Problem #1
UCI. MAE200B, winter 2006. by nasser Abbasi.
Show from Rodrigues formula that the legendre polynomials are orthogonal with unit weight function over and
Solution
Rodrigues formula is
We need to show (for the first part of the problem) that
We start the proof by assuming that (if then we simply switch the indices, and if then this will be shown in part 2 of the problem).
We know is a polynomial in . This can be seen from Rodrigues formula. This poynomial starts with a term which is either a constant, or a constant times depending if is odd or even as can be seen from these few examples
Hence, let us first assume that is odd. Then we can write and if was even, then we write
Now instead of showing the solution for is odd and is even, we show the solution for is even only since the solution in this case will encompase the case for is odd as will be explained more below.
Write LHS of equation (2) as
The strong form of the proof requires then that we show that each term (integral) above is zero. i.e. we need to show that
where in the above can be zero (this covers the case for is odd)
Now since , then the above means we need to show that
equation (3) above means that
(the above relation will be used in the proof for part 2).
Now to proof equation (3), use integration by parts.
Differntiate the term and integrate the terms. Let , . Hence and
But
However, now we see that since the expression (after repeated differentiation) will come out to be a polynomial in , and when we substitute for or we will get zero as the value of each term, hence
Hence the result of integration by parts is as follows
Now we repeate the process. We perform integration by part on the integral in the RHS above. This will result in
And again, we perform integration by parts. We see that each time the exponent of becomes one less than before, and the order of differentiation becomes one less than before. So it is a race between and to reach the value zero. However we started by assuming that , hence by repeated integration by parts we will eventually reach the following form
where
But we have shown above in equation 4 that hence this means that
which is what we wanted to proof. This implies that QED.
Now we start part 2, which is to show that when then
Start by using the recurive formula for Legendre polynomials. For clarity I will write instead of .
The recurive formula to use is
Multiply eq (6) by we obtain
Integrate both sides from
But is a polynomial of degree , as can be seen from expansion of Rodrigues formula , hence the second integral shown above on the LHS side is zero, since it is as if we have which we showed in part 1 that this is zero. Hence eq (7) simplifies to
Now use integration by parts on the integral on the LHS.
But since hence and , hence the above becomes
Now substitute (9) into LHS of (8)
Which is what we are asked to show.