The Binomial Theorem 2

Using the Binomial Expansion in Approximation Only elementary arithmetic operations are used Namely: addition, subtraction, multiplication and division Example Find

3

accurate to three decimal places

Solution 3 = (1 + 2 )

1 2

x =2⇒ x >1

Expansion will not converge

3 = (2 + 1)

1 2

1  = 2 • 1 +  2 

1 2

1 x = ⇒ x <1 2

Expansion will converge but the square root of 2 can not be calculated using elementary arithmetic operations 3 = ( 4 − 1)

1 2

1  = 2 • 1 −  4 

1 2

Expansion will converge

1 x = − ⇒ x <1 4

1 2

1  3 = 2 • 1 −  4    1 1  = 2 1 + −  +  2 4  

 1  1    −  2 1 2 2      −   + 2 •1  4

  1  1   3   − −     3  2  2   2   − 1  +      3 •2 •1  4  

3 = 2(1 − 0.125 + 0.0078125 − 0.00097625 + ) 3 ≅ 1.7324219

The first term neglected in the summation gives an estimate of the error

  1   1  3   5       −  −   −  4 2   2  2   2   1     Error = 2 • −    4 • 3 •2 •1  4      = 0.0003051758

Alternatively 1 1 4 1 3 = ( 3) 2 = ( 48 ) 2 4 4 1 1 7 1  = ( 49 − 1) 2 = 1 −  4 4 49  1 x = << 1 Fast convergence 49

1 2

Example Find

3

340

accurate to four decimal places

Solution 3

340 = (125 + 215 ) 215   = 5 1 +  125  

1 3

1 3

215 x = >1 125

3

340 = ( 216 + 124 ) 124   = 6 1 +  216  

3

340 = ( 343 − 3)

1 3

1 3

124 x = <1 216

1 3

3   = 7 1 −  343  

1 3

3 x = <1 343

Faster convergence and hence fewer terms are needed to achieve the required accuracy

3

3

1 3

3   343 = 7 •  1 −  343     1 3   = 7 1 + − +   3  343   

  1  2     −  2  3  3   − 3  +      2 • 1  343   

343 = 7 • (1 − 0.00291545 − 0.0000084996 − ) = 6.97953235

The multiplier “7” should not affect the fourth decimal place in order to have the required accuracy

Finding the Sum of some Series Now, we have the series and it is required to find its original binomial form Example

1 1• 3 1• 3 • 5 1• 3 • 5 • 7 S= + + + + 3 3 • 6 3 • 6 • 9 3 • 6 • 9 • 11 Solution The numerators factors are divided by “-2” so that they are descending by “1” , while the denominator factors are divided by “3”

1 1 3 1 3 5 2 (−2) • (−2) • • (−2) 3 S = −2 + −2 −2 2 + −2 −2 −23 + 1(3) 1 • 2(3) 1 • 2 • 3(3)

Adding “1” to both sides 1 1 3 (−2) • (−2) 2 S +1 = 1+ − 2 + −2 −2 2 1(3) 1 • 2(3) 1 3 5 • • (−2) 3 + −2 −2 −23 + 1 • 2 • 3(3)

  − 2  S + 1 = 1 +      3 

−1 2

1 =   3

S = 3 −1

−1 2

= 3

Example

1 1• 4 1• 4 • 7 S= − + − +  5 5 • 10 5 • 10 • 15

Solution The numerators factors are divided by “-3” so that they are descending by “1” , while the denominator factors are divided by “5” 1 1 4 1 4 7 2 3 (−3) • (−3) • •  −3 − 3 − 3 − 3 − 3 − 3 − 3 S= − +   − +  2 1(5) 1 • 2(5) 1• 2 • 3  5 

Considering the expansion with a change in the sign of x 1 1 4 1 4 7 2 3 • • • 3 −3 −3 3 −3 −3 −3 3 − 3 S =−  −   −   − 1!  5  2!  5  3! 5

Multiplying by “-1” and adding “1” 1 1 4 2 • 3 −3 −3 3 − 3 − S +1 = 1+  +   1!  5  2!  5  1 4 7 3 • • 3 − 3 − 3 − 3 +   − 3! 5

 3 − S + 1 = 1 +   5

−1 3

8 =  5 3

5 S = 1− 2

−1 3

5 =  8

1 3

Example Find the coefficient of x15 in the expansion of

1 − x    1 + x 

4

Solution 4

1 − x  4 −4 Q=  = (1 + x ) (1 − x ) 1 + x 

(

Q = 1 + 4x + 6x2 + 4x 3 + x 4

)

−4 (−4)(−5)  2 ( − x) + ( − x) • 1 + 1! 2!  (−4)(−5)(−4 − r + 1)  r ( − x ) +  + + r! 

