Category Archives: Problem Solving

Finite Sum & Divisibility

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I wish to discuss a small problem from The USSR Olympiad Problem Book (problem 59) about the finite sum of harmonic series. The problem asks us to prove that

\displaystyle{\sum_{k=2}^{n} \frac{1}{k}}  can never be an  integer for any value of n.

I myself couldn’t think much about how to prove such a statement. So by reading the solution, I realised that how a simple observation about parity leads to this conclusion.

Firstly, observe that among the natural numbers from 2 to n there is exactly one natural number which has the highest power of 2 as its divisor. Now, while summing up the reciprocals of these natural numbers we will get a fraction as the answer. In that fraction, the denominator will be an even number since it’s the least common multiple of all numbers from 2 to n. And the numerator will be an odd number since it’s the sum of (n-2) even numbers with one odd number (corresponding to the reciprocal of the number with the highest power of 2 as the factor). Since under no circumstances an even number can completely divide an odd number, denominator can’t be a factor of the numerator. Hence the fraction can’t be reduced to an integer and the sum can never be an integer.

Prime Consequences

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Most of us are aware of the following consequence of Fundamental Theorem of Arithmetic:

There are infinitely many prime numbers.

The classic proof by Euclid is easy to follow. But I wanted to share the following two analytic equivalents (infinite series and infinite products) of the above purely arithmetical statement:

  • \displaystyle{\sum_{p}\frac{1}{p}}   diverges.

For proof, refer to this discussion: https://math.stackexchange.com/q/361308/214604

  • \displaystyle{\sum_{n=1}^\infty \frac{1}{n^{s}} = \prod_p\left(1-\frac{1}{p^s}\right)^{-1}}, where s is any complex number with \text{Re}(s)>1.

The outline of proof,   when s is a real number, has been discussed here: http://mathworld.wolfram.com/EulerProduct.html

Solving Logarithmic Equations

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While reading John Derbyshire’s Prime Obsession I came across the following statement (clearly explained on pp. 74):

Any positive power of \log(x) eventually increases more slowly than any positive power of x.

It is easy to prove this (existence) analytically, by taking derivative to compare slopes. But algebraically it implies that (for example):

There are either no real solution or two real solutions of the equation
\log(x) = x^\varepsilon
for any given \varepsilon>0.

Now the question that arises is “How to find this x?” I had no idea about how to solve such logarithmic equations, so I took help of Google and discovered this Mathematic.SE post. So, we put \log(x)=y and re-write the equation as:

y=e^{y\varepsilon}

Now to be able to use Lambert W function (also called the product logarithm function) we need to re-write the above equation, but I have failed to do so. 

But using WolframAlpha I was able to solve \log(x)=x^2 to get x=e^{\frac{-W(-2)}{2}} (which is an imaginary number, i.e. no real solution of this equation) but I was not able to figure out the steps involved. So if you have any idea about the general method or the special case of higher exponents, please let me know.

Combinatorial Puzzle

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This is a continuation of previous post:

How many distinct numbers can be formed by using four 2s and the four arithmetic operations +,-,\times, \div.

For example:
1 = \frac{2+2}{2+2}=\frac{2}{2}\times\frac{2}{2}
2 = 2+\frac{2-2}{2}=\frac{2}{2}+\frac{2}{2}
3 = 2+2 - \frac{2}{2}
4 = 2+2+2-2 = (2\times 2) + (2-2)
(note that some binary operations do not make sense without parenthesis)

I have no idea about how to approach this problem (since I am not very comfortable with combinatorics). So any help will be appreciated.

Edit[29 May 2017]: This problem has been solved in the comments below. 

Arithmetic Puzzle

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Following is a very common arithmetic puzzle that you may have encountered as a child:

Express any whole number n using the number 2 precisely four times and using only well-known mathematical symbols.

This puzzle has been discussed on pp. 172 of Graham Farmelo’s “The Strangest Man“, and how Paul Dirac solved it by using his knowledge of “well-known mathematical symbols”:

\displaystyle{n = -\log_{2}\left(\log_{2}\left(2^{2^{-n}}\right)\right) = -\log_{2}\left(\log_{2}\left(\underbrace{\sqrt{\sqrt{\ldots\sqrt{2}}}}_\text{n times}\right)\right)}

This is an example of thinking out of the box, enabling you to write any number using only three/four 2s. Though, using a transcendental function to solve an elementary problem may appear like an overkill.  But, building upon such ideas we can try to tackle the general problem, like the “four fours puzzle“.

This post on Puzzling.SE describes usage of following formula consisting of  trigonometric operation \cos(\arctan(x)) = \frac{1}{\sqrt{1+x^2}} and \tan(\arcsin(x))=\frac{x}{\sqrt{1-x^2}} to obtain the square root of any rational number from 0:

\displaystyle{\tan\left(\arcsin\left(\cos\left(\arctan\left(\cos\left(\arctan\left(\sqrt{n}\right)\right)\right)\right)\right)\right)=\sqrt{n+1}}.

