In this lecture: https://ee227c.github.io/notes/ee227c-lecture6.pdf the author states that he wants to create a polynomial $P_{k}(a)$ in terms of Chebyshev polynomials of the first kind with the properties:

a) the Chebyshev polynomial, $T_k$, has to be rescaled to be in the range $[\alpha, \beta]$

b) at the origin, the desired $P_{k}(a)$ has to be $1$.

So he does a rescaling of the Chebyshev polynomial and states that $$ P_{k}(a) = \frac{T_{k}\left(\frac{\alpha + \beta - 2a}{\beta - \alpha}\right)}{T_{k}\left(\frac{\alpha + \beta}{\beta - \alpha}\right)} $$

I don't understand how he derives this result.


  • $\begingroup$ What do you mean by "have to work"? $\endgroup$ – Robert Israel Feb 19 '18 at 20:19
  • $\begingroup$ The author says: "We need to rescale the Chebyshev polynomials so that they’re supported on this interval", (interval $[\alpha, \beta]$) $\endgroup$ – David Feb 19 '18 at 20:20

The function $f(a) = \frac{\alpha + \beta - 2a}{\beta - \alpha}$ of the affine transformation $P_k(a) = c T_k(f(a))$ maps $[\alpha,\beta]$ to $[-1, 1]$ (in an unsual way because $f(\alpha)=1, f(\beta)=-1$, normally one would use $\alpha - \beta$ in the denominator). The requirement $P_(0)=1$ is equivalent to $1 = cT_k(f(0))$ or $$c = \frac{1}{T_k(f(0))} = \frac{1}{T_{k}\left(\frac{\alpha + \beta}{\beta - \alpha}\right)}$$

  • $\begingroup$ Thanks, but I don't know why you are using this specific $P_k(a) = c T_k(f(a))$. Why $f(a)$ has that specific form? For me, the most natural is a linear tansformation from $[-1, 1]$ to $[\alpha, \beta]$, the $f$ should be $f(a) = \frac{\beta - \alpha}{2}a + \frac{\beta + \alpha}{2}$ $\endgroup$ – David Feb 19 '18 at 21:04
  • 1
    $\begingroup$ I agree with your choice of transformation (that's why I wrote unusal). My answers accept the author's choice and derives the constant. $\endgroup$ – gammatester Feb 19 '18 at 21:16

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