Binomial Series 2nd and 3rd terms

KindofSlow

Junior Member
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Mar 5, 2010
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In the beginning of the binomial series where:
for (1+x)^k, Series = 1 + kx + (k(k-1))/2!*x^2 + (k(k-2))/3!*x^3...

It seems to me that the numerator of the 3rd term of k(k-1) should be in the second term of the series instead of the third.
I think this because k(k-1) is the 2nd term of k! so should be in the 2nd term of series.
Is there a simple or obvious or intuitive reason why my thinking is flawed?
As always, any insight is greatly appreciated.

Thank you
 
Say we have the MacLaurin series for \(\displaystyle (1+x)^{m}\) with m a real number.

Then, we call this series the binomial series.

\(\displaystyle f(0)+f'(0)x+\frac{f''(0)}{2!}x^{2}+........\)

Take the third tem.

\(\displaystyle f''(x)=m(m-1)(1+x)^{m-2}\)

\(\displaystyle f''(0)=m(m-1)\)

There it is.

If m is non negative, then \(\displaystyle (1+x)^{m}\) is a polynomial of degree m, so we get:

\(\displaystyle f^{(m+1)}(0)=f^{(m+2)}(0)=f^{(m+3)}(0)=\cdot \cdot \cdot =0\)

and the binomial series reduces to:

\(\displaystyle (1+x)^{m}=1+mx+\frac{m(m-1)}{2!}x^{2}+\frac{m(m-1)(m-2)}{3!}x^{3}+\cdot\cdot \cdot x^{m}\)

and it is valid as long as x is in the interval \(\displaystyle (-\infty, \infty)\):)
 
Interesting. I don't remember ever hearing of a binomial series, but I just checked out wikipedia. So this is in fact a generalization of the binomial expansion it seems, but I had never made such a connection between it and the coefficients of a taylor expansion. So if the power on a binomial is not a non-negative integer, its binomial expansion has to be represented by a series? I think that is what I read. It makes sense, as differentiation then would not reach a constant.

In a probability class I've taken some time ago, I was made to accept that the choose function could now accept arbitrary real numbers, though I didn;t try to understand it then, and now I see complex. Ughh. Haha.
 
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