Math Methods for Polymer Physics
Lecture 1: Series Representations of Functions
Series analysis is an essential tool in polymer physics and physical sciences, in general. Though other broadly speaking, a series expansion allows
one to analyze an arbitrarily complicated function into the sum of a simpler
set of functions. Though other series expansions exist, two are especially
useful: the Taylor series and Fourier series. In a crude way, we may
think of both series as 2 different ways of approximating, or “fitting”, a
given function to a simpler form. For further reading on Taylor series and
Fourier series see chapters 5 and 14, respectively, of Arken and Weber’s text,
Mathematical Methods for Physicists.
1
Taylor Series
Let’s start with Taylor series expansions. The Taylor expansion is a representation of a function, say f (x), as an infinite power series in the polynomials, (x − x0 )n , where x0 is some reference point for the independent variable,
x. Why are Taylor series useful? Well, let’s say you have a complicated
function:
x 3
f (x) = ln cos x2 + 2 +
.
(1)
3
This function is plotted in Fig. 1.
Often it’s sufficient and useful to have a simpler description of the function in the neighborhood of some point, say x = x0 . For many functions, you may replace f (x) with a power series expansion in polynomials of
∆x = x − x0 distance from the reference point.
f (x) =
∞
X
an (∆x)n
n=0
= a0 + a1 ∆x + a2 (∆x)2 + a3 (∆x)3 + . . .
(2)
What are these coefficients an ? The first one can be deduced from x = x0
and ∆x = 0, so that
f (x0 ) = a0 + a1 ∆x + a2 (∆x)2 + a3 (∆x)3 + . . . = a0
{z
}
|
0
1
(3)
Figure 1: Plot of f (x) in eq. (1), dark solid line. For many applications,
it is often necessary to know behavior of the functionf (x) near some point,
say x0 = 2. The Taylor series expansions for f (x) around x = x0 including
1, 2, 3, and 4 terms only are shown as labelled.
To find the higher-order (larger n) coefficients, take derivatives of both sides.
Note that after this operation the right side is still a Taylor series.
f 0 (x0 ) = a1 + 2a2 ∆x + 3a3 (∆x)2 + . . . = a1
|
{z
}
(4)
0
In general, we may show
1 dn f an =
n! dxn .
(5)
x=x0
In order to find the Taylor series expansion we need only to take derivatives
of f (x) evaluated only at the point of reference, x = x0 . Eqs. (2) and
(5) define the Taylor series expansion of a functions of a single variable.
Functions which can be represented by a Taylor series are known as analytic
functions.
Notice from eq. (2) that as x → x0 and ∆x → 0 the higher order (large
n) terms in the power series expansion go to zero very quickly. Hence, if
one is interested in f (x) sufficiently close to x0 , a Taylor series expansion
truncated to include only a few leading terms may often be sufficient to
approximate the function. Geometrically, we can think of this in terms of a
“local” description of a function near x = x0 .
(∆x)2 00
(∆x)3 000
f (x0 ) +
f (x0 ) + . . .
f (x) = f (x0 ) + ∆xf 0 (x0 ) +
| {z } | {z } | 2! {z
} | 3! {z
}
constant
linear
parabolic
2
cubic
(6)
This shows that sufficiently close a point of interest that analytic functions
are well approximated by constant plus a sloped, linear correction plus a
parabolic correction plus . . .. Further away from a given reference point at
x = x0 , the less and less a function looks like a straight line. In order to get
a better a approximation, you need functions with more wiggles (e.g. higher
order polynomials).
Let’s try some examples.
Example 1: Expand ln(x) in a Taylor series around x0 = 1.
a0 = ln 1 = 0
d
1 a1 =
ln(x)
= =1
dx
x
x=1
x=1
d2
1
d
1
a2 =
ln(x)
=
=
−
= −1
dx2
dx x x2 x=1
x=1
x=1
3
2
d
ln(x)
= 3
=2
a3 =
3
dx
x x=1
In general, an =
(−1)n (n
x=1
− 1)! for n > 1.
(x − 1)2 (x − 1)3 (x − 1)4
+
−
+ ...
2
3
4
∞
X
(−1)n (x − 1)n
=
n
ln(x) = (x − 1) −
(7)
n=1
Example 2: Expand
1
in a Taylor around x = 0.
1−x
a0 = 0
d 1 1
a1 =
=
= 1!
2
dx 1 − x
(1 − x) x=0
x=0
2
d2 1 a2 =
=
= 2!
2
3
dx 1 − x (1 − x) x=0
x=0
d3 1 1 × 2 × 3 a3 =
=
= 3!
dx3 1 − x (1 − x)4 x=0
x=0
In general, an = n!. So
∞
X
1
=
xn
1−x
n=0
3
(8)
which is the well-known geometric series.
