Error in a Maclaurin series I'm having trouble figuring out what I have to do with this question.
"Using Taylor's theorem, determine the largest positive real value $r$ for which we can guarantee that the Maclaurin polynomial $P_6(x)$ of $f(x)$ has an error of less than $0.2$ on the interval $0 < x< r$."
Relevant information:
$f(x) = (\sin(x))^2$
I calculated $P_6(x)$ to be $x^2 - (x^4)/3 + 2x^6/45$
I calculated the error $R_6(x) = -4\sin(2c)*x^7/315$
The little info I gather from this question (may be incorrect) and where I am stuck:
Since $c$ lies between $a$ and $x$ this means that $0 < c < x$ since $0 < x < r$.
 I know I will be solving the error equation for $r$, but I don't know what to do with the $c$ and $x$ in the equation, or what to substitute in their place. 
 A: You are doing fine; now you want to use that $\displaystyle\lvert R_6(x)\rvert=\frac{4\lvert\sin 2c\rvert\lvert x\rvert^7}{315}<\frac{4r^7}{315}$
since $|\sin 2c|\le1$, and $|x|^7=x^7<r^7$ since $0<x<r$.
Now solve $\displaystyle\frac{4r^7}{315}<.2$ for $r$.

(Notice that in this case, $P_6(x)=P_7(x)$, so we could get even a better estimate using $\lvert R_7(x)\rvert$ instead.)
A: i think you can simplify first, for example, 
$$\begin{align}\sin^2x &= \frac12 - \frac12\cos 2x \\
& =\frac12 - \frac12\left(1 - \frac1{2}(2x)^2 + \frac1{4!}(2x)^4  - \frac1{6!}(2x)^6 + \frac1{8!}(2x)^8 - \cdots\right)\\
&=x^2-\frac13x^4 + \frac2{45}x^6-\frac{2^7}{8!}x^8+\cdots\end{align}$$ this is an alternating series so the error committed by stopping at the third term negative and smaller than absolute value.
we can make  $$\frac{2^7}{8!}r^8 \le 0.2 \text{  by demanding }r \le \left(0.2\times 8!/2^7\right)^{1/6} = 1.99475 $$
therefore the sixth degree polynomial $$ P_6(x)= x^2-\frac13x^4 + \frac2{45}x^6$$ approximates $\sin^2 x$ within $0.2$ for $|x| \le 1.99475$
