Which is bigger?

Problem: Which is bigger…\(3^{\pi}\) or \({\pi}^{3}\)?

Solution: Write …\(\pi=3+\delta\), where \(\delta>0\) is the fractional part of …\(\pi\).  Then, in terms of \(\delta\):

\({\pi}^{3}=(3+\delta)^3=27+27 \delta+9 \delta^2+\delta^3\).

On the other hand…

\(3^{\pi}=3^{3+\delta}=27 \times 3^{\delta}=27 e^{\delta \log 3}\).

Now we make the observation that \(\log 3>1\), and use the expansion for the exponential function

\(e^x=1+x+\frac{1}{2} x^2+\frac{1}{6} x^3+\ldots\)

Given these facts, we can then state the following:

\(3^{\pi}>27+27 \delta+\frac{27}{2} \delta^2+\frac{9}{2} \delta^3>27+27 \delta+9 \delta^2+\delta^3\)

And therefore \(3^{\pi}>{\pi}^3\).

An interesting sequence and its limit

Problem: Consider a sequence of numbers \(a_m(j)\) such that \(a_m(j+1) = a_m(j)^2 + 2 a_m(j)\) for all \(j \geq 0\) and \(m \geq 0\) with \(a_m(0)=\frac{d}{2^m}\) for some real number \(d\). Determine \(\displaystyle\lim_{m \to +\infty}a_m(m)\).

Solution:  The above recursive relation can be rewritten as follows:


One of the most underrated tricks in solving recusive relations is to substitute in a new sequence whose pattern has a clear path to solution.  In this case the substitution glares at you. Let \(b_m(j)=1+a_m(j)\).  The new sequence satisfies the recurrence


and has an initial condition \(b_m(0)=1+\frac{d}{2^m}\).  We then can step through the first few cases:



until we get to the desired case:


Using the known limiting relation \(\displaystyle\lim_{m \to +\infty}(1+\frac{z}{m})^m=\exp{z}\), and the relation \(b_m(j)=1+a_m(j)\), we get the desired result:

\(\displaystyle\lim_{m \to +\infty}a_m(m)=\exp{d}-1\).

The Expanding Rail Problem


A railroad track is a mile in length. On a really hot day, the track expanded so that, while its endpoints were fixed in place, the track bowed into a circular arc. If the track expanded by a foot, about how far did the track deviate from its original position?



This problem is a classic illustration in how to do estimates and accurate computation.  I got it out of Forman Acton’s book, “Real Computing Made Real”.  This old chestnut is one that is simple to set up, but difficult to execute.

I take a different approach to Acton, who uses this to challenge the reader in his understanding of preserving significant figures.  His approach is entirely numerical.  I choose, rather, to illustrate the power of analysis to preserve the sig figs and to help feel my way around while doing so.

First, consider a crude estimate to help us gain some perspective.  Instead of the circular arc, imagine the track, originally of length \(a=5280\) is stretched like a violin string at the center to a length \(a+\varepsilon=5281\); it would then look the the straight line segments as in the dotted lines in the figure.  This is not the correct answer, but it puts us in the ballpark.  To find the deviation \(d\), we simply use Pythagoras:


We can then solve for \(d\) as usual.  Nevertheless, note that, as stated, \(a\) is 3 orders of magnitude larger than \(\varepsilon\).  We can therefore neglect the term \((\frac{\epsilon}{2})^2\) and write, with great [numerical] accuracy:

\(d \approx \sqrt{\frac{a \varepsilon}{2}}=51.38093 \ldots\)

The take away here is that \(\varepsilon \ll d\), while at the same time, \(d \ll a\).  More accurately stated, \(d=O[\sqrt{a \varepsilon}]\).  Those who guessed that \(d\) would be about 1 ft would be very wrong.  The neat thing about this problem is that such a small change results in a big result.

That said, we have not solved the correct problem and now need to get on with it.  To do this, we write down the geometric relations algebraically:

\(\sin \frac{\theta}{2}=\frac{a}{2 r}\),

Now that we are guided by orders of magnitude, we can scale our quantities appropriately. Let \(x=\frac{a}{2 r}\), \(y=\frac{d}{r}\), and \(\delta=\frac{\varepsilon}{a}\). We can then rewrite the above equations in a more compact form, eliminating \(\theta\):

\(y^2-2 y+x^2=0\).

We do not know a lot about the magnitude of the quantities \(x\) and \(y\), but we do know from the above analysis that \(y \ll x\), so even though we could solve the second equation exactly in terms of \(x\), we need not even do that. All we need to do is ignore the \(y^2\) term and we get, in a first approximation,


The quantity \(x\) is determined through the first equation and depends on the value of the small quantity \(\delta\). If we think about what to expect as a solution, note that this equation is transcendental and we cannot expect an exact solution. However, if we look at the limiting equation at \(\delta=0\), which is \(\sin{x}=x\). The only solution to that equation lies at \(x=0\). We then conclude that when \(\delta\) is small, so is \(x\).

