Find the Fourier Series of the function

[lesdes]

Homework Statement

Find the Fourier series of the function $f$ given by $f(x) = 1$, $|x| \geq \frac{\pi}{2}$ and $f(x) = 0$, $|x| \leq \frac{\pi}{2}$ over the interval $[-\pi, \pi]$.

Homework Equations

From my lecture notes, the Fourier series is
$f(t) = \frac{a_0}{2}*1 + \sum_{n=1}^\infty a_n cos(nt) + \sum_{n=1}^\infty b_n sin(nt)$. Instead of $\infty$ we use $N\leq20 000$ because that is the limit of the human ear in Hz. (No clue why this is relevant because it is never mentioned or used again throughout the course; it is only mentioned in the lecture notes next to the definition)

The Fourier coefficients are given by
$a_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t)cos(kt) \, dt$, and
$b_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t)sin(kt) \, dt$.

Also there is this neat result that if $f$ is odd, then
$a_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t)cos(kt) \, dt = 0$, and
$b_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t)sin(kt) \, dt = \frac{2}{\pi} \int_{0}^{\pi} f(t)sin(kt) \, dt$.

If $f$ is even, then
$a_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t)cos(kt) \, dt = \frac{2}{\pi} \int_{0}^{\pi} f(t)cos(kt)$, and
$b_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t)sin(kt)\, dt = 0$.

The Attempt at a Solution

The computations of the Fourier coefficients seem simple enough. I am however confused as to what precisely the problem states.

Do I have to calculate the Fourier series for the function $f(x) = 1$ over the interval $[-\pi, \pi]$ with $|x| \geq \frac{\pi}{2}$ and then for a second function $g(x) = 0$ over the interval $[-\pi, \pi]$ with $|x| \leq \frac{\pi}{2}$?
Or does this mean $f(x) = \begin{cases} 1 & \text{if } |x| \geq \frac{\pi}{2}\\ 0 & \text{if } |x | \leq \frac{\pi}{2} \end{cases}$.

If the former, then what do the conditions $|x| \geq \frac{\pi}{2}$ and $|x| \leq \frac{\pi}{2}$ mean? Do I still use the integrals from $a = -\pi$ and $b = \pi$ or does for example $|x| \geq \frac{\pi}{2}$ mean that the integral bounds are $a = -\frac{\pi}{2}$ and $b = \frac{\pi}{2}$?

If the latter, then I have not really an idea on how to proceed.

 

[Mark44]

Or does this mean
$f(x) = \begin{cases} 1 & \text{if } |x| \geq \frac{\pi}{2}\\ 0 & \text{if } |x | \leq \frac{\pi}{2} \end{cases}$.

It means this. Your integrals will need to take into account the three different subintervals: $[-\pi, -\pi/2], [-\pi/2, \pi/2], [\pi/2, \pi]$. There is some simplification, though, because this is an even function, which means there won't be any sine terms.

 

[Mark44]

because that is the limit of the human ear

Apparently, this problem has to do with acoustics, and you don't need to be adding waves that are above the ability of the human ear to hear.

 

[lesdes]

It means this. Your integrals will need to take into account the three different subintervals: $[-\pi, -\pi/2], [-\pi/2, \pi/2], [\pi/2, \pi]$. There is some simplification, though, because this is an even function, which means there won't be any sine terms.

Thank you. I will try and solve it and if I run into any more problems I will come back here or if I arrive at a solution.

Apparently, this problem has to do with acoustics, and you don't need to be adding waves that are above the ability of the human ear to hear.

Well, the course is Linear Algebra and Fourier Series. The prof just said when he explained the topic of Fourier series "We will make a cut here (he referred to $\infty$) and use $N \leq 20000$ (Hz) because that is the limit of the human ear". And the problems are all of the sort "Find the Fourier series of the function...". That is why I find it odd that he mentioned that.

 

[lesdes]

This is what I have done.

$a_0 = \frac{1}{2\pi}(\int_{-\pi}^{-\frac{\pi}{2}} 1 \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 0 \, dx + \int_{\frac{\pi}{2}}^{\pi} 1 \, dx = \frac{1}{2\pi}(x\left. \right|_{-\pi}^{-\frac{\pi}{2}} + 0 + x\left. \right|_{\frac{\pi}{2}}^{\pi}) = \frac{1}{2\pi}(\frac{\pi}{2} + \frac{\pi}{2}) = 2$.

Since $f$ is even, we have $b_n = 0$ for every $n$.
Since $f$ is even, we have $a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)cosnx \, dx = \frac{2}{\pi} \int_{0}^{\pi} f(x)cosnx \, dx$.

Thus,
$a_n = \frac{2}{\pi}(\int_{0}^{\frac{\pi}{2}} 0*cosnx \, dx + \int_{\frac{\pi}{2}}^{\pi} 1*cosnx \, dx) = \frac{2}{\pi}*\frac{1}{n}sinnx\left. \right|_{\frac{\pi}{2}}^{\pi} = \frac{2}{\pi n}(sinn\pi -sinn\frac{\pi}{2}) = -\frac{2}{n\pi}sinn\frac{\pi}{2}$.

For $n$ even, we have $sinn\frac{\pi}{2} = 0$ and for $n$ odd, we have $sinn\frac{\pi}{2} = \pm1$.
The Fourier series of $f$ is
$$S_n(x) = 2 - \frac{2}{\pi}sinx - \frac{2}{3\pi}sin3x - \frac{2}{5\pi}sin5x - ... = 2 + \sum_{n=0}^\infty -\frac{2}{\pi(2n - 1)} sin(x(2n - 1)).$$

 

[Ray Vickson]

This is what I have done.

