# Quick but not simple question. $2^\sqrt2$ or e, which is greater?

$$2^\sqrt2$$ 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?

• What is your question? – Michael Rozenberg Mar 25 at 12:57
• 2^√2 > e or 2^√2 < e ? is my question – J.Bo Mar 25 at 12:59
• 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. – lulu Mar 25 at 13:01
• What's wrong with using a calculator? – fleablood Mar 25 at 13:03
• Oh, that's simple but good idea. THX – J.Bo Mar 25 at 13:04

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, $$e^{\sqrt 2} \approx 4.113250377 > 4$$.

• 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? – Teepeemm Mar 25 at 14:24
• @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 $x=1.41$. – lhf Mar 26 at 1:49

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.

• It's $2^\sqrt2$. See here – YuiTo Cheng Mar 25 at 14:53
• @YuiToCheng: well, I dealt with both cases. – Jack D'Aurizio Mar 25 at 14:58
• +1. I think your answer truly doesn't require any numerical calculation. – YuiTo Cheng Mar 25 at 15:04
• (+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. – Jam Mar 28 at 11:33

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 $$2$$:

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

• Nice solution because this does not require a calculator. – quarague Mar 25 at 14:24

$$2\sqrt{2}^2 = 8$$

$$e^2 < 2.8*2.8 = 7.84$$

You're welcome

• It's $2^\sqrt2$... – YuiTo Cheng Mar 25 at 14:36
• Aww, there is a bug in EE. Yellow formula in startpost is $2\sqrt{2}$ for me. – Felor Mar 25 at 14:47
• I see. It's not your fault. Corrected. It's a careless typo. – YuiTo Cheng Mar 25 at 14:50
• 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. – Felor Mar 25 at 14:52
• Yeah, it's exactly the summary of the first answer. – YuiTo Cheng Mar 25 at 14:57

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.

• 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. – Jam Mar 28 at 15:32