11.3 Fourier Expansions II

In this section we discuss Fourier expansions in terms of the eigenfunctions of Problems 1-4 for Section 11.1.

Fourier Cosine Series

From Exercise 11.1. 20, the eigenfunctions

\[ 1,\, \cos{\pi x\over L}, \, \cos{2\pi x\over L},\dots, \, \cos{n\pi x\over L},\dots \]

of the boundary value problem

\begin{equation} y''+\lambda y=0,\quad y'(0)=0,\quad y'(L)=0 \tag{11.3.1}\end{equation}

(Problem 2) are orthogonal on $[0,L]$. If $f $ is integrable on $[0,L]$ then the Fourier expansion of $f$ in terms of these functions is called the Fourier cosine series of $f$ on $[0,L]$. This series is

\[ a_0+\sum_{n=1}^\infty a_n\cos{n\pi x\over L}, \]

where

\[ a_0={\dst\int_0^Lf(x)\,dx\over\dst\int_0^L\,dx}={1\over L}\int_0^Lf(x)\,dx \]

and

\[ a_n={\dst\int_0^Lf(x)\cos{n\pi x\over L}\,dx\over\dst\int_0^L \cos^2{n\pi x\over L}\,dx}={2\over L}\int_0^Lf(x)\cos{n\pi x\over L}\,dx,\quad n=1,2,3,\dots. \]

Comparing this definition with Theorem 6(a) shows that the Fourier cosine series of $f$ on $[0,L]$ is the Fourier series of the function

\[ f_1(x)= \left\{\begin{array}{cl} f(-x),&-L<x<0,\\f(x),&\phantom{-}0\le x\le L, \end{array}\right. \]

obtained by extending $f$ over $[-L,L]$ as an even function (Figure 11.3.1).

Figure 11.3.1.

Applying Theorem 11.2.4 to $f_1$ yields the next theorem.

Theorem 11.3.1

If $f$ is piecewise smooth on $[0,L]$, then the Fourier cosine series

\[ C(x)=a_0+\sum_{n=1}^\infty a_n\cos{n\pi x\over L} \]

of $f$ on $[0,L]$, with

\[ a_0={1\over L}\int_0^Lf(x)\,dx \mbox{\quad and \quad} a_n={2\over L}\int_0^Lf(x)\cos{n\pi x\over L}\,dx,\quad n=1,2,3,\dots, \]

converges for all $x$ in $[0,L];$ moreover$,$

\[ C(x)= \left\{\begin{array}{cl} f(0+)&\mbox{if }x=0 \\ f(x)&\mbox{if $0<x<L$ and $f$ is continuous at $x $}\\[6pt] \dst{f(x-)+f(x+)\over2}&\mbox{if $0<x<L$ and $f$ is discontinuous at $x $}\\[6pt] f(L-)&\mbox{if }x=L. \end{array}\right. \]

Example 11.3.1

Find the Fourier cosine series of $f(x)=x$ on $[0,L]$.

Solution The coefficients are

\[ a_0={1\over L}\int_0^Lx\,dx={1\over L}{x^2\over2} \lims0L={L\over2} \]

and, if $n\ge1$

\begin{eqnarray*} a_n&=&{2\over L}\int_0^Lx\cos{n\pi x\over L}\,dx ={2\over n\pi}\left[x\sin{n\pi x\over L}\lims0L- \int_0^L \sin{n\pi x\over L}\,dx\right] \\ &=&-{2\over n\pi}\int_0^L \sin{n\pi x\over L}\,dx ={2L\over n^2\pi^2}\cos{n\pi x\over L}\lims0L ={2L\over n^2\pi^2}[(-1)^n-1] \\ &=& \left\{\begin{array}{cl} \dst-{4L\over(2m-1)^2\pi^2}&\mbox{if $n=2m-1$},\\ 0&\mbox{if $n=2m$}. \end{array}\right. \end{eqnarray*}

Therefore

\[ C(x)=\dst{L\over2}-{4L\over\pi^2}\sum_{n=1}^\infty{1\over(2n-1)^2} \cos{(2n-1)\pi x\over L}. \]

Theorem 11.3.1 implies that

\[ C(x)=x,\quad 0\le x\le L. \]

Fourier Sine Series

From Exercise 11.1. 19, the eigenfunctions

\[ \sin{\pi x\over L}, \, \sin{2\pi x\over L},\dots, \, \sin{n\pi x\over L},\dots \]

of the boundary value problem

\[ y''+\lambda y=0,\quad y(0)=0,\quad y(L)=0 \]

(Problem 1) are orthogonal on $[0,L]$. If $f$ is integrable on $[0,L]$ then the Fourier expansion of $f$ in terms of these functions is called the Fourier sine series of $f$ on $[0,L]$. This series is

\[ \sum_{n=1}^\infty b_n\sin{n\pi x\over L}, \]

where

\[ b_n={\dst\int_0^Lf(x)\sin{n\pi x\over L}\,dx\over\dst\int_0^L \sin^2{n\pi x\over L}\,dx}={2\over L}\int_0^Lf(x)\sin{n\pi x\over L}\,dx,\quad n=1,2,3,\dots. \]

Comparing this definition with Theorem 6(b) shows that the Fourier sine series of $f$ on $[0,L]$ is the Fourier series of the function

\[ f_2(x)= \left\{\begin{array}{cl} -f(-x),&-L<x<0,\\f(x),&\phantom{-}0\le x\le L, \end{array}\right. \]

obtained by extending $f$ over $[-L,L]$ as an odd function (Figure 11.3.2).

