Binomial theorem

Visualisation of binomial expansion up to the 4th power

For positive values of a and b, the binomial theorem with n = 2 is the geometrically evident fact that a square of side a + b can be cut into a square of side a, a square of side b, and two rectangles with sides a and b. With n = 3, the theorem states that a cube of side a + b can be cut into a cube of side a, a cube of side b, three a × a × b rectangular boxes, and three a × b × b rectangular boxes.

In calculus, this picture also gives a geometric proof of the derivative ${\displaystyle (x^{n})'=nx^{n-1}:}$[14] if one sets ${\displaystyle a=x}$ and ${\displaystyle b=\Delta x,}$ interpreting b as an infinitesimal change in a, then this picture shows the infinitesimal change in the volume of an n-dimensional hypercube, ${\displaystyle (x+\Delta x)^{n},}$ where the coefficient of the linear term (in ${\displaystyle \Delta x}$) is ${\displaystyle nx^{n-1},}$ the area of the n faces, each of dimension n − 1:

$${\displaystyle (x+\Delta x)^{n}=x^{n}+nx^{n-1}\Delta x+{\binom {n}{2}}x^{n-2}(\Delta x)^{2}+\cdots .}$$

Substituting this into the definition of the derivative via a difference quotient and taking limits means that the higher order terms, ${\displaystyle (\Delta x)^{2}}$ and higher, become negligible, and yields the formula ${\displaystyle (x^{n})'=nx^{n-1},}$ interpreted as

"the infinitesimal rate of change in volume of an n-cube as side length varies is the area of n of its (n − 1)-dimensional faces".

If one integrates this picture, which corresponds to applying the fundamental theorem of calculus, one obtains Cavalieri's quadrature formula, the integral ${\displaystyle \textstyle {\int x^{n-1}\,dx={\tfrac {1}{n}}x^{n}}}$ – see proof of Cavalieri's quadrature formula for details.[14]

The coefficient of x^n−ky^k is given by the formula

$${\displaystyle {\binom {n}{k}}={\frac {n!}{k!\;(n-k)!}},}$$

which is defined in terms of the factorial function n!. Equivalently, this formula can be written

$${\displaystyle {\binom {n}{k}}={\frac {n(n-1)\cdots (n-k+1)}{k(k-1)\cdots 1}}=\prod _{\ell =1}^{k}{\frac {n-\ell +1}{\ell }}=\prod _{\ell =0}^{k-1}{\frac {n-\ell }{k-\ell }}}$$

with k factors in both the numerator and denominator of the fraction. Although this formula involves a fraction, the binomial coefficient ${\displaystyle {\tbinom {n}{k}}}$ is actually an integer.

The binomial coefficient ${\displaystyle {\tbinom {n}{k}}}$ can be interpreted as the number of ways to choose k elements from an n-element set. This is related to binomials for the following reason: if we write (x + y)ⁿ as a product

$${\displaystyle (x+y)(x+y)(x+y)\cdots (x+y),}$$

then, according to the distributive law, there will be one term in the expansion for each choice of either x or y from each of the binomials of the product. For example, there will only be one term xⁿ, corresponding to choosing x from each binomial. However, there will be several terms of the form xⁿ⁻²y², one for each way of choosing exactly two binomials to contribute a y. Therefore, after combining like terms, the coefficient of xⁿ⁻²y² will be equal to the number of ways to choose exactly 2 elements from an n-element set.

The coefficient of xy² in

$${\displaystyle {\begin{aligned}(x+y)^{3}&=(x+y)(x+y)(x+y)\\&=xxx+xxy+xyx+{\underline {xyy}}+yxx+{\underline {yxy}}+{\underline {yyx}}+yyy\\&=x^{3}+3x^{2}y+{\underline {3xy^{2}}}+y^{3}\end{aligned}}}$$

equals ${\displaystyle {\tbinom {3}{2}}=3}$ because there are three x,y strings of length 3 with exactly two ys, namely,

$${\displaystyle xyy,\;yxy,\;yyx,}$$

corresponding to the three 2-element subsets of {1, 2, 3}, namely,

$${\displaystyle \{2,3\},\;\{1,3\},\;\{1,2\},}$$

where each subset specifies the positions of the y in a corresponding string.

