Challenge complex numbers problem for current students (2 Viewers)

[Faera]

haha, i'll say...

but can you use the limit as x ---> -0 ? because when x < 0 you dont get a smooth curve...

i think...

would it matter if it wasnt a smooth curve?

 

[Grey Council]

Really?

[/B]

yes, really. Read:

Originally posted by Grey Council

how do you get x^0 = 1?
x=x^1
divide by x on both sides:
x/x = x^0
= 1

I PROVED that x^0 = 1. why would I ask you something which I already knew how to PROVE? (forget about knowing about it lol)

hehe, btw, Faera, if you got that 0^0 is undefined, than it must be undefined.

hell, David Chan, (turtle_2468) told me its undefined, so I think I'll believe him. ^_^

 

[Xayma]

Originally posted by Grey Council
I PROVED that x^0 = 1. why would I ask you something which I already knew how to PROVE? (forget about knowing about it lol)

But you forgot x<>0 as you divided by x.

 

[Grey Council]

well, not taking the case of x=0 (cause thats what the whole debate is about, right? )

anyway, just stick in 0^0 on your calculator, it says error.
so its undefined. ^_^

 

[Xayma]

yes but if I also do 10¹⁰⁰ it says maths error, does that make 10¹⁰⁰ undefined? Or if I chuck e^-i it also comes up with a maths error (even when its in the mode for complex calculation )

 

[Faera]

e ^ (-i) is not in the form x^x...
nor is 10^100
and the only reason the thing says error calc can't handle that because it's too large a number...
try doing it on your comp...

 

[Xayma]

I know, but just because a calc puts it as undefined doesnt make it so. Of course x^x is probably undefined.

 

[Grey Council]

sigh

Okay, the definitve answer. ^___^

the defn of 0^0 is plain silly. Because the only way you get numbers to the power of zero from basic axioms, power definitions etc is to start from x=x^1 and divide, you get x/x=x^0. But you can't divide by 0 so 0/0 and hence 0^0 is plain silly...
The reason you can get 0^(0.1), for example, is that you can say 0^1=0
and letting the weird number above be m
m^10=0^1=0
so m=0
But doing 0^0 is a whole different ball game...

2. You know probability of k heads in n coins is nCk/2^n
So probability of both ppl getting k heads is (nCk)^2/2^2n
So all you have to prove is that (the sum from 0 to n of (nCk)^2)/2^2n=(2n)! / [2^(2n) * (n!)^2]
ie sum from 0 to n of (nCk)^2)=(2n)!/(n!)^2 (note lhs = no. of ways of getting same no of heads)
So restating problem 2:
Equivalent to -
Show that the number of ways that they get the same number of heads is
(2n)Cn.
Proof: By establishing a bijection (ie correlating one way to get same heads with one way to pick n items from 2n).

Arrange the 2n objects in a row, and draw a line down the middle (separating objects into left and right). The ones you pick, tick them, the ones you don't, cross them.
A tick means heads on the left, and tails on the right. A cross means tails on the left, and heads on the right.
This establishes a relation between the two, basically as it "preserves" the constant k between the two sides, k being (left heads-left tails)-(right heads - right tails). cross decreases k by 1, ticks increases by 1. As cross=ticks, k stays at 0 by the time all the ticks and crosses are converted into heads and tails ie heads-tails is same on both sides ie heads is same on both sides.

 

[Kirsti]

oooh! i took the first one as being z^(n-1) not (z^n) -1
hmm.. that makes it easier

 

[George W. Bush]

axiom: if a/b = c then a = bc

0^0 = 0/0 which can equal every possible number
i.e. indeterminate, not undefined