Coefficient of x15 is obtained by the sum of products of 0

15

Coeff.x ⊗ Ceff.x ⇒ r = 15 1 14 Coeff.x ⊗ Ceff.x ⇒ r = 14 2 13 Coeff.x ⊗ Ceff.x ⇒ r = 13 3 12 Coeff.x ⊗ Ceff.x ⇒ r = 12 4 11 Coeff.x ⊗ Ceff.x ⇒ r = 11

Notice that the coefficients can be written in terms of combinations, e.g.

(− 4)(− 5)(− 4 − r + 1) r (-1) r! 4 • 5 • 6 •  • (r + 4 - 1) = r! (r + 3)! r+3 = = Cr 3! r!

Coeff.x

15

18 3

= (1)C

+ (4)C + (6)C 17 3

16 3

+ (4)C + (1)C 15 3

Coeff.x

15

14 3

18 • 17 • 16 17 • 16 • 15 16 • 15 • 14 = (1) + (4) + (6) 3•2•1 3•2•1 3•2•1 15 • 14 • 13 14 • 13 • 12 + (4) + (1) 3•2•1 3•2•1 = 9080

Example Find the coefficient of x24 in the expansion of

(x Solution

3

(

4

5

+ x + x ++ x 3

4

5

)

8 4

Q = x + x + x ++ x 12

2

)

8 4 5 4

= x (1 + x + x +  + x ) 1 − x = x   1− x 12

6

  

4

4

1 − x  −4 12 6 4   ( ) Q=x  = x 1 − x 1 − x  1 − x   4 4•3  12  6 2 = x  1 + (− x ) + (−6) +  1! 2!   12

6

(

)

−4 (−4)(−5)(−4 − r + 1)   r 1 + ( − x ) +  + ( − x ) +    1! r!  

Coeff.x

24

15 3

= (1)C = 125

9 3

+ (−4)C + (6)(1)

The Multinomial

( x1 + x 2 + x 3 +  + x m )

n

For the case n is a positive integer To find the general term, consider x1 is chosen r1 times, x2 is chosen r2 times, … Thus

r1 + r2 + r3 +  + rm = n

Hence, the general term has the form n r1

n−r1 r2

C C

n−r1 −r2 r3

C

n−r1 −r2 −−rm −1 rm

C

n! (n − r1 )! (n − r1 − r2 )! = • •  r1! (n − r1 )! r2 ! (n − r1 − r2 )! r3 ! (n − r1 − r2 − r3 )! (n − r1 − r2 −  − rm−1 )! • rm! (n − r1 − r2 −  − rm )!

n! = r1! r2 ! rm!

Example

Find the coefficient of

2

3

x y z

5 10

in the expansion of

z   1 + 2 x − 3y +  2 

Solution The term having these powers of x, y, and z is 10! z (1)0 (2x)2 (−3y)3   0! 2! 3! 5! 2

Thus

5

5

Coeff. x2 y 3z5 =

10! 1 (2)2 (−3)3   = −8505 0! 2! 3! 5! 2

Example

Find the coefficient of in the expansion of Solution The general term is

x12

(1 + x + 2x )

3 8

(

8! (1)r1 (x)r2 (2x)3 r1! r2 ! r3 !

r1 + r2 + r3 = 8

The power of x should satisfy

r2 + 3r3 = 12

ri , i = 1, 2, 3 are chosen to satisfy to satisfy these two conditions

)

r3

r1

r2

r3

4

0

4

2

3

3

0

6

2

First choose the highest possible value of r corresponding to highest power of x (r3 in this example) and then those with lower power 8! 8! Coeff. x 7 =

(2)4 +

(2)3

4! 0! 4! 2! 3! 3! 8! + (2)2 = 5712 0! 6! 2!

Example

Find the coefficient of in the expansion of

x7

(1 − 2x − x

3

+x

)

4 6

Solution The general term is

(

6! (1)r1 (−2 x)r2 − x 3 r1! r2 ! r3 ! r4 !

) (x ) r3

4 r4

r1 + r2 + r3 + r4 = 6

The power of x should satisfy

r2 + 3r3 + 4r4 = 7

ri , i = 1, 2, 3,4 are chosen to satisfy to satisfy these two conditions

Coeff. x 7 =

r1

r2

r3

r4

4

0

1

1

2

3

0

1

3

1

2

0

1

4

1

0

6! 6! (−2)0 (−1)1 + (−2)3 (−1)0 4! 0! 1! 1! 2! 3! 0! 1! 6! 6! 1 2 + (−2) (−1) + (−2)4 (−1)1 3! 1! 2! 0! 1! 4! 1! 0!

Coeff. x 7 = −30 − 960 − 120 − 480 = −1590