In the Shor's algorithm, we need to compute the sequence of controlled $U^{2^j}$ operations used by the phase estimation procedure, where $U$ is defined as $$ U|y\rangle=|xy\;(\mod N)\rangle\text{ for } 0\leq y\leq N-1\\ U|y\rangle=|y\rangle\text{ for } N\leq y\leq 2^L-1 $$ That means, we need the transformation $$ |z\rangle|y\rangle\to |z\rangle U^{z_t2^{t-1}}\cdots U^{z_12^0} |y(\mod N)\rangle\\ =|z\rangle|x^{z_t2^{t-1}}\times\cdots\times ^{z_12^{0}}y(\mod N)\rangle=|z\rangle|x^zy(\mod N)\rangle $$
This is done by reversibly computing the function $xz (\mod N )$ of $z$ in a third register, and then by reversibly multiplying the contents of the second register by $xz (\mod N )$, using the trick of uncomputation to erase the contents of the third register upon completion.
We use modular multiplication to compute $x^2 (\mod N )$, by squaring $x$ modulo $N$, then computes $x^4 (\mod N )$ by squaring $x^2 (mod N )$, and continues in this way, computing $x^{2^j}(\mod N )$ for all $j$ up to $t − 1$.
Then it is stated that, we use $\color{red}{t=2L+1+\log(2+1/(2\epsilon))=O(L)}$, so a total of $t−1=O(L)$ squaring operations is performed at a cost of $O(L^2)$ each (this cost assumes the circuit used to do the squaring implements the familiar algorithm we all learn as children for multiplication), for a total cost of $O(L^3)$ for the first stage.
My Understanding
In the phase estimation procedure, the probability of obtaining $m$ such that $|m-b|<e$, where $b$ is the best $t$ bit approximation of the phase $\phi$, is $P(|m-b|<e)\leq 1-\frac{1}{2(e-1)}$.
Suppose we want to approximate the phase $\phi$ to an accuracy $2^{-n}$ then $|m-b|=2^t(\phi'-\phi)< 2^t\times 2^{-n}=2^{t-n}\implies e=2^t(2^{-n}-2^{-t})=2^{t-n}-1$. Therefore, $$ P(|m-b|<e)=P(|m-b|<2^{t-n}-1)\leq 1-\frac{1}{2(e-1)}1-\frac{1}{2(2^{t-n}-2)}=1-\epsilon\\ \epsilon=\frac{1}{2(2^{t-n}-2)}\implies 2^{t-n}-2=\frac{1}{2\epsilon}\implies \boxed{t=n+\log_2\big(2+\frac{1}{2\epsilon}\big)} $$ which is the number of qubits used in the first register to obtain the phase $\phi$ accurate to $n$ bits with probability of success at least $1-\epsilon$.
With this in mind, if I approach the Shor's algorithm how does $2L+1$ comes in $t=2L+1+\log(2+1/(2\epsilon))$ ?
My understanding is that, $L$ is the number of bits needed to specify $N$, ie., second register has $L$ qubits, such that the unitary operator $U$ is $2^L-1\times 2^L-1$.
Please refer to Pages 226,228, Quantum Computation and Quantum Information by Nielsen and Chuang
