Answer:
Step-by-step explanation:
To find the Taylor series of sinc(x) we will use the taylor series of sin(x). We have that
[tex]\sin(x) = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}[/tex]
which is the taylor series expansion based at 0. Then for [tex]x\neq 0[/tex], by dividing both sidex by x, we have that
[tex]\text{sinc}(x) = \frac{\sin(x)}{x}= \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n+1)!}[/tex]
which is the taylor series expansion for the sinc function. Since the series of sine converges for every value of x. Then the taylor series of sinc converges for every value of x, but 0.
