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4 votes
2 answers
73 views

Roots and extrema of the polynomial $P_n(x)=\sum_{k=0}^n\binom{n+k}{2k}(-x)^k$.

Answering a recent question I came across the family of polynomials: $$P_n(x)=\sum_{k=0}^n\binom{n+k}{2k}(-x)^k$$ with numerical evidence of the following interesting properties: $P_n(2)=\begin{cases}...
user's user avatar
  • 26.7k
0 votes
1 answer
57 views

Largest root of a linear combination of Chebyshev polynomials

I wonder if we can say something about roots of a linear combination of Chebyshev polynomials of the first kind. I have an example in my hand: $$(m+1)T_n(x)+(m-3)T_{n-2}(x)=0$$ for some $m>0$. I ...
kswim's user avatar
  • 433
0 votes
1 answer
363 views

Complex argument in Chebyshev polynomials of second kind?

I am looking at Chebyshev polynomials of second kind in order to characterize the spectra of $2$-Toeplitz perturbed matrices (I am not a mathematician myself, just a control theoretician). In all the ...
lyapunov00's user avatar
3 votes
1 answer
896 views

Relationship between Chebyshev polynomials and square roots ($\sqrt{3}+\sqrt{2}=\frac{1}{\sqrt{T_1(5)-\sqrt{T_1(5)^2-1}}}$ etc.)

(If my English is strange, I would appreciate it if you could correct it.) There seems to be a property about the sum of square roots. (This is almost self-explanatory.) let $ a,\ b,\ t \in \mathbb{N}^...
isato's user avatar
  • 426
3 votes
3 answers
121 views

Proving that $\sec\frac\pi{30}=\sqrt{2-\sqrt{5}+\sqrt{15-6\sqrt{5}}}$

I recently saw on this site, the identity $$\sec\frac\pi{30}=\sqrt{2-\sqrt{5}+\sqrt{15-6\sqrt{5}}}$$ which I instantly wanted to prove. I know that I can "reduce" the problem to the evaluation of $\...
clathratus's user avatar
  • 17.3k
2 votes
0 answers
95 views

Vieta's Formula for Chebyshev basis

Let $p(x)=x^d+\sum_{i=0}^{d-1} a_ix^i$. Then Vieta's formula tells us that the $a_i$ can be expressed as signed elementary symmetric polynomials of the roots $\{\alpha_1,\ldots,\alpha_d\}$ of $p(x)$: $...
Taylor's user avatar
  • 161
3 votes
0 answers
101 views

$\cos\frac\pi{n}$ Analytic expression

I recently found out that $$\sin\frac\pi5=\frac12\sqrt{\frac{5-\sqrt5}2}$$ Which means that $$\cos\frac\pi5=\frac{1+\sqrt5}4$$ I also recently found that if $n\in\Bbb N$, $$\sin nx=\sin x\,U_{n-1}(\...
clathratus's user avatar
  • 17.3k
2 votes
0 answers
40 views

Numerical analysis: Chebyshev coefficient representation error.

I am unsure if numerical analysis questions are suitable for this forum, but I have nowhere else to ask, so if this question is inappropriate, tell me and I will delete it. If $x_k$ are the Chebyshev ...
user401855's user avatar
  • 1,077
0 votes
1 answer
112 views

Solving $x+y+z+u+t=0$, $x^3+y^3+z^3+u^3+t^3=0$, $x^5+y^5+z^5+u^5+t^5=-10$

Find all real numbers $x$, $y$, $z$, $t$, $u$ that $x+y+z+u+t=0$ $x^3+y^3+z^3+u^3+t^3=0$ $x^5+y^5+z^5+u^5+t^5=-10$ I'm learning about Chebyshev polynominals but in this case, I still haven't got ...
Uchiha Itachi's user avatar
7 votes
3 answers
3k views

Roots of the Chebyshev polynomials of the second kind.

It is known that the roots of the Chebyshev polynomials of the second kind, denote it by $U_n(x)$, are in the interval $(-1,1)$ and they are simple (of multiplicity one). I have noticed that the roots ...
Math137's user avatar
  • 1,849
2 votes
1 answer
185 views

Building a cubic function with integer coefficients and trigonometric roots

I want to find the answer to the following problem: Construct a cubic polynomial with integer coefficients, whose roots - $\cos{\frac{2 \pi}{7}}$, $\cos{\frac{4 \pi}{7}}$ and $\cos{\frac{6 \pi}{7}}$. ...
Victor's user avatar
  • 21
4 votes
1 answer
120 views

Is it possible to calculate the roots of the difference between successive terms of this polynomial series $\rm{P}_n (x)=x\rm{P}_{n-1}-r\rm{P}_{n-2}$

Consider the polynomial series defined by the following recursion formula: $$ \begin{align} &\mathrm{P}_0 = 1 \\ &\mathrm{P}_1 = x-r \\ &\mathrm{P}_n = x\mathrm{P}_{n-1} - r\mathrm{P}_{n-...
JMK's user avatar
  • 493
5 votes
1 answer
10k views

Calculation of Chebyshev coefficients

The Chebyshev polynomials can be defined recursively as: $T_0(x)=1$; $T_1(x)=x$; $T_{n+1}(x)=2xT_n(x) + T_{n-1}(x)$ The coefficients of these polynomails for a function, $\space f(x)$, under ...
Hasan's user avatar
  • 257