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.
the problems goes like that..
Teacher: Find number of non-negative integer solutions to the equation x1+ x2 + x3 + ....+ x100 = 10
Student: It's just the co - efficient of x10 in the (1 + x + x3+...)100 = (1-x)-100
Teacher: Good, write the next question.In how many...
Student: wait sir! (1 + x + x2+ x3+...) = (1-x) if and only if |x|<1 . But here no such condition is given. So the relation (1 + x + x2+ x3+...) = (1-x) is not always true. So isn't the method wrong ?
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Suppose you are the teacher. what will be your answer?
nt dbt
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5 Answers
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]
21
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!
21yes. 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.