MATH 414 Lecture 8

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Sine and Cosine Series

A regular Fourier series expresses a function in terms of both sines and cosines. For example, on the interval $[-\pi ,\pi ]$,

${\begin{aligned}f(x)&=a_{0}+\sum _{k=1}^{\infty }a_{k}\,\cos {(k\,x)}+b_{k}\sin {(k\,x)}\\a_{0}&={\frac {1}{2\pi }}\,\int _{-\pi }^{\pi }f(x)\,\mathrm {d} x\\a_{n}&={\frac {1}{\pi }}\,\int _{-\pi }^{\pi }f(x)\,\cos {(n\,x)}\,\mathrm {d} x\\b_{n}&={\frac {1}{\pi }}\,\int _{-\pi }^{\pi }f(x)\,\sin {(n\,x)}\,\mathrm {d} x\end{aligned}}$

Sine Series

The Fourier Sine Series of a function $f(x)$ on an interval $\left[0,\pi \right]$ is a fourier series that produces an odd extension of the function $f(x)$ (i.e. for $-\pi <x<0$, we have $-f(-x)$.)

${\begin{aligned}f(x)=\sum _{k=1}^{\infty }b_{k}\,\sin {(k\,x)}b_{n}={\frac {2}{\pi }}\int _{0}^{\pi }f(x)\,\sin {(n\,x)}\,\mathrm {d} x\end{aligned}}$

Cosine Series

The Fourier Cosine Series of a function $f(x)$ on an interval $\left[0,\pi \right]$ is a fourier series that produces an even extension of the function $f(x)$ (i.e. for $-\pi <x<0$, we have $f(-x)$.

${\begin{aligned}f(x)&=a_{0}+\sum _{k=1}^{\infty }a_{k}\,\cos {(k\,x)}\\a_{0}={\frac {1}{\pi }}\,\int _{0}^{\pi }f(x)\,\mathrm {d} x\\a_{n}={\frac {2}{\pi }}\,\int _{0}^{\pi }f(x)\,\cos {(n\,x)}\,\mathrm {d} x\end{aligned}}\,\!$

Complex Fourier Series

Recall

${\begin{aligned}\mathrm {e} ^{i\,n\,x}&=\cos {(n\,x)}+i\,\sin {(n\,x)}\\\mathrm {e} ^{-i\,n\,x}&=\cos {(n\,x)}-i\,\sin {(n\,x)}\end{aligned}}$

Let's add and divide by 2:

${\begin{aligned}\cos {(n\,x)}&={\frac {1}{2}}\left(\mathrm {e} ^{i\,n\,x}+\mathrm {e} ^{-i\,n\,x}\right)\\\sin {(n\,x)}&={\frac {1}{2i}}\left(\mathrm {e} ^{i\,n\,x}+\mathrm {e} ^{-i\,n\,x}\right)\end{aligned}}$

Substituting this into the fourier series definition and simplifying gives

$f(x)=a_{0}+\sum _{k=1}^{\infty }\left({\frac {a_{k}-i\,b_{k}}{2}}\right)\mathrm {e} ^{i\,n\,x}+\sum _{k=1}^{\infty }\left({\frac {a_{k}+i\,b_{n}}{2}}\,\mathrm {e} ^{-i\,n\,x}\right)\,\!$

We'll rewrite this formula a different way. Let $c_{0}:=a_{0}$, $c_{k}:={\frac {a_{k}-i\,b_{k}}{2}}$. If we change the index on our sum by negating, we get

$f(x)=\sum _{n=0}^{\infty }c_{n}\,\mathrm {e} ^{i\,n\,x}+\sum _{n=-\infty }^{-}1\left({\frac {a_{-n}+i\,b_{-n}}{2}}\right)\mathrm {e} ^{i\,n\,x}$

Observe that $\left({\frac {a_{-n}+i\,b_{-n}}{2}}\right)$ is the same as $c_{n}$, so we're left with

${\begin{aligned}f(x)&=\sum _{n=-\infty }^{\infty }c_{n}\,\mathrm {e} ^{i\,n\,x}\\c_{0}&={\frac {1}{2\pi }}\,\int _{-\pi }^{\pi }f(x)\,\mathrm {e} ^{-0x}\,\mathrm {d} x={\frac {1}{2\,\pi }}\,\int _{-\pi }^{\pi }f(x)\,\mathrm {d} x\\c_{n}&={\frac {1}{2\pi }}\,\int {-\pi }^{\pi }f(x)\,\mathrm {e} ^{-i\,n\,x}\,\mathrm {d} x\end{aligned}}$