show that when m=n this is equivalent to ordinary defination x2(A) = ||A||2||A i
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show that when m=n this is equivalent to ordinary defination x2(A) = ||A||2||A inverse||2 show that when m=n this is equivalent to ordinary defination x2(A) = ||A||2||A inverse||2Explanation / Answer
(valid for any elements x,y of a commutative ring), which explains the name "binomial coefficient". Another occurrence of this number is in combinatorics, where it gives the number of ways, disregarding order, that k objects can be chosen from among n objects; more formally, the number of k-element subsets (or k-combinations) of an n-element set. This number can be seen as equal to the one of the first definition, independently of any of the formulas below to compute it: if in each of the n factors of the power (1 + X)n one temporarily labels the term X with an index i (running from 1 to n), then each subset of k indices gives after expansion a contribution Xk, and the coefficient of that monomial in the result will be the number of such subsets. This shows in particular that is a natural number for any natural numbers n and k. There are many other combinatorial interpretations of binomial coefficients (counting problems for which the answer is given by a binomial coefficient expression), for instance the number of words formed of n bits (digits 0 or 1) whose sum is k is given by , while the number of ways to write where every ai is a nonnegative integer is given by . Most of these interpretations are easily seen to be equivalent to counting k-combinations. [edit]Computing the value of binomial coefficients Several methods exist to compute the value of without actually expanding a binomial power or counting k-combinations. [edit]Recursive formula One has a recursive formula for binomial coefficients with initial values The formula follows either from tracing the contributions to Xk in (1 + X)n-1(1 + X), or by counting k-combinations of {1, 2, ..., n} that contain n and that do not contain n separately. It follows easily that when k > n, and for all n, so the recursion can stop when reaching such cases. This recursive formula then allows the construction of Pascal's triangle. [edit]Multiplicative formula A more efficient method to compute individual binomial coefficients is given by the formula where the numerator of the first fraction is expressed as a falling factorial power. This formula is easiest to understand for the combinatorial interpretation of binomial coefficients. The numerator gives the number of ways to select a sequence of k distinct objects, retaining the order of selection, from a set of n objects. The denominator counts the number of distinct sequences that define the same k-combination when order is disregarded. [edit]Factorial formula Finally there is a formula using factorials that is easy to remember: where n! denotes the factorial of n. This formula follows from the multiplicative formula above by multiplying numerator and denominator by (n - k)!; as a consequence it involves many factors common to numerator and denominator. It is less practical for explicit computation unless common factors are first canceled (in particular since factorial values grow very rapidly). The formula does exhibit a symmetry that is less evident from the multiplicative formula (though it is from the definitions) (1) [edit]Generalization and connection to the binomial series The multiplicative formula allows the definition of binomial coefficients to be extended[4] by replacing n by an arbitrary number a (negative, real, complex) or even an element of any commutative ring in which all positive integers are invertible: With this definition one has a generalization of the binomial formula (with one of the variables set to 1), which justifies still calling the binomial coefficients: (2) This formula is valid for all complex numbers a and X with |X| < 1. It can also be interpreted as an identity of formal power series in X, where it actually can serve as definition of arbitrary powers of series with constant coefficient equal to 1; the point is that with this definition all identities hold that one expects for exponentiation, notably If a is a nonnegative integer n, then all terms with k > n are zero, and the infinite series becomes a finite sum, thereby recovering the binomial formula. However for other values of a, including negative integers and rational numbers, the series is really infinite. [edit]Pascal's triangle 1000th row of Pascal's triangle, arranged vertically, with grey-scale representations of decimal digits of the