# Forms of Fourier Series

///Forms of Fourier Series

## Forms of Fourier Series

The following is a summary of the relationships between the three equivalent forms of Fourier series.

## First Form: Sines Only

The first type of form we began with could be described as a “sines only” form:
$f(t)=\sum_{n=0}^{\infty} A_n\sin(2\pi nt+\phi_n).$ Here I’m including $n=0$, which gives a constant term $A_0\sin(\phi_0)$.

## Second Form: Sines and Cosines

If we use the trig identity $\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)$, we can rewrite the shifted sine term as
$\sin(2\pi nt+\phi_n) = \sin(2\pi nt)\cos(\phi_n)+\cos(2\pi nt)\sin(\phi_n).$ Using this, we can rewrite our original Fourier series in a “sines and cosines” form, as
$f(t)=\sum_{n=0}^{\infty} \left(a_n\cos(2\pi nt)+b_n\sin(2\pi nt)\right),$ where $a_n = A_n\sin(\phi_n)$ and $b_n=A_n\cos(\phi_n)$. Notice that the constant term (when $n=0$) is $a_0$. (Sometimes you’ll see this constant term instead denoted $\frac{a_0}{2}$, for reasons that will slightly simplify some formulas we’ll see in the future. For now, let’s not worry about that.)

## Third Form: Complex Exponentials

Our third form uses the famous identity $e^{i\theta}=\cos(\theta)+i\sin(\theta)$. Notice that this equation implies
\begin{align*}
e^{2\pi int}&=\cos(2\pi nt)+i\sin(2\pi nt)\\
e^{-2\pi int}&=\cos(-2\pi nt)+i\sin(-2\pi nt)=\cos(2\pi nt)-i\sin(2\pi nt).
\end{align*}
Adding these two equations and dividing by $2$ (respectively, subtracting them and dividing by $2i$) yields
\begin{align*}
\cos(2\pi nt)&=\frac{1}{2}\left(e^{2\pi int}+e^{-2\pi int}\right),\\
\sin(2\pi nt)&=\frac{1}{2i}\left(e^{2\pi int}-e^{-2\pi int}\right)= -\frac{i}{2}\left(e^{2\pi int}-e^{-2\pi int}\right).
\end{align*}
Substituting these into the “sines and cosines” form and simplifying yields
$f(t)=\sum_{n=0}^{\infty} \left(\frac{1}{2}(a_n-ib_n)e^{2\pi int}+\frac{1}{2}(a_n+ib_n)e^{-2\pi int}\right).$ We can write the above expression even more simply as
$f(t)=\sum_{n=-\infty}^{\infty} c_n e^{2\pi int},$ where $c_0=a_0$ and for $n>0$ we have
\begin{align*}
c_n &= \frac{1}{2}(a_n-ib_n)\\
c_{-n}&= \frac{1}{2}(a_n+ib_n).
\end{align*}
In particular, notice that $c_n$ and $c_{-n}$ are complex conjugates.

What is great about this final form is that it is very nearly a geometric series, with common ratio $e^{2\pi i t}$. (It actually is a geometric series when all of the constants $c_n$ are identical.) This will make certain calculations much simpler, and is why we will often prefer this form.

By |2018-04-24T16:21:20+00:00April 24th, 2018|Categories: Courses, Math 344|0 Comments

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