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Use the zeroes of T3 to construct an interpolating polynomial of degree two for the function x^3 on the interval [-1,1]

Okay, so I have been looking at Finding the zeroes using Chebyshev polynomials which is similar, but not the same type of question.

The zeroes of T3(x)=4x3−3x are 0,±3√2.

Here is what I have tried.

$$\frac{(-\frac{\sqrt{3}}2)^3(x-0)(x-\frac{\sqrt{3}}2)}{(\frac{\sqrt{3}}2)(\frac{\sqrt{3}}2 + \frac{\sqrt{3}}2)} + (0)^3(other stuff) + \frac{(\frac{\sqrt{3}}2)^3(x-0)(x+\frac{\sqrt{3}}2)}{(\frac{\sqrt{3}}2)(\frac{\sqrt{3}}2 + \frac{\sqrt{3}}2)}$$

Am I doing this correctly? When I solve through this, it comes down to $\frac{0}{1.5}$

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We have:

$$f(x) = x^3, ~x \in [-1,1]$$

No translation of the range is required here.

Using the zeros of Chebyshev polynomials for $T_3(x)$, we have:

$$T_3(x) = 4x^3 - 3x = x(4x^2 - 3) = 0 \implies x_0 = -\dfrac{\sqrt{3}}{2} , x_1 = 0, x_3 = \dfrac{\sqrt{3}}{2}$$

We now evaluate our function, $f(x) = x^3$ at those zero values and have:

  • $f(x_0) = f\left(-\dfrac{\sqrt{3}}{2}\right) = -3 \dfrac{\sqrt{3}}{8}$
  • $f(x_1) = f(0) = 0$
  • $f(x_2) = f\left(\dfrac{\sqrt{3}}{2}\right) = 3 \dfrac{\sqrt{3}}{8}$

We can now set up and solve for the interpolating polynomial $q(x) = ax^2 + b x + c$ using the Vandermonde matrix as:

$$\begin{bmatrix}x_0^2 & x_0 & 1\\ x_1^2 & x_1 & 1\\ x_2^2 & x_2 & 1\end{bmatrix} \begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}f(x_0) \\ f(x_1) \\ f(x_2)\end{bmatrix}$$

This gives us:

$$\begin{bmatrix}-\dfrac{\sqrt{3}}{2}^2 & -\dfrac{\sqrt{3}}{2} & 1\\ 0 & 0 & 1\\ \dfrac{\sqrt{3}}{2}^2 & \dfrac{\sqrt{3}}{2} & 1\end{bmatrix} \begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}-3\dfrac{\sqrt{3}}{8} \\ 0 \\ 3\dfrac{\sqrt{3}}{8}\end{bmatrix}$$

This yields:

$$a = 0, b = \dfrac{3}{4}, c = 0$$

So, our interpolating polynomial is:

$$q(x) = \dfrac{3}{4} x$$

We can plot the original and interpolating function and have:

enter image description here

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