Binomial theorem

The binomial coefficients appear as the entries of Pascal's triangle.

In elementary algebra, the binomial theorem describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand the power (x + y)ⁿ into a sum involving terms of the form ax^by^c, where the coefficient of each term is a positive integer, and the sum of the exponents of x and y in each term is n. For example,

$(x+y)^4 \;=\; x^4 \,+\, 4 x^3y \,+\, 6 x^2 y^2 \,+\, 4 x y^3 \,+\, y^4.$

The coefficients appearing in the binomial expansion are known as binomial coefficients. They are the same as the entries of Pascal's triangle, and can be determined by a simple formula involving factorials. These numbers also arise in combinatorics, where the coefficient of x^n−ky^k is equal to the number of different combinations of k elements that can be chosen from an n-element set.

History

This formula and the triangular arrangement of the binomial coefficients are often attributed to Blaise Pascal, who described them in the 17th century, but they were known to many mathematicians who preceded him. The 4th century B.C. Greek mathematician Euclid mentioned the special case of the binomial theorem for exponent 2^[1]^[2] as did the 3rd century B.C. Indian mathematician Pingala to higher orders. A more general binomial theorem and the so-called "Pascal's triangle" were known in the 10th-century A.D. to Indian mathematician Halayudha and Persian mathematician Al-Karaji,^[3] and in the 13th century to Chinese mathematician Yang Hui, who all derived similar results.^[4] Al-Karaji also provided a mathematical proof of both the binomial theorem and Pascal's triangle, using mathematical induction.^[3]

Statement of the theorem

According to the theorem, it is possible to expand any power of x + y into a sum of the form

$\begin{align} (x+y)^n & = {n \choose 0}x^n y^0 + {n \choose 1}x^{n-1}y^1 + {n \choose 2}x^{n-2}y^2 + {n \choose 3}x^{n-3}y^3 + \cdots \\ & {} \qquad \cdots + {n \choose n-1}x^1 y^{n-1} + {n \choose n}x^0 y^n, \end{align}$

where $\tbinom nk$ denotes the corresponding binomial coefficient. Using summation notation, the formula above can be written

$(x+y)^n = \sum_{k=0}^n {n \choose k}x^{n-k}y^k.$

This formula is sometimes referred to as the binomial formula or the binomial identity.

A variant of the binomial formula is obtained by substituting 1 for x and x for y, so that it involves only a single variable. In this form, the formula reads

$(1+x)^n = {n \choose 0}x^0 + {n \choose 1}x^1 + {n \choose 2}x^2 + \cdots + {n \choose {n-1}}x^{n-1} + {n \choose n}x^n,$

or equivalently

$(1+x)^n = \sum_{k=0}^n {n \choose k}x^k.$

Examples

Pascal's triangle

The most basic example of the binomial theorem is the formula for square of x + y:

$(x+y)^2 = x^2 + 2xy + y^2.\!$

The binomial coefficients 1, 2, 1 appearing in this expansion correspond to the third row of Pascal's triangle. The coefficients of higher powers of x + y correspond to later rows of the triangle:

$\begin{align} (x+y)^3 & = x^3 + 3x^2y + 3xy^2 + y^3, \\[8pt] (x+y)^4 & = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4, \\[8pt] (x+y)^5 & = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5, \\[8pt] (x+y)^6 & = x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6, \\[8pt] (x+y)^7 & = x^7 + 7x^6y + 21x^5y^2 + 35x^4y^3 + 35x^3y^4 + 21x^2y^5 + 7xy^6 + y^7. \end{align}$

The binomial theorem can be applied to the powers of any binomial. For example,

$\begin{align} (x+2)^3 &= x^3 + 3x^2(2) + 3x(2)^2 + 2^3 \\ &= x^3 + 6x^2 + 12x + 8.\end{align}$

For a binomial involving subtraction, the theorem can be applied as long as the negation of the second term is used. This has the effect of negating every other term of the expansion:

$(x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3.\!$

Geometrical explanation

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.

The binomial coefficients

Main article: Binomial coefficient

The coefficients that appear in the binomial expansion are called binomial coefficients. These are usually written $\tbinom nk$ , and pronounced “n choose k”.

Formulas

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

${n \choose k} = \frac{n!}{k!\,(n-k)!}$

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

${n \choose k} = \frac{n (n-1) \cdots (n-k+1)}{k (k-1) \cdots 1} = \prod_{\ell=1}^k \frac{n-\ell+1}{\ell}$

with k factors in both the numerator and denominator of the fraction. Note that, although this formula involves a fraction, the binomial coefficient $\tbinom nk$ is actually an integer.

Combinatorial interpretation

The binomial coefficient $\tbinom nk$ 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

$(x+y)(x+y)(x+y)\cdots(x+y),$

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.

Proofs

Combinatorial proof

Example

The coefficient of xy² in

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

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

$xyy, \; yxy, \; yyx,$

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

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

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

General case

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}
${n \choose k}$ (this is either by definition, or by a short combinatorial argument if one is defining ${n \choose k}$ as $\frac{n!}{k!\,(n-k)!}$ ).

This proves the binomial theorem.

Inductive proof

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

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

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

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

If j + k = n + 1, then (j − 1) + k = n and j + (k − 1) = n, so the right hand side is

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

by Pascal's identity. 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

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

and this completes the inductive step.

