Taylor Remainder Calculator

Lagrange error bound

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About Taylor Remainder Calculator

A Taylor remainder calculator computing the Lagrange error bound |Rₙ(x)| ≤ M|x−a|ⁿ⁺¹/(n+1)! where M bounds |f⁽ⁿ⁺¹⁾|. Select from preset functions. Determine how many terms needed for desired accuracy. All calculations are client-side. Essential for numerical analysis.

Taylor Remainder Calculator Features

  • Lagrange bound
  • Term count
  • Error estimate
  • Presets
  • Accuracy
Taylor remainder: Rₙ(x) = f(x) − Tₙ(x). Lagrange form: |Rₙ(x)| ≤ M|x−a|ⁿ⁺¹/(n+1)! where M = max|f⁽ⁿ⁺¹⁾(c)| for c between a and x. For eˣ: M = eˣ. For sin/cos: M = 1.

How to Use

Set up the estimate:

  • f(x): Function
  • n: Degree
  • x: Evaluation point

Applications

  • How many terms for ε accuracy?
  • Verify numerical approximations
  • Error control in computation
  • Convergence rate analysis

Alternating Series

For alternating series, the error is bounded by the first omitted term: |R| ≤ |aₙ₊₁|. This is tighter than Lagrange for sin/cos Maclaurin series.

Step-by-Step Instructions

  1. 1Select a function.
  2. 2Set center a.
  3. 3Choose degree n.
  4. 4Enter evaluation x.
  5. 5Get error bound.

Taylor Remainder Calculator — Frequently Asked Questions

How do I find M (the derivative bound)?+

M = max|f⁽ⁿ⁺¹⁾(c)| for c between a and x. For eˣ at a=0: M = eˣ (increasing). For sin/cos: M = 1 (always). For 1/(1+x): M = (n+1)! (geometric — use ratio test instead). Often, overestimate M for a simpler bound.

How many terms do I need for a given accuracy?+

Solve M|x−a|ⁿ⁺¹/(n+1)! < ε for n. For eˣ near 0: e·|x|ⁿ⁺¹/(n+1)! < ε. The factorial grows fast, so convergence is rapid for |x−a| < 1. For |x−a| > 1, you may need many terms.

Is the Lagrange bound tight?+

Usually not — it's an upper bound using the maximum of the (n+1)th derivative. The actual error is often much smaller. For alternating series, the alternating series estimate is tighter and exact.

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