How does a function represented in bra-ket notation become a vector of only coefficients? I am learning quantum mechanics from the Miller Quantum Mechanics for Scientists and Engineers textbook. On page 97 it states that 
$$f(x)= \sum_{n}c_{n}\psi_{n}(x)$$
becomes
$$|f(x)\rangle = \begin{bmatrix}
c_{1} \\
c_{2} \\
\vdots \\
\end{bmatrix}$$
How is this jump made? The textbook doesn't explain it. I haven't taken linear algebra, so I could just be missing something fundamental. 
 A: It's a sloppiness of notation that is very common in physics. In reality, you have
$$ |f(x)\rangle = \sum_{n} c_n |\psi_n\rangle. $$
If all of the $|\psi_n\rangle$ are linearly independent you can write
$$ |f(x)\rangle \rightarrow \left[ \begin{array}{c}
  c_1 \\
  c_2 \\
  \vdots \end{array}\right], $$
where the arrow means that the vector is represented by the column vector, but they're not the same. This is especially true if the $|\psi_n\rangle$ are not orthonormal - when they are orthonormal, the inner product on the vector space is faithfully reproduced by the matrix multiplication with the conjugate transpose of the coefficient matrices. That is, if
\begin{align}
   |\phi(x)\rangle &= \sum_n p_n|\psi_n\rangle \rightarrow \left[\begin{array}{c} 
   p_1 \\
   p_2 \\
   \vdots \end{array}\right] \Rightarrow\\
   \langle \phi(x)|f(x)\rangle & = \sum_{m,n} p_m^* c_n \langle\psi_m|\psi_n\rangle \tag1\\
   &=\sum_{m,n} p_m^* c_n \delta_{m,n} \\
   &=\sum_{n} p_n^* c_n \\
   &= \left[\begin{array}{ccc} p_1^* & p_2^* & \ldots \end{array}\right] \left[ \begin{array}{c}
  c_1 \\
  c_2 \\
  \vdots \end{array}\right].\tag2
\end{align}
The reason for saying that the matrix representation and the bra/ket representation are not exactly the same is because if you don't have $\langle\psi_m|\psi_n\rangle= \delta_{m,n}$ (orthonormality) then the matrix multiplication in (2) wouldn't equal the sum in (1). Linear independence of the $|\psi_n\rangle$ is the necessary and sufficient condition for all the $c_n$ to be uniquely fixed by $|f(x)\rangle$ and $\left\{|\psi_n\rangle\right\}$, and orthonormality implies linear independence.
