#1 India's Top IT Training Institute
×

About Uncodemy

Know more about who we are and what we stand for.

About Us

About Us

Discover our mission to equip individuals with essential IT skills and industry expertise.

Read More →
UnCodeMy Gallery

UnCodeMy Gallery

Explore highlights from our IT training sessions and success stories.

Read More →
Mentors

Mentors

Meet the industry experts who train and guide you through your course.

Read More →
Free Resources

Free Resources

Free tutorials, interview questions, tools and calculators to help you learn.

Explore →
×

Free Resources

Learn, practice and plan your career — bilkul free.

Tutorials

Tutorials

Step-by-step written tutorials to learn any technology at your own pace.

Read More →
Free Counselling

Free Counselling

Book a free video counselling session and pick the right course for you.

Read More →
Online Compiler

Online Compiler

Write, run and test your code online — no setup required.

Read More →
⚡ FREE · STEP BY STEP

Secant Method Calculator: Find Roots Without Derivatives

Enter f(x) and two starting values to see each secant iteration.

  • ⚡ No derivative needed
  • 📋 Iteration table
  • 🎯 Your tolerance

How to use the Secant Method Calculator

  1. 1. Enter f(x)Type the function whose root you want, for example x^3 − 2x − 5.
  2. 2. Give two starting valuesx₀ and x₁ should be near the root. They do not need to bracket it.
  3. 3. Read the rootEach row shows the two latest values, their f values and the next x.

How the secant method works

The secant method replaces the tangent of Newton-Raphson with the line through the last two points. Starting with x₀ and x₁, the next value is where that line crosses the x-axis:

xₙ₊₁ = xₙ − f(xₙ)(xₙ − xₙ₋₁) / (f(xₙ) − f(xₙ₋₁)).

Then drop the oldest point and repeat. It needs only one new function value per step and no derivative. Its order of convergence is about 1.618, faster than bisection and regula falsi but a little slower than Newton-Raphson. Unlike regula falsi, it does not keep the root bracketed, so it can fail if the starting values are far from the root.

Worked example

Find the root of x³ − 2x − 5 = 0 with x₀ = 2, x₁ = 3. f(2) = −1 and f(3) = 16, so x₂ = 3 − 16(3 − 2)/(16 + 1) = 2.05882. f(2.05882) = −0.39080, so x₃ = 2.05882 + 0.39080 × 0.94118/16.39080 = 2.08126.

Next x₄ = 2.09482, then x₅ = 2.09455 and x₆ = 2.09455. The root is x ≈ 2.0946, reached in 5 iterations for a tolerance of 0.0001.

Tips

If f(xₙ) and f(xₙ₋₁) become equal the line is flat and the method stops; choose other starting values. When you need a guaranteed bracket, use regula falsi or the bisection method.

Frequently asked questions

How is the secant method different from regula falsi?

Both use a chord. Regula falsi keeps one end on each side of the root; the secant method always uses the two latest points, which is faster but can diverge.

Do the starting values have to bracket the root?

No. They only need to be reasonably close to the root.

Why use it instead of Newton-Raphson?

It needs no derivative, which helps when f′(x) is hard to find or expensive to compute.

When does it stop?

When two successive values differ by less than your tolerance, or f(x) becomes exactly 0.

Is this calculator free?

Yes. You fill a short form once when you see your first result, and other Uncodemy tools then open without the form. Everything is calculated in your browser.

More free tools

See all free tools