anonymous
  • anonymous
@ganeshie8
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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zzr0ck3r
  • zzr0ck3r
8eihsenag@
ganeshie8
  • ganeshie8
send meeting id
anonymous
  • anonymous
oh ok. Let me open teamview

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anonymous
  • anonymous
uhm... can not connect? O.o
anonymous
  • anonymous
is it the same ID you game me last time?
anonymous
  • anonymous
m14-003-342 ?
ganeshie8
  • ganeshie8
nope that was finished.. you need to start a fresh meeting.. click on Meeting, then you will see "presentation" icon on left pane
anonymous
  • anonymous
m09-542-741
anonymous
  • anonymous
what do I do now?
anonymous
  • anonymous
@ganeshie8 uhm... hello? :D
ganeshie8
  • ganeshie8
joining..
anonymous
  • anonymous
oh I see your name
ganeshie8
  • ganeshie8
It says you haven't started the presentation...
anonymous
  • anonymous
O.o how do I start it?
anonymous
  • anonymous
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 \(f(x)\equiv g(x) \pmod{p}\)
anonymous
  • anonymous
\((x+jp)^k = x^k + p*(some integer)\)
anonymous
  • anonymous
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
anonymous
  • anonymous
\((x+jp)^k = x^k + p*(some integer) \equiv x^k + 0 \pmod{p}\)

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