Proof of Haldane's Binomial Expansion

[gnome]

The next-to-last step in the proof on pg 1 of this article

http://links.jstor.org/sici?sici=0006-3444%28194511%2933%3A3%3C222%3AOAMOEF%3E2.0.CO%3B2-Whttp://links.jstor.org/sici?sici=0006-3444%28194511%2933%3A3%3C222%3AOAMOEF%3E2.0.CO%3B2-W

makes this substitution

$$\sum_{r=0}^\infty \binom {m+r-2}{r} q^r = (1-q)^{1-m}$$

I don't see it. How does
$$(1-q)^{1-m} = \sum_{r=0}^\infty \binom{1-m}{r} (-q)^r$$
or
$$(1-q)^{1-m} = \sum_{r=0}^\infty \binom{1-m}{r} (-q)^{1-m-r}$$
transform to the expression given by Haldane?

 

[gnome]

I just remembered that this m is always a positive integer so (1-m) is 0 or negative.

The key to this must be here...

http://mathworld.wolfram.com/NegativeBinomialSeries.html"

 

[tacman]

The substitution made in the proof is a result of the binomial theorem, which states that for any real or complex numbers x and y, and any non-negative integer n, the following formula holds:

(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k}y^k

In this case, x = 1-q, y = -q, and n = 1-m. Substituting these values into the binomial theorem, we get:

(1-q-q)^{1-m} = \sum_{k=0}^{1-m} \binom{1-m}{k} (1-q)^{1-m-k}(-q)^k

Since the sum is from k = 0 to 1-m, the term with k = 0 will be 1, and the rest of the terms will have a negative sign because (-q)^k will alternate between positive and negative values. This results in the following simplification:

(1-q-q)^{1-m} = 1 - \sum_{k=1}^{1-m} \binom{1-m}{k} (1-q)^{1-m-k}q^k

Simplifying further, we get:

(1-q)^{1-m} = 1 - \sum_{r=0}^{m-1} \binom{1-m}{m-1-r} (1-q)^r q^{m-1-r}

Since the sum is now from r = 0 to m-1, we can substitute r' = m-1-r to get the desired form:

(1-q)^{1-m} = 1 - \sum_{r'=m-1}^{0} \binom{1-m}{r'} (1-q)^{m-1-r'} q^{r'}

This is equivalent to the expression given by Haldane:

(1-q)^{1-m} = \sum_{r'=0}^{m-1} \binom{1-m}{r'} (-q)^{r'}

which can be rewritten as:

(1-q)^{1-m} = \sum_{r=0}^\infty \binom{1-m}{r} (-q)^r

Therefore, the substitution made by Haldane is a valid result of the binomial theorem and is a key