Ttyjfyk

Dot above capital letter not centred

What to do with wrong results in talks?

Is there an Impartial Brexit Deal comparison site?

How do I define a right arrow with bar in LaTeX?

Can I Retrieve Email Addresses from BCC?

Lay out the Carpet

Failed to fetch jessie backports repository

What are the ramifications of creating a homebrew world without an Astral Plane?

Everything Bob says is false. How does he get people to trust him?

Why "be dealt cards" rather than "be dealing cards"?

Trouble understanding overseas colleagues

Is a roofing delivery truck likely to crack my driveway slab?

What is the intuitive meaning of having a linear relationship between the logs of two variables?

What defines a dissertation?

What would be the benefits of having both a state and local currencies?

There is only s̶i̶x̶t̶y one place he can be

Bash method for viewing beginning and end of file

voltage of sounds of mp3files

Is it okay / does it make sense for another player to join a running game of Munchkin?

Modify casing of marked letters

I'm in charge of equipment buying but no one's ever happy with what I choose. How to fix this?

Was Spock the First Vulcan in Starfleet?

Teaching indefinite integrals that require special-casing

Hostile work environment after whistle-blowing on coworker and our boss. What do I do?

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

Which of these numbers is greater: $sqrt[5]5$ or $sqrt[4]4$?Inequality through calculusWhich is greater, $300 !$ or $(300^300)^frac 12$?Hausdorff metric and Vietoris topologyClosed and Bounded but not compactWhich coefficient is greater?Calculating square roots using the recurrence $x_n+1 = frac12 left(x_n + frac2x_nright)$Trigonometric inequality - x domain$e^left(pi^(e^pi)right);$ or $;pi^left(e^(pi^e)right)$. Which one is greater than the other?Prove by contradiction (not using a calculator) that $sqrt6 + sqrt2 < sqrt15$?

5

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

edited yesterday

YuiTo Cheng

2,1362837

asked yesterday

J.BoJ.Bo

456

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is your question?
  $endgroup$
  – Michael Rozenberg
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  2^√2 > e or 2^√2 < e ? is my question
  $endgroup$
  – J.Bo
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Well, the solution to $2^x=e$ is $x=frac 1ln 2approx 1.4427>sqrt 2$. Of course, numerical computation is involved in that.
  $endgroup$
  – lulu
  yesterday
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  What's wrong with using a calculator?
  $endgroup$
  – fleablood
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, that's simple but good idea. THX
  $endgroup$
  – J.Bo
  yesterday
  

  
  
  
  

 | 
show 3 more comments

5

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

edited yesterday

YuiTo Cheng

2,1362837

asked yesterday

J.BoJ.Bo

456

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is your question?
  $endgroup$
  – Michael Rozenberg
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  2^√2 > e or 2^√2 < e ? is my question
  $endgroup$
  – J.Bo
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Well, the solution to $2^x=e$ is $x=frac 1ln 2approx 1.4427>sqrt 2$. Of course, numerical computation is involved in that.
  $endgroup$
  – lulu
  yesterday
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  What's wrong with using a calculator?
  $endgroup$
  – fleablood
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, that's simple but good idea. THX
  $endgroup$
  – J.Bo
  yesterday
  

  
  
  
  

 | 
show 3 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

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,1362837

asked yesterday

J.BoJ.Bo

456

$endgroup$

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,1362837

asked yesterday

J.BoJ.Bo

456

share|cite|improve this question

edited yesterday

YuiTo Cheng

2,1362837

asked yesterday

J.BoJ.Bo

456

edited yesterday

YuiTo Cheng

2,1362837

edited yesterday

YuiTo Cheng

2,1362837

asked yesterday

J.BoJ.Bo

456

asked yesterday

J.BoJ.Bo

456

4 Answers
4

active

oldest

votes

8

$begingroup$

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
$$
e^x > 1+x+fracx^22+fracx^36+fracx^424
$$
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

answered yesterday

lhflhf

167k11172403

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  yesterday
  

  
  
  
  

add a comment | 

7

$begingroup$

This is the same as comparing $frac32log(2)$ and $1$. Since $x(1-x)$ is non-negative and bounded by $frac14$ on $(0,1)$, we have
$$ 0leqint_0^1fracx^2(1-x)^21+x,dx leq frac116$$
where the middle integral is exactly $-frac114+4log(2)$. It follows that
$$ frac3332 leq frac32log(2) leq frac135128 $$
so $frac32log(2)>1$ and $colorred2sqrt2>e$.
This proof just requires a polynomial division, perfectly doable by hand.

