How Do You Calculate the Fourier Series of a Piecewise Function with Period 4?

[Nallyfish]

Homework Statement

The function f has period 4 and is such that
f(x)=2+x, -2<x\leq0
f(x)=2, 0<x<2

Sketch the graph of f for x∈[−4, 4] and obtain its Fourier series.

Homework Equations

The Attempt at a Solution

Okay so I've pretty much sketched the graph, but I've been thrown by the 2+x and I'm not sure what to do from this point onwards.

So a₀=\frac{1}{2}\int^{2}_{0}2dx=2

a_n=\frac{1}{2}\int^{2}_{0}2*cosnxdx

So
a_n=\frac{sin2n}{n}

And
b_n=\frac{-(cos(2n)-1)}{n}

Am I on the right lines...?

 

[LCKurtz]

Homework Statement

The function f has period 4 and is such that
f(x)=2+x, -2<x\leq0
f(x)=2, 0<x<2Okay so I've pretty much sketched the graph, but I've been thrown by the 2+x and I'm not sure what to do from this point onwards.

So a₀=\frac{1}{2}\int^{2}_{0}2dx=2

a_n=\frac{1}{2}\int^{2}_{0}2*cosnxdx

So
a_n=\frac{sin2n}{n}

And
b_n=\frac{-(cos(2n)-1)}{n}

Am I on the right lines...?

No, you aren't. You are given one period of a function whose period is 4. That is a two piece function defined on (-2,2). If you have graphed it you will know that it is neither even nor odd so you may not use the half-range formulas for the coefficients. The formula for a_n in this problem is

a_n = \frac{1}{2}\int_{-2}^{2}f(x)\cos(\frac{n\pi}{2}x)\, dx

and similarly for b_n. You have to break these intgrals up into (-2,0) and (0,2) and use the appropriate formula for f(x) in each.

 

[Nallyfish]

Oh god, I'm an idiot. Thank you for that! Can't believe I did that.

So I now have

a_n=\frac{1}{2}\int^{0}_{-2}(2+x)cos(\frac{n\pi}{2})xdx+\frac{1}{2}\int^{2}_{0}2cos(\frac{n\pi}{2})xdx

= \frac{4sin(n\pi)}{n\pi}

and b_n= \frac{-2(ncos(n\pi))(\pi-sin(n\pi))}{n²\pi²}

 

[Nallyfish]

Sorry, I forgot to take the sines and cosines out

a_n=\frac{2}{n^{2}\pi^{2}}

and

b_n= 0

Is what I've got which I HOPE is correct.
This would make the Fourier series

f(x)=\frac{3}{2}*\sum^{\infty}_{n=1}\frac{2}{n^{2}\pi^{2}}cos(\frac{(n\pi)x}{2})

 

[LCKurtz]

Sorry, I forgot to take the sines and cosines out

a_n=\frac{2}{n^{2}\pi^{2}}

and

b_n= 0

Is what I've got which I HOPE is correct.
This would make the Fourier series

f(x)=\frac{3}{2}*\sum^{\infty}_{n=1}\frac{2}{n^{2}\pi^{2}}cos(\frac{(n\pi)x}{2})

That can't be correct because your original f(x) is neither even nor odd, and this answer is an even function. You can't have all the b_n be zero.

 

[Nallyfish]

I keep reaching different answers for my a_n and b_n

My latest answers are

a_n= \frac{2(n+1)}{n^{2}\pi^{2}}

and

b_n= \frac{2(n\pi^{2}+\pi^{2}-1)}{n\pi}

 

[LCKurtz]

I keep reaching different answers for my a_n and b_n

My latest answers are

a_n= \frac{2(n+1)}{n^{2}\pi^{2}}

and

b_n= \frac{2(n\pi^{2}+\pi^{2}-1)}{n\pi}

Those are both wrong and you haven't calculated a₀. Here's what you should get for your constants:

a_n=\frac{(-1)^{n+1}+1}{\pi n^2}
b_n=\frac{(-1)^{n+1}}{n}
a_0=2-\frac{\pi}{4}

Keep trying.

 

[vela]

Those aren't correct. You should get
\begin{align*}
a_n &= (1+(-1)^{n+1})\frac{2}{n^2\pi^2} \\
b_n &= (-1)^{n+1}\frac{2}{n\pi}
\end{align*}
I think you dropped a factor of pi somewhere, LCKurtz.

 

[Nallyfish]

I got a₀= \frac{3}{2}

Because I thought the formula was

\frac{1}{2L}\int^{L}_{-L}f(x)dx

and L is 2 so

a₀= \frac{1}{4}[\int^{0}_{-2}(2+x) dx + \int^{2}_{0}2 dx]

Which would give \frac{3}{2}

 

[LCKurtz]

Those aren't correct. You should get
\begin{align*}
a_n &= (1+(-1)^{n+1})\frac{2}{n^2\pi^2} \\
b_n &= (-1)^{n+1}\frac{2}{n\pi}
\end{align*}
I think you dropped a factor of pi somewhere, LCKurtz.

I got a₀= \frac{3}{2}

Because I thought the formula was

\frac{1}{2L}\int^{L}_{-L}f(x)dx

and L is 2 so

a₀= \frac{1}{4}[\int^{0}_{-2}(2+x) dx + \int^{2}_{0}2 dx]

Which would give \frac{3}{2}

Yes, 3/2 is correct for a₀. Vela's formulas for the coefficients are correct. I inadvertently used L = pi instead of L = 2 when I calculated the coefficients.