The quantum circuit construction of the quantum Fourier transform apparently requires gates of exponential precision in the number of qubits used. However, such precision is never required in any quantum circuit of polynomial size. For example, let $U$ be the ideal quantum Fourier transform on $n$ qubits, and $V$ be the transform which results if the controlled-$R_k$ gates are performed to a precision $Δ=1/p(n)$ for some polynomial $p(n)$. Show that the error $E(U, V )≡\max_{|\psi\rangle}(U−V)|\psi\rangle$ scales as $O(n^2 /p(n))$, and thus polynomial precision in each gate is sufficient to guarantee polynomial accuracy in the output state.
This is Exercise $5.6$ in Quantum Computation and Quantum Information by Nielsen and Chuang
We have a quantum circuit of $n$ qubit QFT which requires $n$ Hadamard gates, $n(n+1)/2$ controlled-$R_k$ gates and at most $n/2$ swap gates.
$$ E(R_2^1R_3^1\cdots R_n^1\cdots R_2^{n-1},\mathcal{R}_2^1\mathcal{R}_3^1\cdots \mathcal{R}_n^1\cdots \mathcal{R}_2^{n-1})\leq E(R_2^1,\mathcal{R}_2^1)+E(R_3^1,\mathcal{R}_3^1)+\cdots+E(R_n^1,\mathcal{R}_n^1)+\cdots+E(R_2^{n-1},\mathcal{R}_2^{n-1})\\ =\Delta+\Delta+\cdots+\Delta+\cdots+\Delta=\frac{n(n+1)}{2}\Delta=\frac{n(n+1)}{2p(n)}=O(n^2/p(n)) $$
This is clear.
What does it mean to say that The quantum circuit construction of the quantum Fourier transform apparently requires gates of exponential precision in the number of qubits used ?
What is a quantum circuit with polynomial size ?
I am also confused by the terms polynomial precision and polynomial accuracy ?