The Binomial Theorem 1

Lesson 1

THE BINOMIAL THEOREM (1)

Prof. Hany Abdel-Malek

CONTENTS • Case of positive integer power • Taylor’s Expansion • Expansion for any real power • Applications General Term Approximation of roots Sum of Series • Multinomials

The Case of Positive Integer Power n

n 0

n 1

n 2

2

(1 + x) = C + C x + C x +  n r

r

n n

+ C x ++ C x n! C = r! (n − r)! n r

n r

n r −1

C +C

n+1 r

=C

n 0

n n

C =C

n

Example 4

4 1

4 2

2

(1 − 3x) = 1 + C (−3x) + C (−3x) 4 3

3

4 4

4

+ C (−3x) + C (−3x)

4•3 2 = 1 + 4(−3x) + (−3x) 2 •1 4•3•2 3 4 + (−3x) + (−3x) 3 •2 •1 2 3 4 = 1 − 12 x + 54 x − 108 x + 81x

Taylor’s Expansion (Maclaurin)

f ′(0) f ′′(0) 2 f(x) = f(0) + x+ x + 1! 2! (r) f (0) r + x + r! The function is expanded in terms of its value and derivatives at a point ( it is x = 0 in the expansion above)

Case of any power n f(x) = (1 + x)n f ′(x) = n(1 + x)n−1 f ′′(x) = n(n - 1)(1 + x)n-2 f ′′′(x) = n(n - 1)(n - 2)(1 + x)n-3  f (r)(x) = n(n − 1)(n − 2)(n − r + 1)(1 + x)n−r f(0) = 1 f ′(0) = n f ′′(0) = n(n - 1) f ′′′(0) = n(n - 1)(n - 2)  f (r) (0) = n(n −1)(n − 2)(n − r +1)

Substituting in Taylor’s expansion n n(n − 1) 2 n(n − 1)(n − 2) 3 (1 + x) = 1 + x + x + x + 1! 2! 3! n(n − 1)(n − r + 1) r + x + r! n

Remark: It is an infinite expansion Condition for the convergence of the expansion

x <1

Example Find the first four terms and the condition for convergence for:

(i) (ii) (iii)

−4

(1 + 2x)

1/2

(x + 4)

(3x + 4)

-3 2

(1 + 2x)−4

(i)

(−4)(−5) (1 + 2x) = 1 + (−4)(2x) + (2x)2 2! (−4)(−5)(−6) + (2x)3 +  3! = 1 - 8x + 40x 2 − 160 x 3 +  −4

Condition for the convergence of the expansion

2x < 1 − 1 < 2x < 1 1 1 − <x< 2 2

1 2

(x + 4)

(ii) 1 2

1 2

x  (x + 4) = (4) 1 +  4 

1 2

   1  1   1  1  3  − − −        2 3   1  x   2  2   x  x 2 2 2         = 2 1 +   +   +   +  24 2 •1 4 3 •2 •1   4    

Condition for the convergence of the expansion

x <1 4 −4< x <4

(3x + 4)

(iii) 3 − 2

(3x + 4)

3 − 2

−

-3 2

3 2

3x   = (4)  1 +  4      3  5   3  5  7  − − − − −           2 3   3   3x   2   2   3x  3x  2 2 2         −3 = (2) 1 +  −   +   +   +  2•1 3•2•1    2  4   4   4      −

(3x + 4)

3 2

1 9 135 2 945 3  = 1 − x + x − x +  8 4 128 1024  1 9 135 2 945 3 = − x+ x − x + 8 64 1024 8192

Condition for the convergence of the expansion

3x <1 4 −

4 4 <x< 3 3

Frequently Used Examples 1 (i) = (1 + x)−1 1+ x −1 (−1)(−2) 2 =1+ x+ x 1! 2! (−1)(−2)(−3) 3 + x + 3! 1 ∴ = (1 + x)−1 = 1 − x + x2 − x 3 +  1+ x + (-1)r x r + 

Thus the coefficient of xn for any n can be stated directly from the general term

For example , coefficient of x100 is +1 while the coefficient of x55 is -1 A comparison with the sum of the infinite geometric series 2

3

1− x + x − x + In which the first term a=1 and the base r = -x The sum is given by a 1 1− x + x − x += = 1−r 1+ x 2

3

1 −1 (ii) = (1 − x) 1- x −1 (−1)(−2) 2 =1+ (− x) + (− x) 1! 2! (−1)(−2)(−3) 3 + (− x) +  3! 1 ∴ = (1 − x)−1 = 1 + x + x2 + x 3 +  + x r +  1− x Alternatively: replace x by –x in the previous example

