Math Methods for Polymer Physics

Lecture 1: Series Representations of Functions

Series analysis is an essential tool in polymer physics and physical sci-ences, in general. Though other broadly speaking, a series expansion allowsone to analyze an arbitrarily complicated function into the sum of a simplerset of functions. Though other series expansions exist, two are especiallyuseful: the Taylor series and Fourier series. In a crude way, we maythink of both series as 2 different ways of approximating, or “fitting”, agiven function to a simpler form. For further reading on Taylor series andFourier 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 repre-sentation of a function, say f(x), as an infinite power series in the polynomi-als, (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 complicatedfunction:

f(x) = ln(cosx2 + 2

)+(x

3

)3. (1)

This function is plotted in Fig. 1.Often it’s sufficient and useful to have a simpler description of the func-

tion in the neighborhood of some point, say x = x0. For many func-tions, you may replace f(x) with a power series expansion in polynomials of∆x = x− x0 distance from the reference point.

f(x) =∞∑n=0

an (∆x)n

= 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 + . . .︸ ︷︷ ︸0

= a0 (3)

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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 including1, 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 ′(x0) = a1 + 2a2∆x+ 3a3 (∆x)2 + . . .︸ ︷︷ ︸0

= a1 (4)

In general, we may show

an =1n!

dnfdxn

∣∣∣∣∣x=x0

. (5)

In order to find the Taylor series expansion we need only to take derivativesof 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 analyticfunctions.

Notice from eq. (2) that as x→ x0 and ∆x→ 0 the higher order (largen) terms in the power series expansion go to zero very quickly. Hence, ifone is interested in f(x) sufficiently close to x0, a Taylor series expansiontruncated to include only a few leading terms may often be sufficient toapproximate the function. Geometrically, we can think of this in terms of a“local” description of a function near x = x0.

f(x) = f(x0)︸ ︷︷ ︸constant

+ ∆xf ′(x0)︸ ︷︷ ︸linear

+(∆x)2

2!f ′′(x0)︸ ︷︷ ︸

parabolic

+(∆x)3

3!f ′′′(x0)︸ ︷︷ ︸

cubic

+ . . . (6)

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This shows that sufficiently close a point of interest that analytic functionsare well approximated by constant plus a sloped, linear correction plus aparabolic correction plus . . .. Further away from a given reference point atx = x0, the less and less a function looks like a straight line. In order to geta better a approximation, you need functions with more wiggles (e.g. higherorder polynomials).

Let’s try some examples.Example 1: Expand ln(x) in a Taylor series around x0 = 1.

a0 = ln 1 = 0

a1 =d

dxln(x)

∣∣∣∣∣x=1

=1x

∣∣∣∣∣x=1

= 1

a2 =d2

dx2ln(x)

∣∣∣∣∣x=1

=d

dx1x

∣∣∣∣∣x=1

= − 1x2

∣∣∣∣∣x=1

= −1

a3 =d3

dx3ln(x)

∣∣∣∣∣x=1

=2x3

∣∣∣∣∣x=1

= 2

In general, an = (−1)n(n− 1)! for n > 1.

ln(x) = (x− 1)− (x− 1)2

2+

(x− 1)3

3− (x− 1)4

4+ . . .

=∞∑n=1

(−1)n(x− 1)n

n(7)

Example 2: Expand1

1− xin a Taylor around x = 0.

a0 = 0

a1 =d

dx1

1− x

∣∣∣∣∣x=0

=1

(1− x)2

∣∣∣∣∣x=0

= 1!

a2 =d2

dx2

11− x

∣∣∣∣∣x=0

=2

(1− x)3

∣∣∣∣∣x=0

= 2!

a3 =d3

dx3

11− x

∣∣∣∣∣x=0

=1× 2× 3(1− x)4

∣∣∣∣∣x=0

= 3!

In general, an = n!. So1

1− x=∞∑n=0

xn (8)

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which is the well-known geometric series.

