Repeated substitution and Chebyshev polynomials
Recently, I was experimenting with a process in which we start with the polynomial and obtain by substituting , for some number . By accident, I saw that both for and , . Of course, for , as well. This made me wonder: for which values of do we have for some ?
The first few polynomials
Now we can calculate that means that
so the terms with on the left hand side (these terms amount to ) must match the terms with on the right hand side (they amount to ), the terms with on the left must match the ones on the right and the same for the constant terms. This gives the following equations:
We could have expected the , since and if , , so .
More generally, if a number divides , and for some value , then for as well.
An easier description
Notice that in the equations above, the denominator of matches the numerator of and the term with of the denominator matches the constant term of the numerator. This provides us with an easier way to describe the polynomials .
More formally, we can define a sequence of polynomials by , and , so
We will now claim that
Also, if the claim holds for , then
The second equality is obtained by multiplying numerator and denominator by . This shows that the claim holds for as well and we conclude by induction that the claim holds for all .
Finding a couple more values of
If we want , we must have
For example, iff
which has solutions . It struck me that is exactly the diagonal length of a pentagon, whilst , a root of , is the diagonal length of a square and , a root of , is the side length of a triangle, so again the distance between vertices that are two edges apart. It seems that is always a root of . We will prove this and more in the next two sections.
In this section, we will show that our polynomials are almost equal to the so-called "Chebyshev polynomials of the second kind". Knowing this, and the way that these polynomials are defined, will provide us with a neat description of the for which of the .
Firstly, recall two trigonometric identities. The angle sum identity gives
which we will call (1), and the product-to-sum identity gives
which we will call (2). Combining these, we obtain
which we will call (3).
We will now define polynomials by
These are the so-called "Chebyshev polynomials of the second kind". It is not at all obvious that these are actually polynomials, but we will show that now. We have
and, by (1),
so en . Now we can use (3) to find a recurrence relation for :
We can now claim that for all n,
Indeed, and . Now, suppose the claim holds for all with . Then
The zeroes of the Chebyshev polynomials
By definition, , so if we want and , we must have that , so for some integer . We must also have that is not zero, so .
To recap: we have shown that if some polynomial that we defined, , is zero in . Since , this happens if , which happens exactly when