# Peeter Joot's (OLD) Blog.

• ## Recent Comments peeterjoot on Curious problem using the vari… Tom on Curious problem using the vari… peeterjoot on Cartesian to spherical change… Someone on Cartesian to spherical change… peeterjoot on Cartesian to spherical change…

• 346,727

# Posts Tagged ‘Euler-MacLauren summation’

## Final version of my phy452.pdf notes posted

Posted by peeterjoot on September 5, 2013

I’d intended to rework the exam problems over the summer and make that the last update to my stat mech notes. However, I ended up studying world events and some other non-mainstream ideas intensively over the summer, and never got around to that final update.

Since I’m starting a new course (condensed matter) soon, I’ll end up having to focus on that, and have now posted a final version of my notes as is.

Since the last update the following additions were made

September 05, 2013 Large volume fermi gas density

April 30, 2013 Ultra relativistic spin zero condensation temperature

April 24, 2013 Low temperature Fermi gas chemical potential

## Bernoulli polynomials and numbers and Euler-MacLauren summation

Posted by peeterjoot on May 29, 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.)]

Motivation

In  I saw the Euler-summation formula casually used in a few places, allowing an approximation of a sum with derivatives at the origin. This rather powerful relationship was used in passing, and seemed like it was worth some exploration.

Bernoulli polynomials and numbers

Before tackling Euler summation, we first need to understand some properties of Bernoulli polynomials [], and Bernoulli numbers . The properties of interest required for the derivation of the Euler summation formula appear to follow fairly easily with the following choice for the definition of the Bernoulli polynomials $B_k(x)$ and Bernoulli numbers $B_k$ \begin{aligned}B_m(z) = \sum_{k = 0}^m \binom{m}{k} B_k z^{m - k}\end{aligned} \hspace{\stretch{1}}(1.0.1.1) \begin{aligned}0 = \sum_{k = 0}^{m-1} \binom{m}{k} B_k \frac{1}{{m!}}, \qquad \mbox{m > 1}\end{aligned} \hspace{\stretch{1}}(1.0.1.1)

It is conventional to fix $B_0 = 1$. Eq. 1.0.1.1 provides an iterative method to calculate all the higher Bernoulli numbers. Without calculating the Bernoulli numbers explicitly, we can relate these to the values of the polynomials at the origin \begin{aligned}\boxed{B_m(0) = B_m.}\end{aligned} \hspace{\stretch{1}}(1.0.2)

Now, let’s calculate the first few of these, to verify that we’ve got the conventions right. Starting with $m = 2$ we have \begin{aligned}0 = \sum_{k = 0}^{1} \binom{2}{k} B_k \frac{1}{{2!}}= \frac{1}{{2!}}\left( B_0 + 2 B_1 \right),\end{aligned} \hspace{\stretch{1}}(1.0.2)

or $B_1 = -1/2$. Next with $m = 3$ \begin{aligned}0 &= \sum_{k = 0}^{2} \binom{3}{k} B_k \frac{1}{{3!}} \\ &= \frac{B_0}{6} + \frac{B_1}{2} + \frac{B_2}{2} \\ &= \frac{1}{{2}} \left( \frac{1}{{3}} -\frac{1}{{2}} + B_2 \right)\end{aligned} \hspace{\stretch{1}}(1.0.2)

or $B_2 = 1/6$. Thus the first few Bernoulli polynomials are \begin{aligned}\begin{aligned}B_0(z) &= 1 \\ B_1(z) &= z - \frac{1}{{2}} \\ B_2(z) &= z^2 - z + \frac{1}{{6}}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.5a)

The Bernoulli polynomials have a simple relation to their derivative. Proceeding directly, taking derivatives we have \begin{aligned}B_m'(z) &= \sum_{k = 0}^{m-1} (m - k)\binom{m}{k} B_k z^{m - k -1} \\ &= \sum_{k = 0}^{m-1} \frac{m!}{(m - k - 1)! k!} B_k z^{m - k -1} \\ &= m\sum_{k = 0}^{m-1} \frac{(m - 1)!}{(m - 1 - k)! k!} B_k z^{m - 1 - k},\end{aligned} \hspace{\stretch{1}}(1.0.5a)

or \begin{aligned}\boxed{B_m'(z) = m B_{m-1}(z) }\end{aligned} \hspace{\stretch{1}}(1.0.7)

