Emma Hobbs

2022-11-22

Let $L\left(X\right)=\mathrm{exp}\left(\sqrt{\mathrm{log}X\mathrm{log}\mathrm{log}X}\right)$
Prove that if c>0,$Y=L\left(X{\right)}^{c}$, and $u=\mathrm{log}X/\mathrm{log}Y$ , then
${u}^{u}=L\left(X{\right)}^{\left(1/2c\right)\left(1+o\left(1\right)\right)}$
I've tried to write ${u}^{u}=\left(\mathrm{log}X/\mathrm{log}Y{\right)}^{\mathrm{log}X/\mathrm{log}Y}$
But that doesn't seem to get me anywhere.

kuthiwenihca

Expert

First, we have $\mathrm{ln}{u}^{u}=u\mathrm{ln}u=\frac{1}{\mathrm{ln}Y}\left(\mathrm{ln}X\mathrm{ln}\frac{\mathrm{ln}X}{\mathrm{ln}Y}\right)=\frac{1}{\mathrm{ln}Y}\left(\mathrm{ln}X\left(\mathrm{ln}\mathrm{ln}X-\mathrm{ln}\mathrm{ln}Y\right)\right)$, and Therefore, $\mathrm{ln}{u}^{u}=\frac{1}{2c\sqrt{\mathrm{ln}X\mathrm{ln}\mathrm{ln}X}}\mathrm{ln}X\mathrm{ln}\mathrm{ln}X\cdot \left(1+o\left(1\right)\right).$ On the other hand, we have $\mathrm{ln}L\left(X\right)=\sqrt{\mathrm{ln}X\mathrm{ln}\mathrm{ln}X}$, and so
$\frac{\mathrm{ln}{u}^{u}}{\mathrm{ln}L\left(x\right)}=\frac{1}{2c}\left(1+o\left(1\right)\right).$

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