About $sqrt2log(2)$, we have
$$ log(2)=lim_nto +inftysum_k=n+1^2nfrac1kleqlim_nto+inftysum_k=n+1^2nfrac1sqrtksqrtk-1stackreltextCSleqlim_nto +inftysqrtnsum_k=n+1^2nleft(frac1k-1-frac1kright)$$
and the RHS is exactly $frac1sqrt2$. This is just a slick application of creative telescoping and the Cauchy-Schwarz inequality.

share|cite|improve this answer

edited yesterday

answered yesterday

Jack D'AurizioJack D'Aurizio

292k33284672

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's $2^sqrt2$. See here
  $endgroup$
  – YuiTo Cheng
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YuiToCheng: well, I dealt with both cases.
  $endgroup$
  – Jack D'Aurizio
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1. I think your answer truly doesn't require any numerical calculation.
  $endgroup$
  – YuiTo Cheng
  yesterday
  

  
  
  
  

add a comment | 

5

$begingroup$

If you know that $ln(2)approx0.69$ and $1/sqrt2=sqrt2/2approx1.4.14/2=0.707$, 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 $2$:

$$ln(2)=int_1^2dxover xlt1over6left(1+2cdot3over4+2cdot3over5+1over2 right)=1over6left(1+3over2+6over5+1over2 right)=1over6cdot42over10=7over10$$

share|cite|improve this answer

answered yesterday

Barry CipraBarry Cipra

60.5k655128

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Nice solution because this does not require a calculator.
  $endgroup$
  – quarague
  yesterday
  

  
  
  
  

add a comment | 

0

$begingroup$

$2sqrt2^2 = 8$

$e^2 < 2.8*2.8 = 7.84$

You're welcome

share|cite|improve this answer

answered yesterday

FelorFelor

1

New contributor

Felor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's $2^sqrt2$...
  $endgroup$
  – YuiTo Cheng
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Aww, there is a bug in EE. Yellow formula in startpost is $2sqrt2$ for me.
  $endgroup$
  – Felor
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see. It's not your fault. Corrected. It's a careless typo.
  $endgroup$
  – YuiTo Cheng
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One can start with taylor of $e^x$ then. $1+x+...+fracx^5120$, where x is $1.4 < sqrt2$. Sum of it gives 4.042219. Can be calculated by hand. $1.4 < sqrt2$ is trivial.
  $endgroup$
  – Felor
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, it's exactly the summary of the first answer.
  $endgroup$
  – YuiTo Cheng
  yesterday
  

  
  
  
  

 | 
show 1 more comment

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3161742%2fquick-but-not-simple-question-2-sqrt2-or-e-which-is-greater%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

8

$begingroup$

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
$$
e^x > 1+x+fracx^22+fracx^36+fracx^424
$$
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

answered yesterday

lhflhf

167k11172403

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  yesterday
  

  
  
  
  

add a comment | 

8

$begingroup$

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
$$
e^x > 1+x+fracx^22+fracx^36+fracx^424
$$
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

answered yesterday

lhflhf

167k11172403

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  yesterday
  

  
  
  
  

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
$$
e^x > 1+x+fracx^22+fracx^36+fracx^424
$$
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

answered yesterday

lhflhf

167k11172403

$endgroup$

share|cite|improve this answer

answered yesterday

lhflhf

167k11172403

answered yesterday

lhflhf

167k11172403

answered yesterday

lhflhf

167k11172403

7

$begingroup$

This is the same as comparing $frac32log(2)$ and $1$. Since $x(1-x)$ is non-negative and bounded by $frac14$ on $(0,1)$, we have
$$ 0leqint_0^1fracx^2(1-x)^21+x,dx leq frac116$$
where the middle integral is exactly $-frac114+4log(2)$. It follows that
$$ frac3332 leq frac32log(2) leq frac135128 $$
so $frac32log(2)>1$ and $colorred2sqrt2>e$.
This proof just requires a polynomial division, perfectly doable by hand.

