Correctly scaled bounded product approximation

Set $\delta =\epsilon \mathrm{/}(6M^2)$ and, for $\mathrm{\mid }x\mathrm{\mid },\mathrm{\mid }y\mathrm{\mid }\le M$, put $P_{M,\epsilon }(x,y)=2M^2[Q_{\delta }\text{\hspace{0.17em}⁣}\left(\frac{\mathrm{\mid }x+y\mathrm{\mid }}{2M}\right)-Q_{\delta }\text{\hspace{0.17em}⁣}\left(\frac{\mathrm{\mid }x\mathrm{\mid }}{2M}\right)-Q_{\delta }\text{\hspace{0.17em}⁣}\left(\frac{\mathrm{\mid }y\mathrm{\mid }}{2M}\right)]\mathrm{.}$ All three arguments lie in $[0,1]$. Replacing $Q_{\delta }(u)$ by $u^2$ gives $2M^2[((x+y{)}^2-x^2-y^2)\mathrm{/}(4M^2)]=xy$, so the absolute error is at most $2M^2(3\delta )=\epsilon$.

Since $Q_{\delta }(0)=0$, substituting $x=0$ makes the first and third terms equal and the second zero; substituting $y=0$ makes the first and second terms equal and the third zero. The formula is globally defined, so this cancellation holds for every real value of the other coordinate. Absolute value is represented by $\mathrm{\mid }z\mathrm{\mid }=\rho (z)+\rho (-z)$, and the three square networks run in parallel followed by an affine combination, so the depth, weights, and units are $O(1+\log (1\mathrm{/}\delta ))=O(1+\log (M^2\mathrm{/}\epsilon ))$, since $0<\delta <1\mathrm{/}6$ and $\log (1\mathrm{/}\delta )=\log 6+\log (M^2\mathrm{/}\epsilon )$.