Expanding (x + y)ⁿ yields the sum of the 2ⁿ products of the form e₁e₂ ... e_n where each e_i is x or y. Rearranging factors shows that each product equals x^n−ky^k for some k between 0 and n. For a given k, the following are proved equal in succession:

• the number of copies of x^n−ky^k in the expansion
• the number of n-character x,y strings having y in exactly k positions
• the number of k-element subsets of {1, 2, ..., n}
• ${\displaystyle {\tbinom {n}{k}},}$ either by definition, or by a short combinatorial argument if one is defining ${\displaystyle {\tbinom {n}{k}}}$ as ${\displaystyle {\tfrac {n!}{k!(n-k)!}}.}$

This proves the binomial theorem.

Induction yields another proof of the binomial theorem. When n = 0, both sides equal 1, since x⁰ = 1 and ${\displaystyle {\tbinom {0}{0}}=1.}$ Now suppose that the equality holds for a given n; we will prove it for n + 1. For j, k ≥ 0, let [f(x, y)]_j,k denote the coefficient of x^jy^k in the polynomial f(x, y). By the inductive hypothesis, (x + y)ⁿ is a polynomial in x and y such that [(x + y)ⁿ]_j,k is ${\displaystyle {\tbinom {n}{k}}}$ if j + k = n, and 0 otherwise. The identity

$${\displaystyle (x+y)^{n+1}=x(x+y)^{n}+y(x+y)^{n}}$$

shows that (x + y)ⁿ⁺¹ is also a polynomial in x and y, and

$${\displaystyle [(x+y)^{n+1}]_{j,k}=[(x+y)^{n}]_{j-1,k}+[(x+y)^{n}]_{j,k-1},}$$

since if j + k = n + 1, then (j − 1) + k = n and j + (k − 1) = n. Now, the right hand side is

$${\displaystyle {\binom {n}{k}}+{\binom {n}{k-1}}={\binom {n+1}{k}},}$$

by Pascal's identity.[15] On the other hand, if j + k ≠ n + 1, then (j – 1) + k ≠ n and j + (k – 1) ≠ n, so we get 0 + 0 = 0. Thus

$${\displaystyle (x+y)^{n+1}=\sum _{k=0}^{n+1}{\binom {n+1}{k}}x^{n+1-k}y^{k},}$$

which is the inductive hypothesis with n + 1 substituted for n and so completes the inductive step.

Around 1665, Isaac Newton generalized the binomial theorem to allow real exponents other than nonnegative integers. (The same generalization also applies to complex exponents.) In this generalization, the finite sum is replaced by an infinite series. In order to do this, one needs to give meaning to binomial coefficients with an arbitrary upper index, which cannot be done using the usual formula with factorials. However, for an arbitrary number r, one can define

$${\displaystyle {r \choose k}={\frac {r(r-1)\cdots (r-k+1)}{k!}}={\frac {(r)_{k}}{k!}},}$$

where ${\displaystyle (\cdot )_{k}}$ is the Pochhammer symbol, here standing for a falling factorial. This agrees with the usual definitions when r is a nonnegative integer. Then, if x and y are real numbers with |x| > |y|,[Note 1] and r is any complex number, one has

$${\displaystyle {\begin{aligned}(x+y)^{r}&=\sum _{k=0}^{\infty }{r \choose k}x^{r-k}y^{k}\\&=x^{r}+rx^{r-1}y+{\frac {r(r-1)}{2!}}x^{r-2}y^{2}+{\frac {r(r-1)(r-2)}{3!}}x^{r-3}y^{3}+\cdots .\end{aligned}}}$$

When r is a nonnegative integer, the binomial coefficients for k > r are zero, so this equation reduces to the usual binomial theorem, and there are at most r + 1 nonzero terms. For other values of r, the series typically has infinitely many nonzero terms.

For example, r = 1/2 gives the following series for the square root:

$${\displaystyle {\sqrt {1+x}}=1+{\frac {1}{2}}x-{\frac {1}{8}}x^{2}+{\frac {1}{16}}x^{3}-{\frac {5}{128}}x^{4}+{\frac {7}{256}}x^{5}-\cdots }$$

Taking r = −1, the generalized binomial series gives the geometric series formula, valid for |x| < 1:

$${\displaystyle (1+x)^{-1}={\frac {1}{1+x}}=1-x+x^{2}-x^{3}+x^{4}-x^{5}+\cdots }$$

More generally, with s = −r:

$${\displaystyle {\frac {1}{(1-x)^{s}}}=\sum _{k=0}^{\infty }{s+k-1 \choose k}x^{k}.}$$

So, for instance, when s = 1/2,

$${\displaystyle {\frac {1}{\sqrt {1+x}}}=1-{\frac {1}{2}}x+{\frac {3}{8}}x^{2}-{\frac {5}{16}}x^{3}+{\frac {35}{128}}x^{4}-{\frac {63}{256}}x^{5}+\cdots }$$

The generalized binomial theorem can be extended to the case where x and y are complex numbers. For this version, one should again assume |x| > |y|[Note 1] and define the powers of x + y and x using a holomorphic branch of log defined on an open disk of radius |x| centered at x. The generalized binomial theorem is valid also for elements x and y of a Banach algebra as long as xy = yx, and x is invertible, and ||y/x|| < 1.

