4.5. Other Fourier series

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4.5. Other Fourier series

  1. Fourier series for even and odd functions
  2. $\cos$-Fourier series
  3. $\sin$-Fourier series
  4. $\sin$-Fourier series with half-integers
  5. Fourier series in complex form
  6. Miscellaneous

Fourier series for even and odd functions

In the previous Section 4.4 we proved the completeness of the system of functions \begin{equation} \Bigl\{\frac{1}{2},\qquad \cos (\frac{\pi nx}{l}), \qquad \sin (\frac{\pi nx}{l}) \quad n=1,\ldots\Bigr\} \label{eq-4.5.1} \end{equation} on interval $J:=[x_0,x_1]$ with $(x_1-x_0)=2l$. In other words, we proved that any function $f(x)$ on this interval could be decomposed into Fourier series \begin{equation} f(x)= \frac{1}{2}a_0 + \sum_{n=1}^\infty \bigl( a_n\cos (\frac{\pi nx}{l})+b_n \sin (\frac{\pi nx}{l}) \bigr) \label{eq-4.5.2} \end{equation} with coefficients calculated according to (4.3.7) \begin{align} a_n=& \frac{1}{l}\int_J f(x)\cos (\frac{\pi nx}{l})\,dx \qquad n=0,1,2,\ldots, \label{eq-4.5.3}\\ b_n= & \frac{1}{l}\int_J f(x)\sin (\frac{\pi nx}{l})\,dx \qquad n=1,2,\ldots, \label{eq-4.5.4} \end{align} and satisfying Parseval's equality \begin{equation} \frac{l}{2}|a_0|^2 + \sum_{n=1}^\infty l\bigl( |a_n|^2+|b_n |^2 \bigr)=\int_J |f(x)|^2\,dx. \label{eq-4.5.5} \end{equation}

Now we consider some other orthogonal systems and prove their completeness. To do this we first prove

Lemma 1. Let $J$ be a symmetric interval: $J=[-l,l]$. Then

  1. $f(x)$ is even iff $b_n=0$     $\forall n=1,2,\ldots$.
  2. $f(x)$ is odd iff $a_n=0$     $\forall n=0,1,2,\ldots$.

Proof. (1) Note that $\cos (\frac{\pi nx}{l})$ are even functions and $\sin (\frac{\pi nx}{l})$ are odd functions. Therefore if $b_n=0$     $\forall n=1,2,\ldots$ then decomposition (\ref{eq-4.5.2}) contains only even functions and $f(x)$ is even. Conversely, if $f(x)$ is an even function then integrand in (\ref{eq-4.5.4}) is an odd function and its integral over symmetric interval is $0$.

(2) Statement [2] is proven in the similar way.

$\cos$-Fourier series

Let us consider function $f(x)$ on the interval $[0,l]$. Let us extend it as an even function on $[-l,l]$ so $f(x):=f(-x)$ for $x\in [-l,0]$ and decompose it into full Fourier series (\ref{eq-4.5.2}); however, $\sin$-terms disappear and we arrive to decomposition \begin{equation} f(x)= \frac{1}{2}a_0 + \sum_{n=1}^\infty a_n\cos (\frac{\pi nx}{l}). \label{eq-4.5.6} \end{equation} This is decomposition with respect to orthogonal system \begin{equation} \Bigl\{\frac{1}{2},\qquad \cos (\frac{\pi nx}{l}) \quad n=1,\ldots\Bigr\}. \label{eq-4.5.7} \end{equation}

Its coefficients are calculated according to (\ref{eq-4.5.3}) but here integrands are even functions and we can take interval $[0,l]$ instead of $[-l,l]$ and double integrals: \begin{equation} a_n= \frac{2}{l}\int_0^l f(x)\cos (\frac{\pi nx}{l})\,dx \qquad n=0,1,2,\ldots \label{eq-4.5.8} \end{equation} Also Parseval's equality (\ref{eq-4.5.5}) becomes \begin{equation} \frac{l}{4}|a_0|^2 +\sum_{n=1}^\infty \frac{l}{2} |a_n|^2= \int_0^l |f(x)|^2\,dx. \label{eq-4.5.9} \end{equation}

