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$2^\sqrt2$ vs $e$, which is greater?

$(2^\sqrt2)^\sqrt2 = 4\quad $ & $\quad e^\sqrt2$ = ?

$\log(2^\sqrt2) = \sqrt2\log(2)\quad$ & $\quad \log(e) = 1$

I tried but can't induce comparable form.

Is anybody know how to prove it?

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  • $\begingroup$ What is your question? $\endgroup$ – Michael Rozenberg Mar 25 at 12:57
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    $\begingroup$ 2^√2 > e or 2^√2 < e ? is my question $\endgroup$ – J.Bo Mar 25 at 12:59
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    $\begingroup$ Well, the solution to $2^x=e$ is $x=\frac 1{\ln 2}\approx 1.4427>\sqrt 2$. Of course, numerical computation is involved in that. $\endgroup$ – lulu Mar 25 at 13:01
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    $\begingroup$ What's wrong with using a calculator? $\endgroup$ – fleablood Mar 25 at 13:03
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    $\begingroup$ Oh, that's simple but good idea. THX $\endgroup$ – J.Bo Mar 25 at 13:04
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Instead of comparing $2^{\sqrt 2}$ and $e$, let's raise both to $\sqrt 2$ and compare $2^2$ and $e^{\sqrt 2}$: $$ e^{\sqrt 2} > 2.7^{1.4} \approx 4.017068799 > 4 = 2^2 $$ Or use that $$ e^x > 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24} $$ with $x=1.41$ and get $$ e^{\sqrt 2} > e^{1.41} > 4.03594 > 4 $$ In fact, $e^{\sqrt 2} \approx 4.113250377 > 4$.

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    $\begingroup$ I don't understand how you can estimate $2.7^{1.4}$, $e^{1.41}$, and $e^{\sqrt 2}$ without a calculator. And if you have a calculator, why not find $2^{\sqrt 2}$ from the beginning? $\endgroup$ – Teepeemm Mar 25 at 14:24
  • $\begingroup$ @Teepeemm, you're right. The best approach is the second one, with a polynomial. Unfortunately, it's of degree $4$ and you have to use two decimals in $x=1.41$. $\endgroup$ – lhf Mar 26 at 1:49
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This is the same as comparing $\frac{3}{2}\log(2)$ and $1$. Since $x(1-x)$ is non-negative and bounded by $\frac{1}{4}$ on $(0,1)$, we have $$ 0\leq\int_{0}^{1}\frac{x^2(1-x)^2}{1+x}\,dx \leq \frac{1}{16}$$ where the middle integral is exactly $-\frac{11}{4}+4\log(2)$. It follows that $$ \frac{33}{32} \leq \frac{3}{2}\log(2) \leq \frac{135}{128} $$ so $\frac{3}{2}\log(2)>1$ and $\color{red}{2\sqrt{2}>e}$.
This proof just requires a polynomial division, perfectly doable by hand.

About $\sqrt{2}\log(2)$, we have $$ \log(2)=\lim_{n\to +\infty}\sum_{k=n+1}^{2n}\frac{1}{k}\leq\lim_{n\to+\infty}\sum_{k=n+1}^{2n}\frac{1}{\sqrt{k}\sqrt{k-1}}\stackrel{\text{CS}}{\leq}\lim_{n\to +\infty}\sqrt{n\sum_{k=n+1}^{2n}\left(\frac{1}{k-1}-\frac{1}{k}\right)}$$ and the RHS is exactly $\frac{1}{\sqrt{2}}$. This is just a slick application of creative telescoping and the Cauchy-Schwarz inequality.