$a_0 = \frac{1}{2\pi}(\int_{-\pi}^{-\frac{\pi}{2}} 1 \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 0 \, dx + \int_{\frac{\pi}{2}}^{\pi} 1 \, dx = \frac{1}{2\pi}(x\left. \right|_{-\pi}^{-\frac{\pi}{2}} + 0 + x\left. \right|_{\frac{\pi}{2}}^{\pi}) = \frac{1}{2\pi}(\frac{\pi}{2} + \frac{\pi}{2}) = 2$.

Since $f$ is even, we have $b_n = 0$ for every $n$.
Since $f$ is even, we have $a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)cosnx \, dx = \frac{2}{\pi} \int_{0}^{\pi} f(x)cosnx \, dx$.

Thus,
$a_n = \frac{2}{\pi}(\int_{0}^{\frac{\pi}{2}} 0*cosnx \, dx + \int_{\frac{\pi}{2}}^{\pi} 1*cosnx \, dx) = \frac{2}{\pi}*\frac{1}{n}sinnx\left. \right|_{\frac{\pi}{2}}^{\pi} = \frac{2}{\pi n}(sinn\pi -sinn\frac{\pi}{2}) = -\frac{2}{n\pi}sinn\frac{\pi}{2}$.

For $n$ even, we have $sinn\frac{\pi}{2} = 0$ and for $n$ odd, we have $sinn\frac{\pi}{2} = \pm1$.
The Fourier series of $f$ is
$$S_n(x) = 2 - \frac{2}{\pi}sinx - \frac{2}{3\pi}sin3x - \frac{2}{5\pi}sin5x - ... = 2 + \sum_{n=0}^\infty -\frac{2}{\pi(2n - 1)} sin(x(2n - 1)).$$

How can an even function have sines in its expansion?

 

[lesdes]

How can an even function have sines in its expansion?

Ah yes. I accidentally used the $a_n$ in the definition for the Fourier expression from post 1 as $b_n$. I think I mixed them up. I will look at it tomorrow. It's 4 am now. Probably that is also the reason for such silly mistakes.

 

[lesdes]

So I went over it again and I noticed the following error in $a_0$. It should be $a_0=\frac{1}{2}$ and not $a_0=2$.

This is what I have done.

$a_0 = \frac{1}{2\pi}(\int_{-\pi}^{-\frac{\pi}{2}} 1 \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 0 \, dx + \int_{\frac{\pi}{2}}^{\pi} 1 \, dx = \frac{1}{2\pi}(x\left. \right|_{-\pi}^{-\frac{\pi}{2}} + 0 + x\left. \right|_{\frac{\pi}{2}}^{\pi}) = \frac{1}{2\pi}(\frac{\pi}{2} + \frac{\pi}{2}) = 2$.

Also now I have the Fourier series for $f$ given by

$$S_n(x) = \frac{1}{2} - \frac{2}{\pi}cosx + \frac{2}{3\pi}cos3x - \frac{2}{5\pi}cos5x + ... = \frac{1}{2} + \sum_{n=1}^\infty (-1)^n\frac{2}{\pi(2n - 1)} cos(x(2n - 1)).$$
Is this correct now?

As a side question. How do I graph the function $f$ and the Fourier series in Mathematica? I have tried the following but the Fourier series does not show up.

 

[Mark44]

In your second plot, your Mathematica notation might be wrong. I'm not very knowledgeable about Mathematica, so I might be mistaken.
In the factor right after (-1)^i, you have 2/Pi (2i - 1). I think that you believe that this is $\frac 2 {\pi(2i - 1)}$. It's not.

This would actually be evaluated as $\frac 2 \pi (2i - 1)$
I believe this factor should be 2/(Pi * (2i - 1)).

Also for the cosine factor, you are missing an operator for multiplication. This might not be a problem in Mathematica, but it is a problem in virtually every other programming language. I believe this should be Cos(x * (2i - 1)) or Cos((2i - 1) * x).

 

[lesdes]

In your second plot, your Mathematica notation might be wrong. I'm not very knowledgeable about Mathematica, so I might be mistaken.
In the factor right after (-1)^i, you have 2/Pi (2i - 1). I think that you believe that this is $\frac 2 {\pi(2i - 1)}$. It's not.

This would actually be evaluated as $\frac 2 \pi (2i - 1)$
I believe this factor should be 2/(Pi * (2i - 1)).

Also for the cosine factor, you are missing an operator for multiplication. This might not be a problem in Mathematica, but it is a problem in virtually every other programming language. I believe this should be Cos(x * (2i - 1)) or Cos((2i - 1) * x).

Yes thank you! I totally missed the parentheses there and I also noticed that I butchered the syntax of Mathematica. It should be Cos[] and not Cos(). Now I have
the following plot for $n=10$ and $n=100$ for the Fourier series

So my Fourier series is correct then, I suppose?

 

[Mark44]

Those look right on the money! Well done!

 

[lesdes]

Those look right on the money! Well done!

Great!
There is a follow up problem in my assignment to the one here (like check for uniform convergence, $L^2$-convergence). I will first go and read my notes and book about that, especially $L^2$-convergence since I have never done that. Then attempt so solve that problem. If I will have questions regarding that should I ask them here or open a new thread?

 

[Mark44]

Please start a new thread. That way you'll have the problem statement, relevant definitions/equations (e.g., uniform convergence), and your efforts all together in a thread with a relevant title.