$latex displaystyle binom{n}{m} = t (1)&fg=000000$

with $latex {1 leq m < n}&fg=000000$. Currently, the largest number of solutions that is known to be attainable is eight, with $latex {t}&fg=000000$ equal to

In a previous blogpost we proved the convergence of the Fourier series of a Riemann integrable 1-periodic function to . However convergence doesn’t tell us anything about pointwise convergence.

Exercise 1:

(i.) Find an example of a sequence of functions which converge in norm to a function but does not converge pointwise to

(ii.) Find an example of a sequence of functions which converges pointwise to a function $g$ but does not converge in norm to

(iii.) Let be a sequence of functions which converge pointwise to a function and in norm to a function Is it possible to have

Under some conditions however one can ensure pointwise convergence of the Fourier series of a function; in fact in what follows we discuss one important condition under which we can guarantee uniform convergence of the Fourier series of a function (which in particular would imply its pointwise convergence).

Theorem 2: Let be a continuous, 1-periodic function with Fourier coefficients for each Suppose that Then the Fourier series of converges uniformly to As a consequence, holds for all real numbers

Proof: Notice that if then converges uniformly. Let denote this limit (i.e., ).

Then notice that being the uniform limit of a sequence of continuous functions must be a continuous function itself.

However we know that the Fourier series of must converge to in norm.

Thus, one must have However since both are continuous functions so

This proves the claim.

Exercise 3:

If is continuous, periodic and piecewise continuously differentiable, then prove that the Fourier series of converges uniformly to

In a previous blogpost I proved that the first moments of hitting times for a simple random walk on a finite tree are integers. That answered a part of a more general problem which I was asked by professor Yuval Peres, and in this blogpost I address another part of his question by proving the following:

Proposition 1: If is a finite tree, are two vertices of and denotes the number of steps taken by a simple random walk on starting at to reach vertex then is an integer.

Interestingly, in this case too one can reduce proving Proposition 1 to proving it for the special case when are neighbouring vertices, and in particular when is a leaf of The verification of this reduction is left as an exercise for the readers.

Thus we would only need to prove the following corollary of Proposition 1:

Corollary 2: If is a finite tree, are two neighbouring vertices of and is a leaf of and if denotes the number of steps taken by a simple random walk on starting at to reach vertex then is an integer.

We can do further reduction. Taking help of our conclusions drawn here and the observation that

it suffices to show that is an integer having the same parity as that of

Thus, it suffices to show the following claim:

Lemma 4:9 For a simple random walk on a finite tree and any two adjacent vertices of with being a leaf of the variance is an even integer.

Proof of Lemma 4: We induce on the number of vertices of with the base case being (since is a vacuous case). For the tree just comprises of two vertices joined by an edge and thus (which is an even integer) in this case.

Next we make some observations.

Let be the number of vertices which are adjacent to the vertex in the tree and let the neighbours of in the tree be with being the concerned target (leaf) node.

We think of the Markov chain evolving in the following way: when the particle performing the simple random is at the node it rolls a fair sided dice, and if the outcome is then it moves to and then performs a symmetric simple random walk on starting at the as usual until it reaches upon which we again roll the dice and continue so on and so forth unless the particle hits the state upon which we stop; it is easy to notice that the collection obtained in this way is indeed a symmetric simple random walk on the tree Thus let denote the sequence of outcomes on respective independent rolls of the sided dice and let denote the first time the dice shows up as the outcome. Clearly has the geometric distribution with success probability and thus

Now this allows us to write

where s denote the respective outcome of the tosses and is some integer, thus one can write

for some integer and where ; these s basically correspond to the time difference between two consecutive respective visits to the vertex of , thus are independent and identically distributed and are also independent.

Recently I came across a super cute proof of Napoleon’s triangle theorem using complex numbers in Béla Bollobás’s book “The Art of Mathematics: Coffee time in Memphis”. Realizing the fact that I never knew it till now, and the fact that it is neither discussed in the standard textbooks on Euclidean geometry I read till day nor any online reading source (for instance Wikipedia doesn’t discuss this proof, however the Cut the Knot page on Napolean’s triangle discusses this proof), I think that despite being so beautiful this proof is not very popular (especially among math Olympians and other mathematics enthusiasts at the undergraduate level), and thus I decided to dedicate this blogpost for discussing the proof.

Proposition 1(Napoleon’s triangle theorem) Given any triangle , we construct equilateral triangles erected outwards (i.e., the points lie in the exterior of ) on the sides respectively. Let be the centroids of triangles respectively. Then must be an equilateral triangle.

Here is an illustration of how the setup of Napoleon’s triangle theorem looks like.

Here are equilateral triangles erected outwards on the sides of $\Delta ABC,$ and $A”, B”, C”$ are the centroids of $\Delta A’BC, B’CA, C’AB$ respectively; we wish to show that the dotted triangle (i.e., the triangle with its edges marked by dotted lines

Proof: We notice that a triangle is equilateral if and only if the angle between the sides and is and in other words is equilateral if and only if rotating the side by an angle of gives the side of this triangle.

So, it suffices to show that rotation of side of triangle gives the side of this triangle; in terms of complex numbers a rotation by an angle of basically corresponds to multiplication by the complex number

The idea is to write each side of triangle in terms of complex numbers and then verify the fact that multiplication by rotates the side to yield

Let and Then one has and Also one obtains and Thus we immediately get

whence (since and the product of two non-zero complex numbers must necessarily be non-zero).

Now, notice that Hence, thus proving the claim.

EDIT: (Last updated on 7 June, 2021)

Remark: Some sources believe that the proof discussed above was known much before Béla Bollobás’s discovery of the proof.

Here I prove the following interesting result in a very slick way (this is a proof is beleived to be originally due to Paul Erdős, although I discovered on my own long ago (approximately in 2016-17), and since then it has remained one of my favourite deductions).

Proposition 1: A real number satisfies for all real numbers if and only if (where denotes Euler’s number).

Proof: One part of the claim is fairly straightforward

For proving the converse we use the following important property of Euler’s number

Exercise 2:

Show that for all positive real numbers with equality if and only if

(Hint: Consider the function defined for this is a differentiable function; differentiate it (for this step you may require to consider differentiating ) and find where the derivative changes sign, where it is positive and where it is negative to conclude the result.)

Conversely suppose holds for every real number Let be an arbitrary positive real number. Then putting one has