exercise12.tex

\documentclass[11pt]{article}
% % \def\hidesolutions{}
\input{header}
\begin{document}
\exsheet{12}{5}{December} % parameters are the number of the session and the day






\begin{exercise}
    Consider the following functions with period $T$:
    \begin{itemize}
        \item A function $f$ with period $T = 1$ such that 
        \begin{align*}
            f(x) = 
            \left\{\begin{array}{ll}
                1 & \text{if } 0 \leq x < 0.5 \\
                0 & \text{if } 0.5 \leq x \leq 1
            \end{array}\right.
        \end{align*}
        \item A function $g$ with period $T = 2\pi$ such that 
        \begin{align*}
            g(x) = 
            \left\{\begin{array}{ll}
                x & \text{if } 0 \leq x < \pi \\
                2\pi - x & \text{if } \pi \leq x \leq 2\pi
            \end{array}\right.
        \end{align*}
        \item A function $h$ with period $T = 1$ such that 
        \begin{align*}
            h(x) = -x \text{ if } 0 \leq x < 1.
        \end{align*}
    \end{itemize}
    Find the Fourier coefficients of the Fourier series of these functions. 
    \textit{You can use the Fourier series seen in the lecture to get the coefficients.}
\end{exercise}

\begin{solution}     
    \begin{itemize}
    \item As seen in the lecture we know the Fourier coefficients of a square wave,
    \[
        l(x)= \begin{cases}1, & \text { if } 0 \leq x<0.5 \\ -1, & \text { if } 0.5 \leq x<1\end{cases},
    \]
    are given by:
    \[
        a_n = 0\quad \text{for } n \geq 0, \quad b_n = \begin{cases}0, & \text { if n is even} \\ \frac{4}{n\pi}, & \text{ if n is odd}\end{cases}
    \]
    We can express the function $f(x)$ in terms of $l(x)$ as follows:
    \[
        f(x) = \frac{1}{2} + \frac{1}{2}l(x)
    \]
    therefore the Fourier coefficients of $f$ are given by:
    \[
        a_n = \begin{cases} 1, & \text { if n} = 0 \\0,& \text{ if n }>0\end{cases}, \quad b_n = \begin{cases}0, & \text { if n is odd} \\ \frac{2}{n\pi}, & \text{ if n is even}\end{cases}
    \]
    \item  As seen in the lecture we know the Fourier coefficients of a triangle wave,
    \[
        m(x)= \begin{cases} 2 x, & \text { if } 0 \leq x<0.5 \\ 2(1-x), & \text { if } 0.5 \leq x<1\end{cases}
    \]
    are given by:
    \[
        a_0 = 1, \quad a_n = \begin{cases}-\frac{4}{n^2\pi^2}, & \text { if n is odd} \\ 0, & \text{ if n is even}\end{cases}, \quad b_n = 0 \text{ for } n > 0
    \]
    We can express the function $g(x)$ in terms of $m(x)$ as follows:
    \[
        g(x) = \pi m(\frac{x}{2\pi})
    \]
    therefore the Fourier coefficients of $g$ are given by:
    \[
        a_0 = \pi, \quad a_n = \begin{cases}-\frac{4}{n^2\pi}, & \text { if n is odd} \\ 0, & \text{ if n is even}\end{cases}, \quad b_n = 0 \text{ for } n > 0
    \]
    \item As seen in the lecture we know the Fourier coefficients of a sawtooth wave,
    \[
        n(x)=x \text { for } 0 \leq x<1
    \]
    are given by:   
    \[
        a_0 = 1, \quad a_n = 0 \text{ for } n > 0,  \quad a_n = -\frac{1}{n\pi} \text{ for } n > 0,
    \]
    We can express the function $h(x)$ in terms of $n(x)$ as follows:
    \[
        h(x) = -n(x)
    \]
    therefore the Fourier coefficients of $h$ are given by:
    \[
        a_0 = -1, \quad a_n = 0 \text{ for } n > 0,  \quad a_n = \frac{1}{n\pi} \text{ for } n > 0,
    \]
    \end{itemize}
\end{solution}







