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   Binomial Expansion for Real Exponents
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   Author  Topic: Binomial Expansion for Real Exponents  (Read 6649 times)
knightfischer
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Binomial Expansion for Real Exponents  
« on: Sep 6th, 2007, 6:39am »
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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.
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towr
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Re: Binomial Expansion for Real Exponents  
« Reply #1 on: Sep 6th, 2007, 6:48am »
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on Sep 6th, 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 6th, 2007, 7:10am by towr » IP Logged

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knightfischer
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Re: Binomial Expansion for Real Exponents  
« Reply #2 on: Sep 6th, 2007, 8:19am »
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Note:  I should have stipulated |z|<1, |x|<1.
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Re: Binomial Expansion for Real Exponents  
« Reply #3 on: Sep 6th, 2007, 9:11am »
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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
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knightfischer
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Re: Binomial Expansion for Real Exponents  
« Reply #4 on: Sep 6th, 2007, 9:51am »
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Yes, I did not consider that idea.  That seems to work for all k, complex or real.
 
Thanks for your help.
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