## Archive for November, 2010

### Why, American Media, Why?!

November 30, 2010

I don’t see much mainstream news these days (amongst other things, it’s bad for my blood pressure), but with what I have seen recently, there is a question I am compelled to ask: why is the American media giving so much attention to the upcoming nuptuals of William Windsor and Kate Middleton? I mean, it seemed to me that this … utter irrelevancy… earned more attention that the first awarding of a Medal of Honor to a living recipient since Vietnam. WTF?! I mean, didn’t we make a point, with Article I, Section 9, Clause 8, of breaking away from all this aristocratic nonsense? So, why should I, or anyone else in this nation, care one whit about this wedding?

### Happy Thanksgiving…

November 25, 2010

…and a blogging note: due to the four-day holiday weekend here in the U.S., The usual Friday and Monday posts for this weekend are posponed to next week.

### Monday Math 144

November 22, 2010

Given positive integers a, b, and c which satisfy a2+b2=c2 (a Pythagorean triple), the product abc is divisible by 60.
Proof:

### Physics Friday 144

November 19, 2010

Consider a spherical vessel of (inner) radius r with thin walls of thickness Δr, thin enough so that stress does not vary significantly across that thickness. If we want the tension stress on the material of the walls to not exceed a value of σmax in any direction, then what is the maximum pressure we can have inside the sphere (relative to the pressure outside)?

First, the spherical symmetry tells us that the normal stresses on a small element of the wall must be equal, and there is no shear stress. Next, let us consider an imaginary plane through the center of the sphere, splitting the vessel into two hemispheres. If the (gauge) pressure inside the sphere is P, then the force that pressure exerts on the sphere is πr2P, as shown here. This force on the hemisphere must be countered by the material forces, and so we see that
$\sigma{A}=\pi{r^2}P$,
where σ is the stress and A the cross-sectional area. Since we have a thin-walled sphere, A=2πrΔr, and so
$2\pi{r}\Delta{r}\sigma=\pi{r^2}P$,
so the stress is
$\sigma=\frac{rP}{2\Delta{r}}$;
or solving for pressure,
$P=\frac{2\Delta{r}}{r}\sigma$,
and so our desired pressure range is
$P\le{P}_{max}=\frac{2\Delta{r}}{r}\sigma_{max}$

### Monday Math 143

November 15, 2010

Find .
Solution:

### Physics Friday 143

November 12, 2010

Consider a bead sliding down a straight wire inclined by an angle θ, with a coefficient of kinetic friction μ. One can see that the component of the bead’s weight parallel to the wire is , while the component perpendicular is . The normal force exerted on the bead by the wire is thus , and so the frictional force, acting parallel to the wire in the opposite direction to the parallel component of weight, is .
For the net force to be zero, allowing the bead to slide at constant velocity, we thus need , which means . If , then the friction force is less than the component of the weight, and the bead will accelerate down the wire. If , then the friction force is greater than the component of the weight, and the sliding bead will decelerate, coming to a stop somewhere along the wire.

Now, let us take the bead on a helix from last week, and add a friction force, with a coefficient of kinetic friction μ. Here, our constraint force (normal force) exerted by the wire on the bead has, in the Frenet frame, both normal and binormal components, and has magnitude dependent on velocity of
.
We note that this has a minimum value at v=0 of
, meaning the minimum value of the frictional force is .
Now, the tangential component of the weight is . Thus, we see that if , then the friction will exceed the tangential component of the weight, and the sliding bead will decelerate, coming to a stop somewhere along the helical wire. This condition is

which simplifies to
.
Note, however, that for , the frictional force increases with velocity, meaning that there is a speed where the forces balance, and acceleration is zero; the bead can slide along at that constant speed, and if it starts at a different speed, it will accelerate (or decelerate) toward that speed. Setting friction and tangential component of the weight equal, and solving for the zero-net-force velocity vC, we have
.
We note that this requires that , which is equivalent to .

### Monday Math 142

November 8, 2010

Find $\int_0^1\int_0^1\frac1{1-x^2y^2}\,dx\,dy$:
1). Using an infinite series and the Riemann zeta function
2). Using neither infinite series nor polylogarithms
Solutions:

### Physics Friday 142

November 5, 2010

Consider a small bead of mass m sliding frictionlessly down a wire in the shape of a helix, with a vertical distance d=2πc between coils and a radius R, aligned along a vertical axis. If the bead is released from rest at t=0, find, as functions of time, the distance s(t) travelled along the helix by the bead, the speed v(t) of the bead, and the magnitude F(t) of the (normal) force exerted on the bead by the wire.

