Taylor Series Calculator
Taylor and Maclaurin polynomials with exact coefficients, the general term and an error check
A Taylor series calculator rewrites a function as a polynomial built from its derivatives at one point a: f(a) + f′(a)(x − a) + f″(a)(x − a)²/2! + ⋯. Centered at a = 0 it is called a Maclaurin series, and for ex the order-6 polynomial 1 + x + x²/2 + x³/6 + x⁴/24 + x⁵/120 + x⁶/720 gives 2.718056 at x = 1, within 0.000226 of e. Enter any function and center to get the full Taylor series expansion up to order 10: the coefficients, the general term and the error at each order.
Last updated October 2026
Method: Each coefficient is ck = f(k)(a)/k!, found by differentiating the function symbolically k times and evaluating the result at the center. Rational values, including those at multiples of π such as sin(π/6) = 1/2, stay exact fractions; all other values use double precision. The polynomial is then compared with the true function value at a point you choose.
Included: Taylor and Maclaurin polynomials of order 1 to 10, every derivative and coefficient, a plain-text version of the polynomial, the general term and radius of convergence for ex, sin x, cos x, ln(1 + x), 1/(1 − x), arctan x and (1 + x)k, and a table of the approximation and its error at each order.
Not included: Series in several variables, Laurent series with negative powers, orders above 10, a radius of convergence for functions outside the seven families above, and complex centers.
∑ Enter a function and a center
Use ^ for powers and * or a space for multiplication. Functions: sqrt, ln, log (base 10), sin, cos, tan, sec, arcsin, arctan, sinh, cosh, e and pi. Angles are in radians.
Center 0 gives the Maclaurin series. The center also accepts values such as 1/2 or pi/2.
∑ Maclaurin polynomial of order 6 (center 0)
📘 General term and convergence
This function matches the exponential series, so the whole infinite series has a closed form:
Radius of convergence: infinite. The series converges to the function for every real x.
📋 Step by step: the coefficients
Differentiate repeatedly, evaluate each derivative at a, then divide by k! to get the coefficient ck = f(k)(a)/k!.
| k | f(k)(x) | f(k)(a) | ck |
|---|---|---|---|
| 0 | ex | 1 | 1 |
| 1 | ex | 1 | 1 |
| 2 | ex | 1 | 1/2 ≈ 0.5 |
| 3 | ex | 1 | 1/6 ≈ 0.166667 |
| 4 | ex | 1 | 1/24 ≈ 0.0416667 |
| 5 | ex | 1 | 1/120 ≈ 0.00833333 |
| 6 | ex | 1 | 1/720 ≈ 0.00138889 |
Exact fractions are shown whenever the value at the center is rational; otherwise the coefficient is a decimal rounded to 10 significant digits.
🎯 Approximation vs. actual value at x = 1
| Order | Pn(x) | Actual f(x) | Error |
|---|---|---|---|
| 1 | 2 | 2.718281828 | 0.7183 |
| 2 | 2.5 | 2.718281828 | 0.2183 |
| 3 | 2.666666667 | 2.718281828 | 0.05162 |
| 4 | 2.708333333 | 2.718281828 | 0.009948 |
| 5 | 2.716666667 | 2.718281828 | 0.001615 |
| 6 | 2.718055556 | 2.718281828 | 0.0002263 |
Error = |f(x) − Pn(x)|. Orders whose new coefficient is 0 repeat the previous value, which is why sine and cosine rows come in pairs.
Taylor series calculator: formula, worked examples and convergence
A Taylor polynomial is a local copy of a function. It matches the function's value at the center, then its slope, then its curvature, and so on up to the order you choose, so near the center the polynomial and the function are almost impossible to tell apart. Keep adding terms forever and you get the Taylor series, which for many familiar functions equals the function exactly over a whole interval. This page explains the formula, works through three examples with real numbers, and shows when the series converges and how large the error is.
Every coefficient is a derivative divided by a factorial, so the Derivative Calculator is the natural companion for checking individual derivatives. The Integral Calculator integrates functions, a job that series make easier term by term, the Limit Calculator handles the limits behind convergence questions, and the Graphing Calculator lets you plot a function next to its polynomial and watch them separate away from the center.
