Taylor Series Calculator
The Taylor or Maclaurin polynomial of a function, with each coefficient and how far the approximation is off.
Results
What this tool does
A Taylor series turns any smooth function into a polynomial, which is the only kind of function a computer can really evaluate — every sine your calculator has ever produced came from one. The idea is simple: match the value at a point, then the slope, then the curvature, then the rate of change of curvature, and so on. This page computes the derivatives exactly, builds the polynomial in the classic ascending form with factorial denominators, and shows how far it is from the true function at a point of your choosing.
Formula
f(x) ≈ Σₖ₌₀ⁿ f⁽ᵏ⁾(a) (x − a)ᵏ ÷ k! · a = 0 ⟹ Maclaurin
Variables
| Symbol | Meaning | Unit |
|---|---|---|
expr | The expression, in x | — |
centre | Centre the series at | — |
order | How many terms (order) | — |
at | Compare against the function at | — |
PL | The series | — |
TB | k · the k-th derivative · k! · the coefficient | — |
TV | What the function really is there | — |
AV | What the series gives there | — |
ER | How far the series is off | — |
EP | The same as a share | % |
MC | Is this a Maclaurin series? | — |
OR | How many terms (order) | — |
Worked example
- The expression, in xexp(x)
- Centre the series at0
- How many terms (order)5
- Compare against the function at0.5
- The series1 + x + x^2 / 2 + x^3 / 6 + x^4 / 24 + x^5 / 120
- k · the k-th derivative · k! · the coefficient0 1 1 1 1 1 1 1 2 1 2 0.50000000 3 1 6 0.16666667 4 1 24 0.04166667 5 1 120 0.00833333
- What the function really is there1.648721270700
- What the series gives there1.648697916670
- How far the series is off0.000023354033
- The same as a share0.0014 %
- Is this a Maclaurin series?Yes
- How many terms (order)5
Limitations
- The calculation runs entirely in your browser. The values you type are never sent to a server.
Frequently asked questions
What is the difference between Taylor and Maclaurin?
Only where the series is centred. A Maclaurin series is a Taylor series centred at zero, and that is the whole distinction — the formula is identical. Centring matters because the approximation is best near the centre and gets worse as you move away: the series for exp centred at zero is excellent at x = 0.1 and useless at x = 10. If the point you care about is far from zero, centre the series there instead, and the same number of terms will do far better.
Why is the error sometimes large?
Because a Taylor polynomial is a local approximation, and two things make it worse: distance from the centre, and the function having a singularity nearby. The series for 1/(1−x) centred at zero is exactly 1 + x + x² + … and it converges only for |x| < 1 — at x = 1 it fails completely, because the function itself blows up there. Adding more terms helps inside the radius of convergence and does nothing at all outside it. The error shown compares the polynomial against the true function at the point you chose, so you can see this happen rather than being told about it.