I have been having trouble fully understanding how to use the delta function in finding the Fourier Transform of a function.
Given this function, $$f(x) = \begin{cases} 1 - |x|, & |x| < 1 \\ 0, & |x| > 1 \\ \end{cases} $$
It makes sense to take the derivative: $$f'(x)= \begin{cases} 1, & -1 < x < 0 \\ -1, & 0 < x < 1 \\ 0, & |x| >1 \end{cases} $$ And if we look at the graph of this derivative, we see that there is a jump of $+1$ at $x=-1$, a jump of $-2$ at $x=0$ and a jump of $+1$ at $x=1$. I am not sure how to use the delta function for these jumps.
I am looking to get this answer: $$f''(x)=\delta(x+1) + 2\delta(x) + \delta(x-1)$$
I am just not sure why we take the second derivative (which would be just $0$ everywhere) and why the delta functions are in the second derivative, not the first (where we saw the jumps in the graph).
Thank you for the help!
The distributional derivative of $f(x)$ is a linear combination of Heaviside step functions, not of delta functions: $$f'(x)=\mathbb{I}_{\{[1,\infty)\}}(x)-2\mathbb{I}_{\{[0,\infty)\}}(x)+\mathbb{I}_{\{[-1,\infty)\}}(x)=\mathbb{I}_{\{[0,\infty)\}}(x-1)-2\mathbb{I}_{\{[0,\infty)\}}(x)+\mathbb{I}_{\{[0,\infty)\}}(x+1)$$ We have that $f'(x)=1$ for $x \in [-1,0)$, $f'(x)=-1$ for $x \in [0,1)$ and $f'(x)=0$ for $x \in (-\infty,1)\cup[1,\infty)$. The distributional derivative of the Heaviside step function is the delta function, so $$f''(x)=\delta(x-1)-2\delta(x)+\delta(x+1)$$