# 1 Series, averages, partition functions, DNA melting, atom entropy#

# import all python add-ons etc that will be needed later on
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
from sympy import *
init_printing()                      # allows printing of SymPy results in typeset maths format
plt.rcParams.update({'font.size': 14})  # set font size for plots


## 1.1 Power series#

The series expansions of exponential, sine, and cosine functions have been described in earlier chapters and there is nothing unusual or special about these. Any regular function can be expanded in a similar manner in powers of $$x$$, to produce its unique power series, which is in general

$\displaystyle f=a_0 +a_1x+a_2x^2 +a_3x^3 +\cdots = \sum_n a_nx^n$

where the $$a$$’s are positive or negative constants. The capital $$\sum$$ is used to indicate a summation, with sub- and superscripts as necessary. In the reverse process, some series can be summed to form a function. The power series

$\displaystyle 1+x+x^2 +x^3 +x^4 +\cdots+x^n +$

is a geometric series, as the ratio of consecutive terms is constant, and is summed to the algebraic expression,

$\displaystyle \frac{1}{1-x} =1+x+x^2 +x^3 +x^4 +\cdots+x^n +\cdots \qquad\tag {1}$

provided that $$| x | \lt 1$$ where $$| x |$$ is the absolute values of $$x$$. If this condition is not true then the expansion tends to infinity and gives the wrong answer.

The summation is easily found to a finite number of $$N$$ terms, let

$\displaystyle S_N=1+x+x^2 +x^3 +x^4 +\cdots+x^N$

multiply both sides by $$x$$ and subtract the two expressions

\begin{split}\displaystyle \begin{align}xS_N& = x+x^2 +x^3 +x^4 +\cdots+x^{N+1}\\ (1-x)S_n&=(1+x+x^2 +x^3 +x^4 +\cdots+x^N)-(x+x^2 +x^3 +x^4 +\cdots+x^{N+1}) \\&=(1-x^{N+1})\end{align}\end{split}

and the summation is

$\displaystyle \sum_{n=0}^N x^n=\frac{1-x^{N+1}}{1-x}$

Taken to infinity the summation becomes

$\displaystyle\sum_{n=0}^\infty x^n = \frac{1}{1-x } \qquad\tag{2}$

but only if $$-1\lt x\lt 1$$.

If the summation starts at $$m<N$$ then

$\displaystyle \sum_{n=1}^N x^n=\frac{x^m-x^{N+1}}{1-x}$

The function $$f(x) = 1/(1-x)$$ (eqn. 2) can be represented by the polynomial series equation (1), again this is only true if $$-1\, x \, \lt 1$$. The polynomial is calculated as

$\displaystyle f(0) = 1, \quad f(1/2) = 1 + 1/2 + 1/4 + 1/8 + \cdots$

which looks as though it will converge as terms are becoming increasingly smaller as the series grows. Convergence always needs to be checked. If $$x$$ is greater than 1 then the series obviously increases to infinity, which is clear to see from

$\displaystyle f (2) = 1 + 2 + 4 + 8 + \cdots$

The expansion of $$1/(1 + x)$$ and of $$1/(1 + ax)$$, for example, follows immediately by substitution of $$x \rightarrow -x$$ and $$x \rightarrow ax$$ into the series of equation (1). In the latter case for the series to be valid $$|\, ax \,| < 1$$, where $$a$$ is a constant.

## 1.2 Summation of waves#

The summation of waves occurs in fourier analysis, a wave in its general form is $$e^{-i\omega t}$$ which propagates at frequency $$\omega$$ and time $$t$$. The sum of many waves of different frequency and in phase with one another produces a pulse, this is how a pulse is formed in a mode-locked laser. It is also how diffraction from an array of scatterers, (electrons in atoms) produces a spot in a x-ray diffraction pattern from a single crystal. The sum to $$N$$ terms is formed from waves at frequencies $$n\omega$$,

$\displaystyle S=\sum_{n=-N}^N e^{-in\omega t}$

where $$i=\sqrt{-1}$$. The limits here are $$-N\to N$$ and this changes the summation to $$\displaystyle \frac{-x^{N+1}+x^N}{1-x}$$ as may be worked out following the method above. Substituting $$x\to e^{-i\omega t}$$ gives

$\displaystyle S=\frac{-e^{i(N+1)\omega t}+e^{-iN\omega t}}{1-e^{i\omega t}}$

this can be further simplified by multiplying top and bottom of the fraction by $$e^{-i\omega t/2}$$, which is in effect multiplying by one.

$\displaystyle S = \frac{e^{-i\omega(N+1/2) t} - e^{i\omega (N+1/2)t}}{e^{-i\omega t/2}-e^{+i\omega t/2}}$

and using the definition of a sine in terms of exponentials $$\sin(x)=(e^{ix}-e^{-ix})/2i$$ gives

$\displaystyle S=\frac{\sin((N+1/2)\omega t)}{\sin(\omega t/2)}$

which is a series of spikes separated by $$2\pi/\omega$$, see fig. 21b chapter 9 (Fourier transforms).

## 1.3 Diffraction#

Suppose that there is a line of identical oscillators each emitting waves in the arrangement of figure 0. These could be a row of slits in a transmission grating or reflective strips in a reflection grating but in either case the sources are all in phase and emit over all angles. Thy could also be water waves or scattered off atoms by x-rays. If we observe at some distant point, a distance many times larger than the grating period $$d$$, and at an angle $$theta$$ we would like to know how the intensity of the summed waves varies as $$theta$$ is changed. At some angles the waves will be in phase with one another and so produce a wave of maximum amplitude, less so at another angle and at yet other angles be completely out of phase. Thus we expect to repeatedly see bright and dark regions as the angle is changed.

The calculation to do this means adding up the path lengths from each emitter and seeing what the total path is. The total number of wavelengths does not matter just what the surplus over a whole number of waves is, as this determines the phase and hence amplitude.

Using figure 0, the difference in path-length between $$r_1$$ and $$r_0$$ is $$r_1-r_0=\delta$$ and at the angle drawn $$2\delta=\lambda$$. The distance between $$r_2-r_1=2\delta$$ and so on thus $$n\delta=r_n-r_0$$. Figure 0. An array of $$N$$ coherent oscillators separated by spacing $$d$$. The path difference measured from the first oscillator is $$n\delta$$ where $$n=0,1\cdots N-1$$. At the angle shown $$2\delta =\lambda$$ where $$\lambda$$ is the wavelength.

