Find an upper bound for $|f(x)-P_4(x)|$, for $0 \le x \le 0.4$

Question

So I'm stuck on something that is supposed to be fairly easy...

I was able to work through the Taylor polynomial which I believe to be:

$$x+x^3+\frac{x^5}{2!}+\frac{x^7}{3!}+\frac{x^9}{4!}$$

But I'm stuck on the upper bound. I don't know how to go about finding it. Like something I learned so long ago and can't figure out for the life of me. My teacher told me to put the equation on a graph but when I do I don't know what the bounds would be as I think it would just approach infinity? But I don't know since the problem states a given range.

Graph: https://www.desmos.com/calculator/zy7fjl26zz

Answer 1

You are only supposed to consider the interval $0\leq x\leq 0.4$.

The difference $|f(x)-p_4(x)|$ is bounded by the error term in the Taylor polynomial. In this case, you have $$ |f(x)-p_4(x)|\leq \frac{e^c\,|x|^{11}}{5!}, $$ where $0\leq c\leq 0.4$. So $$ |f(x)-p_4(x)|\leq \frac{e^{0.4}\,0.4^{11}}{5!}\leq \frac{2\times 0.4^{11}}{120}=0.000000699050\overline6\leq10^{-6}. $$