Find the infinite product .
Archive for November, 2009
Find the infinite product .
Consider a pendulum clock with a pendulum whose length is made of aluminum. If the average temperature around the clock is
ΔT=5 C° (=5 K) warmer in the summer than in the winter, then how much faster or slower does the clock run in the summer than in the winter?
Can you construct a regular pentadecagon (15-sided) polygon with compass and straightedge? What’s the restriction on n≥3 such that a regular n-gon is constructible?
Constructing an equilateral triangle, n=3, is very simple, as are constructing a square, n=4, and a hexagon, n=6. The method of constructing a regular pentagon, n=5, has been known since the days of Euclid.
In addition, if one can construct a regular n-gon, then by bisecting its sides one can construct a 2n-gon; by repeating, the polygons are thus constructible for 4n, 8n, etc.
Further, suppose that we can construct a regular n-gon and d≥3 is a divisor of n. Then by connecting every vertices of the n-gon, one produces a d-gon. So, suppose that the regular k-gon is not constructible. Then we can conclude that a polygon with a number of sides that is any multiple of k is also not constructible.
Now, suppose we have that a regular a-gon is constructible, and the regular b-gon is also constructible, with a and b coprime (gcd(a,b)=1). Suppose we construct these two polygons inscribed in the same circle with one (and only one, since a and b are coprime) vertex in common. Then imagine the regular polygon with ab sides. One can form a regular a-gon by connecting every b vertices, and a regular b-gon by connecting every a vertices. Now, Bézout’s identity tells us that for any two non-zero integers a and b, there exist integers u and v such that , which means that for our coprime a and b, there exist integers u and v for which , and thus there exist integer multiples of a and b which differ by exactly one. This means that in our construction of the a-gon and b-gon in the same circle with common vertex, there exist a vertex of the a-gon and a vertex of the b-gon which form one side of the ab-gon, and from that side and the circle in which the polygons are inscribed, one can construct the entire ab-gon. Thus, if the regular a-gon and regular b-gon are both constructible with a and b coprime, then the regular ab-gon is constructible.
So since 15=3*5, and the equilateral triangle and pentagon are both constructible, the regular pentadecagon is thus constructible. To use this as an example of the argument given above, we have that 3*2+5*(-1)=1. Thus, traveling around from the common vertex, if we take two sides of the pentagon and one side of the triangle, the resulting vertices are the sixth and fifth around of the pentadecagon, respectively, and so they form one side of the pentadecagon.
The question of what values of n allow the regular n-gon to be constructible was studied by many mathematicians; Gauss developed a sufficient condition in 1796 which he conjectured was also necessary; Pierre Wantzel later proved that it was so. The proof arises from analytic geometry, and is best expressed in terms of field theory and, in particular, quadratic extensions. The result, however, is that the regular n-gon is constructible if and only if φ(n) is a power of 2, where φ(n) is Euler’s totient function. Our result about ab with coprime a and b thus follows from φ(n) being a multiplicative function.
Since for prime p, we see that , again confirming that if n gives us a constructible polygon, n times any power of two also gives us a constructible polygon. As for prime p>2, we see that is divisible by p for k>1, and is thus not a power of 2 in that case; so, no odd prime may appear as a factor with multiplicity greater than 1; thus the enneagon, n=9=32, is not constructible. Further, a prime factor p will be able to give a power of 2 only if is a power of 2, which means that p is of the form . I showed a proof last week that the only primes of this form are the Fermat primes, primes of the form . Thus, the above is equivalent to Gauss’ phrasing that a regular n-gon is constructible with compass and straightedge if (and only if) n is the product of a power of 2 (including 20=1) and any number (including zero) of distinct Fermat primes.
So we see that the first few constructible regular polygons are
n=3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, …
while the first few non-constructible regular polygons are
n=7, 9, 11, 13, 14, 18, 19, …
Since the only known Fermat primes are F0=3, F1=5, F2=17, F3=257, and F4=65537, their product, 4,294,967,295, is thus the largest odd number of sides for which the regular polygon is known to be constructible.
Consider a set of identical books (or other rectangular objects), stacked flat one atop another. By offsetting the books, one can have a “leaning” tower that angles to one side.
Is it possible to make such a tower where no part of the top book is over any part of the base book, so that the “lean” is larger than the length of a book? What is the maximum possible horizontal offset between the top and bottom books?
Let , n a positive integer. If k is prime (an odd prime, as k>2), show that n must be a power of two.
Suppose we have a cube, of edge length L, with mass M symmetrically distributed so that the geometric center of the cube is the center of mass. Suppose we have a fluid with density ρ, such that (the average density of the cube is less than that of the fluid), so that the cube will float in the fluid. If the cube is floating such that the upper face is entirely outside the fluid, and is tipped toward one edge by an angle θ from the horizontal, then what is the torque on the cube about its center?
A useful polynomial factoring rule is that for positive integer n, the linear term x–y is a factor of . More specifically,
expanding the right hand side of the above gives you the left after cancellation of various terms.
For example, for the first few values of n
and so on.
It is this which gives us the formula for a finite geometric series; letting x=r, y=1, and our power be n+1, then
Further, note that if x and y are integers, then is also an integer, and thus the integer x–y is a factor of the integer .
Now, let n be odd. If we let x=a and y=-b, then
and so on. And we see that if x and y are integers, then is also an integer, and thus the integer x+y is a factor of the integer when n is odd.
Suppose, as in this post, we have a fluid of density ρ, with some arbitrary three-dimensional object immersed in the fluid, occupying a volume V. This object has surface ∂V. We saw that for any point on the surface, the pressure is , and the force on an area element is:
. Thus, the torque on that element is
where r is the coordinate vector to the element. Recalling that for vectors v and w and scalar a,
, so since P(x,y,z) is a scalar, we see
and so the total torque (about the origin) is
and one form of the divergence theorem tells us that for vector field A,
Now, the product rule for the curl of the product of a scalar field ψ and a vector field a is
Now, applying that to , and noting that (as it is a radial vector field), and , we see
Now, since is a constant vector, it can be “factored out” of the integral:
We recall that the total buoyant force on the object is ; plugging this into the above, we see
Now, recall that the center of mass of a region with density function ρ(x,y,z) is . If the density is a constant, it factors out of the integrals, and one gets . Thus, if we consider the volume as if it were filled with the fluid; that is to say, the volume of fluid displaced, its center of mass would be , and then we see that
which is equivalent to the torque if the buoyant force acted entirely on the point rb; this point is called the center of buoyancy. Just as the force of gravity on an extended object can be treated as if it acts entirely on the center of mass, the buoyant force can be treated as if it acts entirely on the center of buoyancy.