coefficients, right-aligned. The left boundary of the image corresponds roughly to the graph of the logarithm of the binomial coefficients, and illustrates that they form a log-concave sequence. Main article: Pascal's rule Main article: Pascal's triangle Pascal's rule is the important recurrence relation (3) which can be used to prove by mathematical induction that is a natural number for all n and k, (equivalent to the statement that k! divides the product of k consecutive integers), a fact that is not immediately obvious from formula (1). Pascal's rule also gives rise to Pascal's triangle: 0: 1 1: 1 1 2: 1 2 1 3: 1 3 3 1 4: 1 4 6 4 1 5: 1 5 10 10 5 1 6: 1 6 15 20 15 6 1 7: 1 7 21 35 35 21 7 1 8: 1 8 28 56 70 56 28 8 1 Row number n contains the numbers for k = 0,…,n. It is constructed by starting with ones at the outside and then always adding two adjacent numbers and writing the sum directly underneath. This method allows the quick calculation of binomial coefficients without the need for fractions or multiplications. For instance, by looking at row number 5 of the triangle, one can quickly read off that (x + y)5 = 1 x5 + 5 x4y + 10 x3y2 + 10 x2y3 + 5 x y4 + 1 y5. The differences between elements on other diagonals are the elements in the previous diagonal, as a consequence of the recurrence relation (3) above. [edit]Combinatorics and statistics Binomial coefficients are of importance in combinatorics, because they provide ready formulas for certain frequent counting problems: There are ways to choose k elements from a set of n elements. See Combination. There are ways to choose k elements from a set of n if repetitions are allowed. See Multiset. There are strings containing k ones and n zeros. There are strings consisting of k ones and n zeros such that no two ones are adjacent. The Catalan numbers are The binomial distribution in statistics is The formula for a Bézier curve. [edit]Binomial coefficients as polynomials For any nonnegative integer k, the expression can be simplified and defined as a polynomial divided by k!: This presents a polynomial in t with rational coefficients. As such, it can be evaluated at any real or complex number t to define binomial coefficients with such first arguments. These "generalized binomial coefficients" appear in Newton's generalized binomial theorem. For each k, the polynomial can be characterized as the unique degree k polynomial p(t) satisfying p(0) = p(1) = ... = p(k - 1) = 0 and p(k) = 1. Its coefficients are expressible in terms of Stirling numbers of the first kind, by definition of the latter: The derivative of can be calculated by logarithmic differentiation: [edit]Binomial coefficients as a basis for the space of polynomials Over any field containing Q, each polynomial p(t) of degree at most d is uniquely expressible as a linear combination . The coefficient ak is the kth difference of the sequence p(0), p(1), …, p(k). Explicitly,[5] (3.5) [edit]Integer-valued polynomials Each polynomial is integer-valued: it takes integer values at integer inputs. (One way to prove this is by induction on k, using Pascal's identity.) Therefore any integer linear combination of binomial coefficient polynomials is integer-valued too. Conversely, (3.5) shows that any integer-valued polynomial is an integer linear combination of these binomial coefficient polynomials. More generally, for any subring R of a characteristic 0 field K, a polynomial in K[t] takes values in R at all integers if and only if it is an R-linear combination of binomial coefficient polynomials. [edit]Example The integer-valued polynomial 3t(3t + 1)/2 can be rewritten as [edit]Identities involving binomial coefficients The factorial formula facilitates relating nearby binomial coefficients. For instance, if k is a positive integer and n is arbitrary, then and, with a little more work, [edit]Powers of -1 A special binomial coefficient is ; it equals powers of -1: [edit]Series involving binomial coefficients The formula (5) is obtained from (2) using x = 1. This is equivalent to saying that the elements in one row of Pascal's triangle always add up to two raised to an integer power. A combinatorial interpretation of this fact involving double counting is given by counting subsets of size 0, size 1, size 2, and so on up to size n of a set S of n elements. Since we count the number of subsets of size i for 0 = i = n, this sum must be equal to the number of subsets of S, which is known to be 2n. That is, Equation 5 is a statement that the power set for a finite set with n elements has size 2n. The formulas (6a) and (6b) follow from (2), after differentiating with