Only two forces act on the bead, its weight and the force exerted by the wire. Only the former has a component tangential to the wire. Now, the wire makes a constant angle θ from the horizontal, with . Then the component of the net force tangent to the wire is , for a tangential acceleration of . This is the time derivative of our speed, so integrating with v(0)=0, we have ,
and integrating again,
.
Note that these are the same as for a straight wire inclined at an angle .

Now, we need only find the normal force. Here, we need to work in three dimensions. For the bead at a given moment, we have three mutually orthogonal directions, the Frenet frame, given by the Frenet-Serret formulas:
We have the tangent vector T, which points along the wire; the normal vector N, which points inward to the helix, and has no vertical component; and the binormal vector , perpendicular to both. Choosing a downward tangent, we then see that B is downward normal to the wire.
The weight has no component along the normal vector; its other components are
 (as found earlier), and .
Similarly, our force of constraint has no tangent component, but we do have both
FN and FB.
The net acceleration is a combination of the tangential (aT) and centripetal (aN) accelerations.
Thus, the net force has no binormal component, and so
.
The centripetal acceleration is
, where v is the speed, and  is the radius of curvature (not the radius R of the helix). As the curvature of a helix is , the radius of curvature is
,
and we have

Thus, the normal component of the net force is this times the mass, and since the weight has no normal component, we have
.
Thus, the magnitude of the force the wire exerts is
,
and plugging in for ,
we get
.

Note that if you take the limit as , then , and , since the helix approaches a horizontal wire in that limit.
Similarly, taking the limit  gives  and , since in that limit, we approach the bead falling down a vertical wire. Lastly, taking  and  while holding the ratio  (and thus the angle ) constant, we get
, which is the result for a straight wire at an angle θ.

### Monday Math 141

November 1, 2010

Let us combine the results from the previous two weeks (here and here). We found that for a regular n-gon with unit sides, the diagonals have lengths $d_k=\frac{\sin\frac{k\pi}n}{\sin\frac{\pi}n}$, k=2,3,…,n-2, with d1=dn-1=1 the length of the sides.
Now, then, we consider the product of two diagonal lengths, dkdm. From the above, this is
$d_kd_m=\frac{\sin\frac{k\pi}n\sin\frac{m\pi}n}{\sin^2\frac{\pi}n}$.
We found here that the numerator product is
$\sin\frac{k\pi}n\sin\frac{m\pi}n=\left(\sin\frac{\pi}n\right)\sum_{i=1}^{m}\sin\frac{(k-m+2i-1)\pi}n$.
Thus,
$\begin{array}{rcl}d_kd_m&=&\frac{\sin\frac{k\pi}n\sin\frac{k\pi}n}{\sin^2\frac{\pi}n}\\&=&\frac1{\sin^2\frac{\pi}n}\left(\sin\frac{\pi}n\right)\sum_{i=1}^{m}\sin\frac{(k-m+2i-1)\pi}n\\&=&\frac1{\sin\frac{\pi}n}\sum_{i=1}^{m}\sin\frac{(k-m+2i-1)\pi}n\\d_kd_m&=&\sum_{i=1}^{m}\frac{\sin\frac{(k-m+2i-1)\pi}n}{\sin\frac{\pi}n}\end{array}$.
Now, if km, then km+2i-1 is a positive integer for all i=1,2,…,m; similarly, if k+mn, then the largest value of km+2i-1 in the sum, k+m-1, is then an integer less than n, and thus the term $\frac{\sin\frac{(k-m+2i-1)\pi}n}{\sin\frac{\pi}n}$ is equal to a diagonal (or side) $d_{k-m+2i-1}$ for all i in the sum, so
$d_kd_m=\sum_{i=1}^{m}d_{k-m+2i-1}$, with 2≤mkn-2, k+mn. This is the “diagonal product formula” named by Peter Steinbach, who outlined a proof using the n-gon formed by the complex n-roots of unity (compare this problem, for example).
Now, we can generalize the formula first by exchanging k and m to see that for mk,
$d_kd_m=\sum_{i=1}^{k}d_{m-k+2i-1}$;
we combine to form
$d_kd_m=\sum_{i=1}^{\min(k,m)}d_{|k-m|+2i-1}$,
which gives us the formula for any 2≤kn-2, 2≤mn-2, k+mn.
For k+mn, we see that one or both of k and m must be greater than $\frac{n}2$. However, recall that dk=dnk. Thus, if $d_k>\frac{n}2$, then $d_{n-k}<\frac{n}2$, and similarly for m; thus, we see we can pick the smaller of k and nk; the latter gives the same terms in the sum, just in opposite order, so only the limit matters. Doing the same with m, we thus find the formula for all k and m in the range 2 to n-2:
$d_kd_m=\sum_{i=1}^{\min(k,m,n-k,n-m)}d_{|k-m|+2i-1}$,

