Newton's Method Calculator

Newton's method (also called Newton-Raphson) is an iterative way to find a root of an equation f(x) = 0. Starting from a guess x_n, it follows the tangent line down to the x-axis and uses that crossing as the next guess: x_{n+1} = x_n - f(x_n) / f'(x_n). Each step usually lands much closer to the root, and the calculator repeats it until f(x) is within your tolerance.

Enter your function in terms of x, such as x^2 - 2 or cos(x) - x, then give a starting guess and a tolerance. Click an example to load a known problem. The calculator runs the iteration and shows a table with x_n, f(x_n), the slope f'(x_n), and the next estimate for every step, plus the final root and its residual.

No. You only enter f(x). The calculator estimates the slope f'(x) it needs at each step using a central difference, the symmetric quotient [f(x + h) - f(x - h)] / (2h) with a tiny h. This matches the exact derivative to several decimal places for smooth functions, so you do not have to differentiate by hand.

Newton's method is fast but not foolproof. It can diverge if the starting guess is far from the root, oscillate between values, or stall when the slope f'(x) is near zero (a flat tangent points nowhere useful). The calculator reports these cases as diverged, flat slope, or max iterations reached. Try a starting guess closer to where you expect the root, or raise the iteration limit.

Tolerance is how close to zero f(x) must get before the result counts as converged; a smaller value like 0.0000001 gives more decimal places. The calculator also stops when the change between successive guesses falls within tolerance. Max iterations caps how many steps it runs (1 to 100) so a non-converging case still ends and shows you the trace.

Near a simple root Newton's method converges quadratically, meaning the number of correct digits roughly doubles each step, so it typically reaches your tolerance in a handful of iterations. Values are rounded to six decimal places for display. The residual f(root) shown next to the answer tells you exactly how close the estimate sits to a true zero.