respect to x (twice in the latter) and then substituting x = 1. The Chu-Vandermonde identity, which holds for any complex-values m and n and any non-negative integer k, is (7a) and can be found by examination of the coefficient of xk in the expansion of (1 + x)m (1 + x)n - m = (1 + x)n using equation (2). When m = 1, equation (7a) reduces to equation (3). A similar looking formula, which applies for any integers j, k, and n satisfying 0 = j = k = n, is (7b) and can be found by examination of the coefficient of xn + 1 in the expansion of using When j = k, equation (7b) gives From expansion (7a) using n = 2m, k = m, and (1), one finds (8) Let F(n) denote the nth Fibonacci number. We obtain a formula about the diagonals of Pascal's triangle (9) This can be proved by induction using (3) or by Zeckendorf's representation (Just note that the lhs gives the number of subsets of {F(2),...,F(n)} without consecutive members, which also form all the numbers below F(n+1)). Also using (3) and induction, one can show that (10) Again by (3) and induction, one can show that for k = 0, ... , n-1 (11) as well as (12) which is itself a special case of the result from the theory of finite differences that for any polynomial P(x) of degree less than n,[citation needed] (13a) Differentiating (2) k times and setting x = -1 yields this for , when 0 = k m by induction on M. Using (8) one can derive (15) and (16) [edit]Identities with combinatorial proofs Many identities involving binomial coefficients can be proved by combinatorial means. For example, the following identity for nonnegative integers (which reduces to (6) when q = 1): (16b) can be given a double counting proof as follows. The left side counts the number of ways of selecting a subset of [n] of at least q elements, and marking q elements among those selected. The right side counts the same parameter, because there are ways of choosing a set of q marks and they occur in all subsets that additionally contain some subset of the remaining elements, of which there are 2n - q. The recursion formula where both sides count the number of k-element subsets of {1, 2, . . . , n} with the right hand side ?rst grouping them into those that contain element n and those that do not. The identity (8) also has a combinatorial proof. The identity reads Suppose you have 2n empty squares arranged in a row and you want to mark (select) n of them. There are ways to do this. On the other hand, you may select your n squares by selecting k squares from among the first n and n - k squares from the remaining n squares. This gives Now apply (4) to get the result. [edit]Sum of coefficients row See also: Combination#Number of k-combinations for all k The number of k-combinations for all k, , is the sum of the nth row (counting from 0) of the binomial coefficients. These combinations are enumerated by the 1 digits of the set of base 2 numbers counting from 0 to 2n - 1, where each digit position is an item from the set of n. [edit]Continuous identities Certain trigonometric integrals have values expressible in terms of binomial coefficients: For and (17) (18) (19) These can be proved by using Euler's formula to convert trigonometric functions to complex exponentials, expanding using the binomial theorem, and integrating term by term. [edit]Generating functions [edit]Ordinary generating functions For a fixed n, the ordinary generating function of the sequence is: For a fixed k, the ordinary generating function of the sequence is: The bivariate generating function of the binomial coefficients is: Another bivariate generating function of the binomial coefficients, which is symmetric, is: [edit]Exponential generating function The exponential bivariate generating function of the binomial coefficients is: [edit]Divisibility properties In 1852, Kummer proved that if m and n are nonnegative integers and p is a prime number, then the largest power of p dividing equals pc, where c is the number of carries when m and n are added in base p. Equivalently, the exponent of a prime p in equals the number of nonnegative integers j such that the fractional part of k/pj is greater than the fractional part of n/pj. It can be deduced from this that is divisible by n/gcd(n,k). A somewhat surprising result by David Singmaster (1974) is that any integer divides almost all binomial coefficients. More precisely, fix an integer d and let f(N) denote the number of binomial coefficients with n = i k) prev (binomial-iter n k (+ i 1) (/ (* (- n i) prev) (+ i 1))))) ;; Use symmetry property C(n,k)=C(n, n-k) (if (Related Questions
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