11. Oct 21, 2009

### Bill Foster

http://en.wikipedia.org/wiki/Binomial_theorem

12. Oct 21, 2009

### Bill Foster

Because if $$k=\frac{l-1}{2}$$, that term is zero. So the sum has to start at $$k=\frac{l+1}{2}$$.

If $$j=k-\frac{l-1}{2}$$, then the first term for j is when $$k=\frac{l+1}{2}$$:

$$j=k-\frac{l-1}{2} = \frac{l+1}{2}-\frac{l-1}{2} = 1$$.

The last term is $$k = l$$

So $$j=k-\frac{l-1}{2} = l-\frac{l-1}{2} = \frac{l+1}{2}$$

13. Oct 21, 2009

### gabbagabbahey

Contrary to popular belief, wikipedia is not always the most accurate source for information.

Edit: After my morning cup of coffee, I realized that there is nothing wrong with only summing up to $k=l$. However, the sum can be extended to infinity since $$\begin{pmatrix}l\\k\end{pmatrix}$$ is zero for $k>l$.

Last edited: Oct 22, 2009
14. Oct 22, 2009

### Bill Foster

I would like to direct your attention to equation 2 in that link, and the text that immediately precedes it.

It would only go to infinity if l goes to infinity.

15. Oct 22, 2009

### gabbagabbahey

Anyways, can you agree that

$$A_l=-\frac{2l+1}{2^l}\sum_{j=0}^{j=\frac{l+1}{2}}\frac{(-1)^j(2j+l-1)!}{(j+\frac{l-1}{2})!(\frac{l+1}{2}-j)!(2j)!}$$

?

16. Oct 23, 2009

### Bill Foster

It looks clearer written in terms of k.

17. Oct 23, 2009

### gabbagabbahey

Since $l$ is odd, make the substitution $l=2n+1$....