NUMERICAL ANALYSIS

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) = (4x – 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 2x , 3x and ex.

Preliminary Results y = 2x is the thicker graph

Preliminary Results y = 3x 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 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

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) = 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). Now put x = 0 in both expressions

Maclaurin • This gives a1 = 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 • 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

Maclaurin • pn(x) = a0 + a1x + a2x2 + a3x3 + …. • f(0) = pn(0) = a0 • Differentiate once so that Because

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

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 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

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

Maclaurin • First get the coefficients:

Maclaurin • This gives the polynomial

NUMERICAL ANALYSIS