# Interpolating polynomial with Chebyshev nodes

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?

-
Yes, those look like the correct nodes for the $[-1,1]$ interval, and yes that looks like the correct way to shift and scale them for the interval $[0,70]$. –  in_wolframAlpha_we_trust Nov 22 '12 at 12:01