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!
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1 Answer
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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: $\text{the coefficients of odd powers of x are 0}\implies \text{the Taylor Series is not alternating}$
is just wrong
