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bahrom7893 Group Title

Zarkon last one!

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

    • 2 years ago
  2. bahrom7893 Group Title
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    mine shutdown like 3 hours ago

    • 2 years ago
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  3. bahrom7893 Group Title
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    Part A's easy

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

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

    • 2 years ago
  6. Pippa Group Title
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    YAY :D

    • 2 years ago
  7. Zarkon Group Title
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    part b looks like induction

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

    • 2 years ago
  9. Zarkon Group Title
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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

    • 2 years ago
  10. FoolForMath Group Title
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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 :)

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

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

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

    • 2 years ago
  14. bahrom7893 Group Title
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    not any time soon, hopefully.

    • 2 years ago
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