Evaluating scalar products of two functions

  • #1
264
37
Homework Statement
Find ##\braket{f|f} , \braket{f|g} , \braket{g|g}##.
Relevant Equations
integral, scalar product
I am to consider a basis function ##\phi_i(x)##, where ##\phi_0 = 1 ,\phi_1 = cosx , \phi_2 = cos(2x) ## and where the scalar product in this vector space is defined by ##\braket{f|g} = \int_{0}^{\pi}f^*(x)g(x)dx##

The functions are defined by ##f(x) = sin^2(x)+cos(x)+1## and ##g(x) = cos^2(x)-cos(x)##
My task is to find ##\braket{f|f} , \braket{f|g} , \braket{g|g}##.

Given that f(x) and g(x) have no imaginary parts, is this problem really as simple as multiplying these two functions and evaluating the integral? If so I have no trouble doing that, I guess I just want to make sure I understand correctly what the question is asking. I am pretty sure my thought process is correct but I am just a bit unsure because of how simple the approach is.
ψ
 
  • #2
Homework Statement:: Find ##\braket{f|f} , \braket{f|g} , \braket{g|g}##.
Relevant Equations:: integral, scalar product

I am to consider a basis function ##\phi_i(x)##, where ##\phi_0 = 1 ,\phi_1 = cosx , \phi_2 = cos(2x) ## and where the scalar product in this vector space is defined by ##\braket{f|g} = \int_{0}^{\pi}f^*(x)g(x)dx##

The functions are defined by ##f(x) = sin^2(x)+cos(x)+1## and ##g(x) = cos^2(x)-cos(x)##
My task is to find ##\braket{f|f} , \braket{f|g} , \braket{g|g}##.

Given that f(x) and g(x) have no imaginary parts, is this problem really as simple as multiplying these two functions and evaluating the integral? If so I have no trouble doing that, I guess I just want to make sure I understand correctly what the question is asking. I am pretty sure my thought process is correct but I am just a bit unsure because of how simple the approach is.
ψ
In four words, you have it right! :)

-Dan
 
  • Like
Likes guyvsdcsniper

Suggested for: Evaluating scalar products of two functions

Replies
16
Views
1K
Replies
2
Views
850
Replies
5
Views
755
Replies
3
Views
790
Replies
4
Views
695
Back
Top