Binomial theorem for fractional exponents?

[Juan Pablo]

I am curious, is there any way to use the binomial theorem for fractional exponents? Is there any other way to expand a binomial with a fractional exponent?
I suppose Newton's theorem is not a way since it requires factorials.

Thanks!

 

[rock.freak667]

You'd work it out in the same basic way. But just that you'd have an infinite number of terms

nC1=n!/(n-1)!1!

and n!=n(n-1)!

so nC1 simplifies to n

Similarly

nC2=n!/(n-2)!2!

=\frac{n(n-1)(n-2)!}{(n-2)!2!}=\frac{n(n-1)}{2!}

and so forth for nC3,nC4,etc

 

[robert Ihnot]

Approximating square roots

One use, or was so before calculators, is to approximate certain square roots. Take this case,

\sqrt{1+a^2} = a +\frac{1}{2a}-\frac{1}{8a^3} +-+

In the case of \sqrt{101} = 10 + 1/20-1/8000 + -

This is just a little less that 10.05 and can be easily carried out.

 

[Surajit93]

I also wish to know, is there any way that I can expand the general

(1+x)^(1/2) = Polynomial_function(x) ??

(1+x)^(n) easily works out for n>0, n<0,n=0...but what about for n = 1/m form ??
m is an Integer.

 

[uart]

I also wish to know, is there any way that I can expand the general

(1+x)^(1/2) = Polynomial_function(x) ??

(1+x)^(n) easily works out for n>0, n<0,n=0...but what about for n = 1/m form ??
m is an Integer.

(1+x)^{a}=1+ax+a(a-1)\,\frac{x^{2}}{2!}+a(a-1)(a-2)\,\frac{x^{3}}{3!}+...

This works for integer and non integer "a". So a=1/m is no problems.