Www.le.ac.uk Binomial Expansion and Maclaurin series Department of Mathematics University of...

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Binomial Expansion and Maclaurin series

Department of MathematicsUniversity of Leicester

Binomial expansion

• The binomial expansion is a series approximation of a function:

• Where n is a real number.

nxxf )1()(

Positive and Integer exponents

• Let n be any positive integer. The the series will always be of finite length.

• Example: Let n=2.

221 xx

2)1( x

Pascal's triangle

• Pascal’s triangle can be used to solve expansions of . Pascal’s triangle has the form

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

nx)1(

Pascal's triangle

• Where every number is the sum of the two numbers above it.

Pascal's triangle

• We first take 1 on its own, then proceed in a triangle below it. The numbers on the diagonals always take the value 1.

1

1 1

Pascal's triangle

• With the 2nd line, we start on the left. The fist number has only one above it, so is again 1. The 2nd has 2 ones above it, so added together equal 2. And the 3rd is again 1. 1

1 1

1 2 1

Pascal's triangle

• This carries on line by line:

111121133114641

Generate Pascal’s Triangle

Order of Pascals Triangle

Generate

Clear

Pascal's triangle

• To solve an expansion ,

we take the line in the triangle where the 2nd and 2nd to last element is n, and multiply each number, from left to right, by and so on.

nx)1(

,,,1 2xx

Pascal's triangle

• Example: for , we take the line 1 3 3 1. and multiply each element by

3)1( x32 ,,,1 xxx

323 .1.3.31.1)1( xxxx

32331 xxx

Positive n

• Any positive integer n will always have a finite number of elements. Thus being a finite sequence.

• This is not the same for negative integers or other real numbers that aren’t positive integers.

• To work these we use the Binomial expansion.

Binomial expansion

• The series is given by, for |x|<1:

k

kk

nn xCx

1

)1(

Binomial expansion

• Where:

...!3

)2)(1(

!2

)1(1 32

x

nnnx

nnnx

!

)1)...(2)(1(

k

knnnnCkn

Binomial expansion

• As any other terms after the first 3 are very small, we can discount them. Making it an approximation.

• Otherwise it would go on forever.

• The expansion only matches the true function if |x|<1, otherwise it is a non-convergent series and therefore can not exist.

Binomial expansion: example

• Example: Let :

• So that n=-2. Rearrange so that it is of the form (1+x).

2)32()( xxf

22 )2

31(

4

1)32(

xx


• Then using the formula:

• Which is valid for |x|< 2/3

...

2

27

4

273

4

1 32 xxx


• This is valid for |x|< because, as

the 2nd term in the equation, here , has to

have a modulus of less than 1.

Therefore:

And:

3

2

2

3x

12

3

2

3 x

x

3

21.

3

2x

Series expansion of rational functions• We can expand more complicated

expressions, now, using the method of partial fractions where needed:

• Example: let:

• We can split this up using partial fractions, into:

)25)(1(

)36()(

xx

xxf

)25(

1

)1(

1

)25)(1(

)36(

xxxx

x

Series expansion of rational functions: example

• Where we can see:

• Then using the binomial series to expand both functions:

11 )25()1()25(

1

)1(

1

xxxx


• Therefore:

• Given: |x|<1

...!3

)3)(2)(1(

!2

)2)(1(

1

)1(1)1(

321

xxx

x

...1 32 xxx


• And:

• Given:|x|<

..!3

2

5)3)(2)(1(

!2

2

5)2)(1(

12

5)1(

12

11

2

5

2

1

32

1xxx

x

...

8

125

4

25

2

51

2

1 32 xxx

5

2


• Therefore:

11 )25()1( xx

...16

141

8

33

4

9

2

3 32

xxx

Maclaurin series

• The Binomial expansion is used to approximate a function of the form:

as a polynomial representation.

• But this doesn’t work for all functions, such as that of sinx or cosx. For these we use Maclaurin series.

nxxf )1()(

Maclaurin series

• Maclaurin series is used to approximate a function as a series around the origin, to see what a function would look like in this area.

• As we do the proof, bare in mind the number of assumptions needed to prove this, and that all must apply for the function to satisfy the expansion.

Maclaurin series

• Maclaurin series is given by:

...!3

)0(''

!2

)0('')0(')0()( 32 x

fx

fxffxf

Maclaurin series: proof

• Take any function f(x) and represent it as a power series:

• Assuming that the function can be written in this way.

