Calculate the Lagrange remainder (Taylor polynomial error bound) |Rₙ(x)| ≤ M·|x−a|ⁿ⁺¹/(n+1)!. Find the maximum error, number of terms needed for desired accuracy, and convergence rate.
The Lagrange error bound — also called the Lagrange remainder or Taylor remainder — gives a rigorous upper limit on the error when a function is approximated by its nth-degree Taylor polynomial. The formula states |Rₙ(x)| ≤ M·|x − a|ⁿ⁺¹ / (n + 1)!, where M is an upper bound on the absolute value of the (n + 1)th derivative of the function on the interval between a and x, and a is the center of the Taylor expansion.
This calculator computes the error bound for any given M, center a, evaluation point x, and polynomial degree n. It also determines the minimum number of terms you need to guarantee a desired level of accuracy — a question that appears constantly in calculus courses, numerical analysis, and scientific computing.
Understanding the Lagrange error bound is essential for tasks ranging from AP Calculus exam problems to real-world engineering applications where truncated series are used to compute functions. This tool provides a convergence table and bar chart so you can visualise how rapidly the error decreases as you add more terms, and it shows the number of correct decimal digits you can expect at each degree. Whether you are studying for an exam or designing an algorithm, this calculator makes Taylor remainder analysis fast and intuitive.
Lagrange Error Bound Calculator helps you solve lagrange error bound problems quickly while keeping each step transparent. Instead of redoing long algebra by hand, you can enter Max |f⁽ⁿ⁺¹⁾(c)| bound (M), Center of expansion (a), Evaluation point (x) once and immediately inspect Lagrange Error Bound |Rₙ(x)|, (n+1)! Factorial, |x − a| to validate your work.
This is useful for homework checks, classroom examples, and practical what-if analysis. You keep the conceptual understanding while reducing arithmetic mistakes in multi-step calculations.
|Rₙ(x)| ≤ M · |x − a|^(n+1) / (n + 1)!, where M = max|f⁽ⁿ⁺¹⁾(c)| for c between a and x.
Result: Lagrange Error Bound |Rₙ(x)| shown by the calculator
Using the preset "sin(0.1) n=3", the calculator evaluates the lagrange error bound setup, applies the selected algebra rules, and reports Lagrange Error Bound |Rₙ(x)| with supporting checks so you can verify each transformation.
This calculator takes Max |f⁽ⁿ⁺¹⁾(c)| bound (M), Center of expansion (a), Evaluation point (x), Polynomial degree (n) and applies the relevant lagrange error bound relationships from your chosen method. It returns both final and intermediate values so you can audit the process instead of treating it as a black box.
Start with the primary output, then use Lagrange Error Bound |Rₙ(x)|, (n+1)! Factorial, |x − a|, |x − a|ⁿ⁺¹ to confirm signs, magnitude, and internal consistency. If anything looks off, change one input and compare the updated outputs to isolate the issue quickly.
A strong workflow is manual solve first, calculator verify second. Repeating that loop improves speed and accuracy because you learn to spot common setup errors before they cost points on multi-step algebra problems.
It is an upper bound on the absolute error when approximating a function by its nth-degree Taylor polynomial: |Rₙ(x)| ≤ M·|x−a|^(n+1) / (n+1)!, where M bounds the (n+1)th derivative.
Compute the (n+1)th derivative and find its maximum absolute value on the interval between a and x. For sin and cos, M = 1. For eˣ on [0, b], M = eᵇ.
They are the same thing — the Lagrange form of the Taylor remainder gives a specific formula for the error using the (n+1)th derivative evaluated at some point c between a and x. This provides clearer practical guidance for reliable use.
Yes, the error bound decreases as n increases (for convergent series). The factorial in the denominator grows much faster than the power term in the numerator.
No. As long as M is a true upper bound on |f⁽ⁿ⁺¹⁾(c)|, the Lagrange remainder guarantees the error cannot exceed the computed bound.
It requires the function to be (n+1) times differentiable on the interval. It also does not apply to asymptotic or divergent series where the Taylor series does not converge to the function.