Approximating Functions
Power Series - Maclaurin
Consider the geometric series (a=1, r=x)
In general a function may be expanded in a
power series defined as;
1 − xn
1 + x + x 2 + x 3 + x 4 + K + x n −1 =
1− x
f ( x) = c0 + c1 x + c2 x 2 + c3 x 3 + K + cn x n + K
But, written the other way round, this is a polynomial
expansion of a function;
Here, all of the polynomial terms are centred
on x=0, it is and expansion about the point
x=0 or a Maclaurin Series.
1− xn
= 1 + x + x 2 + x 3 + x 4 + K + x n −1
1− x
For example : cos(x)
cos( x) = 1 −
Note: x measured in radians
1−
cos( x) = 1 −
1 2 1 4
1 6
x +
x −
x +K
2
24
720
1 2 1 4
x +
x
2
24
1 2 1 4
1 6
x +
x −
x +K
2
24
720
cos(x)
• We’ll see why this is the expansion later
• More terms → the power series becomes
more accurate for a wider range of values of x
1−
Why bother ?
• Approximating an analytic function by its
series expansion often helps us to visualise
and understand its behaviour – eg: <E>(T) in
the problem class.
1−
1 2 1 4
1 6
x +
x −
x
2
24
720
Finding the coefficients: Maclaurin
c0
f ( x) = c0 + c1 x + c2 x 2 + c3 x 3 + K
f (0) = c0 + c1 × 0 + c2 × 0 + c3 × 0 + K = c0
c1
• Series can be used to represent experimental
data when you don’t know the analytic form
(eg: curve fitting, drawing a straight line…).
1 2
x
2
df
= c1 + 2c2 x + 3c3 x 2 + 4c4 x 3 + K
dx
df
= c1
dx x =0
1
c2
The coefficients are the derivatives
d2 f
= 2c2 + 2 × 3c3 + 3 × 4c4 x 2 + K
dx 2
d2 f
dx 2
c0 = f (0)
c1 =
= 2!c2
x =0
c2 =
c3
d3 f
= 2 × 3c3 + 2 × 3 × 4c4 x + K
dx 3
d3 f
dx 3
= 3!c3
df
dx
c3 =
x =0
1 dn f
n! dx n
x =0
K
x=0
1 d2 f
2! dx 2
1 d2 f
3! dx 2
cn =
x =0
That’s it - you can now expand any function !
x =0
The Maclaurin Series - Summary
Maclaurin for cos(x)
cos(0) = 1
f ( x) = f (0) +
df
dx
3
2
x+
x =0
1d f
2! dx 2
x2 +
x=0
1d f
3! dx 3
x3 + K
x=0
Knowledge of all of the derivatives at one
point completely determines any well behaved
function (eventually)
cos( x) = 1 −
1 2 1 4
1 6
x +
x −
x +K
2
24
720
d
cos( x ) = − sin(0) = 0
dx
x =0
: All odd derivs 0
d2
cos( x) = − cos(0) = −1
dx 2
x =0
: Even derivs alternate
-1, +1, -1, +1 etc..
K
cos( x) = 1 −
1 2 1 4 1 6
x + x − x +K
2!
4!
6!
Power Series - Taylor
It may be convenient to expand about some
other point, eg: x=a, then the power series is;
f ( x ) = c0 + c1 ( x − a) + c2 ( x − a) 2 + c3 ( x − a ) 3 + K
It works well for small x – how many terms
would we need to approximate a whole cycle
of cos(x) ?
Expansion about the point x=a is a Taylor
Series.
2
1
1 4
1 6
cos( x) = 1 − x 2 +
x +
x +K
2
24
720
Taylor Series
The analysis is very similar to the Maclaurin
series leading to;
f ( x) = f ( a ) +
1 d3 f
+
3! dx 3
df
dx
( x − a) +
x =a
1 d2 f
2! dx 2
( x − a)2 +
x =a
3
( x − a) + K
x =a
OK but if we want to approximate cos(x) at
x=π we will need a lot of terms – better to
expand about x= π using a Taylor expansion.
cos(x) : Taylor Series
cos( x) = −1 +
1
1
1
( x − π ) 2 − ( x − π )4 + ( x − π )6 − K
2!
4!
6!
cos(π ) = −1
d
cos( x )
= − sin(π ) = 0
dx
x =π
: All odd derivs 0
d2
cos( x)
= − cos(π ) = +1
dx 2
x =π
: Even derivs alternate
+1, -1, +1, -1 etc..
K
So …
cos( x ) = −1 +
1
1
1
( x − π )2 − ( x − π )4 + ( x − π )6 − K
2!
4!
6!
Summary
• Knowledge of the derivatives of a function can be
used to make a polynomial expansion which, if you
use enough terms, reproduces the function exactly
Now an excellent approximation at x=π with a
few terms.
Potential Energy of a Diatomic Molecule
In H2 the potential energy of interaction of the
atoms looks like this
• A Maclaurin series achieves this by expansion
about x=0
• Faster convergence can be achieved away from
x=0 by using a Taylor series which expands about
any point, eg: x=a.
3
The Morse Potential
To a good approximation the potential energy
is give by the Morse form which is;
{
E (r ) = De 1 − e −α (r − a )
Problem Class 2
Compute an Harmonic approximation to the
Morse potential for H2 and thus compute the
vibrations of the molecule.
}
2
with, De=4.79eV, a = 0.074 nm and α=19.3 nm-1.
4