The Taylor series formula
For a function f with derivatives of every order at a point a, the Taylor series and the Taylor polynomial of order n are
f(x) = Σk=0..∞ f(k)(a) / k! · (x − a)k Pn(x) = f(a) + f′(a)(x − a) + f″(a)/2! · (x − a)2 + … + f(n)(a)/n! · (x − a)n The number ck = f(k)(a)/k! is the k-th coefficient. The factorial is there for a precise reason: differentiating (x − a)k k times produces k!, so dividing by k! makes the k-th derivative of the polynomial at a equal to the k-th derivative of the function. Taylor's theorem, equation 1.4.35 in the NIST Digital Library of Mathematical Functions, makes the link exact: f(x) = Pn(x) + Rn, where the remainder is Rn = f(n+1)(c)/(n + 1)! · (x − a)n+1 for some c between a and x (equation 1.4.36). The remainder is the error the calculator's comparison table measures.
Maclaurin series: centering at zero
A Maclaurin series is a Taylor series with a = 0, so every (x − a) becomes plain x. Most tables of series are Maclaurin series, because many functions have simple derivative values at 0: ex gives 1 every time, sin x cycles through 0, 1, 0, −1, and ln(1 + x) produces alternating factorials. Centering at 0 is the right choice when you care about small inputs, such as small angles in physics or small interest rates in finance. When the inputs of interest sit far from 0, a different center gives a far better polynomial for the same number of terms, as the second example shows.
Worked example 1: the Maclaurin polynomial of ex
Every derivative of ex is ex itself, and e0 = 1, so the coefficients are simply ck = 1/k!. The order-6 polynomial is
P6(x) = 1 + x + x2/2 + x3/6 + x4/24 + x5/120 + x6/720 Evaluating at x = 1 approximates e itself. The table lists each order next to the true value e = 2.718281828; it is exactly what the calculator shows for ex, center 0 and x = 1.
| Order n | Term added | Pn(1) | Error |e − Pn(1)| |
|---|---|---|---|
| 1 | x | 2 | 0.7183 |
| 2 | x2/2 | 2.5 | 0.2183 |
| 3 | x3/6 | 2.666666667 | 0.05162 |
| 4 | x4/24 | 2.708333333 | 0.009948 |
| 5 | x5/120 | 2.716666667 | 0.001615 |
| 6 | x6/720 | 2.718055556 | 0.0002263 |
Each new term divides the error by roughly the next whole number, the signature of a factorial in the denominator: from order 5 to order 6 the error falls from 0.001615 to 0.0002263, about a factor of 7. Push the order to 10 and the calculator returns 2.718281801, within 2.731 × 10−8 of e. The same polynomial works for any x, but more slowly as x grows, because the powers xk grow before the factorials catch up.
Worked example 2: approximating √4.1 with a center of 4
A Maclaurin series cannot help with √x at all, because the square root has no derivative at 0. The trick is to center at a nearby perfect square, a = 4, where every derivative value is a fraction. The derivatives of f(x) = √x are 1/(2√x), −1/(4x3/2) and 3/(8x5/2), which at x = 4 equal 1/4, −1/32 and 3/256. Dividing by 1!, 2! and 3! gives
P3(x) = 2 + (x − 4)/4 − (x − 4)2/64 + (x − 4)3/512 With x = 4.1, the step from the center is only 0.1. The linear approximation gives 2 + 0.1/4 = 2.025, off by 0.0001543. Adding the quadratic term subtracts 0.01/64 = 0.00015625 and gives 2.02484375, off by just 0.000001923. The cubic term adds 0.001/512 and brings the result to 2.0248457031, within 3.0 × 10−8 of the true value √4.1 = 2.0248456731. The cubic polynomial delivers seven correct decimals, all from fractions you could compute by hand. Enter sqrt(x), center 4 and x = 4.1 in the calculator to reproduce every number.
