Transcript Binomial expansion and Maclaurin series

Binomial Expansion and Maclaurin
series
Department of Mathematics
University of Leicester
www.le.ac.uk
Binomial expansion
• The binomial expansion is a series
approximation of a function:
f ( x)  (1  x)
• Where n is a real number.
n
Positive and Integer exponents
• Let n be any positive integer. The the series
will always be of finite length.
• Example: Let n=2.
(1  x)
2
 1  2x  x
2
Pascal's triangle
• Pascal’s triangle can be used to solve
n
(
1

x)
expansions of
. Pascal’s triangle has the
form
1
1
1
1 2 1
1 3 3 1
1 4 6 4 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:
1
11
121
1331
14 6 41
Generate Pascal’s Triangle
Order of
Pascals
Triangle
2
Generate
Clear
Pascal's triangle
• To solve an expansion (1  x) ,
n
we take the line in the triangle where the 2nd
and 2nd to last element is n, and multiply each
2
number, from left to right, by 1, x, x , and so
on.
Pascal's triangle
3
(
1

x
)
• Example: for
, we take the line 1 3 3 1.
and multiply each element by
1, x, x , x
(1  x)  1.1  3.x  3.x  1.x
3
2
 1  3x  3x 2  x 3
2
3
3
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:

 
(1  x)   C k x
n
n
k 1
k
Binomial expansion
n(n  1) 2 n(n  1)( n  2) 3
 1  nx 
x 
x  ...
2!
3!
• Where:
n(n  1)( n  2)...( n  k  1)
Ck 
k!
 
n
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 :
f ( x)  ( 2  3 x)
2
• So that n=-2. Rearrange so that it is of the form
(1+x).
1
3x 2
(2  3x)  (1  )
4
2
2
Binomial expansion: example
• Then using the formula:
1
27 2 27 3 
  3x  x  x  ...
4
4
2

• Which is valid for |x|< 2/3
Binomial expansion: example
2
• This is valid for |x|< because, as
3
3x
nd
the 2 term in the equation, here , has to
2
have a modulus of less than 1.
Therefore:
And:
3x 3
 x 1
2 2
2
2
x  .1 
3
3
Series expansion of rational functions
• We can expand more complicated expressions,
now, using the method of partial fractions
where needed:
• Example: let:
(6 x  3)
f ( x) 
(1  x)(5 x  2)
• We can split this up using partial fractions,
into:
(6 x  3)
1
1


(1  x)(5x  2) (1  x) (5x  2)
Series expansion of rational functions:
example
• Where we can see:
1
1
1
1

 (1  x)  (5x  2)
(1  x) (5x  2)
• Then using the binomial series to expand both
functions:
Series expansion of rational functions:
example
• Therefore:
(1) x (1)(2) x (1)(2)(3) x
(1  x)  1 


 ...
1
2!
3!
2
1
 1  x  x 2  x 3  ...
• Given: |x|<1
3
Series expansion of rational functions:
example
• And:
2
3


5 x (1)(2) 5x  (1)(2)(3) 5x 
 
 
 (1)

1
1  5x 
1
2
2


2


 ..
  1  1 

2 2 
2
1
2!
3!




2
3

1  5x 25x 125x
 1  

 ...
2 2
4
8

2
• Given:|x|< 5
Series expansion of rational functions:
example
• Therefore:
1
1
(1  x)  (5x  2) 
2
3
3 9 x 33x 141x
  

 ...
2 4
8
16
Maclaurin series
• The Binomial expansion is used to approximate
a function of the form:
f ( x)  (1  x)
n
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.
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:
f ' ' (0) 2 f ' ' (0) 3
f ( x)  f (0)  f ' (0) x 
x 
x  ...
2!
3!
Maclaurin series: proof
• Take any function f(x) and represent it as a
power series:
f ( x)  a0  a1 x  a 2 x  a3 x  ...
2
3
• Assuming that the function can be written in
this way.
Maclaurin series: proof
• Using this, we can equate this at x=0:
f (0)  a0  a1 0  a2 0  a3 0  ...
• Therefore:
f (0)  a0
Maclaurin series: proof
• This is assuming that f(x) exists at x=0:
• For instance:
1
f ( x) 
x
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.
Maclaurin series: proof
• Now differentiate both sides:
f ' ( x)  a1  2a 2 x  3a3 x  ...
2
• Then again equating at x=0:
f ' (0)  a1
Maclaurin series: proof
• 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.
Maclaurin series: proof
• Then differentiate again:
f ' ' ( x)  2a2  3.2a3 x  ...
Then equating at x=0:
f ' ' (0)
 a2
2!
Maclaurin series: proof
• We can find all of the a i ' s by doing this. And
Thus achieving the formula:
f ' ' (0) 2 f ' ' (0) 3
f ( x)  f (0)  f ' (0) x 
x 
x  ...
2!
3!
Maclaurin series: example
• Example: Let f(x)=
e
x
• Using the formula, and the fact that
which at x=0,
f (0)  e  1
0
d x
x
e e
dx
Maclaurin series: example
• Then:
1 2 1 3
e  1  x  x  x  ...
2!
3!
x
• Which we can simplify to:

