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6 votes
2 answers
251 views

Polynomial that grows faster than any other polynomial outside $[−1,1]^n$

Consider this statement: "Chebyshev polynomials increase in magnitude more quickly outside the range $[−1,1]$ than any other polynomial that is restricted to have magnitude no greater than one ...
Mathews Boban's user avatar
0 votes
0 answers
158 views

Question on convergence of Chebyshev series

I have written a script that plots some function and its truncated Chebyshev expansion in $[-1,1]$, which is given by $$ \sum a_nT_n(x) \quad \text{with} \quad a_n = \int_{-1}^{1}T_n(x)f(x)/\sqrt{1-x^...
user210089's user avatar
2 votes
0 answers
139 views

Approximating Chebyshev Polynomials by Truncated Fourier Series

It is known (see e.g. https://www.math.ucdavis.edu/~bremer/classes/fall2018/MAT128a/lecture9.pdf) that any continuous function $f: [-1,1] \to \mathbb{C}$ admits a Chebyshev expansion $$ f(x) = \sum_{n=...
Lior Eldar's user avatar
3 votes
1 answer
157 views

Is there any subsequence of the sequence $(\frac{\cos(\alpha - n \beta) - \lambda \cos(\alpha + n \beta)}{ \cos^n(\beta) })$ that converges?

Let's take $\alpha$ and $\beta$ two reals in $(0, \frac{\pi}{2})$. Let's take $\lambda \in (0,1)$. Let's define the sequence $(u_n)$ as follows: \begin{eqnarray} u_n & = & \frac{\cos(\alpha - ...
InfiniteMath's user avatar
1 vote
1 answer
220 views

Relation between two extremal properties of Chebyshev polynomials

The Chebyshev polynomials $T_n$ have the following “minimal ∞-norm” property, also known as “Chebyshev's theorem”: (A) Let $P$ be an $n$-the degree polynomial with leading coefficient $2^{n-1}$. Then ...
Martin R's user avatar
  • 117k
1 vote
0 answers
78 views

The envelope for the extremals of $\cos((n+1) \arccos x)-\cos((n-1) \arccos x)$ forms an ellipse.

The Chebyshev polynomial of the first kind is defined on $[-1, 1]$ by $$T_n(x) = \cos(n \arccos x).$$ Prove that the envelope for the extremals of $T_{n+1}(x)-T_{n-1}(x)$ forms an ellipse. The ...
Ho-Oh's user avatar
  • 919
2 votes
0 answers
58 views

Fractional exponent elementary symmetric polynomials.

I am wondering if there is any literature on relations between fractional power symmetric polynomials. For a particular example, with the variables $\textbf{x} = (x_1,x_2,\dots x_n),$, can we ...
dezdichado's user avatar
  • 14.1k
-2 votes
2 answers
153 views

Chebyshev polynomials semigroup property $T_n \circ T_m = T_{nm}$

Consider set of Chebyshev polynomials $T_n(x):\mathbb{R} \to \mathbb{R}$ given by formula $$ T_n(\cos(x)) = \cos(nx) $$ I am interested in elegant way to show that Chebyshev polynomials form a ...
NeoFanatic's user avatar
0 votes
1 answer
82 views

How we generalize the cartesian form of epicycloids?

I have the following parametric form of epicycloids: $x(t)=\frac{a\cdot\cos t+\cos(a\cdot t)}{1+a}$ $y(t)=\frac{a\cdot\sin t+\sin(a\cdot t)}{1+a}$ where $a=2,3,4,\ldots$ is a variable that ...
user avatar
2 votes
1 answer
1k views

Prove the orthogonality relation of Chebyshev polynomials of the first kind

The Chebyshev polynomials of the first kind are obtained from the recurrence relation $$\begin{aligned}T_{0}(x)&=1\\T_{1}(x)&=x\\T_{n+1}(x)&=2x\,T_{n}(x)-T_{n-1}(x)~.\end{aligned}$$ Prove ...
Mark's user avatar
  • 7,880
0 votes
0 answers
64 views

Chebyshev expansion of $f(x)=\frac{1}{1+(x-s)^2}$

The Chebyshev polynomials of the first kind are obtained from the recurrence relation $$\begin{aligned}T_{0}(x)&=1\\T_{1}(x)&=x\\T_{n+1}(x)&=2x\,T_{n}(x)-T_{n-1}(x)~.\end{aligned}$$ I ...
Mark's user avatar
  • 7,880
0 votes
1 answer
43 views

Can we define the generating function for all $x$ and all $t$

The Chebyshev polynomials of second kind are defined for any $x \in \Bbb R$ (or even $x \in \Bbb C$), e.g. via the recurrence relation $$ U_0(x) = 1 \\ U_1(x) = 2x \\ U_{n+1}(x) = 2x U_n(x) - U_{...
Safwane's user avatar
  • 3,854
1 vote
0 answers
125 views

Show that the lowest-norm monic polynomial is of the form $\frac{(b-a)^n}{2^n}\frac{1}{2^n}T_n\left(\frac{2}{b-a}x-\frac{b+a}{b-a}\right)$

Let $T_n\in P_n[-1,1]$ the n-th Chebyshev polynomial. Show that the lowest-norm monic polynomial in $P_n[a,b]$ is of the form $$\frac{(b-a)^n}{2^n}\frac{1}{2^n}T_n\left(\frac{2}{b-a}x-\frac{b+a}{b-a}\...
Sofía Contreras's user avatar
2 votes
1 answer
57 views

Finding zeroeth coefficient of a Chebyshev polynomial expansion

Let $v_\theta = (\cos\theta,\sin\theta)$ be a unit vector in the plane. I have a kernel $p(\theta,\theta') = p(v_\theta\cdot v_{\theta'})$ that satisfies $$\int_0^{2\pi} p(v_\theta\cdot v_{\theta'})\,...
Phil's user avatar
  • 1,088
1 vote
0 answers
36 views

Chebyshev polynom problem.

Consider $P_n = \{Q_n(x) : \deg Q_n = n; \|{Q_n}\| = \max_{[a,b]}|Q_n(x)| = M >0\}$. Now consider $\bar{T}(x) = M T_n(\frac{2x - (b+a)}{b-a}) - $ Chebyshev polynomial (normed on space $P_n)$. We ...
openspace's user avatar
  • 6,550

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