Taylor Series Explorer

Choose a function, a centre a and a degree n. The dashed polynomial matches f and its first n derivatives at a; the orange segment at x shows how far the two have drifted apart, and Lagrange's bound says how far they could have.

Taylor's formula	$P_{n} (x) = k = 0 \sum n \frac{f ^{(k)} ( a )}{k !} (x - a)^{k}$
Lagrange's remainder	$R_{n} (x) = \frac{f ^{(n + 1)} ( ξ )}{( n + 1 )!} (x - a)^{n + 1}, ξ \in (a, x) ∣ R_{n} (x) ∣ \leq \frac{M ∣ x - a ∣ ^{n + 1}}{( n + 1 )!}$
Series to know	$e^{x} = 1 + x + \frac{x ^{2}}{2 !} + \dots, sin x = x - \frac{x ^{3}}{3 !} + \dots cos x = 1 - \frac{x ^{2}}{2 !} + \dots, ln (1 + x) = x - \frac{x ^{2}}{2} + \dots$

Worked example — Taylor Series Explorer

Derivatives at a	0, 1, 0, −1
The polynomial at x	0.833333
The function at x	0.841471
The error	0.0081377
Lagrange's bound	0.035061

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Taylor Series Explorer

The Taylor polynomial of degree n of f about a is the polynomial with the same value and the same first n derivatives as f at a: P_n(x) = Σ f^(k)(a)(x − a)^k/k!, summed from k = 0 to n. About a = 0 it is also called a Maclaurin polynomial.

What each symbol means

a: the centre, where Pₙ and f agree
n: the degree: how many derivatives are matched
Rₙ(x): the remainder, f(x) − Pₙ(x)

Where the formula comes from

Match the value, then the slope. P₀ = f(a) is the height at a. Adding f′(a)(x − a) gives P₁, the tangent line of the derivative page.
$P_{1} (x) = f (a) + f^{'} (a) (x - a)$
Why k! appears. Differentiate c_k(x − a)^k k times and c_k·k! is left, while every other term is 0 at a. To match f^(k)(a), the coefficient must be f^(k)(a)/k!.
$\frac{d ^{k}}{d x ^{k}} c_{k} (x - a)^{k} = c_{k} k! \Rightarrow c_{k} = \frac{f ^{(k)} ( a )}{k !}$
Work it for sin x about 0. The derivatives at 0 cycle 0, 1, 0, −1, so only odd powers survive, with alternating signs.
$sin x = x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - \dots$
Bound the error. Lagrange's form of the remainder puts the (n + 1)th derivative at an unknown ξ between a and x. Replace it by M, its largest size on that interval, and the bound needs nothing from f(x) itself.
$∣ R_{n} (x) ∣ = \frac{∣ f ^{(n + 1)} ( ξ ) ∣}{( n + 1 )!} ∣ x - a ∣^{n + 1} \leq \frac{M ∣ x - a ∣ ^{n + 1}}{( n + 1 )!}$
Build new series from old ones. Substitute into a known series, multiply two, or integrate one term by term, instead of differentiating again: e^−x² comes from e^u with u = −x².
$e^{- x^{2}} = 1 - x^{2} + \frac{x ^{4}}{2 !} - \frac{x ^{6}}{3 !} + \dots$

The chord's miss on the derivative page was a Taylor term 🖖

On the derivative page the chord's slope missed the tangent's by about ½f″(x)·h. Expand f(x + h) = f(x) + f′(x)h + ½f″(x)h² + ⋯, subtract f(x) and divide by h: the chord's slope is f′(x) + ½f″(x)h + ⋯, and the miss is the next term. The same expansion explains the Riemann sum page. Expand f about the middle of a strip and the f′ term is odd about that point, so it integrates to zero across the strip. The midpoint rule is left with an f″ error of order h³ per strip, while the left-endpoint rule keeps an f′ error of order h².

Smooth everywhere, and still not equal to its series 🖖

Lagrange built his 1797 account of calculus on the assumption that every function equals its Taylor series. Cauchy's counterexample of 1823 is f(x) = e^−1/x², with f(0) = 0. Every derivative of it exists, and every one is 0 at x = 0, so its Taylor polynomial about 0 is 0 at every degree, while the function is positive at every x ≠ 0. Adding terms never helps. A function with derivatives of every order is called smooth; one that equals its Taylor series near each point is called analytic, and the first does not imply the second.

