# binomial theorem for all powers

• Aug 31st 2012, 02:12 AM
startanewww
binomial theorem for all powers
My book mentioned that
"If -1<x<1, then (1+x)^n = 1 + nx + n(n-1)(x^2)/2! + n(n-1)(n-2)(x^3)/3! + ... + n(n-1)(n-2)...(n-r+1)(x^r)/r! + ..."
Why -1<x<1? Why is this restriction on x necessary?
• Aug 31st 2012, 02:49 AM
Prove It
Re: binomial theorem for all powers
Quote:

Originally Posted by startanewww
My book mentioned that
"If -1<x<1, then (1+x)^n = 1 + nx + n(n-1)(x^2)/2! + n(n-1)(n-2)(x^3)/3! + ... + n(n-1)(n-2)...(n-r+1)(x^r)/r! + ..."
Why -1<x<1? Why is this restriction on x necessary?

Because it's possible that infinite series might diverge (go to infinity). You need to check the interval for which the series is convergent.
• Aug 31st 2012, 02:58 AM
startanewww
Re: binomial theorem for all powers
In that case, shouldn't the restriction be "x =/= +infinity or -infinity (when n is just a small number)"?
If -1<x<1, will the value of (1+x)^n be very small?
I really don't understand..
• Aug 31st 2012, 03:11 AM
Prove It
Re: binomial theorem for all powers
The series depends on x. But the series itself is only convergent if it follows specific conditions, which means that it will only be convergent for particular values of x (the ones which will satisfy the conditions). I suggest you research convergence tests for infinite series.
• Aug 31st 2012, 04:10 AM
Plato
Re: binomial theorem for all powers
Quote:

Originally Posted by startanewww
My book mentioned that
"If -1<x<1, then (1+x)^n = 1 + nx + n(n-1)(x^2)/2! + n(n-1)(n-2)(x^3)/3! + ... + n(n-1)(n-2)...(n-r+1)(x^r)/r! + ..."
Why -1<x<1? Why is this restriction on x necessary?

I think that you are missing the point.
$\displaystyle {\left( {1 + x} \right)^n} = \sum\limits_{k = 0}^n {{\binom{n}{k} x^k}}$ for any $\displaystyle x\in\mathbb{R}$.
So it would be true for $\displaystyle -1<x<1.$
• Aug 31st 2012, 04:21 AM
Prove It
Re: binomial theorem for all powers
Quote:

Originally Posted by Plato
I think that you are missing the point.
$\displaystyle {\left( {1 + x} \right)^n} = \sum\limits_{k = 0}^n {{\binom{n}{k} x^k}}$ for any $\displaystyle x\in\mathbb{R}$.
So it would be true for $\displaystyle -1<x<1.$

The OP was actually referring to the GENERALISED Binomial Theorem, which is an infinite series.