Eigenvalue/Orthogonal Eigenfunctions

  • Thread starter Thread starter jegues
  • Start date Start date
  • Tags Tags
    Eigenfunctions
Click For Summary

Homework Help Overview

The discussion revolves around a problem related to eigenvalues and orthogonal eigenfunctions, specifically focusing on the concept of orthogonality in the context of integrals involving sine functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore how to derive functions f(x) and g(x) for the orthogonality condition, questioning whether to select specific values for n. There is discussion about the integral definition of orthogonality and attempts to clarify the conditions under which the integral evaluates to zero.

Discussion Status

Some participants have provided guidance on how to approach the integral and suggested using specific trigonometric identities to facilitate the evaluation. There is ongoing exploration of the implications of the integral results and the conditions for orthogonality, with no explicit consensus reached.

Contextual Notes

Participants note confusion regarding the setup of the problem and the implications of selecting values for n and m. There is also mention of potential typographical errors in the expressions used, which may affect understanding.

jegues
Messages
1,085
Reaction score
3

Homework Statement



See figure attached for problem statement.

Homework Equations





The Attempt at a Solution



See figure attached for attempt.

I'm confused as to how to do part B?

I know that the definition of orthogonal is,

[tex]\int_\alpha ^{\beta}f(x)g(x) = 0[/tex]

but how do I obtain f(x) and g(x)

Do I just choose two n values such as n=1, and n=2, plug it in and obtain my 2 functions and preform the integral?

Thanks again!

Whoops! There the figure should be attached now.
 

Attachments

  • 2009Q7.jpg
    2009Q7.jpg
    28.2 KB · Views: 436
Last edited:
Physics news on Phys.org
No picture.
 
LCKurtz said:
No picture.

It's there now, still looking for some help on this one.
 
jegues said:

Homework Statement



See figure attached for problem statement.

Homework Equations


The Attempt at a Solution



See figure attached for attempt.

I'm confused as to how to do part B?

I know that the definition of orthogonal is,

[tex]\int_\alpha ^{\beta}f(x)g(x) = 0[/tex]

but how do I obtain f(x) and g(x)

Do I just choose two n values such as n=1, and n=2, plug it in and obtain my 2 functions and preform the integral?

Thanks again!

Whoops! There the figure should be attached now.

LCKurtz said:
No picture.

jegues said:
It's there now, still looking for some help on this one.

Basically, yes, except you don't plug in numbers for n and m. You need to show

[tex]\int_0^1 \sin(\frac{(2n-1)\pi x}{2})\sin(\frac{(2m-1)\pi x}{2})\, dx = 0,\, m\ne n[/tex]

[Edit] Fixed typo; put x's in the integrals
 
Last edited:
LCKurtz said:
Basically, yes, except you don't plug in numbers for n and m. You need to show

[tex]\int_0^1 \sin(\frac{(2n-1)\pi}{2})\sin(\frac{(2m-1)\pi}{2})\, dx = 0,\, m\ne n[/tex]

Hmmm...

[tex]\text{Since, }\quad m,n \in Z[/tex]

[tex]m \neq n[/tex]

So the integral should always come out in one of two cases,

[tex]\int_0^1 -dx \quad\text{or, } \quad \int_0^1 dx[/tex]

Neither gives 0.

What am I doing wrong/misunderstanding?
 
Try defining the indices like:

2n-1 = k and 2m-1 = j or something. (k=1,3,5,... and j=1,3,5,...). Then try making use of the trig identity

sin(jx)sin(kx) = (1/2)[ cos((j-k)x) - cos((j+k)x) ]

in the integrand.
 
jegues said:
Hmmm...

[tex]\text{Since, }\quad m,n \in Z[/tex]

[tex]m \neq n[/tex]

So the integral should always come out in one of two cases,

[tex]\int_0^1 -dx \quad\text{or, } \quad \int_0^1 dx[/tex]

Neither gives 0.

What am I doing wrong/misunderstanding?

Woops. I inadvertently left the x's out of the sine functions.
They aren't constant and I edited it.
 
haselwhat? said:
Try defining the indices like:

2n-1 = k and 2m-1 = j or something. (k=1,3,5,... and j=1,3,5,...). Then try making use of the trig identity

sin(jx)sin(kx) = (1/2)[ cos((j-k)x) - cos((j+k)x) ]

in the integrand.

How does this translate to the RHS of the equation?

What identity are you using you change to two sin terms into two cos terms?
 
Last edited:
Its just a trig identity I used once to show orthogonality. Hopefully it makes the integration easier and hopefully gives zero, which is your "RHS", right?
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K