Building the coefficients step by step: ln x at a = 1
The coefficient table in the calculator follows the same four columns as the table below. For f(x) = ln x centered at a = 1, each derivative is a power of x with a growing factorial in front, the values at x = 1 alternate in sign, and the division by k! leaves the simple fractions 1/k with alternating signs.
| k | f(k)(x) | f(k)(1) | k! | ck = f(k)(1)/k! |
|---|---|---|---|---|
| 0 | ln x | 0 | 1 | 0 |
| 1 | 1/x | 1 | 1 | 1 |
| 2 | −1/x2 | −1 | 2 | −1/2 |
| 3 | 2/x3 | 2 | 6 | 1/3 |
| 4 | −6/x4 | −6 | 24 | −1/4 |
| 5 | 24/x5 | 24 | 120 | 1/5 |
The result is ln x ≈ (x − 1) − (x − 1)2/2 + (x − 1)3/3 − (x − 1)4/4 + (x − 1)5/5. At x = 1.5 the successive orders give 0.5, 0.375, 0.4166666667, 0.4010416667 and 0.4072916667, zigzagging around the true value ln 1.5 = 0.4054651081 with errors of 0.09453, 0.03047, 0.0112, 0.004423 and 0.001827. Convergence is slower than for ex because the coefficients shrink like 1/k rather than 1/k!.
Common Maclaurin series and where they converge
The calculator recognizes these seven families when the center is 0 and prints the general term and the interval of convergence. The series and their convergence conditions are those in NIST DLMF sections 4.2, 4.6, 4.19 and 4.24.
| Function | First terms | General term | Converges for |
|---|---|---|---|
| ex | 1 + x + x2/2! + x3/3! + … | xn/n! | all x |
| sin x | x − x3/3! + x5/5! − … | (−1)n x2n+1/(2n+1)! | all x |
| cos x | 1 − x2/2! + x4/4! − … | (−1)n x2n/(2n)! | all x |
| ln(1 + x) | x − x2/2 + x3/3 − … | (−1)n+1 xn/n, n ≥ 1 | −1 < x ≤ 1 |
| 1/(1 − x) | 1 + x + x2 + x3 + … | xn | −1 < x < 1 |
| arctan x | x − x3/3 + x5/5 − … | (−1)n x2n+1/(2n+1) | −1 ≤ x ≤ 1 |
| (1 + x)k | 1 + kx + k(k − 1)x2/2! + … | k(k − 1)…(k − n + 1) xn/n! | −1 < x < 1 (all x if k = 0, 1, 2, …) |
The binomial row covers many functions at once. With k = 1/2 it is √(1 + x) = 1 + x/2 − x2/8 + x3/16 − 5x4/128 + …, which at x = 0.2 gives 1.1, 1.095, 1.0955 and 1.0954375 for orders 1 to 4, against √1.2 = 1.095445115. With k = −1 it becomes 1/(1 + x) = 1 − x + x2 − …, and when k is a whole number the series stops after the xk term and reproduces the ordinary binomial expansion. The behavior at the endpoints x = ±1 depends on k.
Radius and interval of convergence
A Taylor polynomial of any finite order is just a polynomial, defined everywhere. The question of convergence is about the infinite series: for which x do the partial sums settle down to the function value? The answer is always a symmetric interval around the center, a − R < x < a + R, where R is the radius of convergence, plus possibly one or both endpoints. Inside the interval, adding terms drives the error to zero. Outside it, the terms grow and the partial sums run away.
The geometric series shows both sides clearly. At x = 0.5, inside the interval, the order-10 polynomial of 1/(1 − x) gives 1.999023438, within 0.0009766 of the true value 2. At x = 2, outside, the partial sums are 3, 7, 15, 31 and 63 for orders 1 to 5, doubling every time, while the function value is −1. Exactly on the boundary convergence can be painfully slow: the arctangent series at x = 1 converges to π/4 = 0.7853981634, but after order 9 it still reads 0.8349206349, and the series for ln(1 + x) at x = 1 still misses ln 2 by 0.04751 after ten orders.