The sum of this Fourier series is $2l$-periodic. Note that even and then periodic continuation does not introduce new jumps.

image

$\sin$-Fourier series

Let us consider function $f(x)$ on the interval $[0,l]$. Let us extend it as an odd function on $[-l,l]$ so $f(x):=-f(-x)$ for $x\in [-l,0]$ and decompose it into full Fourier series (\ref{eq-4.5.2}); however, $\cos$-terms disappear and we arrive to decomposition \begin{equation} f(x)= \sum_{n=1}^\infty b_n\sin (\frac{\pi nx}{l}). \label{eq-4.5.10} \end{equation} This is decomposition with respect to orthogonal system \begin{equation} \Bigl\{ \sin (\frac{\pi nx}{l}) \quad n=1,\ldots\Bigr\}. \label{eq-4.5.11} \end{equation}

Its coefficients are calculated according to (\ref{eq-4.5.4}) but here integrands are even functions and we can take interval $[0,l]$ instead of $[-l,l]$ and double integrals: \begin{equation} b_n= \frac{2}{l}\int_0^l f(x)\sin (\frac{\pi nx}{l})\,dx \qquad n=1,2,\ldots \label{eq-4.5.12} \end{equation} Also Parseval's equality (\ref{eq-4.5.5}) becomes \begin{equation} \sum_{n=1}^\infty \frac{l}{2} |b_n|^2= \int_0^l |f(x)|^2\,dx. \label{eq-4.5.13} \end{equation}

The sum of this Fourier series is $2l$-periodic. Note that odd and then periodic continuation does not introduce new jumps iff $f(0)=f(l)=0$.

image

General case

image

$f(0)=f(l)=0$

$\sin$-Fourier series with half-integers

Let us consider function $f(x)$ on the interval $[0,l]$. Let us extend it as an even with respect to $x=l$ function on $[0,2l]$ so $f(x):=f(2l-x)$ for $x\in [l,2l]$; then we make an odd continuation to $[-2l,2l]$ and decompose it into full Fourier series (\ref{eq-4.5.2}) but with $l$ replaced by $2l$; however, $\cos$-terms disappear and we arrive to decomposition \begin{equation*} f(x)= \sum_{n=1}^\infty b'_n\sin (\frac{\pi nx}{2l}). \end{equation*}

Then $f(2l-x)= \sum_{n=1}^\infty b'_n\sin (\frac{\pi nx}{2l})(-1)^{n+1}$ and since $f(x)=f(2l-x)$ due to original even continuation we conclude that $b'_n=0$ as $n=2m$ and we arrive to \begin{equation} f(x)= \sum_{n=0}^\infty b_n\sin (\frac{\pi(2n+1)x}{2l}) \label{eq-4.5.14} \end{equation} with $b_n:=b'_{2n+1}$ where we replaced $m$ by $n$.

This is decomposition with respect to orthogonal system \begin{equation} \Bigl\{ \sin (\frac{\pi (2n+1)x}{2l}) \quad n=1,\ldots\Bigr\}. \label{eq-4.5.15} \end{equation}

Its coefficients are calculated according to (\ref{eq-4.5.12}) (with $l$ replaced by $2l$) but here we can take interval $[0,l]$ instead of $[0,2l]$ and double integrals: \begin{equation} b_n= \frac{2}{l}\int_0^l f(x)\sin (\frac{\pi (2n+1)x}{2l})\,dx \qquad n=0,2,\ldots \label{eq-4.5.16} \end{equation} Also Parseval's equality (\ref{eq-4.5.13}) becomes \begin{equation} \sum_{n=0}^\infty \frac{l}{2} |b_n|^2= \int_0^l |f(x)|^2\,dx. \label{eq-4.5.17} \end{equation}

