# Taylor Remainder for Even, Bell-shaped Functions

Given a function $$f$$ and its Taylor series. I'm looking for an easy-to-compute upper bound for Taylor's remainder. The missing term in Taylor series serves as an upper bound if they are alternating series. However, when $$f$$ is an even non-negative bell-shaped function, all coefficients of odd powers are 0. It's impossible to have an alternating Taylor series. In this case, I wonder intuitively how the Taylor series approaches $$f$$. Does it have to be true that all coefficients are non-negative? What does it look like geometrically? Does the Taylor Remainder have a convenient upper bound? Thank you so much for your help!

Notice that $$e^{-x^2}$$ is an even, positive, bell-shaped function, but is defined as
$$e^{-x^2}:=\sum\limits_{k=0}^{\infty}\frac{(-1)^kx^{2k}}{k!}$$
and since power series is Taylor series, our function has an alternating Taylor series, with nice upper bounds. The implication: $$\text{the coefficients of odd powers of x are 0}\implies \text{the Taylor Series is not alternating}$$