Answer/Explanation
The complete question is:
Show that the set function {1, cos x, cos 2x, . . .} is orthogonal with respect to given weight on the prescribed interval [- π, π]
Step-by-step explanation:
If we make the identification For ∅° (x) = 1 and ∅n(x) = cos nx, we must show that ∫ lim(π) lim(-π) .∅°(x)dx = 0 , n ≠0, and ∫ lim(π) lim(-π) .∅°(x)dx = 0, m≠n.
Therefore, in the first case, we have
(∅(x), ∅(n)) ∫ lim(π) lim(-π) .∅°(x)dx = ∫ lim(π) lim(-π) cosn(x)dx
This will therefore be equal to :
1/n sin nx lim(π) lim(-π) = 1/n [sin nπ - sin(-nπ)] = 0 , n ≠0 (In the first case)
and in the second case, we have,,
(∅(m) , ∅(n)) = ∫ lim(π) lim(-π) .∅°(x)dx
This will therefore be equal to:
∫ lim(π) lim(-π) cos mx cos nx dx
Therefore, 1/2 ∫ lim(π) lim(-π)( cos (m+n)x + cos( m-n)x dx (Where this equation represents the trigonometric function)
1/2 [ sin (m+n)x / m+n) ]+ [ sin (m-n)x / m-n) ] lim(π) lim(-π) = 0, m ≠ n
Now, to go ahead to find the norms in the given set intervals, we have,
for ∅°(x) = 1 we have:
//∅°(x)//² = ∫lim(π) lim(-π) dx = 2π
So therefore, //∅°(x)//² = √2π
For ∅°∨n(x) = cos nx , n > 0.
It then follows that,
//∅°(x)//² = ∫lim(π) lim(-π) cos²nxdx = 1/2 ∫lim(π) lim(-π) [1 + cos2nx]dx = π
Thus, for n > 0 , //∅°(x)// = √π
It is therefore ggod to note that,
Any orthogonal set of non zero functions {∅∨n(x)}, n = 0, 1, 2, . . . can be normalized—that is, made into an orthonormal set by dividing each function by its norm. It follows from the above equations that has been set.
Therefore,
{ 1/√2π , cosx/√π , cos2x/√π...} is orthonormal on the interval {-π, π}.