A version of the binomial theorem is valid for the following Pochhammer symbol-like family of polynomials: for a given real constant c, define ${\displaystyle x^{(0)}=1}$ and

$${\displaystyle x^{(n)}=\prod _{k=1}^{n}[x+(k-1)c]}$$

for ${\displaystyle n>0.}$ Then[16]

$${\displaystyle (a+b)^{(n)}=\sum _{k=0}^{n}{\binom {n}{k}}a^{(n-k)}b^{(k)}.}$$

The case c = 0 recovers the usual binomial theorem.

More generally, a sequence ${\displaystyle \{p_{n}\}_{n=0}^{\infty }}$ of polynomials is said to be of binomial type if

• ${\displaystyle \deg p_{n}=n}$ for all ${\displaystyle n}$,
• ${\displaystyle p_{0}(0)=1}$, and
• ${\displaystyle p_{n}(x+y)=\sum _{k=0}^{n}{\binom {n}{k}}p_{k}(x)p_{n-k}(y)}$ for all ${\displaystyle x}$, ${\displaystyle y}$, and ${\displaystyle n}$.

An operator ${\displaystyle Q}$ on the space of polynomials is said to be the basis operator of the sequence ${\displaystyle \{p_{n}\}_{n=0}^{\infty }}$ if ${\displaystyle Qp_{0}=0}$ and ${\displaystyle Qp_{n}=np_{n-1}}$ for all ${\displaystyle n\geqslant 1}$. A sequence ${\displaystyle \{p_{n}\}_{n=0}^{\infty }}$ is binomial if and only if its basis operator is a Delta operator.[17] Writing ${\displaystyle E^{a}}$ for the shift by ${\displaystyle a}$ operator, the Delta operators corresponding to the above "Pochhammer" families of polynomials are the backward difference ${\displaystyle I-E^{-c}}$ for ${\displaystyle c>0}$, the ordinary derivative for ${\displaystyle c=0}$, and the forward difference ${\displaystyle E^{-c}-I}$ for ${\displaystyle c<0}$.

The binomial theorem can be generalized to include powers of sums with more than two terms. The general version is

$${\displaystyle (x_{1}+x_{2}+\cdots +x_{m})^{n}=\sum _{k_{1}+k_{2}+\cdots +k_{m}=n}{\binom {n}{k_{1},k_{2},\ldots ,k_{m}}}x_{1}^{k_{1}}x_{2}^{k_{2}}\cdots x_{m}^{k_{m}},}$$

where the summation is taken over all sequences of nonnegative integer indices k₁ through k_m such that the sum of all k_i is n. (For each term in the expansion, the exponents must add up to n). The coefficients ${\displaystyle {\tbinom {n}{k_{1},\cdots ,k_{m}}}}$ are known as multinomial coefficients, and can be computed by the formula

$${\displaystyle {\binom {n}{k_{1},k_{2},\ldots ,k_{m}}}={\frac {n!}{k_{1}!\cdot k_{2}!\cdots k_{m}!}}.}$$

Combinatorially, the multinomial coefficient ${\displaystyle {\tbinom {n}{k_{1},\cdots ,k_{m}}}}$ counts the number of different ways to partition an n-element set into disjoint subsets of sizes k₁, ..., k_m.

