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) = e x and f(x) = ln x.  For e x we can look at the graphs of exponential functions along with their derivatives –  we will consider 2 x, 3 x and e x.

Preliminary Results y = 2 x is the thicker graph

Preliminary Results Notice that the two graphs are almost the same, but not quite y = 3 x

Preliminary Results y = e x This time the two graphs overlap exactly

Preliminary Results  The graphs show that the derivative of e x is e x.  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  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 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  p n (x) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + ……  then we require p n (0) = 0 as well.  p n (0) = a 0 + a a a ……  = a = a 0.  We want this to be 0 so a 0 = 0.

Maclaurin  Now we differentiate both f(x) and p n (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  6a 3 = -1 and so a 3 =  We now have the following coefficients for the polynomial:  a 0 = 0a 1 = 1 a 2 = 0 a 3 =  Giving sin x = 1x

Maclaurin  p n (x) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + ….  f(0) = p n (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 p n (x) approximate the function f(x) near x = 0.

Maclaurin  f(x) = p n (x) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + …  f(0) = p n (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 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