I'm reading this lecture note and I don't understand the bottom part of page 2.
1) What does it mean by "project[ing] the function $f(x)$ onto the basis functions"?
2)How did this equation go from the left hand side to the right hand side?
Wouldn't the RHS of this equation be $$\sum^{\infty}_{n=1} c_n \int^2_1x^{-1}\sin\left(\frac{m\pi \ln x}{\ln 2}\right)\sin\left(\frac{n\pi \ln x}{\ln 2}\right) $$
On
(1) That would mean calculating each of the inner products $\langle f,f_n\rangle,$ where the $f_n$ are the basis functions.
(2) The change that has occurred here is that $f(x)$ has been replaced by $$\sum^{\infty}_{n=1} c_n x^{-1/2}\sin\left(\frac{m\pi \log x}{\log 2}\right),$$ and the series integrated termwise.
On
It is as you wrote. However $$ \int_1^2 x^{-1}\sin\left(\frac{m\pi\ln x}{\ln 2}\right)\sin\left(\frac{n\pi\ln x}{\ln x}\right)dx=0,\;\; n \ne m. $$ You can see this by making the substitution $y=\frac{\pi \ln x}{\ln 2}$. Then $$ dy = \frac{\pi}{x\ln 2}dx \\ \frac{\ln 2}{\pi}dy = \frac{dx}{x}, $$ and $$ \int_1^2 x^{-1}\sin\left(\frac{m\pi\ln x}{\ln 2}\right)\sin\left(\frac{n\pi\ln x}{\ln x}\right)dx= \frac{\ln 2}{\pi}\int_0^{\pi} \sin(my)\sin(ny)dy. $$ So the functions $\{ S_n(x)=\sin\frac{n\pi\ln x}{\ln s}\}_{n=1,\infty}$ form an orthogonal set of elements of $L^2(1,2,1/x)$.
The projection $P_{\mathcal{S}}f$ of $f$ onto the closed subspace $\mathcal{S}$ spanned by $\{S_n\}_{n=1}^{\infty}$ is defined as the closest $s\in\mathcal{S}$ to $f$, meaning that $s\in\mathcal{S}$ and $\|f-s\| \le \|f-s'\|$ for all $s'\in\mathcal{S}$, which is equivalent $s$ being the orthogonal projection onto $\mathcal{S}$, which is to say that $(f-s)\perp \mathcal{S}$. It can be shown that $$ P_{\mathcal{S}}f = \sum_{n=1}^{\infty} \frac{\langle f,S_n\rangle}{\langle S_n, S_n\rangle} S_n $$
You have an inner product space of functions which can be represented as a linear sum of the basis functions. It looks like your basis functions are over the range $1 \le x \le 2$ and the functions are $x^{-1/2}\sin\left(\frac{m\pi \ln x}{\ln 2}\right)$. It is usually convenient to choose basis functions that are orthonormal-that the inner product of any two is $0$ and the square of any one is $1$. The inner product is usually taken with some weight function $w(x)$, so the inner product of $f_m(x)$ and $f_n(x)$ is $\int f_m(x)f_n(x)w(x)dx$. The weight function determines the basis functions if you want them to be orthonormal, that $\int f_m(x)f_n(x)w(x)=\delta_{nm}$
Any function $f(x)$ can be expanded in terms of the basis functions $f(x)=\sum c_nf_n(x)$. If the basis functions are orthonormal, you can evaluate the coefficients by $c_n=\int f(x)f_n(x)w(x)dx$. This is also called projecting them onto the basis functions by analogy with projecting a vector in space onto the $x,y,z$ axes.