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TopNewestUsing the second generalized continued fraction

\( \LARGE \ln {2} \approx \frac {2}{3 - \frac {1^2}{9 - \frac {2^2}{15 - \frac {3^2}{21} }}} = \frac {445}{642} > 0.693 \)

Because \( \ln{2} > 0.693 \Rightarrow ( \ln{2} )^2 > 0.48 = 3 \cdot \frac {4}{25} \Rightarrow \ln{2} > \frac {2\sqrt3}{5} \)

And because \( \pi > \frac {22}{7} - \frac {1}{630} > 3.1 \)

Now suppose that

\( ( \large \frac {e}{2} )^{\sqrt{3}} \geq (\sqrt2)^{\frac {\pi}{2} } \), raise both sides to the power of \(4\)

\( \Rightarrow ( \large \frac {e}{2} )^{4 \sqrt{3}} \geq 2^{\pi} \), log both sides

\( \large \Rightarrow 4 \sqrt{3} \space (1 - \ln 2) \geq \pi \space \ln 2 \)

\( \large \Rightarrow 4 \sqrt{3} \geq \ln {2} \space ( \pi + 4\sqrt{3} ) \)

\( \large \Rightarrow \frac { 4 \sqrt{3} }{ \pi + 4\sqrt{3} } \geq \ln {2} \)

\( \large \Rightarrow \frac { 4 \sqrt{3} }{ \pi + 4\sqrt{3} } \geq \ln {2} > \frac {2\sqrt3}{5} \)

\( \large \Rightarrow 10 - \pi > 4\sqrt{3} \)

\( \large \Rightarrow 10 - 3.1 > 10 - \pi > 4\sqrt{3} \), because \( \pi > 3.1 \)

\( \large \Rightarrow 6.9^2 > 48 \), which is a contradiction.

Hence \( \LARGE ( \frac {e}{2} )^{\sqrt{3}} < (\sqrt2)^{\frac {\pi}{2} } \) – Pi Han Goh · 4 years ago

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– Eric Edwards · 4 years ago

Those are some heavy-duty bounds! I like the integral trick to get \(\pi > 22/7\), and I see now why I was having trouble getting \(\ln 2 > .693\). Where in the world does that crazy continued fraction expansion come from? :-)Log in to reply

Source

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\( \LARGE \log (1 + \frac {x}{y} ) = \frac { 2} { 2y + x - \frac {(1x)^2}{3(2y+x) - \frac {(2x)^2}{5(2y+x) - \frac {(3x)^2}{7(2y+x) - \ldots } } } } \)

Substitute \( \large x = y = 1 \)

Or click here – Pi Han Goh · 4 years ago

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– NurFitri Hartina · 4 years ago

that was very good, that's a lot of storyLog in to reply

(meant to be a reply to George W.'s comment) I think your last step is off since you went \(174<\frac{505}{2}\log 2 < (.69)\frac{505}{2}\), which is backwards. However, if you use \(\log 2 < .7\) (which entails first showing that \(e^7 > 1024\) from the definition of \(e\) (using memorized values is kinda cheaty)) then you get \(174 < 176.75\).

Overall, I think a different approach is called for since showing \(3.14 < \pi\) by hand seems daunting in itself. Getting \(\sqrt{3} < 1.74\) is easy to show once you have it conjectured, but Newton's method is pretty fast by hand if you wanna be a bit more hardcore. All this is beside the point!

This kind of problem is usually solved by comparing to a third number that's between them. Maybe \(\sqrt[3]{5}\)? I thought about expanding \(\pi/4 = 1-\frac{1}{3} + \frac{1}{5}...\), but that convergence is too slow to be helpful. – Eric Edwards · 4 years ago

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– George Williams · 4 years ago

Actually, \( 0.69 < \log 2 \approx 0.6931 \), so \( 0.69 \frac{505}{2} < \frac{505}{2} \log 2 \). Yeah, the memorized value is sort of cheating (although I find it easier to believe than memorizing partial fraction representations, but that's my opinion.) I'll see if I can find a nicer proof which doesn't require knowing \( \log 2 \), although I believe knowing \( \pi \) is quite reasonable.Log in to reply

We will show that the latter is the greater, some easy approximations will do the trick. Note that \( 1.57 < \pi / 2 \), \( \sqrt{3} < 1.74 \), and \( 0.69 < \log 2 \). So we want to show that \( (e/2)^{174/100} < \sqrt{2}^{157/100} \), or \( e^{174} < 2^{157/2 + 174} \), meaning that \( 174 < \frac{505}{2} \log 2 \). Since \( 0.69 < \log 2 \), we have that \( 174 < 174.225 \), which is true.

(Please inform me if I have made a mistake.) – George Williams · 4 years ago

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Wow. Which level is this? – Pola Forest · 4 years ago

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Use logarithms! – Harsa Mitra · 4 years ago

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Let, (e/2)^√3=a×(√2)^π/2

⇒e^√3×2^-√3=a×2^π/4

⇒e^√3=a×2^(π/4+√3)

⇒√3=lna+(π/4+√3)ln2 [ln]

⇒1.73205=lna+(3.1415926536/4+1.73205)×.6931 [ ln2≈2/[3−1^2/{9−2^2/(15−3^2/21)}]=445/642=.69314]

⇒lna=1.73205-1.744843

⇒lna=-.012793

⇒a=1/e^.012793

⇒1/e^.012793<1

so,(e/2)^√3 / (√2)^π/2 <1

so,(e/2)^√3<(√2)^π/2 – Mahmudul Hasan · 4 years ago

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Note that: (1/2e)^3/2=(1/2e)^3^1/2 and (2^1/2)^1/2pi=(2^1/2)^pi^1/2.

Since we are comparing those numbers, we can delete the latter " ^1/2 ". So we got (1/2e)^3 and (2^1/2)^pi.

1/2e ~ 1.36 ; 2^1/2 ~ 1.41 and pi > 3. So, (2^1/2)1/2pi is the greatest. – Leonardo Cidrão · 4 years ago

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– Suhail Sherif · 4 years ago

The first one is (e/2)^(3^1/2). The second is ((2^1/2)^pi)^1/2 You can't remove the ^1/2 at the end of them. (Removing them involves squaring both sides, which gives (e/2)^(2*3^1/2) and (2^1/2)^pi)Log in to reply

– Leonardo Cidrão · 4 years ago

Just for further information about what I had said: http://en.wikipedia.org/wiki/Exponentiation#Rational_exponentsLog in to reply

andproperties – Leonardo Cidrão · 4 years agoLog in to reply

– Ivan Koswara · 4 years ago

The identity is \((a^b)^c = a^{bc}\), not \(a^{b^c} = a^{bc}\). When you do \( \left( \left( \dfrac{e}{2} \right)^{3^{1/2}} \right)^2 = \left( \dfrac{e}{2} \right)^{2 \cdot 3 \cdot 1/2}\), you accidentally made the mistake with \(3^{1/2}\). (The result should have been \(\left( \dfrac{e}{2} \right)^{2 \cdot (3^{1/2})}\); note that the 1/2 is as an exponent of the 3 instead of a multiplicand.)Log in to reply

– Leonardo Cidrão · 4 years ago

Ohh I see. Thanks Ivan for correct me, but I think that there is a solution similar to mine, because it's not necessary to go crazy with logarithms in this problem. I also have to ask for apologies to Suhail: Sorry dude, my bad.Log in to reply

this is just sort of an easy mathematical task. why make things complicated? We are now in the 21st century (computer era). – Mharfe Micaroz · 4 years ago

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