Details
- Interval: —
- M (max |f^(n+1)|): —
Step-by-step
Explanation:
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.
$$|R_n(x)|\le \frac{M}{(n+1)!}|x-a|^{n+1}$$
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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.
Keep exploring Calculus and Mathematics or jump back to Calculators to browse more tools.
Questions About the Taylor Remainder Bound Calculator
Quick answers about Taylor polynomials, Lagrange remainder, and error bounds.
It computes a Taylor polynomial value $T_n(x)$ and an explicit error bound for the remainder $|R_n(x)|$ using the Lagrange remainder formula.
It is the inequality $|R_n(x)|\le \frac{M}{(n+1)!}|x-a|^{n+1}$, where $M=\max_{t\in[a,x]}|f^{(n+1)}(t)|$.
$M$ is an upper bound on the size of the $(n+1)$th derivative over the interval between $a$ and $x$, written as $M=\max_{t\in[a,x]}|f^{(n+1)}(t)|$.
No. The tool reports an upper bound for the error, meaning the true error satisfies $|f(x)-T_n(x)|=|R_n(x)|$ and is guaranteed to be no larger than the bound.
The bound scales like $\frac{M}{(n+1)!}|x-a|^{n+1}$. Increasing $n$ typically decreases the bound, and choosing $x$ closer to $a$ (smaller $|x-a|$) usually makes the bound much smaller.
Because $M$ is computed over the entire interval between $a$ and $x$. For example, $\ln(1+x)$ requires $1+t>0$ on that interval, and $\frac{1}{1-x}$ requires $t\ne 1$ on that interval.