The general form is a wave is $$e^{i(kr-\omega t)}$$ where $$\omega$$ is the frequency and $$k=2\pi/\lambda$$ is the wavevector. If all the emitters have the same amplitude then the total wave is

$\displaystyle E=E_0\left(e^{i(kr_0-\omega t)}+e^{i(kr_1-\omega t)}+\cdots +e^{i(kr_{N-1}-\omega t)}\right)$

which is simplified by factoring out $$e^{-i\omega t}$$ to give

$\displaystyle E=E_0e^{-i\omega t}\left(e^{ikr_0}+e^{ikr_1}+\cdots +e^{ikr_{N-1}}\right)$

The term in brackets can be further simplified by factoring out the first term in $$r_0$$ and using the definition of $$\delta$$ worked out from figure 0. This produces

$\displaystyle E=E_0e^{-i\omega t}e^{ikr_0}\left(1+e^{ik\delta}+e^{2ik\delta} +\cdots +e^{(N-1)ik\delta}\right)$

If we substitute $$x=e^{ik\delta}$$, this series becomes $$\displaystyle \sum_{n=0}^{N-1} x^n=\frac{1-x^N}{1-x}$$ and by substituting back gives

$\displaystyle E=E_0e^{-i\omega t}e^{ikr_0}\frac{e^{ik\delta N}-1}{e^{ik\delta}-1}$

The intensity measure on a detector, such as a CCD, is $$I=E^*E$$ giving

$\displaystyle I=I_0\left(\frac{e^{-ik\delta N}-1}{e^{-ik\delta}-1}\right)\left(\frac{e^{ik\delta N}-1}{e^{ik\delta}-1} \right) =I_0\frac{2-e^{-ik\delta N}-e^{ik\delta N}}{2-e^{-ik\delta}-e^{ik\delta }}$

Converting the exponential to cosines using $$2\cos(x)=e^{-x}+e^{-ix}$$ gives

$\displaystyle I=I_0\frac{1-\cos(k\delta N)}{1-\cos(k\delta)}$

The maximum intensity can be found when $$\delta =0$$ because all the oscillators with be in phase at $$\theta=0$$. Differentiating w.r.t. $$\delta$$, putting $$\delta =0$$ and using l’Hopital’s rule, where the top and bottom is differentiated separately until the ratio is not $$0/0$$, gives

$\displaystyle I(max)\to \frac{N\sin(k\delta N)}{\sin(k\delta)} \to N^2\frac{\cos(k\delta N)}{\cos(k\delta)}\to N^2$

and the maximum intensity is $$I_{max}=N^2I_0$$ as intuition would dictate for $$N$$ identical oscillators. However, by the nature of cosines we expect the waves to come into phase each time the cosine has a value of $$1$$ and this will happen when $$k\delta =2\pi m$$ where $$m=0,\pm 1,\pm 2\cdots$$ where $$m$$ is a positive integer.

The path length difference is related to the angle $$\theta$$ as

$\displaystyle \delta = d\sin(\theta)$

then the intensity becomes

$\displaystyle I=I_0\frac{1-\cos(Nkd\sin(\theta))}{1-\cos(kd\sin(\theta))}$

and has maxima at $$\theta=0,\pm \pi, \pm 2\pi\cdots$$.

This last equation is more often put into a sine form using the identity $$1-\cos(x)=2\sin^2(x/2)$$ which produces

$\displaystyle I=I_0\frac{\sin^2(Nkd\sin(\theta)/2)}{\sin^2(kd\sin(\theta)/2)}$

which is a sinc function, i.e. a central lobe plus decreasing oscillations to either side.

## 2 Making new series#

(i) New series can often be made from existing ones. Suppose each term in the series of $$f(x)=1+x+x^2+\cdots$$ is differentiated, then using eqn 2,

$\displaystyle f^′(x)= \frac{1}{(1 - x)^2} =1+2x+3x^2 +4x^3 +\cdots+nx^{n-1} +\cdots \qquad\tag{3}$

and if $$| \,x\, | \lt 1$$ this is the expansion of $$1/(1 - x)^2$$, which can be written as

$\displaystyle \sum_{n=0}^{\infty} nx^{n-1} = \frac{1}{(1 - x)^2} \qquad\tag{4}$

(ii) Substituting $$x \to -x$$ gives the series

$\displaystyle \frac{1}{(1 + x)^2} =1-2x+3x^2-4x^3 +\cdots \qquad\tag{5}$

The summation expression is

$\displaystyle \sum_{n=1}^\infty (-1)^n nx^{n-1} = \frac{1}{(1+x)^2}$

and the $$\displaystyle (-1)^n$$ ensures that alternate terms are positive and negative.

(iii) The sum

$\displaystyle \sum_{n=0}^\infty (1-x)x^n = 1,\qquad |x|\lt 1$

and the new sum $$\displaystyle \sum_{n=0}^\infty x(1-x)^n$$ can be found with this result by substituting $$y=1-x$$ then

$\displaystyle \sum_{n=0}^\infty y(1-y)^n = 1,\qquad |y-1|\lt 1$

but notice that there are now new limits. We can of course change notation and replace $$y$$ with $$x$$ or $$p$$ or any other symbol.

## 2.1 Table of Summations#

$\begin{split}\displaystyle \begin{array}{l|l|l} \hline \sum_0^\infty\limits x^n = \displaystyle\frac{1}{1-x},\quad |x|<1& \sum_0^\infty\limits (-x)^n = \displaystyle\frac{1}{1+x},\quad |x|<1& \sum_0^\infty\limits nx^n =\displaystyle\frac{x}{(1-x)^2},\quad |x|<1\\ \sum_0^\infty\limits nx(1-x)^n=\displaystyle \frac{1-x}{x},\quad |x-1|<1& \sum_0^\infty\limits (1-x)x^n =1,\quad |x|<1& \sum_0^\infty\limits n(1-x)x^n =\displaystyle\frac{x}{1-x},\quad |x|<1\\ \sum_0^\infty\limits nx(1-x)^{n-1} =\displaystyle\frac{1}{x},\quad |x-1|<1& \sum_0^\infty\limits x(1-x)^{n-1} =\displaystyle\frac{1}{1-x},\quad |x-1|<1& \sum_0^\infty\limits n(1-x)x^{n-1} =\displaystyle\frac{1}{1-x},\quad |x|<1\\ \hline \end{array}\end{split}$

Figure 1, below, shows the function of equation (3) and several approximations which are ever longer series with increasing powers of $$x$$. Expanding the series as far as $$x^6$$ only matches the $$f(x)$$ well up to about $$x = 0.5$$ indicating the rather obvious fact that that many terms may be needed to accurately reproduce a function. The plot also shows haw a function may be approximated by its first couple of terms when the variable $$x$$ is very small. This is a very common procedure to account for a perturbation.

