Is it possible to define the Fourier series for tangent?

Today I want to derive something that feels fundamentally wrong. I want to derive a Fourier series for $\tan(t)$. And I will do it by calculating the Fourier series of $\tan(t+ i a)$ and take the limit as 

$$\lim_{a \rightarrow 0} \tan(t+i a)$$

The result will be a series that converges almost nowhere, but nonetheless represents $\tan(t)$, at least when you consider Cesaro sums to assign a value to the series. 

We are going to make use of a related geometric series we defined previously in the post on neural synchronization. 

$$\sum^N_{k=0} (e^{i \theta})^k =\sum^N_{k=0}x ^k$$ with $x = e^{i \theta}$

$$\frac{1- e^{i (N+1) \theta}}{1- e^{i \theta}}$$

However, we will make two small changes. We will sum up to infinity, that is $k=\infty$ and $\theta = t - i a$

When we do that our sum can converge for $a>0$. Writing this out explicitly gives us

$$\sum^{\infty}_{k=0} (e^{it-a})^k = \frac{1}{1- e^{i t-a}}$$

Now, we can translate between this ratio, and a Fourier series. As 

$\sum^{\infty}_{k=0} (e^{it-a})^k$ is basically a complex valued Fourier series with terms $e^{ak}$ for the $k$-frequency wave.

Now lets see if we can somehow finesse tangent into an expression similar to $\frac{1}{1- e^{i t-a}}$. Recall that 

$$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}=i\frac{e^{-i\theta}-e^{i\theta}}{e^{-i\theta}+e^{i\theta}}$$  

To simplify things a bit, lets call $z=e^{-i\theta}$. That give us 

$$i \frac{z-z^{-1}}{z+z^{-1}}$$  

$$i \frac{z^2-1}{z^2+1}$$

$$i-\frac{2i}{z^2+1}$$

Now we can substitute in $z=e^{-i\theta} =e^{ it -  a}$ and we get.

$$i-\frac{2i}{e^{2(it-a)}+1}$$ .

We can now convert it to a series, we have 

$$\tan(t - i a) = i + 2i\sum^\infty_{k=1} (-1)^k e^{2 k (i t - a)}$$   

Which is a Fourier series with only positive frequency $k$. 

Now if $a \rightarrow 0$ , then 

$$\tan(t) = i + 2i \sum^\infty_{k=1} (-1)^k e^{2 i k t}$$   

However this series doesn't converge at all for any value of $t$. However, we can use Cesaro summation to assign a sign a value to this series. There is a great Mathologer video on this. His "supersum", is the Cesaro summation.

Cesaro summation basically says that if we take the average of the partial sums of a series, and that average converges, then that is the sum of the divergent series. More precisely, we have a series $a_k$

then 

$$s_k = \sum_{i=0}^k a_i$$

$$v_k = \frac{1}{n} \sum_{i=0}^k s_i$$

Lets use a concrete example. Consider the series $1-1+1-1+1-1+\cdots$ 

This its partials sums $s_k$ is

$$s_k = \{1,0,1,0,1,0,\cdots\}$$

Then if we compute the average we get

$$v_k = \{1,\frac{1}{2},\frac{2}{3},\frac{1}{2},\frac{3}{5},\frac{1}{2},\cdots\}$$

which converges to $\frac{1}{2}$.

Interestingly this sum occurs in our fourier series for tangent. When time $t = n \pi$ for any integer $n$ we have 

$$\tan(n \pi) = i + 2i\sum^\infty_{k=1} (-1)^k e^{2 k i n \pi}$$

$$i+2i\sum^\infty_{k=1} (-1)^k$$

$$i +2i(1-1+1-1\cdots)$$

We already know alternating $-1$ and $1$ is $\frac{1}{2}$

Which gives us 

$$\tan(n \pi) = i +2i(1-1+1-1\cdots)=0$$

Now, lets show that for any value of $t$ (except for $t=\frac{\pi}{2}+n \pi$ with integer n) converges. Lets restict ourselves to the term $\sum^\infty_{k=1} (-1)^k e^{2 i k t}$. 