\begin{exercise}
    Explicitly write down the coefficients $a_n$ and $b_n$ and the periods of the following Fourier series:
    \begin{align*}
        f(x) &= \sum_{n=1}^{\infty} \frac{1}{(2n+1)^2}\sin(2\pi (2n+1) x),
        \\
        g(x) &= \sum_{n=1}^{\infty} \frac{(-1)^n}{(2n-1)^3}\cos(2\pi (2n-1) x),
        \\
        h(x) &= \frac \pi 3 + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}\cos(\pi n x).
    \end{align*}
    Determine the Fourier coefficients of the derivatives of those functions. 
\end{exercise}
\begin{solution}     
    \begin{itemize}
    \item We substitute $m = 2n+1$ to obtain 
    \[
            f(x) = \sum_{\substack{m=3\\m \text{ is odd}}}^{\infty} \frac{1}{m^2}\sin(2\pi m x),
    \]

    therefore the Fourier coefficients are given by 
    \[
    a_m = 0 \text{ for } m \geq 0, \quad b_m = \begin{cases}\frac{1}{m^2}, & \text { if m is odd and }m \geq 3 \\ 0, & \text{ otherwise }  \end{cases} 
    \]

    and the period is $T = 1$. The derivative of $f(x)$ is given by:
    \[
    f'(x) = \sum_{\substack{m=3\\m \text{ is odd}}}^{\infty} \frac{2\pi}{m}\sin(2\pi m x),
    \]

    therefore the Fourier coefficients of the derivative of $f(x)$ are given by 

    \[
    b_m = 0 \text{ for } m \geq 0, \quad a_m = \begin{cases}\frac{2\pi}{m}, & \text { if m is odd and }m \geq 3 \\ 0, & \text{ otherwise }  \end{cases} 
    \]

    \item We substitute $m = 2n-1$ to obtain 
    \[
        g(x) = \sum_{\substack{ m=1 \\ m \text{ is odd}}}^{\infty} \frac{(-1)^{\frac{m+1}{2}}}{m^3}\cos(2\pi m x),
    \]

    therefore the Fourier coefficients are given by 
    \[
    b_m = 0 \text{ for } m >0, \quad a_m = \begin{cases}\frac{(-1)^{\frac{m+1}{2}}}{m^3}, & \text { if m is odd and }m \geq 1 \\ 0, & \text{ if m is even and } m \geq 1\\ 0 & \text{ if } m = 0  \end{cases} 
    \]

    and the period is $T = 1$. The derivative of $g(x)$ is given by:
    \[
    g'(x) = \sum_{\substack{m=1\\m \text{ is odd}}}^{\infty} -\frac{2\pi (-1)^{\frac{m+1}{2}}}{m^2}\sin(2\pi m x),
    \]

    therefore the Fourier coefficients of the derivative of $g(x)$ are given by 
    \[
    a_m = 0 \text{ for } m \geq 0, \quad b_m = \begin{cases}-\frac{2\pi (-1)^{\frac{m+1}{2}}}{m^2}, & \text { if m is odd and }m \geq 1 \\ 0, & \text{ if m is even and } m \geq 1\\ 0 & \text{ if } m = 0  \end{cases} 
    \]
    \item 
    the Fourier coefficients are given by 
    \[
    b_n = 0 \text{ for } n >0, \quad a_n = \begin{cases}\frac{2\pi}{3}, & \text { if }n = 0 \\ -\frac{(-1)^n}{n^2}, & \text{ if } n > 1 \end{cases} 
    \]

    and the period $T = 2$. The derivative of $h(x)$ is given by:
    \[
    h'(x) =\sum_{n=1}^{\infty} -\frac{\pi(-1)^{n+1}}{n}\sin(\pi n x),
    \]

    therefore the Fourier coefficients of the derivative of $h(x)$ are given by 
    \[
    a_n = 0 \text{ for } n \geq 0, \quad b_n = -\frac{\pi(-1)^{n+1}}{n},\quad \text{for }n > 0
    \]
    \end{itemize}
\end{solution}




