Sometimes, cosine is unexpected

Revision en4, by adamant, 2023-03-12 05:01:57

Hi everyone!

Consider the sequence of Laurent polynomials (i.e. polynomials, in which negative degrees are allowed)

$$$ A_k = \left(x+\frac{1}{x}\right)^k, $$$

and another sequence of Laurent polynomials

$$$ B_k = x^k + \frac{1}{x^k}. $$$

As it turns out, these two sequences have the same linear span. What it practically means, is that every element of $$$B_k$$$ may be represented as a finite linear combination of the elements of $$$A_k$$$, and vice versa. It is somewhat trivial in one direction:

$$$ A_k = \sum\limits_{i=0}^k \binom{k}{i} x^{k-2i} = \sum\limits_{i=0}^{\lfloor k/2 \rfloor} \binom{k}{i} \left(x^{k-i}+\frac{1}{x^{k-i}}\right). $$$

With some special consideration for even $$$k$$$, as the middle coefficient should be additionally divided by $$$2$$$. Now, the inverse transform is much trickier. To find it, consider the substitution $$$x=e^{i \theta}$$$, then you will notice that

$$$\begin{gather} A_k = 2^k \cos^k \theta, \\ B_k = 2 \cos k \theta. \end{gather}$$$
Tags tutorial, cosinus, polynomials, chebyshev

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  Rev. Lang. By When Δ Comment
en8 English adamant 2023-03-12 21:11:47 0 (published)
en7 English adamant 2023-03-12 21:11:39 13 (saved to drafts)
en6 English adamant 2023-03-12 14:01:40 89
en5 English adamant 2023-03-12 13:55:55 6437 (published)
en4 English adamant 2023-03-12 05:01:57 1032 Tiny change: 'r even $k$' -> 'r even $k$, as it would be additionally divided by $2$.'
en3 English adamant 2023-03-12 04:43:57 84 Tiny change: '.' -> 'Hi everyone!\n\n'
en2 English adamant 2023-03-12 04:42:59 45
en1 English adamant 2023-03-12 04:42:29 50 Initial revision (saved to drafts)