Calculate $\sqrt{5}$ using taylor series to the accuracy of 3 digit after the point $(0.5*10^{-3})$. I need to calculate $\sqrt{5}$ using taylor series to the accuracy of 3 digit after the point $(0.5*10^{-3})$.
I defined : $$f(x)=\sqrt{x}$$
Therefore :
$$f'(x)=\frac{1}{2\sqrt{x}}$$
$$f''(x)=-\frac{1}{4\sqrt{x^3}}$$
$$f'''(x)=\frac{3}{8\sqrt{x^5}}$$
for $x=5$ :
$$|R_2(5)|=\left|\frac{f^{(3)}(c)}{3!}\right|=\left|\frac{\frac{1}{8\sqrt{c^5}}}{3!}\right|\left|\frac{1}{3!*8\sqrt{c^5}}\right|$$
I got stuck now, how can I evaluate: $$???\geq|{3!*8\sqrt{c^5}}|$$
Also, I wonder how many derivatives I need to calculate inorder to reach to the required accuracy, is there a way to find out?
I just took an arbitrary guess to the third derivative.
Any help will be appreciated.
 A: Some comments:


*

*To make an approximation using Taylor series, you're going to want to
choose a point for your expansion.  I would suggest take the Taylor
expansion around $x = 4$, since the square roots will be easy to
take.  This means you have to evaluate the Taylor coefficients directly in the expansion $f(4) + f'(4)(x-4) + \frac{f''(4)}{2}(x-4)^2 + \dots$, before you can plug in $x=5$ to get your estimate.

*Your formula for the third derivative is wrong.  It might be easier
to see why if you write $f''(x)$ in the form $(-1/4)x^{-3/2}$.

*When you bound the error, remember that taking reciprocals of
positive quantities reverses inequalities.  Thus $|1/\sqrt{c^5}| \le
   1/A$ is equivalent to $|\sqrt{c^5}| \ge A$.

*You need to find out which $n$ is large enough such that the $n$-th error is suitably small.One way to find out how many terms are needed for a specified accuracy is to find an expression for a reasonable upper bound on the error as a function of $n$, but guess and check is also feasible as an approach.
$n$-th remainder term as a function of $n$.

*It looks like you're using the Lagrange remainder formula for the error, but without the usual $(x-a)^3$ term.
A: The Taylor series of $x\mapsto\sqrt{1+x}$ is given by
$$\sqrt{1+x}=1+{x\over2}-{x^2\over8}+{x^3\over16}+\ldots\qquad\bigl(|x|<1\bigr),$$
and is alternating when $0<x<1$. Therefore the error after truncating is between $0$ and the first neglected term. Our aim now is to use this with a suitable small $x>0$. We write
$$\sqrt{5}=\sqrt{4.84+0.16}=\sqrt{4.84}\>\sqrt{1+{0.16\over 4.84}}\ .$$
This gives
$$\sqrt{5}=2.2\sqrt{1+x}=2.2\>\left(1+{x\over2}-{x^2\over8}+R\right)$$
whereby
$$0<x:={0.16\over4.84}<0.04$$
and
$$0<R<{0.04^3\over16}=0.000004\ .$$
It follows that
$$\sqrt{5}=2.2\>\left(1+{x\over2}-{x^2\over8}\right)=2.236063110\ldots$$
with an error $<0.00001$. The first decimal places of $\sqrt{5}$ are $2.236067977$.
A: the Taylor series of $y=\sqrt{x+4}$ is
$$y=2+\frac{x}4{}-\frac{x^2}{64}+\frac{x^3}{512}-\frac{5x^4}{16384}+....$$
to get $\sqrt{5}$ put $x=1$ 
