# Help with Taylor Polynomial Estimation Solution.

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

• When you calculate the error bound, it seems like you're using a known value of $e^{0.1}$. Commented Oct 12, 2020 at 2:35

You should finish the error bound something this, to avoid using a pre-computed value of $$e^{.1}$$.
We have$$1.105171 The second inequality gives $$e^{.1}<\frac{1.105171}{1-\frac{10^{-5}}{120}}\approx1.105171092097591$$ so that $$1.105171