The Chebyshev polynomials are a sequence of orthogonal polynomials that are related to De Moivre's formula. They have numerous properties, which make them useful in areas like solving polynomials and approximating functions.
The Chebyshev polynomial of the first kind, denoted by , is defined as
Since we know that we can conclude that
To find , we can equivalently find a function of in terms of .
Using the cosine sum formula, we get
Recall that . Then
To find , we can, just as in the previous example, find a function of in terms of .
Using the cosine sum formula again, we get But then we have to replace with and then manually compute and then...
Forget it. This is turning into a hopeless bash; we can't be doing this for or , and we definitely can't easily generalize this to . We could always use De Moivre's formula, but the calculation is also very extensive.
If only there were an easier way...
How would we obtain a more general formula? In answering the previous question, most people tried to expand
We can easily convert the first 2 terms into the form. However, the issue with this approach is that is not easy to deal with, and will (currently) require much further expansion.
Instead, we will use the fact (from trigonometric sum and product formulas) that
This gives us the recurrence relation:
The following is a table of initial values of :
We can make the following conjectures about these coefficients:
- Coefficients are integers.
- Constant term is for and 0 for .
- Leading term is .
- Linear term of is .
- The non-zero terms have exponents with the same parity as .
- Sum of coefficients is 1.
- Non-zero coefficients alternate in sign.
- The coefficient of the term in is .
- The sum of the absolute value of the coefficients is .
- The roots of are where .
We will prove some of these conjectures. The rest are left as exercises for the reader.
Prove conjecture 2.
We will prove this using induction. First, in the two base cases, we see that and , which satisfies that the constant terms are and , respectively. Now we must prove that given has a coefficient of if and if , then also satisfies this.
Note that the constant term can be evaluated by plugging in . Doing so in the recurrence relation of gives
This means that if , then ; also, if , then , completing the induction.
Prove conjecture 6.
The sum of the coefficients of is just . Recalling that , we see that we want to evaluate when . But this means , so .
Therefore, and we are done.
Prove conjecture 10.
We desire to find the roots of .
Substituting , we want to instead find the roots of
This happens at for .
But this means so we are done.
Because we have a function relating to , it makes sense to suspect that there is such a function for and too. Indeed, the Chebyshev polynomials of the second kind are exactly this:
The Chebyshev polynomial of the second kind, denoted by , is defined by
- Prove that
- Prove that
- Show that the generating function for is given by