\begin{exercise}
    Let \( f(x) \) be a periodic function with period \( T \), represented by its Fourier series:
    \begin{align*}
        f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty \left( a_n \cos\left(\frac{2\pi n x}{T}\right) + b_n \sin\left(\frac{2\pi n x}{T}\right) \right).
    \end{align*}
    As explained in the lecture, 
    \begin{align*}
        \int_{x_0}^{x}
        f(x) 
        dx
        = 
        \frac{a_0}{2} ( x - x_0 )
        + 
        \sum_{n=1}^{\infty} 
        a_n 
        \int_{x_0}^{x}
        \cos\left(\frac{2\pi n x}{T}\right)
        dx 
        +
        b_n
        \int_{x_0}^{x}
        \sin\left(\frac{2\pi n x}{T}\right)
        dx
        .
    \end{align*}
    Complete this discussion and find the Fourier series of a function $g(x)$ such that 
    \begin{align*}
        \int_{x_0}^{x} f(x) = c x + g(x)
    \end{align*}
    for some $c \in \mathbb R$. 
\end{exercise}
\begin{solution}
    We first compute 
    \begin{gather*}
        \int_{x_0}^x \frac{a_0}{2} \, dt = \frac{a_0}{2} (x - x_0).
    \end{gather*}
    We integrate the sign and cosine terms, using the fundamental theorem of calculus:
    \begin{gather*}
        \int_{x_0}^x \cos\left(\frac{2\pi n t}{T}\right) \, dt 
        = 
        \frac{T}{2\pi n} \left( \sin\left(\frac{2\pi n x}{T}\right) - \sin\left(\frac{2\pi n x_0}{T}\right) \right).
    \end{gather*}
    \begin{gather*}
        \int_{x_0}^x \sin\left(\frac{2\pi n t}{T}\right) \, dt 
        = 
        -\frac{T}{2\pi n} \left( \cos\left(\frac{2\pi n x}{T}\right) - \cos\left(\frac{2\pi n x_0}{T}\right) \right).
    \end{gather*}
    Combining all terms, the integral \( F(x) \) becomes:
    \begin{gather*}
        F(x) 
        = 
        \frac{a_0}{2} (x - x_0) 
        + 
        \sum_{n=1}^\infty 
        a_n \frac{T}{2\pi n} \left( \sin\left(\frac{2\pi n x}{T}\right) - \sin\left(\frac{2\pi n x_0}{T}\right) \right) 
        - 
        b_n \frac{T}{2\pi n} \left( \cos\left(\frac{2\pi n x}{T}\right) - \cos\left(\frac{2\pi n x_0}{T}\right) \right).
    \end{gather*}
    We rewrite this once again and obtain
    \begin{align*}
        F(x) 
        &
        = 
        \frac{a_0}{2} x 
        - 
        \frac{a_0 x_0 }{2}
        + 
        \sum_{n=1}^\infty 
        \left( 
        (-a_n) \frac{T}{2\pi n} \sin\left(\frac{2\pi n x_0}{T}\right) 
        + 
        b_n \frac{T}{2\pi n} \cos\left(\frac{2\pi n x_0}{T}\right) 
        \right)
        \\&\qquad\qquad\qquad\qquad 
        + 
        \sum_{n=1}^\infty 
        (-b_n) \frac{T}{2\pi n} \cos\left(\frac{2\pi n x}{T}\right)
        + 
        a_n \frac{T}{2\pi n} \sin\left(\frac{2\pi n x}{T}\right) 
        .
    \end{align*}
    We thus obtain $c = \frac{a_0}{2}$ and the Fourier coefficients of the function $g$:
    \begin{align*}
        \frac{A_0}{2} &:= -\frac{a_0 x_0 }{2}
        + 
        \sum_{n=1}^\infty 
        \left( 
        (-a_n) \frac{T}{2\pi n} \sin\left(\frac{2\pi n x_0}{T}\right) 
        + 
        b_n \frac{T}{2\pi n} \cos\left(\frac{2\pi n x_0}{T}\right) 
        \right)
        \\
        A_n &:= (-b_n) \frac{T}{2\pi n} 
        \\
        B_n &:= a_n \frac{T}{2\pi n} 
    \end{align*}
    This completes the discussion. 
\end{solution}