This is very useful information, because it tells us how to go about approximating the solution to the first, transcendental equation. Because we now know that \(x\) is small, we can Taylor expand the sine term about \(x=0\) and get the following approximate equation:

\(x=x(1+\delta)-\frac{1}{6}x^3 (1+\delta)^3+O[x^5]\)

This can be solved easily for \(x\):

\(x=\sqrt{\frac{6 \delta}{(1+\delta)^3}}(1+O[\delta])\)

[That last factor \(O[\delta]\) comes from considering an expansion to a quadratic equation that would be solved if we considered the \(O[x^5]\) term. More on this later.]

In fact, we can even neglect the factor of \((1+\delta)^3\) in the denominator because it, too, adds a factor of \(O[\delta]\). Then we are simply left with

\(x=\sqrt{6 \delta}(1+O[\delta])\).

If we substitute back into the above equation for \(y\), and use the original definitions of the dimensionless parameters, we find, after a little rearrangement, the following first approximation for the desired deviation:

\(d=\frac{\sqrt{6}}{4}\sqrt{a \varepsilon}(1+O[\frac{\varepsilon}{a}])\).

Using the numbers provided, the deviation resulting from a 1 ft addition in length to the track is about 44.49719 ft, to within some factor of a constant times 0.019% error.

Is this good enough? Acton provides an answer good to 7 significant figures: 44.49845, which means the approximation is good for an error of 0.0028%, or 4 sig figs. This is fine in most cases, but let’s assume we want to reproduce Acton’s result. How do we do this?

We go back to the transcendental equation and Taylor expand out an additional term:

\(x=x(1+\delta)-\frac{1}{6}x^3 (1+\delta)^3+\frac{1}{120}x^5 (1+\delta)^5+O[x^7]\)

Now we can write that \(x=\sqrt{6 \delta}(1+B \delta+O[\delta^2])\) for some \(B\), for which we will solve by plugging into the above equation. When we do this, we note that all terms in \(x\) are of even powers, and terms of \(O[\delta]\) all cancel, so we are left with a coefficient of \(\delta^2\) equaling zero; this gives \(B=-\frac{27}{20}\). This result gets put into the quadratic equation above, where we now assume that \(y\) takes the form \(y=3 \delta (1+C \delta+O[\delta^2])\). Note that we include the \(y^2\) term now. Verifying that the terms of \(O[\delta]\) cancel, and equating the coefficient of \(\delta^2\) to zero, we get that \(C=-\frac{6}{5}\). When we go back to the original parameters, we find that

\(d=y r\),
\(\Rightarrow d=\frac{a}{2}\frac{3\delta(1+C\delta+O[\delta^2])}{\sqrt{6 \delta}(1+B\delta+O[\delta^2])}\),
\(\Rightarrow d=\frac{\sqrt{6}}{4}a\sqrt{\delta}(1+(C-B)\delta+O[\delta^2])\),

which gives us our correction:

\(d=\frac{\sqrt{6}}{4}\sqrt{a \varepsilon}(1+\frac{3}{20}\frac{\varepsilon}{a}+O[(\frac{\varepsilon}{a})^2])\).

This gives us \(d=44.498455 \ldots\), which is not too shabby.

Integrating binomial coefficients

Problem: Let \(C(z)\) = coefficient of \(x^{2010}\) in the expansion of \((1+x)^z\) about \(x=0\). What then is the value of

\(\int_{0}^{1}dy\) \(C(-y-1)\) \(\sum_{m=1}^{2010}\frac{1}{y+m}\) ?


Let \(k=2010\).  The value of the binomial coefficient desired is

\(C(z)=\binom{z}{k}=\frac{z(z-1)(z-2) \ldots (z-k+1)}{k!}\).

It then follows that

\(C(-y-1)=\frac{(-1)^k}{k!}(y+1)(y+2) \ldots (y+k)\).

If we let \(\frac{(-1)^k}{k!}p(y)=C(-y-1)\), then it turns out that



\(I(k)=\int_{0}^{1}dy \: C(-y-1)\sum_{m=1}^{k}\frac{1}{y+m}=\int_{0}^{1}dy \: p'(y)\). \(I(k)=\frac{(-1)^k}{k!}(p(1)-p(0))=\frac{(-1)^k}{k!}(2)(3) \ldots (k+1)\) \(I(k)=\frac{(-1)^k}{k!}(k+1)!=(-1)^k (k+1)\)

The answer is therefore 2011.

Nonlinear system producing periodic functions

Consider the following system:

\(y’\) = \(-z^3\); \(z’\) = \(y^3\); \(y[0]=1\), \(z[0]=0\).