Why is the radius 1 for 1/(1 − x) and ln(1 + x)? Because each function breaks down at distance 1 from the center, at x = 1 and x = −1 respectively. The general rule is that R equals the distance from the center to the nearest point where the function stops being smooth, and that point may be a complex number. That explains the surprise of 1/(1 + x2): it is smooth for every real x, yet its Maclaurin series has radius 1, because the function blows up at the complex points x = ±i, which sit at distance 1 from 0.
How accurate is the polynomial? Taylor's remainder
The remainder formula turns accuracy into a calculation. For ex at x = 1 with order n = 6, the seventh derivative ec is at most e on the interval from 0 to 1, so the error is at most e/7! = 2.718 / 5,040 ≈ 0.000539. The actual error from the table is 0.0002263, comfortably inside the bound. Bounds like this let you choose the order before computing anything: to guarantee an error below 0.000001 for e, you need (n + 1)! larger than e × 1,000,000, about 2.7 million, which first happens at n = 9, since 10! = 3,628,800. The calculator confirms it: the order-8 error is 0.000003059 and the order-9 error is 0.0000003029.
Alternating series give an even simpler bound: when the terms alternate in sign and shrink steadily, the error is smaller than the first term you left out. The order-7 Maclaurin polynomial of sin x at x = 1 gives 0.841468254, while sin 1 = 0.8414709848. The first omitted term is 1/9! ≈ 0.000002756, and the actual error, 0.000002731, sits just under it.
How the error grows with distance from the center
The same order behaves very differently depending on how far x is from the center and whether x lies inside the interval of convergence. This table gives the error of the order-10 Maclaurin polynomial at four distances; each entry is what the calculator reports with order 10, center 0 and the given x.
| Function | x = 0.1 | x = 0.5 | x = 1 | x = 2 |
|---|---|---|---|---|
| ex | < 10−15 | 1.276 × 10−11 | 2.731 × 10−8 | 6.139 × 10−5 |
| sin x | < 10−15 | 1.221 × 10−11 | 2.489 × 10−8 | 5.002 × 10−5 |
| cos x | < 10−15 | 5.09 × 10−13 | 2.076 × 10−9 | 8.366 × 10−6 |
| ln(1 + x) | 8.329 × 10−13 | 3.046 × 10−5 | 0.04751 | 65.92 |
| arctan x | 9.014 × 10−13 | 3.667 × 10−5 | 0.04952 | 43.23 |
| 1/(1 − x) | 1.111 × 10−11 | 0.0009766 | undefined | 2,048 |
The first three rows have an infinite radius, so the error stays small even at x = 2. The last three have radius 1: at x = 2 the polynomial no longer approximates anything, and at x = 1 the function 1/(1 − x) is not even defined. Entries marked < 10−15 are below the resolution of double-precision arithmetic, so the true error is effectively zero.
Choosing a good center
Accuracy depends on the distance from the center far more than on the order, so center the series where you need answers. To approximate sin(0.6), a center of π/6 ≈ 0.5236 is only 0.0764 away, compared with 0.6 for a Maclaurin series. The derivatives of sin x at π/6 are 1/2, √3/2, −1/2, −√3/2 and 1/2, so the calculator shows the exact coefficients 1/2, −1/4 and 1/48 and the decimals 0.8660254038 and −0.1443375673 for the terms involving √3. The errors at x = 0.6 for orders 1 to 4 are 0.001523, 0.00006364, 0.0000007285 and 0.00000001865, so the order-4 polynomial gives 0.5646424547 against sin 0.6 = 0.5646424734. Typing pi/6 as the center is all it takes; the calculator evaluates sine and cosine exactly at multiples of π/6 and π/4 wherever the result is rational.
How to use this calculator
The same tool works as a Maclaurin series calculator (center 0) and as a Taylor polynomial calculator for any other center. Type the function using x as the variable, with ^ for powers and * or a space for multiplication; sqrt, ln, log, sin, cos, tan, arctan, e and pi are all understood, and angles are in radians. Set the center a, which can be 0 for a Maclaurin series or any other number, including fractions such as 1/2 and expressions such as pi/2. Choose the order with the buttons from 1 to 10. If you also enter a comparison point, the calculator evaluates every polynomial from order 1 up to your chosen order at that point and lists each one next to the true function value and the error.
