A Taylor polynomial copies a function at one point: its height, its slope, its bend, and as many more derivatives as you ask for. Scroll down: at each stop a short note explains the next idea, then the picture is yours to play with before the next one arrives.
Everything on this page follows from the assumptions and definitions below. They are granted, not proved, so judge each one for yourself. A step that leans on one is tagged with its number.
The dark curve is ex. The green line is a polynomial built to imitate it near the blue dot, the center, which sits at a = 0.
A degree 0 polynomial is a constant. The best it can do is copy the height of the curve at the center: P0(x) = 1.
A degree 1 polynomial can copy the slope too. The function and its derivative are both 1 at x = 0, so P1(x) = 1 + x. That is the tangent line you already know.
Try itMove the Degree slider between 0 and 1. How far from the center does the tangent line stay close to the curve? On which side does it fail first?
The tangent line is straight and the curve is not. A degree 2 polynomial can bend, so ask it to copy the second derivative at the center as well.
Take P2(x) = 1 + x + cx². Its second derivative is 2c, and it has to equal the function's second derivative at 0, which is 1. So c = 1/2:
P2(x) = 1 + x + x²/2
Adding the x² term did not disturb the height or the slope at 0, because x² and its derivative are both zero there.
Try itThe slider now goes to 2. Step from 1 to 2 and back. Where does the extra term help the most?
Keep going. To copy the third derivative, add a term cx³. Differentiating x³ three times gives 3 · 2 · 1 = 3!, so the coefficient has to be the third derivative divided by 3!.
The same happens at every degree, and at any center a:
Pn(x) = f(a) + f′(a)(x − a) + f″(a)(x − a)²/2! + … + f(n)(a)(x − a)n/n!
That is the Taylor polynomial of degree n. The table on the right builds it one line at a time: a derivative, its value at the center, and that value divided by k!.
Rests on
Try itThe slider now goes to 12. Raise the degree one step at a time. About how far from 0 does the green curve hug the dark one at degree 4? At degree 8?
A menu has appeared, and the function is now sin x. Its derivatives go around in a circle: sin, cos, −sin, −cos, and back to sin. At x = 0 their values are 0, 1, 0, −1, over and over.
Every even-numbered line in the table is zero, so those terms vanish:
sin x ≈ x − x³/6 + x⁵/120 − x⁷/5040
That means the degree 2 polynomial is the same as degree 1, degree 4 is the same as degree 3, and so on.
Try itStep the degree from 1 up to 7 and watch which steps change the picture. Then pick cos x from the menu. Which powers does it keep?
An orange dot has appeared on the axis. It marks a point x, and the orange bar above it is the gap between the function and the polynomial there. The length of that bar is the error:
error = |f(x) − Pn(x)|
The chart on the right shows the error at this x for every degree from 0 to 12 at once. Its scale is in powers of ten: a bar that reaches from 10⁻⁴ up to 10⁻² is a hundred times worse.
Right now the function is sine, the degree is 3 and x = 1.5. The polynomial is off by about 0.06.
Rests on
Try itDrag the orange dot to x = 1. What is the lowest degree with an error under 0.001? Now drag it to x = 4 and ask again.
The polynomial is built from derivatives at the center, so the center is where it knows the function best. The farther x is from a, the worse it does.
You can now drag the blue dot. When the center is not 0 the polynomial is written in powers of (x − a), and its coefficients are decimals, because they come from values like sin 1.5 that have no tidy form.
Try itThe orange dot is at x = 4, where the degree 3 polynomial about 0 is badly off. Leave the degree alone and slide the center toward 4. Watch the error.
You measured the error by comparing with the true value. Usually the true value is the thing you do not have. Lagrange's bound caps the error without it:
|f(x) − Pn(x)| ≤ M · |x − a|n+1/(n + 1)!
Here M is any number at least as large as the size of the next derivative, number n + 1, everywhere between a and x. For sine and cosine every derivative stays between −1 and 1, so M = 1 always works.
The bound is on the right, under the error, and the chart now carries a dark mark for it above each bar. The bar never rises past its mark.
Try itPick cos x, keep the center at 0 and put x at 1. Find the lowest degree whose bound is under 0.0001. You have just guaranteed four decimal places without knowing cos 1.
So far a higher degree has always helped. Here is a function where it does not: ln(1 + x), now on the menu. About 0 its polynomials are
x − x²/2 + x³/3 − x⁴/4 + …
The coefficients shrink slowly, like 1/k. If |x| is less than 1 the powers of x shrink fast enough to make up for it. If x is greater than 1 the powers grow faster than the coefficients shrink, and before long each new term is bigger than the last.
The shaded region is more than 1 away from the center. The orange dot is in it, at x = 1.5.
Try itRaise the degree and watch the chart: the bars climb. Then drag the orange dot to 0.5 and look at the chart again.
The last function on the menu is 1/(1 − x). Its polynomials about 0 are the simplest of all:
1 + x + x² + x³ + … + xn
The function blows up at x = 1, and no polynomial can follow it there. That spot is 1 away from the center, and the polynomials stop improving 1 away from the center in both directions. At x = −1.5 the function is a perfectly ordinary 0.4, and the polynomials still fail.
The distance from the center to the trouble sets the reach. Move the center and the reach moves with it.
Try itDrag the orange dot from −0.5 to −1.5 and compare the charts. Then drag the center to a = −1. How far does the shaded region retreat, and does the polynomial now work at −1.5?
Take a pendulum hanging at an angle θ from straight down, measured in radians. Under the assumptions listed for this page, the pull back toward the middle is proportional to sin θ. That sine makes the motion hard to solve exactly.
