We found that if represent a function as a power series
f(x) = a0+a1(x-a)+...+an(x-a)n+...,
that
an =
f(n)(a)n!
.
Maclaurin series are Taylor series where a = 0.
If f(x) = 1/x then want f(n)(a) to find taylor series generated by
this. x-1 goes to (-1)x-2, which goes to 2!x-3. So
nth derivative is n!(-1)n x-(n+1). This means that
an =
(-1)n n! x-(n+1)n!
= (-1)nx-(n+1).
So if expand about x = 2 then a = 2 so our taylor series is
å
n
(-1)n (x)-(n+1)(x-2)n =
12
-
(x-2)222
+
(x-2)223
+¼.
This is a geometric series with r = -(x-2)/2 and first term 1/2.
Converges for r<1 or | x-2 | < 2. Know what the infinite sum is
for this:
S =
a1-2
=
1/21+(x-2)/2
=
12+x-2
=
1x
.
Taylor Polynomials of order n, Pn(x), are series where we just keep the
first n terms. They provide the best polynomical approximation of the
relevant function possible using polynomials of nth degree!
When can we expect taylor series to converge to the function we are
approximating and how accurate is our approximation?
function is equal to Pn(x)+Rn(x) where Rn(x) is the remainder. It's
absolute value is called the error associated with the approximation.
Taylor's theorem:If f is differentiable through order n+1 in an open
interval I containing a then for each x in I there exists a number c between
x and a such that
Rn(x) =
f(n+1)(c)(n+1)!
(x-1)(n+1).
This is a generalisation of the Mean value theorem. If Rn(x)® 0 as n® ¥
then we say that the series generated by f at a converges to f on I.
Estimating the Remainder:If there are positive constants M and r such that
| f(n+1)(t) | £ Mrn+1 for all t between x and a, inclusive then the
remainer term satisfied the inequality
| Rn(a) | £ M
rn+1 | x-a | n+1(n+1)!
.
Normally what we say the error is O(hn+1) but the derivative term need not
be small which is why we need results like this. In general the derivative being
large doesn't turn out to be a problem.
If have common interval of convergence then can add, subtract and multiply
taylor series term-by-term.
Application of power series
Binomial series:Use for (1+x)3 = x3+3x2+3x+1. With taylor can deal with
(1+x)1/2 by expanding about 0. Then
= 1+
x2
-[1/2(-1/2)]1/2! x2+¼.
Note that this only converges for x < 1 since if radius greater than one would
include 1/0 when the function is no longer continuous. Also do not want to include
negative numbers in our interval of convergence since the square root of negative numbers
are not real number!
Series solutions of differential equations.
If have
dydx
-y = x
with y(0) = 0, as in an initial value problem then we can represent y as a
series as well as y'. This lets us solve the equation.
So we have 2a2-a1 = 1 and a1 = a0, a0 = 1 since y(0) = 1 as well as
nan-an-1 = 0 for the other cases. So a0 = 1 and a1 = a0 = 1. Then
2a2-a1 = 1 so a2 = 1 = 2/2, but 3a3-a2 = 0 so a3 = 1/3, a4 = a3/4 and so on.
Can write as an = 2/n! for n ³ 2.
y = 1+x+x2+x2+2x3/3!+...
which is the same as
y = 1+x+2(x2/2!+x3/3!+...) = 1+x+2(ex-1-x) = 2ex-1-x.
Fourier series
Express functions as
f(x) =
a02
+
å
n
(an cos
npxL
+ bn sin
npxL
).
Need to determine an and bn. This can easily be done since
ó õ
L
-L
cos
npxL
cos
mpxL
dx = L
if n = m or else its zero. Same for sin. sin cos is always zero. So
an =
1L
ó õ
L
-L
f(x) cos
npL
dx.
and
bn =
1L
ó õ
L
-L
f(x) sin
npL
dx.
a0 =
1L
ó õ
L
-L
f(x)dx.
Won't be asked on fourier or application of power series to DE's which you will
see a lot of next year and beyond.
Will bring in sample papers on tuesday and put up solutions for the homework problems
on the web during the week. Remember to ask question in tutorials.
File translated from
TEX
by
TTH,
version 2.70. On 3 May 2002, 13:15.