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The Newton root-finding iteration seeks a zero of a function $F$ by repeatedly solving its tangent line approximation. At a current guess $t_r$ with derivative $F^{\prime }(t_r)\ne 0$, setGeometrically, follow the tangent line to the horizontal axis, then return to the function graph for the next tangent line. Close to a suitable zero of a function, the error becomes dramatically smaller.^[ref]

For a concrete square root calculation, take the polynomial $F(t)=t^2-2$ and start at $t_0=1$. The Newton root-finding iteration becomes $t_{r+1}=(t_r+2/t_r)/2$, givingThe errors use the unrounded iterates. After the first step, each iterate overestimates $\sqrt{2}$, and the exact identityexhibits quadratic convergence: once close, the number of correct decimal digits roughly doubles per step. This speed makes the Newton root-finding iteration useful when accurate derivatives are available.

Global convergence requires additional care. For the polynomial $F(x)=x^3-2x+2$, starting at $x_0=0$ gives $F(0)=2$, $F^{\prime }(0)=-2$, hence $x_1=1$. Then $F(1)=1$, $F^{\prime }(1)=1$ gives $x_2=0$. The sequence repeats $0,1,0,1,\dots$ forever, although a real polynomial root lies in $(-2,-1)$ by the intermediate value theorem. Neither point in the cycle is a zero of a function, and neither has zero derivative. A maintained sign-changing interval and fallback bisection method can protect a root solver from unsuitable Newton steps.

For the closed SIR model with $\gamma >0$, the infectious compartment's fraction approaches $i_{\infty }=0$. One proof outline uses the bounded, increasing removed compartment's fraction: since $r(t)-r_0=\gamma {\int }_0^{t}i(u)du$, the total time-integrated $i$ is finite. The bounded derivative $di/dt$ from the SIR model equations prevents indefinitely recurring spikes, so $i(t)$ tends to zero.

For $s_0>0$, the final size relation for an epidemic follows by dividing the SIR model equations for $s$ and $r$ while $i>0$, giving $ds/dr=-(\beta /\gamma )s$. Integration then yieldsThis also follows from the SIR first integral and conservation of population. For the stipulated initial value problem, solveThe bisection method givesAbout $5.88\%$ of the original population remains in the susceptible compartment. The $94.12\%$ is the final removed compartment's fraction, including the initially $1\%$ in the infectious compartment. The newly acquired infection during the modelled period accounts for $s_0-s_{\infty }\approx 0.93120263520$ of the total population, not $1-s_{\infty }$. Dividing by $s_0$ instead gives the proportion of initially susceptible compartment members who acquire infection; its denominator must be stated.

At the SIR infection peak, the flow into the infectious compartment only balances its outflow; it does not stop. Further infection explains why the final susceptible compartment's fraction is well below $1/3$. Also $s_{\infty }=s_0{e}^{-R_0(r_{\infty }-r_0)}>0$ for $s_0>0$ and finite $R_0$: the closed SIR model does not require everybody to acquire infection before its infectious compartment empties in the long-time limit.

Consider only neutral $\mathrm{HA}$ added to pure water, at fixed temperature, with ideal dilute thermodynamic activities, water thermodynamic activity approximated by one, and no added salts or other chemical species. Normalize every concentration by $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$:Conservation of acid units and zero net electric charge, respectively, requireThe second equation balances positive hydronium against negative conjugate base and hydroxide. It would need all additional ions if salts or a chemical base were added. Together with acid dissociation constant $K_a=hx/y$ and the self-ionization of water relation $K_w=ho$, these giveThis is exact within the stated ideal model, including depletion and self-ionization of water. Multiplication by $h(K_a+h)$ yields the cubic equationFor positive $K_a,K_w,c$, the unmultiplied left-hand side increases strictly from negative to positive values as $h$ increases from zero to infinity; its derivative is $1+K_w/h^2+K_{a}c/(K_a+h{)}^2>0$. There is therefore exactly one positive root of a polynomial. A bisection method can find it without choosing an unphysical root of a polynomial.

If $K_w/h^2\ll 1$, neglect hydroxide, so $h\approx x$ and the quadratic equation retaining acid depletion givesThe last form avoids subtracting nearly equal numbers. If also $x/c\ll 1$, neglect depletion to obtainCheck both conditions after computing: $x/c$ measures the fraction lost from the initial undissociated weak acid inventory, while $K_w/h^2$ measures the omitted hydroxide contribution to electric charge balance. A $5\%$ depletion criterion is a chosen tolerance, not a universal boundary. Keeping depletion does not cure the neglect of self-ionization of water. These weak acid mass and charge balances also explain why the square root estimate cannot be extended to arbitrarily dilute chemical solutions.