Author 
Topic: Binomial Expansion for Real Exponents (Read 6649 times) 

knightfischer
Junior Member
Gender:
Posts: 54


Binomial Expansion for Real Exponents
« on: Sep 6^{th}, 2007, 6:39am » 
Quote Modify

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(k1)(k2)...(kn+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.


IP Logged 



towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730


Re: Binomial Expansion for Real Exponents
« Reply #1 on: Sep 6^{th}, 2007, 6:48am » 
Quote Modify

on Sep 6^{th}, 2007, 6:39am, 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..

« Last Edit: Sep 6^{th}, 2007, 7:10am by towr » 
IP Logged 
Wikipedia, Google, Mathworld, Integer sequence DB



knightfischer
Junior Member
Gender:
Posts: 54


Re: Binomial Expansion for Real Exponents
« Reply #2 on: Sep 6^{th}, 2007, 8:19am » 
Quote Modify

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


IP Logged 



towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730


Re: Binomial Expansion for Real Exponents
« Reply #3 on: Sep 6^{th}, 2007, 9:11am » 
Quote Modify

Can't you deduce c, by looking at lim x 0 g(x) lim x 0 Sum(n 0 to inf) {(kCn)x^n} = 1 lim x 0 c(1+x)^k = c So c=1


IP Logged 
Wikipedia, Google, Mathworld, Integer sequence DB



knightfischer
Junior Member
Gender:
Posts: 54


Re: Binomial Expansion for Real Exponents
« Reply #4 on: Sep 6^{th}, 2007, 9:51am » 
Quote Modify

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


IP Logged 



