2
$\begingroup$

I have obtained as the expression for some quantity the following gargantuan formula:

$$ \frac{k^8 + 3k^7 + 8k^6 + 3k^5 - 16k^4 - 32k^3 + 63k^2 - 34k + 6}{k^6 + 3k^5 + 6k^4 - 24k^2 + 21k - 5}$$.

What I really need is a (very) good lower bound on it, that will hopefully be a more manageable expression.

Is there a systematic way of finding such bounds?

The bound needs to be valid only on $[5,\infty]$.

$\endgroup$
  • $\begingroup$ $k$ is in $\mathbb{R}$? $\endgroup$ – Joseph O'Rourke Oct 24 '13 at 11:29
  • $\begingroup$ this expression ranges from $-\infty$ to $\infty$ (for real $k$, and also for real positive $k$), so there is no lower bound. $\endgroup$ – Carlo Beenakker Oct 24 '13 at 11:31
  • $\begingroup$ @JosephO'Rourke Yes. $\endgroup$ – Felix Goldberg Oct 24 '13 at 11:43
  • $\begingroup$ @CarloBeenakker Sure, that's why I specified in my P.S. that I need actually a "local bound". I'll edit to make this point more prominent. Thanks. $\endgroup$ – Felix Goldberg Oct 24 '13 at 11:44
  • 2
    $\begingroup$ As for a systematic way, your question reminds me of the problem of finding surrogate functions. A surrogate function is a function approximating the original one but cheaper to compute. $\endgroup$ – Waldemar Oct 24 '13 at 12:22
5
$\begingroup$

$$k^2+1.98-2.8/k$$

lower bound, error $<0.02$ for all $k>5$

$\endgroup$
  • $\begingroup$ This looks great, but now I am curious - how did you obtain this result? $\endgroup$ – Felix Goldberg Oct 24 '13 at 13:42
  • $\begingroup$ from the large-$k$ expansion $\endgroup$ – Carlo Beenakker Oct 24 '13 at 13:46
  • $\begingroup$ Do you mean the Taylor series? Or something else? $\endgroup$ – Felix Goldberg Oct 24 '13 at 13:49
  • 1
    $\begingroup$ Taylor series around $1/k=0$ (it works because 1/5 is small enough respect to unity) $\endgroup$ – Carlo Beenakker Oct 24 '13 at 14:40
2
$\begingroup$

The function blows up at approximately $-2.6$ and $+0.5$ and $+0.7$, and otherwise looks something like a parabola. It is unclear what is meant by a "lower bound" for such a function.
     FelixPlot
     FelxiPlot01

$\endgroup$
  • $\begingroup$ Suppose we zoom in on $k\geq 5$ - can you suggest an adequate bound in that case? $\endgroup$ – Felix Goldberg Oct 24 '13 at 11:45
  • $\begingroup$ @FelixGoldberg: Well, then the lower bound is just the value of the expression at $k=5$, which is $\frac{373393}{14125} \approx 26.4$. $\endgroup$ – Joseph O'Rourke Oct 24 '13 at 11:55
  • $\begingroup$ I want a bound which is also a function of $k$... $\endgroup$ – Felix Goldberg Oct 24 '13 at 12:07
  • $\begingroup$ @FelixGoldberg: I'm sorry, Felix, we are talking past one another. Your function is monotonically increasing $k \ge 5$. I am not sure what it means to express this as a function of $k$. $\endgroup$ – Joseph O'Rourke Oct 24 '13 at 12:33
  • $\begingroup$ Now I see, from Waldemar's comment and Carlo's answer. $\endgroup$ – Joseph O'Rourke Oct 24 '13 at 13:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.