Monthly Archives: October 2017

Geometry & Arithmetic

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A couple of weeks ago I discussed a geometric solution to an arithmetic problem. In this post, I will discuss an arithmetical solution to a geometry problem. Consider the following question:

Given a square whose sides are reflecting mirrors. A ray of light leaves a point inside the square and is reflected repeatedly in the mirrors. What is the nature of its paths?

It may happen that the ray passes through a corner of the square. In that case, we assume that it returns along its former path.

arth

In figure, the parallels to the axis are the lines, x = m + \frac{1}{2} and y = n + \frac{1}{2}, where m and n are integers. The thick square, of side 1, around the origin is the square of the problem and E\equiv(a,b) is the starting point. We construct all images of E in the mirrors, for direct or repeated reflection. One can observe that they are of four types, the coordinates of the images of the different types being

  1. (a+2n, b+2m)
  2. (a+2n, -b+2m+1)
  3. (-a+2n+1, b+2m)
  4. (-a+2n+1, -b+2m+1)

where m and n are arbitrary integers. Further, if the velocity at E has direction cosines, \lambda, \mu, then the corresponding images of the velocity have direction cosines

  1. (\lambda, \mu)
  2. (\lambda, -\mu)
  3. (-\lambda, \mu)
  4. (-\lambda, -\mu)

where we suppose (on the grounds of symmetry) that \mu is positive. If we think of the plane as divided into squares of unit side, then interior of a typical square being

\displaystyle{n -\frac{1}{2} < x < n+\frac{1}{2}, \qquad m-\frac{1}{2}<y<m+\frac{1}{2}}

then each squares contains just one image of every point in the original sqaure, given by n=m=0 (shown by the bold points in the figure). And if the image in any of the above squares of any point in the original sqaure is of type (1.), (2.), (3.) or (4.), then the image in any of the above squares of any other point in the original square is of the same type.

We can now imagine E moving with the ray (shown by dotted lines in the figure). When the point E meets the mirror, it coincides with an image and the image of E which momentarily coincides with E continues the motion of E, in its original direction, in one of the squares adjacent to the fundamental square (the thick square). We follow the motion of the image, in this square, until it in its turn it meets a side of the square. Clearly, the original path of E will be continued indefinitely in the same line L (dotted line in the figure), by a series of different images.

The segment of L in any square (for a given n and m) is the image of a straight portion of the path of E in the original square. There is one-to-one correspondence between the segments of L, in different squares, and the portions of the path of E between successive reflections, each segment of L being an image of the corresponding portion of the path of E.

The path of E in the original square will be periodic if E returns to its original position moving in the same direction; and this will happen if and only if L passes through an image of type (1.) of the original E. The coordinates of an arbitrary point of L are x=a+\lambda t, \quad y = b+\mu tf.

Hence the path will be periodic if and only if \lambda t = 2n, \quad \mu t = 2m, for some t and integral n,m, i.e. if  \frac{\lambda}{\mu} is rational.

When \frac{\lambda}{\mu} is irrational, then the path of E approaches arbitrarily near to every point (c,d) of the sqaure. This follows directly from Kronecker’s Theorem in one dimension (see § 23.3 of G H. Hardy and E. M. Wright’s An Introduction to the Theory of Numbers.):

[Kronecker’s Theorem in one dimension] If \theta is irrational, \alpha is arbitrary, and N and \epsilon are positive, then there are integers p and q such that p>N and |p\theta - q-\alpha|<\epsilon.

Here, we have \theta = \frac{\lambda}{\mu} and \alpha = (b-d)\frac{\lambda}{2\mu} - \frac{1}{2}(a-c), with large enough integers p=m and q=n. Hence we can conclude that

[König-Szücs Theorem]Given a square whose sides are reflecting mirrors. A ray of light leaves a point inside the square and is reflected repeatedly in the mirrors. Either the path is closed and periodic or it is dense in the square, passing arbitrarily near to every point of the square. A necessary and sufficient condition for the periodicity is that the angle between a side of the square and the initial direction of the ray should have a rational tangent.

Another way of stating the above Kronecker’s theorem is

[Kronecker’s Theorem in one dimension] If \theta is irrational, then the set of points n\theta - \lfloor n\theta\rfloor is dense in the interval (0,1).

Then with some knowledge of Fourier series, we can try to answer a more general question

Given an irrational number \theta, what can be said about the distribution of the fractional parts of the sequence of numbers n\theta, for n=1,2,3,\ldots?

The answer to this question is called Weyl’s Equidistribution Theorem (see §4.2 of  Elias M. Stein & Rami Shakarchi’s Fourier Analysis: An Introduction)

[Weyl’s Equidistribution Theorem] If \theta is irrational, then the sequence of fractional parts \{n\theta - \lfloor n\theta\rfloor\}_{n=1}^{\infty} is equidistributed in [0,1).

I really enjoyed reading about this unexpected link between geometry and arithmetic (and Fourier analysis). Most of the material has been taken/copied from Hardy’s book. The solution to the geometry problem reminds me of the solution to the Cross Diagonal Cover  Problem.

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Understanding Geometry – 4

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Aleksej Ivanovič Markuševič’s book, “Remarkable Curves” discusses the properties of ellipses, parabolas, hyperbolas, lemniscates, cycloids, brachistochrone, spirals and catenaries.  Among these “lemniscates” are the ones that I encountered only once before starting undergraduate education (all other curves appeared frequently in physics textbooks) and that too just to calculate the area enclosed by this curve. So I will discuss the properties of lemniscates in this post.