# power series calculation

fig1 = plt.figure(figsize=(5, 5))
plt.rcParams.update({'font.size': 14})  # set font size for plots

m    = 8                             # maximum power of series is m
numx = 100                           # number of points to plot
maxy = 15                            # max y axis
s = [0.0 for i in range(numx)]       # make array of zeros
x = np.linspace(0.0,0.999,numx)

sers = lambda x,j: sum( n*x**(n-1) for n in range(1,j+1) )  # sum =1+2x+3x^2 + + nx^(n-1) ... 1/(1-x)^2
j = 1
while j <= m:
for i in range(numx):
s[i]= sers(x[i],j)
plt.plot(x,s)
j = j + 1
pass

plt.plot(x, 1.0/(1.0 - x)**2,color='black')            # exact expression

txt = [ ' n = '+ str(i) for i in range(1,m+1) ]
for i, q in enumerate(txt):
yval = sers(x[numx-1],i+1)
if yval< maxy:
plt.text(1.0,yval,q,verticalalignment='center')
pass
plt.title(r"$f\;'(x)=1+2x+3x^2+\cdots+nx^{n-1}+\cdots$",fontsize=14)  # r"$...$" uses markdown
plt.xlabel(r'$x$')
plt.ylabel(r"$f\;'(x)$")
plt.axis([0.0,1.0,0.0,maxy] )
plt.tight_layout()
plt.show() Figure 1. The function $$1/(1-x)^2$$ (top black line) and its series approximations $$\Sigma_n nx^{n-1}$$ with $$n$$ = 1, 2, $$\cdots$$.

## 3 Convergence#

In summing a series, it is important to ensure that it converges to some sensible expression such as $$1/(1 - x)$$, and is not going to be infinite or undefined. With many series, the summation is infinite and cannot therefore be expressed in a simple form. There are a number of convergence tests to ensure that a series has a finite result; many of these are complex, but a ratio test is a good way of determining if the series will be finite. In the ratio test, the ratio of any term $$w$$ in the series to its preceding term is calculated

$\displaystyle r=\lim_{n=+\infty} \left| \frac{w_{n+1}}{w^n} \right|$

and then the limit is taken as $$n$$ tends to infinity. If the $$(n + 1)$$th term is smaller than the $$n^th$$, then the series is converging. The second step is to check whether $$r \lt 1$$, as $$n$$ goes to infinity.

In the series expansion of $$\displaystyle (1 - x)^{-1} = 1 + x + x^2 \cdots$$ any individual term in the series is $$x^n$$ so the ratio $$\displaystyle r = \lim_{n \rightarrow \infty} \left| \frac{x^{n+1}}{x^n} \right | = |x|$$ is less than 1 only if $$|x| \lt 1$$; therefore when this is true the series converges. The interval of convergence for this sum is 0 $$\le x \lt 1$$, and the radius of convergence $$s = 1$$. For any other values this series diverges.

Should the radius $$s$$ be infinity the series is convergent for all values of $$x$$; for instance, the exponential series is convergent for all $$x$$. Testing for convergence properly is far more complicated than this example leads us to believe and a maths textbook should be consulted to understand how to do this rigorously.

In general, a series converges if $$x$$ is smaller than some number $$s$$, so that $$|x_0 - x| \lt s$$ where $$x_0$$ is a displacement about which the series converges; see Section 6 describing Taylor series for examples of this. The number $$s$$ can be determined from the convergence test, Fig. 2. A useful property is that if a power series converges to some number and if each term of it is differentiated, the resulting series can be shown also to converge.

## 4 Average quantities#

Any quantity, $$x$$, which is one of a set of measurements has an average value

$\displaystyle \langle x \rangle= \frac{\sum xp(x)}{\sum p(x)} \qquad\tag{6}$

and this is also called the first moment of the distribution $$p(x)$$; the second moment is $$\langle x^2 \rangle$$, the fourth $$\langle x^4 \rangle$$, etc. The average of $$x^2$$ is

$\displaystyle \langle x^2 \rangle = \frac{\sum x^2p(x)}{\sum p(x)} \qquad\tag{7}$

The variance or standard deviation squared of $$x$$ is

$\displaystyle \sigma^2 =\langle x^2 \rangle -\langle x \rangle^2$

A more meaningful description is that $$\sigma$$ is the dispersion in the value of $$x$$, meaning it is the spread in its value; experimental values are often quoted as $$\langle x \rangle \pm \sigma$$. The chapter on integration has examples of averaging using integrals rather than summations, and variance is described in Chapter 13 Data analysis.

## 4.1 Average energy of a molecule or atom#

The average energy is described by

$\displaystyle \langle E \rangle = \frac{\sum Ep(E)}{\sum p(E)} \qquad\tag{8}$

where the summation covers all the energy states and $$p(E)$$ is the distribution function, which is very often a thermal or Boltzmann distribution, $$\displaystyle p(E) = e^{-E/k_BT}$$ with temperature $$T$$ and Boltzmann’s constant $$k_B$$. In special situations this might be replaced by the Bose-Einstein or Fermi-Dirac distribution among others.

If the lowest energy is zero and the largest infinity and $$E$$ has discrete integer values, in units of $$k_BT$$, the average energy is, by definition,

$\displaystyle \langle E \rangle=\frac{\sum_{E=0}^\infty\limits Ee^{-E/k_BT}}{\sum_{E=0}^\infty\limits e^{-E/k_BT}}\qquad\tag{8a}$

The denominator is called the partition function (see Section 4) and using the property of exponentials $$\displaystyle e^{nx} = (e^x)^n$$ is written as

$\displaystyle Z=\sum_{E=0}^\infty {\left(e^{-1/k_BT}\right)^E} =\frac{1}{1 - e^{-1/k_BT}}\qquad\tag{8b}$

based on the summation $$\displaystyle \sum_{n=0}^\infty x^n = (1-x)^{-1}$$ and by substituting $$\displaystyle x = e^{-1/k_BT}$$.