\begin{exercise}
    Suppose that $f : \bbR \rightarrow \bbR$ and $g : \bbR \rightarrow \bbR$ are functions. 
    Recall that a function is called \textit{even} if 
    \begin{align*}
        f(-x) = f(x)
    \end{align*}
    and \textit{odd} if 
    \begin{align*}
        f(-x) = - f(x)
    \end{align*}
    \begin{itemize}
     \item Show that if $f$ and $g$ are both odd or both even, then $fg$ is even 
     \item Show that if one of $f$ and $g$ is odd and the other is even, then $fg$ is odd.
     \item Show that the only function that is both odd and even has constant value zero. 
    \end{itemize}
\end{exercise}
\begin{solution}
    \begin{itemize}
     \item If $f$ and $g$ are both even, then 
     \begin{align*}
        f(-x)g(-x) = f(x)g(x),
     \end{align*}
     and if both are odd, then 
     \begin{align*}
        f(-x)g(-x) = (-1)^{2} f(x)g(x) = f(x) g(x),
     \end{align*}
     so $fg$ must be an even function.
     \item 
     Suppose that $f$ is even and that $g$ is odd. Then 
     \begin{align*}
        f(-x) g(-x) = f(x) g(-x) = - f(x)g(x),
     \end{align*}
     showing that $fg$ is an odd function. The same is true if we switch the role of $f$ and $g$.
     \item 
     Suppose that $f$ is both odd and even. Then 
     \begin{align*}
        f(x) = f(-x) = -f(x)
     \end{align*}
     for each $x \in \bbR$. But the only number that equals its negative is zero. So $f(x) = 0$ for each $x$.
    \end{itemize}
\end{solution}









\begin{exercise}
    Compute the Fourier transform of the function
    \begin{gather*}
        f(x) = \left\{\begin{array}{ll}
            x   & \text{ if $0 \leq x < 1$ }
            \\
            0   & \text{ otherwise }
          \end{array}\right.
    \end{gather*}
    You can either directly use the complex exponential, or you can express it in terms of the sine and cosine function. 

