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bahrom7893

  • 4 years ago

Zarkon last one!

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  1. Zarkon
    • 4 years ago
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    I can already feel my brain shutting down ;)

  2. bahrom7893
    • 4 years ago
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    mine shutdown like 3 hours ago

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  3. bahrom7893
    • 4 years ago
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    Part A's easy

  4. Zarkon
    • 4 years ago
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    yes...i get 65/24

  5. bahrom7893
    • 4 years ago
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    1+1+1/2+1/6+1/24

  6. Pippa
    • 4 years ago
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    YAY :D

  7. Zarkon
    • 4 years ago
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    part b looks like induction

  8. FoolForMath
    • 4 years ago
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    part b is definitely induction ... this property could be used to show that 2<e<3

  9. Zarkon
    • 4 years ago
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    for n=1 we have 1/1!=1 and 1/2^(1-1)=1/1=1 \(1\le 1\) so the basis step holds assume \[\frac{1}{k!}\le\frac{1}{2^{k-1}}\] is true for some \(k\) then \[\frac{1}{(k+1)!}=\frac{1}{k!}\frac{1}{k+1}\le\frac{1}{2^{k-1}}\frac{1}{k+1}\] since \(\frac{1}{2}\ge\frac{1}{n+1}\) for \(n\ge1\) we have \[\frac{1}{2^{k-1}}\frac{1}{k+1}\le\frac{1}{2^{k-1}}\frac{1}{2}=\frac{1}{2^{k}}\] QED

  10. FoolForMath
    • 4 years ago
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    Prof Zarkon I need your inputs on another problem, i have pinged you from the relevant thread. Please do check it :)

  11. FoolForMath
    • 4 years ago
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    btw for part c the upper bound is \( 3\).

  12. bahrom7893
    • 4 years ago
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    ty guys, i can die a peaceful death now, knowing that 9.3 is finally done...

  13. FoolForMath
    • 4 years ago
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    when (exactly) are you going to die? :P

  14. bahrom7893
    • 4 years ago
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    not any time soon, hopefully.

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