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\begin{center}

{\bf Fourier Series and their Applications}

\end{center}

${}$

The functions $1,\cos x,\sin x,\cos2x,\sin2x,\cdots$ are
orthogonal over $(-\pi,\pi)$.

$\ds\int_{-\pi}^\pi\cos mx\cos
nxdx=\left\{\begin{array}{cl}0&m\not=n\\ \pi&m=n\not=0\\
2\pi&m=n=0\end{array}\right.$

$\ds\int_{-\pi}^\pi\cos mx\sin nxdx=0$ for all $m,n$

$\ds\int_{-\pi}^\pi\sin mx\sin
nxdx=\left\{\begin{array}{cl}0&m\not=n\\
\pi&m=n\not=0\end{array}\right.$

In fact the functions satisfy these relations over any interval
$(\alpha,\alpha+2\pi)$.  Assuming that $f(x)$, defined and
integrable in $(-\pi,\pi)$, has an expansion.

$\ds\frac{1}{2}a_0+\sum_{n=1}^\infty(a_n\cos mx+b_n\sin nx)$

uniformly convergent over $(-\pi,\pi)$

$\ds\int_{-\pi}^{\pi}f(x)dx=\pi a_0$

$\ds\int_{-\pi}^{\pi}f(x)\cos nxdx=\pi a_n$  therefore $\ds
a_n+ib_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)e^{inx}dx$

$\ds\int_{-\pi}^{\pi}f(x)\sin nxdx=\pi b_n$

These coefficients exist irrespective of whether or not the series
converges and is equal to $f(x)$, and they are called the Fourier
coefficients.

${}$

Sufficient Conditions for convergence

\begin{itemize}
\item[a)]
If $f(x)$ is differentiable at $\xi$ (or if $\exists m$, such that
$\ds\left|\frac{f(x)-f(\xi)}{x-\xi}\right|<m$,
$x\epsilon(\xi-h,\xi+h)$) then the fourier series converges at
$\xi$ to $f(\xi)$.

\item[b)]
If $f(x)$ is monotonic in $\xi<x<\xi+h$ and in $\xi-h<x<\xi$ for
some $h>0$, then the fourier series converges at $\xi$ to the
value

$\frac{1}{2}\{f(\xi-0)+f(\xi+0)\}$.
\end{itemize}

\newpage

General Range

The range $a\leq x\leq b$ is standardised by substituting $\ds
X=\frac{\pi\left(x-\frac{a+b}{2}\right)}
{\left(\frac{b-a}{2}\right)}$

then $-\pi<X<\pi$.

The series $\ds\frac{1}{2}a_0+\sum_{n=1}^\infty a_n\cos nX+b_n\sin
nX$

becomes $\ds\frac{1}{2}a_0+\sum_{n=1}^\infty
a_n\cos2n\pi\left(\frac{2x-(a+b)}{b-a}\right)+
b_n\sin2n\pi\left(\frac{2x-(a+b)}{b-a}\right)$

${}$

Periodicity of $f(x)$

We suppose that $f(x)$ is represented by the series (when cgt.)
for all $x$, hence since the sum function of the series is
periodic, with period $2\pi$, we have $f(x+2\pi)=f(x)$ which
defines $f(x)$ outside the original range.

${}$

Fourier Series for $\frac{1}{2}-t \hspace{0.3in} (0<t<1)$

First consider the identity

$\ds1+\sum_{n=1}^me^{nix}= \frac{e^{(m+1)ix}-1}{e^{ix}-1}=
\frac{e^{\left(m+\frac{1}{2}\right)ix}-e^{-\frac{1}{2}ix}}
{2i\sin\frac{1}{2}x}$

$\ds\hspace{0.5in}=\frac{\cos\left(m+\frac{1}{2}\right)x+
i\sin\left(m+\frac{1}{2}\right)x-
\left(\cos\frac{1}{2}x-i\sin\frac{1}{2}x\right)}
{2i\sin\frac{1}{2}x}$

Hence for $x$ real ($x\not=0,\pm2\pi,\cdots$) taking real and
imaginary parts:

