x means the positive square root.
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Names also exist for higher order
polynomials but are used less frequently.
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|1x1| indicates the modulus or absolute value
of x. If x is positive, |1x1| = x and |1−x1| = x.
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ω is the Greek letter omega.
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We choose a suffix 3 in p3(x) because the
polynomial is of degree 3.
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df
Note that f ′ (0) =  
 dx  x = 0
 d2 f 
f ″ (0) =  2 
 dx  x = 0
and so on.
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Don’t forget that the argument of the sine
function must be a dimensionless variable
or an angle measured in radians.
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exp1(x) is often written as e x.
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Notice that it is quite unnecessary to
calculate a0 , a1 and a2 ; we already know
their values.
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Brook Taylor (1685–1731) was one of the
outstanding British mathematicians of the
early 18th century. What are now referred
to as Taylor polynomials were introduced
by Taylor in 1715. He also pioneered their
application in physics and the theory of
equations.
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Notice particularly that the constant term
disappears when you differentiate.
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Don’t let the algebra put you off. The
central column is merely what you get if
you keep differentiating Equation 17.
p n(x) = a0 + a1x + a2x2 + a3x3 + … + a nx0n
(Eqn 17)
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n! = n(n − 1)(n − 2) … 3 × 2 × 1
and, by convention, 0! = 1.
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 a0 =

 a1 =

 a2 =


 a3 =

a =
 4
M =

 an =

f (0)
f ′ (0)
1!
f ′′ (0)
2!
f (3) (0)
3!
f (4) (0)
4!
M
f (n) (0)
n!
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pn ( x) = f (0) +
f ′ (0)
f ′′ (0) 2
x+
x
1!
2!
f (3) (0) 3
f (n) (0) n
x +…+
x
3!
n!
(Eqn 19)
+
Equation 19 is only valid provided that
f1(0) , f1′(0) , …, f1(n)(0) all exist.
Throughout this module we only consider
functions f1(x) for which this is true.
In Equation 19, 1! is just 1 while 2! is just
2, but this is a neat way to write the
formula (and possibly easier to remember).
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It is assumed that all the derivatives exist.
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A complete discussion of infinite series is
beyond the scope of FLAP. In this module
∞
xn
you may regard f ( x) = ∑ f ( n ) (0)
n!
n=0
as an instruction to take as many terms in
the series as you require in order to obtain
the required degree of accuracy.
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We should emphasize that this is a very
rough approximation and much better
estimates are available, but they properly
belong in a course on mathematical
analysis. For the rule of thumb given here
to be of any use, the series must converge.
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Don’t forget to set your calculator to
radian mode.
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Note that a0 , a1, a2, …, etc. are constant
coefficients whose values are to be
determined, but a is a given value of x.
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Shortly we will see that the working can be
abbreviated considerably in such
calculations.
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The remarks (in Subsection 3.2) about
convergence of series apply here also.
Some authors use the nomenclature,
Maclaurin series, for the expansion near
x = 0 and reserve Taylor series for the
general expansion. We do not make this
distinction within FLAP.
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A Taylor series is simply a power series
where we know the values of the
coefficients in terms of the function f1(x).
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ε is the Greek letter epsilon.
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You will find it useful to memorize these
series.
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R represents the set of real numbers.
So x ∈ R means ‘x is a real number’.
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This is called a binomial expansion.
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This function is usually known as sinh1(x),
and is an example of a hyperbolic function.
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You can omit Example 2 if you are not
familiar with the terms simple harmonic
motion and potential (see the Glossary).
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Don’t forget that V1′(x) = 0 at the
equilibrium position, x0 .
Remember, a function f1(x) will have a
minimum at x − x0 if f1′(x0 ) = 0 and
f1″(x0 ) is positive.
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You may find Question T13 helpful.
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Although a particularly simple example
has been given here, this iterative
technique is especially useful in situations
where f1(x) is a complicated function of x.
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Iteration means ‘repetition with different
starting values’.
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Note that the constant ε in Equation 48
6
 a 12
a 
V (r) = ε   − 2   
 r 
 r 

