Binomial Expansion

5 Key excerpts on "Binomial Expansion"

• 

  eBook - ePub

  Mathematics for Engineering

  • W Bolton(Author)
  • 2012(Publication Date)
  • Routledge

    (Publisher)

  Multiplying out the binomial terms is very laborious. However, if you look at the above results you can see that: 1 The first term is always 1.

  2 The expansion is a series of terms with increasing powers of x, terminating with x

  n

  , where n is the power to which the (1 + x) is raised.

  3 Each term is multiplied by some number (termed a coefficient). The second term is always nx.

  4 The series terminates after (n + 1) terms.

  What we need is the rule to generate the series. The general formula we can use to write the series is:

  Such an expression is valid with any positive whole number value for n and any value of x. The product 1 × 2 is known as factorial 2 and written as 2!; the product 1 × 2 × 3 is factorial 3 and written as 3!. Hence factorial 5, i.e. 5!, is 1 × 2 × 3 × 4 × 5. Many calculators have a function key which allows factorials to be easily obtained; the sequence to find factorial 5 is: key in 5, then select the function x!.

  If we have (a + b)

  n

  , we can put such a binomial expression in the form used above:

  Expanding the bracketed term then gives:

  The above is a more general form of the binomial theorem; however, it is often simpler to just remember the simpler form and multiply the result by the a

  n

  factor. This is illustrated in the following examples.

  Example

  Determine the series expansion of (1 + x)4 .

  Example

  Determine the series expansion of (2y + x)4 .

  Example

  Determine the series expansion of (4 − 3x)3 .

  This can be written as 43 [1 + (−¾x)]3 . Hence:

  Example

  Evaluate 2.54 .

  We can write 2.54 as (1 + 1.5)4 = (1 + x)4 with x = 1.5. Hence, since:

  then 2.54 = 1 + 4 × 1.5 + 6 × 1.52 + 4 × 1.53 + 1.54 = 39.0625.

  Revision

  10 Determine the series expansions of:

  (a) (1 + x)7 ,

  (b) (2 + x)4 ,

  (c) (2x + 3y)4 ,

  (d) (1 − ½x)6 .

  7.4.1 Expansion for a fractional positive index

  We can use the binomial theorem with fractional positive indices. For example:

  The series does not terminate. Suppose we try to use this expansion to determine the square root of 4. Since 4½ = (1 + 3)½

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  eBook - ePub

  Discrete Mathematics with Ducks

  • sarah-marie belcastro(Author)
  • 2018(Publication Date)
  • Chapman and Hall/CRC

    (Publisher)

  Then evaluate it, somehow. 6.7 Binomial Basics Hey! You! Don’t read this unless you have worked through the problems in Section 1.6. I mean it! A binomial is a polynomial with exactly two terms, such as 3 a - 2 b 5. Consider the simple binomial (x + y). We can rewrite (x + y) n as (x + y) · (x + y) · (x + y) · ⋯ · (x + y), where there are n copies of (x + y) in that product. If we expand thebinomial power (x + y) n into a polynomial, we know what all of the variable parts of the terms are—they’re x n, x n - 1 y, x n - 2 y 2, ⋯, x 2 y n - 2, x y n - 1, y n. But we don’tknow what the coefficients are. Well, we know they should be called binomial coefficients because they are coefficients of a Binomial Expansion, but that’s not yet helpful. Notice that each term has variables whose degrees total to n. (Yes, literally go back a few sentences and notice it actively.) We will have one occurrence of a given term in the full expansion for every way there is of forming it, meaning, for example, that there will be only one copy of x n because we can only form x n by multiplying together all n copies of x in (x + y) · (x + y) · (x + y) · ⋯ · (x + y). We can form x n - 1 y by multiplying together all but one of the x s and the remaining y, and we can figure out the number of different x n - 1 y s we can have by counting the number of ways of choosing that single y. Oh, hey! That’s n 1, and in fact, the number of ways of forming x n was n 0. Similarly, the number of ways to form x n - k y k is n k because we choose k copies of y from the n copies of (x + y) (or, equivalently, because we choose n - k copies of x from the n copies of (x + y)). Yup. Conclusion 1: Another name for n k is binomial coefficient. Conclusion 2: We have proven a theorem. Theorem 1.6 [the binomial theorem] (x + y) n = ∑ k = 0 n n k x n - k y k = n 0 x n + n 1 x n - 1 y + ⋯ + n n - 1 x y n - 1 + n n y n. This theorem has lots of cool consequences

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  eBook - ePub

  Combinatorics

  • Nicholas Loehr(Author)
  • 2017(Publication Date)
  • Chapman and Hall/CRC

    (Publisher)

  To see why, rewrite the given identity as (p - q) (r) = 0, where p - q is a polynomial in the formal variable r. If p - q were nonzero, say of degree d, then p - q would have at most d real roots. Since we are assuming the equation holds for infinitely many values of r, we conclude that p - q is the zero polynomial. Thus, p (r) = q (r) must hold for every r. A similar result holds for polynomial identities involving multiple parameters. 2.3 The Binomial Theorem The Binomial Theorem is a famous formula for expanding the n th power of a sum of two terms. Binomial coefficients are so named because of their appearance in this theorem. 2.8. The Binomial Theorem. For all x, y ∈ R and all n ∈ Z ≥ 0, (x + y) n = ∑ k = 0 n n k x k y n - k. For example, (x + y) 4 = 1 y 4 + 4 x y 3 + 6 x 2 y 2 + 4 x 3 y + 1 x 4. We now give a combinatorial proof of the Binomial Theorem under the additional assumption that x and y are positive integers. (Since both sides of the formula are polynomials in x and y, we can deduce the general case using the result mentioned in Remark 2.7.) Let A = { V 1, …, V x, C 1, …, C y } be an alphabet consisting of x + y letters, where A consists of x vowels V 1, …, V x and y consonants C 1, …, C y. Let S be the set of all n -letter words using the alphabet A. By the Word Rule, | S | = (x + y) n. On the other hand, we can classify words in S based on how many vowels they contain. For 0 ≤ k ≤ n, let S k be the set of words in S that contain exactly k vowels (and hence n - k consonants). To build a word in S k, first choose a set of k positions out of n where the vowels will appear (n k ways, by the Subset Rule); then fill these positions with a sequence of vowels (x k ways, by the Word Rule); then fill the remaining positions with a sequence of consonants (y n - k ways, by the Word Rule). By the Product Rule, | S k | = n k x k y n - k

