Taylor polynomials too high or too low? I have used the second order Taylor polynomial for the square root of x about x=100 to approximate the square root of 101, to which I got 80,399/8,000 = 10.049875.
Then I calculated the error and got 1/1,600,000 = 0.000,000,625.
Neither of those were any problem.
However, how can I know if the answer I got from using Taylor is actually higher or lower than the actual value of the square root of 101?
When I use the calculator I get the answer 10.04987562, so I know my answer from using Taylor is too low, but how can I know?
I have to explain why it is either higher or lower without using a calculator.
Also, if someone wants to fix my post to make it look better, that's fine, I haven't quite figured out to write math stuff on a computer yet..
Thanks :)
 A: For any $\alpha \in \mathbb{R}$ and $|x| < 1$, we have the Taylor series expansion:
$$(1+x)^\alpha = \sum_{n=0}^\infty \binom{\alpha}{n} x^n
= 1 + \alpha x + \frac{\alpha(\alpha-1)}{2!} x^2
+ \frac{\alpha(\alpha-1)(\alpha-2)}{3!} x^3 + \cdots
$$
Let's concentrate on the case $\alpha, x \in (0,1)$. 
If one look at the sign of the coefficients of $x^n$,
one find 
$$\binom{\alpha}{n}
\begin{cases}
> 0, & n = 0 \text{ or odd }\\
< 0, & n \ne 0 \text{ and even }
\end{cases}
$$
Furthermore, the magnitude $\displaystyle\;\left|\binom{\alpha}{n}\right|\;$ is monotonic decreasing
in $n$. Aside from the constant term, this is an alternating series and the
limit will be sandwiched by successive partial sums.
More precisely, let $\displaystyle\;S_p = \sum_{n=0}^p \binom{\alpha}{n} x^n$, we have
$$1 + \alpha x + \frac{\alpha(\alpha-1)}{2} x^2 = S_2 < S_4  < \cdots <
\sum_{n=0}^\infty \binom{\alpha}{n} x^n < \cdots 
< S_5 < S_3 < S_1 = 1 + \alpha x$$
Now take $\alpha = \frac12$. This implies
the $2^{nd}$, $4^{th}$, $6^{th}$ even order approximations of square root will be too low while $1^{st}, 3^{rd}, 5^{th}$ odd order approximations will be too high.
A: You want to compare $\frac{80399}{8000}$ and $\sqrt{101}$, but you don't know the exact value of $\sqrt{101}$.  But you can square both sides, and because both are non-negative this won't change the result of the comparison.  Now you can compare $\left(\frac{80399}{8000}\right)^2$ and $101$, which should be much easier.
