The Binomial Theorem.

Transcription

1 The Binomial Theorem

2 The Problem Evaluate (A+B) N as a polynomial in powers of A and B Where N is a positive integer A and B are numbers Example: (A+B) 5 = A 5 +5A 4 B+10A 3 B 2 +10A 2 B 3 +5AB 4 +B 5

3 The Solution Let us approach this as a problem of building words out of an alphabet Let our alphabet consist of only 2 letters: A, B Thus, BABA, AAB, AB, ABBAB are all words using our alphabet List all the 2 letters word can be constructed using this alphabet: AA, AB, BA, BB And all 3 letter words? AAA, ABA, BAA, BBA, AAB, ABB, BAB, BBB

4 If we have all the possible words with 2 letters We can derive all 3 letter words by the following process: Append each 2 letter word with an A Append each 2 letter word with a B The combined output of the 2 steps above is the list of all 3 letter words Lets work this out: The set of all possible words with 2 letters: AA, AB, BA, BB Append with A to get: AAA, ABA, BAA, BBA Append with B to get: AAB, ABB, BAB, BBB Combine the outputs to get: AAA, ABA, BAA, BBA, AAB, ABB, BAB, BBB

5 Let us express this 3 step process, [2 steps to Append and 1 step to Combine] as an operation represented as a between 2 SETS: [P, Q, R,, S] [X, Y]=[PX, QX, RX,, SX, PY, QY, RY,, SY] Here, P, Q, R,, S, X, Y are words of our alphabet [A, B] PX means Appending the word P with the word X So, if P represents ABB and X represents BA, PX represents ABBBA Thus: [AA, ABB, BAAB, AB] [AA,B]=[AAAA, ABBAA, BAABAA, ABAA, AAB, ABBB, BAABB, ABB]

6 The operation is Associative: ([P] [Q]) [R]=[PQ] [R]=[PQR]=P [QR]=P ([Q] [R]) Where P, Q, R are words in our definition The operation is not Commutative : That is P Q is not same as Q P. For example, if P=AAB and Q=BA: [AAB] [BA]=[AABBA] is not [BA] [AAB]=[BAAAB]

7 Using this notation we can derive the set of all 3 letter words, given the set of all 2 letter words as: [List of all 2 letter words] [A, B] [AA, AB, BA, BB] [A,B]=[AAA, ABA, BAA, BBA, AAB, ABB, BAB, BBB] As also, the list of all 4-letter-words is: [List of all 3 letter words] [A, B] [AAA, ABA, BAA, BBA, AAB, ABB, BAB, BBB] [A,B]=[AAAA, ABAA, BAAA, BBAA, AABA, ABBA, BABA, BBBA, AAAB, ABAB, BAAB, BBAB, AABB, ABBB, BABB, BBBB]

8 Clearly, this is true not only for 2, 3 and 4 letter words, but it true for K and (K+1) letter words, K>1 Thus list of all words with (K+1) letters will be: [List of ALL K letter words] [A, B] For example: Thus list of all words with 101 letters will be: [List of ALL 100 letter words] [A, B]

9 Finally, list of all words with N letters will be: [List of ALL (N-1) letter words] [A, B] Thus we can build a list of all N letter words as [All 2 letter words] [A, B]=[All 3 letter words] [All 3 letter words] [A, B]=[All 4 letter words] [All 4 letter words] [A, B]=[All 5 letter words] [All (N-1) letter words] [A, B]=[All N letter words] NOTE: List of all 2 letter words = [A, B] [A, B]

10 Substituting [A, B] [A, B] for the list of all 2 letter words we obtain: [[A, B] [A, B]] [A, B]=[All 3 letter words] [[[A, B] [A, B]] [A, B]] [A, B]=[All 4 letter words] [[[[A, B] [A, B]] [A, B]] [A, B]] [A, B]=[All 5 letter words] =[All N letter words] Note that there are 2 N N letter words using our alphabet since each letter in the N-long word can independently be A or B

11 Using the Associative Property of : [A, B] [A, B] [A, B]=[All 3 letter words] [A, B] [A, B] [A, B] [A, B]=[All 4 letter words] [A, B] [A, B] [A, B] [A, B] [A, B]=[All 5 letter words] [A, B] [A, B] [A, B] {N times}=[all N letter words]

12 Clearly, the operation is indeed analogous to the basic arithmetic operation of multiply & add : In fact, if in the definition of as: [P, Q, R,, S] [X, Y]=[PX, QX, RX,, SX, PY, QY, RY,, SY] We replace, by + and by * we obtain: [P+ Q+ R+ + S]*[X+ Y]=[PX+ QX+ RX+ + SX+ PY+ QY+ RY+ + SY] The expression so obtained is mathematically correct if: P, Q, X, Y are mathematical objects [real numbers, complex numbers etc] The symbol * denotes multiplication and + denotes addition PX is interpreted as P*X, QX is Q*X and so on

