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












6












$begingroup$



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




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



$log(2^sqrt2) = sqrt2log(2)quad$ & $quad log(e) = 1$



I tried but can't induce comparable form.



Is anybody know how to prove it?






real-analysis analysis inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is your question?
    $endgroup$
    – Michael Rozenberg
    Mar 25 at 12:57






  • 1




    $begingroup$
    2^√2 > e or 2^√2 < e ? is my question
    $endgroup$
    – J.Bo
    Mar 25 at 12:59








  • 3




    $begingroup$
    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






  • 3




    $begingroup$
    What's wrong with using a calculator?
    $endgroup$
    – fleablood
    Mar 25 at 13:03






  • 1




    $begingroup$
    Oh, that's simple but good idea. THX
    $endgroup$
    – J.Bo
    Mar 25 at 13:04
















6












$begingroup$



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




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



$log(2^sqrt2) = sqrt2log(2)quad$ & $quad log(e) = 1$



I tried but can't induce comparable form.



Is anybody know how to prove it?






real-analysis analysis inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is your question?
    $endgroup$
    – Michael Rozenberg
    Mar 25 at 12:57






  • 1




    $begingroup$
    2^√2 > e or 2^√2 < e ? is my question
    $endgroup$
    – J.Bo
    Mar 25 at 12:59








  • 3




    $begingroup$
    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






  • 3




    $begingroup$
    What's wrong with using a calculator?
    $endgroup$
    – fleablood
    Mar 25 at 13:03






  • 1




    $begingroup$
    Oh, that's simple but good idea. THX
    $endgroup$
    – J.Bo
    Mar 25 at 13:04














6












6








6


2



$begingroup$



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




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



$log(2^sqrt2) = sqrt2log(2)quad$ & $quad log(e) = 1$



I tried but can't induce comparable form.



Is anybody know how to prove it?






real-analysis analysis inequality







share|cite|improve this question











$endgroup$





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




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



$log(2^sqrt2) = sqrt2log(2)quad$ & $quad log(e) = 1$



I tried but can't induce comparable form.



Is anybody know how to prove it?






real-analysis analysis inequality




real-analysis analysis inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 25 at 14:49









YuiTo Cheng

2,1963937




2,1963937










asked Mar 25 at 12:52









J.BoJ.Bo

506




506












  • $begingroup$
    What is your question?
    $endgroup$
    – Michael Rozenberg
    Mar 25 at 12:57






  • 1




    $begingroup$
    2^√2 > e or 2^√2 < e ? is my question
    $endgroup$
    – J.Bo
    Mar 25 at 12:59








  • 3




    $begingroup$
    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






  • 3




    $begingroup$
    What's wrong with using a calculator?
    $endgroup$
    – fleablood
    Mar 25 at 13:03






  • 1




    $begingroup$
    Oh, that's simple but good idea. THX
    $endgroup$
    – J.Bo
    Mar 25 at 13:04


















  • $begingroup$
    What is your question?
    $endgroup$
    – Michael Rozenberg
    Mar 25 at 12:57






  • 1




    $begingroup$
    2^√2 > e or 2^√2 < e ? is my question
    $endgroup$
    – J.Bo
    Mar 25 at 12:59








  • 3




    $begingroup$
    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






  • 3




    $begingroup$
    What's wrong with using a calculator?
    $endgroup$
    – fleablood
    Mar 25 at 13:03






  • 1




    $begingroup$
    Oh, that's simple but good idea. THX
    $endgroup$
    – J.Bo
    Mar 25 at 13:04
















$begingroup$
What is your question?
$endgroup$
– Michael Rozenberg
Mar 25 at 12:57




$begingroup$
What is your question?
$endgroup$
– Michael Rozenberg
Mar 25 at 12:57




1




1




$begingroup$
2^√2 > e or 2^√2 < e ? is my question
$endgroup$
– J.Bo
Mar 25 at 12:59






$begingroup$
2^√2 > e or 2^√2 < e ? is my question
$endgroup$
– J.Bo
Mar 25 at 12:59






3




3




$begingroup$
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




$begingroup$
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




3




3




$begingroup$
What's wrong with using a calculator?
$endgroup$
– fleablood
Mar 25 at 13:03




$begingroup$
What's wrong with using a calculator?
$endgroup$
– fleablood
Mar 25 at 13:03




1




1




$begingroup$
Oh, that's simple but good idea. THX
$endgroup$
– J.Bo
Mar 25 at 13:04