Generalizations

Newton's generalized binomial theorem

Around 1665, Isaac Newton generalized the formula to allow real exponents other than nonnegative integers, and in fact it can be generalized further, 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 above formula with factorials; however factoring out (n−k)! from numerator and denominator in that formula, and replacing n by r which now stands for an arbitrary number, one can define

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

where $(\cdot)_k$ is the Pochhammer symbol here standing for a falling factorial. Then, if x and y are real numbers with |x| > |y|,^[5] and r is any complex number, one has

$\begin{align} (x+y)^r & =\sum_{k=0}^\infty {r \choose k} x^{r-k} y^k \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(2) \\ & = x^r + r x^{r-1} y + \frac{r(r-1)}{2!} x^{r-2} y^2 + \frac{r(r-1)(r-2)}{3!} x^{r-3} y^3 + \cdots. \end{align}$

When r is a nonnegative integer, the binomial coefficients for k > r are zero, so (2) specializes to (1), and there are at most r + 1 nonzero terms. For other values of r, the series (2) has an infinite number of nonzero terms, at least if x and y are nonzero.

This is important when one is working with infinite series and would like to represent them in terms of generalized hypergeometric functions.

Taking r = −s leads to a particularly handy but non-obvious formula:

$\frac{1}{(1-x)^s} = \sum_{k=0}^\infty {s+k-1 \choose k} x^k \equiv \sum_{k=0}^\infty {s+k-1 \choose s-1} x^k.$

Further specializing to s = 1 yields the geometric series formula.

Generalizations

Formula (2) can be generalized to the case where x and y are complex numbers. For this version, one should assume |x| > |y|^[5] 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.

Formula (2) is valid also for elements x and y of a Banach algebra as long as xy = yx, x is invertible, and ||y/x|| < 1.

The multinomial theorem

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

$(x_1 + x_2 + \cdots + x_m)^n = \sum_{k_1,k_2,\ldots,k_m} {n \choose k_1, k_2, \ldots, k_m} x_1^{k_1} x_2^{k_2} \cdots x_m^{k_m}.$

where the summation is taken over all sequences of nonnegative integer indices k₁ through k_m such the sum of all k_i is n. (For each term in the expansion, the exponents must add up to n). The coefficients $\tbinom n{k_1,\cdots,k_n}$ are known as multinomial coefficients, and can be computed by the formula

${n \choose k_1, k_2, \ldots, k_m} = \frac{n!}{k_1!\, k_2! \cdots k_m!}.$

Combinatorially, the multinomial coefficient $\tbinom n{k_1,\cdots,k_n}$ counts the number of different ways to partition an n-element set into disjoint subsets of sizes k₁, ..., k_n.

Applications

Multiple angle identities

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,

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

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

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

De Moivre's formula tells us that

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

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

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

De Moivre's formula yields

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

In general,

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

and

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

Series for e

The number e is often defined by the formula

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

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

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

The kth term of this sum is

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

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

$\lim_{n\to\infty} {n \choose k}\frac{1}{n^k} = \frac{1}{k!}.$

This indicates that e can be written as a series:

$e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots.$

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 in abstract algebra

Formula (1) is valid more generally for any elements x and y of a semiring satisfying xy = yx. The theorem is true even more generally: alternativity suffices in place of associativity.

The binomial theorem can be stated by saying that the polynomial sequence { 1, x, x², x³, ... } is of binomial type.

Notes

↑ Binomial Theorem
↑ The Story of the Binomial Theorem, by J. L. Coolidge, The American Mathematical Monthly 56:3 (1949), pp. 147–157
↑ ^3.0 ^3.1 O'Connor, John J.; Robertson, Edmund F., "Abu Bekr ibn Muhammad ibn al-Husayn Al-Karaji", MacTutor History of Mathematics archive, University of St Andrews, http://www-history.mcs.st-andrews.ac.uk/Biographies/Al-Karaji.html .
↑ Landau, James A (1999-05-08). "Historia Matematica Mailing List Archive: Re: [HM] Pascal's Triangle" (mailing list email). Archives of Historia Matematica. http://archives.math.utk.edu/hypermail/historia/may99/0073.html. Retrieved 2007-04-13.
↑ ^5.0 ^5.1 This is to guarantee convergence. Depending on r, the series may also converge sometimes when |x| = |y|.

References

Bag, Amulya Kumar (1966). "Binomial theorem in ancient India". Indian J. History Sci 1 (1): 68–74.
Graham, Ronald; Donald Knuth, Oren Patashnik (1994). "(5) Binomial Coefficients". Concrete Matehamtics (2nd ed.). Addison Wesley. pp. 153–256. ISBN 0-201-55802-5. OCLC 17649857.
Solomentsev, E.D. (2001), "Newton binomial", in Hazewinkel, Michiel, Encyclopaedia of Mathematics, Springer, ISBN 978-1556080104, http://eom.springer.de/n/n066500.htm

External links

Binomial Theorem by Stephen Wolfram, and "Binomial Theorem (Step-by-Step)" by Bruce Colletti and Jeff Bryant, Wolfram Demonstrations Project, 2007.

Binomial Theorem Introduction

This article incorporates material from inductive proof of binomial theorem on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.

Binomial theorem

Contents

History

Statement of the theorem

Examples

Geometrical explanation

The binomial coefficients

Formulas

Combinatorial interpretation

Proofs

Combinatorial proof

Example

General case

Inductive proof

Generalizations

Newton's generalized binomial theorem

Generalizations

The multinomial theorem

Applications

Multiple angle identities

Series for e

The binomial theorem in abstract algebra

See also

Notes

References

External links