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 $[5,\infty]$.

  • $\begingroup$ $k$ 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
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    $\begingroup$ 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


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

  • $\begingroup$ 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
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    $\begingroup$ 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

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.

  • $\begingroup$ 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

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