1. ## Orthogonal functions

I have two functions and I want to show that they are orthogonal over a given interval. Do I just multiply them together and show that the integral over the interval is zero?

2. I think it's the definition of orthogonal function, isn't it?
What are the two functions?

3. Originally Posted by billym
I have two functions and I want to show that they are orthogonal over a given interval. Do I just multiply them together and show that the integral over the interval is zero?
If your definition of inner product is <f, g>= $\int_a^b f(x)g(x)dx$. But there are other ways of defining the inner product in function spaces, usually involving a "weighting function", so the inner product is $= \int_a^b f(x)g(x)\mu(x)dx$. For example, the Leguerre polynomials are orthogonal using the inner product defined by $= \int_0^\infty f(x)g(x)e^{-x}dx$. In that case the " $e^-{x}$ is necessary because the interval of definition is infinite. Bessel functions are orthogonal using the inner product defined by $= \int_0^1 x f(x)g(x)dx$. Here, the "x" factor is necessary because the Bessel functions may have a pole of order 1 at 0.

In general, two vectors in an innerproduct space are orthgonal if and only if their inner product is 0. Check to see what inner product you should be using.

4. Mine are

$f(x)=\frac{3x^2-1}{2}$ and $g(x)=\frac{5x^3-3x}{2}$ on $x \in [-1,1]$

Is $\int_a^b f(x)g(x)dx$ the inner product I should use? (I hope so because I already stapled this thing together).

5. actually it turn out quite nice:

$h(x) = f(x) \cdot g(x) = \frac{{x\left( {3x^2 - 1} \right)\left( {5x^2 - 3} \right)}}
{4}$

now you can easily see that h(x) is an odd function since h(x) = -h(-x)
and we all know that the integral of an odd function over a symmetrical interval is 0.