About $sqrt2log(2)$, we have
$$ log(2)=lim_nto +inftysum_k=n+1^2nfrac1kleqlim_nto+inftysum_k=n+1^2nfrac1sqrtksqrtk-1stackreltextCSleqlim_nto +inftysqrtnsum_k=n+1^2nleft(frac1k-1-frac1kright)$$
and the RHS is exactly $frac1sqrt2$. This is just a slick application of creative telescoping and the Cauchy-Schwarz inequality.

share|cite|improve this answer

edited yesterday

answered yesterday

Jack D'AurizioJack D'Aurizio

292k33284672

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's $2^sqrt2$. See here
  $endgroup$
  – YuiTo Cheng
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YuiToCheng: well, I dealt with both cases.
  $endgroup$
  – Jack D'Aurizio
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1. I think your answer truly doesn't require any numerical calculation.
  $endgroup$
  – YuiTo Cheng
  yesterday
  

  
  
  
  

add a comment | 

7

$begingroup$

This is the same as comparing $frac32log(2)$ and $1$. Since $x(1-x)$ is non-negative and bounded by $frac14$ on $(0,1)$, we have
$$ 0leqint_0^1fracx^2(1-x)^21+x,dx leq frac116$$
where the middle integral is exactly $-frac114+4log(2)$. It follows that
$$ frac3332 leq frac32log(2) leq frac135128 $$
so $frac32log(2)>1$ and $colorred2sqrt2>e$.
This proof just requires a polynomial division, perfectly doable by hand.

About $sqrt2log(2)$, we have
$$ log(2)=lim_nto +inftysum_k=n+1^2nfrac1kleqlim_nto+inftysum_k=n+1^2nfrac1sqrtksqrtk-1stackreltextCSleqlim_nto +inftysqrtnsum_k=n+1^2nleft(frac1k-1-frac1kright)$$
and the RHS is exactly $frac1sqrt2$. This is just a slick application of creative telescoping and the Cauchy-Schwarz inequality.

share|cite|improve this answer

edited yesterday

answered yesterday

Jack D'AurizioJack D'Aurizio

292k33284672

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's $2^sqrt2$. See here
  $endgroup$
  – YuiTo Cheng
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YuiToCheng: well, I dealt with both cases.
  $endgroup$
  – Jack D'Aurizio
  yesterday
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1. I think your answer truly doesn't require any numerical calculation.
  $endgroup$
  – YuiTo Cheng
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

This is the same as comparing $frac32log(2)$ and $1$. Since $x(1-x)$ is non-negative and bounded by $frac14$ on $(0,1)$, we have
$$ 0leqint_0^1fracx^2(1-x)^21+x,dx leq frac116$$
where the middle integral is exactly $-frac114+4log(2)$. It follows that
$$ frac3332 leq frac32log(2) leq frac135128 $$
so $frac32log(2)>1$ and $colorred2sqrt2>e$.
This proof just requires a polynomial division, perfectly doable by hand.

About $sqrt2log(2)$, we have
$$ log(2)=lim_nto +inftysum_k=n+1^2nfrac1kleqlim_nto+inftysum_k=n+1^2nfrac1sqrtksqrtk-1stackreltextCSleqlim_nto +inftysqrtnsum_k=n+1^2nleft(frac1k-1-frac1kright)$$
and the RHS is exactly $frac1sqrt2$. This is just a slick application of creative telescoping and the Cauchy-Schwarz inequality.

