$V$ is the vector space of continuous functions $f:[-1,1] \rightarrow \mathbb{R}$, with inner product $\langle f,g \rangle = \int_{-1}^1f(x)g(x)(1-x^2)dx$
Let $p(x) = 1, q(x)=x$ be elements of $V$
Find $||p||, ||q||$
Answer:
$||p||^2=\int_{-1}^1 1(1−x^2)dx=\int_{-1}^1(1−x^2)dx=[x−\frac{x^3}{3}]^1_{−1}=\frac{4}{3}$. Hence, $||p||=\sqrt{||p||^2}=\sqrt{\frac{4}{3}}=\frac{2}{\sqrt3}$
Similarly, $||q||^2=\int_{-1}^1 x^2(1−x^2)dx=\int_{-1}^1(x^2-x^5)dx=[\frac{x^3}{3}−\frac{x^5}{5}]^1_{−1}=\frac{4}{15}$. Hence, $||q||=\sqrt{||q||^2}=\sqrt{\frac{4}{15}}=\frac{2}{\sqrt15}$
Then to find the angle between p and q, we use $\theta = cos^{-1}\frac{\langle p,q \rangle}{||p||||q||}$
Then, $\langle p,q \rangle = \int_{-1}^1 p(x)q(x)(1-x^2)dx = \int_{-1}^1 x(1-x^2)dx = [\frac{x^2}{2}-\frac{x^4}{4}]^1_{-1} = 0$
Giving $cos^{-1}= \frac{0}{\frac{4}{\sqrt3 \sqrt15}} = \frac{\pi}{2}$ radians.