Let us compute $v_k$. We have

$$v_k=\sum^k_{m=1} \frac{(-1)^m e^{2 i m t}}{k}$$  

Here we are only going to show it converges to some constant we will call $S$. 

We will use Dirichlet's Test for convergence. It states when a complex series converges. Specifically we have $\frac{1}{k} \rightarrow 0$ as $k\rightarrow \infty$. The second part says for fixed $t$ we have $|\sum^k_{m=1} (-1)^m e^{2 i m t}|<M$ for all $k$. This basically means that the sum can oscillate between values $M$. This clearly fails for $t=\frac{\pi}{2}+n \pi$ as $$|\sum^k_{m=1} (-1)^m e^{2 i m \frac{\pi}{2}+n}|=|\sum^k_{m=1} (-1)^m (-1)^m |= \sum^k_{m=1} 1 = \infty$$ These will be the normal places where $tan(t)$ is undefined.

However, at all other times there is indeed a maximum value. We know that $$\sum^k_{m=1} (-1)^m e^{2 i m t} = \frac{1-e^{2 i (k+1) t}}{1+ e^{2 i t}}<|\frac{1-e^{2 i (k+1) t}}{1+ e^{2 i t}}|\leq |\frac{2}{1+ e^{2 i t}}| = M$$ .

These two conditions, $\frac{1}{k}\rightarrow 0$, and that $M$ is finite, satisfy the Dirichlet's test. Thus or average partials will converge to some sum $S$ for all times that are not $t=\frac{\pi}{2}+n \pi$.

While that's a long round about way to show that our series in Cesaro summable, now we can make use of some very nice properties of Cesaro sums. Specifically, we have multiplication by a constant or

$$c \sum^\infty_{m=1} a_k =  \sum^\infty_{m=1} c a_k$$

We have shift, that is if $a_0=0$

$$\sum^\infty_{m=1} a_k = \sum^\infty_{m=0} a_k$$

With that we can now write 

$$S =  \sum^\infty_{k=1} (-1)^k e^{2 k (i t)}$$

$$-e^{-2 i t}S  =  \sum^\infty_{k=1} (-1)^{k-1} e^{2 (k-1) (i t)}$$

Using the shift property we have

$$-e^{-2 i t}S  =  1 + \sum^\infty_{k=1} (-1)^k e^{2 k (i t)}$$

Subtract the the equations give

$$(1+e^{-2 i t})S = \sum^\infty_{k=1} (-1)^k e^{2 k (i t)}-\sum^\infty_{k=1} (-1)^k e^{2 k (i t)}-1 =-1$$

$$(1+e^{-2 i t })S = -1$$

Rearranging this gives

$$S =\frac{-1}{1+e^{-2 i t }}$$

Which is a valid value for $S$ for all $t$ except $t=\frac{\pi}{2}+n \pi$. 

Now we know that 

$$\tan(t) = i + 2i S$$

Computing that we get 

$$\tan(t) = i + 2i \frac{-1}{1+e^{-2 i t }}$$

$$\tan(t) = i\frac{1+e^{-2 i t }}{1+e^{-2 i t }} + 2i \frac{-1}{1+e^{-2 i t }}$$

$$\tan(t) = i\frac{1+e^{-2 i t }-2}{1+e^{-2 i t }}$$

$$\tan(t) = i\frac{-1+e^{-2 i t }}{1+e^{-2 i t }}$$

$$\tan(t) = i\frac{e^{it}}{e^{it}}\frac{-1+e^{-2 i t }}{1+e^{-2 i t }}$$

$$\tan(t) = i \frac{-e^{it}+e^{- i t }}{e^{it}+e^{- i t }}$$

Thus we have shown that even though our Fourier series does not define a convergent sum, it none the less converges everywhere to $tan(t)$(except for $t=\frac{\pi}{2}+n \pi$). 

Thus we can say that $tan(t)$ has the Fourier series

$$\tan(t) = i + 2i \sum^\infty_{k=1} (-1)^k e^{2 i k t}$$


Author: Alexander White