The headline result is the polynomial itself, in readable notation, followed by a plain-text version you can paste into a graphing tool or a spreadsheet. Below it, the coefficient table shows every derivative, its value at the center and the resulting coefficient, which is the full step-by-step working. For the seven recognized families at center 0, an extra card shows the general term, the radius of convergence and whether your comparison point lies inside the interval. If a derivative does not exist at the center, as with √x at 0, the calculator stops at the last order that does exist and says so.
Who uses Taylor polynomials
Taylor series are a core topic of AP Calculus BC and of college Calculus II, where students build polynomials, find intervals of convergence and estimate errors, and the coefficient table mirrors the work those courses expect. Physicists rely on low-order polynomials constantly. The small-angle approximation sin θ ≈ θ behind the simple pendulum formula is the order-1 Maclaurin polynomial of sine: at 10°, which is 0.17453293 radians, it overstates sin θ = 0.17364818 by 0.51%, at 5° by only 0.13%, and at 20° by 2.06%. The companion approximation cos θ ≈ 1 − θ2/2 is off by 0.0000386 at 10°.
Engineers linearize nonlinear models around an operating point, which is an order-1 Taylor polynomial in disguise, and control and signal processing courses build on the same idea. Numerical analysts use Taylor expansions to derive finite-difference formulas and to analyze the error of methods for differential equations. In finance, the duration and convexity of a bond are the first two terms of a Taylor expansion of its price in the interest rate.
Key terms
- Taylor polynomial Pn(x): the sum of the terms up to order n; a finite approximation.
- Taylor series: the infinite sum of all terms, which may or may not equal the function.
- Maclaurin series: a Taylor series centered at a = 0.
- Center a: the point where the derivatives are evaluated and where the approximation is best.
- Order n: the highest power of (x − a) kept in the polynomial.
- Coefficient ck: f(k)(a)/k!, the number multiplying (x − a)k.
- Remainder Rn(x): the error f(x) − Pn(x).
- Radius of convergence: the distance from the center within which the series converges.
- Interval of convergence: the set of x values where the series converges, including any endpoints that work.
- Analytic function: a function that equals its Taylor series in some interval around every point of its domain.
Limitations
The calculator handles functions of one variable and orders up to 10. It computes each coefficient from exact symbolic derivatives, but values that are not rational, such as e/k! for ex centered at 1, are shown as decimals rounded to 10 significant digits. The general term and the radius of convergence appear only for the seven recognized families at center 0; for other functions the radius depends on where the function or its complex extension breaks down, which the calculator does not try to locate. The comparison column uses double precision, so errors smaller than about 10−15 are reported as such rather than as exact values. Finally, a Taylor series can converge to the wrong function: the classic example is e−1/x2, whose Maclaurin series is zero although the function is not.
Tips for working with Taylor polynomials
- Center the series as close as possible to the inputs you care about; distance matters more than order.
- Build new series from known ones: replacing x with −x2 in the series for ex gives the series for e−x2 without any differentiation.
- Check the first two coefficients by hand: c0 must equal f(a) and c1 must equal f′(a).
- Watch the error column as the order rises: a shrinking error means you are inside the interval of convergence, a growing one means you are outside.
- For alternating series, use the first omitted term as a quick error bound.
Related calculators
- To check any single derivative with a step-by-step rule list, use the Derivative Calculator.
- To integrate a function, for example term by term after expanding it, use the Integral Calculator.
- For the limits behind convergence tests, use the Limit Calculator.
- To plot a function next to its Taylor polynomial, use the Graphing Calculator.
- For the factorials in every denominator, use the Factorial Calculator.
- For sums of the geometric series 1 + x + x2 + …, use the Geometric Sequence Calculator.
- For the slope of a tangent line, which is the order-1 Taylor polynomial, use the Slope Calculator.
