Express the following functions as Legendre series (as a sum of Legendre polynomials)

a) $-2x^2 + 7$
b) $3x^4+6x^2-2$
c) $\cos^5 \theta$ - the L.P. will be in functions of $\cos \theta$.

If anyone could help me understand the setup process for Legendres, that would be highly appreciated. I have no exposure to Legendre series, and I am using Griffith's Intro. to Electrodynamics which doesn't provide the explanation of Legendre series in any detail.

You may determine the coefficients in the series algebraically or using the orthogonality properties of the Legendre polynomials.

Edit: It seems that the solutions for Legendre polynomial are set, such as the values of $P_0$, $P_1$, $P_2$, etc... But how do I incorporate the given function? Do I make them into a polynomial, then attempt to set that in series that is equal to the Legendre series? For (c), I am assuming to express that in complex terms.. The hint suggests to use orthogonality properties, in which the normalization would be 1?

  • $\begingroup$ @Amzoti it does not. $\endgroup$
    – julesverne
    Mar 14, 2013 at 2:54
  • $\begingroup$ This could help you understand the setup process. $\endgroup$
    – Julien
    Mar 14, 2013 at 3:20
  • $\begingroup$ @julien Thanks for the link. It does help me in understanding the setup of the problem, however I do not understand on what to put in for $P_m$ and $P_n$, and for values of $m$ and its meaning? $\endgroup$
    – julesverne
    Mar 14, 2013 at 3:54

1 Answer 1


Lets do the first one two different ways and then you can try the rest.

The Legendre Polynomials are given by:

$P_0(x) = 1$

$P_1(x) = x$

$P_2(x) = \frac{1}{2}(3x^2 - 1)$

$P_3(x) = \frac{1}{2}(5x^3-3x)$


We are given $f(x) = -2x^2 + 7$

Method 1: (so we can check our answer)

How can we add those $P_i(x)$ to get $f(x)$?

Well, if you add $-\frac{4}{3} P_2(x) + \frac{19}{3}P_0 = f(x)$

Method 2: (the method you are asked to use)

To use the Legendre Series, we put:

$$\tag 1 f(x) = \sum_{i=0}^\infty c_iP_i(x)$$

To solve this, we solve a series of integrals given by:

$$\int_{-1}^1 f(x)P_i(x)~dx = \sum_{i=0}^\infty c_i \int_{-1}^1 (P_i(x))^2 ~dx$$

Because the Legendre polynomials are orthogonal, all the integrals on the right are zero except the one we care about, namely $c_i$. So, lets crank those $c_i$ using this approach.

$i = 0$

$$\int_{-1}^1 f(x)P_0(x)~dx = \sum_{i=0}^\infty c_0 \int_{-1}^1 (P_0(x))^2 ~dx$$

$$\int_{-1}^1 (-2x^2 + 7)(1)~dx = \sum_{i=0}^\infty c_0 \int_{-1}^1 (P_0(x))^2 ~dx = c_0 \int_{-1}^1 (1)^2~dx$$

This yields $\frac{38}{3} = c_0 \cdot 2 \rightarrow c_0 = \frac{19}{3}$

$i = 1$

$$\int_{-1}^1 f(x)P_1(x)~dx = \sum_{i=0}^\infty c_1 \int_{-1}^1 (P_1(x))^2 ~dx$$

$$\int_{-1}^1 (-2x^2 + 7)(x)~dx = \sum_{i=0}^\infty c_1 \int_{-1}^1 (P_1(x))^2 ~dx = c_1 \int_{-1}^1 (x)^2~dx$$

This yields $0 = c_1 \cdot 0 \rightarrow c_1 = 0$

$i = 2$

$$\int_{-1}^1 f(x)P_2(x)~dx = \sum_{i=0}^\infty c_2 \int_{-1}^1 (P_2(x))^2 ~dx$$

$$\int_{-1}^1 (-2x^2 + 7)(\frac{1}{2}(3x^2-1))~dx = \sum_{i=0}^\infty c_2 \int_{-1}^1 (P_2(x))^2 ~dx = c_2 \int_{-1}^1 (\frac{1}{2}(3x^2-1))^2~dx$$

This yields $-\frac{8}{15} = c_2 \cdot \frac{2}{5} \rightarrow c_2 = -\frac{4}{3}$

$\therefore ~ f(x) = -2x^2 + 7 = c_0P_0(x) + c_1P_1(x) + c_2P_2(x) = \frac{19}{3}P_0(x) -\frac{4}{3}P_2(x)$

  • $\begingroup$ thank you for your response. so this explains using algebraic properties, which is SUPERB, because I didn't understand the purpose of the idea. For the orthogonality method, I'm not exactly sure what we are putting in for values of $P_i$. Since we are solving for the coefficient, $a_n$, I'm assuming $P_i$ have some values to put in based on the nature of the problem? $\endgroup$
    – julesverne
    Mar 14, 2013 at 3:45
  • $\begingroup$ Oh I see, you have to do the SUM OF, which means input the values we know of $P_i$ from Legendre, and input that value into the integral. Since it is summed over infinity, how do we know when to stop, as we don't have "infinite" values of $P_i$? $\endgroup$
    – julesverne
    Mar 14, 2013 at 3:48
  • $\begingroup$ @MhenniBenghorbal: thank you - it is really hard with Chrome locking up as I am writing responses - caused me to make about a dozen errors! Appreciate the feedback! Regards $\endgroup$
    – Amzoti
    Mar 14, 2013 at 4:23
  • $\begingroup$ @Amzoti: Maybe you are interested in this problem. $\endgroup$ Mar 14, 2013 at 4:32
  • $\begingroup$ @MhenniBenghorbal: Looks interesting - I will see if I can remember enough (it has been a long time since I've look at this sort of thing - but this is a good refresher) regarding this topic to solve it! :) Thanks! $\endgroup$
    – Amzoti
    Mar 14, 2013 at 4:34

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