Fourier Transformation in exponential form negative infinity

Question

I have a quick question about Fourier series. Given that $$f(x) $$ satisfies 4 conditions required for transformation, we know that f(x) can be expressed in terms of sines and cosines:

$$ f(x) =a_0/2+ \sum^\infty_{n=1} a_n cos(wnx) + b_nsin(wnx), w = \frac{2\pi}{L}$$ Furthermore, using Euler's Identity, $$e^{ix} = cos x + isin x $$ and we can convert the equation to:

$$ f(x)= \sum ^ \infty_{n=-\infty} c_n exp(wnx)$$.

My question is, in the expression involving sines and cosines, note that index n is iterated from 1 and to the infinity. However, in bottom expression, the index n starts from 0 and is iterated in both directions. Of course, we know some results such as a_n = real(c_n + c_{-n})/2 and others from this. However, even before that, why is it that we can't iterate n in positive direction only instead of stretching the summation in both directions? i.e.

$$f(x)= \sum ^ \infty_{n=0} c_n exp(wnx) $$

Is there a reason that we have to go in both direction instead of one direction, say positive direction?

---

sidenote: I am sure that this is a mistake. Please correct me

$$ e^{i\pi} = e^{2(1/2)i\pi} = ((e^{i\pi})^2)^{1/2} = ((-1)^2)^{1/2} = 1^{1/2} = 1. $$

Answer 1

For $x>0$

$(x^a)^b=e^{b\ln(x^a)}=e^{ab\ln(x)}=x^{ab}$

but if $z\in \mathbb C$, in general we don't have

$(z^a)^b=z^{ab}$

caus we could have

$z_1\neq z_2$ and $Log(z_1)=Log(z_2)$.

Answer 2

One answer is that the positive exponentials "are not enough". For example, if $f(x)=e^{-ix}$, then the Fourier coefficients (with respect to the exponential expansion), for $n\geq 0$, are equal to $$c_n=\frac{1}{2\pi}\int_0^{2\pi}e^{-ix}e^{-inx}\,dx=\frac{1}{2\pi}\int_0^{2\pi}e^{-i(n+1)x}\,dx=0.$$ So, if we expressed $f$ in terms of its Fourier series, we would obtain that $f\equiv 0$, which is a contradiction.

On the other hand, the positive together with the negative exponentials are actually "enough" for any reasonable representation to hold. This happens because they form an orthonormal basis of $L^2$.

Answer 3

I am indeed a bit late, but I would like to provide extra information on why the summation becomes from $-\infty$ to $\infty$.

Any periodic signal $s(t)$, with integer frequencies, can be represented with Fourier Series in trigonometric form as

$$\begin{align*} s(t) = \sum\limits_{n=0}^{\infty} [a_n\sin(nt) + b_n\cos(nt)] \end{align*}.$$

We can simplify the summation to realize that the first coefficient will be $b_0$ by expanding the first iteration of the sum, where $n = 0$. However, I will be skipping that since everything that follows will fall down to something more general.

Now, if you do want to change to the complex exponential form, you must realize that both sine and cosine can be expressed in the following way:

$$\begin{align*} \cos(\theta) & = \frac{e^{i\theta} + e^{-i\theta}}{2}\\ \sin(\theta) & = \frac{e^{i\theta} - e^{-i\theta}}{2i}. \end{align*}$$

If we change the summation for an arbitrary $n$ value, it is possible to realize that

$$\begin{align*} a_n\sin(nt) + b_n\cos(nt) & = a_n\left(\frac{e^{int} - e^{-int}}{2i}\right) + b_n\left(\frac{e^{int} + e^{-int}}{2}\right)\\ & = \frac{a_n}{2i}e^{int} - \frac{a_n}{2i}e^{-int} + \frac{b_n}{2}e^{int} + \frac{b_n}{2}e^{-int}, \end{align*}$$

then recall that dividing by $i$ is the same as multiplying by $i^{-1}$, which is the same as $i$ to the power of $3$, $-i$, leading to

$$\begin{align*} \frac{-ia_n}{2}e^{int} - \frac{ia_n}{2}e^{-int} + \frac{b_n}{2}e^{int} + \frac{b_n}{2}e^{-int} & = \left(\frac{b_n - i a_n}{2}\right)e^{-int} + \left(\frac{b_n + i a_n}{2}\right)e^{int}\\ & = c_{-n}e^{-int} + c_ne^{int}, \end{align*}$$ where we define $c_n$ and $c_{-n}$ as

$$\begin{align*} c_n & = \frac{b_n + i a_n}{2}\\ c_{-n} & = \frac{b_n - i a_n}{2}. \end{align*}$$

The sum then becomes

$$\begin{align*} s(t) & = \sum\limits_{n=0}^{\infty} [a_n\sin(nt) + b_n\cos(nt)]\\ & = \sum\limits_{n=0}^{\infty} \left[\left(\frac{b_n - i a_n}{2}\right)e^{-int} + \left(\frac{b_n + i a_n}{2}\right)e^{int}\right]\\ & = \sum\limits_{n=0}^{\infty} \left(c_{-n} e^{-int} + c_ne^{int}\right). \end{align*}$$

Take a minute to realize that $\forall n$, where $n\in\mathbb{N^*}$, the first exponential will make the use of the negative of that $n$, so as $c_{-n}$. Because of that, we can carefully flip the summation limits for the first exponential, leading to

$$\begin{align*} \sum\limits_{n=1}^{\infty} c_{-n}e^{-int} = \sum\limits_{n=-\infty}^{-1} c_ne^{int}. \end{align*}$$

Notice that I took out the iteration 0 from the above sum, since we are only interested in using an iteration once, and I will be using it in a second sum.

Thus, you can rearrange the summation and make sense of the following notation

$$\begin{align*} s(t) & = \sum\limits_{n=0}^{\infty} \left(c_{-n} e^{-int} + c_ne^{int}\right)\\ & = \sum\limits_{n=-\infty}^{-1} c_n e^{int} + \sum\limits_{n=0}^{\infty} c_n e^{int}, \end{align*}$$ which nicely simplifies to

$$\begin{align*} s(t) & = \sum\limits_{n=-\infty}^{\infty} c_ne^{int}, \end{align*}$$ where $$\begin{align*} c_n & = \begin{cases} \frac{b_n - i a_n}{2}, n < 0\\ \frac{b_n + i a_n}{2}, n \ge 0 \end{cases}\\ \; \end{align*}$$