The sum of this Fourier series is $4l$-periodic. Note that odd and then periodic continuation does not introduce new jumps iff $f(0)=0$.

image

General case

image

$f(0)=0$

Fourier series in complex form

Consider (\ref{eq-4.5.2})--(\ref{eq-4.5.5}). Plugging \begin{align*} &\cos(\frac{\pi n x}{l})= \frac{1}{2}e^{\frac{i\pi n x}{l}}+\frac{1}{2}e^{-\frac{i\pi n x}{l}}\\ &\sin(\frac{\pi n x}{l})= \frac{1}{2i}e^{\frac{i\pi n x}{l}}-\frac{1}{2i}e^{-\frac{i\pi n x}{l}} \end{align*} and separating terms with $n$ and $-n$ and replacing in the latter $-n$ by $n=-1,-2,\ldots$ we get \begin{equation} f(x)= \sum_{n=-\infty}^\infty c_n e^{\frac{i\pi nx}{l}} \label{eq-4.5.18} \end{equation} with $c_0=\frac{1}{2}a_0$, $c_n = \frac{1}{2}(a_n -i b_n)$ as $n=1,2,\ldots$ and $c_n = \frac{1}{2}(a_{-n} +i b_{-n})$ as $n=-1,-2,\ldots$ which could be written as \begin{equation} c_n= \frac{1}{2l}\int_J f(x)e^{-\frac{i\pi n x}{l}}\,dx \qquad n=\ldots,-2, -1, 0,1,2,\ldots. \label{eq-4.5.19} \end{equation} Parseval's equality (\ref{eq-4.5.5}) becomes \begin{equation} 2l\sum_{n=-\infty}^\infty |c_n|^2= \int_J |f(x)|^2\,dx. \label{eq-4.5.20} \end{equation} One can see easily that the system \begin{equation} \Bigl\{X_n:=e^{\frac{i\pi nx}{l}} \quad \ldots,-2, -1, 0,1,2,\ldots\Bigr\} \label{eq-4.5.21} \end{equation} on interval $J:=[x_0,x_1]$ with $(x_1-x_0)=2l$ is orthogonal: \begin{equation} \int_J X_n(x)\bar{X_m(x)}\,dx = 2l\delta_{mn}. \label{eq-4.5.22} \end{equation}

Remark 1. All our formulae are due to (Section 4.3) but we need the completeness of the systems and those are due to compleness of the system (\ref{eq-4.5.1}) established in Section 4.4.

Remark 2. Recall that with periodic boundary conditions all eigenvalues $(\frac{\pi n }{l})^2$ are of multiplicity $2$ i.e. the corresponding eigenspace (consisting of all eigenfunctions with the given eigenvalue) has dimension $2$ and $\{\cos (\frac{\pi n x}{l}), \sin (\frac{\pi n x}{l})\}$ and $\{e^{\frac{i\pi n x}{l}}, e^{-\frac{\pi n x}{l}}\}$ are just two different orthogonal basises in this eigenspace.

Remark 3. Fourier series in the complex form are from $-\infty$ to $\infty$ and this means that both sums $\sum_{n=0}^{\pm \infty} c_n e^{\frac{i\pi nx}{l}}$ must converge which is a stronger requirement than convergence of Fourier series in the trigonometric form. For piecewise differentiable function $f$ Fourier series in the complex form converges at points where $f$ is continuous but not at jump points where such series converges only in the sense of principal value: \begin{equation} \lim_{N\to +\infty} \sum_{n=-N}^{n=N} c_n e^{\frac{i\pi nx}{l}}= \frac{1}{2}\bigl( f(x^+) + f(x^-)\bigr). \label{eq-4.5.23} \end{equation}

Miscellaneous

We consider in appendices

  1. Multidimensional Fourier series
  2. Harmonic oscillator and Hermite functions

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