0
$\begingroup$

I was doing some numerical integration when I figured the function I was dealing with (i.e., the function I was integrating) evaluated to big numbers on a tiny portion of the interval (over which I was integrating) and to extremely small numbers on the rest of it. So I decided to find an appropriate interval to perform the integration, with the intention of making the calculations more precise and less costly.

After some rearranging, the problem I ended up with was:

Find a function $f \colon (0, 1) \times (0, 1) \to (0, 1)$ such that $$\begin{cases}x \in (0, 1) \\x^{\alpha}(1-x) > C\end{cases} \;\,\Rightarrow\; x < f(\alpha, \,C) \;\;\;\;\;\;\;\;\forall\;\;\alpha, \, C \in (0, 1).$$

Of course there are many functions that fulfill the condition (for example, $f(\alpha, \, C) \equiv 1$), but the idea is to build a function that makes the result as sharp as possible.

In the context I want to apply this, $\alpha$ is a small number, whereas $C$ is big (close to $1$). If it's helpful, one can make assumptions on that direction.

Any ideas?

Improve this question
$\endgroup$
5
  • $\begingroup$ Isn't $f(\alpha,C)=1-C$ enough for you? $\endgroup$
    – Ilya Bogdanov
    Nov 20, 2020 at 9:26
  • $\begingroup$ Unfortunately, it isn't, @IlyaBogdanov. I got a better result applying AM-GM to $x^{\alpha}(1-x)$, but not even that seems to be good enough. $\endgroup$
    – Valentino
    Nov 20, 2020 at 13:17
  • $\begingroup$ You are not able to formulate an explicit requirement on $f$? I mean, like $f$ must minimize something and maximize some other thing or something like that? $\endgroup$
    – მამუკა ჯიბლაძე
    Nov 23, 2020 at 19:10
  • $\begingroup$ In any case there is no $x$ with $x^\alpha(1-x)>C$ for $C>\frac{\alpha^\alpha}{(1+\alpha)^{1+\alpha}}$. $\endgroup$
    – მამუკა ჯიბლაძე
    Nov 23, 2020 at 19:15
  • $\begingroup$ Using A220883 one may revert the series for $y(1-y)^\alpha$ where $y=1-x$, obtaining a version for $f$ as follows:$$f(\alpha,C)=1-C-\alpha C^2-...-\prod_{k=0}^{n-2}(k+n\alpha)\frac{C^n}{n!}-...$$ $\endgroup$
    – მამუკა ჯიბლაძე
    Nov 23, 2020 at 22:27

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.