I have the following problem:
Given $\space f_1(x)$ and $\space f_2(x)$ continuous real functions defined on the interval $[a;b]$. $\space f_1$ and $\space f_2$ are normalised such that:
$$\int_a^bf_1(x)f_1(x)dx=\int_a^bf_2(x)f_2(x)dx=1 $$
Give the component of $\space f_1(x)$ that is orthogonal to $\space f_2(x)$ on the same interval.
I know that the component of $\space f_1(x)$ in the "direction" of $\space f_2(x)$ is $\langle f_1,f_2\rangle f_2(x)$, where$ \langle f_1,f_2\rangle=\int_a^bf_1(x)f_2(x)dx$. Now if i want the component that is orthogonal to $\space f_2(x)$ should i just multiply by $\space f_2(x)$ and integrate over that interval?
The component is $f_1-\langle f_1 f_2 \rangle f_2$.