NUMERICAL ANALYSIS Maclaurin and Taylor Series Preliminary Results

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:

Preliminary Results ¬We must also remember how to differentiate more complicated expressions: ¬E. g

Preliminary Results ¬We must write in a form suitable for differentiation: ¬f(x) = (4 x – 1)1/2 ¬then we differentiate

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 2 x , 3 x and ex.

Preliminary Results y = 2 x is the thicker graph

Preliminary Results y = 3 x 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

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 18 th 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

Maclaurin ¬We now put x = 0 in each of these.

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) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + …… ¬then we require pn(0) = 0 as well. ¬pn(0) = a 0 + a 10 + a 202 + a 303 + …… ¬= a 0 + 0 = a 0. ¬We want this to be 0 so a 0 = 0.

Maclaurin ¬Now we differentiate both f(x) and pn(x). Now put x = 0 in both expressions

Maclaurin ¬This gives a 1 = 1. ¬Differentiate again to get Put x = 0 again and we get that

Maclaurin ¬To get the cubic polynomial approximation we must differentiate once more. For the last time we put x = 0 to get

Maclaurin ¬ 6 a 3 = -1 and so a 3 = ¬We now have the following coefficients for the polynomial: ¬a 0 = 0 a 1 = 1 a 2 = 0 a 3 = ¬Giving sin x = 1 x

Maclaurin ¬pn(x) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + …. ¬f(0) = pn(0) = a 0 ¬Differentiate once so that Because

Maclaurin ¬This can be written as ¬sin x = x – x 3 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) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + … ¬f(0) = pn(0) = a 0 ¬so a 0 = f(0) ¬Differentiate

Maclaurin ¬Differentiate again

Maclaurin ¬To get a cubic polynomial we must differentiate once more. ¬(If we wanted a higher degree polynomial we would continue. )

Maclaurin ¬We can now write the polynomial as follows: ¬This is called the Maclaurin expansion of f(x).

Maclaurin ¬The numbers 2 and 6 come about from 2 x 1 and 3 x 2(x 1). ¬We can write these in a shorter way as ¬ 2! and 3! – read as factorial 2 and factorial 3. ¬ 4! = 4 x 3 x 2 x 1 = 24 5! = 5 x 4 x 3 x 2 x 1 = 120

Maclaurin ¬This allows us to write the Maclaurin expansion as

Maclaurin ¬Example : obtain the Maclaurin expansion of degree 2 for the function defined by

Maclaurin ¬First get the coefficients:

Maclaurin ¬This gives the polynomial