Help with Taylor Polynomial Estimation Solution.

Question

Estimate $e^{0.1}$ to 6 decimal places using a Taylor polynomial about 0. Use error bounding to prove that your estimate is accurate to at least 6 decimal places.

Is my solution correct?

We know that the derivatives of each term are equal to $e^x,$ as the derivative of $e^x$ is always equal to $e^x.$ Therefore, we have that $f^{(n)} (0) = 1$ for all $n.$ This makes our Taylor Polynomial equal to $$1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ...$$ Since we're estimating $e^{0.1}$ to 6 decimal places, we use a quartic polynomial to get $$1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} = \boxed{1.105171}.$$ We know that the absolute value of the error in estimating $e^{0.1}$ by a Taylor polynomial of degree 4 about $a=0$ is at most $$\left| \frac{M}{5!}(0.1)^{5} \right|,$$ where $M$ is the maximum value of $|f^{5}(x)|$ for $x$ on $[0,0.1].$ This gives us $M = e^{0.1},$ or $$\left| \frac{e^{0.1}}{5!}(0.1)^{5} \right| \approx 0.00000009209 < E = 0.000001.$$ Therefore, our estimate is accurate to at least 6 decimal places

Answer 1

You should finish the error bound something this, to avoid using a pre-computed value of $e^{.1}$.

We have$$1.105171<e^{.1}<1.105171+\frac{10^{-5}}{120}e^{.1}$$ The second inequality gives $$e^{.1}<\frac{1.105171}{1-\frac{10^{-5}}{120}}\approx1.105171092097591$$ so that $$1.105171<e^{.1}<1.1051711$$