There’s a number of difference relations that the polynomials satisfy. The one that we need is \begin{aligned}\boxed{B_m(z + 1) - B_m(z) = m z^{m -1}.}\end{aligned} \hspace{\stretch{1}}(1.0.8)

To prepare for demonstrating this difference in general, let’s perform this calculation for the specific cases of $m = 1$ and $m = 3$ to remove some of the index abstraction from the mix. For $m = 1$ we have \begin{aligned}B_1(z + 1) - B_1(z) &= \sum_{k = 0}^1 \binom{1}{k} B_k \left(\left( z + 1 \right)^{1 - k}- z^{1 - k}\right) \\ &= B_0\left(\left( z + 1 \right)^1- z^1\right)+ 1B_1\left(\left( z + 1 \right)^0- z^0\right) \\ &= B_0 \\ &= 1.\end{aligned} \hspace{\stretch{1}}(1.0.8)

For $m = 3$ (a value of $m > 1$ that is representative) we have \begin{aligned}B_3(z + 1) - B_3(z) &= \sum_{k = 0}^3 \binom{3}{k} B_k \left(\left( z + 1 \right)^{3 - k}- z^{3 - k}\right) \\ &= B_0\left(\left( z + 1 \right)^3- z^3\right)+ 3B_1\left(\left( z + 1 \right)^2- z^2\right)+ 3B_2\left(\left( z + 1 \right)^1- z^1\right)+ B_3\not{{\left(\left( z + 1 \right)^0- z^0\right)}} \\ &= B_0\left(3 z^2 + 3 z + 1\right)+ 3B_1(2 z + 1)+ 3B_2 \\ &= 3 z^2 + z^1 \left( 3 - 3 \right)+ z^0 \left( 1 - \frac{3}{2} + \frac{3}{6} \right) \\ &= 3 z^2.\end{aligned} \hspace{\stretch{1}}(1.0.8)

Evaluating this in general, we see that the term with the highest order Bernoulli number is immediately killed, and we’ll have just one highest order monomial out of the mix. We expect all the remaining monomial terms to be killed term by term. That general difference is, for $m \ge 2$ is \begin{aligned}B_m(z + 1) - B_m(z) &= \sum_{k = 0}^{m - 1}\binom{m}{k} B_k \left(\left( z + 1\right)^{m - k}- z^{m - k}\right) \\ &= \sum_{k = 0}^{m - 1}\binom{m}{k} B_k \sum_{s = 0}^{m - k - 1} \binom{m - k}{s} z^s= m! \sum_{s = 0}^{m - 1}\frac{z^s}{s!}\sum_{k = 0}^{m - s - 1} \frac{1}{\not{{(m -k)!}} k!} \frac{\not{{(m - k)!}}}{(m - k - s)!} B_k \\ &= \frac{m! }{(m -1)!} z^{m - 1}\sum_{k = 0}^{m - m + 1 - 1} \frac{1}{ k! (m - k - m + 1)!} B_k +m! \sum_{s = 0}^{m - 2}\frac{z^s}{s!}\sum_{k = 0}^{m - s - 1} \frac{1}{ k! (m - k - s)!} B_k \\ &= m z^{m - 1}+ m! \sum_{s = 0}^{m - 2}\frac{z^s}{s!}\left( \sum_{k = 0}^{(m-s) - 1} \binom{m - s}{s} B_k \frac{1}{{(m - s)!}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.8)

This last sum up to $m -s -1$ has the form of eq. 1.0.1.1, so is killed off. This proves eq. 1.0.8 as desired.

