Why the function to generate Taylor series need be differentiable on an interval and not just at the point in question instead

Question

The definition of the Taylor series generated by function $f$ at $x=a$,

$$ f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + ... + \frac{f^{n}(a)}{n!}(x-a)^n + ... , $$

requires that the $f$ must be differentiable throughout some interval containing $a$ as an interior point.

However, I fail to see why $f$ need be differentiable to any order on an interval and not just only at the point $x=a$ (and the latter condition could still mean that it's continuous on some interval containing $x=a$ even though it's not differentiable to all orders on that interval)?

Answer 1

A Ha!

If your expansion is just $f(a)+f'(a)(x-a)$, you just need $f$ is differentiable at one point "a".

If your expansion is $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^{2}$, you need $f'$ to be exist in an interval to guarantee the existence of $f''(a)$

.....

If your expansion is $f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+\dots+\frac{f^{(n+1)}(a)}{(n+1)!}$, you need $f^{(n)}$ to be exist in an interval to guarantee the existence of $f^{(n+1)}$ ......

Answer 2

The Taylor series is supposed to give you an approximation of your function for it to be useful.

If you get far away from your starting point, even with differentiability your approximation will get worse.

But imagine a function where in the near vincinity of your point the function is not smooth. Then at the other side of this breaking point, your approximation will fail.

So while you are right, in theory an infinitesimal small interval around your point $a$ would suffice to give you a valid Taylor series, it will not help you with anything.