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# Posts Tagged ‘Taylor expansion’

## A final pre-exam update of my notes compilation for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on April 22, 2013

Here’s my third update of my notes compilation for this course, including all of the following:

April 21, 2013 Fermi function expansion for thermodynamic quantities

April 20, 2013 Relativistic Fermi Gas

April 10, 2013 Non integral binomial coefficient

April 10, 2013 energy distribution around mean energy

April 09, 2013 Velocity volume element to momentum volume element

April 04, 2013 Phonon modes

April 03, 2013 BEC and phonons

April 03, 2013 Max entropy, fugacity, and Fermi gas

April 02, 2013 Bosons

April 02, 2013 Relativisitic density of states

March 28, 2013 Bosons

plus everything detailed in the description of my previous update and before.

## Non integral binomial coefficient

Posted by peeterjoot on April 10, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Updated.  Original generalization of binomial coefficient wasn’t correct for negative exponents.

In [2] appendix section F was the use of binomial coefficients in a non-integral binomial expansion. This surprised me, since I’d never seen that before. However, on reflection, this is a very sensible notation, provided the binomial coefficients are defined in terms of the gamma function. Let’s explore this little detail explicitly.

Taylor series

\begin{aligned}f(x) = (a + x)^b.\end{aligned} \hspace{\stretch{1}}(1.1)

Our derivatives are

\begin{aligned}\begin{aligned}f'(x) &= b (a + x)^{b-1} \\ f''(x) &= b(b-1) (a + x)^{b-2} \\ f^3(x) &= b(b-1)(b-(3-1)) (a + x)^{b-3} \\ \dot{v}s & \\ f^k(x) &= b(b-1)\cdots (b-(k-1)) (a + x)^{b-k}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.2)

Our Taylor series is then

\begin{aligned}(a + x)^b = \sum_{k = 0}^\infty \frac{1}{{k!}} b(b-1)\cdots (b-(k-1)) a^{b-k} x^k.\end{aligned} \hspace{\stretch{1}}(1.0.2)

Note that if $b$ is a positive integer, then all the elements of this series become zero at $b = k - 1$, or

\begin{aligned}(a + x)^b = \sum_{k = 0}^k \frac{1}{{k!}} b(b-1)\cdots (b-(k-1)) a^{b-k} x^k.\end{aligned} \hspace{\stretch{1}}(1.0.2)

Gamma function

Let’s now relate this to the gamma function. From [1] section 6.1.1 we have

\begin{aligned}\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} dt.\end{aligned} \hspace{\stretch{1}}(1.0.2)

Iteratively integrating by parts, we find the usual relation between gamma functions of integral separation

\begin{aligned}\Gamma(z + 1) &= \int_0^\infty t^{z} e^{-t} dt \\ &= \int_0^\infty t^{z} d \left( \frac{ e^{-t}}{-1} \right) \\ &= {\left.{{t^z \frac{e^{-t}}{-1}}}\right\vert}_{{0}}^{{\infty}}- \int_0^\infty z t^{z-1} \frac{e^{-t}}{-1} dt \\ &= z \int_0^\infty t^{z-1} e^{-t} dt \\ &= z (z - 1)\int_0^\infty t^{z-2} e^{-t} dt \\ &= z (z - 1)(z - (3-1))\int_0^\infty t^{z-3} e^{-t} dt \\ &= z (z - 1) \cdots (z - (k-1))\int_0^\infty t^{z-k} e^{-t} dt \\ &= z (z - 1) \cdots (z - (k-1))\int_0^\infty t^{(z + 1 - k) - 1} e^{-t} dt,\end{aligned} \hspace{\stretch{1}}(1.0.2)

or

\begin{aligned}\Gamma(z + 1) = z (z - 1) \cdots (z - (k-1)) \Gamma( z - (k-1) ).\end{aligned} \hspace{\stretch{1}}(1.0.2)

Flipping this gives us a nice closed form expression for the products of a number of positive unit separated values

\begin{aligned}z (z - 1) \cdots (z - k)=\frac{\Gamma(z + 1)}{\Gamma( z - (k-1) )}.\end{aligned} \hspace{\stretch{1}}(1.0.2)

Binomial coefficient for positive exponents

Considering first positive exponents $b$, we can now use this in our Taylor expansion eq. 1.0.2

\begin{aligned}(a + x)^b &= \sum_{k = 0}^\infty \frac{1}{{k!}} \frac{\Gamma(b+1)}{\Gamma(b - k + 1)}a^{b-k} x^k \\ &= \sum_{k = 0}^\infty \frac{\Gamma(b + 1)}{\Gamma(k + 1)\Gamma(b - k + 1)}a^{b-k} x^k.\end{aligned} \hspace{\stretch{1}}(1.0.2)

