Let $n \in \mathbb N$ and consider the polynomial function $f_n \colon \mathbb R \to \mathbb R$ defined by $$f_n(x) = \sum_{k=0}^n (-1)^k \binom {2n+1} {2k+1} (1 - x^2)^{n-k} x^{2k}$$ for any $x \in \mathbb R$. (These functions are related to Chebyshev polynomials, see the update below.)

By plotting the graphs of the functions as $n$ increases, one sees that they exhibit an oscillating behavior in $[-1, 1]$. For example, here are the graphs of $f_3, f_5, f_7$:


As $n \to \infty$, it looks as though the crests of the wave describe the graph of another function. For example, here is the graph of $f_{50}$:

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Let $f \colon D \to \mathbb R$ be defined by $$f(x) = \limsup_{n \to \infty} f_n(x)$$ whenever the limit superior exists and is finite. I would like to find as much information as possible about this function.

So far, I have only been able to show the following (see the update below):

  1. $f$ is an even function, since all of the $f_n$'s are even.

  2. $0 \notin D$. Indeed, $f_n(0) = 2n + 1 \to \infty$ as $n \to \infty$.

  3. $f(\pm 1) = 1$, because $f_n(\pm 1) = (-1)^n$ for any $n \in \mathbb N$.

  4. $f \left (\pm \frac {\sqrt 2} 2 \right ) = 1$. This is because: $$f_n \left ( \pm \frac {\sqrt 2} 2 \right ) = \sum_{k=0}^n (-1)^k \binom {2n+1} {2k+1} \left ( \frac 1 2 \right )^n = (-1)^{\left \lfloor \frac n 2 \right \rfloor} 2^n \left ( \frac 1 2 \right )^n = (-1)^{\left \lfloor \frac n 2 \right \rfloor} \le 1$$ In particular, $f_{4m} \left (\pm \frac {\sqrt 2} 2 \right ) = 1$ for any $m \in \mathbb N$, so $\limsup_{n \to \infty} f_n \left ( \pm \frac {\sqrt 2} 2 \right ) = 1$.

By looking at the definition of $f_n(x)$, it seems as though one should use the binomial theorem to find a better expression to work with, but I'm not sure how.

What else can we say about $f$? Is it possible to find a "simple" expression?

Thank you in advance for any reply.

Update: By looking up the coefficients of the first few polynomials, I found out that they are closely related to the Chebyshev polynomials of the second kind. In fact, it appears that $$f_n(\sin \alpha) = \frac {\sin ((2n+1) \alpha)}{\sin \alpha}$$ for any $\alpha \in \mathbb R \smallsetminus \pi \mathbb Z$, which immediately provides us with many other values of $f$. For instance, $$f_n \left (\sin \frac \pi 6 \right ) = \frac{\sin \left ( (2n+1) \frac \pi 6 \right )}{\sin \frac \pi 6} \le \frac 1 {\frac 1 2} = 2$$ In particular, $$f_{6m+1} \left (\sin \frac \pi 6 \right ) = \frac{\sin \left ( (12 m + 3) \frac \pi 6 \right )}{\sin \frac \pi 6} = \frac{\sin \left ( 2 m \pi + \frac \pi 2 \right )}{\sin \frac \pi 6} = \frac 1 {\frac 1 2} = 2$$ for any $m \in \mathbb N$, and thus $f \left (\pm \frac 1 2 \right ) = 2$.

How can we get a simple expression for $f$ using this information?

  • $\begingroup$ One thing which looks interesting is the behavior close to $x=0$; usinf Taylor,we have $$f_n(x)=(2 n+1)-\frac{2}{3} \left(2 n^3+3 n^2+n\right) x^2+\frac{2}{15} \left(2 n^5+5 n^4-5 n^2-2 n\right) x^4+O\left(x^6\right)$$ $\endgroup$ Jul 20, 2018 at 8:09
  • $\begingroup$ Interested by your observations, I added a few things (of no use, I am afraid). $\endgroup$ Jul 25, 2018 at 4:10

2 Answers 2


This is not an answer since it is just the result from a CAS.

Defining $$u=1-2 x^2-2 \sqrt{x^2 \left(x^2-1\right)} \qquad \text{and}\qquad v=1-2 x^2+2 \sqrt{x^2 \left(x^2-1\right)}$$ a CAS produced

$$f_n(x)=\frac{ \left(u^n+v^n\right)}{2 }+\frac{ \left(u^n-v^n\right)}{2 }\,\frac{\sqrt{x^2 \left(x^2-1\right)} }{ x^2}$$


