the binomial expansion
DESCRIPTION
The Binomial Expansion. Introduction. You first met the Binomial Expansion in C2 In this chapter you will have a brief reminder of expanding for positive integer powers We will also look at how to multiply out a bracket with a fractional or negative power - PowerPoint PPT PresentationTRANSCRIPT
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The Binomial Expansion
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Introductionβ’ You first met the Binomial Expansion in C2
β’ In this chapter you will have a brief reminder of expanding for positive integer powers
β’ We will also look at how to multiply out a bracket with a fractional or negative power
β’ We will also use partial fractions to allow the expansion of more complicated expressions
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Teachings for Exercise 3A
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The Binomial ExpansionYou need to be able to expand expressions of the form (1 + x)n where n is any real
number
3A
Find: (1+π₯ )4
(1+π₯ )π 1ΒΏ +ππ₯+π (πβ1) π₯2
2 !+π (πβ1)(πβ2) π₯
3
3 !β¦β¦+ΒΏππΆπ π₯π ΒΏ
(1+π₯ )4 1ΒΏ +(4 )π₯+4 (3) π₯2
2+(4 )(3)(2) π₯
3
6+(4 )(3)(2)(1) π₯4
24
1ΒΏ +4 π₯+6 π₯2+4 π₯3+π₯4
Every term after this one will contain a (0) so can be ignored
The expansion is finite and exact
Always start by writing out the general form
Sub in:n = 4x = x
Work out each term separately and simplify
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The Binomial ExpansionYou need to be able to expand expressions of the form (1 + x)n where n is any real
number
3A
Find: (1β2 π₯ )3
(1+π₯ )π 1ΒΏ +ππ₯+π (πβ1) π₯2
2 !+π (πβ1)(πβ2) π₯
3
3 !β¦β¦+ΒΏππΆπ π₯π ΒΏ
(1β2 π₯ )31ΒΏ +(3)(β2π₯)+3 (2)(β2π₯ )2
2+(3)(2)(1)
(β2 π₯)3
6
1ΒΏ β6 π₯+12 π₯2β8π₯3
Every term after this one will contain a (0) so can be ignored
The expansion is finite and exact
Always start by writing out the general form
Sub in:n = 3
x = -2xWork out each term separately and
simplifyIt is VERY important to put brackets
around the x parts
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The Binomial ExpansionYou need to be able to expand expressions of the form (1 + x)n where n is any real
number
3A
Find:1
(1+π₯)
(1+π₯ )π 1ΒΏ +ππ₯+π (πβ1) π₯2
2 !+π (πβ1)(πβ2) π₯
3
3 !
(1+π₯ )β11ΒΏ +(β1)(π₯)+(β1)(β2)(π₯)2
2+(β1)(β2)(β3)
(π₯ )3
6
1ΒΏβπ₯+π₯2βπ₯3
Rewrite this as a power of x first
Sub in:n = -1x = x
Work out each term separately and simplify
ΒΏΒΏWrite out the general form (it is very unlikely you will have to go beyond the first 4
terms)
With a negative power you will not get a (0) term
The expansion is infinite It can be used as an approximation for the
original term
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The Binomial ExpansionYou need to be able to expand expressions of the form (1 + x)n where n is any real
number
3A
Find:β1β3 π₯
(1+π₯ )π 1ΒΏ +ππ₯+π (πβ1) π₯2
2 !+π (πβ1)(πβ2) π₯
3
3 !
(1β3 π₯ )12 1ΒΏ+( 12 )(β3 π₯)+( 12 )(β 12 ) (β3 π₯)2
2+(12 )(β 12 )(β 32 ) (β3 π₯)3
6
1ΒΏβ 32 π₯β98 π₯
2β 2716
π₯3
Rewrite this as a power of x first
Sub in:n = 1/2x = -3x
Work out each term separately and simplify You should use your
calculator carefully
ΒΏΒΏWrite out the general form (it is very unlikely you will have to go beyond the first 4
terms)
With a fractional power you will not get a (0) term
The expansion is infinite It can be used as an approximation for the
original term
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The Binomial ExpansionYou need to be able to expand expressions of the form (1 + x)n where n is any real
number
3A
Find the Binomial expansion of:ΒΏ(1+π₯ )π 1ΒΏ +ππ₯+π (πβ1) π₯
2
2 !+π (πβ1)(πβ2) π₯
3
3 !