These particular series, eqs. (7) and (8), do not converge for all valuesof x. When the series does not converge, for some large enough ∆x, thesuccessive terms terms an(∆x)n become larger than that the sum of theprevious terms, meaning that adding more terms in the series expansiondoes not provide a better approximation, and the Taylor series fails to rep-resent the function. For ln(x) around x0 = 1 and 1

1− x around x0 = 0,these only converge for |∆x| < 1.

In general we may define Rc as the radius of convergence of the Taylorseries of f(x) around x = x0. If |∆x| < Rc, then

∑∞n=0 an(∆x)n = f(x).

Otherwise series does not provide a good approximation of f(x) (addingmore terms makes things worse).

There are some functions for which Rc →∞ and the Taylor series alwaysconverges. Important examples include ex, sinx, cosx. These functions arisein many contexts, so it is useful to commit these series to memory.

Example 3: Expand ex around x = 0. Well, first notice

dn

dxnex

∣∣∣∣∣x=0

= ex∣∣∣∣x=0

= 1

From eq. (5) this gives right away the Taylor series coefficient of ex

ex = 1 + x+x2

2!+x3

3!+x4

4!+ . . . =

∞∑n=0

xn

n!. (9)

The Taylor series representation of ex is a particularly useful way to see thatddx(ex) = ex. Indeed, it is reasonable to view

∑∞n=0

xn

n! as the definition ofex.

You should also commit expressions of sinx and cosx to memory. Theseconverge for all x:

sinx = x− x3

3!+x5

5!− x7

7!+ . . . (10)

cosx = 1− x2

2!+x4

4!− x6

6!+ . . . (11)

Notice that these expansions allow you to derive the following importantidentity,

eix = cosx+ i sinx, (12)

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),

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expanded around the point x = x0 and y = y0:

f(x, y) = f(x0, y0) + ∆x∂f

∂x+ ∆y

∂f

∂y

+12!

[(∆x)2

∂2f

∂x2+ 2∆x∆y

∂2f

∂x∂y+ (∆y)2

∂2f

∂y2

]+

13!

[(∆x)3

∂3f

∂x3+ 3(∆x)2∆y

∂3f

∂x2∂y

+3∆x(∆y)2∂3f

∂x∂y2+ (∆y)3

∂3f

∂y3

]+ . . . (13)

where ∆x = x − x0, ∆y = y − y0 and all partial derivatives are evaluatedat (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 physicalsystems, 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 func-tion. By “local”, we mean, sufficiently close to some set of values for theindependent 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 forcesgenerated 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 fieldis simply, U = −1

2p · E, while the energy required to polarize the bead isUpolarization = |p|2/(2α). Therefore, if the polarizable bead is subject to anelectric field E(x) that varies in space (as near the focal point of a laserbeam, the net electrostatic interaction between the bead and the field isdescribed by the potential energy,

U(x) = −α2|E(x)|2. (14)

Hence, the potential energy is lowest in regions where the electric-field in-tensity, |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 rathercomplicated. But, if we are interested only in the behavior very close to thecenter of the trap, the behavior always has the same simple form,

U(∆x) = −U0︸︷︷︸constant

+U1∆x︸ ︷︷ ︸=0

+U2

2(∆x)2︸ ︷︷ ︸

quadratic

+ . . . . (15)

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Figure 2: Top: a schematic depiction of a polarizable bead near to the high-intensity focal point of a laser beam. Bottom: A sketch of U , the potentialenergy of a optically-trapped colloidal particle in terms of ∆x, the deviationfrom the center of the trap.

By definition, is minimum at ∆x = 0, so we knowdUdx

∣∣∣x=0

= U1 = 0. Thismeans that the force on the bead at the center of the trap is zero, because theelectric field intensity is maximal. Local equilibrium (mechanical, dynamic,etc.) always looks like this: constant + quadratic (first non-trivial term inexpression about equilibrium).

What is force if bead is displaced?

