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09:48
they are $\phi$
phase angles
they are related but I did not bother to calculate how because it is irrelavent to the question
"Evert equation f(x) = 0 of degree odd has at least one real root of a sign opposite to that of its last term (if the coefficient of the highest degree term is positive)." Why?
10:29
Ok
Are you there?
Can I help?
Ok
Tell a function with odd degree
any polynomial with odd degree
the fact I stated is true for any general polynomial of odd degree
10:37
Absolutely... First tell me a generalized odd degree polynomial function as an example
We will then prove it for all other polynomials too
?
What happened?
Are you there?
y = a_0 x^n + a_1 x^{n-1} + ... + a_n
n is actually 2k+1 right?
Take it x^n common from the entire polynomial.
You will get : x^n(a0 + a1/x + a2/x^2 +......) = f(x)
10:48
f(+infinity) = x ^ n (a0) [because other terms will be zero] Same for f(-infinity)
Now if a0 is +ve , clearly f(-infinity)=-infinity and f(+infinity)=+infinity
We know a polynomial is a continuous function. So between -infinity and infinity , it must cut the x-axis at some point
Hence there must be atleast one real root
Understood?
:O
understood
what's that obvious theorem called?
intermediate value theorem?
God knows if there is any such theorem or not
In mathematical analysis, the intermediate value theorem states that if a continuous function, f, with an interval, [a, b], as its domain, takes values f(a) and f(b) at each end of the interval, then it also takes any value between f(a) and f(b) at some point within the interval. This has two important corollaries: 1) If a continuous function has values of opposite sign inside an interval, then it has a root in that interval (Bolzano's theorem). 2) The image of a continuous function over an interval is itself an interval. == Motivation == This captures an intuitive property of continuous functions...
Hmm then so... Intermediate theorem
We are still required to prove the last part
Should I ?
I know it
it's obvious
I find it very obvious
10:58
Very good
Ok. Then.
I have a doubt
Can you help?
what is it?
Lets say there are n distinct objects , r1 are of type 1 r2 are of type 2 r3 are of type 3 ... rk are of type k. In how many ways can they arrange among themselves?
Are you there?
Anyone interested in discussing the problem?
(0fcourse a extended version of the above problem)
11:14
I keep disappearing :p sorry
r1 type, r2 type, etc. have to be together?
n distinct objects :/
of type r1, r2, r3 .... ?
I am confused
Let say there are 10 numbers . 1 repeated 1 time 2 repeated 2 times 3 repeated 3 times 4 repeated 4 times.
In how many ways can the digits arrange among themselves to make a number 10 digit long
11:17
(r1 + r2 + r3....)!/(r1! r2! r3!...)?
Absolutely
Now let say we want to make a number 7 digit long. Now?
can do it with cases
but that'd be slow
a lot of cases
not a good idea lol
Yes.. Very very slow and confusing ... And what about a generalized formulae?
I need to think
need to think of some out of the box idea
Think on it.. I bitted my nails while thinking on it.
Now my nails pain.. Lol.
Are you preparing for jee?
11:26
yes
Cbse 2017 ?
Please let me know when you are done thinking
Ok
Let me know if you can find solution
did you attempt KVPY?
12:28
No
You?
12:51
yes
Cleared cut-off?
13:21
Interview?
results not out yet
14:18
did you get your admit card?
Jee. I got today
downloaded?
mine isn't loading
it asks me to enter my roll
it does not show anything after that
Anonymous
Even mine isn't loading :/
Anonymous
Server down
it asks for roll number
I enter it
then the same page reopens -,-
and asks for the roll again
Anonymous
14:28
true^
Anonymous
same happening with me
Anonymous
try it after few days maybe :/
Anonymous
@YashasSamaga Mine just worked
Anonymous
Try again
14:34
Our servers are facing a temporary technical problem.We regret the inconvenience caused.
Please try again after some time
I did @ 3 pm
Anonymous
@MadhuchhandaMandal Where are you from ?
Anonymous
14:51
@YashasSamaga That's very easy toh. Just conditional probability. $$P(\frac{3G3B}{\text{atleast 1 girl}})$$. Probability of atleast 1 girl = $2^6-1$. And you can arrange 3G3B in 6C3 ways. So answer should be $6C3/(2^6-1)$.
Anonymous
Order matters here because GBBBBB is not same as BGBBBB
15:18
WB
Ok. Is anyone active now?
Lemme show you something
If anyone is active
Anonymous
15:48
@MadhuchhandaMandal say
Anonymous
@MadhuchhandaMandal kolkata?

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