This set is due Monday, November 8.
- The goal of this exercise is to use the method of generating functions to verify some results of D.J. Newman and Paul Erdös.
For the purposes of this exercise, an arithmetic progression is a subset of of the form for some fixed naturals with , and we call the common difference of .
Recall that is a partition of a set iff the following hold:
- for all .
- for all .
We want to show the following:
Suppose that is partitioned into finitely many arithmetic progressions . Then either , or at least two of the common differences coincide. Moreover, if denote the common differences, then
To do this, we can use the method of generating functions as follows: Associate to the generating function of :
- Verify that
where the last equality is valid for all . (We will need that it is valid for all complex numbers with . You may assume this.)
- For , define so that . Find a formula for similar to the formula above for , involving and .
- Suppose that and the are all different. Let be the largest of them, and consider
Derive a contradiction by letting .
and use L’Hôpital’s rule to deduce that
- For a positive integer and , let
Suppose that are positive integers and . Show that for any ,
For this, you may want to verify first that the numbers of the form with and run through a complete residue system .
- This problem requires some basic facts about matrices, eigenvalues, and traces. A good reference is Sheldon Axler, Linear algebra done right. Springer, 2nd edition (1997), but feel free to ask me if you are not familiar or do not feel comfortable with linear algebra. Let where is an odd integer. Consider the matrix with entry in row , column ().
- Show that is a permutation matrix, i.e., all its entries are 0s or 1s, and there is exactly one 1 in each column and each row.
- Show that , and conclude that the eigenvalues of are all among .
- Verify that the trace of is 1 to conclude that the eigenvalues of are 1 with multiplicity and with multiplicity .
- Solve exercises 4.6.9–11 from the textbook.
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