Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
BINOMIAL SERIES — PART 1
SERIES 3
INU0114/514 (M ATHS 1)
Dr Adrian Jannetta MIMA CMath FRAS
Binomial Series — Part 1
1 / 22
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
Objectives
The purpose of this session is to introduce power series expansions and
the binomial series in particular.
• The factorial function.
• Pascal’s Triangle.
• Binomial expansions of (1 + x)n and (a + b)n when n is a positive
integer.
• Applications and examples
Binomial Series — Part 1
2 / 22
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
Binomial theorem
The binomial theorem shows us how to quickly represent a power of
a + b as an expanded sum. Using Sigma notation it is like this:
(a + b)n =
n X
n
k=0
k
an−k bk
where n is a positive integer.
Using the binomial theorem we can prove that
(x + y)4 ≡ x4 + 4x3 y + 6x2 y 2 + 4xy 3 + y 4
without the working through the tedious multiplication of
(x + y)(x + y)(x + y)(x + y).
We’ll build our understanding of the binomial theorem with factorials
and Pascal’s triangle.
Binomial Series — Part 1
3 / 22
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
Expanding binomial terms
Expanding binomial terms
Binomial terms
A binomial is the sum (or difference) of two terms. For example, a + b,
3x − 5, 1 + x2 are binomials.
We are often required to express a binomial to some power and then
expand the brackets.
Consider the following expansions:
(1 + x)0
=
1
1
=
1+x
(1 + x)2
=
1 + 2x + x2
(1 + x)3
=
1 + 3x + 3x2 + x3
(1 + x)4
=
1 + 4x + 6x2 + 4x3 + x4
(1 + x)
The coefficients in each row follow a pattern. They are rows in Pascal’s
Triangle.
Binomial Series — Part 1
4 / 22
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
Pascal’s triangle
Pascal’s Triangle
Here is Pascal’s Triangle.
1
ւց
1
ւց
1
ւց
1
ւց
2
ւց
1
ւց
1
3
3
1
ւց
ւց
ւց
ւց
1
4
6
4
1
The next row of the triangle is obtained by adding elements from row
above.
The coefficients of the terms in a binomial expansion can be obtained
from Pascal’s Triangle.
Binomial Series — Part 1
5 / 22
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
Examples
Using Pascal’s Triangle
Use the next row of the triangle to expand the binomial expression
(1 + x)5
The next row of the triangle contains 1, 5, 10, 10, 5 and 1.
The powers of x increase from x0 to x5 :
(1 + x)5 ≡ 1(x0 ) + 5(x1 ) + 10(x2 ) + 10(x3 ) + 5(x4 ) + 1(x5 )
Simplifying this gives:
(1 + x)5 ≡ 1 + 5x + 10x2 + 10x3 + 5x4 + x5
Binomial Series — Part 1
6 / 22
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
Examples
The binomial (a + b)n
Let’s expand the expression (a + b)n where a is a constant.
(a + b)0
(a + b)1
(a + b)2
(a + b)3
(a + b)4
=
=
=
=
=
1
a+b
a2 + 2ab + b2
3
a + 3a2 b + 3ab2 + b3
4
a + 4a3 b + 6a2 b2 + 4ab3 + b4
The coefficients are again from Pascal’s Triangle but there are other patterns too:
• The sum of the powers of a and b in each term is equal to n.
• The powers of a start at the value n and decrease as we go from term to
term.
• The powers of b start at the value 0 and increase as we go from term to term.
• The are n + 1 terms in the expansion.
Binomial Series — Part 1
7 / 22
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
Examples
A binomial expansion
Use Pascal’s Triangle to expand (2 − x)4 .
Using all the patterns to expand the expression:
(2 − x)4 = 1 24 (−x)0 + 4 23 (−x)1 + 6 22 (−x)2
+4 21 (−x)3 + 1 20 (−x)4
= 16 + 32 (−x) + 24 x2 + 8 −x3 + x4
∴ (2 − x)4
Binomial Series — Part 1
=
16 − 32x + 24x2 − 8x3 + x4
8 / 22
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
Factorial function
The factorial function is only defined for non-negative integers.
The factorial of n is denoted by n! and is calculated as follows:
n! = n × (n − 1) × (n − 2) × (n − 3) × . . .× 3 × 2 × 1
(1)
So, for example 5! = 120 (since 5! = 5 × 4 × 3 × 2 × 1 = 120).
In the same way you should be able to show that 7! = 5040.
It is not obvious under this definition what the value of 1! or 0! should be.
By convention the following definitions are used:
1! = 1
0! = 1
and
Most scientific calculators are able to calculate the factorial of a number;
it is usually accessed with a button like n! or x! .
