which is the greater? ( i still dont understand )

without calculator

Note by NurFitri Hartina
6 years, 3 months ago

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14 votes

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Using the second generalized continued fraction

ln22312922153221=445642>0.693 \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 ln2>0.693(ln2)2>0.48=3425ln2>235 \ln{2} > 0.693 \Rightarrow ( \ln{2} )^2 > 0.48 = 3 \cdot \frac {4}{25} \Rightarrow \ln{2} > \frac {2\sqrt3}{5}

And because π>2271630>3.1 \pi > \frac {22}{7} - \frac {1}{630} > 3.1

Now suppose that

(e2)3(2)π2 ( \large \frac {e}{2} )^{\sqrt{3}} \geq (\sqrt2)^{\frac {\pi}{2} } , raise both sides to the power of 44

(e2)432π \Rightarrow ( \large \frac {e}{2} )^{4 \sqrt{3}} \geq 2^{\pi} , log both sides

43 (1ln2)π ln2 \large \Rightarrow 4 \sqrt{3} \space (1 - \ln 2) \geq \pi \space \ln 2

43ln2 (π+43) \large \Rightarrow 4 \sqrt{3} \geq \ln {2} \space ( \pi + 4\sqrt{3} )

43π+43ln2 \large \Rightarrow \frac { 4 \sqrt{3} }{ \pi + 4\sqrt{3} } \geq \ln {2}

43π+43ln2>235 \large \Rightarrow \frac { 4 \sqrt{3} }{ \pi + 4\sqrt{3} } \geq \ln {2} > \frac {2\sqrt3}{5}

10π>43 \large \Rightarrow 10 - \pi > 4\sqrt{3}

103.1>10π>43 \large \Rightarrow 10 - 3.1 > 10 - \pi > 4\sqrt{3} , because π>3.1 \pi > 3.1

6.92>48 \large \Rightarrow 6.9^2 > 48 , which is a contradiction.

Hence (e2)3<(2)π2 \LARGE ( \frac {e}{2} )^{\sqrt{3}} < (\sqrt2)^{\frac {\pi}{2} }

Pi Han Goh - 6 years, 3 months ago

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Those are some heavy-duty bounds! I like the integral trick to get π>22/7\pi > 22/7, and I see now why I was having trouble getting ln2>.693\ln 2 > .693. Where in the world does that crazy continued fraction expansion come from? :-)

Eric Edwards - 6 years, 3 months ago

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Source

Scroll down to

log(1+xy)=22y+x(1x)23(2y+x)(2x)25(2y+x)(3x)27(2y+x) \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 x=y=1 \large x = y = 1

Or click here

Pi Han Goh - 6 years, 3 months ago

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that was very good, that's a lot of story

NurFitri Hartina - 6 years, 3 months ago

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We will show that the latter is the greater, some easy approximations will do the trick. Note that 1.57<π/2 1.57 < \pi / 2 , 3<1.74 \sqrt{3} < 1.74 , and 0.69<log2 0.69 < \log 2 . So we want to show that (e/2)174/100<2157/100 (e/2)^{174/100} < \sqrt{2}^{157/100} , or e174<2157/2+174 e^{174} < 2^{157/2 + 174} , meaning that 174<5052log2 174 < \frac{505}{2} \log 2 . Since 0.69<log2 0.69 < \log 2 , we have that 174<174.225 174 < 174.225 , which is true.

(Please inform me if I have made a mistake.)

George Williams - 6 years, 3 months ago

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(meant to be a reply to George W.'s comment) I think your last step is off since you went 174<5052log2<(.69)5052174<\frac{505}{2}\log 2 < (.69)\frac{505}{2}, which is backwards. However, if you use log2<.7\log 2 < .7 (which entails first showing that e7>1024e^7 > 1024 from the definition of ee (using memorized values is kinda cheaty)) then you get 174<176.75174 < 176.75.

Overall, I think a different approach is called for since showing 3.14<π3.14 < \pi by hand seems daunting in itself. Getting 3<1.74\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 53\sqrt[3]{5}? I thought about expanding π/4=113+15...\pi/4 = 1-\frac{1}{3} + \frac{1}{5}..., but that convergence is too slow to be helpful.

Eric Edwards - 6 years, 3 months ago

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Actually, 0.69<log20.6931 0.69 < \log 2 \approx 0.6931 , so 0.695052<5052log2 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 log2 \log 2 , although I believe knowing π \pi is quite reasonable.

George Williams - 6 years, 3 months ago

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Wow. Which level is this?

Pola Forest - 6 years, 3 months 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 - 6 years, 3 months ago

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Use logarithms!

Harsa Mitra - 6 years, 3 months 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 - 6 years, 3 months ago

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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)

Suhail Sherif - 6 years, 3 months ago

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Just for further information about what I had said: http://en.wikipedia.org/wiki/Exponentiation#Rational_exponents

Leonardo Cidrão - 6 years, 3 months ago

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Sorry dude, but I think you comitted a mistake. I do can remove them, because if we square, for example, (e/2)^(3^1/2), we get: [(e/2)^(3^1/2)]^2= (e/2)^2.3.(1/2)=(e/2)^3 . I used a identity from exponenciation, as you can see at http://en.wikipedia.org/wiki/Exponentiation#Identitiesandproperties

Leonardo Cidrão - 6 years, 3 months ago

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@Leonardo Cidrão The identity is (ab)c=abc(a^b)^c = a^{bc}, not abc=abca^{b^c} = a^{bc}. When you do ((e2)31/2)2=(e2)231/2 \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 31/23^{1/2}. (The result should have been (e2)2(31/2)\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.)

Ivan Koswara - 6 years, 3 months ago

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@Ivan Koswara 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.

Leonardo Cidrão - 6 years, 3 months ago

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this is just sort of an easy mathematical task. why make things complicated? We are now in the 21st century (computer era).

Mharfe Micaroz - 6 years, 3 months ago

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