Some guile is now needed in writing the numerator, which becomes $$\displaystyle e^{-1/k_BT} \sum_{E=0}^\infty E(e^{-1/k_BT})^{E-1}$$. Notice that $$\displaystyle e^{-1/k_BT}$$ is placed outside the summation to get the summation into the mathematical form $$\displaystyle \sum_{E=0}^\infty Ex^{E-1}= (1-x)^{-2}$$, see eqn. (4), which when $$x$$ is substituted gives,

$\displaystyle \sum_{E=0}^\infty Ee^{-E/k_BT}= \frac{e^{-1/k_BT}}{(1-e^{-1/k_BT})^2}\qquad\tag{8c}$

Dividing by the partition function $$Z$$ produces the average energy

$\displaystyle \langle E \rangle = \frac{e^{-1/k_BT}}{1-e^{-1/k_BT}} = \frac{1}{e^{1/{k_BT}}-1}$

This result is interesting and evaluating by expanding the exponential assuming $$k_BT \gg 1$$ produces $$k_BT$$ which is $$\approx 208\;\mathrm{cm^{-1}}$$ at $$300$$ K. This means that every molecule has the same total average energy when $$k_BT \gg 1$$. However, this is not the case if just individual energy levels are considered, such as vibrations or rotations. Rewriting the partition function to sum over an index rather than $$E$$ which means that $$E_i$$ has individual values depending on the particular type of energy level in the molecule or atom, making equation 8b,

$\displaystyle Z = \sum _i e^{-E_i/k_BT}$

and if differentiated with respect to $$T$$,

$\displaystyle \frac{dZ}{dT}=\frac{d}{dT}\sum_i e^{-E_i/k_BT}=\frac{1}{k_BT^2} \sum_i E_ie^{-E_i/k_BT}$

which is related to the average energy from equation 8c; $$\langle E\rangle = \sum_{i} E_ie^{-E_/k_BT}/Z$$ and therefore by substituting with the derivative and rearranging the average energy is

$\displaystyle \langle E\rangle =k_BT^2\frac{1}{Z}\frac{dZ}{dT}=k_BT^2\frac{d\log(Z)}{dT}\qquad\tag{8d}$

If rotational levels are considered typically several tens of levels are populated at room temperature and the summation therefore involves many terms, for example if the rotational constant $$B=1\;\mathrm{cm^{-1}}$$, $$\approx 55$$ levels are needed to produce a constant average energy of $$\approx 14500\;\mathrm{cm^{-1}}$$. A vibration of $$1000\;\mathrm{cm^{-1}}$$ levels only to $$n=2$$ are needed to produce four figure accuracy of $$500.4 \; \mathrm{ cm^{-1}}$$.

## 4.2 Polymers#

In synthesizing polymers by radical or condensation polymerisation, the random nature of the chemistry that adds monomers to an already growing chain dictates that a range of polymer lengths is normally produced; the polymer is poly-disperse. Two different averages are frequently taken to characterize the polymer; one is the number average mass, the other the weight average mass.

The number average mass $$m_n$$ is

$\displaystyle \langle m_n \rangle = \frac{\sum_k m_kp(n_k)}{\sum_k p(n_k)} = \frac{n_1m_1+n_2m_2 +\cdots }{n_1+n_2+\cdots} \qquad\tag{9}$

where $$p$$ is the distribution of the number of polymers of length $$k$$ and $$m_k$$ is the mass of each polymer $$k$$. The sum $$\sum p(n_k) = N$$ the total number of molecules.

The weight (mass)average mass is

$\displaystyle \langle m_w \rangle = \frac{\sum_k \limits m_k\cdot m_kp(n_k)}{\sum_k\limits m_kp(n_k)} = \frac{n_1m_1^2+n_2m_2^2 +\cdots }{n_1m_1+n_2m_2+\cdots} \qquad\tag{10}$

If the number average and mass (weight) average is the same, the polymer is mono-disperse, however, this is not usually the case. To illustrate that these two averages are different, suppose the number distribution, which may not necessarily relate to a particular polymer but has approximately the right shape, is $$\displaystyle p(n_k) = n_ke^{-an_k}$$ where $$a$$ is a positive constant roughly describing the width of the distribution. If $$n_k = k$$ then, by definition, the number of moles of polymers with length $$1$$ is $$1$$ and the number of length $$2$$ is $$2$$ and so forth. The mass of the polymer with $$k$$ segments is $$m_k = km_0$$, if $$m_0$$ is the mass of a mole of monomer units, then the number average, if $$k$$ is taken to infinity, is

$\displaystyle \langle m_n\rangle = m_0 \frac{\sum_{k=1}\limits n_k^2e^{-an_k}}{\sum_{k=1}\limits n_ke^{-an_k}} = m_0\frac{e^a+1}{e^a-1}$

In contrast the weight average is

$\displaystyle \langle m_w\rangle = m_0 \frac{\sum_{k=1}\limits n_k^3e^{-an_k}}{\sum_{k=1}\limits n_k^2e^{-an_k}} = m_0\frac{4e^a+e^{2a}+1}{e^{2a}-1}$

which is clearly not the same as $$\langle m_n \rangle$$. The two averages are approximately linear, for $$a<1$$, if plotted on a log-log plot and have a negative slope, with $$\langle m_w\rangle \gt \langle m_n\rangle$$. When $$a \gt 1$$ the two curves converge to a limiting value of $$m_0$$.