Figure 11.3.2.

Applying Theorem 11.2.4 to $f_2$ yields the next theorem.

Theorem 11.3.2

If $f$ is piecewise smooth on $[0,L]$, then the Fourier sine series

\[ S(x)=\sum_{n=1}^\infty b_n\sin{n\pi x\over L} \]

of $f$ on $[0,L]$, with

\[ b_n={2\over L}\int_0^Lf(x)\sin{n\pi x\over L}\,dx, \]

converges for all $x$ in $[0,L];$ moreover$,$

\[ S(x)= \left\{\begin{array}{cl} 0&\mbox{if }x=0 \\ f(x)&\mbox{if $0<x<L$ and $f$ is continuous at $x $}\\[6pt] \dst{f(x-)+f(x+)\over2}&\mbox{if $0<x<L$ and $f$ is discontinuous at $x $}\\[6pt] 0&\mbox{if }x=L. \end{array}\right. \]

Example 11.3.2

Find the Fourier sine series of $f(x)=x$ on $[0,L]$.

Solution The coefficients are

\begin{eqnarray*} b_n&=&{2\over L}\int_0^Lx\sin{n\pi x\over L}\,dx =-{2\over n\pi}\left[x\cos{n\pi x\over L}\lims0L- \int_0^L \cos{n\pi x\over L}\,dx\right] \\ &=& (-1)^{n+1}{2L\over n\pi}+{2L\over n^2\pi^2}\sin{n\pi x\over L}\lims0L =(-1)^{n+1}{2L\over n\pi}. \end{eqnarray*}

Therefore

\[ S(x)=\dst-{2L\over\pi}\sum_{n=1}^\infty{(-1)^n\over n} \sin{n\pi x\over L}. \]

Theorem 11.3.2 implies that

\[ S(x)= \left\{\begin{array}{cl} x,&0\le x< L,\\0,& x=L. \end{array}\right. \]

Mixed Fourier Cosine Series

From Exercise 11.1. 22, the eigenfunctions

\[ \cos{\pi x\over 2L}, \, \cos{3\pi x\over 2L},\dots, \, \cos{(2n-1)\pi x\over 2L},\dots \]

of the boundary value problem

\begin{equation} y''+\lambda y=0,\quad y'(0)=0,\quad y(L)=0 \tag{11.3.2}\end{equation}

(Problem 4) are orthogonal on $[0,L]$. If $f$ is integrable on $[0,L]$ then the Fourier expansion of $f$ in terms of these functions is

\[ \sum_{n=1}^\infty c_n\cos{(2n-1)\pi x\over2L}, \]

where

\[ c_n={\dst\int_0^Lf(x)\cos{(2n-1)\pi x\over2L}\,dx\over\dst\int_0^L \cos^2{(2n-1)\pi x\over L}\,dx}={2\over L}\int_0^Lf(x)\cos{(2n-1)\pi x\over2L}\,dx. \]

We’ll call this expansion the mixed Fourier cosine series of $f$ on $[0,L]$, because the boundary conditions of (11.3.2) are “mixed” in that they require $y$ to be zero at one boundary point and $y'$ to be zero at the other. By contrast, the “ordinary” Fourier cosine series is associated with (11.3.1), where the boundary conditions require that $y'$ be zero at both endpoints.

It can be shown (Exercise 57) that the mixed Fourier cosine series of $f$ on $[0,L]$ is simply the restriction to $[0,L]$ of the Fourier cosine series of

\[ f_3(x)= \left\{\begin{array}{cl} f(x),&0\le x\le L,\\-f(2L-x),&L< x\le 2L \end{array}\right. \]

on $[0,2L]$ (Figure 11.3.3).

Figure 11.3.3.

Applying Theorem 11.3.1 with $f$ replaced by $f_3$ and $L$ replaced by $2L$ yields the next theorem.

Theorem 11.3.3

If $f$ is piecewise smooth on $[0,L]$, then the mixed Fourier cosine series

\[ C_M(x)=\sum_{n=1}^\infty c_n\cos{(2n-1)\pi x\over2L} \]

of $f$ on $[0,L]$, with

\[ c_n={2\over L}\int_0^Lf(x)\cos{(2n-1)\pi x\over2L}\,dx, \]

converges for all $x$ in $[0,L]; $ moreover$,$

\[ C_M(x)= \left\{\begin{array}{cl} f(0+)&\mbox{if $x=0$} \\ f(x)&\mbox{if $0<x<L$ and $f$ is continuous at $x$ }\\[6pt] \dst{f(x-)+f(x+)\over2}&\mbox{if $0<x<L$ and $f$ is discontinuous at $x$ }\\[6pt] 0&\mbox{if $x=L$.} \end{array}\right. \]

Example 11.3.3

Find the mixed Fourier cosine series of $f(x)=x-L$ on $[0,L]$.