(1βπ₯ )13 1ΒΏ +( 13 )(βπ₯)+( 13 )(β 23 ) (β π₯)2
2+( 13 )(β 23 )(β 53 )(βπ₯)3
6
1ΒΏ β 13 π₯β19 π₯
2β 581
π₯3
Sub in:n = 1/3x = -x
Work out each term separately and simplify
Write out the general formand state the values of x for which it is validβ¦
Imagine we substitute x = 2 into the expansion1ΒΏβ 23β
49β4081
1ΒΏ β0.666β0.444β0.4938
The values fluctuate (easier to see as decimals)
The result is that the sequence will not converge and hence for x = 2, the expansion
is not valid
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The Binomial ExpansionYou need to be able to expand expressions of the form (1 + x)n where n is any real
number
3A
Find the Binomial expansion of:ΒΏ(1+π₯ )π 1ΒΏ +ππ₯+π (πβ1) π₯
2
2 !+π (πβ1)(πβ2) π₯
3
3 !
(1βπ₯ )13 1ΒΏ +( 13 )(βπ₯)+( 13 )(β 23 ) (β π₯)2
2+( 13 )(β 23 )(β 53 )(βπ₯)3
6
1ΒΏ β 13 π₯β19 π₯
2β 581
π₯3
Sub in:n = 1/3x = -x
Work out each term separately and simplify
Write out the general formand state the values of x for which it is validβ¦
Imagine we substitute x = 0.5 into the expansion
1ΒΏβ 16β136β
5648
1ΒΏ β0.166 27 β0.0077
The values continuously get smaller This means the sequence will converge (like an infinite series) and hence for x =
0.5, the sequence IS validβ¦
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The Binomial ExpansionYou need to be able to expand expressions of the form (1 + x)n where n is any real
number
3A
Find the Binomial expansion of:ΒΏ(1+π₯ )π 1ΒΏ +ππ₯+π (πβ1) π₯
2
2 !+π (πβ1)(πβ2) π₯
3
3 !
(1βπ₯ )13 1ΒΏ +( 13 )(βπ₯)+( 13 )(β 23 ) (β π₯)2
2+( 13 )(β 23 )(β 53 )(βπ₯)3
6
1ΒΏ β 13 π₯β19 π₯
2β 581
π₯3
Sub in:n = 1/3x = -x
Work out each term separately and simplify
Write out the general formand state the values of x for which it is validβ¦
How do we work out for what set of values x is valid?The reason an expansion diverges or converges is down to the x
termβ¦If the term is bigger than 1 or less than -1, squaring/cubing etc will accelerate the size of the term, diverging
the sequenceIf the term is between 1 and -1, squaring and cubing cause the terms to become increasingly small, to the
sum of the sequence will converge, and be valid
β1<β π₯<1 ΒΏβπ₯β¨ΒΏ1ΒΏ π₯β¨ΒΏ1
Write using
ModulusThe expansion is valid when
the modulus value of x is less than 1
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The Binomial ExpansionYou need to be able to expand expressions of the form (1 + x)n where n is any real
number
3A
Find the Binomial expansion of: 1ΒΏΒΏ
(1+π₯ )π 1ΒΏ +ππ₯+π (πβ1) π₯2
2 !+π (πβ1)(πβ2) π₯
3
3 !
(1+4 π₯ )β 21ΒΏ +(β2 )(4 π₯)+(β2 ) (β3 )(4 π₯)2
2+(β2 ) (β3 ) (β4 )
(4 π₯)3
6
1ΒΏ β8π₯+48 π₯2β256 π₯3
Sub in:n = -2x = 4x
Work out each term separately and simplify
Write out the general form:
and state the values of x for which it is validβ¦
ΒΏΒΏ
The βxβ term is 4xβ¦
|4 π₯|<1
|π₯|< 14Divide by 4
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The Binomial ExpansionYou need to be able to expand expressions of the form (1 + x)n where n is any real
number
3A
Find the Binomial expansion of:β1β2π₯
(1+π₯ )π 1ΒΏ +ππ₯+π (πβ1) π₯2
2 !+π (πβ1)(πβ2) π₯
3
3 !
