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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.

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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 $\leq \frac{0.1^3}{3!}$.

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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 $x^2$.

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Alright - how's this for a generalization? When a Taylor polynomial expansion P(x) for function f(x) happens to alternate in signs, then both the Alternating Series Estimation Theorem and the Lagrange form of the remainder provide us with upper bound errors between the P(x) and f(x). However, the Alternating Series remainder will always be less than or equal to the Lagrange remainder and therefore will give us a better idea of the accuracy of our estimation.

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