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I have to approximate $\sqrt2$ using Taylor expansion with an error $<10^{-2}$.

I noticed that I can do MacLaurin expansion of $\sqrt{x+1}$ then put $x=1$

So: $$\sqrt{x+1}=1 + \dfrac{x}{2} - \dfrac{x^2}{8} + \dfrac{x^3}{16} + {{\frac1{2}}\choose{n+1}}x^{n+1}(1+\xi)^{-\frac1{2}-n}$$

I have to find the order of the polynomial at which

$\sqrt2 -($the value I found with the polynomial$)<10^{-2}$

I can check the value of the polynomial in $x=1$ order by order until I find that

$\sqrt2 -($the value I found with Taylor$)<10^{-2}$

Or is there a faster way to find the desired order?

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The series$\sum_{k=0}^\infty\binom{1/2}{k}$ is alternating (which is the reason why it converges, by the way). For such series this estimate holds $$\left\lvert\sum_{k=0}^\infty a_k-\sum_{h=0}^n a_h\right\rvert\le \rvert a_{n+1}\rvert$$

See, for instance, the proof of Leibniz criterion.

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The Taylor series of the function $f(x):=(1+x)^{1/2}$ just barely converges for $x:=1$. Evaluate the Taylor expansion of $g(x):=(1+x)^{-1/2}$ at $x:=-{1\over2}$ instead.

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