$\ds Re: \,\,\, 1+\sum_{n=1}^m\cos nx=\frac{1}{2}+
\frac{1}{2}\frac{\sin\left(m+\frac{1}{2}\right)x}
{\sin\frac{1}{2}x}$

or $\ds\frac{1}{2}+\sum_{n=1}^m\cos
nx=\frac{1}{2}\frac{\sin\left(m+\frac{1}{2}\right)x}
{\sin\frac{1}{2}x} \hfill$ (1)

$\ds Im: \,\,\, -\frac{1}{2}\cot\frac{1}{2}x+ \sum_{n=1}^m\sin
nx=-\frac{1}{2}\frac{\cos\left(m+\frac{1}{2}\right)x}
{\sin\frac{1}{2}x} \hfill$ (2)

Integrate (1) and (2) from $x$ to $\pi$

(1):$\ds \,\,\, \frac{1}{2}(\pi-x)-\sum_{n=1}^m\frac{\sin nx}{n}=
\frac{1}{2}\int_x^\pi\frac{\sin\left(m+\frac{1}{2}\right)t}
{\sin\frac{1}{2}t}dt \hfill$ (3)

(2):$\ds \,\,\,
\left[-\log\left(\sin\frac{1}{2}t\right)\right]_x^\pi+
\left[\sum_{n=1}^m\frac{-\cos nt}{n}\right]_x^\pi=
-\frac{1}{2}\int_x^\pi\frac{\cos\left(m+\frac{1}{2}\right)t}
{\sin\frac{1}{2}t}dt \hfill$ (4)

Now suppose $\delta\leq x\leq 2\pi-\delta$.

Then using the Riemann Lebesgue theorem, we have, letting
$m\to\infty$ in (3) and (4)

$\ds\frac{1}{2}(\pi-x)=\sum_{n=1}^\infty\frac{\sin nx}{n} \hfill$
(5)

$\ds\log\sin\frac{1}{2}x-\sum_{n=1}^\infty\frac{(-1)^n}{n}+
\sum_{n=1}^\infty\frac{\cos nx}{n}=0$

Therefore $\ds\log2\sin\frac{1}{2}x= -\sum_{n=1}^\infty\frac{\cos
nx}{n} \hfill$ (6)

${}$

Alternative Proof of (5) and (6)

$\ds\log(1-\xi)=-\int_0^\xi\frac{dt}{1-t}$

where we take a cut along the positive real axis in the $t$-plane
from 1 to $\infty$.  The branch of $\log(1-\xi)$ chosen is that
which is real when $\xi$ is real, and is one-valued in the cut
plane.  In particular this vanishes at $\xi=0$

$\ds\frac{1}{1-t}=1+t+\cdots+t^{m-1}+\frac{t^m}{1-t}$

therefore $\ds\int_0^\xi\frac{dt}{1-t}=
\sum_{n=1}^m\frac{\xi^n}{n}+ \int_0^\xi\frac{t^m}{1-t}dt$

where the path is taken along the radius $0-\xi$.


DIAGRAM


For all $t$ on the radius through $\xi$

$\begin{array}{lccc}
|1-t|&\geq&|\sin\theta|&Re(\xi)>0\\&\geq&1&{\rm
otherwise}\end{array}$

Hence in all cases $|1-t|\geq\sin\delta$ when
$\delta\leq\arg\xi\leq2\pi-\delta \hspace{0.3in} (0<\delta<\pi)$

Therefore $\ds\left|\int_0^\xi\frac{t^m}{1-t}dt\right|=
\left|\int_0^r\frac{(pe^{i\theta})^me^{i\theta}}{1-t}dp\right|$

$\ds\leq\int_0^r\frac{p^m}{|\sin\delta|}dp=
\frac{1}{\sin\delta}\frac{r^{m+1}}{m+1}\leq
\frac{1}{(m+1)\sin\delta} \hfill 0\leq r\leq1$

Hence $\ds\lim_{m\to\infty}
\left|\int_0^\xi\frac{t^mdt}{1-t}\right|=0 \hfill
\delta\leq\arg\xi\leq2\pi-\delta$

$\hfill 0\leq r\leq1$

When $r=1$ the convergence is uniform with respect to $\theta$.
Hence we have

$\ds\log(1-\xi)=-\sum_{n=1}^\infty\frac{\xi^n}{n}$

Where the series converges on $|\xi|=1$ except at $\xi=1$,
uniformly in

$\delta\leq\arg\xi\leq2\pi-\delta$

${}$

$\ds\log(1-\xi)=\log|1-\xi|+i\arg(1-\xi)$

$\ds=\log\left(2\sin\frac{1}{2}\theta\right)-
i\left(\frac{\pi}{2}-\frac{\theta}{2}\right)$

