This community is invite only. If you're enrolled in Mark's PDE class but haven't received an invite, just let him know!
Show that the polynomials
$$1,x,3x^2-1$$
form an orthogonal set over the interval $[-1,1]$.
To show that two functions $f(x)$ and $g(x)$ are orthogonal over an interval $[-L,L],$ we must show that
$$\left\langle f(x),g(x) \right\rangle = \int_{-L}^Lf(x)g(x)dxdx = 0.$$
$$\left\langle 1,x \right\rangle = \int_{-1}^1(1)xdx = \left[\frac{x^2}{2}\right]_{-1}^1 = \frac{1}{2}-\frac{1}{2} = 0.$$
So $1$ and $x$ are orthogonal.
$$\left\langle 1,3x^2-1 \right\rangle = \int_{-1}^1(1)(3x^2-1)dx = \left[x^3-x\right]_{-1}^1 = (1-1) - \left((-1)-(-1)\right) = 0.$$
So $1$ and $3x^2-1$ are orthogonal.
$$\left\langle x,3x^2-1 \right\rangle = \int_{-1}^1x(3x^2-1)dx = \int_{-1}^1(3x^3-x)dx = \left[\frac{3x^4}{4}-\frac{x^2}{2}\right]_{-1}^1 = \left(\frac{3}{4}-\frac{1}{2}\right) - \left(\frac{3}{4}-\frac{1}{2}\right) = 0.$$
So $x$ and $3x^2-1$ are orthogonal. Thus, the polynomials $1$, $x$, $3x^2-1$ form an orthogonal set over the interval $[-1,1].$
Comments
To show that two functions $f(x)$ and $g(x)$ are orthogonal over an interval $[-L,L],$ we must show that
$$\left\langle f(x),g(x) \right\rangle = \int_{-L}^Lf(x)g(x)dxdx = 0.$$
$$\left\langle 1,x \right\rangle = \int_{-1}^1(1)xdx = \left[\frac{x^2}{2}\right]_{-1}^1 = \frac{1}{2}-\frac{1}{2} = 0.$$
So $1$ and $x$ are orthogonal.
$$\left\langle 1,3x^2-1 \right\rangle = \int_{-1}^1(1)(3x^2-1)dx = \left[x^3-x\right]_{-1}^1 = (1-1) - \left((-1)-(-1)\right) = 0.$$
So $1$ and $3x^2-1$ are orthogonal.
$$\left\langle x,3x^2-1 \right\rangle = \int_{-1}^1x(3x^2-1)dx = \int_{-1}^1(3x^3-x)dx = \left[\frac{3x^4}{4}-\frac{x^2}{2}\right]_{-1}^1 = \left(\frac{3}{4}-\frac{1}{2}\right) - \left(\frac{3}{4}-\frac{1}{2}\right) = 0.$$
So $x$ and $3x^2-1$ are orthogonal. Thus, the polynomials $1$, $x$, $3x^2-1$ form an orthogonal set over the interval $[-1,1].$