Mathematics question - Fourier Series

fouadalnoor · Mar 21, 2012

Hi guys,

I am confused about the problem in the attached picture. can someone please explain how he gets the right hand side of that equation?

-> To[1+cos((2*pi*x)/a)]?

Thanks!

MrAl · Mar 22, 2012

Hi,

It looks like they just used the identity:
cos(u)^2=(cos(2*u)+1)/2

You might want to look into reducing trig powers into simpler forms.
Another example would be:
cos(u)^3=(cos(3*u)+3*cos(u))/4

fouadalnoor · Mar 25, 2012

MrAl said:
Hi,

It looks like they just used the identity:
cos(u)^2=(cos(2*u)+1)/2

You might want to look into reducing trig powers into simpler forms.
Another example would be:
cos(u)^3=(cos(3*u)+3*cos(u))/4

Hi, yeah thanks! My friend actually pointed that out a little earlier and it all makes sense now. I am really confused about another thing at this point though, take a look at the attached lecture slides:

How does he get the solution in the red box? I can't seem to use the boundary conditions to get all those exponentials involving x...

MrAl · Mar 27, 2012

Hi again,

This is actually easy...below 'd' stands for partial derivative...

They already give this:
du/dt=0 ==>
d^2u/(dx^2)=u ==>
u=Ae^x+Be^-x

so take that equation:
u=Ae^x+Be^-x [EQ1]

and apply the boundary conditions:
u(0,t)=1 and u(1,t)=0

to that equation EQ1, once for each boundary, and that will give you two equations in x from which you can solve for A and B. Once you have the solutions for A and B you can insert them into EQ1 above and you'll get the solution found within the red box.

To apply boundary condition 1 that means that u=1 when x=0, and to apply boundary condition 2 that means u=0 when x=1.

fouadalnoor · Mar 27, 2012

MrAl said:
Hi again,

This is actually easy...below 'd' stands for partial derivative...

They already give this:
du/dt=0 ==>
d^2u/(dx^2)=u ==>
u=Ae^x+Be^-x

so take that equation:
u=Ae^x+Be^-x [EQ1]

and apply the boundary conditions:
u(0,t)=1 and u(1,t)=0

to that equation EQ1, once for each boundary, and that will give you two equations in x from which you can solve for A and B. Once you have the solutions for A and B you can insert them into EQ1 above and you'll get the solution found within the red box.

To apply boundary condition 1 that means that u=1 when x=0, and to apply boundary condition 2 that means u=0 when x=1.

Yeah that's what I tried doing...

Boundary 1 says: u(0,t)=1
Boundary 2 says: u(1,t)=0

Thus:

1=A+B
0= Ae^(x)+Be(-x)

Now I dont see how you can get how you find out what A and B are... (I am sure I am just not thinking about it properly, I bet there is a simple solution )

MrAl · Mar 27, 2012

fouadalnoor said:
Yeah that's what I tried doing...

Boundary 1 says: u(0,t)=1
Boundary 2 says: u(1,t)=0

Thus:

1=A+B
0= Ae^(x)+Be(-x)

Now I dont see how you can get how you find out what A and B are... (I am sure I am just not thinking about it properly, I bet there is a simple solution )

Hi,

That happens to me too when im tired

You seem to have just missed it...

Boundary 1 says: u(0,t)=1
Boundary 2 says: u(1,t)=0

So using u=Ae^(x)+Be(-x) and those boundary conditions we get these two equations:
1=Ae^(0)+Be(-0)
0=Ae^(1)+Be(-1)

which of course boil down to:
1=A+B
0=A*e+B/e

and you can solve those two equations for A and B, then insert the results into the equation for u: Ae^(x)+Be(-x). Your result should match that found in the red box.

Mathematics question - Fourier Series

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Mathematics question - Fourier Series

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