Solution The coefficients are

\begin{eqnarray*} c_n&=&{2\over L}\int_0^L(x-L)\cos{(2n-1)\pi x\over2L}\,dx \\ &=&{4\over(2n-1)\pi}\left[(x-L)\sin{(2n-1)\pi x\over2L}\lims0L-\int_0^L \sin{(2n-1)\pi x\over2L}\,dx\right] \\ &=&{8L\over(2n-1)^2\pi^2} \cos{(2n-1)\pi x\over2L}\lims0L =-{8L\over(2n-1)^2\pi^2}. \end{eqnarray*}

Therefore

\[ C_M(x)=-{8L\over\pi^2}\sum_{n=1}^\infty{1\over(2n-1)^2} \cos{(2n-1)\pi x\over2L}. \]

Theorem 11.3.3 implies that

\[ C_M(x)= x-L,\quad 0\le x\le L. \]

Mixed Fourier Sine Series

From Exercise 11.1. 21, the eigenfunctions

\[ \sin{\pi x\over 2L}, \, \sin{3\pi x\over 2L},\dots, \, \sin{(2n-1)\pi x\over 2L},\dots \]

of the boundary value problem

\[ y''+\lambda y=0,\quad y(0)=0,\quad y'(L)=0 \]

(Problem 3) are orthogonal on $[0,L]$. If $f$ is integrable on $[0,L]$, then the Fourier expansion of $f$ in terms of these functions is

\[ \sum_{n=1}^\infty d_n\sin{(2n-1)\pi x\over2L}, \]

where

\[ d_n={\dst\int_0^Lf(x)\sin{(2n-1)\pi x\over2L}\,dx\over\dst\int_0^L \sin^2{(2n-1)\pi x\over2L}\,dx}={2\over L}\int_0^Lf(x)\sin{(2n-1)\pi x\over2L}\,dx. \]

We’ll call this expansion the mixed Fourier sine series of $f$ on $[0,L]$.

It can be shown (Exercise 58) that the mixed Fourier sine series of $f$ on $[0,L]$ is simply the restriction to $[0,L]$ of the Fourier sine series of

\[ f_4(x)= \left\{\begin{array}{cl} f(x),&0\le x\le L,\\f(2L-x),&L< x\le 2L, \end{array}\right. \]

on $[0,2L]$ (Figure 11.3.4).

Figure 11.3.4.

Applying Theorem 11.3.2 with $f$ replaced by $f_4$ and $L$ replaced by $2L$ yields the next theorem.

Theorem 11.3.4

If $f$ is piecewise smooth on $[0,L]$, then the mixed Fourier sine series

\[ S_M(x)=\sum_{n=1}^\infty d_n\sin{(2n-1)\pi x\over2L} \]

of $f$ on $[0,L]$, with

\[ d_n={2\over L}\int_0^Lf(x)\sin{(2n-1)\pi x\over2L}\,dx, \]

converges for all $x$ in $[0,L]; $ moreover$,$

\[ S_M(x)= \left\{\begin{array}{cl} 0&\mbox{if $x=0$} \\ f(x)&\mbox{if $0<x<L$ and $f$ is continuous at $x$ }\\[6pt] \dst{f(x-)+f(x+)\over2}&\mbox{if $0<x<L$ and $f$ is discontinuous at $x$ }\\[6pt] f(L-)&\mbox{if $x=L$.} \end{array}\right. \]

Example 11.3.4

Find the mixed Fourier sine series of $f(x)=x$ on $[0,L]$.

Solution The coefficients are

\begin{eqnarray*} d_n&=&{2\over L}\int_0^Lx\sin{(2n-1)\pi x\over2L}\,dx \\ &=&-{4\over(2n-1)\pi}\left[x\cos{(2n-1)\pi x\over2L}\lims0L- \int_0^L \cos{(2n-1)\pi x\over2L}\,dx\right] \\ &=&{4\over(2n-1)\pi} \int_0^L \cos{(2n-1)\pi x\over2L}\,dx \\ &=&{8L\over(2n-1)^2\pi^2}\sin{(2n-1)\pi x\over2L}\lims0L=(-1)^{n+1}{8L\over(2n-1)^2\pi^2}. \end{eqnarray*}

Therefore

\[ S_M(x)=-{8L\over\pi^2}\sum_{n=1}^\infty{(-1)^n\over(2n-1)^2} \sin{(2n-1)\pi x\over2L}. \]

Theorem 11.3.4 implies that

\[ S_M(x)=x,\quad 0\le x\le L. \]

A Useful Observation

In applications involving expansions in terms of the eigenfunctions of Problems 1-4, the functions being expanded are often polynomials that satisfy the boundary conditions of the problem under consideration. In this case the next theorem presents an efficient way to obtain the coefficients in the expansion.

Theorem 11.3.5

(a) If $f'(0)=f'(L)=0$, $f''$ is continuous$,$ and $f'''$ is piecewise continuous on $[0,L],$ then

\begin{equation} f(x)=a_0+\sum_{n=1}^\infty a_n\cos{n\pi x\over L}, \quad 0\le x\le L, \tag{11.3.3}\end{equation}

with

\begin{equation} a_0={1\over L}\int_0^L f(x)\,dx \mbox{\quad and \quad} a_n= {2L^2\over n^3\pi^3}\int_0^L f'''(x)\sin{n\pi x\over L}\,dx, \quad n\ge1. \tag{11.3.4}\end{equation}

