Why are the limits of the fourier cosine/sine series [0,∞) while the limits of the fourier exponential series are (-∞,∞)?

Question

Ok so bear with me here. I am a computer/electrical engineering student who somehow got sucked into math land. Recently, I have been examining the Fourier series and it seems that there are two approaches to the determining coefficients for the Fourier series.

The following summation is referred to as the Fourier series on Wikipedia for a function with period $2\pi$.

1. $f(t) = \frac{1}{2}a_0+\sum_{n=1}^\infty[a_n\cos(nt)+b_n\sin(nt)]$

I have encountered the following two integrals which allow us to extract the coefficients of the Fourier series.

2. $a_n:=\frac{1}{\pi}\int_{-\pi}^\pi f(t)\cos(nt)\;dt$

and

3. $b_n:=\frac{1}{\pi}\int_{-\pi}^\pi f(t)\sin(nt)\;dt$

I have also seen the following summation referred to as the exponential form of the Fourier series

4. $f(t) = \sum_{n=-\infty}^\infty c_n\cdot e^{i nt}$

My question here is why are the limits of the summation in Eq. 1 $[0,\infty)$ whilst $(-\infty, \infty)$ in Eq. 2's summation?

I understand that there is a formula for converting the $c_n$ coefficients of Eq. 4 to the $a_n$ and $b_n$ coefficients of Eq. 1. I have not listed it here as it seems not to pertinent to my question.

I have another hunch; the Eq.1 representation of the Fourier series has two different coefficients, but the summation only spans half of the real line. Eq. 4 only has one coefficient, but the summation spans the entire real line.

It seems that the choice of limits can be justified by the following observations.

1. If the limits of the summation in Eq. 1 were extended to the entire real line, this might be redundant as one could choose to ignore $a_n$ solving instead for $-a_n$.

2. Truncating the limits of the summation for Eq. 2 to only half the real line makes it impossible to represent some signals, for example, real valued signals require that the imaginary parts of the complex exponential cancel out in the summation.

But is there a deeper more satisfying reason for the difference in these two representations?

Answer 1

The complex fourier series is obtained simply from using the formulas $\cos(nx)=\frac{e^{inx}+e^{-inx}}{2}$ and $\sin(nx)=\frac{e^{inx}-e^{-inx}}{2i}$. Just write a partial sum of the real fourier series and then write sine and cosine using exponentials. This way you will get a partial sum of the complex fourier series. As you can see from the formulas we do have negative powers as well.

Answer 2

When you expand a function as a Fourier series, you can choose between two alternative basis. The first is the set of functions $$1, \cos t, \sin t, \cos 2t, \sin 2t, \ldots$$ The second is the set $$1, e^{it}, e^{-it}, e^{2it}, e^{-2it}, \ldots$$

You can write your function $f$ either as a linear combination of functions of the first set, or of functions of the second set. The two sets are related by the formulas: \begin{align} e^{int} &= \cos nt + i\sin nt \\ e^{-int} &= \cos nt - i\sin nt \end{align} or, if you prefer: \begin{align} \cos nt &= \frac{e^{int}+e^{-int}}{2} \\ \sin nt &= \frac{e^{int}-e^{-int}}{2i} \end{align}

As you can see, for each $n>0$, the pair $\{\cos nt, \sin nt\}$ is equivalent to (spans the same set of functions as) the pair $\{e^{int}, e^{-int}\}$. This means that you can take $n=1,2,3,\ldots$, and for each of those $n$ you use both $\cos nt$ and $\sin nt$. Or you could use both $e^{int}$ and $e^{-int}$.

The catch is that if you choose the basis set $\{e^{int}, e^{-int}\}$, then you can pretend that you were taking $n=\ldots,-3,-2,-1,1,2,3,\ldots$, and for each of those $n$ you just use $e^{int}$. And also, if you plug in $n=0$ in $e^{int}$, you end up with $e^0=1$, which is exactly the function that's missing, so that $\{e^{int}:n\in\mathbb Z\}$ is just a very short way to write: $$\{1,e^{it},e^{-it}, e^{2it}, e^{-2it}, \ldots\}$$

As you mention in your question, if you consider the functions $e^{int}$ but allow only $n\geqslant 0$, then you can't represent all functions. That is, the set $\{1, e^{it}, e^{2it}, \ldots\}$ is not powerful enough. In the other hand, if you try to make $\{\cos nt,\sin nt\}$ even more powerful by extending it to allow for negative $n$, then you get a redundant set (or, in the language of linear algebra, a linearly dependent set), because $\cos(-nt)$ is just the same as $\cos nt$, and $\sin(-nt)$ is the same as $-\sin nt$.

You ask for some "deeper reason", but I believe that this is already quite deep: the set $\{e^{int}:n\in\mathbb Z\}$ has exactly the right amount of "power" we need. It generates all functions that you want, without being redundant. Also, each pair $\{e^{int},e^{-int}\}$ is equivalent to $\{\cos nt,\sin nt\}$.

Answer 3

Once you substitute the Fourier coefficients into the series, the truncated Fourier series becomes the symmetrically truncated exponential Fourier series: \begin{align} \frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)dt+\sum_{n=1}^{N}\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos(nt)dt\cos(nx)\\+\sum_{n=1}^{N}\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\sin(nt)dt\sin(nx) \\ =\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)dt+\sum_{n=1}^{N}\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos(n(x-t))dt \\ = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)dt+\sum_{n=1}^{N}\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)(e^{in(x-t)}+e^{-in(x-t)})dt \\ = \sum_{n=-N}^{N}\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)e^{-int}dt\cdot e^{inx} \end{align}