We are given three functions $\{u_1(x) = -3, u_2(x) = -12x, u_3(x) = -12x^2\}$, which form a basis for a subspace of $P_2$ (polynomials of degree up to 2). We need to find an orthogonal basis for this space using the Gram-Schmidt process with respect to the integration inner product defined as:

$\langle f, g \rangle = \int_0^1 f(x)g(x)\,dx$

Let's denote the orthogonal basis we seek as $\{v_1(x), v_2(x), v_3(x)\}$. The Gram-Schmidt process will generate these orthogonal vectors (polynomials) step by step. Here's how we proceed:

Step 1: Set $v_1(x) = u_1(x)$

We start by setting the first orthogonal polynomial to the first function in the original set: $v_1(x) = u_1(x) = -3$ This is already a constant, so it's orthogonal to other polynomials.

Step 2: Find $v_2(x)$

Next, we construct $v_2(x)$ by subtracting from $u_2(x)$ its projection onto $v_1(x)$.

The projection of $u_2(x)$ onto $v_1(x)$ is given by: $\text{proj}_{v_1} u_2 = \frac{\langle u_2, v_1 \rangle}{\langle v_1, v_1 \rangle} v_1$ We compute the inner products:

1. $\langle u_2, v_1 \rangle = \int_0^1 (-12x)(-3)\,dx = \int_0^1 36x\,dx = 18$

2. $\langle v_1, v_1 \rangle = \int_0^1 (-3)(-3)\,dx = \int_0^1 9\,dx = 9$

Now, the projection is: $\text{proj}_{v_1} u_2 = \frac{18}{9}(-3) = -6$

So, we subtract the projection from $u_2(x)$ to get $v_2(x)$: $v_2(x) = u_2(x) - \text{proj}_{v_1} u_2 = -12x - (-6) = -12x + 6$

Step 3: Find $v_3(x)$

Now we need to construct $v_3(x)$ by subtracting from $u_3(x)$ its projections onto $v_1(x)$ and $v_2(x)$.

The projection of $u_3(x)$ onto $v_1(x)$ is: $\text{proj}_{v_1} u_3 = \frac{\langle u_3, v_1 \rangle}{\langle v_1, v_1 \rangle} v_1$ Compute the inner product: $\langle u_3, v_1 \rangle = \int_0^1 (-12x^2)(-3)\,dx = \int_0^1 36x^2\,dx = 12$ Thus, the projection is: $\text{proj}_{v_1} u_3 = \frac{12}{9}(-3) = -4$

Next, the projection of $u_3(x)$ onto $v_2(x)$ is: $\text{proj}_{v_2} u_3 = \frac{\langle u_3, v_2 \rangle}{\langle v_2, v_2 \rangle} v_2$ First, compute $\langle u_3, v_2 \rangle$ and $\langle v_2, v_2 \rangle$:

1. $\langle u_3, v_2 \rangle = \int_0^1 (-12x^2)(-12x + 6)\,dx = \int_0^1 (144x^3 - 72x^2)\,dx = 36 - 24 = 12$

2. $\langle v_2, v_2 \rangle = \int_0^1 (-12x + 6)^2\,dx = \int_0^1 (144x^2 - 144x + 36)\,dx = 48 - 72 + 36 = 12$

Thus, the projection is: $\text{proj}_{v_2} u_3 = \frac{12}{12}(-12x + 6) = -12x + 6$

Now, subtract the projections from $u_3(x)$: $v_3(x) = u_3(x) - \text{proj}_{v_1} u_3 - \text{proj}_{v_2} u_3 = -12x^2 - (-4) - (-12x + 6) = -12x^2 + 12x - 10$

Final Orthogonal Basis

Thus, the orthogonal basis is: $v_1(x) = -3$ $v_2(x) = -12x + 6$ $v_3(x) = -12x^2 + 12x - 10$

Let me know if you would like more details or have any questions.

Here are five related questions to expand on this:

1. What is the Gram-Schmidt process used for in general?
2. How do you normalize an orthogonal basis to make it orthonormal?
3. Why is the inner product in this case defined by an integral?
4. Can you apply the Gram-Schmidt process to a basis that isn't linearly independent?
5. How would you modify the process if the inner product was defined differently?

Tip: Always double-check inner product computations carefully when applying the Gram-Schmidt process, as small errors can propagate through the entire calculation!