Show that the solutions \(y=f[x]\) and \(z=g[x]\) are periodic functions, and evaluate the period.


Clearly, \(y^3 y’\) + \(z^3 z’=0\), so that the quantity \(y^4+z^4\) is constant.  Given the boundary conditions, \(y^4+z^4=1\) for all values of \(x\).  When plotted in phase space, this is a closed curve and is therefore representative of a periodic system.

To compute the period \(p\), we begin with the obvious observation that

\(p=\int_{0}^{p}dx=4 \int_{0}^{p/4}dx\).

Note that


so that

\(p=4 \int_{0}^{1}dz(1-z^4)^{-\frac{3}{4}}\)

This integral is doable by using the substitution \(w=1-z^4\) to reveal that


The true nerd will recognize the above integral is in the form of a Beta integral and is thus equal to

\(p\) = \(\frac{\Gamma[\frac{1}{4}]^2}{\sqrt{\pi}}\) \(\approx 7.4163\)

Solution to the irregular hexagon problem

Three equilateral triangles share a common point which is the center of a circle on which the three sides opposite the point become three sides of a cyclic hexagon [a hexagon inscribed in the circle] ABCDEF, with, say, AB, CD, and EF as the sides of the equilateral triangles. Consider the triangle formed from the midpoints of sides BC, DE, and FA [G, H, and J, respectively]. Prove that triangle GHJ is equilateral.



Let O be the origin of the circle.  Let \(\alpha =\angle BOC\), \(\beta =\angle DOE\), and \(\gamma =\angle FOA\).  Note that \(\alpha +\beta +\gamma =\pi \).

The length of a side of \(\Delta GHJ\), say, \(\overline{GH}\), is determined by the length of the bisectors, in this case \(\left|\overline{OG}\right|\) and \(\left|\overline{OH}\right|\); and the angle \(\angle GOH\) between the bisectors, as follows:

\(\left|\overline{GH}\right|^2\) = \(\left|\overline{OG}\right|^2 \) + \(\left|\overline{OH}\right|^2\) – \(2 \left|\overline{OG}\right|\) \( \left|\overline{OH}\right|\) \(\cos \angle GOH \)

where \(\left|\overline{OG}\right|\) = \(\cos \frac{\alpha}{2}\) , \(\left|\overline{OH}\right|\)= \(\cos \frac{\beta}{2}\), and \(\angle GOH=\frac{\alpha }{2}+\frac{\beta }{2}+\frac{\pi }{3}\).

Then in terms of the parameters \(\alpha\) and \(\beta\), the length of the side \(\overline{GH}\) is

\(\left|\overline{GH}\right|^2\) = \(\cos^2 \frac{\alpha}{2}\) + \(\cos^2 \frac{\beta}{2}\) – \(2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos(\frac{\alpha }{2}+\frac{\beta }{2}+\frac{\pi }{3})\)

The same reasoning applies to the other sides, e.g.,

\(\left|\overline{JG}\right|^2\) = \(\cos^2 \frac{\alpha}{2}\) + \(\cos^2 \frac{\gamma}{2}\) – \(2 \cos \frac{\alpha}{2} \cos \frac{\gamma}{2} \cos(\frac{\alpha }{2}+\frac{\gamma}{2}+\frac{\pi }{3})\)

To prove that \(\Delta GHJ\) is equilateral, we must show that the lengths of its sides are equal.  The problem then reduces to showing that

\(\cos^2 \frac{\alpha}{2}\) + \(\cos^2 \frac{\beta}{2}\) – \(2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos(\frac{\alpha }{2}+\frac{\beta }{2}+\frac{\pi }{3})\) = \(\cos^2 \frac{\alpha}{2}\) + \(\cos^2 \frac{\gamma}{2}\) – \(2 \cos \frac{\alpha}{2} \cos \frac{\gamma}{2} \cos(\frac{\alpha }{2}+\frac{\gamma}{2}+\frac{\pi }{3})\)

where, again,  \(\alpha +\beta +\gamma =\pi \).

Suffice it to say that this is an exercise in manipulating trig identities beyond the patience of most sane people, even those that consider math puzzles for recreation.  I will leave most of the details to the reader, assuring said reader that the identity to be proven does in fact hold true.  Your trusty problem poser has performed said manipulations himself, and can offer the following road map.

First, observe that \(2 \cos(z + \frac{\pi}{3})\) = \(\cos z\) – \(\sqrt{3}\) \(\sin z\).  Note that the term associated with the \(\sqrt{3}\) will be treated separately.

Also note that \(\cos \frac{\gamma }{2}\) = \(\sin \frac{\alpha +\beta }{2}\) and \(\sin \frac{\gamma }{2}\)=\(\cos \frac{\alpha +\beta }{2}\).

Keep organized, and enjoy!