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Topic: Binomial Expansion for Real Exponents (Read 6940 times) |
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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. |
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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..
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| « Last Edit: Sep 6th, 2007, 7:10am by towr » |
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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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towr
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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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