(1β2 π₯ )12 1ΒΏ +( 12 )(β2π₯ )+( 12 )(β 12 ) (β2 π₯)2
2+( 12 )(β 12 )(β 32 ) (β2 π₯)3
6
1ΒΏ βπ₯β 12 π₯2β 12π₯3
Sub in:n = 1/2x = -2x
Work out each term separately and simplify
Write out the general form:
and by using x = 0.01, find an estimate for β2
ΒΏΒΏ
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The Binomial ExpansionYou need to be able to expand expressions of the form (1 + x)n where n is any real
number
3A
Find the Binomial expansion of:β1β2π₯and by using x = 0.01, find an estimate for β2
β1β2π₯ΒΏ1βπ₯β12 π₯
2β 12 π₯3
x = 0.01β0.98ΒΏ1β0.01β0.00005β0.0000005
β 98100ΒΏ0.98994957β 210 ΒΏ0.9899495
7 β2ΒΏ9.899495β 2ΒΏ1.414213571
Rewrite left using a fraction
Square root top and bottom separately
Multiply by 10
Divide by 7
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Teachings for Exercise 3B
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The Binomial ExpansionYou can use the expansion for (1 + x)n to expand (a + bx)n by taking out a as a
factor
3B
Find the first 4 terms in the Binomial expansion of:β 4+π₯ΒΏΒΏΒΏ [4 (1+ π₯4 )]
12
ΒΏ
ΒΏ 412(1+π₯4 )
12
ΒΏ2(1+ π₯4 )12
(1+π₯ )π 1ΒΏ +ππ₯+π (πβ1) π₯2
2 !+π (πβ1)(πβ2) π₯
3
3 !
Write out the general form:
(1+ π₯4 )121ΒΏ +( 12 )( π₯4 )+( 12 )(β 12 )
(π₯4 )2
2+( 12 )(β 12 )(β 32 )
(π₯4 )3
6
(1+ π₯4 )12 1ΒΏ +1
8 π₯β 1128 π₯
2 +11024 π₯
3
2(1+ π₯4 )12 2ΒΏ +1
4 π₯β 164 π₯2+1512 π₯
3
Take a factor 4 out of the brackets
Both parts in the square brackets are to the power 1/2
You can work out the part outside the bracket
Sub in:n = 1/2x = x/4Work out each term
carefully and simplify it
Remember we had a 2 outside the bracket
Multiply each term by 2
|π₯4 |<1|π₯|<4
Multiply by 4
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The Binomial ExpansionYou can use the expansion for (1 + x)n to expand (a + bx)n by taking out a as a
factor
3B
Find the first 4 terms in the Binomial expansion of: 1ΒΏΒΏΒΏΒΏ
ΒΏ [2(1+ 3 π₯2 )]β 2
ΒΏ
ΒΏ2β 2(1+ 3 π₯2 )β2
ΒΏ14 (1+ 3 π₯2 )
β2
(1+π₯ )π 1ΒΏ +ππ₯+π (πβ1) π₯2
2 !+π (πβ1)(πβ2) π₯
3
3 !
Write out the general form:
(1+ 3 π₯2 )β2
1ΒΏ +(β2 )( 3 π₯2 )+(β2 ) (β3 )( 3 π₯2 )
2
2+(β2 ) (β3 ) (β4 )
( 3 π₯2 )3
6
(1+ 3 π₯2 )β2
1ΒΏ β3 π₯+274 π₯2β 272 π₯3
14 (1+ 3π₯2 )
β 214ΒΏ β 34 π₯
+2716 π₯2β 278 π₯3
Take a factor 2 out of the brackets
Both parts in the square brackets are to the power -2
You can work out the part outside the bracket
Sub in:n = -2x = 3x/2
Work out each term carefully and simplify it
Remember we had a 1/4 outside the bracket
Divide each term by 4
|3 π₯2 |<1|π₯|< 23
Multiply by 2, divide by 3
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Teachings for Exercise 3C
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The Binomial Expansion
3C
You can use Partial fractions to simplify the expansions of more difficult expressions
Find the expansion of: up to and including the term in x34β5 π₯
(1+π₯)(2β π₯)
Express as Partial Fractions4β5 π₯
(1+π₯)(2β π₯)ΒΏ
π΄(1+π₯ )
+π΅(2βπ₯)
ΒΏπ΄ (2βπ₯ )+π΅(1+π₯)
(1+π₯ )(2βπ₯)
ΒΏ π΄ (2βπ₯ )+π΅(1+π₯ )4β5 π₯ΒΏ3π΅β6ΒΏπ΅β2ΒΏ3 π΄9ΒΏ π΄3
4β5 π₯(1+π₯)(2β π₯)
ΒΏ3
(1+π₯ )β 2
(2βπ₯ )
Cross-multiply and combine
The numerators must be equal
If x = 2
If x = -1
Express the original fraction as Partial Fractions, using A and B
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The Binomial Expansion
3C
You can use Partial fractions to simplify the expansions of more difficult expressions
Find the expansion of: up to and including the term in x34β5 π₯
(1+π₯)(2β π₯)4β5 π₯
(1+π₯)(2β π₯)ΒΏ
3(1+π₯ )
β 2(2βπ₯ )
ΒΏ3ΒΏβ2ΒΏExpand each term separately
3ΒΏ
(1+π₯ )π 1ΒΏ +ππ₯+π (πβ1) π₯2
2 !+π (πβ1)(πβ2) π₯
3
3 !
