Discriminant and Resultant

Question

Please help to calculate the discriminant: $$D(\cos(n\arccos(x))$$

How to calculate this according to the discriminant formula through the product of the squares of the root differences?

I managed to do this for polynomials, but I have never encountered a trigonometric case.

Explain, please.I really want to understand how to solve such cases

Answer 1

I don't know if there is any definition for the resultant or discriminant of trig functions in general. But notice that your function is a (famous) polynomial in disguise. In fact, $$T_n(x) = \cos(n \cos^{-1}(x))$$ is the $n$-th Chebyshev polynomial (of the first kind)

For example, $$ \cos(3 \cos^{-1}(x)) = 4x^3 -3x$$ and you can use the usual computations for the discriminant and resultant.

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Note that the expression $T_n = \cos(n \cos^{-1}(x))$ gives you a way to compute a pretty formula for the discriminant of $T_n$ without knowing the coefficients of $T_n$ explicitly. Let me outline it as a series of exercises:

Exercise 1: The roots of $T_n$ are $$x_k = \cos\left(\frac{2k - 1}{2n}\pi\right),\qquad k = 1,\dots,n.$$

Recall that the discriminant of a polynomial $f(x)$ with leading coefficients $a$ is given by $$\Delta = (-1)^{n(n-1)/2}a^{n - 2}f'(x_1)\dots f'(x_n),$$ where $f'$ denotes the derivative of $f$ and $x_1,\dots, x_n$ are the roots of $f(x)$.

Exercise 2: With the notation from exercise 1, $$T_n'(x_k) = (-1)^{k- 1}\frac{n}{\sin\left(\frac{(2k - 1)\pi}{2n}\right)}$$

and write $$S_{k, n} = \sin\left(\frac{(2k - 1)\pi}{2n}\right)$$

Exercise 3: $$\prod_{k = 1}^n S_{k, n} = \frac{1}{2^{n - 1}}$$

Hint: Euler's formula $$\sin(\theta) = \frac{i}{2}(e^{-i\theta} - e^{i\theta})$$ might help. You might also want to use the factorization $$x^n + 1 = \prod_{k = 1}^n (x - e^{\pi i (2k - 1)/n}).$$

Using the expression for $\Delta$ above, conclude that:

Exercise 4: The discriminant of $T_n$ is $$\Delta_n = 2^{(n -1)^2}n^n$$