Find a rational approximation $\sqrt e$ and $\cos1$ with accuracy $d=0,001$ using Taylor I know that I can write $\sqrt e=1+\frac{1}{2}+\frac{\frac{1}{2^2}}{2!}+\frac{\frac{1}{2^3}}{3!}+...$ and calculate in which punct I have $d=0,001$. Hovewer I think it is not an elegant way and I can do for example: $$|\sqrt e-(1+\frac{1}{2}+\frac{\frac{1}{2^2}}{2!}+\frac{\frac{1}{2^3}}{3!}+...+\frac{\frac{1}{2^{\alpha}}}{\alpha!}+o(\frac{1}{2})^{\alpha})|<0,001$$Unfortunately then I don't know how I can calculate this $\alpha$.Maybe there is a better way for that?
 A: By Taylor's Theorem, there is an explicit formula for the remainder, i.e. Lagrange form  of the remainder: if $x>0$ then there exists $t\in (0,x)$ such that
$$e^x-\sum_{k=0}^n\frac{x^k}{k!}=\frac{e^t x^{n+1}}{(n+1)!}.$$
Hence, if $x=1/2$ then there is $t\in (0,1/2)$ such that
$$\left|\sqrt{e}-\sum_{k=0}^n\frac{(1/2)^k}{k!}\right|=\frac{e^t (1/2)^{n+1}}{(n+1)!}<\frac{2(1/2)^{n+1}}{(n+1)!}$$
because $e^t<e^{1/2}<4^{1/2}=2$. Now it suffices to find $n$ such that
$$\frac{2(1/2)^{n+1}}{(n+1)!}<0.001.$$
Can you take it from here?
A: A brute-force method:
The series for the cosine is alternating, with decreasing terms, and you can use the property that the error is less than the first omitted term.
You can use the same trick for $\sqrt e$ by evaluating $\dfrac1{\sqrt e}=e^{-1/2}$ instead, which also yields an alternating series.
This might require (a little) too many terms, but you are on the safe side.

Hints:
$$7!>1000,\\5!\,2^5>1000.$$

$$1-\frac12+\frac1{24}-\frac1{720}=\frac{389}{720}.$$
$$\dfrac1{1-\dfrac12+\dfrac18-\dfrac1{48}+\dfrac1{384}}=\frac{384}{233}.$$

