Prove a set is orthogonal.

[roto25]

How would you prove, using the integral product, that the set of {cos x, cos 2x, cos 3x, cos 4x, ...} is an orthogonal set?

 

[tiny-tim]

welcome to pf!

hi roto25! welcome to pf!

i] define the integral product

ii] define orthogonal set

iii] apply i and ii …

what do you get? ​

 

[Bavid]

over the interval -pi to pi, the integral of cos(mx)cos(nx)dx is zero, as long as m and n are integers. Therefore, if you select ANY pair of elements from the set, the 'integral of their product' will be zero, thereby satisfying the condition of orthogonality.

 

[chiro]

over the interval -pi to pi, the integral of cos(mx)cos(nx)dx is zero, as long as m and n are integers. Therefore, if you select ANY pair of elements from the set, the 'integral of their product' will be zero, thereby satisfying the condition of orthogonality.

On top of what Bavid said if you don't know where to start set up the integral and use integration by parts.

 

[blue_raver22]

To prove that a set is orthogonal, we need to show that the inner product of any two distinct elements in the set is equal to 0. In this case, we are dealing with a set of trigonometric functions {cos x, cos 2x, cos 3x, cos 4x, ...}.

The inner product of two functions f(x) and g(x) is defined as:

∫f(x)g(x)dx

To prove that the set is orthogonal, we need to show that for any two distinct elements of the set, the inner product is equal to 0. Let's take two arbitrary elements from the set, cos nx and cos mx, where n and m are distinct positive integers.

∫cos nx cos mx dx

= ∫cos nx cos mx dx

= ∫cos nx cos mx dx

= ∫cos nx cos mx dx (using the trigonometric identity cos(A+B)=cosAcosB-sinAsinB)

= 0 (since n and m are distinct positive integers, nx and mx will never be equal, so cos nx cos mx will always be 0)

Therefore, the inner product of any two distinct elements in the set is equal to 0, proving that the set {cos x, cos 2x, cos 3x, cos 4x, ...} is an orthogonal set.