Find the second Taylor polynomial $\displaystyle P_2(x)$ for the function $\displaystyle e^xcos(x)$ about $\displaystyle x_0=0$

Use $\displaystyle P_2(0.5)$ to approximate f(0.5). Find an upper bound for error $\displaystyle |f(0.5)-P_2(0.5)|$ using the error formula and compare it to the actual error.

Now $\displaystyle f(x)=P_n(x)+R_n(x)$

$\displaystyle P_2(x)=1+x$

$\displaystyle R_n(x)=\frac{f^{n+1}(E(x))}{(n+1)!}(x-x_0)^n$

In mine case:

$\displaystyle R_2(x)=\frac{-2e^{E(x)}(sin(E(x))+Cos(E(x)))}{3!}x^3$

And substitute for $\displaystyle R_2(0.5)$ and I need to find upper bound on $\displaystyle |R_2(0.5)|$ for 0 < E(x) < 0.5

The upper bound is for E(x)=0.5 that is $\displaystyle |R_2(0.5)| \leq 0.093222005$.

But in the solution manual it is 0.0532. I rechecked 10 times and I start to believe that it's their error.

Thanks in advance.