Find p/q using Taylor's Theorem?

Question

For my real analysis class I have to answer the following question:

Find $p,q \in \mathbb{N}$ such that

$$\left|\sqrt{101}-\frac{p}{q}\right|<\frac{1}{1600000}.$$

In the previous question we had to prove that for every $x \in [0, \infty[$,

$$\left|\sqrt{1+x}-(1+\frac{x}{2}-\frac{x^2}{8})\right|\le\frac{1}{16}x^3.$$

For the previous question I used Taylor's Theorem and Remainder. I assume I have to reuse my result of the previous question, so I have tried that for this question too, but I just can't quite wrap my head around this one. It seems like an easy question, but I just can't seem to figure it out. Is there anyone here that is able to give me a hint?

Thanks in advance.

Answer 1

An alternate approach via number theory instead of Taylor series:

The continued fraction expansion of $\sqrt{101}$ is $[10;20,20,20, \dots]$.

The first three convergents are $$10, \frac{201}{20}, \frac{4030}{401}$$ Using the third convergent, $$\sqrt{101}-\frac{4030}{401} \approx 3.09 \times 10^{-7} < \frac{1}{1600000}$$ so $4030/401$ will do.

Answer 2

In fact, you don't need more terms for the Taylor polynomial. You have that $$\left|\sqrt{1+x} - \left(1 + \frac{x}{2} - \frac{x^2}{8}\right)\right| \leq \frac{x^3}{16}$$ Choosing $x = \frac{1}{100}$ gives you $$\left|\sqrt{1 + \frac{1}{100}} -\frac{80399}{80000}\right| \leq \frac{1}{16\;000\;000}$$ Multiplying the inequality by $10$ we get: $$10\left|\sqrt{1 + \frac{1}{100}} -\frac{80399}{80000}\right|=\left|\sqrt{100 + 1} -\frac{80399}{8000}\right| \leq \frac{1}{16\;000\;00}$$ So $(p,q) = (80399,8000)$

Answer 3

Let $f(x) = (100+x)^\frac 12$

We want to find $f(1)$ Find the Taylor series for this function.

$f(x) = 10 - \frac {1}{20} x + \frac {1}{8000} x^2 + \cdots$

(I estimated this off the top of my head. This may not be the correct Talor polynomial for our)

Use your theorems for the remainder of the Talor polynomial to determine how many terms you need for the remainder to be less than the given bound when $x\in [0,1]$