This will not help much, I am afraid, but after your edit, I computed $f_n\left(\sin \left(\frac{\pi k}{12}\right)\right)$ and obtained the (may be) interesting values $$\left( \begin{array}{cc} k & f_n\left(\sin \left(\frac{\pi k}{12}\right)\right) \\ 0 & 2 n+1 \\ 1 & \cos \left(\frac{n \pi }{6}\right)+\left(2+\sqrt{3}\right) \sin \left(\frac{n \pi }{6}\right) \\ 2 & \cos \left(\frac{n \pi }{3}\right)+\sqrt{3} \sin \left(\frac{n \pi }{3}\right) \\ 3 & \cos \left(\frac{n \pi }{2}\right)+\sin \left(\frac{n \pi }{2}\right) \\ 4 & \cos \left(\frac{2 n \pi }{3}\right)+\frac{1}{\sqrt{3}}\sin \left(\frac{2 n \pi }{3}\right) \\ 5 & \cos \left(\frac{5 n \pi }{6}\right)+\left(2-\sqrt{3}\right) \sin \left(\frac{5 n \pi }{6}\right) \\ 6 & (-1)^n \end{array} \right)$$

  • $\begingroup$ Which computer algebra system is this? I would love to see a "real" derivation of this by a general technique. $\endgroup$
    – Hans
    Jul 20, 2018 at 5:28
  • $\begingroup$ @Hans. Me too,but I m too lazy to work the proof. Concerning the CAS, it is an old junk we made over years in my research group to face our needs. Can you access Mathematica (it could probably do it) ? I just checked now on Wolfram Development Platform (which is free) : it gives the same. $\endgroup$ Jul 20, 2018 at 5:39
  • $\begingroup$ I suspected Wolfram would be able to do it. I am not concerned about proof which is just a routine and tedious computation. I am more interested in the algorithm that a CAS uses to derive the function. $\endgroup$
    – Hans
    Jul 20, 2018 at 5:43
  • $\begingroup$ I tried Wolfram with no success. $\endgroup$ Jul 20, 2018 at 5:59
  • $\begingroup$ I've updated my question, maybe you'd like to look at it again. $\endgroup$ Jul 24, 2018 at 18:59

Note that $f_n(0)=2n+1$ and so $f$ is undefined for $x=0$.

Otherwise, for $\alpha \in (0,\pi/2]$ where $\sin(\alpha)>0$ $$f(\sin(\alpha)) = \frac{\sup_n \sin((2n+1)\alpha)}{\sin(\alpha)}.$$ If $\alpha$ is an irrational multiple of $2\pi$ then $(2n+1)\alpha \pmod{2\pi}$ is dense in $[0, 2\pi]$ and therefore $\sup_n \sin((2n+1)\alpha)=1$. So in this case $$f(\sin(\alpha))=\sin^{-1}(\alpha).$$ If $\alpha = 2\pi \frac{p}{q}$ for some coprime integers $p, q$ then a case by case examination on $q \pmod 8$ shows

$$s(q) = \sup_n \sin((2n+1)\alpha) = \begin{cases} \cos(\frac{\pi}{2q}) & \textrm{If $q$ is odd}\\ \cos(\frac{\pi}q) & \textrm{If $q \equiv 2, 6 \pmod 8$}\\ \cos(\frac{2\pi}q) & \textrm{If $q \equiv 0 \pmod 8$}\\ 1 & \textrm{If $q \equiv 4 \pmod 8$} \end{cases}$$ so in this case $$f(\sin(\alpha))=\frac{s(q)}{\sin(\alpha)}.$$ Since you already noted that $f$ is even this completely describes $f$.

  • $\begingroup$ That's not true, because $f \left ( \pm \frac {\sqrt 2} 2 \right ) = 1 \neq \sqrt 2$. Indeed, if $\alpha = \frac \pi 4$, there is no $n \in \mathbb N$ such that $\sin((2n + 1) \alpha) = 1$, because $\sin((2n +1) \alpha) = \pm \frac {\sqrt 2} 2$. $\endgroup$ Jul 24, 2018 at 19:21
  • $\begingroup$ @LucaBressan This is the "obvious" function. It is indeed not pointwise the supremum. I'll edit my answer to reflect this. $\endgroup$
    – WimC
    Jul 24, 2018 at 19:22
  • $\begingroup$ If $\operatorname{asin}(x)$ is not a rational multipe of $\pi$ then the supremum is actually equal to $\lvert x \rvert^{-1}$. $\endgroup$
    – WimC
    Jul 24, 2018 at 19:27
  • $\begingroup$ I agree that $x \mapsto \lvert x \rvert^{-1}$ is a "natural" answer to the problem, but I'm still curious to see if there's an easy way to include the exceptions. $\endgroup$ Jul 25, 2018 at 9:23
  • $\begingroup$ Could you please clarify a bit how you got the expression for $s(q)$ by case examination? I see that if $q \equiv 4 \pmod 8$ then letting $q = 8t + 4$ and $n = [p (2t+1) - 1]/2$ we have $$\sin \left ( (2n+1) \alpha \right ) = \sin \left ( \frac {(2n+1) p} {2t+1} \cdot \frac \pi 2 \right ) = \sin \left ( p^2 \frac \pi 2 \right ) = 1$$ because $p^2 \equiv 1 \pmod 4$ for any odd $p$. But the other cases seem quite harder to prove. $\endgroup$ Jul 25, 2018 at 13:11

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