Use the legendre function to operate on a vector and then examine the format of the output.

Calculate the second-degree Legendre function values of a vector.

deg = 2;
x = 0:0.1:0.2;
P = legendre(deg,x)

P = 3×3

   -0.5000   -0.4850   -0.4400
         0   -0.2985   -0.5879
    3.0000    2.9700    2.8800

The format of the output is such that:

The equation for the second-degree associated Legendre function ${\mathit{P}}_2^{\mathit{m}}$ is

Therefore, the value of ${\mathit{P}}_2^0\left(0\right)$ is

This result agrees with P(1,1) = -0.5000.

Calculate the associated Legendre function values with several normalizations.

Calculate the first-degree, unnormalized Legendre function values ${\mathit{P}}_1^{\mathit{m}}$. The first row of values corresponds to $\mathit{m}=0$, and the second row to $\mathit{m}=1$.

x = 0:0.2:1;
n = 1;
P_unnorm = legendre(n,x)

P_unnorm = 2×6

         0    0.2000    0.4000    0.6000    0.8000    1.0000
   -1.0000   -0.9798   -0.9165   -0.8000   -0.6000         0

Next, compute the Schmidt seminormalized function values. Compared to the unnormalized values, the Schmidt form differs when $\mathit{m}>0$ by the scaling

For the first row, the two normalizations are the same, since $\mathit{m}=0$. For the second row, the scaling constant multiplying each value is -1.

P_sch = legendre(n,x,'sch')

P_sch = 2×6

         0    0.2000    0.4000    0.6000    0.8000    1.0000
    1.0000    0.9798    0.9165    0.8000    0.6000         0

C1 = (-1) * sqrt(2*factorial(0)/factorial(2))

C1 = -1

Lastly, compute the fully normalized function values. Compared to the unnormalized values, the fully normalized form differs by the scaling factor

This scaling factor applies for all values of $\mathit{m}$, so the first and second rows have different scaling factors.

P_norm = legendre(n,x,'norm')

P_norm = 2×6

         0    0.2449    0.4899    0.7348    0.9798    1.2247
    0.8660    0.8485    0.7937    0.6928    0.5196         0

Cm0 = sqrt((3/2))

Cm0 = 1.2247

Cm1 = (-1) * sqrt((3/2)/2)

Cm1 = -0.8660

Spherical harmonics arise in the solution to Laplace's equation and are used to represent functions defined on the surface of a sphere. Use legendre to compute and visualize the spherical harmonic for ${\mathit{Y}}_3^2$.

The equation for spherical harmonics includes a term for the Legendre function, as well as a complex exponential:

First, create a grid of values to represent all combinations of $0\le \theta \le \pi$ (colatitude angle) and $0\le \varphi \le 2\pi$ (azimuthal angle). Here, the colatitude $$\theta$$ ranges from 0 at the North Pole, to $$\pi /2$$ at the Equator, and to $$\pi$$ at the South Pole.

dx = pi/60;
col = 0:dx:pi;
az = 0:dx:2*pi;
[phi,theta] = meshgrid(az,col);

Calculate ${\mathit{P}}_{\mathit{l}}^{\mathit{m}}\left(\cos \theta \right)$ on the grid for $\mathit{l}=3$.

l = 3;
Plm = legendre(l,cos(theta));

Since legendre computes the answer for all values of $\mathit{m}$, Plm contains some extra function values. Extract the values for $\mathit{m}=2$ and discard the rest. Use the reshape function to orient the results as a matrix with the same size as phi and theta.

m = 2;
if l ~= 0
    Plm = reshape(Plm(m+1,:,:),size(phi));
end

Calculate the spherical harmonic values for ${\mathit{Y}}_3^2$.

a = (2*l+1)*factorial(l-m);
b = 4*pi*factorial(l+m);
C = sqrt(a/b);
Ylm = C .*Plm .*exp(1i*m*phi);

Convert the spherical coordinates to Cartesian coordinates. Here, $$\pi /2-\theta$$ becomes the latitude angle that ranges from $$\pi /2$$ at the North Pole, to 0 at the Equator, and to $$-\pi /2$$ at the South Pole. Plot the spherical harmonic for ${\mathit{Y}}_3^2$ using both the positive and negative real values.

[Xm,Ym,Zm] = sph2cart(phi, pi/2-theta, abs(real(Ylm)));
surf(Xm,Ym,Zm)
title('$Y_3^2$ spherical harmonic','interpreter','latex')