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  • $\begingroup$ It's $2^\sqrt2$. See here $\endgroup$ – YuiTo Cheng Mar 25 at 14:53
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    $\begingroup$ @YuiToCheng: well, I dealt with both cases. $\endgroup$ – Jack D'Aurizio Mar 25 at 14:58
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    $\begingroup$ +1. I think your answer truly doesn't require any numerical calculation. $\endgroup$ – YuiTo Cheng Mar 25 at 15:04
  • $\begingroup$ (+1), slick answer as always. For posterity, and since the last step had confused me for a while, rewrite the sum as $\sum\frac{1}{\sqrt{k}\sqrt{k-1}}=\sum1\cdot\sqrt{\frac{1}{k(k-1)}}$ before applying CS. $\endgroup$ – Jam Mar 28 at 11:33
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If you know that $\ln(2)\approx0.69$ and $1/\sqrt2=\sqrt2/2\approx1.414/2=0.707$, then you have $\ln(2)\lt1/\sqrt2$, in which case $\ln(2^\sqrt2)=\sqrt2\ln2\lt1=\ln(e)$, hence $2^\sqrt2\lt e$.

It's not hard to show that $\sqrt2\gt1.4$, since $1.4^2=1.96\lt2$. It's a little trickier to show that $\ln(2)\lt0.7$, but this can be done by comparing the area beneath the curve $y=1/x$ to the areas of the trapezoids containing it with endpoints at $x=1$, $4/3$, $5/3$, and $2$:

$$\ln(2)=\int_1^2{dx\over x}\lt{1\over6}\left(1+2\cdot{3\over4}+2\cdot{3\over5}+{1\over2} \right)={1\over6}\left(1+{3\over2}+{6\over5}+{1\over2} \right)={1\over6}\cdot{42\over10}={7\over10}$$

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    $\begingroup$ Nice solution because this does not require a calculator. $\endgroup$ – quarague Mar 25 at 14:24
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$2\sqrt{2}^2 = 8$

$e^2 < 2.8*2.8 = 7.84$

You're welcome

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  • $\begingroup$ It's $2^\sqrt2$... $\endgroup$ – YuiTo Cheng Mar 25 at 14:36
  • $\begingroup$ Aww, there is a bug in EE. Yellow formula in startpost is $2\sqrt{2}$ for me. $\endgroup$ – Felor Mar 25 at 14:47
  • $\begingroup$ I see. It's not your fault. Corrected. It's a careless typo. $\endgroup$ – YuiTo Cheng Mar 25 at 14:50
  • $\begingroup$ One can start with taylor of $e^x$ then. $1+x+...+\frac{x^5}{120}$, where x is $1.4 < \sqrt{2}$. Sum of it gives 4.042219. Can be calculated by hand. $1.4 < \sqrt{2}$ is trivial. $\endgroup$ – Felor Mar 25 at 14:52
  • $\begingroup$ Yeah, it's exactly the summary of the first answer. $\endgroup$ – YuiTo Cheng Mar 25 at 14:57
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Let $f(x)=\ln(x),\,g(x)=x^{-1/2}$. By the Taylor series of $e^x$, we have

$$e^{-0.35}>1-0.35+\frac12\left(0.35\right)^2-\frac16\left(0.35\right)^3=0.704>0.7$$

Hence, $g(e^{0.7})-f(e^{0.7})=e^{-0.35}-0.7>0$. Applying Taylor series again shows

$$e^{0.7}>1+0.7+\frac{1}{2}(0.7)^2+\frac{1}{6}(0.7)^3=2.002>2$$

Observe that $f$ and $g$ are respectively strictly increasing and decreasing over $(0,\infty)$, so $g>f$ holds for all $x$ in $(0,2.002)$. Therefore $g(2)>f(2)$, which rearranges to $e>2^{\sqrt{2}}$. These calculations are perfectly feasible to do by hand.

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  • $\begingroup$ Addendum: it's not feasible to use the series expansion of $\ln(x)$ vs. $x^{-1/2}$, nor $\ln(x)$ vs. the reciprocal of the series expansion of $x^{1/2}$ to compare their values at $x=2$ as you'd have to go to at least $37$ and $23$ terms, respectively. $\endgroup$ – Jam Mar 28 at 15:32

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