Taylor series of $\frac{\sin(x)}{x}$ is given by $1 – \frac{x^2}{6} + \frac{x^4}{120} + … $
We know that the Fourier Transform of this sinc function is a scaled Rect function. Why is it incorrect to simply take a Fourier Transform of each term of the Taylor Series shown above and sum them up?
Taking the definition of FT as $\int_{-\infty}^{\infty} f(t) e^{-i 2 \pi\omega t} dt$,
$$FT(1) = \delta(\omega)$$
$$FT(x^2) = – \frac{\delta^{"}(\omega)}{4\pi^2}$$
And so on.
The sum doesn't resemble a Rect function. I do realize that $\delta(x)$ is not a function in traditional sense etc. So at least I was hoping to use it as an operator under integral sign for doing something like $FT(\frac{\sin(x)}{x} \frac{\sin(x)}{x}) = FT(\frac{\sin(x)}{x}) \star FT(\frac{\sin(x)}{x})$, where $\star$ is convolution.I wanted to use Rect function for the first term on the right hand side of this equation, but for the second term, instead of using another Rect function, I wanted to use the term wise FT of the Taylor series and then sum the infinite set of convolutions. Note, I already know that the FT of this particular squared sinc function is a scaled unit triangle function in the frequency domain. So my question is: why doesn't the term-wise FT give correct result (either directly for sinc or as an operator under integral sign for this squared sinc)?
Best Answer
Let $f(x)=\frac{\sin(x)}{x}$. Then, we have for $\phi\in\mathbb{S}$ (i.e., $\phi$ is a Schwartz function)
$$\begin{align} \pi \int_{-1/2\pi}^{1/2\pi}\phi(k)\,dk&=\langle \mathscr{F}\{f\},\phi\rangle\\\\ &=\langle f,\mathscr{F}\{\phi\}\rangle\\\\ &=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\int_{-\infty}^\infty x^{2n}\int_{-\infty}^\infty \phi(k) e^{i2\pi kx}\,dk\,dx \\\\ &=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \int_{-\infty}^\infty \int_{-\infty}^\infty \phi(k) \frac1{(i2\pi )^{2n}}\frac{d^{2n}}{dk^{2n}}e^{i2\pi k x}\,dk\,dx\\\\ &=\sum_{n=0}^\infty \frac{1}{(2\pi)^{2n}(2n+1)!}\int_{-\infty}^\infty \int_{-\infty}^\infty \phi^{(2n)}(k)e^{i2\pi kx}\,dk\,dx\\\\ &=\sum_{n=0}^\infty \frac{\phi^{(2n)}(0)}{(2\pi)^{2n}(2n+1)!} \end{align}$$
Alternatively, we can expand $\phi$ in a Taylor series to arrive at the same result. Proceeding, we see that
$$\begin{align} \pi \int_{-1/2\pi}^{1/2\pi}\phi(k)\,dk&=\pi \int_{-1/2\pi}^{1/2\pi} \sum_{n=0}^\infty \frac{\phi^{(n)}(0)}{n!} k^n\,dk\\\\ &=2\pi \sum_{n=0}^\infty \frac{\phi^{(2n)}(0)}{(2n)!}\int_0^{1/2\pi} k^{2n}\,dk\\\\ &=2\pi \sum_{n=0}^\infty \frac{\phi^{(2n)}(0)}{(2\pi)^{2n+1}(2n+1)!}\\\\ &=\sum_{n=0}^\infty \frac{\phi^{(2n)}(0)}{(2\pi)^{2n}(2n+1)!} \end{align}$$
Therefore, in distribution we see that
$$\begin{align} \pi \text{rect}(\pi k)&=\mathscr{F}\{f\}(k)\\\\ &= \sum_{n=0}^\infty \frac{\delta^{2n}(k)}{(2\pi)^{2n}(2n+1)!} \end{align}$$
as was to be shown!