When working in more dimensions, it is often useful to deal with products of binomial expressions. By the binomial theorem this is equal to

$${\displaystyle (x_{1}+y_{1})^{n_{1}}\dotsm (x_{d}+y_{d})^{n_{d}}=\sum _{k_{1}=0}^{n_{1}}\dotsm \sum _{k_{d}=0}^{n_{d}}{\binom {n_{1}}{k_{1}}}x_{1}^{k_{1}}y_{1}^{n_{1}-k_{1}}\dotsc {\binom {n_{d}}{k_{d}}}x_{d}^{k_{d}}y_{d}^{n_{d}-k_{d}}.}$$

This may be written more concisely, by multi-index notation, as

$${\displaystyle (x+y)^{\alpha }=\sum _{\nu \leq \alpha }{\binom {\alpha }{\nu }}x^{\nu }y^{\alpha -\nu }.}$$

The general Leibniz rule gives the nth derivative of a product of two functions in a form similar to that of the binomial theorem:[18]

$${\displaystyle (fg)^{(n)}(x)=\sum _{k=0}^{n}{\binom {n}{k}}f^{(n-k)}(x)g^{(k)}(x).}$$

Here, the superscript (n) indicates the nth derivative of a function. If one sets f(x) = e^ax and g(x) = e^bx, and then cancels the common factor of e^{(a + b)x} from both sides of the result, the ordinary binomial theorem is recovered.[19]

For the complex numbers the binomial theorem can be combined with de Moivre's formula to yield multiple-angle formulas for the sine and cosine. According to De Moivre's formula,

$${\displaystyle \cos \left(nx\right)+i\sin \left(nx\right)=\left(\cos x+i\sin x\right)^{n}.}$$

Using the binomial theorem, the expression on the right can be expanded, and then the real and imaginary parts can be taken to yield formulas for cos(nx) and sin(nx). For example, since

$${\displaystyle \left(\cos x+i\sin x\right)^{2}=\cos ^{2}x+2i\cos x\sin x-\sin ^{2}x,}$$

De Moivre's formula tells us that

$${\displaystyle \cos(2x)=\cos ^{2}x-\sin ^{2}x\quad {\text{and}}\quad \sin(2x)=2\cos x\sin x,}$$

which are the usual double-angle identities. Similarly, since

$${\displaystyle \left(\cos x+i\sin x\right)^{3}=\cos ^{3}x+3i\cos ^{2}x\sin x-3\cos x\sin ^{2}x-i\sin ^{3}x,}$$

De Moivre's formula yields

$${\displaystyle \cos(3x)=\cos ^{3}x-3\cos x\sin ^{2}x\quad {\text{and}}\quad \sin(3x)=3\cos ^{2}x\sin x-\sin ^{3}x.}$$

In general,

$${\displaystyle \cos(nx)=\sum _{k{\text{ even}}}(-1)^{k/2}{n \choose k}\cos ^{n-k}x\sin ^{k}x}$$

and

$${\displaystyle \sin(nx)=\sum _{k{\text{ odd}}}(-1)^{(k-1)/2}{n \choose k}\cos ^{n-k}x\sin ^{k}x.}$$

The number e is often defined by the formula

$${\displaystyle e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}.}$$

Applying the binomial theorem to this expression yields the usual infinite series for e. In particular:

$${\displaystyle \left(1+{\frac {1}{n}}\right)^{n}=1+{n \choose 1}{\frac {1}{n}}+{n \choose 2}{\frac {1}{n^{2}}}+{n \choose 3}{\frac {1}{n^{3}}}+\cdots +{n \choose n}{\frac {1}{n^{n}}}.}$$

The kth term of this sum is

$${\displaystyle {n \choose k}{\frac {1}{n^{k}}}={\frac {1}{k!}}\cdot {\frac {n(n-1)(n-2)\cdots (n-k+1)}{n^{k}}}}$$

As n → ∞, the rational expression on the right approaches 1, and therefore

$${\displaystyle \lim _{n\to \infty }{n \choose k}{\frac {1}{n^{k}}}={\frac {1}{k!}}.}$$

This indicates that e can be written as a series:

$${\displaystyle e=\sum _{k=0}^{\infty }{\frac {1}{k!}}={\frac {1}{0!}}+{\frac {1}{1!}}+{\frac {1}{2!}}+{\frac {1}{3!}}+\cdots .}$$

Indeed, since each term of the binomial expansion is an increasing function of n, it follows from the monotone convergence theorem for series that the sum of this infinite series is equal to e.

The binomial theorem is closely related to the probability mass function of the negative binomial distribution. The probability of a (countable) collection of independent Bernoulli trials ${\displaystyle \{X_{t}\}_{t\in S}}$ with probability of success ${\displaystyle p\in [0,1]}$ all not happening is

${\displaystyle P\left(\bigcap _{t\in S}X_{t}^{C}\right)=(1-p)^{|S|}=\sum _{n=0}^{|S|}{|S| \choose n}(-p)^{n}.}$

An upper bound for this quantity is ${\displaystyle e^{-p|S|}.}$[20]