Binomial theorem

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In mathematics, the binomial theorem is an important formula giving the expansion of powers of sums. Its simplest version says

(x+y)^n=\sum_{k=0}^n{n \choose k}x^{n-k}y^{k}\quad\quad\quad(1)

whenever n is any non-negative integer, the number

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

is the binomial coefficient (using the choose function), and n! denotes the factorial of n.

This formula and the triangular arrangement of the binomial coefficients are often attributed to Blaise Pascal, who described them in the 17th century. However, it was known to many mathematicians who preceded him; 13th-century Chinese mathematician Yang Hui, 11th-century Persian mathematician Omar Khayyám, and 3rd-century BC Indian mathematician Pingala all derived similar results. [1]

For example, here are the cases where 2 ≤ n ≤ 5:

(x + y)^2 = x^2 + 2xy + y^2\,
(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3\,
(x + y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4\,
(x + y)^5 = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 +5xy^4 + y^5\,

Formula (1) is valid for all real or complex numbers x and y, and more generally for any elements x and y of a semiring as long as xy = yx (the theorem is true even more generally: note that associativity is not required, just alternativity).

Contents

Consider a = (x + y)n . a can be written as a product of sums, a=s_1s_2 \cdots s_n , where each si = x + y . The expansion of a is the sum of all products involving one term -- either x or y -- from each si . For example, the term xn in the expansion of a is had by picking x in each si .

The coefficient of each term in the expansion of a is determined by how many different ways there are to pick terms from the si such that their product is of the same form as the term (excluding the coefficient). Consider t = xn − 1y. t can be formed from a by picking y from one of the si and x in the rest of them. There are n ways to pick a si to provide the y; t is thus formed in n different ways in the expansion of a, making its coefficient n. In general, for t = xn − kyk, there are

{n \choose k}

different ways to pick the si that provide the ys (since k ys are picked from the n si), and thus this must be the coefficient for t. The binomial theorem follows naturally from here.

Isaac Newton generalized the formula to other exponents by considering an infinite series:

{(x+y)^r=\sum_{k=0}^\infty {r \choose k} x^{r-k} y^k \quad\quad\quad(2)}

where r can be any complex number (in particular r can be any real number, not necessarily positive and not necessarily an integer), and the coefficients are given by

\begin{align} {r \choose k} &{}= {1 \over k!}\prod_{n=0}^{k-1}(r-n)=\frac{r(r-1)(r-2)\cdots(r-(k-1))}{k!}. \end{align}

In case k = 0, this is a product of no numbers at all and therefore equal to 1, and in case k = 1 it is equal to r, as the additional factors (r − 1), etc., do not appear.

Another way to express this quantity is

{r \choose k}=\frac{(r)_k}{k!},

which is important when one is working with infinite series and would like to represent them in terms of generalized hypergeometric functions. The notation (\cdot)_k is the Pochhammer symbol. This form is vital in applied mathematics, for example, when evaluating the formulas that model the statistical properties of the phase-front curvature of a light wave as it propagates through optical atmospheric turbulence.

A particularly handy but non-obvious form holds for the reciprocal power:

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

For a more extensive account of Newton's generalized binomial theorem, see binomial series.

The sum in (2) converges and the equality is true whenever the real or complex numbers x and y are "close together" in the sense that the absolute value | y/x | is less than one.

The geometric series is a special case of (2) where we choose x = 1 and r = −1.

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

The binomial theorem can be stated by saying that the polynomial sequence

\left\{\,x^k:k=0,1,2,\dots\,\right\}\,

is of binomial type.

One way to prove the binomial theorem (1) is with mathematical induction. When n = 0, we have

(a+b)^0 = 1 = \sum_{k=0}^0 { 0 \choose k } a^{0-k}b^k.

