Deriving fourier series using complex numbers - introduction

Question

So this is the follow up thread to the one I asked before but you don't need to read the other one for this to make sense. If you want to, read PZZ's answer: link to the thread.

So I know that there exist a basis in $L^2$ which is a set of functions in the form $e^{inx}$. It turns out that this is an orthonormal basis. Also given any $f \in L^2$, there exists a sequence of complex numbers $(c_n)$ such that $$ f = \sum_{n \in \mathbb{Z}}{c_ne^{inz}}$$ It turns out these sequences lie in the vector space $l^2$.

What I am confused about is that in my lecture notes, I have the following derivation. Suppose $ f = \sum_{n \in \mathbb{Z}}{c_ne^{inz}}$ is the fourier series of $f :[-\pi,\pi] \rightarrow \mathbb{C}$ Then by definition: $$ \begin{align} c_n &= \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx} dx \\ c_{-n} &= \overline{c_n} = \frac{1}{2\pi} \int_{-\pi}^{\pi}f(x)e^{inx} dx \end{align} $$

And so we can rewrite the fourier series as following: $$\sum_{n=\infty}^{\infty}c_n e^{int} = c_0 + \sum_{n=1}^{\infty}\left( c_ne^{int}+c_{-n}e^{-int} \right)$$

And this is where my problem is. If someeone could explain how the one single sum series is equal to a constant + sum of series with starting n = 1, it would be appreciated. I am basically looking for a "summary" of what fourier series are. I've googled countless pdfs and lecture but they all start of with the definition that $$f(t) = c_n + \sum a_n \cos{nt} + b_n \sin{nt} $$. I can derive that using my above definition (sub $c_n = a_n + ib_n$) but I'd like to know where its coming from and what fourier series actually are.

Answer 1

I'm not quite sure what you're asking...but...first, in the world of complex variables trigonometric functions and exponential functions all unify. Thus by moving to complexes we have a nice uniform way of dealing with sine and cosine.

If $c_n=a_n+ib_n$, then $$ c_ne^{int}+c_{-n}e^{-int} = c_ne^{int}+\bar{c_{n}}e^{-int} =$$ $$ (a_n+ib_n)(\cos(nt)+i\sin(nt))+(a_n-ib_n)(\cos(nt)-i\sin(nt)) = 2a_n\cos(nt)-2b_n\sin(nt) $$

So dealing with trig functions or exponentials is just a matter of notation/taste.

Now to address your question as to what these series are for...

Recall that an analytic function is equal to its Taylor series:

$$ f(x) = \sum\limits_{k=0}^\infty \frac{f^{(k)}(a)}{k!}(x-a)^k $$

If we stop this sum, say at $N$, we have

$$ f(x) \approx \sum\limits_{k=0}^N \frac{f^{(k)}(a)}{k!}(x-a)^k $$

which is a polynomial approximation of $f(x)$. Polynomials are nice. Taylor polynomials give us easily understood approximations of our function.

Now Fourier series do essentially the same thing for periodic functions. If we stop the summation at some $N$, we get a Fourier polynomial. This is a nice approximation made up of "easy" to understand trig functions.

From a theoretic viewpoint, Taylor series are wonderful because you can treat analytic functions sort of like polynomials. Fourier series allow one to treat nice periodic functions sort of like trig functions.

Answer 2

Let us start with the Fourier series of a periodic function with period T: $$a{\left[k\right]}=\frac{2}{T}\int_{T}x_T{\left(t\right)}\cos{\left(k\frac{2\pi}{T}t\right)}, a{\left[0\right]}=1,$$ $$b{\left[k\right]}=\frac{2}{T}\int_{T}x_T{\left(t\right)}\sin{\left(n\frac{2\pi}{T}t\right)}, \ b{\left[0\right]}=0, $$ $$ x_T{\left(t\right)}=\ \sum_{k=0}^{\infty}a{\left[k\right]}\cos{\left(n\frac{2\pi}{T}t\right)}+b{\left[k\right]}\sin{\left(n\frac{2\pi}{T}t\right)} $$ We set $\omega_0=\frac{2\pi}{T}$

$$a{\left[k\right]}=\frac{2}{T}\int_{T}x_T{\left(t\right)}\cos{\left(k\omega_0t\right)}, a{\left[0\right]}=1,$$ $$ b{\left[k\right]}=\frac{2}{T}\int_{T}x_T{\left(t\right)}\sin{\left(k\omega_0t\right))},$$ $$ b{\left[0\right]}=0, x_T{\left(t\right)}=\ \sum_{k=0}^{\infty}a{\left[k\right]}\cos{\left(k\omega_0t\right)} $$

We define $a{\left[k\omega_0\right]} = Ta{\left[k\right]}$

$$x_T{\left(t\right)}=\frac{1}{T}\ \sum_{k=0}^{\infty}a{\left[k\omega_0\right]}\cos{\left(k\omega_0t\right)}+b{\left[k\omega_0\right]}\sin{\left(k\omega_0t\right)}$$