Now, let us look at a few examples. The smallest n to give us a diagonal is n=4, the square. For square of unit side, the length of the diagonal is $\sqrt2$, and so the only diagonal product is
$d_2^2=\sum_{i=1}^{2}d_{2i-1}=d_1+d_3$, which is
$(\sqrt2)^2=2=1+1$
For the pentagon, the diagonals have length $\phi=\frac{1+\sqrt5}2$, the golden ratio, so the only unique diagonal product, since $d_2^2=d_2d_3=d_3^2$, is
$d_2^2=\sum_{i=1}^{2}d_{2i-1}=d_1+d_3$, which is
$\phi^2=1+\phi$,
a classic equation for the golden ratio.
For the hexagon, we have $d_2=d_4=\sqrt{3}$ and $d_3=2$.
Thus we have three unique products
$d_2^2=\sum_{i=1}^{2}d_{2i-1}=d_1+d_3$, which is
$(\sqrt3)^2=3=1+2$;
$d_2d_3=\sum_{i=1}^{2}d_{2i}=d_2+d_4$, which is
$2(\sqrt3)=\sqrt{3}+\sqrt{3}$;
and
$d_3^2=\sum_{i=1}^{3}d_{2i-1}=d_1+d_3+d_5$, which is
$2^2=4=1+2+1$.

Further, this can be used to prove some interesting relations of the (non-constructible) diagonal lengths of the heptagon, and thus the sides of the heptagonal triangle. Letting the heptagon have unit side, labelling the shorter and longer diagonal lengths as $a=\frac{\sin\frac{2\pi}7}{\sin\frac{\pi}7}$ and $b=\frac{\sin\frac{3\pi}7}{\sin\frac{\pi}7}$ respectively, then our diagonal product formula tells us (with $d_1=d_6=1$, $d_2=d_5=a$, and $d_3=d_4=b$):
$d_2^2=\sum_{i=1}^{2}d_{2i-1}=d_1+d_3$, and thus
$a^2=1+b$;
$d_2d_3=\sum_{i=1}^{2}d_{2i}=d_2+d_4$, and thus
$ab=a+b$;
and
$d_3^2=\sum_{i=1}^{3}d_{2i-1}=d_1+d_3+d_5$, which is
$b^2=1+a+b$.
Via simple algebra on these equations, we can find formula for the quotients of these numbers; dividing $ab=a+b$ by a or b and solving for the quotient left, we see
$\frac{b}{a}=b-1$ and $\frac{a}{b}=a-1$;
and dividing by the product ab gives
$\frac{1}{a}+\frac{1}{b}=1$,
while further algebra with these formulas lets us find the reciprocals of a and b as linear combinations of a, b, and 1:
$\begin{array}{rcl}b^2&=&1+a+b\\b&=&\frac{1}b+\frac{a}{b}+1\\b&=&\frac{1}b+a-1+1\\b&=&\frac{1}b+a\\\frac{1}b&=&b-a\end{array}$
and
$\begin{array}{rcl}a^2&=&1+b\\a&=&\frac{1}a+\frac{b}{a}\\a&=&\frac{1}a+b-1\\\frac{1}a&=&a-b+1\end{array}$.

These relations appear, for example, in the substitution rules for Danzer’s 7-fold quasiperiodic tiling and Maloney’s 7-fold quasiperiodic tiling