## 4.3 Sequence repeats in DNA#

Our DNA is packaged into $$46$$ chromosomes (22 pairs and 2 sex chromosomes) but each chromosome is still too vast to analyse and so it is broken into more manageable pieces by mechanical or chemical or enzymatic means. If we want to identify a particular sequence, say of $$n$$ base pairs we will want to estimate what is the average separation of these is so that a suitable number of fragments are present containing the sequence. It is assumed that the DNA is initially so long, $$\sim 10^6$$ base pairs, that there are equal amounts of GCAT bases. If the sequence to identify is $$n=8$$ bases long and is GGCATGGA then the chance of finding the first G is $$1/4$$ and of the second adjacent G is $$1/16$$ and as each event is independent of any other the probabilities multiply to give $$\displaystyle p=\frac{1}{4}\frac{1}{4}\frac{1}{4}\cdots= \frac{1}{4^n}$$, thus for $$n=8$$ the chance of finding any specified consecutive sequence is very small $$1/4^8=1/65536$$. This is true whatever the sequence is as long as the bases are in a fixed order. To detect an 8-mer the fragment will, of course, have to have on average at least these many bases since we do not know before hand if the sequence is present. What then is the average distance from one sequence to the next? The probability of not seeing the next sequence in $$m$$ moves is $$(1-1/4^n)^{m-1}$$ and then seeing one in the next move is the product

$\displaystyle p(m)=\left(1-\frac{1}{4^n}\right)^{m-1}\frac{1}{4^n}$

With the substitution $$p=1/4^n$$ the average is therefore

$\mu= \frac{\sum_{m=0}^\infty\limits m\left(1-p\right)^{m-1}p}{\sum_{m=0}^\infty\limits \left(1-p\right)^{m-1}p}$

The normalisation (denominator) is the sum $$\displaystyle \frac{1}{1-p}\sum_{m=0}^\infty\limits \left(1-p\right)^{m}p= \frac{1}{1-p}\to 1$$ because $$p\lt\lt 1$$. The numerator is $$\displaystyle \frac{1}{1-p}\sum_{m=0}^\infty\limits mp\left(1-p\right)^m= \frac{1}{p}$$ so the average is

$\displaystyle \mu = \frac{1}{p}=4^n$

which means that when the sequence is long a huge number of base pairs are needed, on average, if the sequence is to be present more than once. This is important in PCR where a sequence of perhaps $$15 \to 30$$ primer base pairs are used to identify and so amplify the DNA. This primer sequence is so long that the probability of this occurring by random is tiny in fragmented DNA with only a few thousand base pairs. This makes the PCR amplification very specific indeed, i.e if the target sequence is not there, amplification is highly unlikely to occur by random chance.

## 5 Partition Functions#

One of the commonest uses of the summation of a series of terms is in the calculation of the partition function. These are met formally in statistical mechanics but are also used in kinetic theory and in spectroscopy.

The partition function is the sum over states of the distribution of energy levels usually assumed to follow Boltzmann’s exponential law (or distribution), which describes the probability

$\displaystyle pdE = ge^{-E/k_BT}dE$

that an atom or molecule has energy in the range $$E$$ to $$E + dE$$ and where $$g$$ is the degeneracy of the energy level, $$k_B$$ is the Boltzmann constant ($$1.3805 \cdot 10^{-23}\,\mathrm{ J K^{-1} }$$) and $$T$$ the temperature. This distribution is obtained by considering the most probable way of arranging a fixed number of particles among all the available energy levels in a constant volume and with a constant total energy. Related distributions, but applicable only in certain circumstances, are the Fermi-Dirac and Bose-Einstein.

The German word for the partition function, Zustandssumme, translates as ‘sum over states’. In English, it is called the partition function because it describes how much energy is in one level compared to the total, i.e. how energy is partitioned when the system, an ensemble of molecules for example, is in contact with a heat bath. A distribution based on being in thermal equilibrium with a heat bath is also called a canonical distribution. Once the partition function is calculated all the various thermodynamic properties, such as heat capacity, entropy and free energy, can be calculated.

The partition function is the summation over all the energy levels

$\displaystyle Z = \sum_{i=0}^m g_ie^{-E_i/k_BT} \qquad\tag{11}$

which extends over all energy levels from $$i = 0 \cdots m$$. The energy of the $$i^th$$ level is $$E_i$$, its degeneracy $$g_i$$. A state is degenerate if it has more than one energy level, the number of these levels is called the multiplicity. The vibrational levels in molecules are singly or non-degenerate, the rotational levels are degenerate and have a multiplicity that increases with the quantum number and is $$g_J = 2J + 1$$ where $$J$$ is the quantum number. In some situations, such as the harmonic oscillator, there is an infinite number of levels to sum in the partition function and $$m = \infty$$; in others, such as when an atom is in a magnetic field, there is a finite number.

The fractional population in any one level is the ratio

$\displaystyle f_i = \frac{n_i}{N}=\frac{g_ie^{-E_i/k_BT}}{\sum_{i=0}^m\limits g_ie^{E_i/k_BT}} \qquad\tag{12}$

and the partition function normalizes the expression. $$N$$ is the total number of particles. The partition function assumes that the energy is based on a thermal scale starting at zero, if this is not the case plausible answers can be obtained, which on scrutiny may prove to be wrong. The way to avoid this is always to make the energy scale start at zero; Q10 illustrates this effect.

We can derive the fraction in another way. As the total probability that a ‘particle’ has to be in one of possibly many states must be one, $$\sum_i p_i = 1$$ where $$p_i$$ is the probability of being in state $$i$$ and is proportional to the statistical weight from that state and hence

$p=\alpha g_i e^{-E_i/k_BT}$

where $$\alpha$$ is the normalisation constant. The total probability is therefore

$\sum_i p_i=\alpha \sum_i g_ie^{-E_i/k_BT}=1$

and therefore

$\alpha = \frac{1}{\sum_i g_ie^{-E_i/k_BT} }$

Conventionally we write $$\alpha=1/Z$$ and the probability that the particle is in level $$i$$ becomes

$\displaystyle p_i = \frac{g_ie^{-E_i/k_BT}}{Z}$

which is equation 12. The partition function, $$Z$$, is the normalization term and determines what fraction of the total energy is in each level or how energy is partitioned among the various level

## 5.1 Bond rotation#

The use of the word ‘particle’ is rather general, it might represent a molecule in a given vibrational or rotational energy level, a nuclear spin state in an NMR experiment, or the torsional energy of a restricted rotor as in the alkyl chain of hydrocarbons (Jackson 2006). In this example, consider butane, which has trans and gauche configurations as shown in figure 1a. These have different energies due to interactions between the protons on carbon $$1$$ and $$4$$ as bond C2 to C3 rotates. Figure 1a Bond angles in butane. The angles, $$0,120,240$$ represent the position of $$C_4$$.