Taking real and imaginary parts gives

$\ds\frac{1}{2}\pi-\theta=\sum_1^\infty\frac{\sin n\theta}{n}$

$\ds\log\left(2\sin\frac{1}{2}\theta\right)=
-\sum_1^\infty\frac{\cos n\theta}{n} \hfill 0<\theta<2\pi$

Convergence being uniform in $\delta\leq\theta\leq2\pi-\delta$.

${}$

Fourier Expansion of the Bernoulli polynomials in $0\leq t\leq1$.
Values of the Bernoulli numbers.

Put $x=2\pi t$ in $\ds\frac{1}{2}(\pi-x)= \sum_1^\infty\frac{\sin
nx}{n}$

Therefore $\ds t-\frac{1}{2}=
-\frac{1}{\pi}\sum_1^\infty\frac{\sin2\pi nt}{n}$

$\ds\hspace{0.5in}=-2\sum_1^\infty\frac{\sin2\pi nt}{2\pi n}
\hfill (1) \hspace{0.5in} 0<t<1$

Therefore $\ds P_1(t)=-2\sum_1^\infty\frac{\sin2\pi nt}{2\pi n}$

$P'_2(t)=P_1(t) \hspace{0.5in} P_2(0)=0$

Therefore $\ds P_2(t)=\int_0^t P_1(s)ds$

The series (1) converges uniformly in $\epsilon\leq
t\leq1-\epsilon$

$\ds\int_\epsilon^t
P_1(s)ds=-2\sum_1^\infty\int_\epsilon^t\frac{\sin2n\pi s}{2n\pi}ds
\hfill \epsilon\leq t\leq1-\epsilon$

$\ds=-2\sum_1^\infty\frac{\cos2n\pi\epsilon-\cos2n\pi
t}{(2n\pi)^2}$

The series on the right converges absolutely and uniformly since

$\ds\left|\frac{\cos(2n\pi
t)}{(2n\pi)^2}\right|\leq\frac{1}{(2n\pi)^2}$ and
$\ds\sum\frac{1}{n^2}$ converges.

Hence $\ds\int_0^t P_1(s)ds=-2\sum_1^\infty\frac{1-\cos2n\pi
t}{(2n\pi)^2} \hfill 0\leq t\leq 1$

(using continuity)

Hence $\ds P_2(t)=\bar{P_2}+2\sum_1^\infty\frac{\cos2n\pi
t}{(2n\pi)^2} \hfill (2)$

$\ds\bar{P_2}=-2\sum_1^\infty\frac{1}{(2n\pi)^2}=
\frac{-2}{(2\pi)^2}S_2$

Next $P'_3(t)=P_2(t)-\bar{P_2}$

Therefore $\ds P_3(t)=2\sum_1^\infty\frac{\sin2n\pi t}{(2n\pi)^3}$

and generally we have

$\ds P_{2m}(t)-\bar{P_{2m}}=
(-1)^{m-1}2\sum_{m=1}^\infty\frac{\cos2n\pi t}{(2n\pi)^{2m}}$

$\ds P_{2m+1}(t)=(-1)^{m-1}2\sum_{m=1}^\infty\frac{\sin2n\pi
t}{(2n\pi)^{2m+1}}$

$\ds\bar{P_{2m}}=(-1)^m\frac{2S_{2m}}{(2\pi)^{2m}}$

We also have

$\ds\frac{\phi_m(t)}{m!}=P_m(t) \hspace{0,5in} m=2,3,\cdots$

$\ds\frac{B_m}{(2m)!}=(-1)^m\bar{P_{2m}}$

For $k\geq2$ it can be shown that

$\ds 1\leq S_k\leq1+\frac{1}{2^{k-2}}(S_2-1)$

Therefore $S_k=1+o(k)$

Also $\ds P_{2m+1}(t)\sim(-1)^{m-1}\frac{2}{(2\pi)^{2m+1}}\sin2\pi
t$

$\ds P_{2m}(t)\sim(-1)^{m-1}\frac{2}{(2\pi)^m}(1-\cos2\pi t)$


${}$

Fourier Series of the Square Wave

We have $\ds\frac{1}{2}(\pi-x)=\sum_{n=1}^\infty\frac{\sin nx}{n}
\hfill 0<x<2\pi$