n
x
e 
n 0 n!
x
for all x
Maclaurin series: example
• Example: Let f(x)= cos x
• Using a table of differentiation:
Differentiated:
at x=0:
cos x
 sin x
1
0
 cos x
sin x
1
0
Maclaurin series: example
• Then. Using the formula:
1 2
1 4
cos x  1  0  x  0  x  0  ...
2!
4!
• Which we can simplify to:
(1) 2n
cos x  
x
n 0 (2n)!

n
for all x
Maclaurin series: example
• Example: Let f(x)=
sin x
• Using a table of differentiation:
Differentiated:
at x=0:
sin x
cos x
 sin x
0
1
0
 cos x
1
Maclaurin series: example
• Then. Using the formula:
1 3
1 5
sin x  0  x  0  x  0  x  0  ...
3!
5!
• Which we can simplify to:
(1) ( 2n1)
sin x  
x
n 0 (2n  1)!

n
for all x
Maclaurin series: example
• Example: Let f(x)=
log( 1  x), for  1  x  1
• Using a table of differentiation:
Differentiated:
log( 1  x)
1

1 x
1

(1  x) 2
1
2
(1  x) 3
at x=0:
0
1
1
2
Maclaurin series: example
• Then. Using the formula:
2
3
x x
log( 1  x)  0  x    ...
2 3
• Which we can simplify to:

n
x
log( 1  x)  
n 1 n
Maclaurin series: example
• Example: Let f(x)=
log( 1  x), for  1  x  1
• Using a table of differentiation:
Differentiated:
log( 1  x)
1
1 x
1

(1  x) 2
1
2
(1  x) 3
at x=0:
0
1
1
2
Maclaurin series: example
• Then. Using the formula:
2
3
x x
log( 1  x)  0  x    ...
2 3
• Which we can simplify to:
n 1
(1) n
log( 1  x)  
x
n
n 1

Maclaurin series: Chain rule example
• Example: Let f(x)= e
• Let
yx
x2
2
y
• Then we know that e , using Maclaurin series,
equals:
1 2 1 3
e  1  y  y  y  ...
2!
3!
y
Maclaurin series: example
• Then, substituting back in
yx
2
, we get:
1 2 2 1 2 3
e  1  ( x )  ( x )  ( x )  ...
2!
3!
x2
2
1 4 1 6
 1  x  x  x  ...
2!
3!
2
Maclaurin series: example
• Which we can then simplify to:

2n
x
e 
n 0 n!
x
2
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:
2
4
x x
cos( x)  1    ...
2! 4!
• The integral of f(x) equals sin(x).
Using integration
• If we integrate the expansion, we get:
2
4
x x
cos(
x
)
dx

1



...
dx

 2! 4!
3
5
3
5
x
x
x x
 x

 ...  x    ...
3.2! 5.4!
3! 5!
Using integration
• Then we add the integral of f(x) at x=0, which
is: sin(0)=0.
• Therefore:
3
5
x x
 x    ...
3! 5!
• Which we know is the series expansion for
sin(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.
Using differentiation
• For example:
• Let f(x)=cos(x). The series expansion is:
2
4
x x
cos( x)  1    ...
2! 4!
• The differential of f(x) equals -sin(x).
Using differentiation
• If we differentiate the series expansion, we
get:
2
4


d
d
x x
cos( x)  1    ...
dx
dx  2! 4!

3
5
2.x 4 x 6 x
 0


 ...
2! 4!
6!
Using differentiation
• Which equals:
3
5
x x
  x    ...
3! 5!
• Which we know is the series expansion for
–sin(x).
Conclusion
• Binomial expansion and Maclaurin series are
used to represent functions as a polynmial
series.
Conclusion
• Binomial expansion:

 
(1  x)   C k x
n
n
k
k 1
• Maclaurin seires:
f ' ' (0) 2 f ' ' (0) 3
f ( x)  f (0)  f ' (0) x 
x 
x  ...
2!
3!