Now suppose $f'$ is continuous and $f''$ is piecewise continuous on $[0,L].$

(b) If $f(0)=f(L)=0$, then

\[ f(x)=\sum_{n=1}^\infty b_n\sin{n\pi x\over L}, \quad 0\le x\le L, \]

with

\begin{equation} b_n=-{2L\over n^2\pi^2}\int_0^L f''(x)\sin{n\pi x\over L}\,dx. \tag{11.3.5}\end{equation}

(c) If $f'(0)=f(L)=0$, then

\[ f(x)= \sum_{n=1}^\infty c_n\cos{(2n-1)\pi x\over2L}, \quad 0\le x\le L, \]

with

\begin{equation} c_n=-{8L\over(2n-1)^2\pi^2}\int_0^L f''(x)\cos{(2n-1)\pi x\over2L} \,dx. \tag{11.3.6}\end{equation}

(d) If $f(0)=f'(L)=0$, then

\[ f(x)= \sum_{n=1}^\infty d_n\sin{(2n-1)\pi x\over2L}, \quad 0\le x\le L, \]

with

\begin{equation} d_n=-{8L\over(2n-1)^2\pi^2}\int_0^L f''(x)\sin{(2n-1)\pi x\over2L} \,dx. \tag{11.3.7}\end{equation}

Proof We’ll prove (a) and leave the rest to you (Exercises 35, 42, and 50). Since $f$ is continuous on $[0,L]$, Theorem 11.3.1 implies (11.3.3) with $a_0$, $a_1$, $a_2$,... as defined in Theorem 11.3.1. We already know that $a_0$ is as in (11.3.4). If $n\ge1$, integrating twice by parts yields

\begin{eqnarray*} a_n&=& {2\over L}\int_0^L f(x)\cos{n\pi x\over L}\,dx \\ &=&{2\over n\pi}\left[f(x)\sin{n\pi x\over L}\lims0L -\int_0^Lf'(x)\sin{n\pi x\over L}\,dx\right] \\ &=&-{2\over n\pi} \int_0^Lf'(x)\sin{n\pi x\over L}\,dx \mbox{ (since $\sin0=\sin n\pi=0$)} \\ &=&{2L\over n^2\pi^2}\left[f'(x)\cos{n\pi x\over L}\lims0L -\int_0^Lf''(x)\cos{n\pi x\over L}\right]\,dx \\ &=& -{2L\over n^2\pi^2}\int_0^Lf''(x)\cos{n\pi x\over L}\,dx \mbox{ (since $f'(0)=f'(L)=0$)} \\ &=&-{2L^2\over n^3\pi^3}\left[f''(x)\sin{n\pi x\over L}\lims0L -\int_0^Lf'''(x)\sin{n\pi x\over L}\,dx\right] \\ &=& {2L^2\over n^3\pi^3}\int_0^Lf'''(x)\sin{n\pi x\over L}\,dx \mbox{ (since $\sin0=\sin n\pi=0$).} \end{eqnarray*}

(By an argument similar to one used in the proof of Theorem 8.3.1, the last integration by parts is legitimate in the case where $f'''$ is undefined at finitely many points in $[0,L]$, so long as it’s piecewise continuous on $[0,L]$.) This completes the proof.

Example 11.3.5

Find the Fourier cosine expansion of $f(x)=x^2(3L-2x)$ on $[0,L]$.

Solution Here

\[ a_0={1\over L}\int_0^L(3Lx^2-2x^3)\,dx={1\over L}\left(Lx^3-{x^4\over2} \right)\lims0L={L^3\over2} \]

and

\[ a_n={2\over L}\int_0^L(3Lx^2-2x^3)\cos{n\pi x\over L}\,dx,\quad n\ge1. \]

Evaluating this integral directly is laborious. However, since $f'(x)=6Lx-6x^2$, we see that $f'(0)=f'(L)=0$. Since $f'''(x)=-12$, we see from (11.3.4) that if $n\ge1$ then

\begin{eqnarray*} a_n&=&-{24L^2\over n^3\pi^3}\int_0^L\sin{n\pi x\over L}\,dx ={24L^3\over n^4\pi^4}\cos{n\pi x\over L}\lims0L={24L^3\over n^4\pi^4}[(-1)^n-1] \\ &=& \left\{\begin{array}{cl} \dst-{48L^3\over(2m-1)^4\pi^4}&\mbox{if $n=2m-1$},\\ 0&\mbox{if $n=2m$.} \end{array}\right. \end{eqnarray*}

Therefore

\[ C(x)={L^3\over2}-{48L^3\over\pi^4}\sum_{n=1}^\infty{1\over (2n-1)^4}\cos{(2n-1)\pi x\over L}. \]

Example 11.3.6

Find the Fourier sine expansion of $f(x)=x(x^2-3Lx+2L^2)$ on $[0,L]$.

Solution Since $f(0)=f(L)=0$ and $f''(x)=6(x-L)$, we see from (11.3.5) that

\begin{eqnarray*} b_n&=&- {12L\over n^2\pi^2}\int_0^L(x-L)\sin{n\pi x\over L}\,dx \\ &=&{12L^2\over n^3\pi^3}\left[(x-L)\cos{n\pi x\over L}\lims0L -\int_0^L\cos{n\pi x\over L}\,dx\right] \\ &=&{12L^2\over n^3\pi^3}\left[L-\frac{L}{n\pi}\sin\frac{n\pi x}{L}\lims0L\right] ={12L^3\over n^3\pi^3}. \end{eqnarray*}

Therefore

\[ S(x)=\dst{12L^3\over\pi^3}\sum_{n=1}^\infty{1\over n^3} \sin{n\pi x\over L}. \]

Example 11.3.7

Find the mixed Fourier cosine expansion of $f(x)=3x^3-4Lx^2+L^3$ on $[0,L]$.