These particular series, eqs. (7) and (8), do not converge for all values
of x. When the series does not converge, for some large enough ∆x, the
successive terms terms an (∆x)n become larger than that the sum of the
previous terms, meaning that adding more terms in the series expansion
does not provide a better approximation, and the Taylor series fails to rep1 around x = 0,
resent the function. For ln(x) around x0 = 1 and 1 −
0
x
these only converge for |∆x| < 1.
In general we may define Rc as the radius of convergence
of the Taylor
P
n = f (x).
series of f (x) around x = x0 . If |∆x| < Rc , then ∞
a
(∆x)
n=0 n
Otherwise series does not provide a good approximation of f (x) (adding
more terms makes things worse).
There are some functions for which Rc → ∞ and the Taylor series always
converges. Important examples include ex , sin x, cos x. These functions arise
in many contexts, so it is useful to commit these series to memory.
Example 3: Expand ex around x = 0. Well, first notice
dn x x
=1
e
=
e
dxn x=0
x=0
From eq. (5) this gives right away the Taylor series coefficient of ex
∞
X xn
x2 x3 x4
e =1+x+
+
+
+ ... =
.
2!
3!
4!
n!
x
(9)
n=0
The Taylor series representation of ex is a particularly
way to see that
P∞ xuseful
n
d
x ) = ex . Indeed, it is reasonable to view
(e
as
the
definition of
n=0 n!
dx
x
e .
You should also commit expressions of sin x and cos x to memory. These
converge for all x:
sin x = x −
x3 x5 x7
+
−
+ ...
3!
5!
7!
(10)
x2 x4 x6
+
−
+ ...
(11)
2!
4!
6!
Notice that these expansions allow you to derive the following important
identity,
eix = cos x + i sin x,
(12)
cos x = 1 −
which is used heavily in Fourier analysis.
It is reasonably straightforward to generalize the Taylor series expansion
for a function of a single variable to a mutli-variable function, say f (x, y),
4
expanded around the point x = x0 and y = y0 :
f (x, y) = f (x0 , y0 ) + ∆x
∂f
∂f
+ ∆y
∂x
∂y
1h
∂2f
∂2f
∂2f i
(∆x)2 2 + 2∆x∆y
+ (∆y)2 2
2!
∂x
∂x∂y
∂y
3f
3f
1h
∂
∂
+ (∆x)3 3 + 3(∆x)2 ∆y 2
3!
∂x
∂x ∂y
3
3 i
∂ f
3∂ f
+3∆x(∆y)2
+ ...
+
(∆y)
∂x∂y 2
∂y 3
+
(13)
where ∆x = x − x0 , ∆y = y − y0 and all partial derivatives are evaluated
at (x0 , y0 ). This expansion can be confirmed by taking first, second, third
(etc.) derivatives of both sides of the equation above.
Why is the Taylor expansion a useful description? In many physical
systems, the full expression for a function may be impossible to write down
(i.e. PE of strongly interacting mixtures of charged particles). But often,
equilibrium and dynamic behavior depends only on local properties of function. By “local”, we mean, sufficiently close to some set of values for the
independent variable.
As a concrete example, consider a colloidal bead in a laser trap (Fig.
2) , an experimental tool which has been exploited to measure the forces
generated by single macromolecules. If the bead has a polarizability, α,
then when it is subject to an electric field, E, it obtains a dipole moment,
p = αE. The potential energy of a polarized object in an electric field
is simply, U = − 21 p · E, while the energy required to polarize the bead is
Upolarization = |p|2 /(2α). Therefore, if the polarizable bead is subject to an
electric field E(x) that varies in space (as near the focal point of a laser
beam, the net electrostatic interaction between the bead and the field is
described by the potential energy,
α
U (x) = − |E(x)|2 .
2
(14)
Hence, the potential energy is lowest in regions where the electric-field intensity, |E(x)|2 , is highest. This explains why a small polarizable object,
like colloidal beads, are drawn into the focal point of a high-intensity laser
(shown schematically in Fig. 2).
In general, the pattern of electric field intensity, |E(x)|2 , may be rather
complicated. But, if we are interested only in the behavior very close to the
center of the trap, the behavior always has the same simple form,
U2
U (∆x) = −U0 + U1 ∆x +
(∆x)2 + . . . .
|{z}
| {z }
2
| {z }
constant
=0
quadratic
5
(15)
Figure 2: Top: a schematic depiction of a polarizable bead near to the highintensity focal point of a laser beam. Bottom: A sketch of U , the potential
energy of a optically-trapped colloidal particle in terms of ∆x, the deviation
from the center of the trap.
dU By definition, is minimum at ∆x = 0, so we know
= U1 = 0. This
dx x=0
means that the force on the bead at the center of the trap is zero, because the
electric field intensity is maximal. Local equilibrium (mechanical, dynamic,
etc.) always looks like this: constant + quadratic (first non-trivial term in
expression about equilibrium).
