numerical analysis maclaurin and taylor series. preliminary results in this unit we require certain...
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Preliminary Results
In this unit we require certain knowledge from higher maths.
You must be able to DIFFERENTIATE.Remember the general rule:
1
1
( ) so ( )
so
n n
n n
f x x f x nx
dyor y x nx
dx
Preliminary Results
We must also remember how to differentiate more complicated expressions:
E.g
( ) (4 1)f x x
Preliminary Results
We must write in a form suitable for differentiation:
f(x) = (4x – 1)1/2 then we differentiate
1
2
1
2
1( ) 4 (4 1)
2
( ) 2(4 1)
f x x
f x x
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
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
ExampleFind a polynomial expansion of degree 3
for sin x near x=0. AnswerFirst we must differentiate sin x three times
( ) sin( ) ( ) cos( ) ( ) sin( )
( ) cos( )
f x x f x x f x x
f x x
Maclaurin
We now put x = 0 in each of these.
(0) sin(0) 0
(0) cos(0) 1
(0) sin(0) 0
(0) cos(0) 1
f
f
f
f
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 = a0 .
We want this to be 0 so a0 = 0.
Maclaurin
Now we differentiate both f(x) and pn(x).
( ) cosf x x 2
1 2 3( ) 2 3 ......np x a a x a x
Now put x = 0 in both expressions
(0) cos0 1f
1(0) 0 0......np a
Maclaurin
This gives a1 = 1.
Differentiate again to get( ) sinf x x
2 3( ) 2 3 2 .......np x a a x
Put x = 0 again and we get that
2
2
(0) 0 and (0) 2
so a 0.nf p a
Maclaurin
To get the cubic polynomial approximation we must differentiate once more.
3( ) cos and ( ) 6 other terms in xnf x x p x a
For the last time we put x = 0 to get
3(0) 1 and (0) 6nf p a
Maclaurin
6a3 = -1 and so a3 =
We now have the following coefficients for the polynomial:
a0 = 0 a1 = 1 a2 = 0 a3 =
Giving sin x = 1x
1
6
1
6
31
6x
Maclaurin
pn(x) = a0 + a1x + a2x2 + a3x3 + ….
f(0) = pn(0) = a0
Differentiate once so that2
n 1 2 3p ( ) 2 3 ......x a a x a x
Because
2
2
(0) (0) we see that (0) 2
(0)
2
nf p f a
fa
Maclaurin
This can be written assin 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)
Differentiate2
n 1 2 3
n 1
f (x) = p (x) a + 2a x + 3a x .......
so f (0) = p (0) a
Maclaurin
Differentiate again
n 2 3
n 2
f (x) = p (x) = 2a + 3 2a x + ......
so f (0) = p (0) = 2a
2
f (0)giving a =
2
Maclaurin
To get a cubic polynomial we must differentiate once more.
(If we wanted a higher degree polynomial we would continue.)
n 3
n 3
f (x) = p (x) = 3 2a + other terms in x
so f (0) = p (0) = 6a
3
f (0)finally a =
6
Maclaurin
We can now write the polynomial as follows:
This is called the Maclaurin expansion of f(x).
2 3n
f (0) f (0)p (x) = f(0) + f (0) x + x + x + .....
2 6
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
2 3(0) (0)( ) ( ) (0) (0) ...
2! 3!n
f ff x p x f f x x x
Maclaurin
Example : obtain the Maclaurin expansion of degree 2 for the function defined by
( ) 1f x x
Maclaurin
First get the coefficients:1
2( ) 1 (1 ) (0) 1f x x x f
1
21 1
( ) (1 ) (0)2 2
f x x f
3
21 1 1
( ) (1 ) (0)2 2 4
f x x f
0 1 2
1 (0) 1so (0) 1, (0) and
2 2! 8
fa f a f a