Taylor Series Calculator
Exact coefficients, and the error you actually get.
Work out Taylor Series. Exact coefficients, and the error you actually get. Names the misconception directly.
exp(x), sin(x), cos(x) and ln(x) all have well-known expansions.
Taylor polynomial about x = 0
1 + x + 0.5x^2 + 0.166667x^3 + 0.0416667x^4 + 0.00833333x^5
At x = 1: approximates 2.7166667 against an actual 2.7182818
Each coefficient is f⁽ⁿ⁾(a) divided by n!, with the derivative taken symbolically. Finite differences lose precision fast past the second derivative and are worthless by the fifth, which is exactly the range where a Taylor series becomes interesting. The error column is the thing to watch. Taylor accuracy falls off with distance from the expansion point, and adding terms helps enormously near the centre and barely at all far from it — which is why expanding about a point close to where you actually need the value matters more than the order does. Some functions have no Taylor series at a given point at all. ln(x) about 0 is the standard example, and it is refused rather than answered, because the function and every derivative are undefined there.
How the Taylor Series Calculator works
Taylor coefficients about a point, from repeated symbolic differentiation, with the approximation and its actual error at a value you choose. Finite differences are useless past the second derivative, which is exactly where a series gets interesting.
Also known as: maclaurin series calculator · series expansion of e^x · taylor polynomial approximation · how many terms do i need
Frequently asked questions
What is a Taylor series?
A polynomial that matches a function's value and derivatives at a chosen point. Each coefficient is f⁽ⁿ⁾(a) divided by n!, and the more terms you keep the better it matches — near that point.
What is a Maclaurin series?
A Taylor series expanded about zero. It is the same construction with a specific centre, and it is what you get here by leaving the expansion point at 0.
Why is my approximation bad far from the centre?
Because Taylor accuracy falls off with distance from the expansion point. Adding terms helps enormously nearby and barely at all far away, so choosing a centre close to the value you need matters more than the order does.
Why does ln(x) have no Taylor series about 0?
Because the function and every one of its derivatives are undefined there. Expanding ln(1+x) about 0 works instead, which is why that form appears in every textbook.
What are Taylor series used for?
Everywhere a hard function needs replacing by an easy one: sin x ≈ x for small angles in physics, series approximations inside numerical libraries, and error analysis throughout applied mathematics.
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