Binomial Series — Part 1
9 / 22
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
Calculating the coefficients
Pascal’s triangle isn’t great in some situations. Suppose we needed some
of the values from the 100th row?
The k th coefficient in row n of Pascal’s triangle can be can be calculated
using the formula
n
n!
=
k!(n − k)!
k
This formula is related to Pascal’s triangle like this:
n=0
n=1
n=2
n=3
n=4
n=5
n=6
k −→
Binomial Series — Part 1
1
1
1
1
1
1
1
0
1
2
3
4
5
6
1
1
3 1
6 4 1
10 10 5
15 20 15
2 3 4
10 / 22
1
6
5
1
6
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
Calculating coefficients
Calculate
4
3
.
(This is the 4th row of Pascal’s Triangle and the 3rd coefficient.)
It’s pretty simple to do this:
 ‹
4!
24
4!
4
=
=
=4
=
3!(4 − 3)! 3!1! 6 × 1
3
It usually isn’t necessary to use this formula.
Scientists need to use binomial coefficients so often that most modern scientific
calculators have the formula built into them; look for a button marked with
n Cr (or equivalent) to access the function.
For example: to evaluate
10
5
Input to the calculator 10 →
Binomial Series — Part 1
n Cr
→ 5 to get the answer 252.
11 / 22
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
Simplifying binomial coefficients
The formula for calculating the coefficients is:
n
k
=
n!
k!(n − k)!
Each coefficient can be simplified to reveal a simple pattern depending
on n.
n
n!
n!
= =1
=
1!(n − 0)! n!
0
n
1
n
2
n
3
=
=
=
n!
n × (n − 1)!
=
=n
1!(n − 1)!
(n − 1)!
n!
n × (n − 1) × (n − 2)! n(n − 1)
=
=
2!(n − 2)!
2! × (n − 2)!
2!
n × (n − 1) × (n − 2) × (n − 3)! n(n − 1)(n − 2)
n!
=
=
3!(n − 3)!
3! × (n − 3)!
3!
...and so on.
Binomial Series — Part 1
12 / 22
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
Expansion of (a + b)n for positive integer n
The results obtained in the previous sections and examples suggest that the expansion of
(a + b)n has the following form:
(a + b)n =
where
n
k
n
0
an b0 +
n
1
an−1 b1 +
n
2
an−2 b2 + . . . +
n n
a1 bn−1 +
a0 bn
n−1
n
is a function which gives the binomial coefficient.
The coefficients of the binomial expansion formulae can be written in terms of n.
Doing that makes them more convenient to write down and manipulate.
The expansion written above becomes
(a + b)n = an + nan−1 b1 +
Binomial Series — Part 1
n(n − 1) n−2 2
a b + . . . + bn
2!
13 / 22
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
Binomial expansion
4
Write down the full expansion of 4 + 32 x .
The expansion we require is:
4
4
4
4
=
(4)4 ( 23 x)0 +
(4)3 ( 32 x)1 +
(4)2 ( 23 x)2
4 + 32 x
0
1
2
4
4
1 3 3
+
(4) ( 2 x) +
(4)0 ( 23 x)4
3
4
Simplifying the coefficients and powers:
4 + 23 x
4
=
81 4
3
+(4)(4)( 27
8 x ) + (1)(1)( 16 x )
=
Binomial Series — Part 1
(1)(256)(1) + (4)(64)( 23 x) + (6)(16)( 49 x2 )
81 4
256 + 384x + 216x2 + 54x3 + 16
x
14 / 22
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
Expansion of (1 + x)n for positive integer n
If we let a = 1 and replace b with x in the previous binomial expansion
formula we get a useful variation:
n
n n
n
n
xn
xn−1 +
x2 + . . . +
x1 +
x0 +
(1 + x)n =
n
n−1
2
1
0
n
n
n = 1+
x+
x2 + . . . +
xn−1 + xn
1
2
n−1
And this can also be written in the following form:
(1 + x)n = 1 + nx +
Binomial Series — Part 1
n(n − 1) 2 n(n − 1)(n − 2) 3
x +
x . . . + xn
2!
3!
15 / 22
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
Evaluate binomial coefficients
Given that (1 + x)50 ≈ a + bx + cx2 + dx3 , find the values of a, b, c and d.
We must calculate the following coefficients:
50
50
50
50
(1 + x)50 =
x0 +
x1 +
x2 +
x3
0
1
2
3
Therefore
(1 + x)50 ≈ 1 + 50x + 1225x2 + 19600x3
.
