# Binomial series vs Binomial theorem, scratching my head for three days on this

1. Aug 15, 2010

### gretun

In my book, it says that the Binomial Series is

$$\sum_{n=0}^{\infty }\binom{n}{r} x^n$$

Where $$\binom{n}{r} = \frac{n(n-1)...(n-r+1)}{n!}$$ for $$r\geq1$$ and $$\binom{n}{0} = 1$$

Now here is where it got to be, I know that the $$\binom{n}{r} = \frac{n(n-1)...(n-r+1)}{n!}$$ were derived through the power series, but how does it explained the original statement that $$\sum_{n=0}^{\infty }\binom{n}{r} x^n$$?

Since $$\binom{n}{r}$$ does not actually change. I mean it is still $$\frac{n!}{(n-r)!r!}$$ and so how can you put negative n into n!?

2. Aug 15, 2010

### Mute

One way to think of it is to rewrite the binomial coefficient in terms of Gamma functions:

$$\left(\begin{array}{c}n \\ r \end{array}\right) = \frac{\Gamma(n+1)}{\Gamma(n-r+1)\Gamma(r+1)}$$

The Gamma function $\Gamma(z)$ has poles at $z = 0, -1, -2, \dots$, or equivalently, $1/\Gamma(z)$ has zeroes at these points, so when $n-r+1 \leq 0$ the denominator of the binomial coefficient becomes infinite and so the coefficient vanishes. Hence the series terminates and you get a finite expansion for integer powers.

It is only at the negative integers that the gamma function has poles, so when n - r + 1 is not an integer, the Gamma function doesn't vanish, and you get an infinite series.

(Note: I don't think this is a rigorous argument, but I think it can be made rigorous by defining the properties of $1/\Gamma(z)$ precisely such that it has zeros where $\Gamma(z)$ has poles. I'll leave it to any more mathematician-minded posters to fill in the gaps in rigor).

3. Aug 15, 2010

### Gib Z

I think the main thing the OP is simply missing is that for Binomial series, n is not necessarily an integer, which is why they write it out in the longer form, as the one with factorials only works with integers.

4. Aug 15, 2010

### gretun

Did they flip the formula of combinatorics?

5. Aug 15, 2010

### g_edgar

This is incorrect. It should be

$$\binom{n}{r} = \frac{n(n-1)...(n-r+1)}{r!}$$

6. Aug 15, 2010

### gretun

Oh right sorry lol, my tex was wrong.

7. Aug 17, 2010

### ross_tang

I think your original binomial series is wrong too.

The correct one should be summing about r, not n. i.e.

$$(1+x)^n = \sum_{r=0}^{\infty} \binom{n}{r} x^r$$

This formula is valid, even for complex n. You are right that this formula can be derived using Taylor series.