Fourier transform derivation

Question

I'm reading Hassani's Mathematical Methods book specifically the chapter on Integral Transforms. He derives the fourier transform starting with the concept that the fourier transform has a kernel of the form $e^{itx}$, everything goes well until he stated on the bottom of page 694,

"In other words, as $n$ changes by one unit, $k_n$ changes only slightly. This suggests that the terms in the sum in Equation (29.2) can be lumped together in $j$ intervals of width $\Delta n_j$"

That means as $\Lambda \rightarrow \infty $, $k_n \rightarrow 0$, so becomes almost continuous. The sentence "This suggests that the terms in the sum in Equation (29.2) can be lumped together in $j$ intervals of width $\Delta n_j$" is what I don't understand. Can anyone clarify what he meant?

Answer 1

"The terms in the sum can be lumped together" means that it can be possible to sum every $\Delta n_j$ consecutive terms (the resulting sum can be called lumped term) and then consider the series of such lumped terms as equivalent to the original series. In building such a lumped term some approximation can be made of the kind here illustrated.

In the first diagram I plotted as an example the sequence $$\frac{1}{\sqrt{L+\Lambda}}f_{\Lambda, n}e^{2i\pi nx/(L+\Lambda)}$$ on $36$ points only, at $x=1$, with the hypotheses that $L+\Lambda=1$ and $f_{\Lambda, n}\equiv 1$.

Supposing to lump together terms every $\Delta n_j = 3$, the corresponding piece of sequence that results is displayed in the second diagram. You see that for instance the three consecutive red terms (all with unit modulus) sum up to a term whose modulus is a bit less than $3$.

This lumped term can be substituted for by an approximated lumped term given by $\Delta n_j$ times one of the consecutive terms to be lumped as if all such consecutive terms had the same phase. In the third diagram, such an approximation is displayed, where the approximated lumped term is taken as three times the central term of the consecutive terms to be lumped.

Note that such an approximation can be affected heavely by $f_{\Lambda, n}$ that I've been considering constant in the diagrams. If it takes values too different the lumped terms could no longer be approximated by one of the original terms multiplied by the number of consecutive terms, because such terms would have very different moduli and phases. But this is not a problem because whatever is the level of approximation you want there will always be a real value such that, whatever is $\Lambda$ greater than this value, that level of approximation is attained (as you can see by $(29.3)$).

Answer 2

It should be something like

This suggests that the terms in the sum in Equation (29.2) can be divided into pieces, where in each piece the quantity $k_n$ is constant. Denote the length of the $j$-th piece by $\Delta n_j$, and put $n=n_j$ throughout the $j$-th piece.

In other words, he is writing $$ \sum_{n}f_{\Lambda,n}e^{ik_nx}=\sum_j\sum_{n\in I_j}f_{\Lambda,n}e^{ik_nx}\approx\sum_jf_{\Lambda,n_j}e^{ik_{n_j}x}\Delta n_j $$ where $\{I_j\}$ is a disjoint decomposition of $\mathbb{Z}$ into intervals, $\Delta n_j=|I_j|$ is the number of elements of $I_j$, and $n_j\in I_j$ is an element of $I_j$ chosen by some rule.

Answer 3

I have a slightly different book that defines this differently. The motivation for the Fourier Transform can be seen from the following. This is from Richard Haberman - Applied Partial Differential Equations with Fourier Series and Boundary Problems

In solving boundary value problems on a finite interval $-L < x < L$ with periodic boundary conditions we can use the complex form of the Fourier Series

$$ \frac{f(x+)+f(x-)}{2}= \sum_{n=-\infty}^{\infty} c_{n}e^{\frac{-in \pi x}{L}}$$

Where $f(x)$ represents a linear combination of all the sinusoids that are periodic with period $2L$ then we have the Fourier coefficients as

$$c_{n} = \frac{1}{2L}\int_{-L}^{L} f(x)e^{\frac{-in \pi x}{L}}dx $$

now then we have $ -L < x < L $ as our region of integration. So we extend it to $ - \infty < x< \infty $

$$ \frac{f(x+)+f(x-)}{2}= \sum_{n=-\infty}^{\infty} \left[ \frac{1}{2L}\int_{-L}^{L} f(\bar{x})e^{\frac{-in \pi \bar{x}}{L}}d\bar{x} \right]e^{\frac{-in \pi x}{L}}$$

For periodic functions $ -L < x < L $ the number of waves $ \omega $ in a distance of $ 2 \pi $ are then

$$ \omega = \frac{n \pi}{L} = 2\pi \frac{n}{2L}$$

giving us the distance between waves $$ \Delta \omega = \frac{(n+1) \pi}{L} - \frac{n \pi}{L} = \frac{\pi}{L}$$

$$ \frac{f(x+)+f(x-)}{2}= \sum_{n=-\infty}^{\infty} \left[ \frac{\Delta \omega}{2\pi}\int_{-L}^{L} f(\bar{x})e^{i \omega \bar{x}}d\bar{x} \right]e^{-i \omega x}$$

Then the fourier transform here is as $ L \to \infty $. The values $ \omega $ are the square roots of the eigenvalues so they get closer and closer $ \Delta \omega \to 0$

$$ \frac{f(x+)+f(x-)}{2}= \frac{1}{2 \pi}\int_{-\infty}^{\infty} \left[ \int_{-\infty}^{\infty} f(\bar{x})e^{i \omega \bar{x}} \right]e^{-i \omega x} d\omega$$

Then the fourier transform is

$$F(\omega) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} f(\bar{x}) e^{i \omega \bar{x}} d \bar{x} $$

The notation is that our interval $ - L < x < L$ has extended to infinity as you see. This is commonly what happens when you have some Riemann sum and take the discrete intervals and you let them go infinitly small.