Estimating the value of $\sqrt{5}$ using Taylor Polynomial

Question

I'm trying to estimate the value of $\sqrt{5}$ using the Taylor Polynomial with the accuracy of $\frac{1}{10^3}$.

What I did was to calculate the third derivative of $\sqrt{x}$, which is $\frac{3}{8x^2\sqrt{x}}$. I calculate the third derivative, because I want accuracy on 3 digits.

Next I calculate the degree $n$ in Taylor polynomial for which I need to calculate to get the required accuracy: $|\frac{8x^2*\sqrt{x}}{3!}|\Leftrightarrow \frac{1}{16x^2*\sqrt{x}}\leq\frac{1}{10^3} \Rightarrow n \geq 6 $.

So I have to calculate Taylor polynomial for $\sqrt{x+1}$ of degree $6$ and plug $4$ for $x$.

Am I doing it correctly?

Thanks!

Answer 1

It is better to center at values such as $x_0=4.84=2.2^2$, so that the error decays fast like the powers of $h=x-x_0=0.16$. Note that the series will be an alternating series, meaning that the error can be estimated by the first omitted term.

Answer 2

No, it is not correct: the Taylor series of $\sqrt{1+x}$ converges only for $|x|<1$. But you can write $$\sqrt 5=\sqrt{4\Bigl(1+\frac14\Bigr)}=2\sqrt{1+\frac14}$$ Can you continue?

Answer 3

There are various approaches using Taylor series, rescaling as Bernard did, choice of expansion point, use of the mean value theorem for estimates, etc.

The following is based on the fact that the Taylor series for $x \mapsto \sqrt{x+k}$ is alternating (as Timur notes).

Let $f(x) = \sqrt{4+x}$. Note that the Taylor series converges for $|x|<4$, and is alternating.

In particular, if $P_n$ is the Taylor polynomial of degree $n$, then $f(x) < P_n(x)$ if $n$ is odd and $f(x) > P_n(x)$ if $n$ is even.

In particular, if $|P_{n+1}(x) -P_n(x)| < \epsilon$, then $|f(x)-P_n(x)| < \epsilon$.

If we set $x=1$, then the first six terms of the Taylor expansion are $2,{1 \over 4}, -{1 \over 64}, {1 \over 512}, -{5 \over 16384}$,$ {7 \over 131072}$ and since the last two terms differ by less than ${1 \over 10^3}$ we see that $|f(1)-P_4(1)| < {1 \over 10^3}$.

Hence $f(1) \approx P_4(1) = {36635 \over 16384} \approx 2.2360$, $\sqrt{5} \approx 2.2361$.