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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.

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  • $\begingroup$ Is there an $x$ that makes the right-hand sides match? $\endgroup$ Commented Apr 1, 2023 at 21:50
  • $\begingroup$ Yeah... $\sqrt[3]{\frac{1}{100000}}$... I don't really know what to do with that $\endgroup$
    – Hans
    Commented Apr 1, 2023 at 21:53
  • $\begingroup$ Simplify it? … Then think a minute. $\endgroup$ Commented Apr 1, 2023 at 21:58
  • $\begingroup$ Oh. Are you sure there aren’t 6 zeroes in the problem? There should be. $\endgroup$ Commented Apr 1, 2023 at 22:01
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    $\begingroup$ Wait... if I just solve it for 6 zeroes like you said, the found p and q will also work for 5 zeroes, so it doesn't even need to be a typo $\endgroup$
    – Hans
    Commented Apr 1, 2023 at 22:08

3 Answers 3

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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.

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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 $(p,q) = (80399,8000)$

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  • $\begingroup$ Oh, thanks! Well, that was embarassingly easy hahaha... $\endgroup$
    – Hans
    Commented Apr 2, 2023 at 4:54
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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 $x\in [0,1]$

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  • $\begingroup$ Thanks! How did I not even think of adding more terms to the Taylor polynomial within the left hand side... $\endgroup$
    – Hans
    Commented Apr 1, 2023 at 22:04

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