Fourier Transform (in 1D)

Projecting the function $f(x)$ onto the sine and cosine functions simultaneously is done by projecting it onto $e^{i k x}$:

$\begin{array}{rcl} F(k) & = & \displaystyle { \int_0^{2\pi} f(t)\ e^{-i k t}\ dt \over \int_0^{2\pi} e^{ikt}\ e^{-ikt}\ dt } \\ & = & \displaystyle {1 \over 2 \pi } \int_0^{2\pi} f(t)\ e^{-i k t}\ dt \\ & = & c_k \end{array}$

Recall from the axioms of the dot product that an inner product of complex functions must use the complex conjugate of the second function, so projecting onto $e^{ikt}$ uses the complex conjugate, $e^{-ikt}$, as the second factor.

So $F(k)$ provides the parameters of the shifted, modulated sinusoid of frequency ${ k \over 2\pi }$ onto which $f(x)$ projects.

The integrals above are taken over $[0,2\pi]$ since we assume that the period of the function is $2\pi$.

More generally, let the period of $f$ be $T$ and define the $\omega = {2 \pi \over T}$. $\omega$ is the "angular frequency" measured in radians per second, rather than the "regular" frequency, $1 \over T$, which is measured in cycles per second. Then

$\begin{array}{rcl} F(k) & = & \displaystyle { \int_0^{T} f(t)\ e^{-i k \omega t}\ dt \over \int_0^{T} e^{i k \omega t}\ e^{-i k \omega t}\ dt } \\ & = & \displaystyle {1 \over T } \int_0^{T} f(t)\ e^{-i k \omega t}\ dt \\ & = & c_k \end{array}$

From the coefficients $c_k$, we can reconstruct $f$ by multiplying the $c_k$ by the $e^{i k \omega t}$:

$\displaystyle f(x) = \sum_{k=-\infty}^\infty c_k\ e^{i k \omega t}$

for integer values of $k$ in the sum.