Solution Since $f'(0)=f(L)=0$ and $f''(x)=2(9x-4L)$, we see from (11.3.6) that

\begin{eqnarray*} c_n&=& -{16L\over(2n-1)^2\pi^2} \int_0^L(9x-4L)\cos{(2n-1)\pi x\over2L}\,dx \\ &=&-{32L^2\over(2n-1)^3\pi^3}\left[(9x-4L)\sin{(2n-1)\pi x\over2L} \lims0L-9\int_0^L\sin{(2n-1)\pi x\over2L}\right]\,dx \\ &=&-{32L^2\over(2n-1)^3\pi^3} \left[(-1)^{n+1}5L+{18L\over(2n-1)\pi}\cos{(2n-1)\pi x\over2L} \lims0L\right] \\ &=&{32L^3\over(2n-1)^3\pi^3} \left[(-1)^n5+{18\over(2n-1)\pi}\right]. \end{eqnarray*}

Therefore

\[ C_M(x)=\dst{32L^3\over\pi^3}\sum_{n=1}^\infty{1\over(2n-1)^3}\left[ (-1)^n5+{18\over(2n-1)\pi}\right]\cos{(2n-1)\pi x\over2L}. \]

Example 11.3.8

Find the mixed Fourier sine expansion of

\[ f(x)=x(2x^2-9Lx+12L^2) \]

on $[0,L]$.

Solution Since $f(0)=f'(L)=0$, and $f''(x)=6(2x-3L)$, we see from (11.3.7) that

\begin{eqnarray*} d_n&=&-\dst{48L\over(2n-1)^2\pi^2}\int_0^L(2x-3L) \sin{(2n-1)\pi x\over2L}\,dx \\ &=&\dst{96L^2\over(2n-1)^3\pi^3}\left[(2x-3L)\cos{(2n-1)\pi x\over2L}\lims0L-2\int_0^L\cos{(2n-1)\pi x\over2L}\,dx\right] \\ &=&\dst{96L^2\over(2n-1)^3\pi^3}\left[3L-{4L\over(2n-1)\pi} \sin{(2n-1)\pi x\over2L}\lims0L\right] \\ &=&\dst{96L^3\over(2n-1)^3\pi^3}\left[3+(-1)^n{4\over(2n-1)\pi}\right]. \end{eqnarray*}

Therefore

\[ S_M(x) =\dst{96L^3\over\pi^3}\sum_{n=1}^\infty {1\over(2n-1)^3}\left[3+(-1)^n{4\over(2n-1)\pi}\right]\sin{(2n-1)\pi x\over2L}. \]

11.3 Exercises

In exercises marked by C graph $f$ and some partial sums of the required series. If the interval is $[0,L]$, choose a specific value of $L$ for the graph.

In Exercises 1-10 find the Fourier cosine series.

$f(x)=x^2$; $[0,L]$

Show answer

$C(x)=\dst{L^2\over3}+{4L^2\over\pi^2}\sum_{n=1}^\infty{(-1)^n\over n^2}\cos{n\pi x\over L}$
C $f(x)=1-x$; $[0,1]$

Show answer

$C(x)=\dst{1\over2}+{4\over\pi^2}\sum_{n=1}^\infty{1\over(2n-1)^2} \cos(2n-1)\pi x$
C $f(x)=x^2-2Lx$; $[0,L]$

Show answer

$C(x)=\dst-{2L^2\over3}+{4L^2\over\pi^2}\sum_{n=1}^\infty{1\over n^2}\cos{n\pi x\over L}$
$f(x)=\sin kx$ ($k\ne$ integer); $[0,\pi]$

Show answer

$C(x)=\dst{1-\cos k\pi\over k\pi}-{2k\over\pi}\sum_{n=1}^\infty {[1-(-1)^n\cos k\pi]\over n^2-k^2}\cos nx$.
C $f(x)= \left\{\begin{array}{cl} 1,&0\le x\le{L\over2}\\0,&{L\over2}<x<L; \end{array}\right.$ $[0,L]$

Show answer

$C(x)=\dst{1\over2}-{2\over\pi}\sum_{n=1}^\infty {(-1)^n\over2n-1}\cos{(2n-1)\pi x\over L}$
$f(x)=x^2-L^2$; $[0,L]$

Show answer

$C(x)=\dst-{2L^2\over3}+{4L^2\over\pi^2}\sum_{n=1}^\infty{(-1)^n\over n^2}\cos{n\pi x\over L}$
$f(x)=(x-1)^2$; $[0,1]$

Show answer

$C(x)=\dst{1\over3}+{4\over\pi^2}\sum_{n=1}^\infty{1\over n^2}\cos n\pi x$
$f(x)=e^x$; $[0,\pi]$

Show answer

$C(x)=\dst{e^\pi-1\over\pi}+{2\over\pi}\sum_{n=1}^\infty {[(-1)^ne^\pi-1]\over(n^2+1)}\cos nx$
C $f(x)=x(L-x)$; $[0,L]$

Show answer

$\dst C(x)={L^2\over6}-{L^2\over\pi^2}\sum_{n=1}^\infty{1\over n^2}\cos{2n\pi x\over L}$
C $f(x)=x(x-2L)$; $[0,L]$

Show answer

$C(x)=\dst-{2L^2\over3}+{4L^2\over\pi^2}\sum_{n=1}^\infty{1\over n^2}\cos{n\pi x\over L}$

In Exercises 11-17 find the Fourier sine series.

