Proving one function is greater than other?

2 次查看（过去 30 天）
Fatima Majeed
Fatima Majeed 2024-6-9
评论： Matt J 2024-6-10
I want to see where is this inequality true
where x in (e^e,∞).
  6 个评论
显示 4更早的评论
Matt J
Matt J 2024-6-9
The notation is ambiguous. We must know whether is to be interpreted as or as .

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Matt J
Matt J 2024-6-10
编辑：Matt J 2024-6-10
Make the change of variables and rearrange the inequality as,
Since is a convex function on such that >0 and , it readily follows that for all . QED.
  10 个评论
显示 8更早的评论
Matt J
Matt J 2024-6-10
I want to thank both of you for your efforts.
You're welcome, but please Accept-click the answer if it has resolved your question.

请先登录，再进行评论。

更多回答（1 个）

Torsten
Torsten 2024-6-9
编辑：Torsten 2024-6-10
Ran in:
syms x y
f = 53.989/21.233 * (log(log(x))).^(4/3)./(log(x)).^(1/3);
%x is solution where f starts getting greater than 1
xstart = vpasolve(f==1,x,5)
xstart = 
5.8933180867157239497727768797558
log(log(xstart))/(21.233*log(xstart))
ans = 
0.01521725041175722892398806828762
1/(53.989*log(xstart)^(2/3)*log(log(xstart))^(1/3))
ans = 
0.01521725041175722892398806828762
ftrans = subs(f,x,exp(exp(y)));
%exp(exp(y)) is solution where f ends being greater than 1
yend = vpasolve(ftrans==1,y,13)
yend = 
13.085773977624348668463856655909
log(log(exp(exp(yend))))/(21.233*log(exp(exp(yend))))
ans = 
0.0000012785232351771168083804508742678
1/(53.989*log(exp(exp(yend)))^(2/3)*log(log(exp(exp(yend))))^(1/3))
ans = 
0.0000012785232351771168083804508742678
  3 个评论
显示 1更早的评论
Sam Chak
Sam Chak 2024-6-10
Hi Fatima, can you provide the paper or link, or a cropped section for study purposes? Sounds like an interesting problem.
While I know what a log function is, I never use log(log(x)) or exp(exp(x)) in this approach.
Fatima Majeed
Fatima Majeed 2024-6-10
https://doi.org/10.48550/arXiv.2402.04272
In the end of page 13

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Calculus 的更多信息

标签

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by