1

How is the following identity derived, and how is it related to the stars and bars problem here?

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

I saw this identity in the forum here without justification given.

Community
  • 1
Cokes
  • 155
  • I added the link. I noticed that the coefficient is the count in the stars and bars combinatoric problem.

    At first I thought it had something to do with a generating function, but the negative exponent negated that idea.

    Can we interpret it in the context of the binomial theorem somehow?

     – Cokes Sep 30 '16 at 00:48

2 Answers2

1

We have $$(1-x)^{-n} = \left( \sum_{k \geq 0} x^k\right)^n$$ The coefficient of $x^r$ here is the number of ordered $n$ tuples of non negative integers which sums to $r$, which by the stars and bars problem is $\binom{n+r-1}{n-1}$

Nick R
  • 1,513
1

From geometric series, we know that

$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \dots$$

You can see this by using a telescoping series or any other number of standard techniques.

What we want is an expression for this, raised to the $n$'th power. I.e.,

\begin{align} (1-x)^{-n} &= \underbrace{(1 + x + x^2 + x^3 + \dots) \cdot (1 + x + x^2 + x^3 + \dots) \cdot \dots \cdot (1 + x + x^2 + x^3 + \dots)}_{n~\text{times}} \\ &= c_0 x^0 + c_1 x^1 + c_2 x^2 + c_3 x^3 + \dots, \end{align}

where the coefficients $c_r$ are found by multiplying out these infinite series up to a given order and collecting terms. (See Cauchy product)

  • Zeroth order terms ($x^0=1$) can only be created by choosing the $1$ term from each of these multiplicands within the product, so $c_0=1$.

  • First order terms ($x^1$) can be created by choosing $x$ from one of the multiplicands, and ones from the rest. Since there are $n$ multiplicands, there are $n$ possible choices, so $c_1 = n$.

  • Second order terms ($x^2$) can be created by either choosing two $x^1$'s from two diferent multiplicands, or by choosing one $x^2$ from one of the multiplicands.

More generally, for the $r$'th coefficient, $c_r$, we have $r$ copies of $x$, and distribute them into the $n$ multiplicands. If we distribute, say, 3 copies of $x$ into the second bin, that means we are considering a term associated with choosing $x^3$ from the second multiplicand. Each way of distributing these $r$ copies of $x$ corresponds to a distinct term in the series when multiplied out.

But the number of ways to distribute $r$ items into $n$ bins is precisely the complicated choosing formula in the original post, from which the formula follows.

Nick Alger
  • 18,844