C $f(x)=1$; $[0,L]$

Show answer

$S(x)=\dst{4\over\pi}\sum_{n=1}^\infty{1\over(2n-1)} \sin{(2n-1)\pi x\over L}$
C $f(x)=1-x$; $[0,1]$

Show answer

$S(x)=\dst{2\over\pi}\sum_{n=1}^\infty{1\over n} \sin n\pi x$
$f(x)=\cos kx$ ($k\ne$ integer); $[0,\pi]$

Show answer

$S(x)=\dst{2\over\pi}\sum_{n=1}^\infty [1-(-1)^n\cos k\pi]{n\over n^2-k^2}\sin nx$
C $f(x)= \left\{\begin{array}{cl} 1,&0\le x\le{L\over2}\\0,&{L\over2}<x<L; \end{array}\right.$ $[0,L]$

Show answer

$S(x)=\dst{2\over\pi}\sum_{n=1}^\infty{1\over n} \left[1-\cos{n\pi\over2}\right]\sin{n\pi x\over L}$
C $f(x)= \left\{\begin{array}{cl} x,&0\le x\le{L\over2},\\L-x,&{L\over2}\le x\le L; \end{array}\right.$ $[0,L]$.

Show answer

$S(x)=\dst{4L\over\pi^2}\sum_{n=1}^\infty{(-1)^{n+1}\over(2n-1)^2} \sin{(2n-1)\pi x\over L}$
C $f(x)=x\sin x$; $[0,\pi]$

Show answer

$S(x)=\dst{\pi\over2}\sin x-{16\over\pi}\sum_{n=1}^\infty{n\over(4n^2-1)^2}\sin2nx$
$f(x)=e^x$; $[0,\pi]$

Show answer

$S(x)=\dst-{2\over\pi}\sum_{n=1}^\infty {n[(-1)^ne^\pi-1]\over(n^2+1)}\sin nx$

In Exercises 18-24 find the mixed Fourier cosine series.

C $f(x)=1$; $[0,L]$

Show answer

$C_M(x)=\dst-{4\over\pi}\sum_{n=1}^\infty{(-1)^n\over2n-1} \cos{(2n-1)\pi x\over2L}$
$f(x)=x^2$; $[0,L]$

Show answer

$C_M(x)=\dst-{4L^2\over\pi}\sum_{n=1}^\infty{(-1)^n\over2n-1}\left[ 1-{8\over(2n-1)^2\pi^2}\right]\cos{(2n-1)\pi x\over2L}$
C $f(x)=x$; $[0,1]$

Show answer

$C_M(x)= -\dst{4\over\pi}\sum_{n=1}^\infty\left[(-1)^n+{2\over(2n-1)\pi}\right] \cos{(2n-1)\pi x\over2}$.
C $f(x)= \left\{\begin{array}{cl} 1,&0\le x\le{L\over2}\\0,&{L\over2}<x<L; \end{array}\right.$ $[0,L]$

Show answer

$C_M(x)=\dst-{4\over\pi}\sum_{n=1}^\infty{1\over2n-1}\cos{(2n+1)\pi\over4} \cos{(2n-1)\pi x\over 2L}$
$f(x)=\cos x$; $[0,\pi]$

Show answer

$C_M(x)= \dst{4\over\pi}\sum_{n=1}^\infty(-1)^n{2n-1\over(2n-3)(2n+1)} \cos{(2n-1) x\over2}$
$f(x)=\sin x$; $[0,\pi]$

Show answer

$C_M(x)= -\dst{8\over\pi}\sum_{n=1}^\infty{1\over(2n-3)(2n+1)} \cos{(2n-1) x\over2}$
C $f(x)=x(L-x)$; $[0,L]$

Show answer

$C_M(x)=-\dst{8L^2\over\pi^2}\sum_{n=1}^\infty{1\over(2n-1)^2}\left[ 1+{4(-1)^n\over(2n-1)\pi}\right]\cos{(2n-1)\pi x\over2L}$

In Exercises 25-30 find the mixed Fourier sine series.

C $f(x)=1$; $[0,L]$

Show answer

$S_M(x)=\dst{4\over\pi}\sum_{n=1}^\infty{1\over(2n-1)} \sin{(2n-1)\pi x\over2L}$
$f(x)=x^2$; $[0,L]$

Show answer

$S_M(x)=\dst-{16L^2\over\pi^2}\sum_{n=1}^\infty{1\over(2n-1)^2}\left[ (-1)^n+{2\over(2n-1)\pi}\right]\sin{(2n-1)\pi x\over2L}$
C $f(x)= \left\{\begin{array}{cl} 1,&0\le x\le{L\over2}\\0,&{L\over2}<x<L; \end{array}\right.$ $[0,L]$

Show answer

$S_M(x)=\dst{4\over\pi}\sum_{n=1}^\infty{1\over2n-1} \left[1-\cos{(2n-1)\pi)\over4}\right] \sin{(2n-1)\pi x\over 2L}$
$f(x)=\cos x$; $[0,\pi]$

Show answer

$S_M(x)=\dst{4\over\pi}\sum_{n=1}^\infty{2n-1\over(2n-3)(2n+1)}\sin{(2n-1) x\over2}$
$f(x)=\sin x$; $[0,\pi]$

Show answer

$S_M(x)=\dst{8\over\pi}\sum_{n=1}^\infty{(-1)^n\over(2n-3)(2n+1)}\sin{(2n-1) x\over2}$
C $f(x)=x(L-x)$; $[0,L]$.

