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        general >> complex analysis >> Binomial Expansion for Real Exponents
        
(Message started by: knightfischer on Sep 6^th, 2007, 6:39am)

Title: Binomial Expansion for Real Exponents
Post by knightfischer on Sep 6^th, 2007, 6:39am

I wonder if someone can help me with a proof of the Binomial expansion for real and complex exponents.

(1+z)^k = Sum(n 0 to inf) {(kCn)z^n},
where z is complex, k is real, n is an integer,
kCn = {k(k-1)(k-2)...(k-n+1)}/n!

The proof I found where x is real is as follows:

let g(x) = Sum(n 0 to inf) {(kCn)x^n},
then g'(x) = k(g(x)/(1+x)

define h(x) = {(1+x)^(-k)}g(x)
then h'(x) = 0

Deduce that g(x) = (1+x)^k

The steps of the proof may be a bit laborious to actually carry out, but I managed to demonstrate that h'(x) is identically 0. My questions are as follows:

1. Since h'(x) = 0, then h(x) = c, where c is some constant. Therefore, g(x) = c(1+x)^k. I was able to demonstrate for k rational, c=1; so, in fact, the binomial expansion is valid for k rational. However, if k is irrational, I cannot see how to demonstrate that c must equal 1 in this case also.

For rational k, I applied g(x) = c(1+x)^k as follows:

{c(1+x)^(-k)}^k = 1+x, simply multiplying out shows c must equal 1.

However, for k irrational, this "multiplying out" step does not make sense until I already know the theorem works. I cannot figure out how to demonstrate that c=1 in this case also.

2. Does a similar proof work for z, k complex? Again, the final step of "multiplying out" to demonstrate c=1 is not clear to me, without assuming the theorem is already true.

Any help with this would be greatly appreciated.

Title: Re: Binomial Expansion for Real Exponents
Post by towr on Sep 6^th, 2007, 6:48am

on 09/06/07 at 06:39:24, knightfischer wrote:
1. Since h'(x) = 0, then h(x) = c, where c is some constant. Therefore, g(x) = c(1+x)^k. I was able to demonstrate for k rational, c=1; so, in fact, the binomial expansion is valid for k rational. However, if k is irrational, I cannot see how to demonstrate that c must equal 1 in this case also.
But isn't k always an integer?

For that matter; shouldn't we have
(1+z)^k = Sum(n 0 to k) {(kCn)z^n},
instead of
(1+z)^k = Sum(n 0 to inf) {(kCn)z^n}
at the start.

Nevermind; I should have read http://mathworld.wolfram.com/BinomialTheorem.html first..

Title: Re: Binomial Expansion for Real Exponents
Post by knightfischer on Sep 6^th, 2007, 8:19am

Note: I should have stipulated |z|<1, |x|<1.

Title: Re: Binomial Expansion for Real Exponents
Post by towr on Sep 6^th, 2007, 9:11am

Can't you deduce c, by looking at lim x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif 0 g(x)
lim x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif 0 Sum(n 0 to inf) {(kCn)x^n} = 1
lim x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif 0 c(1+x)^k = c
So c=1

Title: Re: Binomial Expansion for Real Exponents
Post by knightfischer on Sep 6^th, 2007, 9:51am

Yes, I did not consider that idea. That seems to work for all k, complex or real.

Thanks for your help.