Calculate approximate value using Taylor series.

Question

I need to calculate the approximate value for $e^{-\frac{1}{4}}$ with a precision of $10^{-3}$. I understand that I have to find the expansion order to which remainder is smaller than $10^{-3}$, but as $x$ in this case is $-\frac{1}{4}$ then those values are fluctuating depending on $n$ being an odd or even number. Also I don't understand whether I have to simply guess what the $n$th order be, where the remainder is small enough or is there some general way how to do it?

Answer 1

So using the Taylor series expansion we have $$e^{-\frac14}\approx\sum_{k=0}^n\frac{(-\frac14)^k}{k!}$$ For an accuracy within $10^{-3}$ we require $n$ such that $$\Bigg|e^{-\frac14}-\sum_{k=0}^n\frac{(-\frac14)^k}{k!}\Bigg|\lt10^{-3}$$ $$\Bigg|\sum_{k=0}^\infty\frac{(-\frac14)^k}{k!}-\sum_{k=0}^n\frac{(-\frac14)^k}{k!}\Bigg|\lt10^{-3}$$ $$\Bigg|\sum_{k=n+1}^\infty\frac{(-\frac14)^k}{k!}\Bigg|\lt10^{-3}$$ But as the terms of the above series are absolutely decreasing and alternating in sign, we have $$\Bigg|\sum_{k=n+1}^\infty\frac{(-\frac14)^k}{k!}\Bigg|\lt\Bigg|\frac{(-\frac14)^{n+1}}{(n+1)!}\Bigg|=\frac{(\frac14)^{n+1}}{(n+1)!}$$ So we can instead find $n$ such that $$\frac{(\frac14)^{n+1}}{(n+1)!}\lt10^{-3}$$ Testing different values of $n$ shows that the smallest valid solution is $n=3$. So we can accurately approximate $e^{-\frac14}$ by using the first $4$ terms of the expansion to an accuracy $10^{-3}$ $$e^{-\frac14}=1-\frac14+\frac1{2!}\Big(\frac14\Big)^2-\frac1{3!}\Big(\frac14\Big)^3$$ $$=\frac{299}{384}\approx 0.7786$$

Answer 2

The alternating signs are a help rather than a problem, because you immediately know (assuming the series converges) that the error is less than the first absolute value of the first omitted term. (If the terms were all the same sign, you would need to do a bit more work to prove you error estimate.)

Since the general term for $e^{-\frac14}$ is $$ (-1)^n \left( \frac14\right)^n\frac1{n!} = (-1)^n \frac1{4^nn!} $$ we just have to find some $n$ such that $$\frac1{4^nn!} < 10^{-3} \implies 4^nn! > 1000 $$ Although there is no clean way to solve this equation, knowing that $4^4 = 256$ you can immediately see that $4^44! > 1000$, so you only have to try $n=3$ to see if that also works (it does not).

So the first omitted term han be the $n=4$ term, and this gives a value of $$ e^{-\frac14} \approx 1 - \frac14 + \frac1{2\cdot 16} - \frac1{6\cdot 64} = \frac{299}{384} \approx 0.779 $$ and this is correct to about 2 parts in $10^4$.