What is force if bead is displaced?
Fx = −
dU
= −U2 ∆x = −k∆x
dx
(16)
The linear force response is identical to a “Hooke’s Law” elastic spring,
and k is spring constant. For all interest and purpose (near equilibrium or
steady state), we are often interested in expression up to harmonic order.
Therefore, if one “calibrates” the strength of optical trapping (the value of
k) and carefully measure ∆x, you can measure magnitude of external forces
that pull a bead from the center of the trap, generated, say, by a strand of
DNA chemically tethered to the bead.
2
Fourier Series
The second important series representation of functions is the Fourier series.
A simple way to describe this series is to contrast it with the Taylor series
6
described in the previous section:
Taylor Series - decompose f (x) into infinite series of polynomials (∆x)n
Fourier Series - decompose f (x) into infinite series of sines and cosines
Why are Fourier series (and transforms) useful?
1. Fourier analysis is necessary to understand interaction between matter
and radiation/waves (i.e. scattering) and spectral analysis
2. Sines and cosines are “harmonic functions”, which means they form a
complete basis of solutions to certain PDE’s common to the study of
physical systems
Indeed, properties 1 and 2 are intimately related as the wave equation is harmonic, and therefore, radiation (light, x-rays, etc.) is sinusoidal in nature.
In addition, you’ll likely see how property 2 can be used to solve problems
in continuum elasticity and polymer dynamics. For example, in the study
of polymer dynamics, we come across equations like,
d2
R(n) + kR(n) = 0,
dn2
(17)
where k > 0 describes a relaxation rate chain motion, and R(n) specifies
√ d2
the position of the bead n along a polymer chain. Since, dn
kn =
2 sin
√ −k sin
kn , sines and cosines form a natural set of solutions to this equation. For the purposes of this review, a Fourier series is the unique decomposition of an arbitrary function (in some domain) into an infinite series of
sines and cosines.
Let’s say we are interested in a function f (x) in the domain x ∈ [0, L]
(see Fig. 3). In this domain we can write Fourier series as:
∞
a0 X
f (x) =
+
an cos
2
n=1
X
∞
2πn
2πn
x +
bn sin
x
L
L
(18)
n=1
an and bn are coefficients. Just as the coefficients of the Taylor series are
related uniquely to the given function, an and bn are uniquely determined
by properties of f (x) on this domain.
How are an and bn related to f (x)? This relationship derives
from
an
2πn
important properties of sines and cosines. In particular, sin L x and
cos 2πn
x
are orthogonal functions on this domain. This means that if a
L
7
Figure 3: Plot of f (x), in the range of [0, L].
multiply any two of these elementary functions and integrate over the domain x ∈ [0, L], the resulting integral is zero unless these functions
are iden
2πn
tically. Consider two produce of two sine functions sin L x sin 2πm
L x :
L
2πn
2πm
dx sin
x sin
x
L
L
0
Z
1 L
2πx
2πx
=
dx cos
(n − m) − cos
(n + m) .
2 0
L
L
Z
(19)
This integral is only non-zero ifn = m for which the first term in the integrand becomes cos 2πx
L (n − m) = 1. From this we can show the following
for the orthogonality between sines,
( L
Z L
2πn
2πm
if n = m
2
dx sin
(20)
x sin
x =
L
L
0
if n 6= m
0
Similarly, for the cosines,
Z
L
dx cos
0
( L
2πn
2πm
2
x cos
x =
L
L
0
Sines and cosines are always orthogonal,
Z L
2πn
2πm
dx sin
x cos
x =0
L
L
0
if n = m 6= 0
if n =
6 m
for all m, n.
