Emma Hobbs

Answered

2022-11-22

Let $L(X)=\mathrm{exp}(\sqrt{\mathrm{log}X\mathrm{log}\mathrm{log}X})$

Prove that if c>0,$Y=L(X{)}^{c}$, and $u=\mathrm{log}X/\mathrm{log}Y$ , then

${u}^{u}=L(X{)}^{(1/2c)(1+o(1))}$

I've tried to write ${u}^{u}=(\mathrm{log}X/\mathrm{log}Y{)}^{\mathrm{log}X/\mathrm{log}Y}$

But that doesn't seem to get me anywhere.

Prove that if c>0,$Y=L(X{)}^{c}$, and $u=\mathrm{log}X/\mathrm{log}Y$ , then

${u}^{u}=L(X{)}^{(1/2c)(1+o(1))}$

I've tried to write ${u}^{u}=(\mathrm{log}X/\mathrm{log}Y{)}^{\mathrm{log}X/\mathrm{log}Y}$

But that doesn't seem to get me anywhere.

Answer & Explanation

kuthiwenihca

Expert

2022-11-23Added 23 answers

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

$\frac{\mathrm{ln}{u}^{u}}{\mathrm{ln}L(x)}=\frac{1}{2c}(1+o(1)).$

$\frac{\mathrm{ln}{u}^{u}}{\mathrm{ln}L(x)}=\frac{1}{2c}(1+o(1)).$

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