1 −2 (iii) = (1 + x) 2 (1 + x) −2 (−2)(−3) 2 =1+ x+ x 1! 2! (−2)(−3)(−4) 3 + x + 3! 1 2 3 ∴ = 1 − 2 x + 3x − 4 x +  2 (1 + x) r

r

+ (-1) (r + 1)x +  e.g. coefficient of x100 is +101 while the coefficient of x55 is -56

1 −2 (iv) = (1 − x) 2 (1 − x) −2 (−2)(−3) 2 =1+ (− x) + (− x) 1! 2! (−2)(−3)(−4) 3 + (− x) +  3! 1 2 3 ∴ = 1 + 2 x + 3 x + 4 x + 2 (1 − x) + (r + 1)x r + 

Which can be also obtained from the previous example by replacing x by –x

Finding the general term (Coefficient of xn ) Example: Find the expansion and its condition for convergence of 2 + x    2 − x 

2

Hence find the coefficient of xn

2

−2

2

x  x 2 + x     = 1 +  1 −  2  2 2 − x  

2

 x  x  x  = 1 + 2 2  + 3 2  1 + x + 4         r  x  +  + (r +1)  +   2   2

The condition for convergence

-2< x < 2

In order to find the coefficient of xn , the product of the two brackets is considered

  x  x x x 1 + x + 1 + 2  + 3  +  + (r + 1)  +   4  2 2 2   2

2

×x

n −2

r

× x n−1

× xn n

1 1 Coeff.x = (1)(n + 1)  + (1)(n)  2 2 n

n

n−1

1 1 +  (n − 1)  4 2

1 1 Coeff.x =   ((n + 1) + 2n + (n − 1)) = 4n  2 2

n

n

This expression is valid only for n > 2 since we have considered all terms up to x2 in the first bracket

n−2

For the cases n = 0 and n = 1 , not all the terms in the first bracket are considered   x  x x x 1 + x + 1 + 2  + 3  +  + (r + 1)  +   4  2 2 2   2

2

r

Coeff. x 0 = (1)(1) = 1   x  x x x 1 + x + 1 + 2  + 3  +  + (r + 1)  +   4  2 2 2   2

2

Coeff. x1 = (1)(1) + (1)(1) = 2

r

Example: Find the first six terms in the expansion of 1 − 3x 2 1+ x + x

Also find the coefficient of xn

Solution: 1 − 3x 1 − 3x 1− x Q= = • 2 2 1− x 1+ x + x 1+ x + x 1 − 4 x + 3x 2 = (1 − x 3 )

(

)

1 − 4 x + 3x 2 2 3 −1 Q= = 1 − 4 x + 3 x ( 1 − x ) 3 (1 − x )

(

)(

= 1 − 4x + 3x 2 1 + x 3 + (x 3 )2 + (x 3 )3 + 

)

+ (x 3 )r + 

Q = 1 − 4 x + 3x 2 + x 3 − 4 x 4 + 3x 5 +  Coeff. x 3r = 1 Coeff. x 3r +1 = −4 Coeff. x

3r + 2

=3

The value of n can be one of three cases: n= = 3r , n = 3r + 1 , n = 3r +2 e.g. Coeff. x100= -4 since 100=3(33)+1 Coeff. x50 = 3 since 50=3(16)+2

Example: Find the first six terms in the expansion of 1− x (1 + x + x2 )2

Also find the coefficient of xn

Solution: 1−x 1− x 1 − x  Q= = •  2 2 2 2 (1 + x + x ) (1 + x + x )  1 − x  1 − 3x + 3x 2 − x 3 = (1 − x 3 )2

2

( = (1 − 3x + 3x

) )(1 + 2x

Q = 1 − 3x + 3x2 − x 3 (1 − x 3 )−2 2

+ x3

3

+ 3(x 3 )2 + 4(x 3 )3

)

+  + (r + 1)(x 3 )r +  2

3

4

5

Q = 1 − 3 x + 3x + x − 6 x + 6 x +  Coeff. x 3r = (1)(r + 1) + (−1)(r) = 1 Coeff. x 3r +1 = (−3)(r + 1) = −3r − 3 Coeff. x 3r +2 = (3)(r + 1) = 3r + 3

Notice that the case of n is a multiple of 3 can be obtained by considering the first and last terms of the first bracket