Maclaurin series to evaluate $\sqrt{1.2}$

Question

How do I evaluate $\sqrt{1.2}$ using the Maclaurin series?

I tried using the general form of Maclaurin series but I can't get a correct answer, all the derivatives are 0.

Answer 1

Hint: use the function $y=\sqrt{1+x}$ to find the Maclaurin series and then let $x=0.2$

Answer 2

Hint. From the Maclaurin series expansion $$ f(x)=f(0)+\frac {f'(0)}{1!} x+ \frac{f''(0)}{2!} x^2+\frac{f^{(3)}(0)}{3!}x^3+ \cdots, $$ one gets, for $x$ near $0$, $$ f(x)=\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}+O(x^4) $$ then one may put $x=0.2$ in the above identity.

Answer 3

$$\sqrt{1.2}=\sqrt{1.21-0.01}$$

$$1.1\sqrt{1-\frac{0.01}{1.21}}$$

and now expand $\sqrt{1+x}$ with $x=\frac{-1}{121}$ to get a better precision.

Answer 4

According to the binomial expansion theorem:

$$(1+0.2)^{1/2}=1+\frac{0.2}2-\frac{0.2^2}8+\frac{0.2^3}{16}-\mathcal O(0.2^4)$$