While exploring Taylor series with numerical and graphing tools, I noticed a very peculiar and interesting fact: the partial sums of the Maclaurin series expansion of $\sin(x)$ approximate it better than their hyperbolic counterparts approximate $\sinh(x)$. To be specific, I noticed that for every $x\in\mathbb{R}$ and $k\in\mathbb{N}$, the following inequality seems (haven't proven it) to hold
$$\left|\sin(x)-\sum_{n=0}^{k}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}\right|\leq\left|\sinh(x)-\sum_{n=0}^{k}\frac{x^{2n+1}}{(2n+1)!}\right|$$
For all $x\neq 0$, the inequality is strict, showing that $\sum_{n=0}^{k}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}$ does indeed approximate $\sin(x)$ better than $\sum_{n=0}^{k}\frac{x^{2n+1}}{(2n+1)!}$ approximates $\sinh(x)$. But why do they provide better approximations? The remainders
$$\sin(x)-\sum_{n=0}^{k}(-1)^n\frac{x^{2n+1}}{(2n+1)!},\text{ }\sinh(x)-\sum_{n=0}^{k}\frac{x^{2n+1}}{(2n+1)!}$$
obviously can't be equal everywhere, but given the similarity between the partial sums for each function, it's unclear to me why the $\sin$ remainder should be bounded by an inequality as "uniform" as the one above. Why isn't it the other way around, with $\sinh$'s remainder being bounded by $\sin$'s? Why is the inequality true for all $x\in\mathbb{R}$, and not just some disjoint intervals? Looking at the partial sums for each series, it can be seen that the only difference is the $(-1)^n$ factor for the $\sin$'s series, which doesn't seem like the kind of thing that can make the difference between
$$\left|\sin(x)-\sum_{n=0}^{k}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}\right|\leq\left|\sinh(x)-\sum_{n=0}^{k}\frac{x^{2n+1}}{(2n+1)!}\right|$$
and some other inequality relating the two remainders. Can someone give an intuitive explanation for why this happens?
Best Answer
The series for $\sin x$ is alternating. So for any given partial sum $T_n(x)$ of that series, the absolute value of the next term is an overestimate of the error in using $T_n(x)$ to estimate $\sin x$.
In contrast, the terms in the series for $\sinh x$ all have the same sign. So for any given partial sum $H_n(x)$ of that series, the absolute value of the next term is an underestimate of the error in using $H_n(x)$ to estimate $\sinh x$.
But for both series, the absolute value of the $n+1$st term is $\dfrac{|x|^{2n+3}}{(2n+3)!}$.