Details
- Tₙ(x) polynomial: —
Step-by-step
Explanation:
Taylor and Maclaurin Polynomial Calculator
This tool computes a Maclaurin polynomial (the Taylor polynomial centered at $a=0$) and evaluates it at your chosen point $x$. Select a function, enter $x$ and the degree $n$, and the calculator returns the polynomial $T_n(x)$ plus the numeric value $T_n(x)$, with steps. For more tools in this topic, explore Sequences and Series.
Compute the Maclaurin polynomial and its value at x.
$$T_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}x^k$$
What This Calculator Computes
A Taylor polynomial approximates a function near a chosen center $a$. A Maclaurin polynomial is the special case with $a=0$: $$T_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}x^k$$ This tool builds that polynomial up to degree $n$, then evaluates it at your input $x$. The result is a practical approximation you can use for quick estimates, checking work, or understanding how a function behaves near 0.
If you also need a guaranteed error bound for the approximation at a point, use the Taylor Remainder Bound Calculator.
How the Steps Are Built
The calculator follows the definition of the Maclaurin polynomial. For each $k$ from 0 to $n$, it finds the derivative value $f^{(k)}(0)$, divides by $k!$, and uses that as the coefficient of $x^k$. Then it adds the terms to form $T_n(x)$ and finally computes $T_n(x)$ at your chosen $x$.
To keep the output readable, the steps highlight the early coefficients and the final polynomial. The numeric value shown for $T_n(x)$ is formatted using the site’s configured decimal precision.
Domain Notes and Limits
The tool checks basic domain rules before computing the polynomial value. For $\ln(1+x)$, you must have $1+x>0$ at your input $x$. For $\frac{1}{1-x}$, you must have $x\ne 1$. If an input is outside the valid domain, the calculator will stop and report an error instead of producing a misleading result.
There is also a degree limit ($n$ up to 25) to keep calculations stable. Higher degrees make factorials and intermediate values grow very quickly, which can cause overflow or unreliable floating point behavior. If you need convergence behavior or a broader “does this series converge” decision, use the Series Convergence Test.
Worked Examples With Steps
Example 1: $e^x$ at $x=1$
Compute the Maclaurin polynomial of degree $n=3$ for $e^x$ and evaluate it at $x=1$.
Choose $f(x)=e^x$, $x=1$, $n=3$.
The calculator builds $T_3(x)$ from derivatives at 0, then reports the polynomial and the numeric value $T_3(1)$.
Example 2: $\sin(x)$ at $x=0.5$
Sine has alternating odd-power terms around 0, so the polynomial shows the classic alternating pattern.
Choose $f(x)=\sin(x)$, $x=0.5$, $n=5$.
The tool outputs $T_5(x)$ and the approximation $T_5(0.5)$.
Example 3: $\ln(1+x)$ at $x=0.5$
Logarithms require domain care. This example stays inside the valid domain $1+x>0$.
Choose $f(x)=\ln(1+x)$, $x=0.5$, $n=3$.
The calculator constructs the degree-3 Maclaurin polynomial and evaluates it at $x=0.5$.
Example 4: $\frac{1}{1-x}$ at $x=0.2$
This function expands cleanly around 0, but it is undefined at $x=1$, so the input must stay away from that point.
Choose $f(x)=\frac{1}{1-x}$, $x=0.2$, $n=4$.
The tool outputs the polynomial $T_4(x)$ and the numeric approximation $T_4(0.2)$.
Keep exploring Calculus and Mathematics or jump back to Calculators to browse more tools.
Questions About Taylor and Maclaurin Polynomials
Quick answers about Maclaurin polynomials, degree n, and interpreting T_n(x).
It computes the Maclaurin polynomial $T_n(x)$, which is the Taylor polynomial centered at $a=0$, and evaluates it at your chosen input $x$.
The calculator uses $T_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}x^k$ and builds the polynomial term by term up to degree $n$.
$T_n(x)$ is a polynomial approximation of $f(x)$ near 0. The value shown as $T_n(x)$ is the calculator’s approximation to $f(x)$ at your chosen $x$.
Higher degrees make factorials and intermediate terms grow very quickly, which can cause overflow or unstable floating point results. This tool limits $n$ to keep calculations reliable.
For $\ln(1+x)$, it requires $1+x>0$ at your input $x$. For $\frac{1}{1-x}$, it requires $x\ne 1$. If an input violates the domain, the tool returns an error instead of producing a result.
This tool outputs the polynomial and its numeric value, but it does not compute an error bound. For a guaranteed bound at a point, use the Taylor remainder bound tool.
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