...)( 33

2210 xaxaxaaxf


• Using this, we can equate this at x=0:

• Therefore:

...000)0( 3210 aaaaf

0)0( af


• This is assuming that f(x) exists at x=0:

• For instance:

does NOT exist as x=0, and therefore cannot be used as part of this series.

• This also assumes that both f(x) and the polynomial representation are differentiable.

xxf

1)(


• Now differentiate both sides:

• Then again equating at x=0:

...32)(' 2321 xaxaaxf

1)0(' af


• This again assumes that f’(x) exists as 0.

• For all successive derivatives of f(x) in this series we will assume exists at x=0.

• We will also assume from now on that all successive differentials and their polynomial representation are differentiable.


• Then differentiate again:

Then equating at x=0:

...2.32)('' 32 xaaxf

2!2

)0(''a

f


• We can find all of the by doing this. And Thus achieving the formula:

...!3

)0(''

!2

)0('')0(')0()( 32 x

fx

fxffxf

sai '

Maclaurin series: example

• Example: Let f(x)=

• Using the formula, and the fact that

which at x=0,

xe

xx eedx

d

1)0( 0 ef


• Then:

• Which we can simplify to:

for all x

...!3

1

!2

11 32 xxxe x

0 !n

nx

n

xe



• Using a table of differentiation:

Differentiated: at x=0:

xcos

0sin

1cos

0sin

1cos

x

x

x

x


• Then. Using the formula:


for all x

...0!4

10

!2

101cos 42 xxx

n

n

n

xn

x 2

0 )!2(

)1(cos





xsin

1cos

0sin

1cos

0sin

x

x

x

x




for all x

...0!5

10

!3

100sin 53 xxxx

)12(

0 )!12(

)1(sin

n

n

n

xn

x





11),1log( xforx

2)1(

12

1)1(

1

11

1

0)1log(

3

2

x

x

x

x




...32

0)1log(32

xx

xx

1

)1log(n

n

n

xx





11),1log( xforx

2)1(

12

1)1(

1

11

1

0)1log(

3

2

x

x

x

x




...32

0)1log(32

xx

xx

n

n

n

xn

x

1

1)1()1log(

Maclaurin series: Chain rule example• Example: Let f(x)=

• Let

• Then we know that , using Maclaurin series, equals:

2xe

ye

2xy

...!3

1

!2

11 32 yyye y


• Then, substituting back in , we get:

...)(!3

1)(

!2

1)(1 322222

xxxe x

2xy

...!3

1

!2

11 642 xxx


• Which we can then simplify to:

0

2

!

2

n

nx

n

xe

Using integration and differentiation to solve other functions

• If we know a function, we can take it’s series expansion and use it to find the expansion for it’s integral or differential.

Using integration

• For finding the series expansion for an integral of f(x), we use the same expansion for f(x), but we integrate the expansion. Then add the integral of f(x) at x=0 to the series.

Using integration

• For example:

• Let f(x)=cos(x). The series expansion is:

• The integral of f(x) equals sin(x).

...!4!2

1)cos(42

xx

x

Using integration

• If we integrate the expansion, we get:

dxxx

dxx ...!4!2

1)cos(42

...!5!3

...!4.5!2.3

5353

xx

xxx

x

Using integration

• Then we add the integral of f(x) at x=0, which is: sin(0)=0.

• Therefore:

• Which we know is the series expansion for sin(x).

...!5!3

53

xx

x

Using differentiation

• For finding the series expansion for a differential of f(x), we use the same expansion for f(x), but we differentiate the expansion.


• For example:

• Let f(x)=cos(x). The series expansion is:

• The differential of f(x) equals -sin(x).

...!4!2

1)cos(42

xx

x


• If we differentiate the series expansion, we get:

...!4!2

1)cos(42 xx

dx

dx

dx

d

...!6

6

!4

4

!2

.20

53

xxx


• Which equals:

• Which we know is the series expansion for

–sin(x).

...!5!3

53

xx

x

Conclusion

• Binomial expansion and Maclaurin series are used to represent functions as a polynmial series.

Conclusion

• Binomial expansion:

• Maclaurin seires:

k

kk

nn xCx

1

)1(

...!3

)0(''

!2

)0('')0(')0()( 32 x

fx

fxffxf

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