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A2-Level Maths: Core 4for Edexcel

C4.3 Sequences and series

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Binomial expansion for negative and fractional indices

Approximations

Use of partial fractions

Examination-style question

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Negative and fractional indices


Binomial expansion

Previously in the course we found that, when n is a positive whole number,

If n is negative or fractional then, provided that |x| < 1, the infinite series

will converge towards (1 + x)n.

This is a finite series with n + 1 terms.

2 3( 1) ( 1)( 2)(1+ ) =1+ + + +...+

2! 3!n nn n n n n

x nx x x x

2 3( 1) ( 1)( 2)1+ + + +...

2! 3!

n n n n nnx x x


Binomial expansion

In general, for negative and fractional n and |x| < 1,

Start by writing this as (1 + x)–1.

2 3( 1) ( 1)( 2)(1+ ) =1+ + + +...

2! 3!n n n n n n

x nx x x

2 3 4( 1)( 2) ( 1)( 2)( 3) ( 1)( 2)( 3)( 4)1+ ( 1) + + + ...

2! 3! 4!x x x x

Expand up to the term in x4.1

1+ x

This is equal to (1 + x)–1 provided that |x| < 1.

The expansion is then:

2 3 41 ...x x x x


=1

Binomial expansion

12 2 3

31 1 1 12 2 2 2 21

2

( )( ) ( )( )( )(1+ 2 ) =1+ ( )(2 )+ (2 ) + (2 ) +...

2! 3!x x x x

Expand up to the term in x3.1+ 2x

Start by writing this as . 12(1+ 2 )x Here x is replaced by 2x.

+ x 212 x 31

2+ ...x

This converges towards provided that |2x| < 1. 1+ 2x

That is when |x| < . 12


Binomial expansion

2 31 ( 2)( 3) ( 2)( 3)( 4)

= 1+ ( 2) + + +...9 3 2! 3 3! 3

x x x

Find the first four terms in the expansion of (3 – x)–2.

When the first term in the bracket is not 1, we have to factorize it first. For example:

2

2(3 ) = 3 13

xx

22= 3 1

3

x

2 31 2 4= 1+ + + +...

9 3 3 27

x x x


Binomial expansion

That is, when |x| < 3.

Therefore

This expansion is valid for < 1. 3x

2 32 1 2 4

(3 ) = + + + +...9 27 27 243

x x xx

In general, if we are asked to expand an expression of the form (a + bx)n where n is negative or fractional we should start by writing this as:

( + ) = 1+n

n bxa bx a

a

The corresponding binomial expansion will be valid for |x| < . a

b


Binomial expansion

Expand up to the term in x2 giving the range of values for which the expansion is valid.

12

12

3(4 + 3 ) = 4 1+

4

xx

12(4 + 3 )x

12

12

3= 4 1+

4

x

2312 2( )( )1 1 3 3

= 1 + +...2 2 4 2! 4

x x

1 3 27= 1 + +...

2 8 128

x x


Binomial expansion

Therefore

This expansion is valid for < 1.34x

12

1 3 27(4 + 3 ) = + +...

2 16 256

x xx

43

That is, when |x| < . 43


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Approximations


Approximations




Approximations

In general, when the index is negative or fractional, we only have to find the first few terms in a binomial expansion.

For example, it can be shown that:

This is because, as long as x is defined within a valid range, the terms get very small as the series progresses.

By only using the first few terms in an expansion we can therefore give a reasonable approximation.

If x is equal to 0.1 we have:

1 2 3 4(1 ) =1+ + + + +... for 1x x x x x x

1(1 0.1) =1+ 0.1+ 0.01+ 0.001+ 0.0001+...1(0.9) =1.1111...


Approximations


Approximations

If we only expand up to the term in x it is called a linear approximation. For example:

If we expand up to the term in x2 it is called a quadratic approximation. For example:

Binomial expansions can be used to make numerical approximations by choosing suitable values for x.

1(1 ) 1+x x (for |x| < 1)

1 2(1 ) 1+ +x x x (for |x| < 1)

Write a quadratic approximation to and use this to find a rational approximation for .3

12(4 )x


Approximations

Let x = 1:

12

12(4 ) = 4 1

4

xx

12

= 2 14

x

21 12 2( )( )1

= 2 1+ + +...2 4 2! 4

x x 2

= 2 1+ + +...8 128

x x

12

1 1(4 1) 2 1 +

8 128

1133

64

(for |x| < 4)


Approximations

12 2

1 12 2( )( )1

(1+ ) =1+ + +...2 2!

x x x

Expand up to the term in x2 and substitute x = to obtain a rational approximation for

12(1+ )x 2

911.