Sources
Taylor series are exact mathematical results rather than statistics. The formulas and convergence conditions on this page match the NIST Digital Library of Mathematical Functions, the U.S. government's reference for special functions:
- NIST DLMF §1.4, Calculus of One Variable: Taylor's theorem (1.4.35) and the remainder (1.4.36).
- NIST DLMF §4.2, Logarithm, exponential and powers: definitions: the exponential series (4.2.19).
- NIST DLMF §4.6, Power series: ln(1 + z) for |z| ≤ 1, z ≠ −1 (4.6.1) and the binomial series for |z| < 1 (4.6.7).
- NIST DLMF §4.19, Maclaurin series of the trigonometric functions: sin z and cos z (4.19.1, 4.19.2).
- NIST DLMF §4.24, Inverse trigonometric functions: the arctangent series for |z| ≤ 1 (4.24.3).
Every polynomial, coefficient and error figure on this page was computed with the same engine the calculator uses.
⚠️ Common mistakes & edge cases
Forgetting to divide by k!
The coefficient is the derivative value divided by k!, not the derivative value itself. For ex every derivative equals 1 at 0, but the coefficient of x4 is 1/24, not 1. Leaving out the factorials turns a fast-converging series into one that does not converge at all.
Writing xk instead of (x − a)k
With a center other than 0, every power must be a power of (x − a). The polynomial for √x at a = 4 is 2 + (x − 4)/4 − …, and writing 2 + x/4 − … instead gives 3.025 at x = 4.1 rather than 2.025.
Using a series outside its interval of convergence
Polynomials are defined everywhere, so nothing warns you when x is outside the interval. At x = 2 the order-10 polynomial of ln(1 + x) is off by 65.9, even though ln 3 is a perfectly ordinary number. Check the radius first, or watch whether the error shrinks as the order rises.
Plugging in degrees
The series for sin x and cos x assume radians. Plugging in 30 for 30° makes the polynomial try to approximate sin 30 (radians), about −0.988, instead of 0.5. Convert first: 30° is π/6 ≈ 0.5236 radians.
Confusing the order with the number of terms
An order-5 polynomial keeps powers up to 5, but some of those coefficients can be zero. For sin x it has three nonzero terms, and for cos x the order-5 polynomial is the same as the order-4 one. Exam questions that ask for the first three nonzero terms of sin x want order 5.
Expanding where a derivative does not exist
ln x cannot be expanded around 0, and √x has no first derivative there, so a Maclaurin series does not exist for either. Shift the center to a point inside the domain, such as a = 1 for ln x or a = 4 for √x, or expand ln(1 + x) instead.
❓ Frequently asked questions
What is the difference between a Taylor series and a Maclaurin series?
A Maclaurin series is a Taylor series centered at a = 0. The Taylor series of f around a point a is f(a) + f′(a)(x − a) + f″(a)(x − a)²/2! + ⋯, and setting a = 0 turns every (x − a) into x. The Maclaurin series of eˣ, for example, is 1 + x + x²/2! + x³/3! + ⋯. Centering at another point, such as a = 4 for √x, gives a polynomial that is most accurate near that point instead of near 0.
How do you find the Taylor polynomial of degree n?
Differentiate the function n times, evaluate the function and each derivative at the center a, divide the value of the k-th derivative by k!, multiply by (x − a) to the k-th power, and add the terms. For √x at a = 4 the values for k = 0 to 3 are 2, 1/4, −1/32 and 3/256; dividing by 0!, 1!, 2! and 3! gives the coefficients 2, 1/4, −1/64 and 1/512, so P₃(x) = 2 + (x − 4)/4 − (x − 4)²/64 + (x − 4)³/512.
What are the Taylor series of eˣ, sin x and cos x?
Centered at 0, eˣ = 1 + x + x²/2! + x³/3! + ⋯, sin x = x − x³/3! + x⁵/5! − ⋯ and cos x = 1 − x²/2! + x⁴/4! − ⋯. All three converge for every real x. The order-6 Maclaurin polynomial of eˣ gives 2.718056 at x = 1, within 0.000226 of e, and the order-7 polynomial of sin x gives 0.841468 at x = 1, within 0.0000027 of sin 1.