Since $\omega_0=\frac{2\pi}{T}\Longleftrightarrow\frac{\omega_0}{2\pi}=\frac{1}{T}$ we can write: $$x_T{\left(t\right)}=\frac{\omega_0}{2\pi}\ \sum_{k=0}^{\infty}a{\left[k\omega_0\right]}\cos{\left(k\omega_0t\right)}+b{\left[k\omega_0\right]}\sin{\left(k\omega_0t\right)}$$

Now we stretch the period T till infinity: $T\longrightarrow\ \infty\ \Longrightarrow\ \omega_0\longrightarrow\ 0$

$$x{\left(t\right) \equiv \lim_{T\to \infty}{x_T{\left(t\right)}}}=\lim_{\omega_0\to 0}{x_T{\left(t\right)}}=\lim_{\omega_0\to 0}\frac{1}{2\pi}\ \sum_{k=0}^{\infty}\left(a{\left[k\omega_0\right]}\cos{\left(k\omega_0t\right)}+b{\left[k\omega_0\right]}\sin{\left(k\omega_0t\right)}\right)\omega_0$$

$$x\left(t\right)=\lim_{\omega_0\to 0}\frac{1}{2\pi}\ \sum_{k=0}^{\infty}\left(a{\left[k\omega_0\right]}\cos{\left(k\omega_0t\right)}+b{\left[k\omega_0\right]}\sin{\left(k\omega_0t\right)}\right)\omega_0$$

Since we move between frequencies using the step $\omega_0$,
we can consider $\omega_0$ being a frequency interval: $\omega_0\equiv\Delta\omega$

We use the following property: $$\lim_{\Delta\omega\to 0}{\sum_{k=0}^{\infty}f{\left(k\Delta\omega\right)}}\Delta\omega=\int_{0}^{\infty}f\left(\omega\right){d\omega)}$$ This leads to:

$$x{\left(t\right)=\ \lim_{T\to\infty}{x_T{\left(t\right)}}}=\lim_{\omega_0\to0}{x_T{\left(t\right)}}=\frac{1}{2\pi}\int_{0}^{\infty}\left(a{\left(\omega\right)}\cos{\left(\omega t\right)}+b{\left(\omega\right)}\sin{\left(\omega t\right)}\right)\ d\omega$$ We now do the same trick for the coefficients: $$a{\left(\omega\right)}=\lim_{T\to\infty}a{\left[k\omega_0\right]}=\lim_{T\to\infty}Ta{\left[k\right]}=\lim_{T\to\infty}2\int_{T}x_T{\left(t\right)}\cos{\left(k\omega_0t\right))}$$

$$a{\left(\omega\right)}=\lim_{T\to\infty}2\int_{T}x_T{\left(t\right)}\cos{\left(k\omega_0t\right)}=2\int_{0}^{\infty}x{\left(t\right)}\cos{\left(\omega t\right)}$$

We finally arrive at the 'real Fourier transform': $$a{\left(\omega\right)}=2\int_{0}^{\infty}x{\left(t\right)}\cos{\left(\omega t\right)} \ and \ b{\left(\omega\right)}=2\int_{0}^{\infty}x{\left(t\right)}\sin{\left(\omega t\right)}$$ The 'inverse real transform' is: $$x{\left(t\right)}=\frac{1}{2\pi}\int_{0}^{\infty}\left(a{\left(\omega\right)}\cos{\left(\omega t\right)}+b{\left(\omega\right)}\sin{\left(\omega t\right)}\right)\ d\omega$$

We can also express the inverse transform as a decomposition of the function in phase-shifted cosines: $$x{\left(t\right)}=\frac{1}{2\pi}\int_{0}^{\infty}{c{\left(\omega\right)}\cos{\left(\omega t+\varphi\left(\omega\right)\right)}}\ d\omega$$ with $c{\left(\omega\right)}=\sqrt{{a{\left(\omega\right)}}^2+{b{\left(\omega\right)}}^2},\ \varphi\left(\omega\right)=atan2{\left(b{\left(\omega\right)},a{\left(\omega\right)}\right)}$

Now we jump to complex numbers by using the complex exponential notation for cosine: $$x{\left(t\right)}=\frac{1}{2\pi}\int_{0}^{\infty}{c{\left(\omega\right)}\frac{\left(e^{j\left(\omega t+\varphi\left(\omega\right)\right)}+e^{-j\left(\omega t+\varphi\left(\omega\right)\right)}\right)}{2}}\ d\omega$$

$$x{\left(t\right)}=\frac{1}{2\pi}\int_{-\infty}^{\infty}{c{\left(\omega\right)}\frac{\left(e^{j\left(\omega t+\varphi\left(\omega\right)\right)}\right)}{2}}\ d\omega$$

$$x{\left(t\right)}=\frac{1}{2\pi}\int_{-\infty}^{\infty}{c{\left(\omega\right)e^{j\varphi\left(\omega\right)}}\ \frac{e^{j\omega t}}{2}}\ d\omega$$

$$x{\left(t\right)}=\frac{1}{2\pi}\int_{-\infty}^{\infty}{\left(a{\left(\omega\right)}+j\ b{\left(\omega\right)}\ \right)\ \frac{e^{j\omega t}}{2}}\ d\omega$$

$c\left(\omega\right)e^{j\varphi\left(\omega\right)}$ or $a\left(\omega\right)+j\ b\left(\omega\right)$ is the Fourier transform of $x\left(t\right)$.