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geerky42

  • one year ago

How can I verify if this is true for \(n\in\mathbb N\)? \[\Large \dfrac1{n^n}=\sum_{i=0}^n\sum_{k=0}^i\dfrac{(-1)^kn^{k-i}}{k!(n-i)!}\]

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  1. geerky42
    • one year ago
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    It is true for \(n=1\) and \(n=2\).

  2. Kainui
    • one year ago
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    This site is terrible \[\Large \dfrac1{n^n}=\sum_{i=0}^n\sum_{k=0}^i\dfrac{(-1)^kn^{k-i}}{k!(n-i)!}\]

  3. anonymous
    • one year ago
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    yes @Kainui i completely agree :P

  4. geerky42
    • one year ago
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    I stumped this while solving big kid club lol @Kainui

  5. anonymous
    • one year ago
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    wait no lol i read ur question in some other way :P sorry its not zeta

  6. rational
    • one year ago
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    seems to be true http://jsfiddle.net/ganeshie8/ykrpkweu/

  7. dan815
    • one year ago
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    is it enuff if i show it to be true for the integral analog of it

  8. dan815
    • one year ago
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    |dw:1441227726264:dw|

  9. dan815
    • one year ago
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    |dw:1441227959575:dw|

  10. dan815
    • one year ago
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    |dw:1441228057588:dw|

  11. dan815
    • one year ago
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    part of e^-n i shud say in this case

  12. dan815
    • one year ago
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    so its like uare adding higher and higher orders of e^-n together

  13. dan815
    • one year ago
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    summation of 1/factorials have to with gamma function again

  14. dan815
    • one year ago
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    or u can think about it like adding different orders of e^1 now

  15. dan815
    • one year ago
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    higher and higher degeree approximations of e^1 being adding together

  16. dan815
    • one year ago
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    g2g mow the lawn

  17. anonymous
    • one year ago
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    :) @dan815

  18. geerky42
    • one year ago
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    Not sure how approximation of \(e^n\) helps us?

  19. anonymous
    • one year ago
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    where have u seen this formula @geerky42 ? or u just created it ?

  20. geerky42
    • one year ago
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    I kind of created it. I tried to solve @Kainui 's "the big kid's club" and I got stuck on close-to-last step, then I saw solution, and bam! I got this formula.

  21. geerky42
    • one year ago
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    I applied Binomial Theorem and took tedious path, so I sincerely hope this formula is true, because my solution is like 2 whole pages long and otherwise it'd be hard to hunt for mistake lol...

  22. anonymous
    • one year ago
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    :O thats cool

  23. geerky42
    • one year ago
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    @dan815 ?

  24. geerky42
    • one year ago
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    Not sure how approximation of \(e^n\) helps us?

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