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# which is the greater? ( i still dont understand )

without calculator

Note by NurFitri Hartina
4 years ago

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

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 $$\LARGE ( \frac {e}{2} )^{\sqrt{3}} < (\sqrt2)^{\frac {\pi}{2} }$$ · 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? :-) · 4 years ago

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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 · 4 years ago

that was very good, that's a lot of story · 4 years ago

(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. · 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. · 4 years ago

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.) · 4 years ago

Wow. Which level is this? · 4 years ago

Use logarithms! · 4 years ago

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 · 4 years ago

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. · 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) · 4 years ago

Just for further information about what I had said: http://en.wikipedia.org/wiki/Exponentiation#Rational_exponents · 4 years ago

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 · 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.) · 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. · 4 years ago