Fx = −dUdx

= −U2∆x = −k∆x (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 orsteady state), we are often interested in expression up to harmonic order.Therefore, if one “calibrates” the strength of optical trapping (the value ofk) and carefully measure ∆x, you can measure magnitude of external forcesthat pull a bead from the center of the trap, generated, say, by a strand ofDNA 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

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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 matterand radiation/waves (i.e. scattering) and spectral analysis

2. Sines and cosines are “harmonic functions”, which means they form acomplete basis of solutions to certain PDE’s common to the study ofphysical systems

Indeed, properties 1 and 2 are intimately related as the wave equation is har-monic, 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 problemsin continuum elasticity and polymer dynamics. For example, in the studyof polymer dynamics, we come across equations like,

d2

dn2R(n) + kR(n) = 0, (17)

where k > 0 describes a relaxation rate chain motion, and R(n) specifiesthe position of the bead n along a polymer chain. Since, d2

dn2 sin(√

kn)

=

−k sin(√

kn)

, sines and cosines form a natural set of solutions to this equa-tion. For the purposes of this review, a Fourier series is the unique decom-position of an arbitrary function (in some domain) into an infinite series ofsines 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:

f(x) =a0

2+∞∑n=1

an cos(

2πnL

x

)+∞∑n=1

bn sin(

2πnL

x

)(18)

an and bn are coefficients. Just as the coefficients of the Taylor series arerelated uniquely to the given function, an and bn are uniquely determinedby properties of f(x) on this domain.

How are an and bn related to f(x)? This relationship derives from animportant properties of sines and cosines. In particular, sin

(2πnL x

)and

cos(

2πnL x

)are orthogonal functions on this domain. This means that if a

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Figure 3: Plot of f(x), in the range of [0, L].

multiply any two of these elementary functions and integrate over the do-main x ∈ [0, L], the resulting integral is zero unless these functions are iden-tically. Consider two produce of two sine functions sin

(2πnL x

)sin(

2πmL x

):∫ L

0dx sin

(2πnL

x

)sin(

2πmL

x

)=

12

∫ L

0dx[cos(

2πxL

(n−m))− cos

(2πxL

(n+m))]

. (19)

This integral is only non-zero if n = m for which the first term in the inte-grand becomes cos

(2πxL (n−m)

)= 1. From this we can show the following

for the orthogonality between sines,∫ L

0dx sin

(2πnL

x

)sin(

2πmL

x

)=

{L2 if n = m

0 if n 6= m(20)

Similarly, for the cosines,∫ L

0dx cos

(2πnL

x

)cos(

2πmL

x

)=

{L2 if n = m 6= 00 if n 6= m

(21)

Sines and cosines are always orthogonal,∫ L

0dx sin

(2πnL

x

)cos(

2πmL

x

)= 0 for all m, n. (22)

The orthogonality relations, eqs. (20) - (22), are important becausethey allow one to invert the Fourier series, to determine the unique set of

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coefficients, an and bn, the correspond to the function f(x). Operationally,the coefficients of the Fourier series are determined by “projecting out” theterm in series proportional to, say, sin

(2πnL x

), by multiplying both sides

of eq. (18) by sin(

2πnL x

)and integrating the product over the domain

x ∈ [0, L]:∫ L

0dx f(x) sin

(2πmL

x

)=∫ L

0dx sin

(2πmL

x

)[a0

2+∞∑n=1

an cos(

2πnL

x

)+∞∑n=1

bn sin(

2πnL

x

)]Carrying out the integration, a0 and all cosine terms in sum will be zerodue to orthogonality conditions, eqs. (21) and (22). Likewise, all sine termsin sum except n = m term are zero too. Thus, the only term from theright-hand side of eq. (18) that survives this “projection” operation is fromn = m: ∫ L

0dx f(x) sin

(2πmL

x

)=L

2bm

and,

bn =2L

∫ L

0dx sin

(2πnL

x

)f(x) (23)

By performing the same operation with the cosine functions we can alsoderive,

an =2L

∫ L

0dx cos

(2πnL

x

)f(x) (24)