Let’s begin with the well-known curve, ellipse. An ellipse is the locus of points whose sum of distances from two fixed points (called foci) is constant. My favourite fact about ellipses is that we can’t find a general formula for the perimeter of an ellipse, and this little fact leads to the magical world of elliptic integrals. This, in turn, leads to the mysterious elliptic functions, which were discovered as inverse functions of elliptic integrals. Further, these functions are needed in the parameterization of certain curves, now called elliptic curves. For more details about this story, read the paper by Adrian Rice and Ezra Brown, “Why Ellipses are not Elliptic curves“.

Lemniscate is defined as the locus of points whose product of distances from two fixed points F_1 and F_2 (called foci) is constant. Lemniscate means, “with hanging ribbons” in Latin.  If the length of the segment \overline{F_1F_2} is c then for the midpoint of this line segment will lie on the curve if the product constant is c^2/4. In this case we get a figure-eight lying on its side.

Lemniskate_bernoulli2.svg

Lemniscate of Bernoulli; By Kmhkmh (Own work) [CC BY 4.0], via Wikimedia Commons

The attempt to calculate the perimeter of the above curve leads to elliptic integral, hence can’t derive a general formula for its perimeter. Just like an ellipse!

If we equate the value of the constant product not to c^2/4 but to another value, the lemniscate will change its shape. When the constant is less than c^2/4, the lemniscate consists of two ovals, one of which contains inside it the point F_1, and the other the point F_2.

oval

Cassini oval (x^2+y^2)^2−2c^2(x^2−y^2)=a^4−c^4; Source: https://www.encyclopediaofmath.org/legacyimages/common_img/c020700b.gif

When the product constant is greater than c^2/4 but less than c^2/2, the lemniscate has the form of a biscuit. If the constant is close to c^2/4, the “waist” of the biscuit is very narrow and the shape of the curve is very close to the figure-eight shape.

biscuit

Cassini oval (x^2+y^2)^2−2c^2(x^2−y^2)=a^4−c^4; Source: https://www.encyclopediaofmath.org/legacyimages/common_img/c020700b.gif

If the constant differs little from c^2/2, the waist is hardly noticeable, and if the constant is equal or greater than c^2/2 the waist disappears completely, and the lemniscate takes the form of an oval.

oval.gif

Cassini oval (x^2+y^2)^2−2c^2(x^2−y^2)=a^4−c^4; Source: https://www.encyclopediaofmath.org/legacyimages/common_img/c020700a.gif

We can further generalize this whole argument to get lemniscate with an arbitrary number of foci, called polynomial lemniscate.

Arithmetic & Geometry

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After a month of the unexpected break from mathematics, I will resume the regular weekly blog posts. It’s a kind of relaunch of this blog, and I  will begin with the discussion of an arithmetic problem with a geometric solution.  This is problem 103 from  The USSR Olympiad Problem Book:

Prove that
\displaystyle{t_1+t_2+\ldots + t_n = \left\lfloor\frac{n}{1}\right\rfloor + \left\lfloor\frac{n}{2}\right\rfloor + \left\lfloor\frac{n}{3}\right\rfloor + \ldots + \left\lfloor\frac{n}{n}\right\rfloor}
where t_k is the number of divisors of the natural number k.

One can solve this problem by using the principle of mathematical induction or using the fact that the number of integers of the sequence 1,2,3,\ldots , n which are divisible by any chosen integer k is equal to \left\lfloor \frac{n}{k}\right\rfloor. The second approach of counting suggests an elegant geometric solution.

Consider the equilateral hyperbola \displaystyle{y = \frac{k}{x}} , (of which we shall take only the branches in the first quadrant):
ss
We note all the points in the first quadrant which have integer coordinates as the intersection point of the dotted lines. Now, if an integer x is a divisor of the integer k, then the point (x,y) is a point on the graph of the hyperbola xy=k. Conversely, if the hyperbola xy=k contains an integer point, then the x-coordinate is a divisor of k. Hence the number of integers t_k is  equal to the number of the integer points lying on the hyperbola xy=k. The number t_k is thus equal to the number of absciassas of integer points lying on the hyperbola xy=k. Now, we make use of the fact that the hyperbola xy=n lies “farther out” in the quadrant than do xy=1, xy=2, xy=3, \ldots, xy=n-1. The following implication hold:

The sum t_1+t_2\ldots + t_n is equal to the number of integer points lying under or on the hyperbola xy=n. Each such point will lie on a hyperbola xy=k, where k\leq n. The number of integer points with abscissa k located under the hyperbola is equal to the integer part of the length of the segment \overline{AB} [in figure above k=3]. That is \left\lfloor\frac{n}{k}\right\rfloor, since \displaystyle{|\overline{AB}|=\frac{n}{k}}, i.e. ordinate of point A on hyperbola xy=n for abscissa k. Thus, we obtain

\displaystyle{t_1+t_2+\ldots + t_n = \left\lfloor\frac{n}{1}\right\rfloor + \left\lfloor\frac{n}{2}\right\rfloor + \left\lfloor\frac{n}{3}\right\rfloor + \ldots + \left\lfloor\frac{n}{n}\right\rfloor}

Caution: Excess of anything is harmful, even mathematics.