The energy of the trans state at $$0^\text{o}$$ is taken as zero, $$E_0 = 0$$. The energy of the other two conformations, measured relative to this, have minima at $$120^\text{o}$$ and $$240^\text{o}$$ and are $$E_{120}$$ and $$E_{240}$$. The partition function is $$\displaystyle Z = 1 + e^{-E_{120}/ k_BT} + e^{-E_{240}/ k_BT}$$. The probability of being in the trans state is $$1/Z$$ and of being in the $$120^\text{o}$$ gauche state is $$\displaystyle e^{-E_{120}/k_BT}/Z$$ which is less than at $$0^\text{o}$$ but changes with temperature. By symmetry, the energy of the $$120$$ and $$240$$ states are the same and the probability of being in either is $$\displaystyle 2e^{-E_g / k_BT}/Z$$ where $$E_g$$ is an abbreviation for $$E_{120}$$ and $$E_{240}$$. These two levels are accidentally degenerate. Once the partition function has been found, the average energy follows easily because by definition $$\langle E \rangle = \sum_i E_i p_i$$ then

$\displaystyle \langle E \rangle =\frac{2e_ge^{-E_g/k_BT}}{Z}$

because $$E_0 = 0$$. The average length and the average of the square of the length (which will lead to the standard deviation in the length for different conformers) are

\begin{split}\displaystyle \begin{align}\langle L \rangle &=\sum_i L_ip_i=\frac{1}{Z}\left(L_0e^{-E_0/k_BT} +2L_ge^{-E_g/k_BT} \right),\\ \langle L^2 \rangle &=\sum_i L_i^2p_i=\frac{1}{Z}\left(L_0^2e^{-E_0/k_BT} +2L_g^2e^{-E_g/k_BT} \right)\end{align}\end{split}

If the alkyl chain is longer then the probability of the final state being a trans or gauche configuration is multiplied by however many repeat units there are, the partition function becoming $$Z^n$$ for $$n$$ units.

As an example, if $$E_g /k_BT = 1$$, which would be the case for a small barrier between trans to gauche or at a high temperature, then the partition function $$Z = 1.736$$. If there are $$10$$ repeat units in an alkane then the chance of an all-trans chain is $$1/Z^{10} = 0.004$$ compared to $$0.58$$ for butane. If $$E_g /k_BT$$ is larger, which it will be at a low temperature, then the probability of being in the all-trans state is almost one, even for a long alkyl chain because $$Z \to 1$$ at a low temperature. Figure 1b. Probability of an all trans chain vs the ratio $$E_g/k_BT$$ for butane and then for a ten fold longer chain of butane like repeat units. Large $$E_g/k_BT$$ corresponds to low temperature or small energy $$E_g$$.

## 5.2 DNA melting: Zipper model#

DNA can melt if the temperature is raised a little above room temperature, $$60 \to 70^\text{o}$$C would be typical depending on the composition. As a very simple model(see sketch) suppose that as the double stranded molecule unfolds the separated residues become free to move about on the backbone, with degeneracy $$g$$, but linked residues remain fixed in place. One end of the double stranded molecule is fixed thus the molecule can only unfold from the other. Link $$n$$ can only open if all those before it are already open, i.e. links $$n=0,1,2,\cdots, N-1$$ and the energy to do this is $$\epsilon$$ so that $$n$$ open links have energy $$n\epsilon$$. Closed links have energy zero and the energy needed to open a link other than that next to already open ones is infinite. This model was first examined by C. Kittel Am. J. Physics, v37, p917, 1969.

To open $$n$$ links requires energy $$n\epsilon$$ and those open links have degeneracy $$g^n$$. The partition function (equation 11) is therefore

$\displaystyle Z = \sum_{n=0}^{N-1} g^ne^{-n\epsilon/k_BT}$

The limit extends to $$N-1$$ as the last link remains intact. To evaluate this expression let $$x\equiv e^{-\epsilon /k_BT}$$ then

$\displaystyle Z=\sum_{n=0}^{N-1} (gx)^n$

which is the summation of a geometric series $$\sum_{i=0}^n r^i= (1-r^{n+1})/(1-r)$$ thus the partition function evaluates to

$\displaystyle Z=\frac{1-(ge^{-\epsilon/k_BT})^N}{1-ge^{-\epsilon/k_BT}}\qquad\tag{12 a}$

As the average energy is $$\langle E\rangle = \epsilon \langle n\rangle$$ the number of open links can be found via $$\langle e\rangle$$. The relationship between average energy (in thermodynamic terms the internal energy $$U$$) and partition function is equation 8d, using $$d\log(Z)\equiv dZ/Z$$

$\displaystyle \langle E\rangle =\frac{k_BT^2}{Z} \frac{d Z}{dT}$

thus the average number of open links is

$\displaystyle \langle n\rangle =\frac{k_BT^2}{\epsilon Z} \frac{d Z}{dT}$

Using python Sympy to do the algebra gives

g, N, epsilon, k, T = symbols('g, N, epsilon, k, T')

Z   = (1 - (g**N*exp(-N*epsilon/(k*T))))/(1 - g*exp(-epsilon/(k*T)))
avn = (1/epsilon)*simplify( ( k*T**2*diff(Z,T)/Z) )
avn which can be simplified by dividing common factors

$\displaystyle \langle n\rangle = \frac{Ng^N}{g^N-e^{N\epsilon/k_BT}} -\frac{g}{g-e^{\epsilon/k_BT}}=\frac{N}{1-(e^{\epsilon/k_BT}/g)^N} -\frac{1}{1-e^{\epsilon/k_BT}/g}\qquad\tag{12 b}$

whose behaviour is easier to see in a plot of $$\langle n\rangle$$ vs. $$\epsilon/k_BT$$. Figure 1c. Average length of free links in the zipper model $$\langle n\rangle$$ vs. dimensionless energy $$\epsilon/k_BT$$. The vertical line is at $$\ln(g)=\epsilon/k_BT$$.

