this dynamic figure shows the graph of y = cos(x) and graphs of maclaurin polynomials from t0(x) to t40(x)…

Explanation:

Step1: Recall Maclaurin series property

The Maclaurin series of $\cos(x)=\sum_{k = 0}^{\infty}\frac{(- 1)^{k}}{(2k)!}x^{2k}=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\cdots$. The Maclaurin polynomial $T_n(x)$ is a partial - sum of this series. The error term $R_n(x)=\cos(x)-T_n(x)$ is bounded by the next - non - included term of the series according to the Alternating Series Estimation Theorem (since the Maclaurin series of $\cos(x)$ is an alternating series for non - zero $x$).

Step2: Analyze the absolute - value of the error

We know that $|\cos(x)|\leq1$ and $|T_n(x)|$ is also a well - behaved polynomial. The error $|\cos(x)-T_n(x)|$ is bounded. In fact, by the Alternating Series Estimation Theorem, for an alternating series $\sum_{k = 0}^{\infty}(-1)^{k}a_k$ ($a_k\gt0$, $a_{k + 1}\leq a_k$ and $\lim_{k\rightarrow\infty}a_k = 0$), the error $|R_N|=\left|\sum_{k = N}^{\infty}(-1)^{k}a_k\right|\leq a_N$. For the Maclaurin series of $\cos(x)$, the terms $a_k=\frac{|x|^{2k}}{(2k)!}$. The function $\cos(x)$ and its Maclaurin polynomials $T_n(x)$ satisfy $|\cos(x)-T_n(x)|\leq1$. The values of $|\cos(x)-T_n(x)|$ are not alternately positive and negative for all $n>1$ in general. And $|\cos(x)-T_n(x)|$ is bounded.

Answer:

For any $n > 1$, the values of $|\cos(x)-T_n(x)|$ are never larger than 1.