Question

Consider the set \( \{ 1, x, x^2 \} \). An orthonormal basis in \( x \in [-1, 1] \) is formed from these three terms, where the normalization of a function \( f(x) \) is defined via \[ \int_{-1}^{1} \left[ f(x) \right]^2 \, dx = 1. \] If the orthonormal basis set is \( \left( \frac{\sqrt{3}}{2}, \frac{\sqrt{5}}{2} x, \frac{1}{2} \sqrt{\frac{21}{N}} (5x^2 - 3) \right) \), then the value of \( N \) (in integer) is:

Hint:

When normalizing functions in an orthonormal basis, carefully compute the integrals of the squared terms to find the correct scaling factor for each function.

Answer 1

Step 1: The given orthonormal basis set is: \[ \left( \frac{\sqrt{3}}{2}, \frac{\sqrt{5}}{2} x, \frac{1}{2} \sqrt{\frac{21}{N}} (5x^2 - 3) \right). \] For the functions to be orthonormal, the third function needs to be normalized. We focus on the third function: \[ f_3(x) = \frac{1}{2} \sqrt{\frac{21}{N}} (5x^2 - 3). \] To normalize this function, we compute the following integral: \[ \int_{-1}^{1} \left( \frac{1}{2} \sqrt{\frac{21}{N}} (5x^2 - 3) \right)^2 dx = 1. \] Step 2: Expand the integrand: \[ \left( \frac{1}{2} \sqrt{\frac{21}{N}} (5x^2 - 3) \right)^2 = \frac{21}{4N} (5x^2 - 3)^2. \] Now expand \( (5x^2 - 3)^2 \): \[ (5x^2 - 3)^2 = 25x^4 - 30x^2 + 9. \] So the integral becomes: \[ \frac{21}{4N} \int_{-1}^{1} (25x^4 - 30x^2 + 9) \, dx. \] Step 3: Compute the individual integrals: \[ \int_{-1}^{1} x^4 \, dx = \frac{2}{5}, \quad \int_{-1}^{1} x^2 \, dx = \frac{2}{3}, \quad \int_{-1}^{1} dx = 2. \] Substitute these values into the integral: \[ \int_{-1}^{1} (25x^4 - 30x^2 + 9) \, dx = 25 \times \frac{2}{5} - 30 \times \frac{2}{3} + 9 \times 2 = 10 - 20 + 18 = 8. \] Step 4: Now, substitute this value into the normalization equation: \[ \frac{21}{4N} \times 8 = 1 \quad \Rightarrow \quad \frac{168}{4N} = 1 \quad \Rightarrow \quad 42 = N. \] Thus, the value of \( N \) is \( \boxed{3} \).