Find Taylor series polynomial that gives uniform bound on error The problem comes in two parts:


*

*Find an $\epsilon > 0$ such that for every $x\in[0,1]$ $$\left\lvert \sqrt{x}-\sqrt{x+\epsilon}\right\rvert  \le \frac{1}{200}$$


We can show that $\left\lvert \sqrt{x}-\sqrt{x+\epsilon}\right\rvert $ decreases as $x$ increases to 1, so it's enough to consider the case where $x=0$ and we see that we can use $\epsilon = 1/200^2$.


*(The part I didn't get or at least am not satisfied with.) Find a polynomial $P(x)$ such that for every $x\in[0,1]$ $$\left\lvert \sqrt{x}-P(x)\right\rvert \le \frac{1}{100}$$


Hint: Use the power series expansion of $\sqrt{x+\epsilon}$ around $x=1$.
Here's my work so far
First note that $$\left\lvert \sqrt{x}-P(x)\right\rvert  \le \left\lvert \sqrt{x}-\sqrt{x+\epsilon}\right\rvert  + \left\lvert \sqrt{x+\epsilon} - P(x)\right\rvert $$
If we can find a P(x) such that $\left\lvert \sqrt{x+\epsilon} - P(x)\right\rvert  \le 1/200$, then we can use part 1 and be done. So then I tried using the hint:
$$\sqrt{x+\epsilon} = \sqrt{1+\epsilon} + \frac{1}{2}(1+\epsilon)^{-1/2}(x-1)- \frac{1}{8}(1+\epsilon)^{-3/2}(x-1)^2 + \ldots$$
The tricky part here is that we can't get a nice uniform bound on $f^{(k)}(\xi), \xi\in[0,1]$, where $f(x) = \sqrt{x+\epsilon}$, because around $x=0$ it blows up. Note for the remainder term we have
$$\left\lvert \frac{f^{k}(\xi)}{k!}(x-1)^k\right\rvert \le \left\lvert \frac{f^{k}(\xi)}{k!}\right\rvert $$
In other words, the remainder is largest in abs. value when $x=0$, so we can just focus on the Taylor series expansion after substituting $x=0$. The expansion becomes
$$\sqrt{1+\epsilon} - \frac{1}{2}(1+\epsilon)^{-1/2} - \frac{1}{8}(1+\epsilon)^{-3/2} - \ldots$$
Note that $$\frac{1}{(1+\epsilon)^{(2k-1)/2}}\le 1$$
so
$$\sqrt{1+\epsilon} - \frac{1}{2}(1+\epsilon)^{-1/2} - \frac{1}{8}(1+\epsilon)^{-3/2} - \ldots \ge \sqrt{1+\epsilon} - \frac{1}{2}-\frac{1}{8} - \ldots$$
and the coefficients are monotonically decreasing. The form of the coefficients is $$a_k=\frac{1}{2^k k!}\prod_{n=1}^{k-1} (2n-1)$$ Recall that we should be converging to 0, so each new term we add moves us closer to 0.
That's as far as I've gotten. I don't know if that was the right route, and also how to proceed from here. Maybe I've just stared at this for too long, but either way I'd like to get some feedback. Maybe there's a better way to go about this.
Thanks!
Update
I tried out my solution with Mathematica, and it seems to work. The problem is that I had to expand the Taylor series to ~12800th order. This question should be answerable without Mathematica (it is taken from an old exam).
 A: Your approach looks good and the only missing piece is to bound the remainder of the Taylor polynomial to get an explict bound on $N$ (the degree of the polynomial) such that the bound $|\sqrt{x+\epsilon} - P(x)| < \frac{1}{200}$ holds. This is what I will show here.

