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I have a physical problem with a solution component that involves an infinite series of Legendre polynomials. I'm trying to compute the coefficients but am getting some odd behaviour. As I understand it, provided $-1 \leq x \leq x$, then for $f(x)$ I can get my coefficients by

$$A_{n} = \frac{2n+1}{2} \int_{-1}^{1} P_{n}(x)f(x)\space dx.$$

where $P_{n}$ is the legendre function of degree $n$. My functions $f(x)$ uses cosine functions (which satisfy $|x| \leq 1$). Writing $x = \cos\theta$, then an example of the type of function we need to solve is

$$f(x) = \frac{100}{\sqrt{25 + x^2}}.$$

However, when I compute these, I get some odd behavior - for all odd values, $A_{n} = 0$, which is fine. Intially, it seems like the series converges (as expected) with the first few non-zero terms gradually getting smaller and alternating sign;

$A_{2} = -0.26$, $A_{4} = 0.26$, $A_{6} = -2.596 \times 10^{-4}$, $A_{8} = -1.266 \times 10^{-6}$....

But this does not continue -after this, the values rapidly increase, as shown in the figure below, eventually exploding to huge values as $n \rightarrow \infty$ enter image description here

My question is why does this occur, and where am I going wrong? I've seen conflicting things about whether $f((x)$ has to be a polynomial for this to work, so if that's the problem, how can I express the function $f(x)$ if it is non-polynomial, or is there something more subtle (or obvious) I'm missing?

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  • $\begingroup$ Try looking at the limits of your integral. Legendre polynomials are orthogonal for $ P_{n) = P_{n+1}, so integrating between -1 and 1 will provide a null answer as an inverse function $\frac{1}{x} ~ P_{n}(x). $\endgroup$ – Cppg Sep 18 '17 at 13:03
  • $\begingroup$ That's interesting, though I'm not sure I follow? Would you mind expanding ? $\endgroup$ – DRG Sep 18 '17 at 13:26
  • $\begingroup$ If the see here en.m.wikipedia.org/wiki/Legendre_polynomials, the polynomials converge for $ |x| < 1 $ so for an ill-behaved function in a region of -1 to 1, the polynomial is useful for creating a discernable convergence. Note that the series of $ P_{n}(x) $ is of odd parity around the interval, so summing them up within -1 and 1 will give zero. $\endgroup$ – Cppg Sep 18 '17 at 16:26
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So part of the problem was apparently numerical instability - changing the language I used and the errors bounds made the function behave much more predictably. It seems the integrals were blowing up in the original code, which disappeared when they were solved in Mathematica. Though if there are any conditions I should know about for $f(x)$, I'd still be grateful to hear them!

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