Observe that when $b$ is a positive integer we have

\begin{aligned}\frac{\Gamma(b + 1)}{\Gamma(k + 1)\Gamma(b - k + 1)} &= \frac{b!}{k!(b-k)!} \\ &= \binom{b}{k}.\end{aligned} \hspace{\stretch{1}}(1.0.10)

So for positive values of $b$, even non-integer values, we see that is then very reasonable to define the binomial coefficient \index{binomial coefficient} explicitly in terms of the gamma function

\begin{aligned}\binom{b}{k} \equiv\frac{\Gamma(b + 1)}{\Gamma(k + 1)\Gamma(b - k + 1)}.\end{aligned} \hspace{\stretch{1}}(1.0.11)

If we do that, then the binomial expansion for non-integral values of $b$ is simply

\begin{aligned}(a + x)^b = \sum_{k = 0}^\infty \binom{b}{k} a^{b-k} x^k.\end{aligned} \hspace{\stretch{1}}(1.0.11)

Binomial coefficient for negative integer exponents

Using the relation eq. 1.0.11 blindly leads to some trouble, since $\Gamma(-\left\lvert {m} \right\rvert)$ goes to infinity for integer values of $m > 0$. We have to modify the definition of the binomial coefficient. Let’s rewrite eq. 1.0.2 for negative integer values of $b = -m$ as

\begin{aligned}(a + x)^{-m} &= \sum_{k = 0}^\infty \frac{1}{{k!}} (-m)(-m-1)\cdots (-m-(k-1)) a^{-m-k} x^k \\ &= \sum_{k = 0}^\infty \frac{1}{{k!}} (-1)^km(m+1)\cdots (m+(k-1)) a^{-m-k} x^k.\end{aligned} \hspace{\stretch{1}}(1.0.11)

Let’s also put the ratio of gamma functions relation of eq. 1.0.2, in a slightly more general form. For $u, v > 0$, where $u - v$ is an integer, we can write

\begin{aligned}u (u - 1) \cdots (v) = \frac{\Gamma(u + 1)}{\Gamma( v )}.\end{aligned} \hspace{\stretch{1}}(1.0.11)

Our Taylor series takes the form

\begin{aligned}(a + x)^{-m} &= \sum_{k = 0}^\infty \frac{1}{{k!}} (-1)^k(m+k-1) (m+k-2) \cdots (m)a^{-m-k} x^k \\ &= \sum_{k = 0}^\infty (-1)^k \frac{ \Gamma(m + k) }{\Gamma(m)\Gamma(k+1)}a^{-m-k} x^k.\end{aligned} \hspace{\stretch{1}}(1.0.16)

We can now define, for negative integers $-m$

\begin{aligned}\binom{-m}{k}\equiv(-1)^k \frac{ \Gamma(m + k) }{ \Gamma(m)\Gamma(k+1) }.\end{aligned} \hspace{\stretch{1}}(1.0.16)

With such a definition, our Taylor series takes the tidy form

\begin{aligned}(a + x)^{-m} = \sum_{k = 0}^\infty \binom{-m}{k} a^{-m-k} x^k.\end{aligned} \hspace{\stretch{1}}(1.0.16)

For negative integer values of $b = -m$, this is now consistent with eq. 1.0.11.

Observe that we can put eq. 1.0.16 into the standard binomial form with a bit of manipulation

\begin{aligned}\binom{-m}{k} &= (-1)^k \frac{ \Gamma(m + k) }{ \Gamma(m)\Gamma(k+1) } \\ &= (-1)^k \frac{ (m + k -1)! }{ (m-1)! k! } \\ &= (-1)^k \frac{ m (m + k)! }{ (m + k) m! k! },\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}\binom{-m}{k}=(-1)^k \frac{m}{m + k} \binom{m+k}{m}.\end{aligned} \hspace{\stretch{1}}(1.0.19)

Negative non-integral binomial coefficients

TODO. There will be some ugliness due to the changes of sign in the products $b(b-1)\cdots (b -k + 1)$ since $b$ and $b -k + 1$ may have different sign. A product of two ratios of gamma functions will be required to express this product, which will further complicate the definition of binomial coefficient.

# References

[1] M. Abramowitz and I.A. Stegun. \emph{Handbook of mathematical functions with formulas, graphs, and mathematical tables}, volume 55. Dover publications, 1964.

[2] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.