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I should find $\sqrt{1.1}$ using Taylor series of the function $\sqrt{x+1}$ in $x^{}_0=1$ with error smaller than $10^{-4}$. The first derivatives are $$f'(x)=\frac{1}{2\sqrt{x+1}}$$ $$f''(x)=\frac{-1}{4\sqrt{x+1}^ 3}$$ $$f'''(x)=\frac{3}{8\sqrt{x+1}^5}$$ Applying $x^{}_0$ we have: $$f(1)=\sqrt{2}$$ $$f'(1)=\frac{1}{2\sqrt{2}}$$ $$f''(1)=\frac{-1}{4\sqrt{2}^ 3}$$ $$f'''(1)=\frac{3}{8\sqrt{2}^5}$$ And we can build the Taylor polynomial $$T(x)=\sqrt2 + \frac{1}{2\sqrt{2}}(x+1)+\frac{-1}{2!·4\sqrt{2}^3}(x+1)^2+\frac{3}{3!·8\sqrt{2}^5}(x+1)^3+R(\xi)$$ Is everything right until here?
What I don't understand is how can I check that $R(\xi) > 10^{-4}$

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    https://en.wikipedia.org/wiki/Taylor%27s_theorem#Estimates_for_the_remainder – 2017-01-27
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    you need to find f(0), f'(0), f''(0) - which then involves no square roots - then consider f(x + h) where x = 0 and h = 0.1 – 2017-01-27
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    with your working, h would be -0.9 - which is not ideal - the '1' in 'x+1' already gives you the 1 you need for 1.1 – 2017-01-27

3 Answers 3

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$f(x) = \sqrt{1 + x}$

$f'(x) = \frac{1}{2\sqrt{1+x}}$

$f(0) = 1$

$f'(0) = 1/2$

$f(x + h) = f(x) + h f'(x) +.....$

$\sqrt{1.1} = f(.1) = f(0 + .1) \approx1 + 0.1 / 2 = 1.05$

you got confused by the fact they are using x + 1 in the function - i made a start for you, with two terms it is getting closer to the answer

do you see the important part for you? $\sqrt{1.1} = \sqrt(1 + .1) = f(.1)$

since it is easy to workout f(0), f'(0),f''(0) without any square -roots, you then need to centre it around 0 - i.e. f(0 + .1)

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The Taylor Theorem tells us that the estimation error after $n$ terms is given by $f'(c)\frac{(x-a)^{n+1}}{(n+1)!}$, for som $c \in (x-a)$. Now you should be able to find an upper bound on the derivative in that interval, which should give you an upper bound on the error.

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Hint:$$f(x_0+h)=f(x_0)+f'(x_0)h+f''(x_0).\dfrac{h^2}{2!}+f^3(x_0).\dfrac{h^3}{3!}+f^4(x_0).\dfrac{h^4}{4!}+o(h^5)\\h=0.1 \\ \to f(x_0+h)=f(x_0)+f'(x_0)(0.1)+f''(x_0).\dfrac{(0.1)^2}{2!}+f^3(x_0).\dfrac{(0.1)^3}{3!}+f^4(x_0).\dfrac{(0.1)^4}{4!}+o((0.1)^5)$$ so in your case $f^4(x_0).h^4$ is in order of $10^{-4}$ and the next terms are in order of $10^{-5},10^{-6},10^{-6},...$ so , it is obvious to do it before term $f^5(x_0).\dfrac{h^5}{5!}=f^5(1).\dfrac{(10^{-5})}{5!}$