At low values of $$\epsilon/k_BT$$, which means at high temperature or zero energy to unfold, almost the whole zipper is unfolded, as intuition would suggest Conversely at large $$\epsilon/k_BT$$, low temperature or large unfolding energy almost all bonds are intact. The transition between the two is quite sharp and occurs at $$\ln(g)=\epsilon/k_BT$$ as can be seen in equation 12b. The heat capacity which is $$d\langle E\rangle/dT$$ shows a sharp peak at this value. This overall behaviour somewhat unexpected, the model does not immediately suggest that a ‘phase transition’ should occur. In effect it means that either the zipper is closed or it is fully open and hardly ever in between.

## 5.3 Entropy of atoms and molecules#

In the gas phase the entropy of atoms can be calculated knowing the contribution from translation and electronic terms. The calculation is in two parts, first the partition function must be evaluated and then used to obtain the entropy.

The electronic part of the entropy is not zero if there are low lying electronic states as there are in carbon atoms. These are populated according the the Boltzmann distribution and increasingly so with temperature increase. The partition function and therefore the entropy due to translational motion similarly increases as the temperature increases. This occurs because more energy levels become populated and thus there are more ways to arrange the available energy.

To calculate the entropy we start with the partition function as in eqn. 11. The partition function for electronic energy levels is straightforward and is

$\displaystyle Z = \sum_J (2J+1)e^{-E_J/k_BT}\qquad\tag{12c}$

where $$J$$ is the total angular momentum of the electrons consisting of both spin and orbital parts and the degeneracy is $$g = 2J+1$$. The term symbol contains the information and has the forms $$\displaystyle ^{(2S+1)}A_J$$ where $$S$$ is the total spin, so $$2S+1$$ is the spin degeneracy as singlet ($$S=0$$), doublet ($$S=1/2$$), triplet ($$S=1$$) etc. The letter $$A$$ indicates the orbital angular momentum $$L$$ and if $$L=0$$ then $$A=S$$, if $$L=1$$ $$A=P$$ etc. and the subscript $$J$$ is the total which ranges as $$J=|L+S|\cdots \to \cdots |L-S|$$ in unit steps. Thus $$^1D_2$$ is a state that has total spin $$S=0$$, orbital angular momentum $$L=2$$ and total angular momentum $$J=2$$.

The spectroscopic data for carbon atoms can be found on the NIST WebBook website and the lowest few levels in cm$$^{-1}$$ are

$\begin{split}\displaystyle \begin{array}\\ \hline \text{config} & \text{term symbol} & \text{energy } /\mathrm{cm^{-1}}\\ \hline 2s^2.2p^2& ^3P_0& 0.0000000 \\ 2s^2.2p^2& ^3P_1& 16.4167130\\ 2s^2.2p^2& ^3P_2& 43.4134567\\ 2s^2.2p^2& ^1D_2& 10192.657 \\ 2s^2.2p^2& ^1S_0& 21648.030 \\ 2s^1.2p^3& ^5S_2& 33735.121 \\ \hline \end{array}\end{split}$

In calculating the partition function for ‘internal’ energy, the levels at several thousand wavenumbers above the ground state contribute very little to the summation due to the exponential becoming very small. At $$300$$ K, $$k_BT = 207\,\mathrm{cm^{-1}}$$ therefore the energy level at $$10192.6\,\mathrm{cm^{-1}}$$, and also those at higher energy, contributes insignificantly, e.g $$(2\times 2+1)e^{-10192.6/207} \approx 2\cdot 10^{-21}$$ so is utterly negligible. Only the first three terms need be included making the partition function $$Z_{int}= 18.24$$. This is a small value compared to the translational value.

In a molecule, there are in addition to electronic states, vibrational and rotational energy levels and these, particularly the latter, can be very numerous. If there are $$N$$ atoms there are $$3N-6$$ vibrational normal modes if the molecule is non-linear and $$3N-5$$ if it is. Each vibrational level has a stack of rotational levels these being $$2J+1$$ degenerate with rigid rotor energy $$2B(J+1)$$ where $$J$$ is the rotational quantum number. The rotational constant $$B$$ is often small, just a fraction to a few a wavenumbers being typical. The vibrational frequency and rotational constant for I$$_2$$ and CO molecules are respectively $$(214.57, 0.03735)$$ and $$(2214.24,1.9313)\,\mathrm{cm^{-1}}$$.

The translational part of the partition function is found by examining the energy levels of a particle in a box. As there is no potential energy these levels give the kinetic energy. The box has infinitely high walls with zero potential between them. We shall suppose that the atoms are in boxes whose size is very large compared to atomic dimensions so that the energy levels effectively become continuous. In a box these levels are

$\displaystyle \epsilon_x= \frac{h^2}{8m}\left(\frac{n_x}{a^2}+\frac{n_y}{b^2}+\frac{n_z}{c^2}\right)$

where $$a,b,c$$ is the lengths of the sides, $$m$$ the atom’s mass and $$n_x=n_y=n_z =1,2,\cdots$$ are the quantum numbers.

The partition function from eqn. 11 is the triple summation over the quantum numbers in $$x,y,z$$

$\displaystyle Z=\sum_{n_x}^\infty\sum_{n_y}^\infty\sum_{n_z}^\infty \large {e^{-(h^2/8mk_BT)(n_x^2/a^2+n_y^2/b^2+n_z^2/c^2) }}$

which, because the summation in the exponential can be split into the product of three similar summations, form the product $$Z=z_xz_yz_z$$ where

$\displaystyle z_x=\sum_{n_x}^\infty e^{-(h^2/8mk_BT)(n_x^2/a^2) }$

and similarly for $$z_y,z_z$$. The energy levels in translational motion are too close together to be observable and thus the summation can be replaced with an integration (from zero) without introducing any significant error. With the substitution $$w=h^2/8ma^2k_BT$$ the change is

$\displaystyle z_x=\sum_{n_x}^\infty e^{-wn_x^2 } \qquad \longrightarrow \qquad \int_0^\infty e^{-wn^2} dn=\frac{1}{2}\sqrt{\frac{\pi}{w}}$

The final partition function is the product of three similar terms and substituting back gives

$\displaystyle Z= V\left(\frac{2\pi m k_BT}{h^2}\right)^{3/2}\qquad\tag{12d}$

where the volume is $$V=abc$$. Substituting in values for the partition function produces a very large number. The mass must be in kg, $$k_B$$ in J/K and temperature in degrees Kelvin. The volume can be found using the ideal gas law with pressure for $$1$$ atm. expressed in Pascals; $$V=RT/p$$ where $$p=101325$$ Pa. The result for carbon atoms is $$Z = 10^{30}$$. As we shall see the entropy has terms in $$\ln(Z)$$ so the difference in the electronic and translational partition functions is not so great as it first seems.