How accurate is a Taylor polynomial approximation?
Taylor's theorem bounds the error: after the order-n term the remainder equals f⁽ⁿ⁺¹⁾(c)(x − a)ⁿ⁺¹/(n + 1)! for some c between a and x. For eˣ at x = 1 with n = 6, that derivative is at most e, so the error is at most e/7! ≈ 0.000539, and the actual error is 0.000226. For alternating series such as sin x, the error is smaller than the first term left out. The comparison table in the calculator shows the real error at every order.
What is the radius of convergence of a Taylor series?
It is the distance from the center within which the infinite series converges to the function. For eˣ, sin x and cos x the radius is infinite. For ln(1 + x), arctan x, 1/(1 − x) and the binomial series (1 + x)ᵏ it is 1, unless k is a whole number, in which case the binomial series stops and works for every x. In general the radius equals the distance from the center to the nearest point where the function stops being smooth, counting complex points, which is why 1/(1 + x²) has radius 1 even though it is defined for every real x.
Why does the Taylor series of 1/(1 − x) fail at x = 2?
Because x = 2 lies outside the interval of convergence, −1 < x < 1. The partial sums of 1 + x + x² + ⋯ at x = 2 are 3, 7, 15, 31, 63 and keep doubling, while the function value is 1/(1 − 2) = −1, so every added term makes the error larger. Inside the interval the series works: at x = 0.5 the order-10 polynomial gives 1.999023, within 0.000977 of the true value 2.
Can I center the series at π/2 or at another point?
Yes. The center can be any real number, including fractions such as 1/2 and multiples of π such as pi/2 or pi/6. The calculator keeps coefficients exact whenever the derivative values are rational: sin x centered at π/6 has the exact coefficients 1/2, −1/4 and 1/48 for its even terms, while the odd terms involve √3 and appear as decimals. Pick a center close to the x values you care about, because accuracy falls off with distance from the center.
What does the order of a Taylor polynomial mean?
The order is the highest power of (x − a) that is kept, so an order-5 polynomial uses derivatives up to the fifth. Some coefficients can be zero: the order-5 Maclaurin polynomial of sin x has only three nonzero terms, x − x³/6 + x⁵/120, because every even derivative of sin x is 0 at x = 0. That is why the comparison table repeats the same value for orders 1 and 2, then 3 and 4, and so on.
What are Taylor series used for in practice?
They turn hard functions into polynomials, which are easy to evaluate, differentiate and integrate. Physics uses sin θ ≈ θ for small angles, which is off by only 0.51% at 10°. Engineers linearize nonlinear systems around an operating point, numerical software evaluates functions such as eˣ from polynomial approximations after shrinking the input range, and integrals with no elementary antiderivative, such as the integral of e^(−x²), are computed by integrating the series term by term.
Does every function have a Taylor series?
No. The function needs derivatives of every order at the center, so ln x has no series at 0, and √x has no Taylor polynomial beyond order 0 there. Even when every derivative exists, the series may not equal the function: e^(−1/x²), defined as 0 at x = 0, has every derivative equal to 0 at that point, so its Maclaurin series is identically 0 even though the function is positive everywhere else. Functions that do equal their Taylor series near every point are called analytic.
💡 Good to know
The tangent line is a Taylor polynomial
The order-1 Taylor polynomial f(a) + f′(a)(x − a) is exactly the tangent line at a, and the order-2 polynomial adds the curvature. Each extra order makes the polynomial hug the curve for a little longer before the two separate.
Calculators shrink the input first
A raw Maclaurin polynomial would need many terms for e20 or sin(100). Numerical libraries first reduce the input to a small range, using identities such as e20 = (e5)4 or the periodicity of sine, and only then apply a short polynomial where it is most accurate.
e itself comes from a series
Setting x = 1 in the series for ex gives e = 1 + 1 + 1/2! + 1/3! + …, one of the fastest-converging ways to compute the constant. Eleven terms, through 1/10!, already fix it to within 2.731 × 10−8.