0
$\begingroup$

I have the following equation:

$$ n^2 = \frac{49000000000000\epsilon^2-2814154000000\epsilon+40405422121}{1000000000000\epsilon^2+5656854000000\epsilon-705671} $$

where $n$ is a positive integer, and $\epsilon$ is a positive real number with $0 \le \epsilon < .0000001247461938191918$.

I’m trying to find an upper limit on $n$. Am I justified in simply plugging the maximum value for $\epsilon$, finding

$$ n^2 < 1919254381004.712 \implies n \le 1385371,$$

or are there subtleties I need to consider?

$\endgroup$
6
  • 2
    $\begingroup$ Can you see what might happen if the denominator was zero (or very near it) for some value of epsilon between 0 and .000000012347... ? $\endgroup$ Commented Nov 18 at 0:18
  • $\begingroup$ @JohnHughes: Good point. Doing some very quick “needle-drop” calculations, it looks like the zero might happen somewhere between $.9999999701$ and $.9999999702$ of the max value. Looks like maybe it goes asymptotic there, which would mean there is no “max value” for $n$…? $\endgroup$ Commented Nov 18 at 0:34
  • $\begingroup$ @JohnHughes: Setting the denominator equal to zero and solving for $\epsilon$ gives [the positive root] $\epsilon \simeq 0.000000124746193177998$. $\endgroup$ Commented Nov 18 at 0:42
  • 1
    $\begingroup$ Summarizing, the right-hand side is unbounded on $[0, 0..0000001247461938191918]$ (so has neither maximum nor minimum on that interval), so $n$ is unbounded. $\endgroup$ Commented Nov 18 at 0:52
  • $\begingroup$ @EricTowers: <sigh> Well, please make that an answer, and I’ll accept it. $\endgroup$ Commented Nov 18 at 0:55

1 Answer 1

Reset to default
3
$\begingroup$

The denominator of the right-hand side, $$ 1000000000000\epsilon^2+5656854000000\epsilon-705671 $$ has roots at \begin{align*} \epsilon_1 &= -2 \sqrt{2} - \frac{1\,828\,427}{1\,000\,000} \\ &= -5.656\,854\,124\,746{\dots} \\ \epsilon_2 &= 2 \sqrt{2} - \frac{1\,828\,427}{1\,000\,000} \\ &= 1.247\,461\,900\,976{\dots} \times 10^{-7} \end{align*}

The permissible range of $\epsilon$ is $[0,1.247\,461\,938\,191\,918{\dots} \times 10^{-7})$, which contains $\epsilon_2$. (The minimum of the denominator occurs at $\epsilon = \frac{2\,828\,427}{1\,000\,000}$, so the denominator changes sign at $\epsilon_2$).

At $\epsilon_2$, the numerator, \begin{align*} &49000000000000\epsilon_2^2 − 2814154000000 \epsilon_2 + 40405422121 \\ &= 792 \times 10^{12} - 560 \sqrt{2} \times 10^{12} \\ &= 4.04{\dots} \times 10^{10} \end{align*} is not zero, so the right-hand side exhibits arbitrarily large values in the neighborhood of $\epsilon = \epsilon_2$. Since the right-hand side is unbounded, $n$ is unbounded, and has no maximum (or supremum).

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.