# Lagendre Polynomials - using the recursion relation

1. May 24, 2014

### rwooduk

(1) $$P_{l}(u)$$ is normalised such that $$P_{l}(1) = 1$$. Find $$P_{0}(u)$$ and $$P_{2}(u)$$

We have the recursion relation:

$$a_{n+2} = \frac{n(n+1) - l(l+1)}{(n+2)(n+1)}a_{n}$$

I'm going to include a second similar question, which I'm hoping is solved in a similar way, so I can relate it to the above question:

(2) Show that the functions $$H_{0}(x) = 1$$, $$H_{1}(x) = 2x$$, $$H_{2}(x) = 4x^{2} - 2$$, $$H_{3}(x) = 8x^{3} - 12x$$ have coeffiecients that obey the recursion relation

We have the recursion relation:

$$a_{n+2} = \frac{2(p-n)}{(n+2)(n+1)}a_{n}$$

I have the answer for (2) and it makes no sense to me, i tried to tabulate why:

For P = 0, 1, 2 we have Even $$a_{0}$$ values of 1, 0, -2 respectively

For P = 0, 1, 2 we have Odd $$a_{1}$$ values of 0, 2, 0 respectively

And for P = 0, 1, 2 we have n values of 0, 1, 0 respectively

Where do these come from? I've looked for how to solve the above problems, and cant find a step by step example anywhere, it follows no logical pattern.

Any help would be VERY much appreciated.

Thanks.

2. May 24, 2014

### vela

Staff Emeritus
$a_0$ and $a_1$ are arbitrary constants. The rest of the coefficients can be expressed in terms of them through the recurrence relation. In both of these problems, you're looking for polynomial solutions to the differential equation. That means that the series needs to terminate for some value of $n$. Looking at the recurrence relation, you can see this will happen when $p=n$.

Let's consider the case when $p=2$. The even coefficients will be given by
\begin{eqnarray*}
a_2 &= -\frac{2(2-0)}{(0+2)(0+1)}a_0 = -2a_0 \\
a_4 &= -\frac{2(2-2)}{(2+2)(2+1)}a_2 = 0 \\
a_6 &= -\frac{2(2-4)}{(4+2)(4+1)}a_4 = 0
\end{eqnarray*} (Your recurrence relation has a sign error.) The remaining even coefficients will vanish as well. The odd coefficients will be proportional to $a_1$, but the series will never terminate because you'll never have $p=n$ because $p=2$ is even while $n$ will always be odd. To avoid an infinite series solution, we set $a_1=0$.

At this point, we have one polynomial solution, namely $H_2(x) = a_0 + a_2 x^2 = a_0(1-2x^2)$. What value to choose for $a_0$? It depends on how you want them normalized.

3. May 24, 2014

### rwooduk

That's a very clear explanation, many thanks!!

I'll try work through the questions with this knowledge and see how i do.

Thanks again!!

4. May 25, 2014

### rwooduk

hm why does $a_{6} = 0$ ?

and where does the $x^{2}$ come from?

thanks

Last edited: May 25, 2014
5. May 25, 2014

### vela

Staff Emeritus
Because $a_4=0$.

6. May 25, 2014

### rwooduk

yep just realised thanks and the $x^{2}$?

also for p = 1, do i just use odd values of n? n=1 would not be possible, but then for n=3 i get $a_{5} = \frac{2(1-3)}{(3+2)(3+1)} = \frac{-4}{20}$ not sure what to do with that

if possible please could someone do a couple more, i still cant see a logical process :-/

Last edited: May 25, 2014
7. May 27, 2014

### rwooduk

or if someone could direct me to a similar problem somewhere?

8. May 27, 2014

### Ray Vickson

First of all, spend some time really trying to understand what the problem is asking for; then you can worry about how to solve the problem. As I read it, you have polynomials $H_p(x)$ for $p = 0,1,2,3,\ldots$, and it looks like:
(1) For even $p$, $H_p(x)$ contains only the even powers $x^0 (=1), x^2, \ldots, x^{p-2}, x^p$.
(2) For odd $p$, $H_p(x)$ contains only the odd powers $x, x^3, \ldots, x^{p-2}, x^p$
In a court of law you could argue that the question did not contain enough information to allow you to say this with certainty, but more-or-less standard usage and common sense implies it with little room for doubt.

So, you are given a few examples of $H_p(x)$ for small $p$ and are asked to verify that the coefficients obey a certain, given, recursion. What is preventing you from just checking if those coefficients really do obey the recursion?

Oh...I see that there might be one source of confusion: the question did not say it, but the coefficients $a_i$ also depend on $p$; that is, $a_0$ is different for different values of $p$, as is also the case for $a_1$, etc. So, perhaps the question should really have given the recursion as
$$a_{n+2}(p) = \frac{2(p-n)}{(n+2)(n+1)}a_{n}(p)$$

Harder questions would be: how on earth did the recursion arise? Who thought of it, and why? But those are outside the actual scope of the question, assuming you quoted the question accurately.

Last edited: May 27, 2014
9. May 28, 2014

### rwooduk

Many thanks for the reply! I'm going to give it one last go with all the information presented in this thread (and my lecture notes, which cover more the orthogonality property so aren't that useful) and see where I get with it. Least I now understand where the x comes from in the solution!

Thanks again for all the help!

10. May 30, 2014

### rwooduk

update: finally figured this out!

i also found a very useful video for those interested:

start at 12:50 to relate to op

Last edited by a moderator: Sep 25, 2014