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In some approximation problems I'm working on, the errors turned out to be polynomials of various degrees whose graphs on the interval $[-1,1]$ look like this: enter image description here

As you can see, these things look a bit like Chebyshev polynomials. The one shown has degree 6. The degree $n$ polynomial in this family has the following defining properties:

  • Double zeros at $t=-1$ and $t=1$
  • The other $n-4$ zeros lie in $[-1,1]$
  • There are $n-3$ extrema with values alternating between $-1$ and $+1$.

My questions: has this family of polynomials been studied before. If so, do they have a name, or can you provide references.

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    $\begingroup$ the properties seem kind of arbitrary, but you can easily give a general formula for these polynomials, I'm not sure that giving them a name will give you more relevant information. $\endgroup$
    – Carlo Beenakker
    Commented Sep 12, 2022 at 10:26
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    $\begingroup$ I guess $(1-\zeta^2)^2P_{n-4}^{(\alpha,\beta)}(\zeta)$ works with $P_{n}^{(\alpha,\beta)}$ normalized Jacobi Polynomials. For symmetrical Polynomials you can take Gegenbauer's. $\endgroup$
    – Jorge Zuniga
    Commented Sep 12, 2022 at 13:47
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    $\begingroup$ Please define "these polynomials,". $\endgroup$
    – Max Alekseyev
    Commented Sep 12, 2022 at 14:23
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    $\begingroup$ Use Rodrigues' formula to show that the zeroes of $(1-\zeta^2)P_{n+1}^{(1,1)}(\zeta)$ are the locations of extrema of $(1-\zeta^2)^2P_{n}^{(2,2)}(\zeta)$. If you sort them these extremes alternate in sign. This is easily proved since 2nd derivatives $\frac{d^2}{d\zeta^2} (1-\zeta^2)^2P_{n}^{(2,2)}(\zeta)$ are proportional to Legendre Polynomials $L_{n+2}(\zeta)$ and signs interlace at these points. $\endgroup$
    – Jorge Zuniga
    Commented Sep 13, 2022 at 13:47
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    $\begingroup$ @JorgeZuniga, the extremes alternate in sign but they don't have equal absolute values. $\endgroup$
    – Peter Taylor
    Commented Sep 13, 2022 at 14:03

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