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ParthKohli
 one year ago
Cool solution.
ParthKohli
 one year ago
Cool solution.

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Problem: Prove that\[\left(\frac{1 + a}{1+b}\right)^{\frac{1}{ab}}< e\]for distinct reals \(a,b\).

freckles
 one year ago
Best ResponseYou've already chosen the best response.2How many seconds can I have to think about it before you give the solution? :p

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Try it out. Just meant to share this problem.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[(\frac{1+a}{1+b})^{\frac{1}{ab}} \\ =(\frac{1+b+ab}{1+b})^\frac{1}{ab} \\ =(1+\frac{ab}{1+b})^{\frac{1}{ab}}\] \[=(1+\frac{1}{\frac{1+b}{ab}})^\frac{1}{ab} \\ =(1+\frac{\frac{1}{1+b}}{\frac{1}{ab}})^\frac{1}{ab}\] \[\lim_{ab \rightarrow \infty}(1+\frac{\frac{1}{1+b}}{\frac{1}{ab}})^{\frac{1}{ab}}=\lim_{ab \rightarrow \infty} e^{\frac{1}{1+b}} \\ ...\] if ab goes to infinity I don't think we can say anything about the limit of b from that so that limit should be just exp(1/(1+b)) ..I don't know if this is the right route For some reason this is the first thing that popped in my head

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Ganeshie is here  he'll easily crack this one. :

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I did something wrong

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I should have said 1/(ab) goes to infinity

freckles
 one year ago
Best ResponseYou've already chosen the best response.2which would mean ab goes to 0

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1`ganeshie8 is typing a reply...` Well, here we go.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$\left(\frac{1 + a}{1+b}\right)^{\frac{1}{ab}}< e\\\frac1{ab}\log\left(\frac{1+a}{1+b}\right)<1\\\frac1{ab}\left(\log(1+a)\log(1+b)\right)<1\\\log(1+a)\log(1+b)<(1+a)(1+b)$$... is just a restatement of the fact that \(\log\) is concave by Jensen's inequality

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0all those steps are reversible, so the proof just follows from the fact that \(\log\) is concave via Jensen's inequality

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I don't think my way is good enough it only shows for a close to b we have exp(1/(1+b)) but I think this right here is <e for any real b

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[e^{xy} \gt 1+(xy)+(xy)^2/2 \gt 1+xy = 1+\dfrac{xy}{1+y} = \frac{1+x}{1+y}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0haha that one is great

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but hmm @ganeshie8 both you and i have only proven it for \(ab>0\) technically

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0unless you can rigorously show that the even terms outweigh the odd ones in the expansion \(e^x=1+x+\frac12 x^2+\dots\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0https://en.wikipedia.org/wiki/Taylor%27s_theorem#Estimates_for_the_remainder

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2looks the taylor series one also requires \(y\gt 0\) \[e^{xy} \gt 1+(xy)\gt 1+\dfrac{xy}{1+y} = \frac{1+x}{1+y}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@parthkohli share your answer

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1@ganeshie8 That's exactly my answer. I think I got the incomplete question.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Hey, your next project is to implement LaTeX in the drawing tool. That looks good.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2looks the inequality is true only when \(a,b\) have same signs dw:1437673377179:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1So yes, I did get an incomplete question. You can't expect much from Facebook math groups.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Haha okay but my/your solution is incomplete too even after restricting to first quadrant because it requires \(ab\gt 0\) as pointed out by oldrin earlier
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