Write $y=x-\pi$

$\ds-\frac{1}{2}y=\sum_{n=1}^\infty(-1)^n\frac{\sin ny}{n} \hfill
-\pi<y<\pi$

$\ds x=2\sum_{n=1}^\infty(-1)^{n-1}\frac{\sin nx}{n} \hfill
-\pi<x<\pi$

Write $\ds f(x)=2\sum_{n=1}^\infty(-1)^{n-1}\frac{\sin nx}{n}$

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Graph of $f(x)$ is shown by solid lines.

Graph of $f(x+\pi)$ is shown by broken lines.

Graph of $f(x)-f(x+\pi)$ is shown by dotted lines.

The fourier series of $f(x)-f(x+\pi)$ is then

$\ds2\sum_{n=1}^\infty(-1)^n\frac{\sin
nx}{n}-2\sum_{n=1}^\infty(-1)^{n-1}(-1)^n\frac{\sin nx}{n}$

$\ds=4\sum_{n=0}^\infty\frac{\sin(2n+1)x}{2n+1}$

$\ds\frac{4}{\pi}\sum_{n=0}^\infty\frac{\sin(2n+1)x}{2n+1}=
\left\{\begin{array}{cr} +1&0<x<\pi\\
-1&-\pi<x<0\end{array}\right.$

Find coefficients by direct integration.

${}$

Gibbs' Phenomenon

Write $\ds
S_m(x)=\frac{4}{\pi}\sum_{n=0}^m\frac{\sin(2n+1)x}{2n+1}$

$\ds\frac{\pi}{4}\frac{d}{dx}S_m(x)=\sum_{n=0}^m\cos(2n+1)x=
\frac{\sin(2m+2)x}{2\sin x}$

This vanishes in $o<x<\pi$ at $\ds
x=\frac{\pi}{2m+2}\cdots\frac{(2m+1)\pi}{2m+2}$

$S_m(x)$ is symmetrical about $\frac{\pi}{2}$.

Hence consider the value of $S_m$ for $0<x<\frac{\pi}{2}$, and in
particular at

$\ds x=\frac{\pi}{2m+2}$, the first max.

$\ds\frac{\pi}{4}S_m\left(\frac{\pi}{2m+2}\right)=
\int_0^\frac{\pi}{2m+2}\frac{\sin(2m+2)t}{2\sin t}dt$

Put $\ds t=\frac{s}{2m+2}$ then we have

$\ds\frac{\pi}{4}\sin\left(\frac{\pi}{2m+2}\right)=
\frac{1}{2}\int_0^\pi\frac{\sin sds}{(2m+2)\sin\frac{s}{2m+2}}=
\frac{1}{2}\int_0^\pi\frac{sin
s}{s}\phi\left(\frac{s}{2m+2}\right)ds$

where $\ds\phi(u)=\frac{u}{\sin u}$.

Now $1\leq\phi(u)\leq\phi(\delta) \hspace{0.2in} 0\leq
u\leq\delta<\pi \hspace{0.2in}$ and
$\ds0\leq\frac{s}{2m+2}\leq{\pi}{2m+2}$

So $\ds1\leq\phi\left(\frac{s}{2m+2}\right)\leq
\phi\left(\frac{\pi}{2m+2}\right)$

$\ds1\geq\frac{\sin s}{s}\geq0 \hspace{0.5in}$ in $0\leq s\leq\pi$

Hence $\ds\int_0^\pi\frac{\sin s}{s}ds\leq\int_0^\pi\frac{\sin
s}{s}\phi\left(\frac{s}{2m+2}\right)ds$

$\ds\leq\phi\left(\frac{\pi}{2m+2}\right)\int_0^\pi\frac{\sin
s}{s}ds$

Since $\ds\lim_{m\to\infty}\phi\left(\frac{\pi}{2m+2}\right)=1$ we
have

$\ds\lim_{m\to\infty}\int_0^\pi\frac{\sin
s}{s}\phi\left(\frac{s}{2m+2}\right)ds= \int_0^\pi\frac{\sin
s}{s}ds$