TAYLOR SERIES — HOW FAR FROM THE CENTRE YOU CAN STILL TRUST IT

Which Taylor Case Are You In?

A Taylor polynomial matches a function's value and its first n derivatives at one point, and then drifts away as you move off. Two things decide how fast: how quickly the coefficients shrink, and whether the function has a singularity nearby. Adding terms helps inside the radius of convergence and does nothing at all outside it — and moving the centre is often worth more than adding terms.

Good near the centre, useless far from it ε ∝ (x − a)ⁿ⁺¹

Factorials win — the series converges everywhere xⁿ/n! → 0, R = ∞

A radius of convergence — beyond it, more terms do not help ln(1 + x): R = 1

Move the centre — cheaper than adding terms Δa ≠ 0, Δn = 0

Good near the centre, useless far from it

What you know: A low-degree expansion of a well-behaved function. Accurate close to the centre, and the error grows fast as you move away.

Series: ε ∝ (x − a)ⁿ⁺¹

Worked example: sin x ≈ x − x³/6 at x = 1 gives 0.8333 against the true 0.8415 — an error of 0.008. At x = 3 it gives −1.5 against 0.141

Open this case: sin x, degree 3

$Good near the centre, useless far from it. The cubic tracks the sine for about a radian, then leaves in the wrong direction entirely. A low-degree expansion of a well-behaved function. Accurate close to the centre, and the error grows fast as you move away.$

The cubic tracks the sine for about a radian, then leaves in the wrong direction entirely.

Factorials win — the series converges everywhere

What you know: The coefficients carry a factorial in the denominator, so they shrink faster than any power grows. The radius of convergence is infinite.

Series: xⁿ/n! → 0, R = ∞

Worked example: eˣ ≈ 1 + x + x²/2 + x³/6 + x⁴/24 + x⁵/120 at x = 1 gives 2.71667 against e = 2.71828: six terms, three correct digits

Open this case: eˣ, degree 5

Factorials win — the series converges everywhere. Degree 5 already traces the exponential across the whole visible range. The coefficients carry a factorial in the denominator, so they shrink faster than any power grows. The radius of convergence is infinite. — Degree 5 already traces the exponential across the whole visible range.

A radius of convergence — beyond it, more terms do not help

What you know: The function has a singularity a finite distance from the centre. Inside that distance the series converges; outside it, it diverges no matter how many terms you take.

Series: ln(1 + x): R = 1

Worked example: ln(1 + x) with six terms at x = 0.8 gives 0.5701 against 0.5878 — still converging, slowly. At x = 1.2 the terms grow and the sum runs away

Open this case: ln(1 + x) near its edge

A radius of convergence — beyond it, more terms do not help. Inside the radius the polynomial tracks the curve; past it, it leaves and never comes back. The function has a singularity a finite distance from the centre. Inside that distance the series converges; outside it, it diverges no matter how many terms you take. — Inside the radius the polynomial tracks the curve; past it, it leaves and never comes back.

Move the centre — cheaper than adding terms

What you know: You need accuracy at some point far from where the series is centred. Re-expanding around a nearer point costs nothing extra in degree.

Series: Δa ≠ 0, Δn = 0

Worked example: Degree 4 for cos x at x = 2: centred at 0 it gives −0.333 against −0.416, centred at 1 it gives −0.4086 — an error eleven times smaller

Open this case: cos x about 1

Move the centre — cheaper than adding terms. Same degree, different centre: the expansion around 1 hugs the curve at x = 2. You need accuracy at some point far from where the series is centred. Re-expanding around a nearer point costs nothing extra in degree. — Same degree, different centre: the expansion around 1 hugs the curve at x = 2.

Problems solved in full

Approximating sin 1 with the degree-3 Maclaurin polynomial and bounding the error 5 steps
Approximate sin 1 with the degree-3 Maclaurin polynomial, and bound the error before looking up the true value. An error bound you can only compute once you know the answer is worth nothing.