Show answer

$S_M(x)=\dst{8L^2\over\pi^2}\sum_{n=1}^\infty{1\over(2n-1)^2}\left[ (-1)^n+{4\over(2n-1)\pi}\right]\sin{(2n-1)\pi x\over2L}$

In Exercises 31-34 use Theorem 11.3.5(a) to find the Fourier cosine series of $f$ on $[0,L]$.

$f(x)=3x^2(x^2-2L^2)$

Show answer

$C(x)=\dst-{7L^4\over5}-{144L^4\over\pi^4}\sum_{n=1}^\infty{(-1)^n\over n^4}\cos{n\pi x\over L}$
$f(x)=x^3(3x-4L)$

Show answer

$C(x)=\dst-{2L^4\over5}-{48L^4\over\pi^4}\sum_{n=1}^\infty{1+(-1)^n2\over n^4}\cos{n\pi x\over L}$
$f(x)=x^2(3x^2-8Lx+6L^2)$

Show answer

$C(x)=\dst{3L^4\over5}-{48L^4\over\pi^4}\sum_{n=1}^\infty{2+(-1)^n\over n^4}\cos{n\pi x\over L}$
$f(x)=x^2(x-L)^2$

Show answer

$\dst C(x)={L^4\over30}-{3L^4\over\pi^4}\sum_{n=1}^\infty{1\over n^4}\cos{2n\pi x\over L}$
1. Prove Theorem 11.3.5(b).
2. In addition to the assumptions of Theorem 11.3.5(b), suppose $f''(0)=f''(L)=0$, $f'''$ is continuous, and $f^{(4)}$ is piecewise continuous on $[0,L]$. Show that
  
  \[ b_n={2L^3\over n^4\pi^4}\int_0^L f^{(4)}(x)\sin{n\pi x\over L}\,dx, \quad n\ge1. \]

In Exercises 36-41 use Theorem 11.3.5(b) or, where applicable, Exercise $11.1.$ 35(b), to find the Fourier sine series of $f$ on $[0,L]$.

C $f(x)=x(L-x)$

Show answer

$S(x)=\dst{8L^2\over\pi^3}\sum_{n=1}^\infty{1\over(2n-1)^3} \sin{(2n-1)\pi x\over L}$
C $f(x)=x^2(L-x)$

Show answer

$S(x)=\dst-{4L^3\over\pi^3}\sum_{n=1}^\infty{(1+(-1)^n2)\over n^3} \sin{n\pi x\over L}$
$f(x)=x(L^2-x^2)$

Show answer

$S(x)=\dst-{12L^3\over\pi^3}\sum_{n=1}^\infty{(-1)^n\over n^3} \sin{n\pi x\over L}$
$f(x)=x(x^3-2Lx^2+L^3)$

Show answer

$S(x)=\dst{96L^4\over\pi^5}\sum_{n=1}^\infty{1\over(2n-1)^5}\sin{(2n-1)\pi x\over L}$
$f(x)=x(3x^4-10L^2x^2+7L^4)$

Show answer

$S(x)=\dst-{720L^5\over\pi^5}\sum_{n=1}^\infty{(-1)^n\over n^5}\sin{n\pi x\over L}$
$f(x)=x(3x^4-5Lx^3+2L^4)$

Show answer

$S(x)=\dst-{240L^5\over\pi^5}\sum_{n=1}^\infty{1+(-1)^n2\over n^5}\sin{n\pi x\over L}$
1. Prove Theorem 11.3.5(c).
2. In addition to the assumptions of Theorem 11.3.5(c), suppose $f''(L)=0$, $f''$ is continuous, and $f'''$ is piecewise continuous on $[0,L]$. Show that
  
  \[ c_n={16L^2\over(2n-1)^3\pi^3}\int_0^L f'''(x)\sin{(2n-1)\pi x\over2L} \,dx,\quad n\ge1. \]

In Exercises 43-49 use Theorem 11.3.5(c) or, where applicable, Exercise $11.1$. 42(b), to find the mixed Fourier cosine series of $f$ on $[0,L]$.

