Use $x=-\frac{1}{2}$ in the MacLaurin series for $e^x$ to approximate $1/\sqrt{e}$ to four decimal places.
I'm not sure where to start or how to use $x=-1/2$. Does that mean find the Taylor series at $x=-1/2$ (doesn't make sense since I thought the MacLaurin series was for at $x=0$). I also don't get how to do the approximation, like what should I set to $\le 10^{-4}$
Thanks
On
The Taylor formula allows you to compute unknown function values from known ones (as well as derivatives), by elementary operations (powers and divisions by factorials).
So in the case at hand, it would make no sense to expand around $x=-1/2$ as you don't know the value there (which will be theoutput of the exercise).
On the other hand, you can easily develop around $0$ (indeed giving a so-called Maclaurin series) and use the development to evaluate elsewhere.
To ensure the requested accuracy, you need to estimate the magnitude of the remainder as a function of the number of terms, so that you can make this remainder less than $10^{-4}$.
Hint:
$$\frac{x^n}{n!}+\frac{x^{n+1}}{(n+1)!}+\frac{x^{n+2}}{(n+2)!}+\cdots<\frac{x^n}{n!}\left(\frac1{0!}+\frac x{1!}+\frac{x^2}{2!}+\cdots\right)=\frac{x^ne^x}{n!}.$$
On
For each $x\in\mathbb R$ and each natural $n$,$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots+\frac{x^n}{n!}+\frac{e^\eta}{(n+1)!}x^{n+1},$$for some $\eta$ between $0$ and $x$. Now, take $x=-\frac12$. Then $-\frac12<\eta<0$ and therefore $e^\eta<e^0=1$. So, all you have to do is to find some $n\in\mathbb N$ such that $\frac1{(n+1)!2^{n+1}}\leqslant\frac1{10^4}$, that is, such that $(n+1)!2^{n+1}\geqslant10^4$. Just take $n=5$. Then, the approximation of $1/\sqrt e$ that you want is$$1-\frac12+\frac1{2^22!}-\frac1{2^33!}+\frac1{2^44!}-\frac1{2^55!}=\frac{2\,329}{3\,840}\simeq0.60651$$
The Maclaurin series for $e^x$ is $$1+x+\frac{x^2}2+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots.$$ What happens when you stick in $x=-1/2$?