Open this problem in the tool
1. Maclaurin needs the derivatives at 0, and sine cycles through them with period four. Two of the first four are zero, which is why the polynomial ends up so short.
  
  $f (x) = sin x : f (0) = 0, f^{'} (0) = 1, f^{''} (0) = 0, f^{'''} (0) = - 1$
2. So P₃ has two terms, and the x³/6 is the whole correction. It moves 1 down to 0.83333 — a cut of nearly 17%, which is already a hint that the next term will not be negligible either.
  
  $P_{3} (x) = x - \frac{x ^{3}}{6} ⟹ P_{3} (1) = 1 - \frac{1}{6} = 0.83333$
3. This is the step that earns the section. The series for sine alternates with terms that shrink, so the truncation error is smaller than the first term left out. That term is x⁵/120, and computing it requires knowing nothing whatsoever about sin 1.
  
  $∣ R_{3} ∣ \leq \frac{∣ x ∣ ^{5}}{5 !} = \frac{1}{120} = 0.0083333$
4. Now check the promise. The true value is 0.84147, so the actual error is 0.0081377 — and it does sit under the 0.0083333 that was guaranteed in advance.
  
  $sin 1 = 0.84147 ⟹ ∣ sin 1 - P_{3} (1) ∣ = 0.0081377$
5. The bound is tight, and you can see why. The error is the entire omitted tail, and the tail alternates as well: 1/5! overshoots, subtracting 1/7! overshoots the other way, and 1/9! brings it home. Three terms already reproduce the printed error digit for digit.
  
  $\frac{1}{5 !} - \frac{1}{7 !} + \frac{1}{9 !} - \dots = 0.0081377, \frac{0.0083333}{0.0081377} = 1.0240$
Answer

P₃(1) = 0.833333 against a true 0.841471, an error of 0.0081377. The order the numbers arrived in is the point: 1/120 = 0.0083333 was available before sin 1 was, and it overshot the real error by only 2.4%. An approximation that comes with a certificate like that is worth more than one that merely happens to be close. The page's Lagrange row says 0.035061 at degree 3, because it treats P₃ as a degree-3 polynomial; set the degree to 4, which for sine is the same polynomial, and Lagrange gives 1/120 too. Raise the degree to 5 and the same argument hands you 1/5040 (the Lagrange row shows it at degree 6, the same polynomial), three decimal places guaranteed before a single digit of the answer is computed.

$P_{3} (1) = 0.83333, ∣ R_{3} ∣ \leq \frac{1}{120} first, then 0.0081377$
Estimating √e, with a bound found first 4 steps
Use the degree-4 Maclaurin polynomial of eˣ to estimate √e, and bound the error without using the value of e^0.5.

Open this problem in the tool
Show solution
1. Every derivative of eˣ is eˣ, which is 1 at 0, so every coefficient is 1/k!:
  
  $P_{4} (x) = 1 + x + \frac{x ^{2}}{2} + \frac{x ^{3}}{6} + \frac{x ^{4}}{24}$
2. At x = 0.5:
  
  $P_{4} (0.5) = 1 + 0.5 + 0.125 + \frac{0.125}{6} + \frac{0.0625}{24} = \frac{211}{128} = 1.6484375$
3. Lagrange's bound needs M, the largest value of the fifth derivative eᵗ for t between 0 and 0.5. Without knowing e^0.5, use e < 3, so e^0.5 < √3 < 2:
  
  $∣ R_{4} (0.5) ∣ \leq \frac{M ( 0.5 ) ^{5}}{5 !} < \frac{2 \cdot 0.03125}{120} = 0.00052083$
4. The page uses the exact M = e^0.5, which gives a tighter bound, and then the true error, which sits inside both:
  
  $\frac{1.648721 \cdot 0.03125}{120} = 0.00042935, e - P_{4} (0.5) = 0.00028377$
Answer

√e ≈ 1.648438, with an error below 0.00052 guaranteed from e < 3 alone. The true error is 0.00028377. A bound found from a cruder M is still a bound; it only promises less.
A Taylor polynomial about a point other than 0 4 steps
Find the degree-2 Taylor polynomial of √x about 4, use it to estimate √4.1, and bound the error.

