The Binomial Expansion
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Transcript of The Binomial Expansion
The Binomial Expansion
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
Teachings for Exercise 3A
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
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
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
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
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
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¦
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
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
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
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
Teachings for Exercise 3B
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
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
Teachings for Exercise 3C
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
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
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!
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
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)
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