    (Interpretation: the function $f(x)$ describes a localized signal: it is zero at $x=0$, then it rises linearly up to $1$, and then it jumps back to zero and remains zero from there on. The signal is not periodic.)
\end{exercise}
\begin{solution}     
    We write down the solution in two different ways, either using the complex exponential directly, or writing it as a sum of sine and cosine.
    \begin{align*}
        \mathfrak{F}(f)(\alpha)&=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(x) e^{-i \alpha x} d x\\
        &=\frac{1}{\sqrt{2 \pi}} \int_{0}^{1} x e^{-i \alpha x} d x\\
        &=\frac{1}{\sqrt{2 \pi}} \left[ \frac{x}{-i\alpha} e^{-i \alpha x} \right]_0^1 - \frac{1}{\sqrt{2 \pi}} \int_{0}^{1} \frac{1}{-i\alpha} e^{-i \alpha x} d x\\
        &=\frac{1}{\sqrt{2 \pi}} \left[ \frac{x}{-i\alpha} e^{-i \alpha x} \right]_0^1 + \frac{1}{\sqrt{2 \pi}} \int_{0}^{1} \frac{1}{i\alpha} e^{-i \alpha x} d x\\
        &=\frac{1}{\sqrt{2 \pi}} \left[ \frac{x}{-i\alpha} e^{-i \alpha x} \right]_0^1 + \frac{1}{\sqrt{2 \pi}} \left[ \frac{1}{\alpha^2} e^{-
        i \alpha x} \right]_{0}^1\\
        &=\frac{1}{\sqrt{2 \pi}} \frac{1}{-i\alpha} e^{-i \alpha } + \frac{1}{\sqrt{2 \pi}} \left( \frac{1}{\alpha^2} e^{-i \alpha} -\frac{1}{\alpha^2}\right)\\
        &=\frac{1}{\sqrt{2 \pi}}\left( \frac{i}{\alpha} e^{-i \alpha } + \frac{1}{\alpha^2} e^{-i \alpha} -\frac{1}{\alpha^2}\right)\\ 
        &=\frac{1}{\sqrt{2 \pi}}\left( \frac{i}{\alpha} \cos\alpha  + \frac{1}{\alpha} \sin\alpha + \frac{1}{\alpha^2} \cos\alpha - \frac{i}{\alpha^2} \sin\alpha -\frac{1}{\alpha^2}\right)\\ 
    \end{align*}
    Next we do it in terms of sine and cosine functions.
    \[
    \begin{aligned}
    \mathfrak{F}(f)(\alpha)&=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(x) e^{-i \alpha x} d x\\
    &=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(x)(\cos \alpha x - i\sin\alpha x) d x\\
    & = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(x) \cos\alpha x d x - i\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(x) \sin\alpha x d x
    \end{aligned}
    \]
    We evaluate each integral seperately.
    \begin{align*} 
        \int_0^1 x \cos (\alpha x) d x 
        & =
        \left[\frac{x}{\alpha} \sin \alpha x\right]_0^1-\int_0^1 \frac{1}{\alpha} \sin \alpha x d x 
        \\ & 
        =\left[\frac{x}{\alpha} \sin \alpha x\right]_0^1+\left[\frac{1}{\alpha^2} \cos \alpha x\right]_0^1 
        \\ & 
        =\frac{1}{\alpha} \sin \alpha+\frac{1}{\alpha^2} \cos \alpha-\frac{1}{\alpha^2}
    \end{align*}
    \begin{align*} 
        \int_0^1 x \sin (\alpha x) d x 
        & 
        =\left[-\frac{x}{\alpha} \cos (\alpha x)\right]_0^1+\int_0^1 \frac{1}{\alpha} \cos (\alpha(x) d x 
        \\ & 
        =\left[-\frac{x}{\alpha} \cos (\alpha x)\right]_0^1+\left[\frac{1}{\alpha^2} \sin \alpha x\right]_0^1 
        \\ & 
        =-\frac{1}{\alpha} \cos \alpha+\frac{1}{\alpha^2} \sin \alpha
    \end{align*}
    All together we have:
    \begin{align*}
        \mathfrak{F}(f)(\alpha)
        &
        =
        \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(x) e^{-i \alpha x} d x 
        \\&
        = 
        \frac{1}{\sqrt{2\pi}}\left(\frac{1}{\alpha} \sin \alpha+\frac{1}{\alpha^2} \cos \alpha-\frac{1}{\alpha^2} +i\frac{1}{\alpha} \cos \alpha-i\frac{1}{\alpha^2} \sin \alpha\right)
    \end{align*}
\end{solution}