(Eqn 48)
is not the same as the ε used earlier.
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Because β = a, so we are really evaluating
the bracketed term at r = a.
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Note that x represents the difference
between the separation r of the atoms and
their minimum energy separation β. Epot(x)
represents the instantaneous potential
energy of the atoms when their separation
is r = β + x.
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Some important examples of Taylor
expansions are given in Subsection 3.5.
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✧
Only (a) is a polynomial in x.4❏
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✧
The coefficient of x in the two expressions
is identical. In other words the
approximation is improved by adding an
extra term and not by changing the existing
one.4❏
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✧
The coefficients of x and x3 in the two
expressions are identical. So once again
the approximation is improved by leaving
the existing terms unchanged, and adding
an extra term.4❏
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✧
q 1 (0.1) = 1.1 and exp1(0.1) ≈ 1.10511709,
so q1 (x) is a good approximation to exp1(x)
at x = 0.1.
The gradient of y = exp1(x) is
dy
= exp( x).
dx
So at x = 0 the gradient is exp1(0) = 1.
The tangent line at x = 0 has the equation
y = 1 + x (and this is why q1 (x) = 1 + x is a
good approximation to exp1(x) near
x = 0).4❏
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✧
q 2 (0.1) = 1.105 which is a better
approximation to exp1(0.1) ≈ 1.10511709
than q1(0.1) = 1.1.4❏
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✧
If g(x) = exp1(x) then g(3)(0) = exp1(0) = 1
while q3(3) (0) = 3 × 2 × a3 = 6a3 and these
derivatives are equal if 6a3 = 1, and
therefore ☞
q3 ( x) = 1 + x +
x2 x3
+
2
6
(16)4❏
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✧
q 3 (0.1) = 1.1051166167 which is a better
approximation to exp1(0.1) ≈ 1.10511709
than q2(0.1) = 1.105 and q1(0.1) = 1.1.4❏
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✧
p′(x) = 3 + 10x + 21x02 + 36x03,
p″(x) = 10 + 42x + 108x02
and p″(0) = 10. ☞4❏
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✧
(a)
a 0 = 1, a3 = 1/(1.3), a10 = 1/2
and an = 1/(1 + n × 0.1).
(b
pn′ ( x) =
1 2 x 3x 2
nx n−1
+
+
+K+
(1 + n × 0.1)
1.1 1.2 1.3
and
pn′′( x) =
2 3× 2x
n(n − 1)x n−2
+
+ K+
(1 + n × 0.1)
1.2
1.3
(c)
pn′′(0) =
2
≈ 1.6667 4❏
1. 2
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✧
pn′ ( x)= a1 + 2a 2 x + 3a 3 x2 + 4a 4 x3 +
… + nan x0n 1−11
pn′′( x)= 2a2 + 3 × 2a3 x + 4 × 3a4 0x2 +
… + ann(n − 1)x0n 1−12
and
pn′′(0)= 2a2.4❏
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✧
The Taylor series contain an infinite
number of terms, the Taylor polynomial of
degree n only contains n + 1 terms with
coefficients a0, a1, … an. For a given
function f1(x), the Taylor polynomial of
degree n is given by the first n + 1 terms of
the Taylor series of f1(x).4❏
☞
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✧
Since
[x − π0/4]x = 0.8 ≈ 0.8 − 0.7854 = 0.0146
we obtain
p 3 (π0/4) ≈ [1 + 0.0146 − (0.0146)2 /2
− (0.0146)3 /6 ]/ 2 ≈ 0.7174
A calculator gives
sin1(0.8) ≈ 0.7171356.4❏
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✧
First we determine the derivatives of sin1(x)
evaluated at x = π
[sin ( x)]x=π = sin (π) = 0
 d sin ( x) 
= cos(π) = −1

 dx
x=π
 d2

= − sin (π) = 0
 2 sin ( x) 
 dx
 x=π
 d3

= − cos(π) = 1
 3 sin ( x) 
dx

 x=π
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Substituting these values into Equation 32
pn ( x) =
f (m) (a)( x − a) m
m!
m=0
n
∑
(Eqn 32)
we obtain
sin ( x) ≈ −( x − π) +
( x − π) 3
3!
(33)
Since 3 − π ≈ −0.141159 we have, from
Equation 33,
sin (3) ≈ −(−0.14159) +
(−0.14159)3
6
≈ 0.1411
A calculator gives the same value correct
to four decimal places.4❏
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✧
To find a Taylor polynomial of degree 3
you only need to evaluate the function and
its first three derivatives at x = a; but to
find the Taylor series near x = a you have
to evaluate all the derivatives at x = a, in
other words you need to find a formula for
f1(n)(a).4❏
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✧
1
=
L + 3x
1
3x
L1 + 

L
=
1  3x 
1+
L
L
from Equation 42 using x =
−1
−3x
L
1
= 1+ x + x2 + x3 + K −1 < x < 1
1− x
(Eqn 42)
1
1
3x
≈ 1 − 
L + 3x L 
L
=
1 3x
−
L L2
The condition that x is small compared to L
ensures that 3x/L lies between −1 and 1, so
that we are justified in using
Equation 42.4❏
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