(21)
(22)
The orthogonality relations, eqs. (20) - (22), are important because
they allow one to invert the Fourier series, to determine the unique set of
8
coefficients, an and bn , the correspond to the function f (x). Operationally,
the coefficients of the Fourier series are determined
by “projecting out” the
2πn
term in series proportional to, say, sin L x , by multiplying both sides
x
and integrating the product over the domain
of eq. (18) by sin 2πn
L
x ∈ [0, L]:
Z L
2πm
dx f (x) sin
x
L
0
"
X
#
Z L
∞
∞
2πm
a0 X
2πn
2πn
dx sin
=
x
+
an cos
x +
bn sin
x
L
2
L
L
0
n=1
n=1
Carrying out the integration, a0 and all cosine terms in sum will be zero
due to orthogonality conditions, eqs. (21) and (22). Likewise, all sine terms
in sum except n = m term are zero too. Thus, the only term from the
right-hand side of eq. (18) that survives this “projection” operation is from
n = m:
Z L
2πm
L
dx f (x) sin
x = bm
L
2
0
and,
2
bn =
L
Z
L
dx sin
0
2πn
x f (x)
L
(23)
By performing the same operation with the cosine functions we can also
derive,
Z
2 L
2πn
an =
dx cos
x f (x)
(24)
L 0
L
Example 4 Consider a function f (x) = A + Bx (see Fig. ??). Compute
coefficients an and bn for a Fourier series in the domain x ∈ [0, L]:
From eq. (24) we compute
the coefficients to the cosine terms by mul2πn
tiplying f (x) by cos L x and integrating over the domain. For m = 0,
this is easy,
Z
2 L
2
BL2
dx (A + Bx) =
a0 =
AL +
= 2A + BL
(25)
L 0
L
2
Now consider bn ,
L
2πm
x (A + Bx)
dx sin
L
0
Z
2B L
2πm
=
dx x sin
x
L 0
L
2
bn =
L
Z
(26)
How do you do this integral? Let’s review a useful trick for evaluating
integrals.
9
Figure 4: Plot of f (x), in the domain of [0, L].
Aside: Integrations by parts Z
Let’s say you want to compute
L
dx u(x)v 0 (x), and you don’t know the
0
anti-derivative of v 0 (x). The chain rule of differentiation gives you,
d
(u(x)v(x)) = u0 (x)v(x) + u(x)v 0 (x)
dx
(27)
or
d
(u(x)v(x)) − u0 (x)v(x).
dx
Substituting this expression for the integrand,
Z L
Z L
d
0
0
dx u(x)v (x) =
dx
(u(x)v(x)) − u (x)v(x)
dx
0
0
L Z L
dx u0 (x)v(x).
= u(x)v(x) −
u(x)v 0 (x) =
0
(28)
0
Colloquially, we say that this operation “flips” the derivative from v(x) to
u(x). (Hopefully, the remaining integrand is known!)
Applying integration by parts to our case in eq. (26):
2πm
0
u=x
v = sin
x
L
L
2πm
0
cos
x
u =1
v=−
2πn
L
10
and,
Z
L
dx x sin
0
2πn
x
L
xL
=−
cos
2πn
2πn
x
L
L
Z L
2πn
L
dx cos
x
+
2πn 0
L
0
L2
=−
2πn
thus
L
.
πn
Applying integration by parts, we can also show an = 0 for a 6= 0.
All together, we have
X
∞
BL
2πn
BL
−
sin
x .
f (x) = A +
2
πn
L
bn = −
(29)
(30)
n=1
This result is plotted in Fig. 5, where the series has been truncated after
including a finite number of terms. It is quite clear, that additional terms
improve the quality of the Fourier expansion, and the series will ultimately
converge to f (x). It is common to refer to the individual terms contributing
L
to the Fourier sum as “Fourier modes”. From the result bn = − πn
and from
Fig. 5, it is clear that the contribution, or amplitude, of the higher-order
modes decreases as the “mode-number” n increases, explaining why this
sum converges to a reasonable approximation to the function f (x) after a
finite number of terms.
Three final notes on Fourier series. First, domain of Fourier series can
be chosen arbitrarily. It is commonly convenient to shift domain to be
symmetric about x = 0:
L L
x∈ − ,
2 2
In this case, form of Fourier series looks the same. Only formula for coefficients changes.
Z L
2
2
2πn
an =
x f (x),
(31)
dx cos
L −L
L
2
and similarly for the bn .
Second, notice that all terms in Fourier series are periodic under x →
x + nL (shift by length of domain). For this reason Fourier series especially
useful as a general representation of any periodic function. For example,
one may calculate the Fourier coefficients for a given function based on the
“projection operation” within a single domain, say from x − 0 to x = L
in Fig. 6. In crystalline materials, for example, the electron density is a
periodic function that is naturally described in as a Fourier spectrum, and
the modes of non-zero amplitude represent regions of strong-scattering by
diffraction.
11
Figure 5: Plot of f (x), in the domain of [0, L], with the Fourier series expansion truncated after including a different number of terms. Clearly, the
inclusion of higher order terms improves the overall approximation.
Figure 6: Plot of an infinitely periodic function. The Fourier series for
a single domain describes an infinite array of periodic copies of the same
function, translated by one domain length, L.
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Finally, recall that for a discontinuous (non-analytic) function, the Taylor series near to the point of discontinuity does not converge, providing
a poor “fit” to a discontinuous function. However, the convergence of the
Fourier serious does not require a function to be analytic. Any function,
even discontinuous functions, can be decomposed into a Fourier series that
converges as the number of terms included in the series goes to ∞.
13