Taylor-approximation of square root Can someone please guide me with this question: Use the Taylor polynomial of $$f(x)=\sqrt{1+x}$$ around $0$ to get an estimate for $\sqrt{2}$ correct to
one decimal places (that is, remainder less than $0.05$). State clearly the polynomial you are using. First give the solution as a fraction, then use a calculator to write the solution with one decimal precision. Use a calculator to verify that the solution is correct.
 A: Just for your curiosity.
Following Dr. MV's answer, let us consider the more general case where you would like a remainder less that $\epsilon$. Then, as Dr. MV wrote,  you are looking for $n$ such that $$\frac{(2n)!}{4^n(n!)^2(2n-1)}<\epsilon\tag 1$$ For approximating the lhs, use Stirling approximation which is $$m!\sim \sqrt{2 \pi m} \left(\frac m e\right)^m$$ This makes $(1)$ much simpler since it becomes $$\frac{1}{\sqrt{\pi } \sqrt{n} (2 n-1)}<\epsilon\tag 2$$ and we can write  $$\frac{1}{2n\sqrt{\pi } \sqrt{n} }<\frac{1}{\sqrt{\pi } \sqrt{n} (2 n-1)}<\epsilon\tag 3$$ from which $$n >\frac 1{(2 \sqrt{\pi}\epsilon)^{2/3}}\approx \frac{0.43}{\epsilon ^{2/3}}\tag 4$$ Since $n$ is an integer, then select $$n=\left\lceil \frac{0.43}{\epsilon ^{2/3}}\right\rceil\tag 5$$
For your case where $\epsilon=0.05$, the above formula would lead to $n=4$. To $\epsilon=0.005$ would correspond $n=15$. As Dr. MV mentioned, it is possible that one less term could be resuired. For the last considered remainder, summing up to $n=15$ leads to $$\frac{95064943}{67108864}\approx 1.41658$$ but $n=14$ leads to $$\frac{47365319}{33554432}\approx 1.41160$$ which is also within the prescribed tolerance of $0.005$.
A: The Taylor series for $f(x)=\sqrt{1+x}$ is given by
$$f(x)=1+\sum_{n=1}^\infty \frac{(-1)^{n+1} (2n)!x^n}{4^n(n!)^2(2n-1)}$$
Inasmuch as the series is an alternating series, the error of the $N-1$'th partial sum for $x=1$ is bounded by $\frac{(2N)!}{4^N(N!)^2(2N-1)}$.  We want to find $N$ such that the error is less than $0.05$.  Thus, we set
$$\frac{(2N)!}{4^N(N!)^2(2N-1)}<0.05$$
which is satisfied for $N\ge 4$.   Therefore, the square root of $2$ can be approximated to one decimal place as 
$$\sqrt{2}\approx 1+\frac12-\frac18+\frac1{16}=\frac{23}{16}\approx 1.4375$$
However, we can check numerically that for $N=3$, the approximation 
$$\sqrt{1+x}\approx 1+\frac12-\frac18 \approx 1.375$$
is also within the prescribed tolerance of $0.05$.
