I need to find the kinds of Taylor polynomial for $\sqrt x$ at the point $a = 1$, $x$ lies in $<1,2>$.
I calculated the second derivative: $-\frac{1}{4}x^\frac{-3}{2}$
and the third: $\frac{3}{8}x^\frac{-5}{2}$
I calculated Taylor's polynomial:
$$ {1+\frac{1}{2}(x-1)-\frac{1}{8}(x-1)^2}$$
I need advice on estimating the error.
I calculated something like:
$$R \leq \frac{3/(8\cdot2^{5/2})}{3!}\cdot{(x-1)^3}$$
But I'm not sure if that's right.
Best Answer
As another contributor has commented, you can use the binomial theorem by writing $$\sqrt x = (1+(x-1))^{1/2}.$$ You might also note that for $1 \leqslant x \leqslant 2$ the infinite series converges and the terms decrease with alternating sign so the error is always less or equal to the the first neglected term. Your attempt using derivatives is also perfectly legitimate and should give the same result (but note you have mis-typed the powers of $x$ in your calculation of the derivatives).
So you have obtained $$ \sqrt{x} = 1 + \frac{1}{2}(x-1)-\frac{1}{8}(x-1)^2 + \frac{1}{16}(x-1)^3+\cdots. $$ whence truncating after the $(x-1)^2$ term, $$\lvert R \rvert \leqslant \frac{1}{16}(x-1)^3 \leqslant \frac{1}{16} \quad (x \in [1,2]).$$ If you are interested in the error over the full range ($0 < x \leqslant 2$) then you will need to use one of the forms of the Taylor series error, for example the Lagrange estimate, which should give the same conclusion for $x \geqslant 1$.