Hence $\ds\lim_{m\to\infty}S_m\left(\frac{\pi}{2m+2}\right)=
\frac{2}{\pi}\int_0^\pi\frac{\sin s}{s}ds\approx1.179>1$

${}$

Dirichlet's Formula (sufficient conditions for convergence)

Assume that $f(x)$ is bounded and integrable over $[-\pi,\pi]$,
and

$f(x+2\pi)=f(x)$

Write $\ds S_m(x)=\frac{1}{2}a_0+\sum_{n=1}^m(a_n\cos nx+b_n\sin
nx)$

$\ds=\frac{1}{\pi}\int_{-\pi}^\pi
f(t)\left[\frac{1}{2}+\sum_{n=1}^m(\cos nt\cos nx+\sin nt\sin
nx)\right]dt$

$\ds=\frac{1}{\pi}\int_{-\pi}^\pi
f(t)\left[\frac{1}{2}+\sum_{n=1}^m\cos n(t-x)\right]dt$

$\ds=\frac{1}{2\pi}\int_{-\pi}^\pi
f(t)\frac{\sin\left(m+\frac{1}{2}\right)(t-x)}
{\sin\frac{1}{2}(t-x)}dt$

$\ds=\frac{1}{2\pi}\int_{-\pi-x}^{\pi-x}
f(x+s)\frac{\sin\left(m+\frac{1}{2}\right)s}{\sin\frac{1}{2}s}ds$

$\ds=\frac{1}{2\pi}\int_{-\pi}^\pi
f(x+s)\frac{\sin\left(m+\frac{1}{2}\right)s}{\sin\frac{1}{2}s}ds
\hfill$ by periodicity

$\ds=\frac{1}{2\pi}\int_0^\pi[f(x+t)+f(x-t)]
\frac{\sin\left(m+\frac{1}{2}\right)t}{\sin\frac{1}{2}t}dt \hfill$
as $\ds\frac{\sin\left(m+\frac{1}{2}\right)t}{\sin\frac{1}{2}t}$
is even.

Since $\ds\frac{1}{2}+\sum_{n=1}^m\cos nx=
\frac{\sin\left(m+\frac{1}{2}\right)x}{2\sin\frac{1}{2}x}$,

$\ds\frac{1}{2}\pi=\int_0^\pi\frac{\sin\left(m+\frac{1}{2}\right)x}
{2\sin\frac{1}{2}x}dx$

Therefore $\ds\frac{1}{2}[f(x+0)+f(x-0)]=
\frac{1}{2\pi}\int_0^\pi[f(x+0)+f(x-0)]
\frac{\sin\left(m+\frac{1}{2}\right)t}{\sin\frac{1}{2}t}$

Therefore $\ds S_m(x)-\frac{1}{2}[f(x+0)+f(x-0)]$

$\ds\hspace{0.2in}=\frac{1}{2\pi}\int_0^\pi[f(x+t)-f(x+0)]
\frac{\sin\left(m+\frac{1}{2}\right)t}{\sin\frac{1}{2}t}dt$

$\ds\hspace{0.4in}+\frac{1}{2\pi}\int_0^\pi[f(x-t)-f(x-0)]
\frac{\sin\left(m+\frac{1}{2}\right)t}{\sin\frac{1}{2}t}dt \hfill
(1)$


When $f(x+0)=f(x-0)=f(x) \hspace{0.3in} (1)$ becomes

$\ds S_m(x)-f(x)=\frac{1}{2\pi}\int_0^\pi[f(x+t)+f(x-t)-2f(x)]
\frac{\sin\left(m+\frac{1}{2}\right)t}{\sin\frac{1}{2}t}dt \hfill$
(1a)


The integrals appearing in (1) and (1a) are all of the form

$\ds\int_a^b\phi(t)\sin\lambda tdt$ where $a=0, \,\, b=\pi, \,\,
\lambda=m+\frac{1}{2}$

$\ds\phi(t)=\frac{f(x+t)-f(x+0)}{\sin\frac{1}{2}t}, \,\,\,
\frac{f(x-t)-f(x-0)}{\sin\frac{1}{2}t}, \,$ or

$\ds\frac{f(x+t)+f(x-t)-2f(x)}{\sin\frac{1}{2}t}$

Hence if $\phi(t)$ is bounded and integrable over $[0,\pi]$, then
by the Riemann Lebesgue theorem, $\ds\int_0^\pi\phi(t)\sin\lambda
t\to0$ as $\lambda\to\infty$.

