Quick but not simple question. $2^sqrt2$ or e, which is greater?
$begingroup$
$2^sqrt2$ 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
edited Mar 25 at 14:49
YuiTo Cheng
2,1963937
asked Mar 25 at 12:52
J.BoJ.Bo
506
$endgroup$
|
show 4 more comments
$begingroup$
$2^sqrt2$ 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
edited Mar 25 at 14:49
YuiTo Cheng
2,1963937
asked Mar 25 at 12:52
J.BoJ.Bo
506
$endgroup$
$begingroup$
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
|
show 4 more comments
$begingroup$
$2^sqrt2$ 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
edited Mar 25 at 14:49
YuiTo Cheng
2,1963937
asked Mar 25 at 12:52
J.BoJ.Bo
506
$endgroup$
$2^sqrt2$ 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
edited Mar 25 at 14:49
YuiTo Cheng
2,1963937
asked Mar 25 at 12:52
J.BoJ.Bo
506
edited Mar 25 at 14:49
YuiTo Cheng
2,1963937
asked Mar 25 at 12:52
J.BoJ.Bo
506
edited Mar 25 at 14:49
YuiTo Cheng
2,1963937
edited Mar 25 at 14:49
YuiTo Cheng
2,1963937
edited Mar 25 at 14:49
YuiTo Cheng
2,1963937
2,1963937
asked Mar 25 at 12:52
J.BoJ.Bo
506
asked Mar 25 at 12:52
J.BoJ.Bo
506
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
|
show 4 more comments
$begingroup$
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
|
show 4 more comments
5 Answers
5
active
oldest
votes
$begingroup$
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, $e^{sqrt 2} approx 4.113250377 > 4$.
answered Mar 25 at 13:18
lhflhf
167k11172404
$endgroup$
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
add a comment |
$begingroup$
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.
answered Mar 25 at 14:49
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $2$:
$$ln(2)=int_1^2{dxover x}lt{1over6}left(1+2cdot{3over4}+2cdot{3over5}+{1over2} right)={1over6}left(1+{3over2}+{6over5}+{1over2} right)={1over6}cdot{42over10}={7over10}$$
answered Mar 25 at 14:00
Barry CipraBarry Cipra
60.6k655129
$endgroup$
2
$begingroup$
Nice solution because this does not require a calculator.
$endgroup$
– quarague
Mar 25 at 14:24
add a comment |
$begingroup$
$2sqrt{2}^2 = 8$
$e^2 < 2.8*2.8 = 7.84$
You're welcome
answered Mar 25 at 14:35
FelorFelor
1
$endgroup$
$begingroup$
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
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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
|
show 1 more comment
$begingroup$
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.
answered Mar 28 at 14:06
JamJam
5,01921432
$endgroup$
$begingroup$
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
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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, $e^{sqrt 2} approx 4.113250377 > 4$.
answered Mar 25 at 13:18
lhflhf
167k11172404
$endgroup$
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
add a comment |
$begingroup$
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, $e^{sqrt 2} approx 4.113250377 > 4$.
answered Mar 25 at 13:18
lhflhf
167k11172404
$endgroup$
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
add a comment |
$begingroup$
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, $e^{sqrt 2} approx 4.113250377 > 4$.
answered Mar 25 at 13:18
lhflhf
167k11172404
$endgroup$
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, $e^{sqrt 2} approx 4.113250377 > 4$.
answered Mar 25 at 13:18
lhflhf
167k11172404
answered Mar 25 at 13:18
lhflhf
167k11172404
answered Mar 25 at 13:18
lhflhf
167k11172404
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
add a comment |
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
add a comment |
$begingroup$
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.
answered Mar 25 at 14:49
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Mar 25 at 14:49
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Mar 25 at 14:49
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
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.
answered Mar 25 at 14:49
Jack D'AurizioJack D'Aurizio
292k33284672
edited Mar 25 at 14:58
answered Mar 25 at 14:49
Jack D'AurizioJack D'Aurizio
292k33284672
answered Mar 25 at 14:49
Jack D'AurizioJack D'Aurizio
292k33284672
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 $2$:
$$ln(2)=int_1^2{dxover x}lt{1over6}left(1+2cdot{3over4}+2cdot{3over5}+{1over2} right)={1over6}left(1+{3over2}+{6over5}+{1over2} right)={1over6}cdot{42over10}={7over10}$$
answered Mar 25 at 14:00
Barry CipraBarry Cipra
60.6k655129
$endgroup$
2
$begingroup$
Nice solution because this does not require a calculator.
$endgroup$
– quarague
Mar 25 at 14:24
add a comment |
$begingroup$
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 $2$:
$$ln(2)=int_1^2{dxover x}lt{1over6}left(1+2cdot{3over4}+2cdot{3over5}+{1over2} right)={1over6}left(1+{3over2}+{6over5}+{1over2} right)={1over6}cdot{42over10}={7over10}$$
answered Mar 25 at 14:00
Barry CipraBarry Cipra
60.6k655129
$endgroup$
2
$begingroup$
Nice solution because this does not require a calculator.
$endgroup$
– quarague
Mar 25 at 14:24
add a comment |
$begingroup$
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 $2$:
$$ln(2)=int_1^2{dxover x}lt{1over6}left(1+2cdot{3over4}+2cdot{3over5}+{1over2} right)={1over6}left(1+{3over2}+{6over5}+{1over2} right)={1over6}cdot{42over10}={7over10}$$
answered Mar 25 at 14:00
Barry CipraBarry Cipra
60.6k655129
$endgroup$
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 $2$:
$$ln(2)=int_1^2{dxover x}lt{1over6}left(1+2cdot{3over4}+2cdot{3over5}+{1over2} right)={1over6}left(1+{3over2}+{6over5}+{1over2} right)={1over6}cdot{42over10}={7over10}$$
answered Mar 25 at 14:00
Barry CipraBarry Cipra
60.6k655129
edited Mar 28 at 12:49
answered Mar 25 at 14:00
Barry CipraBarry Cipra
60.6k655129
answered Mar 25 at 14:00
Barry CipraBarry Cipra
60.6k655129
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
add a comment |
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
add a comment |
$begingroup$
$2sqrt{2}^2 = 8$
$e^2 < 2.8*2.8 = 7.84$
You're welcome
answered Mar 25 at 14:35
FelorFelor
1
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
$2sqrt{2}^2 = 8$
$e^2 < 2.8*2.8 = 7.84$
You're welcome
answered Mar 25 at 14:35
FelorFelor
1
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
$2sqrt{2}^2 = 8$
$e^2 < 2.8*2.8 = 7.84$
You're welcome
answered Mar 25 at 14:35
FelorFelor
1
$endgroup$
$2sqrt{2}^2 = 8$
$e^2 < 2.8*2.8 = 7.84$
You're welcome
answered Mar 25 at 14:35
FelorFelor
1
answered Mar 25 at 14:35
FelorFelor
1
answered Mar 25 at 14:35
FelorFelor
1
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
$begingroup$
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.
answered Mar 28 at 14:06
JamJam
5,01921432
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Mar 28 at 14:06
JamJam
5,01921432
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Mar 28 at 14:06
JamJam
5,01921432
$endgroup$
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.
answered Mar 28 at 14:06
JamJam
5,01921432
answered Mar 28 at 14:06
JamJam
5,01921432
answered Mar 28 at 14:06
JamJam
5,01921432
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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