13 Since [A+B] N = [A+B]*[A+B]* *[A+B] {N times} And is analogous to multiply and add IF we compute the end result of: [A, B] [A, B] [A, B] {N times} We should get the same result as if [A+B] N is expanded repeatedly multiplying [A+B] by itself N times The only difference will be that instead of + we will have, separating the individual terms The individual terms will be N letter long strings of A and B

14 Since [A, B] [A, B] [A, B] {N times} Gives us the list of ALL N-letter words made of the 2 letter alphabet, A and B [A+B] N =[A+B]*[A+B]* *[A+B] {N times}, due to the analogous nature of and multiply and add, should lead to a summation of terms such that: the set of terms in the summation will be same as the set of all N-long-words of the alphabet A, B Each N-long-word should be interpreted as a * of the A s and B s in it Since there are 2 N of these N-long-words, there will be 2 N terms in the summation So, if we can somehow enumerate ALL the N-long-words using A and B [and we know there will be 2 N of them], we can then add them all up to obtain an expression for [A+B] N Wherein each of the N-long-word should be treated as a product of A s and B s We now find a way to form all these N-long-words using A and B

15 Consider the N letter word as being formed by filling up a row of N positions, with either A or B: A A B B A A B (N-2) (N-1) N N Positions Any N letter word will have k A s and [N-k] B s, where 0<=k<=N B

16 Of all the N letter words, how many will have k A s and [N-k] B s? [0<=k<=N] This is the same as asking, in how many different ways can we choose k positions of the total N positions Assume for now, we have a formula that gives this number We write this formula as: N C k

17 Then the set of ALL N letter words can be partitioned as: Set of N Letter words with N A s = N C N Set of N Letter words with [N-1] A s = N C (N-1) Set of N Letter words with [N-2] A s = N C (N-2) Set of N Letter words with k A s = N C k Set of N Letter words with 1 A s = N C 1 Set of N Letter words with 0 A s = N C 0 Note that all the above sets are mutually disjoint and their union gives the set of ALL N letter words

18 We thus conclude that if we were to expand [A+ B] N using the distribution of * over + then it will lead to the addition of terms [words] such that: N C N of them have N A s and 0 B s N C (N-1) of them will have (N-1) A s and 1 B N C (N-2) of them will have (N-2) A s and 2 B s N C k of them will have k A s and (N-k) B s N C 1 of them will have 1 A and (N-1) B s N C 0 of them will have 0 A s and N B s Note that the word AABAB [for example] = A*A*B*A*B and so on (as mentioned earlier) where * is multiplication

19 Hence, a word with k A s and [N-k] B s is the same as a mathematica term which is the product of k A s and [N-k] B s Such a word can be represented as a mathematical term A k B (N-k) Here we are invoking the fact that: PQ is to be interpreted as P*Q And the * operation IS Commutative So that: AAB=ABA=BAA=A 2 B

20 We thus can say that [A+ B] N when expanded consists of the summation of N C N terms of value A N B 0 N C (N-1) terms of value A (N-1) B 1 N C (N-2) terms of value A (N-2) B 2 N C k terms of value A k B (N-k) N C 1 terms of value A 1 B (N-1) N C 0 terms of value A 0 B N

21 Thus: [A+ B] N = N C N *A N *B 0 + N C (N-1) *A (N-1) *B 1 + N C (N-2) *A (N-2) *B N C k *A k *B (N-k) + + N C 1 *A 1 *B (N-1) + N C 0 *A 0 *B N More concisely N k=0 [A + B] N = N CkA (N k) B k Also, we know from basic Combinatorics that N C k = N! N k!

22 Thus the famous result: [A + B] N = N k=0 N! N k! A(N k) B k For N being a positive integer, and A, B numbers [real or complex]

23 Fractional Exponents Applying the binomial theorem of A = 1, B = x: [1 + x] N = Note that N! N k! (1 + x) n = 1 + nx1 n n 1 n [k 2] N N! k=0 N k! xk = N N 1 (N k 2 )(N k 1 ) + n(n 1) 1! 2! n [k 1] x 2 + n(n 1)(n 2) 3! x k + and so: x Note that for k (n + 1) the coefficients of x k become 0 since they contain a 0 in the numerator. Hence: n n 1 (n [k 1]) (1 + x) n = k=0 x k where the upper limit of summation is changed to instead of N.