For the inductive step, assume the theorem holds when the exponent is m. Then for n = m + 1

(a+b)^{m+1} = a(a+b)^m + b(a+b)^m \,
= a \sum_{k=0}^m { m \choose k } a^{m-k} b^k + b \sum_{j=0}^m { m \choose j } a^{m-j} b^j

by the inductive hypothesis

= \sum_{k=0}^m { m \choose k } a^{m-k+1} b^k + \sum_{j=0}^m { m \choose j } a^{m-j} b^{j+1}

by multiplying through by a and b

= a^{m+1} + \sum_{k=1}^m { m \choose k } a^{m-k+1} b^k + \sum_{j=0}^m { m \choose j } a^{m-j} b^{j+1}

by pulling out the k = 0 term

= a^{m+1} + \sum_{k=1}^m { m \choose k } a^{m-k+1} b^k + \sum_{k=1}^{m+1} { m \choose k-1 }a^{m-k+1}b^{k}

by letting j = k − 1

= a^{m+1} + \sum_{k=1}^m { m \choose k } a^{m-k+1}b^k + \sum_{k=1}^{m} { m \choose k-1 }a^{m+1-k}b^{k} + b^{m+1}

by pulling out the k = m + 1 term from the right hand side

= a^{m+1} + b^{m+1} + \sum_{k=1}^m \left[ { m \choose k } + { m \choose k-1 } \right] a^{m+1-k}b^k

by combining the sums

= a^{m+1} + b^{m+1} + \sum_{k=1}^m { m+1 \choose k } a^{m+1-k}b^k

from Pascal's rule

= \sum_{k=0}^{m+1} { m+1 \choose k } a^{m+1-k}b^k

by adding in the m + 1 terms.

A binomial number is a number in the form of \scriptstyle x^n \,\pm\, y^n (for n at least 2). When the sign is minus or n is odd these binomial numbers can be factored algebraically:

x^n\pm y^n=(x\pm y)(x^{n-1} \mp x^{n-2}y + \cdots \mp xy^{n-2} + y^{n-1}).\,

Examples:

x^2-y^2=(x-y)(x+y)\,
x^3-y^3=(x-y)(x^2+xy+y^2)\,
x^3+y^3=(x+y)(x^2-xy+y^2)\,
x^8-y^8=(x-y)(x+y)(x^2+y^2)(x^4+y^4)\,

To factorise \scriptstyle x^n\,-\,y^n simply, use

x^n-y^n=(x-y) \left( \sum_{k=0}^{n-1}x^ky^{n-1-k} \right).

To quickly expand binomials of the form

(x+y)^n \,

The first term is

x^n \,

(this follows directly from the generalized binomial theorem) and the coefficient of each subsequent term is the current coefficient multiplied by the current exponent of x, divided by the current term number. Exponents of x decrease each term, while exponents of y increase each term (from 0 in the first term) until the exponent of x is 0.

Example:

(x+y)^{10} \,

The first term is

x^{10} \,

To find the coefficient of the second term, multiply 1 (the current coefficient) by 10 (the current exponent of x), and divide by the current term number (1, since this is the first term) to get 10. The exponent of x decrements, and the exponent of y increments. The next term is therefore

10x^9y \,

Similarly, the next coefficient is 10×9/2, which gives 45. After that, it is (10×9×8)/(3×2×1). This continues until (10×9×8×7×6)/(5×4×3×2×1), after which, the coefficients are symmetrical. The whole thing is

x10 + 10x9y + 45x8y2 + 120x7y3 + 210x6y4 + 252x5y5 + 210x4y6 + 120x3y7 + 45x2y8 + 10xy9 + y10.

Notice that the coefficients are perfectly symmetrical. This will happen when the coefficients of x and y within the parentheses of the original expression are the same. Recognizing this can save even more time.

If the original expression instead was

(2x+y)^{10} \,

then the resulting expansion would be the same, except with (2x) in place of x in every place. The factor of 2 must get raised to the power of x in each term. The same holds true if either x or y is raised to a power inside the parentheses of the original expression.

  1. Amulya Kumar Bag. Binomial Theorem in Ancient India. Indian J.History Sci.,1:68-74,1966.
  1. ^ Landau, James A (1999-05-08). Historia Matematica Mailing List Archive: Re: [HM] Pascal's Triangle (mailing list email). Archives of Historia Matematica. Retrieved on 2007-04-13.

This article incorporates material from inductive proof of binomial theorem on PlanetMath, which is licensed under the GFDL.

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