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  eBook - ePub

  Principles and Techniques in Combinatorics

  • Chen Chuan-Chong, Koh Khee-Meng(Authors)
  • 1992(Publication Date)
  • WSPC

    (Publisher)

  B|, as claimed.

  We thus conclude that the coefficient of (2.8.6) in the expansion is given by

  Combining this with identity (2.8.5), we arrive at the following generalization of the binomial theorem, that was first formulated by G.W. Leibniz (1646-1716) and later on proved by Johann Bernoulli (1667-1748).

  Theorem 2.8.1 (The Multinomial Theorem). For n, m ∈ N,

  where the sum is taken over all m-ary sequences (n1 , n2 , …, n

  m

  ) of nonnegative integers with , and

  Example 2.8.1. For n = 4 and m = 3, we have by Theorem 2.8.1 ,

  Because of Theorem 2.8.1 , the numbers of the form (2.8.2) are usually called the multinomial coefficients. Since multinomial coefficients are generalizations of binomial coefficients, it is natural to ask whether some results about binomial coefficients can be generalized to multinomial coefficients. We end this chapter with a short discussion on this.

  1° The identity for binomial coefficients may be written as (here of course n1 + n2 = n). By identity (2.8.5), it is easy to see in general that

  where {α(l), α(2), …, α(m)} = {1, 2 m}.

  2° The identity for binomial coefficients may be written:

  In general, we have:

  3° For binomial coefficients, we have the identity By letting x1 = x2 = ··· = x

  m

  = 1 in the multinomial theorem, we have

  where the sum is taken over all m-ary sequences (n1 , n2 , …, n

  m

  ) of nonnegative integers with .

  Identity (2.8.9) simply says that the sum of the coefficients in the expansion of (x1 + x2 +…+ x

  m

  )

  n

  is given by m

  n

  . Thus, in Example 2.8.1 , the sum of the coefficients in the expansion of (x1 + x2 + x3 )4 is 81, which is 34 .

  4° In the Binomial Expansion the number of distinct terms is n + 1. How many distinct terms are there in the expansion of (x1 + x2 +…+ x

  m

  )

  n

  ? To answer this question, let us first look at Example 2.8.1 . The distinct terms obtained in the expansion of (x1 + x2 + x3 )4

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  eBook - ePub

  Mathematical Foundations of Computer Science

  • Bhavanari Satyanarayana, T.V. Pradeep Kumar, Shaik Mohiddin Shaw(Authors)
  • 2019(Publication Date)
  • CRC Press

    (Publisher)

  C HAPTER - 14 Binomial Theorem LEARNING OBJECTIVES ♦ to understand the basic concepts of Binomial and Multinomial Coefficients ♦ to know some Generating functions of permutations and combinations ♦ to develop problem solving skills related to the Principle of Inclusion & Exclusion If x and a are real numbers then for all n ∈ N we have. that (x+a) n = n C 0 x n a 0 + n C 1 x n − 1 a 1 + n C 2 x n − 2 a 2 + … + n C r x n − r a r + … + n C n − 1 x 1 a n − 1 + n C n x 0 a n That is, (x +. a) n = ∑ r = 0 n C n r x n − r a r and it is called as Binomial Theorem. 14.1  Binomial and Multinomial Coefficients 14.1.1  Pascal’s Triangle We know that (x + a) 0 =. 1 (x+a) 1 = x + a = 1. x + 1. a (x+a) 2 = x 2 + 2 ax+a 2 = 1. x 2 + 2. ax+1.a 2 (x+a) 3 = x 3 + 3 x 2 a + 3 xa 2 + a 3 = 1. x 3 + 3. x 2 a+3.xa 2 + 1. a 3 We observe that the coefficients in the above expansions follow a particular pattern as shown in the Table 14.1. Table 14.1 We observe that each row is bounded by 1 on both sides. Any entry, except the first and last, in a row is the sum of two entries in the preceding row, one on the immediate left and the other on immediate right. The above pattern is known as Pascal’s triangle. It has been checked that the above pattern also holds good for the coefficients in the expansions of the binomial expressions having index (exponent) greater than 3 as shown in the Table 14.2. Table 14.2 Index of the binomial Coefficients of various terms 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 Using the above Pascal’s triangle, we may express (x + a) n for n = 1, 2, 3,. … (x + a) 1 = x + a (x+a) 2 = x 2 + 2 ax + a 2 (x + a) 3 = x 3 + 3 ax 2 + 3 a 2 x + a 3 (x + a) 4 = x 4 + 4 ax 3 + 6 a 2 x 2 + 4 a 3 x + a 4 (x+a) 5 = x 5 + 5 ax 4 + 10 a 2 x 3 + 10 a 3 x[-. -=PLGO-SEPARATOR=--]2 + 5a 4 x + a 5 and so on. 14.1.2  Properties of Binomial Coefficients or Combinatorial Identities An identity that results from some counting process is called a combinatorial identity

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