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

[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!?

 

[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).

 

[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.

 

[gretun]

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.

Did they flip the formula of combinatorics?

 

[g_edgar]

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

This is incorrect. It should be

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

 

[gretun]

This is incorrect. It should be

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

Oh right sorry lol, my tex was wrong.

 

[ross_tang]

I think your original binomial series is wrong too.

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

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.