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.