From this difference result we find for $m > 1$ \begin{aligned}B_m(1) &= \sum_{k = 0}^m \binom{m}{k} B_k \\ &= m! \sum_{k = 0}^{m-1} \binom{m}{k} B_k \frac{1}{{m!}}+B_m \\ &= B_m,\end{aligned} \hspace{\stretch{1}}(1.0.8)

and for $m = 1$ \begin{aligned}B_1(1) = 1 + B_1(0) = 1 - 1/2 = -B_1.\end{aligned} \hspace{\stretch{1}}(1.0.8)

we find that either of the end points in the $[0, 1]$ interval provide us (up to a sign) with the Bernoulli numbers \begin{aligned}\boxed{B_m(1) = \left\{\begin{array}{l l}B_m & \quad m > 1 \\ -B_1 & \quad m = 1 \end{array}\right.}\end{aligned} \hspace{\stretch{1}}(1.0.14)

Integrating eq. 1.0.7 after an $m \rightarrow m + 1$ substitution, and comparing to the difference equation, we have \begin{aligned}(m + 1) z^m &= B_{m + 1}(z + 1) - B_{m + 1}(z) \\ &= (m + 1)\int_z^{z+1} B_m(z) dz,\end{aligned} \hspace{\stretch{1}}(1.0.15)

or \begin{aligned}\boxed{\int_z^{z+1} B_m(z) dz = z^m.}\end{aligned} \hspace{\stretch{1}}(1.0.16)

Evaluating this at $z = 0$ shows that our polynomials are odd functions around the center of the $[0, 1]$ interval, or \begin{aligned}\boxed{\int_0^{1} B_m(z) dz = 0.}\end{aligned} \hspace{\stretch{1}}(1.0.17)

We also obtain Bernoulli’s sum of powers result \begin{aligned}\int_0^n B_m(z) dz &= \int_0^1 B_m(z) dz+\int_1^2 B_m(z) dz+\cdots+\int_n^{n-1} B_m(z) dz \\ &= 0 + 1^m + 2^m + \cdots (n-1)^m,\end{aligned} \hspace{\stretch{1}}(1.0.17)

or \begin{aligned}\boxed{\sum_{k = 1}^{n-1} k^m = \int_1^n B_m(z) dz.}\end{aligned} \hspace{\stretch{1}}(1.0.19)

We don’t need this result for the Euler summation formula, but it’s cool!

To arrive at some of these results I’ve followed, in part, portions of the approach outlined in []. That treatment however, starts by deriving some difference calculus results and uses associated generating functions for a more abstract difference equation related to the Bernoulli polynomials. In this summary of relationships above, I’ve attempted to avoid any requirement to first study the difference equation formalism (although that is also cool too, and not actually that difficult).

Euler-MacLauren summation

Following wikipedia , we utilize the simple boundary conditions for the Bernoulli polynomials in the $[0, 1]$ interval. We can exploit these using integration by parts if we do a periodic extension of these polynomials in that interval.

Writing $\lfloor {x} \rfloor$ for the largest integer less than or equal to $x$, our periodical extension of the $[0, 1]$ interval Bernoulli polynomial is \begin{aligned}P_m(x) = B_m\left( x = \lfloor {x} \rfloor \right).\end{aligned} \hspace{\stretch{1}}(1.0.20)

From eq. 1.0.2 and eq. 1.0.14, our end points are \begin{aligned}P_m(1) = \left\{\begin{array}{l l}B_m(0) = B_m & \quad m > 1 \\ -B_1(0) = -B_1 & \quad m = 1 \end{array}\right.\end{aligned} \hspace{\stretch{1}}(1.0.21)

Utilizing eq. 1.0.7 we can integrate by parts in a specific unit interval \begin{aligned}\int_k^{k+1} f(x) dx &= \int_k^{k+1} f(x) P_0(x) dx \\ &= \int_k^{k+1} f(x) d \left( \frac{P_1(x)}{1} \right) \\ &= {\left.{{\left( f(x) P_1(x) \right)}}\right\vert}_{{k}}^{{k+1}}-\int_k^{k+1} f'(x) P_1(x) dx \\ &= - B_1 f(k+1) - B_1 f(k)-\int_k^{k+1} f'(x) P_1(x) \\ &= \frac{1}{{2}} \left( f(k+1) + f(k) \right)-\int_k^{k+1} f'(x) P_1(x)\end{aligned} \hspace{\stretch{1}}(1.0.21)

