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So I just finished learning the standard binomial theorem and I've just come across the extended (newtons binomial theorem).

As expected I am completely baffled about how it works I do not understand how the following holds :

Suppose $n \in \mathbb{R}$ then for every $x$ satisfying $|x| < 1$, we have $$ (1+x)^{n} = \sum_{r=0}^{\infty} \dbinom{n}r x^r \space \space \space \space ; \space \space r = 0,1,2...$$ and the extended binomial coefficient is given as $$\dbinom{n}r = \dfrac{n!}{(n-r)! \space r!}$$

There are a lot of things that trouble me here first I've read online that factorial are not defined for negative integer values yet somehow we are using them to compute the coefficients in the expansion example:.$$\begin{align*} (b + a)^n &= b^n\left(1 + \frac{a}{b}\right)^n\\ &= b^n \left(\sum_{k = 0}^{n}{n \choose k}\left(\frac{a}{b}\right)^k\right)\\ &= b^n \left(1 + n \left(\frac{a}{b}\right) + \frac{(n)(n-1)}{2}\left(\frac{a}{b}\right)^2 + \cdots\right) \end{align*}$$

How is it that say $-4!$ is undefined yet in the formula we are using it is it then not the same as negative factorial but instead some kind of entirely different method that happens to work?.

I would like some basic intuition on how this binomial theorem actually works if it is possible as the book i am reading just states the formula without any explanation as to how it works if there is no basic explanation maybe a list of topics I need to learn before to understand it would be helpful.

Thanks In advance.

The homeschooler
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1 Answers1

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In the extended binomial theorem, the definition of $^nC_r$ is not as simple as it is for the 'vanilla' binomial theorem. If we define $$ n!=n \cdot(n-1) \cdot (n-2)\cdot\dots\cdot3\cdot 2\cdot 1 $$ then the formula you have provided is indeed meaningless, as $n!$ only makes sense when $n$ is a natural number. The simplest way to address this is to define $^nC_r$ slightly differently when $n$ is a real number: $$ \binom{n}{r}=\frac{n(n-1)(n-2)\cdots(n-r+1)}{r!} \, , $$ meaning that the problem of defining things such as $(-4)!$ does not arise. Notice that this definition is equivalent to the one you gave when $n$ is a natural number. By the way, the numerator can be written as $n^{\underline{r}}$ using falling factorials.

Joe
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  • so in newtons binomial theorem it does not use the gamma function but the general one you defined above correct?. If so how does this actually compute the coefficient? – The homeschooler Mar 22 '21 at 18:38
  • @Thehomeschooler: I deleted the stuff about the gamma function because I wasn't sure that it was correct. To compute the coefficient you should use the alternative formula I wrote about. Please let me know if you have any questions. – Joe Mar 22 '21 at 18:40
  • thanks for the help and im still a bit confused on why choosing from negative numbers gives us the coefficient after all isn't the top half of the equation still using factorial and again factorial aren't defined for negative integers? – The homeschooler Mar 22 '21 at 18:46
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    @Thehomeschooler: I don't exactly understand what you mean, but the formula $$^nC_r=\frac{n(n-1)(n-2)\cdots(n-r+1)}{r!}$$only has factorials on the denominator (which is unproblematic because $r$ is a natural number). – Joe Mar 22 '21 at 18:49
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    @Thehomeschooler as an example, $\binom{-3}{4} = \dfrac{(-3)(-4)(-5)(-6)}{4!}=15$ and $\binom{\pi}{2} = \dfrac{(\pi)(\pi-1)}{2}$ and so on... – JMoravitz Mar 22 '21 at 18:54
  • @joe oh ok so the numerator is just a decreasing multiplication that happens to give the coefficient when divided by $k!$ – The homeschooler Mar 22 '21 at 18:59
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    @Thehomeschooler: Yes, decreasing multiplication avoids the problem. There are two things to note: (1) this definition of $^nC_r$ is exactly the same as the one you gave you in your question provided that both $n$ and $r$ are natural numbers; (2) it's possible to prove why the formula works, e.g. using Taylor series. Historically, it was discovered by Newton who just tried using the binomial theorem for negative numbers and fractions, and it still worked. He then used this to compute pi to astounding numerical accuracy. – Joe Mar 22 '21 at 19:08
  • @joe thanks you have cleared my confusion – The homeschooler Mar 22 '21 at 19:12
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    @Thehomeschooler: No worries! I'm glad I could help. – Joe Mar 22 '21 at 19:26
  • Related to this: We also cannot use the factorial formula for indefinite summation, even though in that case we only need $\binom{n}{k}$ defined for $n∈ℤ$ and $k∈ℕ$. – user21820 Oct 02 '21 at 04:14