The $N$th derivative of $\sqrt{x}$ is given by
$$\frac{d^n}{dx^n}\sqrt{x} = x^{\frac{1}{2}-n}\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\cdots \left(-\frac{2n-3}{2}\right) = x^{\frac{1}{2}-n}\frac{(2n)!}{n!4^n}\frac{(-1)^{n-1}}{2n-1}$$
If we let $P(x)$ be the $N$th order Taylor polynomial of $\sqrt{x+\epsilon}$ about $x=1$ then using the result above we get 
$$\sqrt{x+\epsilon} - P(x) = \sqrt{1+\epsilon}\sum_{n=N}^\infty {2n\choose n}\frac{(-1)^{n+1}}{2n-1}\left(\frac{x-1}{4(1+\epsilon)}\right)^n$$
Since $\epsilon > 0$ and $x\in[0,1]$ we therefore have
$$|\sqrt{x+\epsilon} - P(x)| < \sqrt{1+\epsilon}\sum_{n=N}^\infty {2n\choose n}\frac{1}{(2n-1)4^n}$$
From Stirlings approximation it follows that
$${2n \choose n}\frac{1}{4^n} = \frac{1}{\sqrt{\pi n}}\left(1 - \frac{1}{8n} + O(n^{-2})\right) < \frac{1}{\sqrt{\pi n}}$$
so the sum above is bounded by
$$|\sqrt{x+\epsilon} - P(x)| < \frac{\sqrt{1+\epsilon}}{\sqrt{\pi}}\sum_{n=N}^\infty \frac{1}{(2n-1)\sqrt{n}} < \frac{\sqrt{1+\epsilon}}{\sqrt{\pi}}\sum_{n=N-1}^\infty \frac{1}{2n^{3/2}}$$
Since $n^{-3/2}$ is monotonically decreasing we have 
$$\sum_{n=N-1}^\infty \frac{1}{2n^{3/2}} < \int_{N-2}^\infty\frac{{\rm d}x}{2x^{3/2}} = \frac{1}{\sqrt{N-2}}$$
which gives us
$$|\sqrt{x+\epsilon} - P(x)| < \left(\frac{1+\epsilon}{\pi(N-2)}\right)^{1/2}$$
which is less than $\frac{1}{200}$ when $N > \frac{200^2(1+\epsilon)}{\pi}+2 \approx 12735$. This is a fairly large value of $N$ - and such an approximation would not be very useful in practice - however it does solve the problem as stated and the approach is in line with the given hint. This is largely an effect of us trying to approximating a function that has a very non-polynomial behavior at a point in the interval (in this case $x=0$). There should however exist lower order polynomials that does the job here so I guess another take-home message here is that Taylor polynomials is not always the best ones to use to approximate a function over a larger interval.
A: Take any $k\geq \sqrt 2$ and  $e= k^{-2}.\;$ For $x\in [0,1] $  let $y= x+e-1.$  Then $$\sqrt {x+e} =\sqrt {1+y} .$$ And for all $x\in [0,1]$ we have $$|\sqrt x-\sqrt{1+y}|=|\sqrt x -\sqrt {x+e}|\leq 1/k.$$ $$\text {Now we have }\quad  e-1\leq y\leq e$$ $$\text {so }\quad  |y|\leq 1-e<1 \text { because } 0<e\leq 1/2.$$ $$ \text {Since } |y|<1 \text  {  we have }\quad  \sqrt {1+y}=\sum_{n=0}^{\infty}y^n \binom {1/2}{n}.$$ Since $|\binom {1/2}{n}|\leq 1$ (by induction on  $n$), an upper bound for the absolute value of the remainder of the infinite series, after the term of degree $n,$ is $$U_{n.k}=\sum_{j=n+1}^{\infty}|y|^j.$$ $$\text { We have }\quad  U_{n,k}=|y|^{-n-1}/(1-|y|)\leq |y|^{-n-1}/e\leq (1-e)^{-n-1}/e.$$ The partial sum of the series ,up to the term of degree $n,$ is a polynomial $P_{n,k}(y) ,$ which is a polynomial $Q_{n,k}(x).$ If $n$ is large enough that $(1-e)^{-n-1}/e<1/k,$ then for all $x\in [0,1]$ we have $|Q_{n.k}(x)-\sqrt {x+e}|<1/k$ and  also  $|\sqrt {x+e}-\sqrt x|\leq 1/k , $    so  $$|\sqrt x- Q_{n,k}(x)|\leq 2/k.$$ And $k$ can be arbitrarily large.
Remark:$\binom {1/2}{0}=1,$ and $\binom {1/2}{1}=1/2, $ and for $n\geq 1:\;$  $\;\binom {1/2}{n+1}=\binom {1/2}{n}(1/2-n)/(n+1).$ 
