Eigenvalue/Orthogonal Eigenfunctions

[jegues]

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,

\int_\alpha ^{\beta}f(x)g(x) = 0

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]

No picture.

 

[jegues]

No picture.

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

 

[LCKurtz]

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,

\int_\alpha ^{\beta}f(x)g(x) = 0

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.

No picture.

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

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

[Edit] Fixed typo; put x's in the integrals

 

[jegues]

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

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

Hmmm...

\text{Since, }\quad m,n \in Z

m \neq n

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

\int_0^1 -dx \quad\text{or, } \quad \int_0^1 dx

Neither gives 0.

What am I doing wrong/misunderstanding?

 

[haselwhat?]

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.

 

[LCKurtz]

Hmmm...

\text{Since, }\quad m,n \in Z

m \neq n

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

\int_0^1 -dx \quad\text{or, } \quad \int_0^1 dx

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.

 

[jegues]

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?

 

[haselwhat?]

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?