q_{0} = 1 , q_{1} = x , q_{2} = x ^ 2 - 1/3 where {q_{i}} is the set of orthogonal monomials in the interval [-1,1] with respect to the standard square norm of the differentiable func- tions. Compute q_{3} and then express x ^ 3 as a q_{0} to q_{3}

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.

Learn how to compute orthogonal polynomials q0, q1, q2, and q3 on the interval [-1, 1] and express x^3 as a linear combination of these polynomials. Detailed step-by-step solution involving integral calculus and orthogonal polynomial theory.

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