In fact in the above cases $\phi(t)$ is bounded and integrable
over $[h,\pi] \,\,\, h>0$ and so the convergence depends only on
the behaviour of the function in a sufficiently small interval
$[0,h]$.

${}$

Integration of a Fourier Series

If $f(x)$ is bounded and integrable in $[-\pi,\pi]$ and $\ds
F(x)=\int_{-\pi}^{\pi}\left(f(t)-\frac{1}{2}a_0\right)dt$

Where $\frac{1}{2}a_0=\bar{f}$ is the constant term in the Fourier
series for $f$, then $F(x)$ has a Fourier series, convergent
everywhere to $F(x)$, obtained by integrating the Fourier series
for $f(x)-\frac{1}{2}a_0$ term by term.  [This holds even if the
Fourier series for $f$ does not converge.]

$F(x)$ is an absolutely continuous function and hence possesses a
Fourier series converging everywhere to $F(x)$.

$\ds F(x)=\frac{1}{2}A_0+\sum_{n=1}^\infty(A_n\cos nx+B_n\sin nx)$

Assuming that $f$ is continuous on $(-\pi,\pi)$ ensures the
existence of $F'(x)$, and

$\ds A_n=\frac{1}{\pi}\int_{\pi}^\pi F(x)\cos nxdx \hspace{0.5in}
n=1,2,\cdots$

$\ds\hspace{0.3in}=\frac{1}{\pi}\left[F(x)\frac{\sin
nx}{n}\right]_{-\pi}^\pi-\frac{1}{n}\frac{1}{\pi}\int_{-\pi}^\pi
F'(x)\sin nxdx$

$\ds\hspace{0.3in}=0-\frac{1}{n\pi}\int_{-\pi}^\pi
\left(f(x)-\frac{1}{2}a_0\right)\sin nxdx \hfill
[F(\pi)=F(-\pi)=0]$

$\ds\hspace{0.3in}=-\frac{1}{n\pi}\int_{-\pi}^\pi f(x)\sin
nxdx=-\frac{b_n}{n}$

Similarly $\ds B_n=\frac{a_n}{n}$

Therefore $\ds
F(x)=\frac{1}{2}A_0+\sum_{n=1}^\infty-\frac{b_n}{n}\cos
nx+\frac{a_n}{n}\sin nx$

Putting $x=\pi$ gives
$\ds\frac{1}{2}(a_0)=\sum_{n=1}^\infty\frac{b_n}{n}(-1)^n$

Therefore $\ds F(x)=\sum_{n=1}^\infty\frac{a_n\sin
nx+b_n((-1)^n-\cos nx)}{n}$

$\ds\hspace{0.7in}=\sum_{n=1}^\infty\int_{-\pi}^x\{a_n\cos
nt+b_n\sin nt\}dt$

${}$

Differentiation of a Fourier Series

This is not always valid.

e.g. $\ds\sum_{n=1}^\infty\frac{\sin nx}{n}=\frac{1}{2}(\pi-x)
\hspace{0.5in} 0\leq x\leq2\pi$

$\ds\sum_{n=1}^\infty\frac{d}{dx}\frac{\sin nx}{n}=
\sum_{n=1}^\infty\cos nx \hspace{0.5in}$ which does not converge.

${}$

Sufficient Conditions

If $f(x)$ is continuous and $f'(x)$ exists except at a finite
number of points, and both $f(x)$ and $f'(x)$ have Fourier series
which converge, then the series for $f'(x)$ is obtained by term by
term differentiation of the Fourier series for $f(x)$.

i.e. $\ds f(x)=\frac{1}{2}a_0+\sum_{n=1}^\infty a_n\cos nx+b_n\sin
nx$

$\ds\Rightarrow\frac{1}{2}[f'(x+0)+f'(x-0)]=\sum_{n=1}^\infty
nb_n\cos nx-na_n\sin nx$

[This is really just the same as the result for integration, with
slightly weaker conditions.]

