OpenStudy (anonymous):

@ganeshie8

2 years ago
OpenStudy (zzr0ck3r):

8eihsenag@

2 years ago
ganeshie8 (ganeshie8):

send meeting id

2 years ago
OpenStudy (anonymous):

oh ok. Let me open teamview

2 years ago
OpenStudy (anonymous):

uhm... can not connect? O.o

2 years ago
OpenStudy (anonymous):

is it the same ID you game me last time?

2 years ago
OpenStudy (anonymous):

m14-003-342 ?

2 years ago
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

2 years ago
OpenStudy (anonymous):

m09-542-741

2 years ago
OpenStudy (anonymous):

what do I do now?

2 years ago
OpenStudy (anonymous):

@ganeshie8 uhm... hello? :D

2 years ago
ganeshie8 (ganeshie8):

joining..

2 years ago
OpenStudy (anonymous):

2 years ago
ganeshie8 (ganeshie8):

It says you haven't started the presentation...

2 years ago
OpenStudy (anonymous):

O.o how do I start it?

2 years ago
OpenStudy (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 $$f(x)\equiv g(x) \pmod{p}$$

2 years ago
OpenStudy (anonymous):

$$(x+jp)^k = x^k + p*(some integer)$$

2 years ago
OpenStudy (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

2 years ago
OpenStudy (anonymous):

$$(x+jp)^k = x^k + p*(some integer) \equiv x^k + 0 \pmod{p}$$

2 years ago