Using this we can write n using two 2s:

\displaystyle{n = (\underbrace{\tan\arcsin\cos\arctan\cos\arctan}_{n-4\text{ times}}\,2)^2}

or even with only one 2:

\displaystyle{n = \underbrace{\tan\arcsin\cos\arctan\cos\arctan}_{n^2-4\text{ times}}\,2}

Understanding Geometry – 2

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If you want to brush up your high school geometry knowledge, then KhanAcademy is a good place to start. For example, I learned a new proof of Pythagoras Theorem (there are 4 different proofs on KhanAcademy) which uses scissors-congruence:

In this post, I will share with you few theorems from L. I. Golovina and I. M. Yaglom’s “Induction in Geometry ”  which I learned while trying to prove Midpoint-Polygon Conjecture.

Theorem 1: The sum of interior angles of an n-gon is 2\pi (n-2).

Theorem 2: The number of ways in which a convex n-gon can be divided into triangles by non-intersecting diagonals is given by

\displaystyle{\frac{2(2n-5)!}{(n-1)!(n-3)!}}

Theorem 3: Given a \triangle ABC, with n-1 straight lines CM_1, CM_2, \ldots CM_{n-1} drawn through its vertex C, cutting the triangle into n smaller triangles \triangle ACM_1, \triangle M_1CM_2, \ldots, \triangle M_{n-1}CB. Denote by r_1, r_2, \ldots r_n and \rho_1, \rho_2, \ldots, \rho_n respectively the radii of the inscribed and circumscribed circles of these triangles (all the circumscribed circles are inscribed within the angle C of the triangle) and let r and \rho be the radii of the inscribed and circumscribed circles (respectively) of the \triangle ABC itself. Then

\displaystyle{\frac{r_1}{\rho_1} \cdot\frac{r_2}{\rho_2} \cdots\frac{r_n}{\rho_n} =\frac{r}{\rho} }

Theorem 4: Any convex n-gon which is not a parallelogram can be enclosed by a triangle whose sides lie along three sides of the given n-gon.

Theorem 5 (Levi’s Theorem): Any convex polygon which is not a parallelogram can be covered with three homothetic polygons smaller than the given one.

The above theorem gives a good idea of what “combinatorial geometry” is all about. In this subject, the method of mathematical induction is widely used for proving various theorems. Combinatorial geometry deals with problems, connected with finite configurations of points or figures. In these problems, values are estimated connected with configurations of figures (or points) which are optimal in some sense.

Theorem 6 (Newton’s Theorem): The midpoints of the diagonals of a quadrilateral circumscribed about a circle lie on one straight line passing through the centre of the circle.

Theorem 7 (Simson’s Theorem): Given a \triangle ABC inscribed in the circle S with an arbitrary point P on this circle. Then then feet of the perpendiculars dropped from the point P to the sides of the \triangle ABC are collinear.

We can extend the above idea of Simson’s line to any n-gon inscribed in a circle.

Theorem 8: A 3-dimensional space is divided into \frac{(n+1)(n^2-n+6)}{6} parts by n planes, each three of which intersect and no four of which have a common point.

Theorem 9: Given n spheres in 3-dimesnional space, each four of which intersect. Then all these spheres intersect, i. e. there exists a point belonging to all the spheres.

Theorem 10 (Young’s Theorem): Given n points in the plane such that each pair of them are at a distance of at most 1 from each other. Then all these points can be enclosed in a circle of radius 1/\sqrt{3}.

I won’t be discussing their proofs since the booklet containing the proofs and the detailed discussion is freely available at Mir Books.

Also, I would like to make a passing remark about the existence of a different kind of geometry system, called “finite geometry“. A finite geometry is any geometric system that has only a finite number of points. The familiar Euclidean geometry is not finite because a Euclidean line contains infinitely many points. A geometry based on the graphics displayed on a computer screen, where the pixels are considered to be the points, would be a finite geometry. While there are many systems that could be called finite geometries, attention is mostly paid to the finite projective and affine spaces because of their regularity and simplicity. You can learn more about it here: http://www.ams.org/samplings/feature-column/fcarc-finitegeometries

Huntington’s Red-Blue Set

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While reading Lillian Lieber’s book on infinity, I came across an astonishing example of infinite set (on pp. 207). Let’s call the property of existence of a rational number between given two rational number to be “beauty” (a random word introduced by me to make arguments clearer).

The set of rational numbers between 0 and 1 are arranged in ascending order of magnitude, and all of them are coloured blue. This is clearly a beautiful set. Then another another set of rational numbers between 0 and 1 is taken and arrange in ascending order of magnitude, but all of them are coloured red. This is also a beautiful set. Now, put these two sets together in such a way that each blue number is immediately followed by the corresponding red number. For example, 1/2 is immediately followed by 1/2 etc.  It appears that if we interlace two beautiful sets, the resulting set should be even more beautiful. But since each blue number has an immediate successor, namely the corresponding red number, so that between these two we can’t find even a single other rational number, red or blue, the resulting set is NOT beautiful.

The set created above is called Huntington’s Red-Blue set. It is an ingenious invention, where two beautiful sets combined together lead to loss of beauty. For more details, read the original paper:

Huntington, Edward V. “The Continuum as A Type of Order: An Exposition of the Modern Theory.” Annals of Mathematics, Second Series, 7, no. 1 (1905): 15-43. doi:10.2307/1967192.