Recent content by zack7

Thank you for all the help, I got it solved, got my teacher to explain.

I used the same boundary condition for the integral below (0<x<pi) \frac{2}{pi}\int(12sin(2x)*sin(\frac{2pix}{pi}) This gave me 12 Is this wrong ?

Using this formula \frac{2}{L}\int(f(x)*sin(\frac{npix}{L}) for f(x) = 2sin(x) n=1 since cx=x which gives me one L=pi from the conditions of the wave equation where x goes from 0 to pi This gives me \frac{2}{pi}\int(2sin(x)*sin(\frac{1pix}{pi}) \frac{2}{pi}\int(2sin(x)*sin(x))...

Using this formula \frac{2}{L}\int(f(x)*sin(\frac{npix}{L}) I obtain An= 2 and bn= 12 what should I do next ? Is this the way you were talking about, now I would need to find Bn

This was the equation that my teacher used when he found Bn , I do not know any other way to find it. (\frac{n*pi*a}{l})*b2=Fourier Coefficient of g

Looking at the function 12sin(2x) I get n= 2 L= pi (initial condition) Fourier coefficient = 12 a= 2 ( previous answer (a1) Using the formula I get \frac{12pi}{2pi*2}=B2 which gives me three which is wrong since the answer should be two

That formula was given by my teacher after forming the formula for u(x,t) An= Fourier Coefficient of f (\frac{n*pi*a}{l})*b2=Fourier Coefficient of g The u(x,t) formula is \sum(An*cos\frac{npiat}{l}+Bn*sin\frac{npiat}{l})*sin\frac{npix}{L} where a infinite series is not needed for this...

Using the formula (\frac{n*pi*a}{l})*b2=12 where a=2, ;L=pi, and n=2, Therefore I get three, I think I am making a mistake with my a.

Basically I was taught the the first U(x,0) is f and the second is g, so I go on to name the Fourier coefficient of f coefficient as a and the second as b.

For the wave equation I managed to get the coefficient of f: a1=2 and the coefficient of g: \frac{12pi}{2pi*2}=B2 Is these answers right, since my B2 does not match the answer I was given. Thank you

Yup got it, totally forgot the trig identity cos(A+ B)= cos(A)cos(B)- sin(A)sin(B) Thank you

What I don't get is how to I simplify from \frac{-(cos(n*pi+pi)}{(n+1)}+\frac{-(cos(n*pi-pi)}{(n-1)} to \frac{(cos(n*pi)}{(n+1)}-\frac{(cos(n*pi)}{(n-1)} to \frac{-(2 cos(pi n)}{(n^2-1)}

I have manged to get my answer down to the first line in the picture but I have tried all ways and can't seem to simplify it to the second line. Thank you

Yup used Matlab and got the same answer. Thank you

For the right hand side I got 0.397137375v2 for the left I got 7.5-6.250498975-2.499999995=-1.250498 used A=0.25ft, R=(1/3)ft ,I=0.0222