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:

\[

\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

\[

\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

\[

\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

\[

\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.

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