Taylor Remainder Bound Calculator

This tool computes a Taylor polynomial value $T_n(x)$ and an explicit error bound for the approximation, shown step by step using the Lagrange remainder. Choose a function, set the center $a$, the point $x$, and the degree $n$, then the calculator reports $T_n(x)$ and a bound for $|R_n(x)|$. For more tools in this topic, explore Sequences and Series.

Compute a Taylor polynomial value and a Lagrange remainder bound.

$$|R_n(x)|\le \frac{M}{(n+1)!}|x-a|^{n+1}$$

f(x)

a (center)

x (point)

n (degree)

Content updated: 2026-01-17

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What the Taylor Remainder Bound Means

A Taylor polynomial approximates a function near a center point $a$. The remainder term $R_n(x)$ is the difference between the true value and the approximation: $$f(x)=T_n(x)+R_n(x)$$ This calculator gives a numerical upper bound for the size of that error, so you can say not only “here is the approximation,” but also “here is how far off it can be.”

Note: this is a guaranteed upper bound, not necessarily the exact error. The true error can be smaller than the bound.

$$|R_n(x)|\le \frac{M}{(n+1)!}|x-a|^{n+1},\quad M=\max_{t\in[\min(a,x),\max(a,x)]}|f^{(n+1)}(t)|$$

How the Calculator Computes the Bound

The tool does two things:

Compute $T_n(x)$: it evaluates the Taylor polynomial at $x$ using derivatives at $a$.
Compute an error bound: it finds an $M$ that upper-bounds $|f^{(n+1)}(t)|$ on the interval between $a$ and $x$, then plugs it into the Lagrange remainder formula.

In practice, $M$ is chosen in a way that stays reliable and fast. For example, for $e^x$ the maximum occurs at the interval endpoint with the larger $t$ value. For $\sin(x)$ and $\cos(x)$, the relevant derivative magnitudes stay within 1. For $\ln(1+x)$, the maximum is achieved at the smaller endpoint of the interval. For $\frac{1}{1-x}$, the maximum comes from the endpoint closest to $t=1$ (as long as the interval does not cross the singularity).

The output is designed to be practical: you get $T_n(x)$, the interval used, the value of $M$, and the final bound for $|R_n(x)|$.

Input Rules and Domain Notes

Different functions have different domain requirements, and the bound depends on the interval between $a$ and $x$. The calculator checks these conditions before computing a bound.

Degree $n$: choose a nonnegative integer. Larger $n$ usually improves the approximation, but factorials grow fast.
Interval: the tool uses $t\in[\min(a,x),\max(a,x)]$ when computing $M$.
General rule: if the interval crosses a point where the function is undefined or the required derivative does not exist, a valid bound cannot be computed.
$\ln(1+x)$: requires $1+t>0$ on the whole interval (so you cannot cross $t=-1$).
$\frac{1}{1-x}$: requires $t\ne 1$ on the whole interval (so the interval cannot cross $t=1$).

If you want the Taylor or Maclaurin series itself (not just a bound at a point), use the Taylor and Maclaurin Series tool. If you need to determine convergence behavior for an infinite series, use the Series Convergence Test tool.

Worked Examples With Steps

Example 1: $\ln(1+x)$ near 0

Approximate $\ln(1+x)$ at $x=0.5$ with a Maclaurin polynomial and get a numerical error bound.

Choose $f(x)=\ln(1+x)$, $a=0$, $x=0.5$, $n=3$.

The tool computes $T_3(0.5)$ and then uses: $$|R_3(0.5)|\le \frac{M}{4!}|0.5-0|^4$$ with $M=\max_{t\in[0,0.5]}|f^{(4)}(t)|$.

It then reports the numeric value of $T_3(0.5)$ and a concrete bound for $|R_3(0.5)|$.

Example 2: $e^x$ with a bound on an interval

For $e^x$, the derivatives stay as $e^x$, so the maximum on the interval is easy to interpret.

Choose $f(x)=e^x$, $a=0$, $x=1$, $n=3$.

The bound is computed from: $$|R_3(1)|\le \frac{M}{4!}|1|^4,\quad M=\max_{t\in[0,1]}e^t$$

The calculator then outputs $T_3(1)$ and the final numeric bound for $|R_3(1)|$.

Example 3: $\sin(x)$ with a simple $M$

For $\sin(x)$ and $\cos(x)$, derivatives stay within $[-1,1]$, so $M$ is often just 1.

Choose $f(x)=\sin(x)$, $a=0$, $x=0.5$, $n=5$.

The tool uses: $$|R_5(0.5)|\le \frac{M}{6!}|0.5|^6,\quad M=\max_{t\in[0,0.5]}|\sin^{(6)}(t)|$$

Since the relevant derivatives are sine/cosine values, $M$ is bounded by 1 on the interval, producing a clean numerical bound.

Example 4: $\frac{1}{1-x}$ away from the singularity

Rational functions can be approximated too, but the interval must stay away from where the function is undefined.

Choose $f(x)=\frac{1}{1-x}$, $a=0$, $x=0.2$, $n=4$.

The calculator computes $T_4(0.2)$ and then evaluates a bound using: $$|R_4(0.2)|\le \frac{M}{5!}|0.2|^5$$ where $M$ is taken as the maximum of $|f^{(5)}(t)|$ on $t\in[0,0.2]$.

Because the interval stays far from $t=1$, the bound remains stable and meaningful.

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