Hello everyone. Recently, I decided to upload some course materials, which I have prepared based on lectures I have been a part of, on the course materials webpage .
These notes will be updated regularly.
In some instances, I may also share some sections from my unpublished book, or other course related documents I have prepared.
Let for any positive integers we denote the number of solutions to the equation
In 1770, Edward Waring posed the following problem :
For every positive integer there exists some positive integer such that, for every positive integer
It remained unanswered for a long time, until building up on a partial progress due to Hurwitz, David Hilbert proved the assertion to be true, thereby solving the infamous Waring’s problem. We shall not discuss Hilbert’s proof in this blog post, since it happens to be well-known and can easily be found on the internet.
Consider the number In following the circle method, we write this number of solutions as an integral.
Here, I shall be demonstrating this basic instance of the circle method by writing the number as an integral in this case.
We begin by recalling the notation where as usual is the imaginary unit, and is Euler’s infamous number.
For any suppose
where is a positive integer.
We may write as
where represents the number of solutions to with This can be seen by multiplying the expression of (in ) many times, and observing that in the power of exponential we would get sum of th powers, and then taking it as .
If then, would be the same as
Now, multiplying both sides of equation with and integrating it over (and by exploiting the orthogonality of complex exponential integral) yields
The expression in above allows us to write the number of solutions to the equation as an integral.
Now, we make a claim the proof of which is left as an easy exercise for the reader :
Exercise : Prove that in the asymptotic formula given for above, the main term is of order(Hint : Try to prove this combinatorially)
This observation allows us to neglect all those which contribute an amount of order strictly lower than to the integral in
We consider and regard the absolute value of the integrand as
Then, an appeal to Hua’s inequality allows us to neglect all those for which where as usual denotes the Vinogradov asymptotic symbol.
A general plan in working with Waring’s problem and other similar problems is to divide the values of into two sets : the major arcs whose contribution shapes the main term, and the minor arcs the contribution of which goes to the error term. The distinction between these sets depend very much on the available results about the main term.
Loosely speaking, while powerful (although involved) methods are available for dealing with the major arcs, the crux of the problem lies with the minor arcs. Once a method to deal with minor arcs is known, one tries to extend the method as much as the setup of the problem permits, to ease the work of dealing with major arcs (though dealing with major arcs are usually more straightforward).
Around every rational number (in lowest form), we put an interval given by the conditions
and we do this for and
It is easy to notice that the distance between the centres of these intervals is and this being greater than their length implies that these intervals do not overlap. Moreover by rounding and translation, we can assume that these intervals are contained in
Invoking Weyl’s inequality, one can see that these intervals gives the major arcs, and their complement relative to give the minor arcs, the totality of which we denote by
Lemma : If then we have
where is a positive number depending upon
Proof : Invoking a classical result due to Dirichlet on rational approximations, we get that : for every one can find a rational number satisfying and
Furthermore, if then we shall have : Since gives a stronger criteria than so it follows that : we should have if In other words, this means for every we must have
Since, so applying Weyl’s inequality to the exponential sum and using the fact that we get :
Now, applying Hua’s inequality to this we get :
where and depends on
This concludes the proof of the above lemma.
We now turn our attention to the major arcs
Lemma : For we have
Proof : To begin with, we collect those values of in the sum defining which are in the same residue class modulo To do this, we write with Here, runs through an interval depending upon corresponding to Now, doing this and making use of the fact that for any integer we conclude that :
Now, we attempt to replace the discrete variable by a continuous variable in a quest to replace the by an integral over After this, we can make a change of variable from to This would allow us to replace the summation over by the integral
So, all that remains now, is to estimate the error in going from the summation to the integral.
Exercise : If is a differentiable function, then prove that for we must have(Hint : Use Lagrange’s Mean Value Theorem)
From the above exercise, we see that if we divide any interval into intervals of length together with two possibly broken intervals, then we obtain :
In this case, we have : and hence and
Also, in this case Furthermore, from the very definition of we would have :
Hence the error in replacing the sum by the integral is
Finally we notice that and hence multiplying by from outside gives us the error bound we had in
This completes the proof of the lemma.
Remark : The above proof of the lemma demonstrates a way to replace summation by integrals. There are other effective ways to do this. Such techniques of replacing summations by integrals and bounding the error term play a crucial role when one deals with asymptotic behaviour of terms as we are doing here.
In a previous lemma, we dealt with the minor arcs. Now, we consider dealing with the major arcs.
Now, we define the singular series for Waring’s problem as :
Lemma : If denotes the totality of the major arcs, then
We notice that, the above quantities allows us to deal with
In fact we shall have :
Theorem : If then
for some fixed and where
We shall gloss over the proof of this theorem, because it is quite technical in nature. One may refer to a paper by Landau for the proof.
Finally we notice that, the above theorem would give us what we aimed to prove.
There are several other equations, where the circle method finds a prominent place. For a survey on the remarkable circle method, one may refer to the book on Hardy-Littlewood method authored by Vaughan. The interested reader is suggested to take a look at this wonderful blog post on circle methods and its limitations (heuristics) authored by Prof. Terence Tao.
Hi everyone !
Last year around this time I was introduced to the immensely powerful technique of combinatorial nullstellensatz (devised by prof. Noga Alon) by prof. Fedor Petrov.
I wanted to share a note on this remarkable technique with the audience here. I had two options : make a blog post and use the LaTeX feature of wordpress to type everything out here, or upload a file for the audience to read. I chose the latter one, since in my opinion it is more convenient.
Hope you enjoy reading this note. Feel free to comment below.
Just today morning, after solving a problem here using the infamous Erdős–Szekeres theorem, I felt like posting some details of it over here, which I wrote earlier, and edited recently to add some examples.
So, I wrote this post, to share the handout I prepared on these contents. This is a part of the write-up on the notes from the advanced combinatorics course at St. Petersburg State University delivered during the 2019-2020 session.
For viewing the images, please click on the image to view it better.