${}$

Half-Range Series

Let $f(x)$ be bounded and integrable in $[0,\pi]$

\begin{itemize}
\item[(1)]
Cosine Series

define $f_c(x)=\left\{\begin{array}{lr} f(x)&0\leq x\leq\pi\\
f(-x)&-\pi\leq x\leq0\end{array}\right.$

Then $f_c(x)$ is an even function, which has a Fourier series in
which $b_n\equiv0$

$\ds a_n=\frac{1}{\pi}\int_{-\pi}^\pi f_c(x)\cos nxdx$

$\ds\hspace{0.3in}=\frac{2}{\pi}\int_0^\pi f_c(x)\cos
nxdx=\frac{2}{\pi}\int_0^\pi f(x)\cos nxdx$

\item[(2)]
Sine Series

define $f_s(x)=\left\{\begin{array}{lr} f(x)&0<x<\pi\\
-f(-x)&-\pi<x<0\end{array}\right.$

If $f(0)\not=0 \,\,\,\, f_s$ is discontinuous at 0.

If $f(\pi)\not=0 \,\,\,\, f_s$ is discontinuous at $\pi$.

Then $f_s(x)$ is an odd function, and has a Fourier series in
which $a_n\equiv0$

$\ds b_n=\frac{1}{\pi}\int_{-\pi}^\pi f_s(x)\sin nxdx$

$\ds\hspace{0.3in}=\frac{2}{\pi}\int_0^\pi f_s(x)\sin
nxdx=\frac{2}{\pi}\int_0^\pi f(x)\sin nxdx$
\end{itemize}

${}$

Order of magnitude of Fourier coefficients

$\ds a_n-ib_n=\frac{1}{\pi}\int_{-\pi}^\pi f(x)e^{-inx}dx=c_n
\hfill$ (1)

Suppose $f(x)$ and all its derivatives are bounded and continuous
in

$(-\pi,\alpha_1), (\alpha_1,\alpha_2)\cdots(\alpha_k,\pi)$

Write $\ds c_n^m=\frac{1}{\pi}\int_{-\pi}^\pi
f^{(m)}(x)e^{-inx}dx$

$\ds\hspace{0.2in}=\frac{1}{\pi}\sum_0^k\int_{-\alpha_r}^{\alpha_{r+1}}
f^{(m)}(x)e^{-inx}dx \hfill$ (2)

Integrating (1) by parts gives

$\ds\pi c_n=\pi c_n^0=\sum_{r=0}^k
\left[-\frac{f(x)}{in}e^{-inx}\right]_{\alpha_r}^{\alpha_{r+1}}+
\frac{1}{in}\int_{\alpha_r}^{\alpha_{r+1}}f'(x)e^{inx}dx$

$\ds\hspace{0.3in}=\frac{1}{in}\left[\sum_{r=0}^k
f(\alpha_r+0)e^{-in\alpha r}-f(\alpha_{r+1}-0)e^{-in\alpha_{r+1}}+
\int_{\alpha_r}^{\alpha_{r+1}}f'(x)e^{-inx}dx\right]$

$\ds\hspace{0.3in}=
\frac{1}{in}\left[\frac{}{}f(-\pi+0)e^{-in\pi}-f(\pi-0)e^{-in\pi}\right.$

$\ds\hspace{0.6in}+
\left.\sum_{r=1}^k\{f(\alpha_r+0)-f(\alpha_r-0)\}e^{-in\alpha_r}+
\pi c_n^{(1)}\right]$

$f(-\pi+0)=f(\pi+0)$ by periodicity

Therefore $f(-\pi+0)e^{in\pi}-f(\pi-0)e^{-in\pi}$

$\hspace{0.5in}=[f(\pi+0)-f(\pi-0)]e^{-in\pi}=
[f(\alpha_{k+1}+0)-f(\alpha_{k+1}-0)]e^{in\alpha_{k+1}}$

Hence we have

$\ds\pi c_n^{(0)}=\frac{\pi c_n^{(1)}}{ni}+
\frac{1}{ni}\sum_{r=1}^{k+1}\{f(\alpha_r+0)-f(\alpha_r-0)\}
e^{-in\alpha_r}$

Write $\ds J_n^{(m)}=\frac{1}{\pi}\sum_{r=1}^{k+1}
\{f^{(m)}(\alpha_r+0)-f^{(m)}(\alpha_r-0)\} e^{-in\alpha_r}$

