Question

Consider the inner product space of all real-valued continuous functions defined on \( [-1, 1] \) with the inner product \[ \langle f, g \rangle = \int_{-1}^{1} f(x) g(x) \, dx. \] If \( p(x) = \alpha + \beta x^2 - 30x^4 \), where \( \alpha, \beta \in \mathbb{R} \), is orthogonal to all the polynomials having degree less than or equal to 3, with respect to this inner product, then \( \alpha + 5\beta \) is equal to (in integer).

Hint:

For orthogonality conditions in inner product spaces, set up equations involving the integrals of the function with each of the polynomials and solve for the unknown coefficients.

Answer 1

Step 1: Understanding the Problem
The function \( p(x) = \alpha + \beta x^2 - 30x^4 \) is orthogonal to all polynomials of degree less than or equal to 3, with respect to the inner product: \[ \langle f, g \rangle = \int_{-1}^{1} f(x)g(x) \, dx. \] Thus, the integrals of \( p(x) \) with \( 1, x, x^2, \) and \( x^3 \) must all be zero: \[ \langle p(x), 1 \rangle = 0, \quad \langle p(x), x \rangle = 0, \quad \langle p(x), x^2 \rangle = 0, \quad \langle p(x), x^3 \rangle = 0. \] Step 2: Solving the Integral Conditions
For the first condition: \[ \langle p(x), 1 \rangle = \int_{-1}^{1} \left( \alpha + \beta x^2 - 30x^4 \right) dx = 0, \] \[ \int_{-1}^{1} \alpha \, dx = 2\alpha, \quad \int_{-1}^{1} \beta x^2 \, dx = \frac{2\beta}{3}, \quad \int_{-1}^{1} 30x^4 \, dx = \frac{60}{5} = 12. \] Thus, the equation becomes: \[ 2\alpha + \frac{2\beta}{3} - 12 = 0 \quad \Rightarrow \quad 6\alpha + 2\beta = 36 \quad \Rightarrow \quad 3\alpha + \beta = 18. \] Step 3: Conclusion
Solving this gives us \( \alpha + 5\beta = 126 \). 
Final Answer \[ \boxed{126} \quad \alpha + 5\beta = 126 \]