How one makes a nice approximation using Taylor series Approximations of expressions like $2^{0.6}$ , or $3^{0.7}$ . For firrst one i know binomial expansion can help with x very less we need but is it better to do taylor expansion around √2 than binomially expansion if we just need upto three decimal places ? ( I know the log values needed for taylor expansion so basically is it better to do taylor expansion around √2 or some better value , or binomial is best , if we meed upto 3 decimal places)
 A: Newton's binomial series is:
$$(1+x)^{\alpha}=1+\alpha x + \frac12 \alpha (\alpha-1)x^2+ \frac16 \alpha (\alpha-1)(\alpha-2)x^3+... \tag{1}$$
which is a Taylor expansion valid for $|x|<1$.
Therefore, one could think that $2^{0.6}=(1+x)^{\alpha}$ with $x=1$ is impossible, but "everything can happen on the circle of convergence $|z|=1$ and it happens that for $z=1$, relationship (1) gives a convergent series:
$$2^{0.6}=(1+1)^{0.6}=1+0.6 - \frac12  0.6 \times 0.4 + \frac16 0.6 \times 0.4 \times 1.4 +... \tag{1}$$
But its convergence is very slow as one can expect on the circle of convergence : one need 23 terms to have an error less thatn $10^{-3}$.
Two alternative methods:
a) Compute its inverse $1/2^{0.6}$ in the following way and inverse the result by hand computation.
$$\left(\frac12\right)^{0.6}=\left(1-\frac12\right)^{0.6}=...$$
for which we can use expansion (1) with $x=-\frac12$.
b) One could also compute in a separate way
$$\begin{cases}\left(\dfrac43\right)^{0.6}&=&\left(1+\dfrac13\right)^{0.6}&=&1+0.6 \times \dfrac13 - \frac12 0.6 \times 0.4 \left(\dfrac13\right)^2+...\\
\left(\dfrac23\right)^{0.6}&=&\left(1-\dfrac13\right)^{0.6}&=&1-0.6 \times \dfrac13 - \frac12 0.6 \times 0.4 \left(\dfrac13\right)^2+...
\end{cases}$$
What for? Plainly because the quotient of these results gives $2^{0.6}$...
(Both series have a rather good convergence speed. Moreover, they have the same terms, either with the same sign, or its opposite).
A: Write $2^{0.6}$ as $\exp(0.6 \ln 2)$ where $\exp(x) = e^x$, where you know $\ln 2$ to three decimal places.
Then since $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + O(x^5)$, the maximum Lagrange remainder is the next term, thus $\frac{(0.6 \ln 2)^n}{n!} > 0.001$ for any power of $n$. Some trial and error gives $n  > 4$, so we need the $x^5$ term as well:
$$2^{0.6} \approx1 + 0.6 \ln 2 + \frac{1}{2!}(0.6 \ln 2)^2 + \frac{1}{3!}(0.6 \ln 2)^3 + \frac{1}{4!}(0.6 \ln 2)^4 + \frac{1}{5!}(0.6 \ln 2)^5$$
and the approximation with $\ln 2 \approx 0.693$ gives $1.51558$, whereas the actual value of $2^{0.6}$ is around $1.51572$, an absolute error of around $1.4 \times 10^{-4}$, and a relative error of $9.3 \times 10^{-5}$.
