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Binomial theorem

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Pascal's triangle. Also called Tartaglia's triangle, and Yang Hui's Triangle, after others who invented the triangle.

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

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

for any real or complex numbers x and y, and any non-negative integer n. The binomial coefficient appearing in (1) may be defined in terms of the factorial function n!:

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

Contents

[edit] Examples

Taking n to be 2, 3, 4, or 5 in the binomial theorem yields

\begin{align} (x + y)^2 & = x^2 + 2xy + y^2 \\[6pt] (x + y)^3 & = x^3 + 3x^2y + 3xy^2 + y^3 \\[6pt] (x + y)^4 & = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4 \\[6pt] (x + y)^5 & = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 +5xy^4 + y^5. \end{align}

[edit] Combinatorial proof

[edit] Example

The coefficient of xy2 in

(x + y)3 = (x + y)(x + y)(x + y) = xxx + xxy + xyx + xyy + yxx + yxy + yyx + yyy = x3 + 3x2y + 3xy2 + y3.

equals \binom{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.

[edit] General case

Expanding (x + y)n yields the sum of the 2n products of the form e_1 e_2 \cdots e_n where each ei is x or y. Rearranging factors shows that each product equals xn − kyk for some k between 0 and n. For a given k, the following are proved equal in succession:

  • the number of copies of xn − kyk 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,\ldots,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.

[edit] 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 xjyk 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.

[edit] A quick way to expand binomials

Given a term cxiyj in the binomial expansion of (x + y)n, the next term can be obtained by decreasing i by 1, increasing j by 1, multiplying by the old i, and dividing by the new j. This lets one quickly calculate the whole expansion by hand, one term at a time, starting from the leading term xn = 1xny0. For example, the term following 45x8y2 in the expansion of (x + y)10 is

\frac{8}{3} 45 x^7 y^3 = 120 x^7 y^3.

[edit] Newton's generalized binomial theorem

Around 1665, Isaac Newton generalized the formula to allow exponents other than nonnegative integers. In this generalization, the finite sum is replaced by an infinite series. Namely, if x and y are real numbers with x > |y|,[1] and r is any complex number, then

\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.

The coefficients can also be written

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

where (\cdot)_k is the Pochhammer symbol. This is important when one is working with infinite series and would like to represent them in terms of generalized hypergeometric functions. This form is used 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.[citation needed]

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.

[edit] Generalizations

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

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.

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

[edit] 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 \left\{1,x,x^2,x^3,\dots\right\} is of binomial type.

[edit] 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 knew a special case of the binomial theorem up to the second order,[2][3] 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 to the 10th-century A.D. Indian mathematician Halayudha, the 11th-century A.D. Persian mathematician Omar Khayyám, and 13th-century Chinese mathematician Yang Hui, who all derived similar results.[4]

[edit] The binomial theorem in popular culture

[edit] See also

Sister project The Wikibook Combinatorics has a page on the topic of

[edit] Notes

  1. ^ a b This is to guarantee convergence. Depending on r, the series may also converge sometimes when |x| = |y|.
  2. ^ Binomial Theorem
  3. ^ The Story of the Binomial Theorem, by J. L. Coolidge, The American Mathematical Monthly 56:3 (1949), pp. 147–157
  4. ^ 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 on 2007-04-13. 
  5. ^ Cajori, Florian (1985). A History of Mathematics. New York: Chelsea Publishing Company. pp. 205. ISBN 0-8284-1303-X. 
  6. ^ Mikhail Bulgakov, translated by Michael Glenny. "The Master and Margarita". http://lib.meta.ua/book/1115/. 
  7. ^ Kevin Boss, Middlebury College. "The Master and Margarita, Chapter 18". http://cr.middlebury.edu/public/russian/Bulgakov/public_html/C18.html. 

[edit] References

  • Amulya Kumar Bag. Binomial Theorem in Ancient India. Indian J.History Sci.,1:68-74,1966.

[edit] External links

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

Retrieved from "http://en.wikipedia.org/wiki/Binomial_theorem"
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