# Taylor remainder approximation of $\sqrt{1+2\sin(x)}$

I need help with the following question:

Using Taylor expansion we can find the approximation:

$$f(x)=\sqrt{1+2\sin(x)}\approx 1+\frac{2\sin(x)}{2}-\frac{(2\sin(x))^2}{8}$$

Approximate the reminder for $$|x|\le 0.001$$

My attempt:

I think I need to find $$R_2(0.001)$$. Using the remainder formula I get that:

$$R_2(0.001)=\bigg|\frac{f^{(3)}(x)}{3!}(0.001)^3\bigg|$$

I found that $$f^{(3)}(x)=\frac{\cos (x)(-\sin (x)+\cos ^2(x)+1))}{(2\sin (x)+1)^{\frac{5}{2}}}$$

I tried getting to a number from here but failed.

Any help will be appreciated.

• To estimate the remainder, consider $2\sin(x)<2x$ for $x>0$ Dec 10, 2017 at 17:13
• @Peter I tried that. but what about cos(x)? Dec 10, 2017 at 17:23
• What I mean : First estimate the value of $2\sin(x)$ and use this estimation for the expansion of $\sqrt{1+x}$. This does not give an "optimal" estimation, but it is much easier to determine. The argument is between $0$ and $0.002$, so it is enough to estimate the error of the $\sqrt{1+x}$-expandion at $x=0.002$. Dec 10, 2017 at 17:26

The expansion in the question is not the Taylor expansion of $f(x)$. It is the Taylor expansion of $g(t)=\sqrt{1+t}$ where the substitution $t=2\sin x$ is used. Thus, you need to take the Lagrange's remainder for $g$ $$g(t)=1+\frac{t}2-\frac{t^2}8+\underbrace{\frac{t^3}{16(1+\theta t)^{5/2}}}_{R_3},\qquad 0\le\theta\le 1$$ and do the substitution $$R_3=\frac{(2\sin x)^3}{16(1+\theta (2\sin x))^{5/2}}.$$ You should be able to estimate $|R_3|$ if you use e.g. $|\sin x|\le |x|\le 0.001$.