I am interested in constructing an polynomial that interpolates some known arbitrary function $f(x)$ over the domain $x \in [0,70]$. I want the polynomial to have degree 14 and so need 15 points.
Since I'm free to choose which 15 points I use it would probably be wise to use points indicated by the Chebyshev polynomial of order 15, call it $T_{15}(x)$ where: $T_{15}(x)=16384x^{15}-61440x^{13}+92160x^{11}-70400x^9+28800x^7-6048x^5+560x^3-15x$
I solved this which gives the points:
- -0.9945218954
- -0.9510565163
- -0.8660254038
- -0.7431448255
- -0.5877852523
- -0.4067366431
- -0.2079116908
- 0
- 0.2079116908
- 0.4067366431
- 0.5877852523
- 0.7431448255
- 0.8660254038
- 0.9510565163
- 0.9945218954
It's my understanding that to calculate the roots that I'd use for the domain $[0,70]$ I put each of these roots $x$ into the formula: $p=\frac{70}{2} + \frac{70}{2}x$
Then I interpolate using the points $(p,z(p))$. Will this give the intended interpolant?