Given 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 linear combination of q_{0} to q_{3}

The problem involves finding the next orthogonal polynomial \( q_3(x) \) given the first three orthogonal polynomials \( q_0(x) = 1 \), \( q_1(x) = x \), and \( q_2(x) = x^2 - \frac{1}{3} \). We will use the Gram-Schmidt process to ensure the orthogonality of the polynomials over the interval \([-1, 1]\) with respect to the standard square norm:

\[
\langle f, g \rangle = \int_{-1}^{1} f(x) g(x) \, dx
\]

### Step 1: Find \( q_3(x) \)

We assume \( q_3(x) \) is a linear combination of monomials up to \( x^3 \) because the degree of \( q_3(x) \) should be 3. Hence, let:

\[
q_3(x) = x^3 + A x^2 + B x + C
\]

To ensure orthogonality, \( q_3(x) \) must satisfy:

\[
\langle q_3, q_0 \rangle = 0, \quad \langle q_3, q_1 \rangle = 0, \quad \langle q_3, q_2 \rangle = 0
\]

#### Orthogonality with \( q_0(x) \):

\[
\langle q_3, q_0 \rangle = \int_{-1}^{1} (x^3 + A x^2 + B x + C) \cdot 1 \, dx = \int_{-1}^{1} (x^3 + A x^2 + B x + C) \, dx
\]

\[
= \left[ \frac{x^4}{4} \right]_{-1}^{1} + A \left[ \frac{x^3}{3} \right]_{-1}^{1} + B \left[ \frac{x^2}{2} \right]_{-1}^{1} + C \left[ x \right]_{-1}^{1}
\]

Since all odd powers of \( x \) vanish when integrated over \([-1, 1]\), we get:

\[
\langle q_3, q_0 \rangle = 0 + 0 + 2B \cdot \frac{1}{2} + 0 = B
\]

Thus, \( B = 0 \).

#### Orthogonality with \( q_1(x) \):

\[
\langle q_3, q_1 \rangle = \int_{-1}^{1} (x^3 + A x^2 + C) \cdot x \, dx = \int_{-1}^{1} (x^4 + A x^3 + C x) \, dx
\]

\[
= \left[ \frac{x^5}{5} \right]_{-1}^{1} + A \left[ \frac{x^4}{4} \right]_{-1}^{1} + C \left[ \frac{x^2}{2} \right]_{-1}^{1}
\]

Again, odd powers vanish, and we are left with:

\[
\langle q_3, q_1 \rangle = 0 + A \cdot \frac{2}{4} + 0 = \frac{A}{2}
\]

Thus, \( A = 0 \).

#### Orthogonality with \( q_2(x) \):

\[
q_2(x) = x^2 - \frac{1}{3}
\]
\[
\langle q_3, q_2 \rangle = \int_{-1}^{1} (x^3 + C) \cdot (x^2 - \frac{1}{3}) \, dx = \int_{-1}^{1} (x^5 - \frac{1}{3} x^3 + Cx^2 - \frac{C}{3}) \, dx
\]

This simplifies to:

\[
\langle q_3, q_2 \rangle = 0 + \frac{C}{3} \cdot 2 - 0 = \frac{2C}{3}
\]

Thus, \( C = 0 \).

### Conclusion for \( q_3(x) \):

Therefore, \( q_3(x) = x^3 \), but we need to normalize it to ensure it satisfies the orthogonality conditions correctly. Let’s calculate the norm of \( x^3 \):

\[
\| q_3 \|^2 = \langle x^3, x^3 \rangle = \int_{-1}^{1} x^6 \, dx = \left[ \frac{x^7}{7} \right]_{-1}^{1} = \frac{2}{7}
\]

Thus, the normalized polynomial is:

\[
q_3(x) = \sqrt{\frac{7}{2}} x^3
\]

### Step 2: Express \( x^3 \) as a linear combination of \( q_0, q_1, q_2, q_3 \)

We can now express \( x^3 \) as a linear combination of the orthogonal polynomials:

\[
x^3 = c_0 q_0 + c_1 q_1 + c_2 q_2 + c_3 q_3
\]

Where \( q_0 = 1 \), \( q_1 = x \), \( q_2 = x^2 - \frac{1}{3} \), and \( q_3 = \sqrt{\frac{7}{2}} x^3 \).

We now need to compute the coefficients \( c_0, c_1, c_2, c_3 \) by projecting \( x^3 \) onto each of the polynomials.

Given 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 differentiable functions, compute q_3, and then express x^3 as a linear combination of q_{0} to q_{3}.

Explore how to compute the third orthogonal polynomial q3 using the Gram-Schmidt process and express x^3 as a linear combination of orthogonal polynomials q0 to q3. This problem covers advanced mathematical concepts, including orthogonal polynomials, linear algebra, and integration. Suitable for advanced high school or college-level students.

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