C $f(x)=x^2(L-x)$

Show answer

$C_M(x)=-\dst{64L^3\over\pi^3}\sum_{n=1}^\infty{1\over(2n-1)^3}\left[ (-1)^n+{3\over(2n-1)\pi}\right]\cos{(2n-1)\pi x\over2L}$
$f(x)=L^2-x^2$

Show answer

$C_M(x)=-\dst {32L^2\over\pi^3}\sum_{n=1}^\infty{(-1)^n\over (2n-1)^3}\cos{(2n-1)\pi x\over2L}$
$f(x)=L^3-x^3$

Show answer

$C_M(x)=-\dst{96L^3\over\pi^3}\sum_{n=1}^\infty{1\over(2n-1)^3}\left[ (-1)^n+{2\over(2n-1)\pi}\right]\cos{(2n-1)\pi x\over2L}$
$f(x)=2x^3+3Lx^2-5L^3$

Show answer

$C_M(x)=\dst{96L^3\over\pi^3}\sum_{n=1}^\infty{1\over(2n-1)^3}\left[ (-1)^n3+{4\over(2n-1)\pi}\right]\cos{(2n-1)\pi x\over2L}$
$f(x)=4x^3+3Lx^2-7L^3$

Show answer

$C_M(x)=\dst{96L^3\over\pi^3}\sum_{n=1}^\infty{1\over(2n-1)^3}\left[ (-1)^n5+{8\over(2n-1)\pi}\right]\cos{(2n-1)\pi x\over2L}$
$f(x)=x^4-2Lx^3+L^4$

Show answer

$C_M(x)=-\dst{384L^4\over\pi^4}\sum_{n=1}^\infty{1\over(2n-1)^4}\left[ 1+{(-1)^n4\over(2n-1)\pi}\right]\cos{(2n-1)\pi x\over2L}$
$f(x)=x^4-4Lx^3+6L^2x^2-3L^4$

Show answer

$C_M(x)=-\dst{768L^4\over\pi^4}\sum_{n=1}^\infty{1\over(2n-1)^4}\left[ 1+{(-1)^n2\over(2n-1)\pi}\right]\cos{(2n-1)\pi x\over2L}$
1. Prove Theorem 11.3.5(d).
2. In addition to the assumptions of Theorem 11.3.5(d), suppose $f''(0)=0$, $f''$ is continuous, and $f'''$ is piecewise continuous on $[0,L]$. Show that
  
  \[ d_n=-{16L^2\over(2n-1)^3\pi^3}\int_0^L f'''(x)\cos{(2n-1)\pi x\over2L} \,dx,\quad n\ge1. \]

In Exercises 51-56 use Theorem 11.3.5(d) or, where applicable, Exercise 50(b), to find the mixed Fourier sine series of the $f$ on $[0,L]$.

$f(x)=x(2L -x)$

Show answer

$S_M(x)=\dst {32L^2\over\pi^3}\sum_{n=1}^\infty{1\over (2n-1)^3}\sin{(2n-1)\pi x\over2L}$
$f(x)=x^2(3L-2x)$

Show answer

$S_M(x) =\dst-{96L^3\over\pi^3}\sum_{n=1}^\infty {1\over(2n-1)^3}\left[1+(-1)^n{4\over(2n-1)\pi}\right]\sin{(2n-1)\pi x \over2L}$
$f(x)=(x-L)^3+L^3$

Show answer

$S_M(x) =\dst{96L^3\over\pi^3}\sum_{n=1}^\infty {1\over(2n-1)^3}\left[1+(-1)^n{2\over(2n-1)\pi}\right]\sin{(2n-1)\pi x\over2L}$
$f(x)=x(x^2-3L^2)$

Show answer

$S_M(x)=\dst{192L^3\over\pi^4}\sum_{n=1}^\infty{(-1)^n\over(2n-1)^4} \sin{(2n-1)\pi x\over2L}$
$f(x)=x^3(3x-4L)$

Show answer

$S_M(x)=\dst{1536L^4\over\pi^4}\sum_{n=1}^\infty{1\over(2n-1)^4} \left[(-1)^n+{3\over(2n-1)\pi}\right]\sin{(2n-1)\pi x\over2L}$
$f(x)=x(x^3-2Lx^2+2L^3)$

Show answer

$S_M(x)=\dst{384L^4\over\pi^4}\sum_{n=1}^\infty{1\over(2n-1)^4} \left[(-1)^n+{4\over(2n-1)\pi}\right]\sin{(2n-1)\pi x\over2L}$
Show that the mixed Fourier cosine series of $f$ on $[0,L]$ is the restriction to $[0,L]$ of the Fourier cosine series of

\[ f_3(x)= \left\{\begin{array}{cl} f(x),&0\le x\le L,\\-f(2L-x),&L< x\le 2L \end{array}\right. \]

on $[0,2L]$. Use this to prove Theorem 11.3.3.
Show that the mixed Fourier sine series of $f$ on $[0,L]$ is the restriction to $[0,L]$ of the Fourier sine series of

\[ f_4(x)= \left\{\begin{array}{cl} f(x),&0\le x\le L,\\f(2L-x),&L< x\le 2L \end{array}\right. \]

on $[0,2L]$. Use this to prove Theorem 11.3.4.
Show that the Fourier sine series of $f$ on $[0,L]$ is the restriction to $[0,L]$ of the Fourier sine series of

\[ f_3(x)= \left\{\begin{array}{cl} f(x),&0\le x\le L,\\-f(2L-x),&L< x\le 2L \end{array}\right. \]

on $[0,2L]$.
Show that the Fourier cosine series of $f$ on $[0,L]$ is the restriction to $[0,L]$ of the Fourier cosine series of

\[ f_4(x)= \left\{\begin{array}{cl} f(x),&0\le x\le L,\\f(2L-x),&L< x\le 2L \end{array}\right. \]

on $[0,2L]$.