24 Let us define f n for any positive rational number n: f n = n n 1 (n [k 1]) k=0 x k Note that for a non-integral value of n the coefficients never become 0. In fact they can attain negative values as well. For n = 3 for example: 2 f 3 = x + 2 (3 2 1) x ! 3 2 (3 2 1)(3 2 2) f 3 2 = x x x3 + 3! x 3 +, so that: m m 1 (m [k 1]) k=0 x k Then we have f m =, m being any positive rational number And of course for positive integral values of n and m, f n = (1 + x) n and f m = (1 + x) m

25 Now if we consider the product f n. f m for n, m positive rational numbers: 1 + nx1 1! m m 1 2! + x 2 + n n 1 2! m m 1 m 2 x 2 + 3! n n 1 n 2 3! x 3 + ) x 3 + (1 + mx1 1! For fractional n, m the coefficients do not got to 0 and hence the product is an infinite series multiplied as: f n. f m f n. m m 1 m 2 = f n. 1 + f n. mx1 3! x 3 + 1! + f n. m m 1 2! + x 2 +

26 This leads to another infinite series in powers of x k, k 0, where the coefficients are functions of n and m f n. f m = C 0 + C 1 x 1 + C 2 x 2 + C 3 x 3 For example, for x 2 the coefficient C 2 is: n n 1 + nm m 1 (n+m) n+m 1 +m = 2! 1! 2! 2! And for x 3 the coefficient is C 3 is: n n 1 n 2 n n 1 m m m 1 m m 1 m ! 2! Which simplifies to: 1! + n 1! 2! (n+m) n+m 1 n+m 2 3! 3!

27 Is there an expression we can derive for C k? Expressing as C kn and C km the coefficients of x k in f n and f m respectively. Examining the product f n. f m = 1 + C 1n x 1 + C 2n x 2 + C 3n x 3 + C kn x k +. (1 + C 1m x 1 + C 2m x 2 + C 3m x C km x k + ) C k = C kn + C (k 1)n C 1m + C (k r)n C rm + +C 1n C (k 1)m +C km Note: (k + 1) terms add up to form C k So while the product f n. f(m) has infinite terms [since f n and f(m) have infinite terms], each of the coefficients C k therein is the addition of (k + 1) terms as above Remember: C nk = n n 1 (n [k 1]) and C mk = m m 1 (m [k 1])

28 Now in the case of n and m being positive integers, f(n) and f(m) actually represent (1 + x) n and (1 + x) m respectively [as per the binomial theorem for positive integral exponents] Hence: f n. f m = (1 + x) n. (1 + x) m = (1 + x) (n+m) and now since (n + m) is also a positive integer we know that [again using the binomial theorem]: (1 + x) n+m = 1 + (n+m)x1 1! (n+m)(n+m 1)(n+m 2) 3! (n+m) n+m 1 n+m [k 2] + (n+m)(n+m 1) 2! x n+m [k 1] x 2 + x k +

29 Hence in the case of n, m being positive integers we can derive that C k, the coefficient of x k in f n. f(m) is same as: (n+m) n+m 1 n+m [k 2] n+m [k 1] and thus: C k = C kn + C (k 1)n C 1m + C (k r)n C rm + +C 1n C (k 1)m +C km (n+m) n+m 1 n+m [k 2] n+m [k 1] = We now have a value for C k in terms of n, m assuming their positive integer values Of course for k > n + m, the C k values will be 0 since the numerator will have a 0 term

30 Now comes a key argument! C k which is given in its SUM FORM [for integral or fractional n, m]: C kn + C (k 1)n C 1m + C (k r)n C rm + +C 1n C (k 1)m +C km is, for integral n, m, (n+m) n+m 1 n+m [k 2] n+m [k 1] [FACTOR FORM] Each term of C k, in the Sum Form is as C (k r)n C rm, which is a finite polynomial in n and m So one can imagine that using successive steps of algebraic manipulations the Sum Form of C k can be transformed to the Factor Form of C k for integral values of n and m These successive steps will be using Commutative, Associative and Distributive Laws of numbers NOW: The Commutative, Associative and Distributive laws are exactly the same for rational numbers as for integers and so if the same successive steps are applied on the Sum Form of C k for fractional values of n and Thus C k = m, the result will be the SAME Factor Form of C k (n+m) n+m 1 n+m [k 2] n+m [k 1] for fractional n, m

31 Hence for n, m positive rational numbers: f n. f m = k=0 (n+m) n+m 1 n+m [k 2] n+m [k 1] Note that for k=0, the numerator is taken as 1 And in fact this can be extended to: f n. f m f(p) = k=0 (n+m +p) n+m +p 1 n+m +p [k 2] x k n+m +p [k 1] x k

