Orthogonality, Fourier series and Kronecker delta

[TobyDarkeness]

Homework Statement

Show that the orthogonality relation for the "cosine basis functions" used in the Fourier series is

1/L$$\int$$cos[(n*pi*x)/L)]cos[(m*pi*x)/L)]dx = {Sin([n-m]*pi)}/[(n-m)*pi] + {Sin([n+m]*pi)}/[(n+m)*pi]

By considering the different integer n and m, show that the right hand side is equal to $$\delta$$(nm)

Homework Equations

the limits of integration are -L to L

The Attempt at a Solution

we haven't been taught how to handle this type of integral i know what the solution looks like but i don't know the process to get there. Our task is to find out. Could anyone help me? thanks in advance.

 

[Dick]

The first step is to use a trig identity. See the Product-to-sum rules in http://en.wikipedia.org/wiki/List_of_trigonometric_identities

 

[TobyDarkeness]

ah i see so i should use, 1/2{cos[(n*pi*x)-(m*pi*x)]+cos[(m*pi*x)+(n*pi*x)]} substitute that into my integral and proceed with the integration and Fourier series?

 

[Dick]

ah i see so i should use, 1/2{cos[(n*pi*x)-(m*pi*x)]+cos[(m*pi*x)+(n*pi*x)]} substitute that into my integral and proceed with the integration and Fourier series?

Sure. I hope you haven't been waiting for my reply to proceed.

 

[TobyDarkeness]

Oh no, I was just checking I was heading in the correct direction thanks for the help haven't quite finished but it's coming together. A further question,

Show that the Fourier series for the square wave defined as
f(x)=-1 for -L<=x<=0
f(x)= 1 for 0<=x<=L
is given by the following equation:
f(x)=$$\sum$$ m=1 to infinity of [2/(m*pi)]*[1-(-1)^m)sin[(m*pi*x)/L]

i know this is something to do with the gibs function and (-1)^m is a cos function but i have no idea how to get here. Thanks again.

 

[Dick]

Oh no, I was just checking I was heading in the correct direction thanks for the help haven't quite finished but it's coming together. A further question,

Show that the Fourier series for the square wave defined as
f(x)=-1 for -L<=x<=0
f(x)= 1 for 0<=x<=L
is given by the following equation:
f(x)=$$\sum$$ m=1 to infinity of [2/(m*pi)]*[1-(-1)^m)sin[(m*pi*x)/L]

i know this is something to do with the gibs function and (-1)^m is a cos function but i have no idea how to get here. Thanks again.

You integrate f(x) times the sin function from -L to L to get the coefficient. Just break it into the two integrals from -L to 0 (where f(x)=(-1)) and 0 to L (where f(x)=1).

 

[TobyDarkeness]

thanks again for the help, they worked out pretty well. just another quick question if you can help.

[6] A semi-infinite bar 0 < x < ∞ is subject to periodic heating at x = 0 ; the
temperature at x = 0 is T0 cosωt and is zero at x = ∞. By solving the heat equation

∂T/∂t= 1/2(∂2T/∂x2) ,
show that

T (x,t) = T_0 exp(α x)cos(ωt − x sqrtω ),
where α is a constant to be determined.

Ok I know I need to separate variables and Iv'e been advised to solve for the time dependence first. This is what Iv'e done so far but I'm a little stuck for the next steps.

∂T/∂t= 1/2*(∂^2T/∂x^2)

T(x,t)=X(x)T(t)

∂/∂t*[X(x)T(t)]=1/2*[(∂^2)/(∂x^2)]*(X(x)T(t))

X(x)*[∂T(t)/∂t]=1/2*T(t)*[∂^2X(x)]/[∂x^2]

dividing through by 1/[X(x)T(t)]


1/[T(t)]*[∂T(t)/∂t]=1/2*[1/X(x)]*(∂^2 X(x))/∂x^2


2/T(t)*∂T(t)/∂t=1/X(x)*[(∂^2X(x))/(∂x^2)]


T(x,t) =T_0exp(αx)cos(ωt − x sqrtω)

T(0,t)=T_0cos(ωt)

T(infinity,0)=0

this is what i have not sure where to go next...

thanks again.

 

[TobyDarkeness]

ok i think i have to make the constants complex but I'm not sure what I should sub in.