\begin{exercise}
    Find the Fourier transform of 
    \begin{align*}
        f(x) = \left\{\begin{array}{ll}
                \sin(x) & \text{ if } 0 \leq x \leq 2 \pi 
                \\
                0 & \text{ otherwise }
               \end{array}
               \right.
    \end{align*}
\end{exercise}
\begin{solution} 
We try to perform integration by parts 
\begin{align*} 
    \mathcal{F}(f(x))(\alpha) 
    & =
    \frac{1}{\sqrt{2 \pi}} \int_0^{2 \pi} \sin x e^{-i \alpha x} d x 
    \\ & 
    =
    \frac{1}{\sqrt{2 \pi}}\left[\sin x \frac{e^{-i \alpha x}}{-i \alpha}\right]_0^{2 \pi}
    +
    \frac{1}{\sqrt{2 \pi}} \int_0^{2 \pi} \frac{\cos x}{i \alpha} e^{-\alpha i x} d x 
    \\ & 
    =
    \frac{1}{\sqrt{2 \pi}} \int_0^{2 \pi} \frac{\cos x}{i \alpha} e^{-\alpha i x} d x 
    \\ & 
    =
    \frac{1}{\sqrt{2 \pi}}\left[\frac{\cos x}{a^2} e^{-i \alpha x}\right]_0^{2 \pi}
    +
    \frac{1}{\sqrt{2 \pi}} \int_0^{2 \pi} \frac{\sin x}{\alpha^2} e^{-i \alpha x} d x 
    \\ & 
    =
    \frac{1}{\sqrt{2 \pi} \alpha^2}\left(e^{-2 \pi i \alpha}-1\right)+\frac{1}{\alpha^2} \frac{1}{\sqrt{2 \pi}} \int_0^{2 \pi} \sin x e^{-i \alpha x} d x
\end{align*}
It seems we have obtained the term that we started with. However, 
\begin{align*}
    \frac{1}{\sqrt{2 \pi}} \int_0^{2 \pi} \sin x e^{-i \alpha x} d x 
    =
    \frac{1}{\sqrt{2 \pi} \alpha^2}\left(e^{-2 \pi i \alpha}-1\right)+\frac{1}{\alpha^2} \frac{1}{\sqrt{2 \pi}} \int_0^{2 \pi} \sin x e^{-i \alpha x} d x
\end{align*}
can be rearranged to 
\begin{align*}
    \left( 1 - \frac{1}{\alpha^2} \right)
    \frac{1}{\sqrt{2 \pi}} \int_0^{2 \pi} \sin x e^{-i \alpha x} d x 
    =
    \frac{1}{\sqrt{2 \pi} \alpha^2}\left(e^{-2 \pi i \alpha}-1\right)
    .
\end{align*}
It follows that 
\begin{align*}
    \frac{1}{\sqrt{2 \pi}} \int_0^{2 \pi} \sin x e^{-i \alpha x} d x 
    &
    =
    \left( 1 - \frac{1}{\alpha^2} \right)^{-1}
    \frac{1}{\sqrt{2 \pi} \alpha^2}\left(e^{-2 \pi i \alpha}-1\right)
    \\&
    =
    \left( \frac{\alpha^2-1}{\alpha^2} \right)^{-1}
    \frac{\left(e^{-2 \pi i \alpha}-1\right)}{\sqrt{2 \pi} \alpha^2} 
    \\ & 
    =\frac{\alpha^2}{\alpha^2-1} \frac{\left(e^{-2 \pi i \alpha}-1\right)}{\alpha^2 \sqrt{2 \pi}} 
    \\ & 
    =
    \frac{\left(e^{-2 \pi i \alpha}-1\right)}{\sqrt{2 \pi}\left(\alpha^2-1\right)}
\end{align*}
which is the desired Fourier transform.
\end{solution}

















\begin{exercise}[Extra]
    We have introduced the Fourier transform 
    \begin{align*}
        \frakF(f)(\alpha) := \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x) e^{-i \alpha x} dx
    \end{align*}
    Different authors define the Fourier transform alternatively by:
    \begin{align*}
        \frakF_2(f)(\xi) 
        := \int_{-\infty}^{\infty} f(x) e^{-i 2\pi \xi x} dx
        ,
        \quad 
        \frakF_3(f)(\omega) 
        := \int_{-\infty}^{\infty} f(x) e^{-i \omega x} dx
        .
    \end{align*}
    Express $\frakF_{2}(f)$ and $\frakF_{3}(f)$ in terms of $\frakF(f)$.
\end{exercise}
\begin{solution}    
    Obviously, 
    \begin{align*}
        \frakF_{3}(f)(\omega) = \sqrt{2\pi} \frakF(f)(\omega).
    \end{align*}
    For the other transformation, we write:
    \begin{align*}
        \frakF_2(f)(\xi) 
        &
        = 
        \int_{-\infty}^{\infty} f(x) e^{-i 2\pi \xi x} dx
        = 
        \sqrt{2\pi} 
        \cdot 
        \frac{1}{\sqrt{2\pi}}
        \int_{-\infty}^{\infty} f(x) e^{-i 2\pi \xi x} dx
        = 
        \sqrt{2\pi} 
        \cdot 
        \frakF(f)( 2\pi \xi )
        .
    \end{align*}
\end{solution}




\end{document}