32 Let each of n, m, p be = l h for a positive integers l, h and let us multiply f( l h) by itself h times: f( l h). f( l h). f( l h) [h times] = k=0 ( l h +l h +l h ) l h +l h +l h 1 l h +l h +l h [k 2] l h + l h +l h [k 1] Now the left hand side of the above equation = f( l h) h The right side contains the term l h + l h + l hwherein l h occurs h times so that l h + l h + l h = l giving the right hand side as: k=0 (l) l 1 l [k 2] l [k 1] x k x k

33 Note that the term (l) l 1 l [k 2] l [k 1] becomes 0 for k = l + 1 so we can write the RHS as: l k=0 (l) l 1 l [k 2] l [k 1] x k In the above, since l is a positive integer, we know by the binomial theorem that it represents (1 + x) l. Thus: f( l h) h =(1 + x) l and so: f l h = (1 + x) l h

34 k=0 r r 1 (r [k 1]) Thus: (1 + x) r = f r = for a positive rational number r. Equivalently: (1 + x) r = 1 + rx + r(r 1) x 2 + 2! r(r 1)(r 2) x 3 + r(r 1)(r 2)(r 3) x 4 3! 4! r r 1 r 2 (r k 1 ) + x k + where r is a positive rational number (including positive integers, of course!) x k

35 Lets try it out! x and r both positive integers: x = 2; r = 3 (1 + 2) 3 = (3 1) 3.(3 1)(3 2)(3 3) 4! 2! (3 1)(3 2) 3! (3 1)(3 2)(3 3)(3 4) 5! = = 27 Works of course this is the same as binomial theorem for positive integer exponents Also, in the infinite series expansion, the coefficients after the 4 th terms are all 0.

36 Lets try it out! x and r both positive fractions: x = 1 3 ; r = 1 2 ( )1 2= ! ( 1 3 )4 + ( 1 3 ) ! 3! ! ( 1 3 )3 + ( 1 3 )5 + = = Check: ( )1 2= ( ) 1 2 = So, taking 6 terms of the infinite series we get pretty close to the real answer

37 Lets try it out! x and r both positive fractions: x = 11 ( )1 2= ! ( ) ; r = ( 11 2! 10 )2 + 3! = = ! ( )3 + ( )5 + Check: ( )1 2= (2.1) 1 2 = ? So, taking 6 terms of the infinite series we are close but not as close!

38 Taking 20 terms instead of 6 and then we reach Taking 50 terms instead of 6 and then we reach now we are farther from what we had with 20 terms! Lets try 100 terms: !! Worse still, 101 terms gives: ! Lets look at how the sum behaves as we take more and more values!

39 Sum Value vs Number of Terms!! Series

40 r r 1 3! r 2 (r k 1 ) 4! Lesson Our infinite series formula: (1 + x) r = 1 + rx + r(r 1) x 2 + r(r 1)(r 2) 2! x 3 + r(r 1)(r 2)(r 3) x 4 + x k + Works always if r is a positive integer Works for x < 1 for if r is a positive fraction And this stems from the fact that the infinite sum will converge if and only if x < 1

41 Utilising the Series So, how do we use this infinite sum in real life to evaluate (a + b) r where a, b are real number and r is a positive fraction? If a < b: (a + b) r =(b a b + 1 )r = b r ( a b + 1)r and now since a <1 the infinite series will converge and can b be used. Similarly, if a > b then (a + b) r =a r ( b a + 1)r

42 What if r is a negative fraction? Notice that up until the derivation of: f n. f m f(p) = k=0 (n+m +p) n+m +p 1 n+m +p [k 2] n+m +p [k 1] There is nothing in the argument that is specific to n and m being positive rational numbers In fact the argument is valid for n, m being negative rational numbers, real numbers or even complex numbers since they all follow the same rules of Commutation, Association, Distribution etc. In fact as long as n, m are members of a Commutative Ring the above derivation is true Of course the question is whether the infinite series f(n) converges or not, for a class of inputs E.g. for x < 1, the f(n) converges for any positive rational value of n x k

43 So for a rational number r > 0, we can say: f r. f r = k=0 (r+[ r]) r+[ r] 1 r+[ r] [k 2] n+m [k 1] And remembering that for k = 0, the numerator = 1 The series sum then reduces to 1 since there is a factor (r + [ r]) in each numerator but for when k = 0 when the numerator is 1 So we get: f r. f r = 1 We proved that f r = (1 + x) r for r > 0 Hence: f r = (1 + x) r for r > 0 Thus for all rational values of r, f r = (1 + x) r, x < 1 x k