Error Bound Estimator

Math Calculus • Infinite Series and Sequences

Written by STEM Calculators Team Published December 30, 2025 Updated February 24, 2026

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Inputs

Bound type

Target \(\varepsilon\) (optional)

If provided, finds a smallest \(n\) (up to a cap) so the bound \(\le \varepsilon\).

Options

Prefer symbolic derivatives (if math.js exists) Show steps Auto-fit graph after solve

Taylor mode

Function \(f(x)\)

Use variable x. Constants: pi, e. Functions: sin cos tan asin acos atan sqrt abs exp ln/log. Implicit multiplication (e.g., 2x) allowed.

Center \(a\)

Order \(n\)

Evaluate at \(x_0\)

Plot \(x_{\min}\)

Plot \(x_{\max}\)

How to get \(M\)

Auto samples the interval; uses symbolic derivative if possible, else numeric.

Manual \(M\)

Presets

Click a preset to load and solve.

Ready

Graph

Drag to pan • wheel/pinch to zoom.

Toggles

Plot \(f(x)\) Plot \(P_n(x)\) Shade bound band Legend

x: 0, y: 0

Results & steps

Click Calculate.

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3. Error Bound Estimator — Theory

Taylor remainder (Lagrange form)

Let \(P_n(x)\) be the Taylor polynomial of \(f\) about \(a\): \[ P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k. \] The remainder is \(R_n(x)=f(x)-P_n(x)\). If \(f^{(n+1)}\) is continuous on an interval containing \(a\) and \(x\), then there exists \(\xi\) between \(a\) and \(x\) such that \[ R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}. \] Therefore, if \(\left|f^{(n+1)}(t)\right|\le M\) on the interval between \(a\) and \(x\), then \[ \left|R_n(x)\right|\le \frac{M\,|x-a|^{n+1}}{(n+1)!}. \]

Quick bound for sine/cosine: since \(|\sin(\cdot)|\le 1\) and \(|\cos(\cdot)|\le 1\), if \(f^{(n+1)}\) is \(\pm\sin\) or \(\pm\cos\) on the interval, you may take \(M=1\).

Alternating series error bound

Consider an alternating series \[ \sum_{k=0}^{\infty}(-1)^k b_k, \qquad b_k\ge 0. \] If \(b_k\) is decreasing and \(b_k\to 0\), then the truncation error after \(n\) terms satisfies \[ \left|R_n\right|=\left|S-S_n\right|\le b_{n+1}. \]

How many terms for a target \(\varepsilon\)

In Taylor mode, solve for \(n\) so that \[ \frac{M\,|x_0-a|^{n+1}}{(n+1)!}\le \varepsilon, \] while in alternating mode it is enough to find \(n\) such that \[ b_{n+1}\le \varepsilon. \]

Frequently Asked Questions

What is the Taylor remainder bound in Lagrange form?

If |f^(n+1)(t)| <= M on the interval between a and x0, then the truncation error satisfies |R_n(x0)| <= M|x0-a|^(n+1)/(n+1)!. This gives a guaranteed upper bound on the approximation error.

How do I choose the value of M for the Taylor bound?

M should bound the absolute value of the (n+1)th derivative on the interval between a and x0. The calculator can auto-estimate M by sampling that interval, or you can enter a manual M if you already know a safe bound.

When can I use the alternating series error bound |R_n| <= b_(n+1)?

It applies to series of the form sum (-1)^k b_k when b_k is decreasing and b_k -> 0. Under these conditions, the absolute truncation error after n terms is at most the first omitted term b_(n+1).

What does the target epsilon option do?

It searches for the smallest n (within a maximum cap) such that the selected error bound is less than or equal to epsilon. This helps pick how many Taylor terms or alternating-series terms are needed for a desired accuracy.

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Frequently Asked Questions

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