# Orthogonality of Two Functions

1. Apr 15, 2008

### White Ink

1. The problem statement, all variables and given/known data

Show that:

$$\varphi_{0}(x) = f_{0}(x)$$

and

$$\varphi_{1}(x) = f_{1}(x) - \frac{\left\langle\right\varphi_{0},f_{1}\rangle}{\left\|\varphi_{0}\right\|^{2}}\varphi_{0}(x)$$

are orthogonal on the interval [a,b].

2. Relevant equations

Orthogonal functions satisfy:

$$\left\langle\right\varphi_{m},\varphi_{n}\rangle = \int^{b}_{a}\varphi_{m}(x)\varphi_{n}(x)dx = g(m)\delta_{mn}$$

Where, $$\delta_{mn}$$ is the Delta Kronecker.

Also:

$$\left\langle\right\varphi_{m},\varphi_{m}\rangle = \left\|\varphi_{m}\right\|^{2}$$

3. The attempt at a solution

Since m and n (0 and 1) are not equal, the Delta Kronecker is zero and therefore the proof is a matter of proving that:

$$\left\langle\right\varphi_{0},\varphi_{1}\rangle = 0$$

Having substituted the functions into the inner product formula in 2:

$$\left\langle\right\varphi_{0},\varphi_{1}\rangle = \int^{b}_{a}\varphi_{0}(x)\varphi_{1}(x)dx = \int^{b}_{a}f_{0}(x)\left[f_{1}(x) - \frac{\left\langle\right\varphi_{0},f_{1}\rangle}{\left\|\varphi_{0}\right\|^{2}}\varphi_{0}(x)\right]dx$$

Because the Delta Kronecker is zero, all I have to do is show that:

$$\int^{b}_{a}f_{0}(x)\left[f_{1}(x) - \frac{\left\langle\right\varphi_{0},f_{1}\rangle}{\left\|\varphi_{0}\right\|^{2}}\varphi_{0}(x)\right]dx = 0$$

I'm unsure as to whether I should use integration by parts to do the resulting integral because there is another integral embedded in the $$\varphi_{1}(x)$$ function; which (because it is a definite integral) would be tricky to differentiate or integrate.

2. Apr 15, 2008

### D H

Staff Emeritus
You really don't need to bring the Kronecker delta stuff into the picture. It is completely superfluous. You do need to show that $\langle \varphi_0, \varphi_1 \rangle = 0$ as that is the definition of orthogonality.

Why integrate by parts? Use $\langle f, g \rangle \equiv \int_a^b f(x)g(x)\,dx$ and $\varphi_0(x) = f_0(x)$