Power series as approximation

Question

I have to estimate the error when I approximate the function $$e^{\sin x}$$ to $$1+x+x^{2}+x^{3}$$ when $|x|<0.1$.

I really don't know how to do because my teacher didn't teach me. But what I did it was: I found the Taylor's polynom of $e^{\sin x}$ for $a=0$: $$T_{0}(x)=f(0)+f'(0)(x-0)+\frac{f''(0)}{2!}(x-0)^{2}+\frac{f'''(0)}{3!}(x-0)^{3}$$ then $$T_{0}(x)=1+x+\frac{1}{2}x^{2}$$ My question is: until here, is that the right way to solve it? If yes, what should I do now? How I use the fact that $|x|<0.1$? If it is wrong, where can I find some solved example to understand it? I didn't find anything good on google.

Answer 1

The Taylor expansion of $\exp\sin x$ around zero is $1+x+x^{2}/2+O(x^{4})$. Therefore, the error is \begin{align*} \left|\exp\sin x-(1+x+x^{2}+x^{3})\right| & =\left|-x^{2}/2+O(x^{3})\right|\\ & \leq|x^{2}|/2+|O(x^{3})|\\ & \approx|x^{2}|/2 & \text{for }|x|\text{ small}. \end{align*}

Answer 2

As you noted the Taylor series for $f(x) = e^x$ is $f(x) \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$. Now plug in $\sin x$ and use the fact that for values close to $x$ we have $\sin x \approx x$. So we have:

$$e^{\sin x} \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$$

Now the error of the first approximation is around $\frac{x^2}{2} + \frac{5x^3}{6}$. Use the fact that $\mid x \mid \le 0.1$ and plug it into the estimation to get the numerical approximation of the error.