Taylor Series Approximator

Math Algebra • Sequences and Series

Written by STEM Calculators Team Published December 19, 2025 Updated February 24, 2026

7. Taylor Series Approximator

Build a Taylor/Maclaurin polynomial \(P_n(x)\) about \(x=a\), plot \(f(x)\) vs. \(P_n(x)\), and estimate the Lagrange remainder.

Overlay graph

x-axis: x • y-axis: y. Drag to pan • wheel/pinch to zoom • Auto-fit uses separate x/y scales.

x: 0, y: 0 sx: 40, sy: 40 \(f(x)\) \(P_n(x)\)

Choose a function and click Compute & plot.

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Theory: Taylor series approximation

1) Taylor polynomial about \(x=a\)

If a function \(f\) is differentiable many times near \(x=a\), its Taylor polynomial of order \(n\) is

\[ P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k. \]

The special case \(a=0\) is the Maclaurin polynomial.

2) Lagrange remainder and an error bound idea

The exact error is \(R_n(x)=f(x)-P_n(x)\). When \(f^{(n+1)}\) exists, the Lagrange form says:

\[ R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1} \quad \text{for some } \xi \text{ between } a \text{ and } x. \]

So if you can bound \(\lvert f^{(n+1)}(t)\rvert \le M\) for all \(t\) between \(a\) and \(x\), then:

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

In this calculator, \(M\) is estimated numerically by sampling an approximation of \(f^{(n+1)}\) on the interval. This is a practical estimate (not a formal proof for every input).

3) Solved example: \(\sin(x)\) at \(a=0\), order \(5\)

Find the Maclaurin polynomial \(P_5(x)\) for \(f(x)=\sin(x)\).

Step 1: derivatives at \(0\)

\[ \begin{aligned} f(x)&=\sin x &\Rightarrow&\ f(0)=0\\ f'(x)&=\cos x &\Rightarrow&\ f'(0)=1\\ f''(x)&=-\sin x &\Rightarrow&\ f''(0)=0\\ f^{(3)}(x)&=-\cos x &\Rightarrow&\ f^{(3)}(0)=-1\\ f^{(4)}(x)&=\sin x &\Rightarrow&\ f^{(4)}(0)=0\\ f^{(5)}(x)&=\cos x &\Rightarrow&\ f^{(5)}(0)=1 \end{aligned} \]

Step 2: build \(P_5(x)\)

\[ \begin{aligned} P_5(x) &=\sum_{k=0}^{5}\frac{f^{(k)}(0)}{k!}x^k \\ &=0+\frac{1}{1!}x+0+\frac{-1}{3!}x^3+0+\frac{1}{5!}x^5 \\ &=x-\frac{x^3}{6}+\frac{x^5}{120}. \end{aligned} \]

Step 3: remainder bound (Lagrange form)

For \(n=5\), the next derivative is \(f^{(6)}(x)=-\sin x\), so \(\lvert f^{(6)}(x)\rvert \le 1\). Therefore, for any \(x\),

\[ |R_5(x)| \le \frac{1}{6!}|x|^6 = \frac{|x|^6}{720}. \]

You can reproduce this instantly with the calculator’s “Fill example: \(\sin x\) at 0, order 5”.

4) Notes on domains and numerical differentiation

Some presets have domain restrictions (e.g., \(\ln(1+x)\) needs \(x>-1\); \(\sqrt{1+x}\) needs \(x\ge -1\)). If the graph breaks, reduce the x-range or change the center \(a\).
Derivatives are estimated using finite differences with a step size \(h\). Higher orders can be sensitive; if results look unstable, reduce the order \(n\) or try a different \(h\).
The overlay graph helps you see where the Taylor polynomial is accurate (usually best near \(x=a\)).

Frequently Asked Questions

What does a Taylor polynomial P_n(x) represent?

P_n(x) is an order-n polynomial that approximates f(x) near x = a using derivatives at a. It has the form P_n(x) = sum_{k=0}^n (f^(k)(a)/k!) (x-a)^k.

How is a Maclaurin series different from a Taylor series?

A Maclaurin polynomial is a Taylor polynomial centered at a = 0. The same derivative-and-coefficient formula is used, but (x-a) becomes x.

How does the calculator estimate the Lagrange remainder error?

It uses the bound |R_n(x)| <= (M/(n+1)!) |x-a|^(n+1), where M is an estimated maximum of |f^(n+1)(t)| between a and x. The calculator estimates M numerically by sampling an approximation of the (n+1)th derivative on the interval.

Why does the approximation get worse farther from the center a?

Taylor polynomials are built from local derivative information at a, so accuracy is typically best near a and can degrade as |x-a| grows. The remainder term includes a factor |x-a|^(n+1), which can increase quickly away from the center.

What does the derivative step h (optional) affect?

When derivatives are approximated numerically, h controls the finite-difference spacing. A very large or very small h can change the stability of higher-order derivative estimates, which can affect coefficients and the remainder estimate.

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Frequently Asked Questions

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