$begingroup$
Oh, that's simple but good idea. THX
$endgroup$
– J.Bo
Mar 25 at 13:04










5 Answers
5






active

oldest

votes


















6












$begingroup$

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






share|cite|improve this answer









$endgroup$









  • 8




    $begingroup$
    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



















8












$begingroup$

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
$$ 0leqint_{0}^{1}frac{x^2(1-x)^2}{1+x},dx leq frac{1}{16}$$
where the middle integral is exactly $-frac{11}{4}+4log(2)$. It follows that
$$ frac{33}{32} leq frac{3}{2}log(2) leq frac{135}{128} $$
so $frac{3}{2}log(2)>1$ and $color{red}{2sqrt{2}>e}$.
This proof just requires a polynomial division, perfectly doable by hand.



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






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It's $2^sqrt2$. See here
    $endgroup$
    – YuiTo Cheng
    Mar 25 at 14:53








  • 1




    $begingroup$
    @YuiToCheng: well, I dealt with both cases.
    $endgroup$
    – Jack D'Aurizio
    Mar 25 at 14:58






  • 1




    $begingroup$
    +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 $sumfrac{1}{sqrt{k}sqrt{k-1}}=sum1cdotsqrt{frac{1}{k(k-1)}}$ before applying CS.
    $endgroup$
    – Jam
    Mar 28 at 11:33



















6












$begingroup$

If you know that $ln(2)approx0.69$ and $1/sqrt2=sqrt2/2approx1.414/2=0.707$, then you have $ln(2)lt1/sqrt2$, in which case $ln(2^sqrt2)=sqrt2ln2lt1=ln(e)$, hence $2^sqrt2lt e$.



It's not hard to show that $sqrt2gt1.4$, since $1.4^2=1.96lt2$. 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 $2$:



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






share|cite|improve this answer











$endgroup$









  • 2




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



















0












$begingroup$

$2sqrt{2}^2 = 8$



$e^2 < 2.8*2.8 = 7.84$



You're welcome






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 $2sqrt{2}$ 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



















0












$begingroup$

Let $f(x)=ln(x),,g(x)=x^{-1/2}$. By the Taylor series of $e^x$, we have



$$e^{-0.35}>1-0.35+frac12left(0.35right)^2-frac16left(0.35right)^3=0.704>0.7$$



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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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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5 Answers
5






active

oldest

votes








5 Answers
5






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

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






share|cite|improve this answer









$endgroup$









  • 8




    $begingroup$
    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
















6












$begingroup$

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






share|cite|improve this answer









$endgroup$









  • 8




    $begingroup$
    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














6












6








6





$begingroup$

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






share|cite|improve this answer









$endgroup$



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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 25 at 13:18









lhflhf

167k11172404




167k11172404








  • 8




    $begingroup$
    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














  • 8




    $begingroup$
    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








8




8




$begingroup$
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




$begingroup$
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




$begingroup$
@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











8












$begingroup$

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
$$ 0leqint_{0}^{1}frac{x^2(1-x)^2}{1+x},dx leq frac{1}{16}$$
where the middle integral is exactly $-frac{11}{4}+4log(2)$. It follows that
$$ frac{33}{32} leq frac{3}{2}log(2) leq frac{135}{128} $$
so $frac{3}{2}log(2)>1$ and $color{red}{2sqrt{2}>e}$.
This proof just requires a polynomial division, perfectly doable by hand.



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






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It's $2^sqrt2$. See here
    $endgroup$
    – YuiTo Cheng
    Mar 25 at 14:53








  • 1




    $begingroup$
    @YuiToCheng: well, I dealt with both cases.
    $endgroup$
    – Jack D'Aurizio
    Mar 25 at 14:58






  • 1




    $begingroup$
    +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 $sumfrac{1}{sqrt{k}sqrt{k-1}}=sum1cdotsqrt{frac{1}{k(k-1)}}$ before applying CS.
    $endgroup$
    – Jam
    Mar 28 at 11:33
















8












$begingroup$

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
$$ 0leqint_{0}^{1}frac{x^2(1-x)^2}{1+x},dx leq frac{1}{16}$$
where the middle integral is exactly $-frac{11}{4}+4log(2)$. It follows that
$$ frac{33}{32} leq frac{3}{2}log(2) leq frac{135}{128} $$
so $frac{3}{2}log(2)>1$ and $color{red}{2sqrt{2}>e}$.
This proof just requires a polynomial division, perfectly doable by hand.



