4.5. Other Fourier series


4.5. Other Fourier series

Fourier series for even and odd functions

In the previous Section 4.4 we proved the completeness of the system of functions $$\Bigl\{\frac{1}{2},\qquad \cos (\frac{\pi nx}{l}), \qquad \sin (\frac{\pi nx}{l}) \quad n=1,\ldots\Bigr\} \label{eq-4.5.1}$$ 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 $$f(x)= \frac{1}{2}a_0 + \sum_{n=1}^\infty \bigl( a_n\cos (\frac{\pi nx}{l})+b_n \sin (\frac{\pi nx}{l}) \bigr) \label{eq-4.5.2}$$ 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 $$\frac{l}{2}|a_0|^2 + \sum_{n=1}^\infty l\bigl( |a_n|^2+|b_n |^2 \bigr)=\int_J |f(x)|^2\,dx. \label{eq-4.5.5}$$

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 $$f(x)= \frac{1}{2}a_0 + \sum_{n=1}^\infty a_n\cos (\frac{\pi nx}{l}). \label{eq-4.5.6}$$ This is decomposition with respect to orthogonal system $$\Bigl\{\frac{1}{2},\qquad \cos (\frac{\pi nx}{l}) \quad n=1,\ldots\Bigr\}. \label{eq-4.5.7}$$

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: $$a_n= \frac{2}{l}\int_0^l f(x)\cos (\frac{\pi nx}{l})\,dx \qquad n=0,1,2,\ldots \label{eq-4.5.8}$$ Also Parseval's equality (\ref{eq-4.5.5}) becomes $$\frac{l}{4}|a_0|^2 +\sum_{n=1}^\infty \frac{l}{2} |a_n|^2= \int_0^l |f(x)|^2\,dx. \label{eq-4.5.9}$$

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

$\sin$-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 $$f(x)= \sum_{n=1}^\infty b_n\sin (\frac{\pi nx}{l}). \label{eq-4.5.10}$$ This is decomposition with respect to orthogonal system $$\Bigl\{ \sin (\frac{\pi nx}{l}) \quad n=1,\ldots\Bigr\}. \label{eq-4.5.11}$$

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: $$b_n= \frac{2}{l}\int_0^l f(x)\sin (\frac{\pi nx}{l})\,dx \qquad n=1,2,\ldots \label{eq-4.5.12}$$ Also Parseval's equality (\ref{eq-4.5.5}) becomes $$\sum_{n=1}^\infty \frac{l}{2} |b_n|^2= \int_0^l |f(x)|^2\,dx. \label{eq-4.5.13}$$

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

General case

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

$\sin$-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 $$f(x)= \sum_{n=0}^\infty b_n\sin (\frac{\pi(2n+1)x}{2l}) \label{eq-4.5.14}$$ with $b_n:=b'_{2n+1}$ where we replaced $m$ by $n$.

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

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: $$b_n= \frac{2}{l}\int_0^l f(x)\sin (\frac{\pi (2n+1)x}{2l})\,dx \qquad n=0,2,\ldots \label{eq-4.5.16}$$ Also Parseval's equality (\ref{eq-4.5.13}) becomes $$\sum_{n=0}^\infty \frac{l}{2} |b_n|^2= \int_0^l |f(x)|^2\,dx. \label{eq-4.5.17}$$

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

General case

$f(0)=0$

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 $$f(x)= \sum_{n=-\infty}^\infty c_n e^{\frac{i\pi nx}{l}} \label{eq-4.5.18}$$ 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 $$c_n= \frac{1}{2l}\int_J f(x)e^{-\frac{i\pi n x}{l}}\,dx \qquad n=\ldots,-2, -1, 0,1,2,\ldots. \label{eq-4.5.19}$$ Parseval's equality (\ref{eq-4.5.5}) becomes $$2l\sum_{n=-\infty}^\infty |c_n|^2= \int_J |f(x)|^2\,dx. \label{eq-4.5.20}$$ One can see easily that the system $$\Bigl\{X_n:=e^{\frac{i\pi nx}{l}} \quad \ldots,-2, -1, 0,1,2,\ldots\Bigr\} \label{eq-4.5.21}$$ on interval $J:=[x_0,x_1]$ with $(x_1-x_0)=2l$ is orthogonal: $$\int_J X_n(x)\bar{X}_m(x)\,dx = 2l\delta_{mn}. \label{eq-4.5.22}$$

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

Remark 2. Recall that with periodic boundary conditions all eigenvalues $(\frac{\pi n }{l})^2$, $\ne 0$ 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: $$\lim_{N\to +\infty} \sum_{n=-N}^{n=N} c_n e^{\frac{i\pi nx}{l}}= \frac{1}{2}\bigl( f(x^+) + f(x^-)\bigr). \label{eq-4.5.23}$$

Miscellaneous

We consider in appendices