49I wud say bachchhe.. don't control your emotion! [6]
We are only worried about the coefficient of x...so no problem if we consider x to be a fraction! :D
I feel I m wrong.. par fir bhi itna feel-good factor huwa ki post kar hi diya! [3]
21but how can even the coefficient be the equal in LHS and RHS for any xn. We will put any x:|x|>1 and RHS≠LHS
We use (1-x)-2 = 1+ 2x + 3x2+..... but for |x|>1 this is invalid.
1ok, look,
we are using 'x' and the binomial expansion as a 'tool' to get to the final answer. If the kid still isn't able to convince himself,
see this:
to get the coefficient, (OK, assuming your x>1),
we don't need to sum up (1+x+x2+x3...till infinity)100
if you think, you'll notice, (1+x+x2+x3+..+xn)100 will do, where n≥100
this works well for us, because we know summing up a diverging series up to infinite terms will not be finite.
so we have (xn+1-1)100/(x-1)100
=(xn+1-1)100*(x-1)-100
=(-1)-100(100C0(-1)100+100C0(-1)99xn+1.....)(1-x)-100
no good is going to come off all the terms of this second bracket apart from the first term,
as n+1>100
so we are reduced to (-1)-100*100C0(-1)100*(1-x)-100
=(1-x)-100
this works well for any k, where we need coeff of xk
so you see, it ultimately reduces to the same expression.
cheers!
21
Shubhodip
·2011-06-13 03:13:14
yes. thanks gordo for your explanation. my thought on the same was like this.
Let A= {2,3,4,5,6,7,8} be a sequence. PA(x)= 2x + 3x2+ 4x3+ ..+ 8x8 is called the generating function (of the 1st kind) of the sequence A. i.e. co-efficient of xn is the n-th term of the sequence.
.evaluating value of generating functions at some x is meaningless. By rules of generating function one can prove that (1-x)(1+ x2+ x3+ ....) = 1. Convergence has nothing to do in the actual proof.