Binomial Series — Part 1
16 / 22
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
Binomial expansion formulae
The coefficients of the binomial expansion formulae can be written in
terms of n.
Doing that makes them more convenient to write down and manipulate.
For example, the expansion
n
n
n
an−2 b2 + . . .
an−1 b1 +
an b 0 +
(a + b)n =
2
1
0
n
n a0 b n
a1 bn−1 +
+
n
n−1
becomes
(a + b)n = an + nan−1 b1 +
n(n − 1) n−2 2
a b + . . . + bn
2!
The expansion of (1 + x)n becomes
(1 + x)n = 1 + nx +
Binomial Series — Part 1
n(n − 1) 2 n(n − 1)(n − 2) 3
x +
x . . . + xn
2!
3!
17 / 22
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
Binomial expansion
Write down the expansion of (2 + x)4 .
Using the expansion formula for (a + b)n
(2 + x)4
4(3) 2 2 4(3)(2) 1 3 4(3)(2)(1) 4
(2 )x +
(2 )x +
x
2!
3!
4!
16 + 32x + 24x2 + 8x3 + x4
24 + 4(23 )x1 +
=
=
Binomial expansion
Write down the expansion of (1 − 12 x)3 .
Using the expansion formula for (1 + x)n
1 − 12 x
Binomial Series — Part 1
3
=
=
3(2) 1 2 3(2)(1) 1 3
(− 2 x) +
(− 2 x)
2!
3!
3
3 2
1 3
1− 2x+ 4x − 8x
1 + (3)(− 12 )x +
18 / 22
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
Approximating powers
Write down the expansion of (2 + x)4 up to and including the x2 term. Then use the
expansion to evaluate:
(a) (2.01)4
(b) (1.99)4
Expanding the brackets using the binomial theorem gives:
(2 + x)4 ≈ 16 + 32x + 24x2
(a) Substitute x = 0.01
(2.01)4
=
(2 + 0.01)4
≈
16 + 32(0.01) + 24(0.01)2
≈
16 + 0.32 + 0.0024
≈
16.3224
(b) Substitute x = −0.01
(1.99)4
Binomial Series — Part 1
=
(2 − 0.01)4
≈
16 + 32(−0.01) + 24(−0.01)2
≈
16 − 0.32 + 0.0024
≈
15.6824
19 / 22
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
The expansion of (1 + x)n can also be useful whatever binomial we are expanding.
Binomial expansion using (1 + x)n
Expand (2 + x)8 up to and including the x3 term.
Rather than use (a + b)n , let’s factorise and use the other expansion:
8
(2 + x)8 =
2(1 + 21 x)
8
= 28 1 + 12 x
8
= 256 1 + 12 x
Now expand the brackets using the rule for (1 + x)n :
‹
‹
‹



˜
‹
•
8
8
8
8
( 12 x)1 +
( 21 x)2 +
( 12 x)3 + . . .
( 21 x)0 +
= 256
1
2
3
0
= 256 (1)(1) + 8( 12 x) + 28( 41 x2 ) + 56( 18 x3 ) + . . .
= 256 1 + 4x + 7x2 + 7x3 + . . .
Therefore (2 + x)8 = 256 + 1024x + 1792x2 + 1792x3 + . . .
Binomial Series — Part 1
20 / 22
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
Summary
(a + b)n
=
n
or (a + b)n
=
an + nan−1 b1 +
0
an b0 +
n
1
The version in Sigma notation: (a + b)n =
an−1 b1 +
n
2
an−2 b2 + . . . + bn
n(n − 1) n−2 2
a b + . . . + bn
2!
n X
n
k=0
k
an−k bk .
Similarly, to expand (1 + x)n we can use
(1 + x)n
=
or (1 + x)n
=
n
n
x2 + . . . + xn
2
n(n − 1) 2 n(n − 1)(n − 2)
x +
. . . + xn
1 + nx +
2!
3!
1+
1
x+
These formulae only apply when n is a positive integer.
Binomial Series — Part 1
21 / 22
Adrian Jannetta
Introduction
Binomial theorem
Binomial coefficients
Formulae
Summary
Test
Test yourself...
You should be able to solve the following problems if you have
understood everything in these notes.
1
2
Write down the first five rows of Pascal’s triangle in less than 1
minute! 7
30
Evaluate
and
.
3
3
3
Expand (3 + 2x)4
4
Expand (1 − 12 x)3
Answers:
1 See the example given earlier in the notes.
2 35 and 4060.
3 81 + 216x + 216x2 + 96x3 + 16x4 .
3
3
1
4 1 − 2 x + 4 x2 − 8 x3 .
Binomial Series — Part 1
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Adrian Jannetta