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






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It's $2^sqrt2$. See here
    $endgroup$
    – YuiTo Cheng
    Mar 25 at 14:53








  • 1




    $begingroup$
    @YuiToCheng: well, I dealt with both cases.
    $endgroup$
    – Jack D'Aurizio
    Mar 25 at 14:58






  • 1




    $begingroup$
    +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 $sumfrac{1}{sqrt{k}sqrt{k-1}}=sum1cdotsqrt{frac{1}{k(k-1)}}$ before applying CS.
    $endgroup$
    – Jam
    Mar 28 at 11:33














8












8








8





$begingroup$

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
$$ 0leqint_{0}^{1}frac{x^2(1-x)^2}{1+x},dx leq frac{1}{16}$$
where the middle integral is exactly $-frac{11}{4}+4log(2)$. It follows that
$$ frac{33}{32} leq frac{3}{2}log(2) leq frac{135}{128} $$
so $frac{3}{2}log(2)>1$ and $color{red}{2sqrt{2}>e}$.
This proof just requires a polynomial division, perfectly doable by hand.



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






share|cite|improve this answer











$endgroup$



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
$$ 0leqint_{0}^{1}frac{x^2(1-x)^2}{1+x},dx leq frac{1}{16}$$
where the middle integral is exactly $-frac{11}{4}+4log(2)$. It follows that
$$ frac{33}{32} leq frac{3}{2}log(2) leq frac{135}{128} $$
so $frac{3}{2}log(2)>1$ and $color{red}{2sqrt{2}>e}$.
This proof just requires a polynomial division, perfectly doable by hand.



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







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 25 at 14:58

























answered Mar 25 at 14:49









Jack D'AurizioJack D'Aurizio

292k33284672




292k33284672












  • $begingroup$
    It's $2^sqrt2$. See here
    $endgroup$
    – YuiTo Cheng
    Mar 25 at 14:53








  • 1




    $begingroup$
    @YuiToCheng: well, I dealt with both cases.
    $endgroup$
    – Jack D'Aurizio
    Mar 25 at 14:58






  • 1




    $begingroup$
    +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 $sumfrac{1}{sqrt{k}sqrt{k-1}}=sum1cdotsqrt{frac{1}{k(k-1)}}$ before applying CS.
    $endgroup$
    – Jam
    Mar 28 at 11:33


















  • $begingroup$
    It's $2^sqrt2$. See here
    $endgroup$
    – YuiTo Cheng
    Mar 25 at 14:53








  • 1




    $begingroup$
    @YuiToCheng: well, I dealt with both cases.
    $endgroup$
    – Jack D'Aurizio
    Mar 25 at 14:58






  • 1




    $begingroup$
    +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 $sumfrac{1}{sqrt{k}sqrt{k-1}}=sum1cdotsqrt{frac{1}{k(k-1)}}$ before applying CS.
    $endgroup$
    – Jam
    Mar 28 at 11:33
















$begingroup$
It's $2^sqrt2$. See here
$endgroup$
– YuiTo Cheng
Mar 25 at 14:53






$begingroup$
It's $2^sqrt2$. See here
$endgroup$
– YuiTo Cheng
Mar 25 at 14:53






1




1




$begingroup$
@YuiToCheng: well, I dealt with both cases.
$endgroup$
– Jack D'Aurizio
Mar 25 at 14:58




$begingroup$
@YuiToCheng: well, I dealt with both cases.
$endgroup$
– Jack D'Aurizio
Mar 25 at 14:58




1




1




$begingroup$
+1. I think your answer truly doesn't require any numerical calculation.
$endgroup$
– YuiTo Cheng
Mar 25 at 15:04




$begingroup$
+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 $sumfrac{1}{sqrt{k}sqrt{k-1}}=sum1cdotsqrt{frac{1}{k(k-1)}}$ before applying CS.
$endgroup$
– Jam
Mar 28 at 11:33




$begingroup$
(+1), slick answer as always. For posterity, and since the last step had confused me for a while, rewrite the sum as $sumfrac{1}{sqrt{k}sqrt{k-1}}=sum1cdotsqrt{frac{1}{k(k-1)}}$ before applying CS.
$endgroup$
– Jam
Mar 28 at 11:33











6












$begingroup$

If you know that $ln(2)approx0.69$ and $1/sqrt2=sqrt2/2approx1.414/2=0.707$, then you have $ln(2)lt1/sqrt2$, in which case $ln(2^sqrt2)=sqrt2ln2lt1=ln(e)$, hence $2^sqrt2lt e$.