The usual move is to replace sin θ with its degree 1 Taylor polynomial, which is just θ. How much does that cost? The degree 2 polynomial for sine is the same as degree 1, so Lagrange's bound with n = 2 applies:
|sin θ − θ| ≤ |θ|³/6
At θ = 0.2, about 11.5°, that is 0.0013, less than 1% of the angle. The picture is zoomed in, with sine on the graph and the orange dot at 0.2.
Rests on
Try itRaise the degree from 1 to 2. The polynomial does not change, but the bound gets much tighter. Then drag the orange dot out to 1 radian, about 57°. How good is the small-angle rule there?
Everything on this page came from one idea:
What was left out is the limit. Let the degree grow without end and the polynomial becomes a Taylor series, an infinite sum. Whether that sum settles on a value, and whether the value is the function, are separate questions with their own tests. Stops 8 and 9 were a first look: for those two functions the answer is yes inside the reach and no beyond it.
Everything is unlocked now. Play as long as you like, then scroll on to finish.
Try itPick ex and put the orange dot at x = 3. How many degrees does it take to get the error under 0.001? Is there any x where raising the degree stops helping?
| Claim | Status | Rests on | Detail |
|---|---|---|---|
| The derivative patterns behind the table, for all five functions and any center. | ✓ Proved | — | For each of the five functions, derivatives 0 to 13 match the closed forms the page uses at any center. |
| P₀ = 1, P₁ = 1 + x and P₂ = 1 + x + x²/2 for eˣ. The kth coefficient is the kth derivative at the center divided by k!, and the polynomial has the function’s value and first n derivatives there (stops 1 to 3). | ✓ Proved | D1 | P0 = 1, P1 = 1 + x and P2 = 1 + x + x^2/2 for e^x; the kth derivative of x^k is k!; a degree 6 polynomial built this way has the function's value and first 6 derivatives at the center. |
| sin x keeps only odd powers about 0, so each even degree repeats the odd one below it. cos x keeps only even powers (stop 4). | ✓ Proved | D1 | sin x has only odd powers about 0 (x - x^3/6 + x^5/120 - x^7/5040), so each even degree repeats the odd one below it; cos x has only even powers. |
| The error figures quoted: 0.06 for sine at x = 1.5, degree 3; degree 5 is the first under 0.001 at x = 1; degree 7 guarantees cos 1 to 0.0001 (stops 5 and 7). | ≈ Checked numerically | D2 | sin at x = 1.5, degree 3: error 0.0600. At x = 1 the error is 0.00814 at degree 3 and 0.000196 at degree 5, the first under 0.001. sin at x = 2, degree 5: error 0.0240, bound 2^6/6! = 0.0889. For cos at x = 1 the bound 1/(n + 1)! first drops under 0.0001 at n = 7. |
| Lagrange’s bound, with the M the page uses, is never smaller than the true error. | ≈ Checked numerically | D2 | At 579 settings across the five functions the error never exceeds Lagrange's bound (closest: 1.000 of the bound). |
| ln(1 + x) about 0: coefficients (−1)ᵏ⁻¹/k. More terms help at x = 0.5 and hurt at x = 1.5 (stop 8). | ≈ Checked numerically | D1, D2 | ln(1 + x) about 0 has coefficients (-1)^(k-1)/k; successive terms have ratio tending to |x|. At x = 1.5 the error grows with every degree from 4 (0.682) to 12 (6.3); at x = 0.5 it shrinks with every degree. |
| 1/(1 − x) about 0: Pₙ = 1 + x + … + xⁿ, with error exactly |x|ⁿ⁺¹/|1 − x|. It fails at x = −1.5 about 0 and works there about a = −1, where the reach is 2 (stop 9). | ✓ Proved | D1, D2 | 1/(1 - x) about 0: P_n = 1 + x + ... + x^n and the error is exactly |x|^(n+1)/|1 - x|, which shrinks only when |x| < 1. At x = -1.5 the function is 0.4 and the error grows with every degree; about a = -1 the coefficients are 1/2^(k+1) (reach 2) and the error at -1.5 shrinks with every degree. |
| |sin θ − θ| ≤ |θ|³/6: about 0.0013 at 0.2 radians, under 1% of the angle (stop 10). | ≈ Checked numerically | D2 | sin's degree 2 polynomial about 0 is x, so |sin t - t| <= |t|^3/6. At t = 0.2 (11.46 degrees) the miss is 0.001331, under the bound 0.001333 and 0.67% of the angle; at t = 1 (57.3 degrees) it is 0.1585, 15.9% of the angle. |
| The numbers the tool shows as you drag. | ✓ Checked independently | — | The page's own functions, run at 380 settings across the five functions, match SymPy's coefficients and values to 1.1e-15 (relative), and the whole-number derivatives it shows at a = 0 are exact. |
| Lagrange’s bound as a statement for every function with enough derivatives (stop 7), and that these five functions equal their Taylor series inside the reach (stop 11). | Reviewed | D1, D2 | Both are standard theorems (Taylor’s theorem with the Lagrange remainder). They are quoted here, not proved; the two rows above test them at the settings the page uses. |
| The sentences on the cards. | Reviewed | — | Each card was read against the formulas, looking for anything assumed but not listed. No objection stands. |
| A1, D1, D2 | Assumed | — | Granted, not proved. Whether they suit a real pendulum is yours to judge. |
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: the page’s own functions were run against SymPy’s exact answers. Checked numerically: evidence, not proof. Reviewed: a model or a person read it. Assumed: an assumption or a definition, granted and not proved. Checks run 2026-11-11 with SymPy 1.14.0; rerun them with interactives/checks/taylor_polynomials.py. Cards read by Claude Fable 5.2, 2026-11-11. How it’s checked
Part of Calc Practice: free Calculus 2 practice. Try the taylor and maclaurin polynomials problems or see more interactives.