Given the set of orthogonal polynomials $\{q_i\}$ defined on the interval $[-1,1]$ with respect to the standard square norm, the polynomials $q_0, q_1, q_2$ are already provided:

$q_0 = 1, \quad q_1 = x, \quad q_2 = x^2 - \frac{1}{3}$

We need to compute $q_3$ and then express $x^3$ as a linear combination of $q_0, q_1, q_2, q_3$.

Step 1: Compute $q_3$

Orthogonal polynomials $q_3(x)$ will have the general form:

$q_3(x) = x^3 + a_2 x^2 + a_1 x + a_0$

We require $q_3(x)$ to be orthogonal to $q_0(x), q_1(x), q_2(x)$ over the interval $[-1, 1]$.

Orthogonality Conditions:

1. $\int_{-1}^{1} q_3(x) \cdot q_0(x) \, dx = 0$
2. $\int_{-1}^{1} q_3(x) \cdot q_1(x) \, dx = 0$
3. $\int_{-1}^{1} q_3(x) \cdot q_2(x) \, dx = 0$

Given these, we calculate the coefficients $a_2, a_1, a_0$.

Orthogonality with $q_0(x) = 1$:

$\int_{-1}^{1} \left( x^3 + a_2 x^2 + a_1 x + a_0 \right) dx = 0$

Calculating the integrals:

$\int_{-1}^{1} x^3 \, dx = 0, \quad \int_{-1}^{1} x^2 \, dx = \frac{2}{3}, \quad \int_{-1}^{1} x \, dx = 0, \quad \int_{-1}^{1} dx = 2$

Thus:

$a_0 \cdot 2 + a_2 \cdot \frac{2}{3} = 0 \implies a_0 + \frac{a_2}{3} = 0 \implies a_0 = -\frac{a_2}{3}$

Orthogonality with $q_1(x) = x$:

$\int_{-1}^{1} \left( x^3 + a_2 x^2 + a_1 x + a_0 \right) x \, dx = 0$

This simplifies to:

$\int_{-1}^{1} x^4 \, dx + a_2 \int_{-1}^{1} x^3 \, dx + a_1 \int_{-1}^{1} x^2 \, dx + a_0 \int_{-1}^{1} x \, dx = 0$

Given $\int_{-1}^{1} x^4 \, dx = \frac{2}{5}$:

$\frac{2}{5} + a_1 \cdot \frac{2}{3} = 0 \implies a_1 = -\frac{3}{5}$

Orthogonality with $q_2(x) = x^2 - \frac{1}{3}$:

$\int_{-1}^{1} \left( x^3 + a_2 x^2 + a_1 x + a_0 \right) \left( x^2 - \frac{1}{3} \right) dx = 0$

This expands to:

$\int_{-1}^{1} \left( x^5 + a_2 x^4 + a_1 x^3 + a_0 x^2 - \frac{1}{3}x^3 - \frac{a_2}{3}x^2 - \frac{a_1}{3}x - \frac{a_0}{3} \right) dx = 0$

Matching coefficients and solving, we get $a_2 = 0$, $a_1 = -\frac{3}{5}$, $a_0 = 0$.

Thus:

$q_3(x) = x^3 - \frac{3}{5}x$

Step 2: Express $x^3$ in terms of $q_0, q_1, q_2, q_3$

We need to find constants $c_0, c_1, c_2, c_3$ such that:

$x^3 = c_0 q_0(x) + c_1 q_1(x) + c_2 q_2(x) + c_3 q_3(x)$

Using the known polynomials:

$x^3 = c_0 (1) + c_1 (x) + c_2 \left( x^2 - \frac{1}{3} \right) + c_3 \left( x^3 - \frac{3}{5} x \right)$

Comparing coefficients, we find:

$c_3 = 1, \quad c_1 - \frac{3}{5}c_3 = 0 \implies c_1 = \frac{3}{5}, \quad c_2 = 0, \quad c_0 = 0$

So:

$x^3 = q_3(x) + \frac{3}{5}q_1(x)$

Final Expression:

$x^3 = q_3(x) + \frac{3}{5}q_1(x)$

Would you like more details on the calculations? Here are 5 related questions to consider:

1. What are the properties of orthogonal polynomials in general?
2. How can we derive higher-order orthogonal polynomials for different intervals?
3. What role does the weight function play in orthogonal polynomials?
4. Can you explain the Gram-Schmidt process in the context of generating orthogonal polynomials?
5. How does orthogonality influence the convergence of polynomial approximations?

Tip: Orthogonal polynomials are useful in numerical analysis for approximating functions, especially in quadrature methods.