Find the Fourier sin series for the function $f(x) = x^3$ on the interval $0\leq x \leq L$. the Legendre series for the same function. One representation involves an infinite number of terms, while the other has only a finite number of terms. In the context of separation of variables, why is it important to understand both of these very different-looking series representations of a function?

I'm not exactly sure what the problem is stating and how to go about using the Legendre series. I started with the equation for generating the Legendre polynomial inside the integral, but I do not understand how to incorporate the information of the finite and infinite terms?


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  • $\begingroup$ The procedure is the same for both: write $f$ as a sum of basis functions with unknown coefficients, multiply by some other basis function $f_n$, integrate, and use the orthogonality properties to get an expression for the corresponding coefficient $c_n$. The generating function won't help you much. $\endgroup$ – E.P. Mar 13 '13 at 23:46
  • $\begingroup$ See here. $\endgroup$ – Mhenni Benghorbal Mar 14 '13 at 4:09

We are given:

$$\tag 1 f(x) = x^3$$


Note: see my response here for the Legendre approach.

Using the method from the referenced approach, we find:

$$\tag 2 f(x) = x^3 = c_0P_0(x) + c_1P_1(x) + c_2P_2(x) = \frac{3}{5}P_1(x) + \frac{2}{5} P_3(x)$$

Please note that $(2)$ only has a finite number of terms as mentioned in the problem statement.

Fourier Sine Series

Note that since $f(-x) = -f(x)$, $(1)$ is an odd function and that is very helpful!

If a function is odd, then $a_n = 0$ and the Fourier sin series collapses to:

$$f(x) = \sum_{n=1}^\infty b_n~\sin(n x)$$


$$b_n = \frac{2}{\pi} \int_0^{\pi} f(x)~\sin(n x)~dx$$

However, the question wants us to extend the range to $L$, so we have:

$$b_n = \frac{2}{L} \int_0^{L} f(x)~\sin(\frac{n \pi x}{L})~dx$$

Lets calculate these terms:

$\displaystyle b_1 = \frac{2}{L} \int_0^{L} x^{3}~\sin(\frac{1 \pi x}{L})~dx = \frac{2 (\pi^2-6) L^3}{\pi^3}$

$\displaystyle b_2 = \frac{2}{L} \int_0^{L} x^{3}~\sin(\frac{2 \pi x}{L})~dx = -\frac{(2 \pi^2-3) L^3}{2 \pi^3}$

$\displaystyle b_3 = \frac{2}{L} \int_0^{L} x^{3}~\sin(\frac{3 \pi x}{L})~dx = \frac{2 (3 \pi^2-2) L^3}{9 \pi^3}$


$\displaystyle b_n = \frac{2}{L} \int_0^{L} x^{3}~\sin(\frac{n \pi x}{L})~dx = -\frac{2 L^3 (\pi n (\pi^2 n^2-6) \cos(\pi n)-3 (\pi^2 n^2-2) \sin(\pi n))}{(\pi^4 n^4)}$

Please note that the Fourier sin series has an infinite number of terms as mentioned in the problem statement.

In the context of separation of variables, why is it important to understand both of these very different-looking series representations of a function?

Update to the last question

We have now written two very different ways to solve this problem (function). The first, based on the Legendre polynomials, provided a closed form solution, while the second is based on an infinite series Fourier analysis. These can aid us with different perspectives to solve separation of variable problems.

These two solutions are merely two facets of the same solution. The Fourier series formula shows how every piece-wise component of the solution can be decomposed into its constituent parts, while the Legendre approach demonstrates how all the components combine into a single solution.

Mathematically, both of these provide us with the ability to look at behaviors from a different perspective and are both very useful to analyze the behavior of the function (globally or component wise) and provide another tool in our tool-box for qualitative and quantitative analyses (do they converge, how fast, how large is the error $\ldots$).


  • $\begingroup$ It sure as heck did, thank you so much. I was able to tackle most of my other problems for class because of your guide! Clear, concise, and well derived steps. Thank you again. $\endgroup$ – julesverne Mar 14 '13 at 18:34
  • $\begingroup$ Really nice and thorough answer, and it seemed to really help both the OP, and future visitors too! $\endgroup$ – Namaste Apr 20 '13 at 0:16
  • $\begingroup$ Oh dear! Here at MSE? I have felt that way at times... :-( Just reread the feedback you received from the OP on this answer...that might help rebuild your morale!! $\endgroup$ – Namaste Apr 20 '13 at 2:04

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