The entropy calculated by Statistical Mechanics is, for an ideal monoatomic gas of $$N$$ atoms,

$\displaystyle S= Nk_B\ln \left(\frac{Z}{N} \right)+ \frac{U}{T} + Nk_B \qquad\tag{12e}$

where $$U$$ is the internal energy where

$\displaystyle U = Nk_BT^2 \left(\frac{\partial \ln(Z)}{\partial T} \right)_V\qquad\tag{12f}$

Differentiating to obtain $$U$$ gives

$\displaystyle U= Nk_BT^2\frac{3}{2T}= \frac{3}{2}Nk_BT$

which is the same result as from the kinetic theory of gases for a monoatomic gas and at equilibrium produces energy of $$k_BT/2$$ per degree of freedom.

After substituting and some rearranging the entropy is

$\displaystyle S=Nk_B\left(\frac{3}{2}\ln(T)+ \ln\left(\frac{V}{N} \right)+\frac{3}{2}\ln\left(\frac{2\pi mk_B}{h^2} \right) +\frac{5}{2}\right)\qquad\tag{12g}$

and in terms of pressure substitute $$V=Nk_BT/p$$.

If we take a mole of gas then $$N\to N_A$$ Avogadro’s number and $$Nk_B=R$$ the gas constant, then

$\displaystyle S=R\left(\frac{5}{2}\ln(T)+ \ln\left(\frac{k_B}{p} \right)+\frac{3}{2}\ln\left(\frac{2\pi mk_B}{h^2} \right) +\frac{5}{2}\right)\qquad\tag{12h}$

The thermodynamic calculation gives the entropy as

$\displaystyle S_{thermo}= \frac{3}{2}R\ln(T) +R\ln(V) +S_0$

which is the same as derived above but with the Statistical Mechanics calculation the constant $$S_0$$ is now determined. Additionally the dependence on mass is explained. This dependence is because the energy levels in the particle in a box depend inversely on the mass. As the mass increases more energy levels are available to be filled at a given temperature which produces more ways of distributing the energy among these levels which increases the entropy.

The translational entropy for carbon atoms is $$S_{trans} = 139.8$$ J/mol/K.

The calculation of the internal entropy, i.e. that due to energy levels within the atom or molecule is given by a related formula to 12e which is

$\displaystyle S= Nk_B\ln(Z)+ \frac{U}{T} \qquad\tag{12i}$

The internal energy (12f) is found most easily using $$\displaystyle \frac{d \ln(Z)}{dT}=\frac{1}{Z}\frac{dZ}{dT}$$

$\displaystyle U = Nk_BT \frac{\sum_{J=0}^\infty\limits E_J(2J+1)e^{-E_J/k_BT}}{\sum_{J=0}^\infty\limits (2J+1)e^{-E_J/k_BT}}$

The ratio of summations has the form of an average, see eqn. 8. This is to be expected in hindsight as we assume thermal equilibrium and so the energy over many levels should be the average value.

The entropy/mole becomes

$\displaystyle S_{int}= R\ln\left(\sum_{J=0}^\infty (2J+1)e^{-E_J/k_BT}\right) + \frac{N_A}{T} \frac{\sum_{J=0}^\infty\limits E_J(2J+1)e^{-E_J/k_BT}}{\sum_{J=0}^\infty\limits (2J+1)e^{-E_J/k_BT}}$

which more concisely is

$\displaystyle S_{int}= R\ln(Z) + \frac{N_A}{T} \langle E\rangle$

The values for the electronic energies of C atoms are given above making the contribution to the entropy at $$300$$ K from these states $$S_{int}= 18.24\; \mathrm{J/mol/K}$$ and the total entropy is $$S= 158.12$$ J/mol/K which is acceptably close to the experimental, calorimetric value of $$158.10$$. The calculation is shown below.

The translational entropy of a molecule will increase only as $$\ln(m)$$ eqn. 12h which is a relatively small change with increasing mass, however, the internal entropy changes by far more largely because of the many, possibly hundreds, of closely spaced rotational energy levels needed to be included in the calculation. In the case of CO the calculated and experimental values agree well, both at $$197.6$$ J/mol/K, but the case of I$$_2$$ where the internal entropy is $$88.7$$ J/mol/K and the translational one $$177.9$$ make the total $$266.5$$. This is quite a bit larger than the experimental value of $$260.7$$ J/mol/K. Possibly this is due to the formulas describing the energy levels not being sufficiently accurate for higher vibrational and rotational levels, since the same trend also occurs with chlorine but to a smaller extent.

# entropy of atoms
h   = 6.62607e-34  # J s
N_A = 6.02214e23   # 1/mol
kB  = 1.38065e-23  # J/K
amu = 1.66054e-27  # kg
cmJ = 1.96e-23     # wavenumber to joules
p   = 101325.0     # pascal
R   = N_A*kB       # J/mol/K

T = 300

# carbon from NIST WebBook
m = 12.011
En= [0.0000000 ,16.4167130,43.4134567]
J = [0,1,2]

V = R*T/p
Ztrans = V*(2*np.pi*m*amu*kB*T/h**2)**(3/2)
#print('{:s}{:10.4g}'.format('Z trans',Ztrans))

S_trans= R*(5/2*np.log(T) + np.log(kB/p) + 3/2*np.log(2*np.pi*m*amu*kB/h**2)  + 5/2 )
print('{:s} {:8.4f} {:s}'.format('S translation',S_trans, 'J/mol/K') )

ZE = 0.0
Z  = 0.0
for i in range(len(J)):
temp = (2*J[i]+1)*np.exp(-En[i]*cmJ/(kB*T))
Z  = Z + temp
ZE = ZE+ En[i]*cmJ*temp  # average energy

S_int = R*np.log(Z) + N_A*ZE/(T*Z)

print('{:s} {:8.4f} {:s}'.format('S internal', S_int,  ' J/mol/K') )
print('{:s} {:8.4f} {:s}'.format('S total', S_int+S_trans,' J/mol/K') )

S translation 139.8770 J/mol/K
S internal  18.2438  J/mol/K
S total 158.1208  J/mol/K