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anonymous

  • one year ago

Find the upper-bound for the summation, such that: (using the lowest possible integer)

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  1. amistre64
    • one year ago
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    it feels like there is something missing in the post

  2. anonymous
    • one year ago
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    oops

  3. anonymous
    • one year ago
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    \[\sum_{n=4200}^{x}n ^{1.5} \ge 1.0 \times 10^{8}\]

  4. anonymous
    • one year ago
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    not sure if possible. I could solve it if it was something like 5n

  5. amistre64
    • one year ago
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    does this have anything to do with a remainder thrm?

  6. anonymous
    • one year ago
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    I want to solve it for a game. Putting in random numbers, the closest integer I could get to have the same as close as possible to 100 million is 4563

  7. anonymous
    • one year ago
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    *starting at 4220 >_>

  8. anonymous
    • one year ago
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    Wolfram gives me weird zeta functions or harmonic numbers

  9. amistre64
    • one year ago
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    10^8, or 10^7?

  10. anonymous
    • one year ago
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    100 million, so 10^8 (if I did that right)

  11. amistre64
    • one year ago
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    \[\int_{4200}^{x}x^{1.5}dx=x^{2.5}/2.5-(4200)^{2.5}/2.5\] \[x^{2.5}/2.5-(4200)^{2.5}/2.5=10^8\] \[x^{2.5}-(4200)^{2.5}=2.5*10^8\] \[x^{2.5}=(4200)^{2.5}+2.5*10^8\] \[x=((4200)^{2.5}+2.5*10^8\])^{1/}

  12. amistre64
    • one year ago
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    ugh ... firefox crashed at the end of that

  13. amistre64
    • one year ago
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    \[x=((4200)^{2.5}+2.5*10^8)^{1/2.5}=4545.ddd\]

  14. anonymous
    • one year ago
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    I tried integrals, I must have done it wrong.

  15. amistre64
    • one year ago
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    http://www.wolframalpha.com/input/?i=sum%28n%3D4200+to+4545%29n^%281.5%29

  16. amistre64
    • one year ago
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    letting x=4545 seems to be fair

  17. anonymous
    • one year ago
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    oh, I see what I did wrong. Thank you :D

  18. amistre64
    • one year ago
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    yw

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