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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.

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  • $\begingroup$ To estimate the remainder, consider $2\sin(x)<2x$ for $x>0$ $\endgroup$ – Peter Dec 10 '17 at 17:13
  • $\begingroup$ @Peter I tried that. but what about cos(x)? $\endgroup$ – segevp Dec 10 '17 at 17:23
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    $\begingroup$ 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 $x=0.002$. $\endgroup$ – Peter Dec 10 '17 at 17:26
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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. $|\sin x|\le |x|\le 0.001$.

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