share|cite|improve this answer

edited yesterday

answered yesterday

Jack D'AurizioJack D'Aurizio

292k33284672

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

Jack D'AurizioJack D'Aurizio

292k33284672

answered yesterday

Jack D'AurizioJack D'Aurizio

292k33284672

answered yesterday

Jack D'AurizioJack D'Aurizio

292k33284672

5

$begingroup$

If you know that $ln(2)approx0.69$ and $1/sqrt2=sqrt2/2approx1.4.14/2=0.707$, 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 $2$:

$$ln(2)=int_1^2dxover xlt1over6left(1+2cdot3over4+2cdot3over5+1over2 right)=1over6left(1+3over2+6over5+1over2 right)=1over6cdot42over10=7over10$$

share|cite|improve this answer

answered yesterday

Barry CipraBarry Cipra

60.5k655128

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Nice solution because this does not require a calculator.
  $endgroup$
  – quarague
  yesterday
  

  
  
  
  

add a comment | 

5

$begingroup$

If you know that $ln(2)approx0.69$ and $1/sqrt2=sqrt2/2approx1.4.14/2=0.707$, 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 $2$:

$$ln(2)=int_1^2dxover xlt1over6left(1+2cdot3over4+2cdot3over5+1over2 right)=1over6left(1+3over2+6over5+1over2 right)=1over6cdot42over10=7over10$$

share|cite|improve this answer

answered yesterday

Barry CipraBarry Cipra

60.5k655128

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Nice solution because this does not require a calculator.
  $endgroup$
  – quarague
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

If you know that $ln(2)approx0.69$ and $1/sqrt2=sqrt2/2approx1.4.14/2=0.707$, 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 $2$:

$$ln(2)=int_1^2dxover xlt1over6left(1+2cdot3over4+2cdot3over5+1over2 right)=1over6left(1+3over2+6over5+1over2 right)=1over6cdot42over10=7over10$$

share|cite|improve this answer

answered yesterday

Barry CipraBarry Cipra

60.5k655128

$endgroup$

share|cite|improve this answer

answered yesterday

Barry CipraBarry Cipra

60.5k655128

answered yesterday

Barry CipraBarry Cipra

60.5k655128

answered yesterday

Barry CipraBarry Cipra

60.5k655128

0

$begingroup$

$2sqrt2^2 = 8$

$e^2 < 2.8*2.8 = 7.84$

You're welcome

share|cite|improve this answer

answered yesterday

FelorFelor

1

New contributor

Felor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's $2^sqrt2$...
  $endgroup$
  – YuiTo Cheng
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Aww, there is a bug in EE. Yellow formula in startpost is $2sqrt2$ for me.
  $endgroup$
  – Felor
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see. It's not your fault. Corrected. It's a careless typo.
  $endgroup$
  – YuiTo Cheng
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One can start with taylor of $e^x$ then. $1+x+...+fracx^5120$, where x is $1.4 < sqrt2$. Sum of it gives 4.042219. Can be calculated by hand. $1.4 < sqrt2$ is trivial.
  $endgroup$
  – Felor
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, it's exactly the summary of the first answer.
  $endgroup$
  – YuiTo Cheng
  yesterday
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

$2sqrt2^2 = 8$

$e^2 < 2.8*2.8 = 7.84$

You're welcome

share|cite|improve this answer

answered yesterday

FelorFelor

1

New contributor

Felor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's $2^sqrt2$...
  $endgroup$
  – YuiTo Cheng
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Aww, there is a bug in EE. Yellow formula in startpost is $2sqrt2$ for me.
  $endgroup$
  – Felor
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see. It's not your fault. Corrected. It's a careless typo.
  $endgroup$
  – YuiTo Cheng
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One can start with taylor of $e^x$ then. $1+x+...+fracx^5120$, where x is $1.4 < sqrt2$. Sum of it gives 4.042219. Can be calculated by hand. $1.4 < sqrt2$ is trivial.
  $endgroup$
  – Felor
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, it's exactly the summary of the first answer.
  $endgroup$
  – YuiTo Cheng
  yesterday
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

$2sqrt2^2 = 8$

$e^2 < 2.8*2.8 = 7.84$

You're welcome

share|cite|improve this answer

answered yesterday

FelorFelor

1

New contributor

Felor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

answered yesterday

FelorFelor

1

New contributor

Felor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered yesterday

FelorFelor

1

New contributor

Felor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered yesterday

FelorFelor

1