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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 :)

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  • $\begingroup$ It basically just goes down to inspecting the taylor polynomial and trying to understand what partial sums are above and what partial sums are below the final value $\endgroup$ – RGS Nov 19 '16 at 23:49
  • $\begingroup$ Have you ever graphed a Taylor expansion? They are quite...intuitive. $\endgroup$ – Simply Beautiful Art Nov 20 '16 at 1:27
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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.

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  • $\begingroup$ Okay, I think I got it. Thanks! $\endgroup$ – Mariatorg Nov 20 '16 at 0:28
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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.

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