Suppose f is twice continuously differentiable near a root r with f′(r)=0. Then there is a neighborhood of r such that, if the Newton iteration starts inside it, the iterates converge to r and satisfy ∣xn+1−r∣≤C∣xn−r∣2 for a constant C=2min∣f′∣max∣f′′∣ (the maximum and minimum taken over that neighborhood).
Why is it true?
The tangent line is such a good approximation to a smooth curve that the error it makes is proportional to the square of the distance already traveled, so each correct digit roughly doubles the number of correct digits at the next step.
Proof sketch
Step 1 (Taylor expand around the current guess). Since f is twice differentiable, Taylor's theorem with the Lagrange remainder gives f(r)=f(xn)+f′(xn)(r−xn)+21f′′(ξn)(r−xn)2 for some ξn between xn and r. This is exact, not an approximation, because the remainder term carries the full second-order error.
Step 2 (substitute the root condition). Since f(r)=0, the left side vanishes, leaving 0=f(xn)+f′(xn)(r−xn)+21f′′(ξn)(r−xn)2. Dividing through by f′(xn) (nonzero near r since f′(r)=0 and f′ is continuous) isolates r−xn: 0=f′(xn)f(xn)+(r−xn)+2f′(xn)f′′(ξn)(r−xn)2.
Step 3 (recognize the Newton step). Rearranging, r−(xn−f′(xn)f(xn))=−2f′(xn)f′′(ξn)(r−xn)2. The left-hand parenthesis is exactly the Newton update xn+1, so with en=xn−r this reads r−xn+1=−2f′(xn)f′′(ξn)en2, i.e. en+1=−2f′(xn)f′′(ξn)en2.
Step 4 (bound the constant). Taking absolute values and bounding ∣f′′(ξn)∣ and 1/∣f′(xn)∣ by their extreme values max∣f′′∣ and 1/min∣f′∣ on the neighborhood gives ∣xn+1−r∣≤C∣xn−r∣2 with C=2min∣f′∣max∣f′′∣. Once C∣x0−r∣<1, this recursion forces ∣xn−r∣ to shrink to 0, proving convergence, and the squaring in ∣xn+1−r∣≤C∣xn−r∣2 is exactly the quadratic rate.