Summing gives us \begin{aligned}\int_0^{n} f(x) dx \\ &= \sum_{k = 0}^{n-1}\int_k^{k+1} f(x) dx \\ &= \frac{1}{{2}} f(0) + \sum_{k = 1}^{n-1} f(k) + f(n)-\int_0^{n} f'(x) P_1(x) dx,\end{aligned} \hspace{\stretch{1}}(1.0.21)

or \begin{aligned}\sum_{k = 0}^{n} f(k)=\int_0^{n} f(x) dx+\frac{1}{{2}} \left( f(0) + f(n) \right)-\int_0^{n} f'(x) P_1(x) dx.\end{aligned} \hspace{\stretch{1}}(1.0.26)

Continuing the integration by parts we have \begin{aligned}\int_0^{n} f'(x) P_1(x) dx \\ &= \sum_{k = 0}^{n-1}\int_k^{k+1} f'(x) P_1(x) dx \\ &= \sum_{k = 0}^{n-1}\int_k^{k+1} f'(x) d \left( \frac{P_2(x)}{2} \right) \\ &= \sum_{k = 0}^{n-1}\frac{B_2}{2} \left( f'(k+1) - f'(k) \right)-\sum_{k = 0}^{n-1}\int_k^{k+1} f''(x) \frac{P_2(x)}{2} dx \\ &= \frac{B_2}{2} \left( f'(n) - f'(0) \right)-\int_0^{n} f''(x) \frac{P_2(x)}{2} dx \\ &= \frac{B_2}{2} \left( f'(n) - f'(0) \right)-\frac{B_3}{3!} \left( f''(n) - f''(0) \right)+\int_0^{n} f'''(x) \frac{P_3(x)}{3!} dx \\ &= \sum_{s = 1}^{m}(-1)^{s-1}\frac{B_{s+1}}{(s+1)!} \left( f^s(n) - f^s(0) \right)+(-1)^{m-1}\int_0^{n} f^m(x) \frac{P_m(x)}{m!} dx,\end{aligned} \hspace{\stretch{1}}(1.0.21)

or \begin{aligned}\boxed{\begin{aligned}\sum_{k = 0}^{n} f(k)&=\int_0^{n} f(x) dx+\frac{1}{{2}} \left( f(0) + f(n) \right)\\ &+\sum_{s = 1}^{m}(-1)^{s}\frac{B_{s+1}}{(s+1)!} \left( f^s(n) - f^s(0) \right)+(-1)^{m}\int_0^{n} f^m(x) \frac{P_m(x)}{m!} dx.\end{aligned}}\end{aligned} \hspace{\stretch{1}}(1.0.26)

# References

\bibitem[Behnke et al.(1974)Behnke, Gerike, and Gould]behnke1974fundamentalsV3Heinrich Behnke, Helmuth Gerike, and Sydney Henry Gould. Fundamentals of mathematics, volume 3. MIT Press, 1974.

 RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.

 Wikipedia. Bernoulli number — wikipedia, the free encyclopedia, 2013\natexlab{a}. URL http://en.wikipedia.org/w/index.php?title=Bernoulli_number&oldid=556109551. [Online; accessed 28-May-2013].

 Wikipedia. Bernoulli polynomials — wikipedia, the free encyclopedia, 2013\natexlab{b}. URL http://en.wikipedia.org/w/index.php?title=Bernoulli_polynomials&oldid=548729909. [Online; accessed 28-May-2013].

 Wikipedia. Euler-maclaurin formula — wikipedia, the free encyclopedia, 2013\natexlab{c}. URL http://en.wikipedia.org/w/index.php?title=Euler%E2%80%93Maclaurin_formula&oldid=552061467. [Online; accessed 28-May-2013].