Example 4 Consider a function f(x) = A+Bx (see Fig. ??). Computecoefficients 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 mul-tiplying f(x) by cos

(2πnL x

)and integrating over the domain. For m = 0,

this is easy,

a0 =2L

∫ L

0dx (A+Bx) =

2L

[AL+

BL2

2

]= 2A+BL (25)

Now consider bn,

bn =2L

∫ L

0dx sin

(2πmL

x

)(A+Bx)

=2BL

∫ L

0dxx sin

(2πmL

x

)(26)

How do you do this integral? Let’s review a useful trick for evaluatingintegrals.

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Figure 4: Plot of f(x), in the domain of [0, L].

Aside: Integrations by parts

Let’s say you want to compute∫ L

0dxu(x)v′(x), and you don’t know the

anti-derivative of v′(x). The chain rule of differentiation gives you,

ddx

(u(x)v(x)) = u′(x)v(x) + u(x)v′(x) (27)

oru(x)v′(x) =

ddx

(u(x)v(x))− u′(x)v(x).

Substituting this expression for the integrand,∫ L

0dxu(x)v′(x) =

∫ L

0dx[

ddx

(u(x)v(x))− u′(x)v(x)]

= u(x)v(x)∣∣∣∣L0

−∫ L

0dxu′(x)v(x). (28)

Colloquially, we say that this operation “flips” the derivative from v(x) tou(x). (Hopefully, the remaining integrand is known!)

Applying integration by parts to our case in eq. (26):

u = x v′ = sin(

2πmL

x

)

u′ = 1 v = − L

2πncos(

2πmL

x

)

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and,∫ L

0dxx sin

(2πnL

x

)= − xL

2πncos(

2πnL

x

) ∣∣∣∣∣L

0

+L

2πn

∫ L

0dx cos

(2πnL

x

)= − L2

2πn

thusbn = − L

πn. (29)

Applying integration by parts, we can also show an = 0 for a 6= 0.All together, we have

f(x) =(A+

BL

2

)−∞∑n=1

BL

πnsin(

2πnL

x

). (30)

This result is plotted in Fig. 5, where the series has been truncated afterincluding a finite number of terms. It is quite clear, that additional termsimprove the quality of the Fourier expansion, and the series will ultimatelyconverge to f(x). It is common to refer to the individual terms contributingto the Fourier sum as “Fourier modes”. From the result bn = − L

πn and fromFig. 5, it is clear that the contribution, or amplitude, of the higher-ordermodes decreases as the “mode-number” n increases, explaining why thissum converges to a reasonable approximation to the function f(x) after afinite number of terms.

Three final notes on Fourier series. First, domain of Fourier series canbe chosen arbitrarily. It is commonly convenient to shift domain to besymmetric about x = 0:

x ∈[−L

2,L

2

]In this case, form of Fourier series looks the same. Only formula for coeffi-cients changes.

an =2L

∫ L2

−L2

dx cos(

2πnL

x

)f(x), (31)

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 especiallyuseful 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 = Lin Fig. 6. In crystalline materials, for example, the electron density is aperiodic function that is naturally described in as a Fourier spectrum, andthe modes of non-zero amplitude represent regions of strong-scattering bydiffraction.

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Figure 5: Plot of f(x), in the domain of [0, L], with the Fourier series ex-pansion truncated after including a different number of terms. Clearly, theinclusion of higher order terms improves the overall approximation.

Figure 6: Plot of an infinitely periodic function. The Fourier series fora single domain describes an infinite array of periodic copies of the samefunction, translated by one domain length, L.

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Finally, recall that for a discontinuous (non-analytic) function, the Tay-lor series near to the point of discontinuity does not converge, providinga poor “fit” to a discontinuous function. However, the convergence of theFourier serious does not require a function to be analytic. Any function,even discontinuous functions, can be decomposed into a Fourier series thatconverges as the number of terms included in the series goes to ∞.

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