Fourier Cosine Series for f(t)=1-t; 0≤t≤π | Homework Solution and Equations

[icystrike]

Homework Statement

f(t)=1-t ; 0≤t≤\pi

Homework Equations

The Attempt at a Solution

$$a_{0}=\frac{1}{2}\int_{0}^{2}\left( 1-t\right) dt=0$$
$$a_{n}=\int_{0}^{2}\left(1-t\right ) cos(\frac{n\pi\left t \right}{2}) dt = \frac{4}{(n\pi)^2}(1-(-1)^n)$$
Ans that is given to me is:
$$\frac{2}{\pi} + \sum_{n=1}^{\infty} \frac{4}{(n\pi)^2}(1-(-1)^n) cos(\frac{n\pi\left t \right}{2})$$

I'm wondering where the $$\frac{2}{\pi}$$ from...

 

[icystrike]

bump~

 

[icystrike]

~Bump ( Sorry guys):shy:

 

[LCKurtz]

Homework Statement

f(t)=1-t ; 0≤t≤\pi

Homework Equations

The Attempt at a Solution

$$a_{0}=\frac{1}{2}\int_{0}^{2}\left( 1-t\right) dt=0$$
$$a_{n}=\int_{0}^{2}\left(1-t\right ) cos(\frac{n\pi\left t \right}{2}) dt = \frac{4}{(n\pi)^2}(1-(-1)^n)$$
Ans that is given to me is:
$$\frac{2}{\pi} + \sum_{n=1}^{\infty} \frac{4}{(n\pi)^2}(1-(-1)^n) cos(\frac{n\pi\left t \right}{2})$$

I'm wondering where the $$\frac{2}{\pi}$$ from...

I'm wondering where your upper limit of 2 in your integrals came from. I'm wondering if you calculated a₀ correctly. Did you use the half range formulas for the coefficients?

 

[HallsofIvy]

A "bump" after 39 minutes and then a second after another 1 hour 10 minutes? How close to banning does that get you?

 

[icystrike]

Hi LCKurtz!

I have applied the formula for coeff:

$$a_{0}=\frac{1}{L}\int_{0}^{L}f\left( x\right) dx$$

$$a_{n}=\frac{2}{L}\int_{0}^{L}f\left( x\right) \cos \left( \frac{n\pi x}{L}\right) dx$$

Given $$x\in \left[ 0,L\right]$$Fourier Sine series:

$$b_{n}=\int_{0}^{2}\left( 1-t\right) \sin \left( \frac{n\pi t}{2}\right) dt$$

$$=\frac{2}{n\pi }\left( 1+\left( -1\right) ^{n}\right)$$Fourier Sine Series:

$$\sum_{n=1}^{\infty }\frac{2}{n\pi }\left( 1+\left( -1\right) ^{n}\right)<br /> \sin \left( \frac{n\pi t}{2}\right)$$

 

[LCKurtz]

But isn't L = pi given?

 

[icystrike]

Oh gee.. My bad.. Let me check again