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anonymous

  • one year ago

@ganeshie8

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  1. zzr0ck3r
    • one year ago
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    8eihsenag@

  2. ganeshie8
    • one year ago
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    send meeting id

  3. anonymous
    • one year ago
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    oh ok. Let me open teamview

  4. anonymous
    • one year ago
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    uhm... can not connect? O.o

  5. anonymous
    • one year ago
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    is it the same ID you game me last time?

  6. anonymous
    • one year ago
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    m14-003-342 ?

  7. ganeshie8
    • one year ago
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    nope that was finished.. you need to start a fresh meeting.. click on Meeting, then you will see "presentation" icon on left pane

  8. anonymous
    • one year ago
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    m09-542-741

  9. anonymous
    • one year ago
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    what do I do now?

  10. anonymous
    • one year ago
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    @ganeshie8 uhm... hello? :D

  11. ganeshie8
    • one year ago
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    joining..

  12. anonymous
    • one year ago
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    oh I see your name

  13. ganeshie8
    • one year ago
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    It says you haven't started the presentation...

  14. anonymous
    • one year ago
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    O.o how do I start it?

  15. anonymous
    • one year ago
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    Let : \[f(x) = \sum\limits_{k=0}^{n}a_kx^k\] \[g(x) = \sum\limits_{k=0}^{n}a_k(x+jp)^k\] show that \(f(x)\equiv g(x) \pmod{p}\)

  16. anonymous
    • one year ago
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    \((x+jp)^k = x^k + p*(some integer)\)

  17. anonymous
    • one year ago
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    let me skip typing the sum, so it's x^i (jp)^(k-i), i from 0 to k ? oh, x^0 (jp)^(k-0) + x^1 (jp)^(k-1) + x^2 (jp)^(k-2) + ... + x^(k-1)(jp)^1 + x^k (jp)^0 yeah I see p as a factor of every term except the last term p*(some integer) + x^k yes

  18. anonymous
    • one year ago
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    \((x+jp)^k = x^k + p*(some integer) \equiv x^k + 0 \pmod{p}\)

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spraguer (Moderator)
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