Numerical Equation Solver & Optimizer

What it is for

Many equations have no formula for their solution: cos(x) = x, e^(−x) = x² − 1, or a pair like x² + y² = 4 and x·y = 1. This tool solves them numerically, to about 12 significant digits, and shows how it got there.

Pick a Problem. Equation finds roots, either all of them in an interval (Brent's method) or one by bisection or Newton's method with every iteration listed. System of nonlinear equations solves up to six equations in as many unknowns with a damped Newton method. Minimum and Maximum scan an interval and refine every local extremum.

If you need exact answers with algebraic steps (fractions, square roots), use the step-by-step equation solver instead. This tool is the numeric counterpart.

Worked example

Enter exp(-x) = x^2 - 1, keep Brent, and set the interval from −3 to 3. The tool reports: One root of exp(-x) = x^2 - 1 between -3 and 3: x = 1.14775763214. The table shows the residual f(x) = −2.6·10⁻¹⁵, which means the two sides agree to the last digit a double can hold. Brent needed 5 iterations after the scan found the sign change.

Now switch to Newton's method for x^3 - 2x - 5 = 0 with start value 2. The iteration table is the classic textbook one: x goes 2 → 2.1 → 2.094568121 → 2.094551482, f(x) drops from −1 to 0.061, 0.00019, 1.7·10⁻⁹ and finally −8.9·10⁻¹⁶. The number of correct digits roughly doubles each step, which is what quadratic convergence looks like. Bisection on the same equation over [2, 3] with tolerance 10⁻⁶ needs 19 halvings.

For the system x^2 + y^2 = 4 and x*y = 1 with start values 2, 0.5, Newton's method converges in 4 iterations to x = 1.93185165258, y = 0.517638090205.

How to read the result

• Metrics show the root(s), the residual and the work done (iterations, function evaluations).
• Roots in the interval lists every root with its type: sign change or touches zero. A touching root, like x = 1 in (x − 1)², has no sign change and is found by minimising f²; it is accurate to about half as many digits.
• Bisection steps / Newton iterations are the full tables, ready to copy into homework or a report.
• For systems, Residual of each equation confirms the answer, and the convergence plot shows the largest residual per step on a log scale.
• The graph marks the roots or extrema on f.

Tips and common mistakes

• ln is the natural logarithm and log is base 10. Angles are in radians.
• Brent samples the interval at 401 points. Two roots closer than 1/400 of the interval can hide between samples, so narrow the interval when you suspect clustered roots.
• Sign changes across a pole (tan x at ±π/2) are rejected, because |f| stays large there.
• Newton's method can jump to a different root, or diverge, from a poor start value. Use Brent to see where the roots are, then Newton for the steps.
• A system needs as many equations as unknowns. If Newton stops with "no step reduces the residuals", try other start values; there may be no real solution nearby.
• A minimum flagged at the interval end means the function keeps falling beyond your interval.

FAQ

How accurate are the roots?

The default tolerance is 10⁻¹², so roots of well-behaved functions are correct to about 12 digits. The residual column shows how close f(x) really is to zero.

Why does bisection need so many more steps than Newton?

Bisection gains one binary digit per step, about 3.3 steps per decimal digit. Newton doubles the correct digits each step once it is close, but it needs a derivative and a good start.

Can it find complex roots?

No. It works with real numbers only. For complex roots of polynomials use the step-by-step solver or the math workbench.

Is the derivative computed symbolically?

No. Newton's method uses a central difference with a step of about 6·10⁻⁶ times |x|. That is accurate enough for fast convergence on smooth functions.