NUMERICAL ANALYSIS
Maclaurin and Taylor Series
Preliminary Results
In this unit we require certain knowledge
from higher maths.
You must be able to DIFFERENTIATE.
Remember the general rule:
f ( x)  x so f ( x)  nx
dy
n
n 1
or y  x so
 nx
dx
n
n 1
Preliminary Results
We must also remember how to
differentiate more complicated
expressions:
E.g
f ( x)  (4x 1)
Preliminary Results
We must write in a form suitable for
differentiation:
f(x) = (4x – 1)1/2
then we differentiate
1
f ( x)  4  (4 x  1)
2
f ( x)  2(4 x  1)

1
2

1
2
Preliminary Results
There are 2 new derivatives that we need
for this unit,
f(x) = ex and f(x) = ln x.
For ex we can look at the graphs of
exponential functions along with their
derivatives –
we will consider 2x , 3x and ex.
Preliminary Results
dy
dx
y = 2x is the thicker graph
Preliminary Results
y = 3x
dy
dx
Notice that the two graphs are almost
the same, but not quite
Preliminary Results
y = ex
This time the two graphs
overlap exactly
Preliminary Results
The graphs show that the derivative of ex is
ex.
We will not show the derivative of ln x but
you need to remember that it is
1
x
Maclaurin
We are now in a position to start looking at
Maclaurin series.
These are polynomial approximations to
various functions close to the point where
x = 0.
Historical Note
Colin Maclaurin was one of the
outstanding mathematicians of the 18th
century.
Born Kilmodan Argyll 1698, went to
Glasgow University at the age of 11.
Obtained an MA when 15, in 1713.
In 1717 became professor at Aberdeen.
Historical Note
In 1725 joined James Gregory as professor
of maths at Edinburgh.
Helped the Glasgow excisemen find a way
of getting the volume of the contents of
partially filled rum casks arriving from the
West Indies.
Also set up the first pension fund for
widows and orphans.
Historical note
In 1745 fled from the Jacobite uprising and
went to York where he died in 1746.
Colin Maclaurin
Maclaurin
Example
Find a polynomial expansion of degree 3
for sin x near x=0.
Answer
First we must differentiate sin x three times
f ( x)  sin( x) f ( x)  cos( x)
f ( x)   cos( x)
f ( x)   sin( x)
Maclaurin
We now put x = 0 in each of these.
f (0)  sin(0)  0
f (0)  cos(0)  1
f (0)   sin(0)  0
f (0)   cos(0)  1
Maclaurin
We can now build up the polynomial:
we choose the coefficients of the
polynomial so that the values of f and its
derivatives are the same as the values of p
and its derivatives at x = 0.
For example we know that f(0) = 0, and so
if our polynomial is
Maclaurin
pn(x) = a0 + a1x + a2x2 + a3x3 + ……
then we require pn(0) = 0 as well.
pn(0) = a0 + a10 + a202 + a303 + ……
 = a0 + 0 = a 0 .
We want this to be 0 so a0 = 0.
Maclaurin
Now we differentiate both f(x) and pn(x).
f ( x)  cos x
pn ( x)  a1  2a2 x  3a3 x 2  ......
Now put x = 0 in both expressions
f (0)  cos 0  1
pn (0)  a1  0  0......
Maclaurin
This gives a1 = 1.
Differentiate again to get
f ( x)   sin x
pn ( x)  2a2  3  2a3 x  .......
Put x = 0 again and we get that
f (0)  0 and pn (0)  2a2
so a 2  0.
Maclaurin
To get the cubic polynomial approximation
we must differentiate once more.
f ( x)   cos x and pn ( x)  6a3  other terms in x
For the last time we put x = 0 to get
f (0)  1 and pn (0)  6a3
Maclaurin
6a3 = -1 and so a3 =

1
6
We now have the following coefficients for
the polynomial:
a0 = 0 a1 = 1 a2 = 0 a3 =
Giving sin x = 1x
1 3
 x
6
1

6
Maclaurin
pn(x) = a0 + a1x + a2x2 + a3x3 + ….
f(0) = pn(0) = a0
Differentiate once so that
p n ( x)  a1  2a2 x  3a3 x 2  ......
Because
f (0)  pn (0) we see that f (0)  2a2
f (0)
a2 
2
Maclaurin
This can be written as
sin x = x – x3
6
It is possible to generalise this process as
follows:
let the polynomial pn(x) approximate the
function f(x) near x = 0.
Maclaurin
f(x) = pn(x) = a0 + a1x + a2x2 + a3x3 + …
f(0) = pn(0) = a0
so a0 = f(0)
Differentiate
f (x) = pn (x)  a1 + 2a 2 x + 3a 3 x 2  .......
so f (0) = pn (0)  a1
Maclaurin
Differentiate again
f (x) = pn (x) = 2a 2 + 3  2a 3 x + ......
so f (0) = pn (0) = 2a 2
f (0)
giving a 2 =
2
Maclaurin
To get a cubic polynomial we must
differentiate once more.
(If we wanted a higher degree polynomial
we would continue.)
f (x) = pn (x) = 3  2a 3 + other terms in x
so f (0) = pn (0) = 6a 3
f (0)
finally a 3 =
6
Maclaurin
We can now write the polynomial as
follows:
f (0) 2 f (0) 3
p n (x) = f(0) + f (0) x +
x +
x + .....
2
6
This is called the Maclaurin expansion of
f(x).
Maclaurin
The numbers 2 and 6 come about from 2x1
and 3x2(x1).
We can write these in a shorter way as
2! and 3! – read as factorial 2 and factorial
3.
4! = 4x3x2x1 = 24
5! = 5x4x3x2x1 = 120
Maclaurin
This allows us to write the Maclaurin
expansion as
f (0) 2 f (0) 3
f ( x)  pn ( x)  f (0)  f (0) x 
x 
x  ...
2!
3!
Maclaurin
Example : obtain the Maclaurin expansion
of degree 2 for the function defined by
f ( x)  1  x
Maclaurin
First get the coefficients:
f ( x)  1  x  (1  x)
1
2
f (0)  1
1

1
f ( x)  (1  x) 2
2
f (0) 
3

1
1
f ( x)    (1  x) 2
2
2
so a0  f (0)  1,
1
2
f (0)  
1
4
1
f (0)
1
a1  f (0)  and a2 

2
2!
8
Maclaurin
This gives the polynomial
1
1 2
1 x  1 x  x
2
8
for values of x near 0