Write out the general form:
(1+π₯ )β11ΒΏ +(β1)(π₯)+(β1)(β2)(π₯)2
2+(β1)(β2)(β3)
(π₯ )3
6
1ΒΏβπ₯+π₯2βπ₯33 (1+π₯ )β1 3ΒΏβ3 π₯+3 π₯2β3 π₯3
Both fractions can be rewritten
Sub in:x = xn = -1Work out each term carefully
Remember that this expansion is to be multiplied
by 3
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The Binomial Expansion
3C
You can use Partial fractions to simplify the expansions of more difficult expressions
Find the expansion of: up to and including the term in x34β5 π₯
(1+π₯)(2β π₯)4β5 π₯
(1+π₯)(2β π₯)ΒΏ
3(1+π₯ )
β 2(2βπ₯ )
ΒΏ3ΒΏβ2ΒΏExpand each term separately
2ΒΏ
Both fractions can be rewritten
3 (1+π₯ )β1 3ΒΏβ3 π₯+3 π₯2β3 π₯3
2[2 (1β π₯2 )]β1
2[2β 1(1β π₯2 )β 1]
2[ 12 (1β π₯2 )β 1]
(1β π₯2 )
β1
Take a factor 2 out of the brackets (and keep the current 2 separateβ¦)
Both parts in the square brackets are raised to -1
Work out 2-1
This is actually now cancelled by the 2 outside the square
bracket!
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The Binomial Expansion
3C
You can use Partial fractions to simplify the expansions of more difficult expressions
Find the expansion of: up to and including the term in x34β5 π₯
(1+π₯)(2β π₯)4β5 π₯
(1+π₯)(2β π₯)ΒΏ
3(1+π₯ )
β 2(2βπ₯ )
ΒΏ3ΒΏβ2ΒΏExpand each term separately
2ΒΏ
Both fractions can be rewritten
3 (1+π₯ )β1 3ΒΏβ3 π₯+3 π₯2β3 π₯3
ΒΏ (1β π₯2 )β 1
(1+π₯ )π 1ΒΏ +ππ₯+π (πβ1) π₯2
2 !+π (πβ1)(πβ2) π₯
3
3 !
Write out the general form:
(1β π₯2 )
β1
1ΒΏ +(β1)(β π₯2 )+(β1)(β2)(β π₯2 )
2
2+(β1)(β2)(β3)
(β π₯2 )3
6
1ΒΏ +π₯2
+π₯24
+π₯38
Sub in:x = -x/2n = -1Work out each term carefully
(1β π₯2 )β1
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The Binomial Expansion
3C
You can use Partial fractions to simplify the expansions of more difficult expressions
Find the expansion of: up to and including the term in x34β5 π₯
(1+π₯)(2β π₯)4β5 π₯
(1+π₯)(2β π₯)ΒΏ
3(1+π₯ )
β 2(2βπ₯ )
ΒΏ3ΒΏβ2ΒΏ
Both fractions can be rewritten
3 (1+π₯ )β1 3ΒΏβ3 π₯+3 π₯2β3 π₯3
1ΒΏ +π₯2
+π₯24
+π₯38(1β π₯2 )
β1
ΒΏ (3β3 π₯+3 π₯2β3 π₯3)β(1+ π₯2 +π₯24
+π₯38 )
ΒΏ2β 72 π₯+114 π₯2β 258 π₯3
Replace each bracket with its expansion
Subtract the second from the first (be wary of double negatives in
some questions)
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Summaryβ’ We have been reminded of the Binomial Expansion
β’ We have seen that when the power is a positive integer, the expansion is finite and exact
β’ With negative or fractional powers, the expansion is infinite
β’ We have seen how to decide what set of x-values the expansion is valid for
β’ We have also used partial fractions to break up more complex expansions