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Maths C4
Binomial Theorem
Three quick questions from C2
• Expand the following:• 1) (1+x)4
• 2) (1-2x)3
• 3) (1+3x)4
• Here these expansions are finite (n+1) terms and exact.
2 3( 1) ( 1)( 2)(1+ ) =1+ + + +...+
2! 3!n nn n n n n
x nx x x x
Expand the following:
• 1) (1+x)4=1+4x+6x2+4x3+x4
• 2) (1-2x)4=1+4(-2x) + 6(-2x)2+4(-2x)3+(-2x)4.
• 3) (1+3x)4 =
Three steps
• 1) write coefficients :
• 2) Make first term = 1• 3) second term ascending
...!3
)2)(1(!2
)1(1
)1(32
xnnn
xnnnx
x n
Binomial expansionPreviously 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
What happens is n is a fraction or negative number?
• Click here to investigate
For what values of x is this expansion valid?
...!3
)2)(1(!2
)1(1
)1(32
xnnn
xnnnx
x n
....0331
)1(432
3
xxxx
x
1(1 )x
n is positive integer This expansion is finite and all other terms will be zero after this
n is negative fractional This expansion will be infinite.
What happens if x >1? What happens if it is a fraction?
What happens when n is negative or fractional?
• Use the binomial expansion to find the first four terms of
• This expansion is infinite-this goes on forever and convergent when [x] < 1 or
• -1<x<1. This is very important condition!
1)1(1
1
xx
....1
...!3
)3)(2)(1(
!2
)2)(1()1(1
32
32
xxx
xxx
Infinite and Convergent
Using our applet can you see why this series goes on forever and forever?
Can you see that if x is a fraction i.e. |x| < 1 that this series converges?
x x 1
21 + 2 (1 2 )For this infinite series |2x| < 1.
That is when |x| < . 12
2 3( 1) ( 1)( 2)(1+ ) =1+ + + +...
2! 3!n n n n n n
x nx x x
Binomial expansionIn 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
A summary of the key points
• Look at this summary and take notes accordingly
The condition for convergence
3
1
)1( x
2
1
)31(31 xx
)21( x
For these expansions to be infinite and converge then [x}<1, [2x]<1 or [3x]<1
Binomial expansionExpand 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 expansionTherefore
This expansion is valid for < 1.34x
12
1 3 27(4 + 3 ) = + +...
2 16 256
x xx
43
That is, when |x| < . 43
ApproximationsIn 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...
ApproximationsIf 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
Using partial fractionsWe 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
Using partial fractionsWhen 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
Using partial fractions
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.
Using partial fractions
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
Problem A
• Find the binomial expansions of up to and including the term in x3 stating the range of values for which the expansions are valid.
3
1
)1( x
...!3
)2)(1(!2
)1(1
)1(32
xnnn
xnnnx
x n
Problem B
• Find the binomial expansions of up to and including the term in x3 stating the range of values for which the expansions are valid.
2)41(
1
x
...!3
)2)(1(!2
)1(1
)1(32
xnnn
xnnnx
x n
Problem C
• Find the binomial expansions of up to and including the term in x3 stating the range of values for which the expansions are valid.
)21( x
Problem D
• Find the binomial expansions of up to and including the term in x3 stating the range of values for which the expansions are valid.
• Hint make sure the first term is 1!!!
)4( x
Problem E
• Find the binomial expansions of up to and including the term in x3 stating the range of values for which the expansions are valid. Hint make sure the first term is 1!!!
2)32(
1
x
Problem A
• Find the binomial expansions of up to and including the term in x3 stating the range of values for which the expansions are valid.
3
1
)1( x
...!3
)2)(1(!2
)1(1
)1(32
xnnn
xnnnx
x n
Problem B
• Find the binomial expansions of up to and including the term in x3 stating the range of values for which the expansions are valid.
2)41(
1
x
...!3
)2)(1(!2
)1(1
)1(32
xnnn
xnnnx
x n
Problem C
• Find the binomial expansions of up to and including the term in x3 stating the range of values for which the expansions are valid.
)21( x
Problem D
• Find the binomial expansions of up to and including the term in x3 stating the range of values for which the expansions are valid.
• Hint make sure the first term is 1!!!
)4( x
Problem E
• Find the binomial expansions of up to and including the term in x3 stating the range of values for which the expansions are valid. Hint make sure the first term is 1!!!
2)32(
1
x
Let’s use our Partial Fractions knowledge!
• Example 1 Express the following a partial fractions first and then expand up to the x3 term.
)2)(1(
54
xx
x
Example 2
• Express as partial fractions,
• hence expand the first three terms in ascending powers of x . State the set of values of x for which the expansion is valid.
)2)(1(
48
xx
x
)2)(1(
48
xx
x
Ex 3C Q1 Anom
