Lagrange remainder vs. Alternating Series Estimation Theorem: do they always give you the same error bound?

Question

Given a function and its nth degree Taylor series approximation, we can use the Lagrange form of the remainder to get a maximum value of the error of approximation. If the series is also an alternating series, we can also use the Alternating Series Estimation Theorem to get a maximum value of the error of approximation. Will the two maximums always be the same?

I have tested a few cases and they turn out to be the same. I'm looking for a proof that they are always the same, or else a counter-example.

Answer 1

Suppose your series is $f(x) = \sum_{n=0}^\infty a_n x^n$ with the signs of $a_n$ alternating. If $|a_n| r^n$ is decreasing to $0$, this is an alternating series for $0 < x < r$.

The alternating series bound for the remainder after the $x^n$ term is then $|a_{n+1}| x^{n+1}$. The Lagrange form for the remainder is $\dfrac{f^{(n+1)}(c)}{(n+1)!} x^{n+1}$, where $0 < c < x$, and to get a bound we want to maximize $|f^{(n+1)}(c)|$ on this interval.

Now $$\dfrac{f^{(n+1)}(c)}{(n+1)!} = \sum_{j=n+1}^\infty {j \choose n+1} a_j c^{j-n-1}$$

The bound is the same as the alternating series bound if the maximum occurs at $c=0$. Now the derivative of this is

$$ \dfrac{f^{(n+2)}(c)}{(n+1)!} = \sum_{j=n+2}^\infty (j-n-1) {j \choose n+1} a_j c^{j-n-2} = (n+2)\sum_{j=n+2}^\infty {j \choose n+2} a_j c^{j-n-2} $$

If it weren't for that ${j \choose n+2}$ factor, this would still be an alternating series, and $f^{(n+2)}(c)$ would have the same sign as $a_{n+2}$, which is opposite to the sign of $a_{n+1}$ and $f^{(n+1)}(c)$, implying that the maximum is at $c=0$. But that factor can mess things up.

Consider e.g. a series that starts $1 - x + x^2 - x^3 + x^4$, with the remaining terms very small (but still alternating for $0 < x < 1$). You want to estimate the error in the linear approximation $1 - x$. Then $$\dfrac{f''(c)}{2} \approx 1 - 3 c + 6 c^2$$ If $ 1/2 < x < 1$, the maximum of this is not at $c=0$ but rather at $c=x$. The Lagrange bound is then approximately $(1 - 3 x + 6 x^2) x^2$, which is different from the alternating series bound of $x^2$.

Answer 2

They will not be the same. One counterexample is enough. Use $\cos(x)=1-\frac{x^2}{2!}+...$ Suppose you want to evaluate $\cos(0.1)$, error using first dropped term $\leq \frac{0.1^4}{4!}$, and using LaGrange Error is $\leq \frac{0.1^3}{3!}$.