Orthogonality of functions
Task number: 2693
For the scalar product \(\langle f|g \rangle=\int_{-1}^1 f(x)g(x)dx\) show that the polynomials \(1, x, 3x^2-1, 5x^3-3x\) are pairwise orthogonal.
Resolution
\(\langle 1|x \rangle =\int_{-1}^1 x dx = [\frac{1}{2}x^2]_{-1}^1 = \frac{1}{2} - \frac{1}{2} = 0\)
\(\langle 1|3x^2-1 \rangle =\int_{-1}^1 (3x^2-1) dx = [x^3-x]_{-1}^1 = 0 - 0 = 0\)
\(\langle 1|5x^3-3x \rangle =\int_{-1}^1 (5x^3-3x) dx = [\frac{5}{4}x^4 - \frac{3}{2}x^2]_{-1}^1 = -\frac{1}{4} + \frac{1}{4} = 0\)
\(\langle x|3x^2-1 \rangle =\int_{-1}^1 (3x^3-x) dx = [\frac{3}{4}x^4 - \frac{1}{2}x^2]_{-1}^1 = \frac{1}{4} - \frac{1}{4} = 0\)
\(\langle x|5x^3-3x \rangle =\int_{-1}^1 (5x^4-3x^2) dx = [x^5-x^3]_{-1}^1 = 0 - 0 = 0\)
\(\langle 3x^2-1|5x^3-3x \rangle=\int_{-1}^1 (15x^5-14x^3+3x) dx = [\frac{15}{6}x^6-\frac{7}{2}x^4+\frac{3}{2}x^2]_{-1}^1 = 0\)
Observe that in the first, the third, the fourth and the last case we have integrated an odd function, so the result had to be zero.
