- #1

roldy

- 237

- 2

## Homework Statement

I need to determine the first three terms in the Fourier series pictured in the attachment.

Did I define the peace-wise functions correctly?

I'm re-posting this with the tex code instead of the attached document.

## Homework Equations

[tex]

a_o=\frac{1}{2L}\int_{-L}^Lf(t)dt

[/tex]

[tex]

a_n=\frac{1}{L}\int_{-L}^Lf(t)cos\left(\frac{n\pi{t}}{L}\right)dt

[/tex]

## The Attempt at a Solution

The plot of the series is symmetric, so therefore I am only going to find what [tex]a_o[/tex] and [tex]a_n[/tex].

[tex]

a_o=\frac{2}{3T} \left[\int_\frac{-3T}{4}^\frac{-T}{4}(-1)dt+\int_\frac{-T}{4}^\frac{T}{4}(1)dt+\int_\frac{T}{4}^\frac{3T}{4}(-1)dt\right]

[/tex]

[tex]

=\frac{2}{3T}\left[-\left(-T/4-(-3T/4)\right)+\left(T/4-(-T/4)\right)+\left(3T/4-T/4\right)\right]

=\frac{2}{3T}\left(T/4-3T/4+T/4+T/4+3T/4-T/4\right)=1/3

[/tex]

[tex]

a_n=\frac{4}{3T}\left[\int_\frac{-3T}{4}^\frac{-T}{4}(-1)cos\left(\frac{n\pi{t}}{3T/4}\right)dt+\int_\frac{-T}{4}^\frac{T}{4}(1)cos\left(\frac{n\pi{t}}{3T/4}\right)dt+\int_\frac{T}{4}^\frac{3T}{4}(-1)cos\left(\frac{n\pi{t}}{3T/4}\right)dt\right]

[/tex]

[tex]

=\frac{4}{3T}\left[\left[-sin\left(\frac{n\pi{\left(-T/4\right)}}{3T/4}\right)+sin\left(\frac{n\pi{\left(-3T/4\right)}}{3T/4}\right)\right]\frac{3T}{4n\pi}+\left[sin\left(\frac{n\pi{\left(T/4\right)}}{3T/4}\right)-sin\left(\frac{n\pi{\left(-T/4\right)}}{3T/4}\right)\right]\frac{3T}{4n\pi}

[/tex]

[tex]

-\left[sin\left(\frac{n\pi{\left(3T/4\right)}}{3T/4}\right)-sin\left(\frac{n\pi{\left(T/4\right)}}{3T/4}\right)\right]\frac{3T}{4n\pi}

[/tex]

[tex]

\frac{1}{n\pi}\left[\left[-sin\left(\frac{-n\pi}{3}\right)+sin\left(-n\pi\right)\right]+\left[sin\left(\frac{n\pi}{3}\right)-sin\left(\frac{-n\pi}{3}\right)\right]-\left[sin\left(n\pi\right)-sin\left(\frac{n\pi}{3}\right)\right]\right]

[/tex]

Distributing the signs through and simplifying:

[tex]

\frac{1}{n\pi}\left[4sin\left(\frac{n\pi}{3}\right)-2sin\left(n\pi\right)\right]

[/tex]

So for n=1,2,3 I get

[tex]

a_1=\frac{2\sqrt{3}}{2\pi}

[/tex]

[tex]

a_2=\frac{\sqrt{3}}{2\pi}

[/tex]

[tex]

a_3=0

[/tex]