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ParthKohli

  • one year ago

Cool solution.

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  1. ParthKohli
    • one year ago
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    Problem: Prove that\[\left(\frac{1 + a}{1+b}\right)^{\frac{1}{a-b}}< e\]for distinct reals \(a,b\).

  2. freckles
    • one year ago
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    How many seconds can I have to think about it before you give the solution? :p

  3. ParthKohli
    • one year ago
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    Try it out. Just meant to share this problem.

  4. freckles
    • one year ago
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    \[(\frac{1+a}{1+b})^{\frac{1}{a-b}} \\ =(\frac{1+b+a-b}{1+b})^\frac{1}{a-b} \\ =(1+\frac{a-b}{1+b})^{\frac{1}{a-b}}\] \[=(1+\frac{1}{\frac{1+b}{a-b}})^\frac{1}{a-b} \\ =(1+\frac{\frac{1}{1+b}}{\frac{1}{a-b}})^\frac{1}{a-b}\] \[\lim_{a-b \rightarrow \infty}(1+\frac{\frac{1}{1+b}}{\frac{1}{a-b}})^{\frac{1}{a-b}}=\lim_{a-b \rightarrow \infty} e^{\frac{1}{1+b}} \\ ...\] if a-b 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

  5. ParthKohli
    • one year ago
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    Ganeshie is here - he'll easily crack this one. :|

  6. freckles
    • one year ago
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    I did something wrong

  7. freckles
    • one year ago
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    I should have said 1/(a-b) goes to infinity

  8. freckles
    • one year ago
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    which would mean a-b goes to 0

  9. ParthKohli
    • one year ago
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    `ganeshie8 is typing a reply...` Well, here we go.

  10. anonymous
    • one year ago
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    $$\left(\frac{1 + a}{1+b}\right)^{\frac{1}{a-b}}< e\\\frac1{a-b}\log\left(\frac{1+a}{1+b}\right)<1\\\frac1{a-b}\left(\log(1+a)-\log(1+b)\right)<1\\\log(1+a)-\log(1+b)<(1+a)-(1+b)$$... is just a restatement of the fact that \(\log\) is concave by Jensen's inequality

  11. anonymous
    • one year ago
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    all those steps are reversible, so the proof just follows from the fact that \(\log\) is concave via Jensen's inequality

  12. freckles
    • one year ago
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    I 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

  13. ganeshie8
    • one year ago
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    \[e^{x-y} \gt 1+(x-y)+(x-y)^2/2 \gt 1+x-y = 1+\dfrac{x-y}{1+y} = \frac{1+x}{1+y}\]

  14. anonymous
    • one year ago
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    haha that one is great

  15. anonymous
    • one year ago
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    but hmm @ganeshie8 both you and i have only proven it for \(a-b>0\) technically

  16. ganeshie8
    • one year ago
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    Ah right, I see..

  17. anonymous
    • one year ago
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    unless you can rigorously show that the even terms outweigh the odd ones in the expansion \(e^x=1+x+\frac12 x^2+\dots\)

  18. anonymous
    • one year ago
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    https://en.wikipedia.org/wiki/Taylor%27s_theorem#Estimates_for_the_remainder

  19. ganeshie8
    • one year ago
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    looks the taylor series one also requires \(y\gt 0\) \[e^{x-y} \gt 1+(x-y)\gt 1+\dfrac{x-y}{1+y} = \frac{1+x}{1+y}\]

  20. anonymous
    • one year ago
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    @parthkohli share your answer

  21. ParthKohli
    • one year ago
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    @ganeshie8 That's exactly my answer. I think I got the incomplete question.

  22. ParthKohli
    • one year ago
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    Hey, your next project is to implement LaTeX in the drawing tool. That looks good.

  23. ganeshie8
    • one year ago
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    looks the inequality is true only when \(a,b\) have same signs |dw:1437673377179:dw|

  24. ParthKohli
    • one year ago
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    So yes, I did get an incomplete question. You can't expect much from Facebook math groups.

  25. ganeshie8
    • one year ago
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    Haha okay but my/your solution is incomplete too even after restricting to first quadrant because it requires \(a-b\gt 0\) as pointed out by oldrin earlier

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