Therefore $\ds
c_n^{(0)}=\frac{c_n^{(1)}}{ni}+\frac{J_n^{(0)}}{ni}$

Similarly $\ds
c_n^{(1)}=\frac{c_n^{(2)}}{ni}+\frac{J_n^{(1)}}{ni}$

Therefore $\ds
c_n^{(0)}=\frac{J_n^{(0)}}{ni}+\frac{J_n^{(1)}}{(ni)^2}+
\frac{c_n^{(2)}}{(ni)^2}$

Since $\ds c_n^{(2)}=\frac{1}{\pi}\int_{-\pi}^\pi
f''(x)e^{-inx}dx$ is bounded,

if $J_n^{(0)}=0$ for all $n\geq m$ then $c_n=c_n^{(0)}$ is
$O\left(\frac{1}{n^2}\right)$ as $n\to\infty$

If also $J_n^{(1)}=0$ for all $n\geq m$ then $c_n=c_n^{(0)}$ is
$O\left(\frac{1}{n^3}\right)$ as $n\to\infty$

In particular if $f,f^{(1)},\cdots f^{(r)}$ are continuous but
$f^{(r+1)}$ is not continuous then
$c_n=O\left(\frac{1}{n^{r+2}}\right)$ as $n\to\infty$

In fact $J_n^{(0)}$ vanishes only if $f$ is continuous for if we
write

$f(\alpha_r+0)-f(\alpha_r-0)=j_r$ then $\ds\pi
J_n^{(0)}=\sum_{r=1}^{k+1}j_re^{-in\alpha_r}$.

If $J_n^{(0)}=0$ for $n\geq m$ then

$\ds\sum_{r=1}^{k+1}j_re^{-in\alpha_r}=0 \hspace{0.5in}
n=m,m+1,\cdots$

Taking $n=m,m+1,\cdots,m+k,$ we write $e^{-i\alpha_r}=z_r$

Therefore $\left(\begin{array}{cccc}
z_1^m&z_2^m&\cdots&z_{k+1}^m\\ \vdots\\
z_1^{m+k}&z_2^{m+k}&\cdots&z_{k+1}^{m+k}\end{array}\right)
\left(\begin{array}{c} j_1\\ \vdots\\ j_{k+1}\end{array}\right)=0$

The determinant of the matrix is

$\ds(z_1z_2\cdots z_{k+1})^m \left|\begin{array}{ccc} 1&\cdots&1\\
z_1&\cdots&z_{k+1}\\ \vdots\\
z_1^k&\cdots&z_{k+1}^k\end{array}\right|=(z_1z_2\cdots
z_{k+1})^m\prod_{r>s}(z_r-z_s)$

$z_r-z_s\not=0$ for $r\not=s$

Therefore the determinant is non zero.

Therefore $j_1=j_2=\cdots=j_{k+1}=0$

Therefore $f$ is continuous.

${}$

Parseval's Theorem

If $f(x)$ is bounded and integrable in $(-\pi,\pi)$

then $\ds\frac{1}{\pi}\int_{-\pi}^\pi(f(x))^2dx=
\frac{1}{2}a_0^2+\sum_{n=1}^\infty(a_n^2+b_n^2)$.

[Note that this is true even though $f(x)$ does not equal the sum
of its Fourier series.]

If we assume that $f(x)$ is continuous and the Fourier series
converges to $f(x)$,

$\ds\frac{1}{n}\int_{-\pi}^\pi f^2(x)=\frac{1}{\pi}\int_{-\pi}^\pi
f(x)\left[\frac{1}{2}a_0+\sum_{n=1}^\infty(a_n\cos nx+b_n\sin
nx)\right]dx$

Since uniformity of convergence is not affected by multiplying by
$f(x)$ we can integrate term by term

RHS=$\ds\frac{1}{2}a_0\frac{1}{\pi}\int_{-\pi}^\pi f(x)dx+
\frac{1}{\pi}\sum_{n=1}^\infty a_n\int_{-\pi}^\pi f(x)\cos nxdx+
b_n\int_{-\pi}^\pi f(x)\sin nxdx$

$\ds\hspace{0.5in}=\frac{1}{2}a_0^2+\sum_{n=1}^\infty a_n^2+b_n^2$

\end{document}