If I have $f(4)=1$ and $f^{(n)}(4)=\frac{\left(-1\right)^n n!}{3^n \left(n+1\right)}$, I know I can show that the fifth degree taylor polynomial will approximate $f(5)$ within $0.0002$ accuracy.
Since this would make an alternating series, I could do the "alternating series estimation theorem" but I want to try the Lagrange remainder and Taylor's inequality as well. I know this isn't necessary since the series is alternating, but I'd want to see if I can verify my results in different ways.
So, from what I understand, I need to find a $M$ (maximum) where $|f^{\left(n+1\right)}\left(x\right)|\leq M$. Since the $n$th derivative is given, I could use that and so it's $f^6 \left(4\right) = \frac{(-1)^6 6!}{3^6 \cdot 7}$ … but this $M$ when multiplied by $\frac{1}{\left(n+1\right)!}\left(x-a\right)^{n+1}$ will give me $\frac{\left(-1\right)^6}{ 3^6 \cdot 7}\left(5-4\right)^6$ and that's literally the same expression from the Alternating series estimation theorem (which is the first omitted term).
So, I'm just wondering if the alternating series estimation theorem has any special properties about it, or is it just a special/simple case?
Best Answer
Short Answer: The alternating series estimation is not a special case of Taylor's estimation. It's better, but only applies to alternating series (obviously). So If the series is an alternating series, use the Alternating series estimation, otherwise use Taylor's estimation.
In more detail: If you have an alternating series corresponding to a function whose $(n+1)$-st derivative happens to be maximized at the center, $a$, then you're in the rare case where Taylor's estimate and the alternating series estimate agree. E.g., $f^{(n+1)}(x)=\cos(x)$, centered at $a=0$.
In general, however, the alternating series estimate is better because it's rarely the case that the derivative is maximized at the center (oftentimes it's one of the endpoints of the interval). So $M$ is usually greater than $|f^{(n+1)}(a)|$, making Taylor's estimate larger (i.e., worse).
Moreover, Taylor's estimate requires more work on your part to actually maximize the $(n+1)$-st derivative (i.e., find $M$). For the Alternating series error, you simply use the next term in the series.
Here's a helpful comparison: $$E_{A}\leq\frac{\left|f^{\left(n+1\right)}\left(a\right)\right|}{\left(n+1\right)!}\left|x-a\right|^{n+1}$$ while $$ E_{T} \leq\frac{\left|f^{\left(n+1\right)}\left(x^{\ast}\right)\right|}{\left(n+1\right)!}\left|x-a\right|^{n+1}$$ where $x^\ast$ maximizes $f^{(n+1)}(x)$ on the desired interval about $a$.