1 12 8=1+ +...x x (for |x| < 1)

29When x = we have:

122

9

1 2 1 4(1+ ) =1+ +...

2 9 8 81

11 1 1=1+ +...

9 9 162

11 162 18 1= + +...

3 162 162 162


Approximations

We can check the accuracy of this approximation using a calculator.

11 = 3.317 (to 3 d.p.)

11 179=

3 162

53711 =

162

179=

54

179= 3.315 (to 3 d.p.)

54

Our approximation is therefore correct to 2 decimal places.

If a greater degree of accuracy is required we can extend the expansion to include more terms.

Therefore


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Approximations





We can use partial fractions to carry out more complex binomial expansions.

When x = –1:

For example, we can expand by expressing it in

partial fractions as follows:

5 1

( +1)( 2)

x

x x

Let5 1

+( +1)( 2) +1 2

x A B

x x x x

Multiplying through by (x + 1)(x – 2) gives:

5 1 ( 2)+ ( +1)x A x B x

5 1= 3A

= 2A



When x = 2:

So5 1 2 3

+( +1)( 2) +1 2

x

x x x x

We can now expand 2(1 + x)–1 and 3(–2 + x)–1 :

1 12( +1) + 3( 2)x x

10 1= 3B= 3B

1 12(1+ ) + 3( 2 + )x x

1 2 3( 1)( 2) ( 1)( 2)( 3)2(1+ ) = 2 1+ ( 1) + + +...

2! 3!x x x x

2 3= 2 1 + +...x x x

2 3= 2 2 + 2 2 +...x x x



This is valid for |x| < 2.

11 13( 2 + ) = 3( 2) 1

2

xx

2 33 ( 1)( 2) ( 1)( 2)( 3)

1+ ( 1) + + +...2 2 2! 2 3! 2

x x x

This expands to give:

2 33= 1+ + + +...

2 2 4 8

x x x

2 33 3 3 3

= ...2 4 8 16

x x x

This is valid for |x| < 1.



This is valid when both |x| < 1 and |x| < 2.

2 31 11 13 35= ...

2 4 8 16

x x x

2 3

2 32 3 3 3 3 3+ = 2 2 + 2 2 +... + ...

+1 2 2 4 8 16

x x xx x x

x x

We can now add the two expansions together:

From the number line we can see that both inequalities hold when |x| < 1.

–2 –1 0 1 2


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Approximations





The function f is given by

a) Express f(x) in partial fractions.

b) Expand f(x) as a series in ascending powers of x as far as the term in x2.

c) State the set of values of x for which the expansion is valid.

f 12

6( ) = , , , 1

(1+ 2 )(1 )x x x x

x x

a) Let6

+(1+ 2 )(1 ) 1+ 2 1

A B

x x x x

Multiplying through by (1 + 2x)(1 – x) gives:

6 (1 )+ (1+ 2 )A x B x



When x = 1:

So6 4 2

+(1+ 2 )(1 ) 1+ 2 1x x x x

6 = (1+ 2)B

= 2B

When x = : 126 = (1+ )A

= 4A

12

b) 1 14 2+ = 4(1+ 2 ) + 2(1 )

1+ 2 1x x

x x



Expanding 4(1 + 2x)–1 gives:

Expanding 2(1 – x)–1 gives:

1 2 3( 1)( 2) ( 1)( 2)( 3)4(1+ 2 ) = 4 1+( 1)(2 )+ (2 ) + (2 ) +...

2! 3!x x x x

2 3= 4 1 2 + 4 8 +...x x x

2 3= 4 8 +16 32 +...x x x

1 2 3( 1)( 2) ( 1)( 2)( 3)2(1 ) = 2 1+ ( 1)( )+ ( ) + ( ) +...

2! 3!x x x x

2 3= 2 1+ + + +...x x x

2 3= 2 + 2 + 2 + 2 +...x x x



Therefore f(x) can be expanded as:

2 3 2 34 2+ = (4 8 +16 32 +...)+ (2 + 2 + 2 + 2 +...)

1+ 2 1x x x x x x

x x

2 3= 6 6 +18 30 +...x x x

c) The expansion of (1 + 2x)–1 is valid for |x| < .12

The expansion of (1 – x)–1 is valid for |x| < 1.

The expansion of f(x) is therefore valid for |x| < .12

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