It's not hard to show that $sqrt2gt1.4$, since $1.4^2=1.96lt2$. 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 $2$:



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






share|cite|improve this answer











$endgroup$









  • 2




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
















6












$begingroup$

If you know that $ln(2)approx0.69$ and $1/sqrt2=sqrt2/2approx1.414/2=0.707$, then you have $ln(2)lt1/sqrt2$, in which case $ln(2^sqrt2)=sqrt2ln2lt1=ln(e)$, hence $2^sqrt2lt e$.



It's not hard to show that $sqrt2gt1.4$, since $1.4^2=1.96lt2$. 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 $2$:



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






share|cite|improve this answer











$endgroup$









  • 2




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














6












6








6





$begingroup$

If you know that $ln(2)approx0.69$ and $1/sqrt2=sqrt2/2approx1.414/2=0.707$, then you have $ln(2)lt1/sqrt2$, in which case $ln(2^sqrt2)=sqrt2ln2lt1=ln(e)$, hence $2^sqrt2lt e$.



It's not hard to show that $sqrt2gt1.4$, since $1.4^2=1.96lt2$. 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 $2$:



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






share|cite|improve this answer











$endgroup$



If you know that $ln(2)approx0.69$ and $1/sqrt2=sqrt2/2approx1.414/2=0.707$, then you have $ln(2)lt1/sqrt2$, in which case $ln(2^sqrt2)=sqrt2ln2lt1=ln(e)$, hence $2^sqrt2lt e$.



It's not hard to show that $sqrt2gt1.4$, since $1.4^2=1.96lt2$. 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 $2$:



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







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 28 at 12:49

























answered Mar 25 at 14:00









Barry CipraBarry Cipra

60.6k655129




60.6k655129








  • 2




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














  • 2




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








2




2




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




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











0












$begingroup$

$2sqrt{2}^2 = 8$



$e^2 < 2.8*2.8 = 7.84$



You're welcome






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 $2sqrt{2}$ 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
















0












$begingroup$

$2sqrt{2}^2 = 8$



$e^2 < 2.8*2.8 = 7.84$



You're welcome






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 $2sqrt{2}$ 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














0












0








0





$begingroup$

$2sqrt{2}^2 = 8$



$e^2 < 2.8*2.8 = 7.84$



You're welcome






share|cite|improve this answer









$endgroup$



$2sqrt{2}^2 = 8$



$e^2 < 2.8*2.8 = 7.84$



You're welcome







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 25 at 14:35









FelorFelor

1




1












  • $begingroup$
    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 $2sqrt{2}$ 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


















  • $begingroup$
    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 $2sqrt{2}$ 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
















$begingroup$
It's $2^sqrt2$...
$endgroup$
– YuiTo Cheng
Mar 25 at 14:36




$begingroup$
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 $2sqrt{2}$ for me.
$endgroup$
– Felor
Mar 25 at 14:47






$begingroup$
Aww, there is a bug in EE. Yellow formula in startpost is $2sqrt{2}$ 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






$begingroup$
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






$begingroup$
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




$begingroup$
Yeah, it's exactly the summary of the first answer.
$endgroup$
– YuiTo Cheng
Mar 25 at 14:57











0












$begingroup$

Let $f(x)=ln(x),,g(x)=x^{-1/2}$. By the Taylor series of $e^x$, we have



$$e^{-0.35}>1-0.35+frac12left(0.35right)^2-frac16left(0.35right)^3=0.704>0.7$$



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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















0












$begingroup$

Let $f(x)=ln(x),,g(x)=x^{-1/2}$. By the Taylor series of $e^x$, we have



$$e^{-0.35}>1-0.35+frac12left(0.35right)^2-frac16left(0.35right)^3=0.704>0.7$$



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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












0








0





$begingroup$

Let $f(x)=ln(x),,g(x)=x^{-1/2}$. By the Taylor series of $e^x$, we have



$$e^{-0.35}>1-0.35+frac12left(0.35right)^2-frac16left(0.35right)^3=0.704>0.7$$



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.






share|cite|improve this answer









$endgroup$



Let $f(x)=ln(x),,g(x)=x^{-1/2}$. By the Taylor series of $e^x$, we have



$$e^{-0.35}>1-0.35+frac12left(0.35right)^2-frac16left(0.35right)^3=0.704>0.7$$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 28 at 14:06









JamJam

5,01921432




5,01921432












  • $begingroup$
    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




















  • $begingroup$
    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


















$begingroup$
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






$begingroup$
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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Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
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Can I use Tabulator js library in my java Spring + Thymeleaf project?

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0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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