Open this problem in the tool
Show solution
1. The derivatives at 4:
  
  $f (4) = 2, f^{'} (4) = \frac{1}{2} \cdot 4^{- 1/2} = \frac{1}{4}, f^{''} (4) = - \frac{1}{4} \cdot 4^{- 3/2} = - \frac{1}{32}$
2. Divide each by k!:
  
  $P_{2} (x) = 2 + \frac{x - 4}{4} - \frac{( x - 4 ) ^{2}}{64}$
3. At x = 4.1, x − 4 = 0.1:
  
  $P_{2} (4.1) = 2 + 0.025 - 0.00015625 = 2.02484375$
4. The third derivative, (3/8)x^−5/2, is largest at the left end of [4, 4.1], where it is 3/256:
  
  $∣ R_{2} (4.1) ∣ \leq \frac{\frac{3}{256} ( 0.1 ) ^{3}}{3 !} = 1.9531 \times 1 0^{- 6}$
Answer

√4.1 ≈ 2.024844, with an error below 1.9531 × 10⁻⁶. The true error is 1.9231 × 10⁻⁶, close to the bound because the third derivative barely changes between 4 and 4.1. The expansion is about 4 because √x is easy to evaluate there; about 0, √x has no Taylor polynomial beyond degree 0, because it has no derivative there.
A series by substitution, then integrated 4 steps
Find the Maclaurin series of e^−x² up to x⁶, and use it to estimate the integral of e^−x² from 0 to 0.5.

Open this problem in the tool
Show solution
1. Differentiating e^−x² six times is slow. Substitute u = −x² into the series for eᵘ instead:
  
  $e^{u} = 1 + u + \frac{u ^{2}}{2} + \frac{u ^{3}}{6} + \dots \Rightarrow e^{- x^{2}} = 1 - x^{2} + \frac{x ^{4}}{2} - \frac{x ^{6}}{6} + \dots$
2. Check it against the page at x = 0.5:
  
  $P_{6} (0.5) = 1 - 0.25 + 0.03125 - 0.0026042 = 0.778646$
3. Integrate term by term from 0 to 0.5:
  
  $\int_{0}^{0.5} (1 - x^{2} + \frac{x ^{4}}{2} - \frac{x ^{6}}{6}) d x = 0.5 - \frac{0.125}{3} + \frac{0.03125}{10} - \frac{0.0078125}{42} = 0.461272$
4. The terms alternate and shrink, so the error is below the first one left out, which comes from x⁸/4!:
  
  $\int_{0}^{0.5} \frac{x ^{8}}{4 !} d x = \frac{( 0.5 ) ^{9}}{216} = 0.0000090$
Answer

The integral is about 0.461272, within 0.0000090. The true value is 0.461281. This integral has no formula in elementary functions, which is why a series, or a Riemann sum, is how it gets computed.
A limit by series 2 steps
Find the limit of (sin x − x)/x³ as x → 0.
Show solution
1. Replace sin x by its series, and the x cancels:
  
  $\frac{sin x - x}{x ^{3}} = \frac{( x - \frac{x ^{3}}{6} + \frac{x ^{5}}{120} - \dots ) - x}{x ^{3}}$
2. Divide each remaining term by x³:
  
  $= - \frac{1}{6} + \frac{x ^{2}}{120} - \dots x \to 0 - \frac{1}{6}$
Answer

The limit is −1/6. L'Hôpital's rule gets there too, after three rounds of differentiation. The series gives the answer at once, and also how fast the quotient approaches it: the next term is x²/120.
A high derivative read off a series 2 steps
Find the fifth derivative of f(x) = x²eˣ at x = 0.

Open this problem in the tool
Show solution
1. Multiply the series for eˣ by x²:
  
  $x^{2} e^{x} = x^{2} + x^{3} + \frac{x ^{4}}{2} + \frac{x ^{5}}{6} + \dots$
2. The coefficient of x⁵ is f⁽⁵⁾(0)/5!, so:
  
  $\frac{f ^{(5)} ( 0 )}{5 !} = \frac{1}{6} \Rightarrow f^{(5)} (0) = \frac{120}{6} = 20$
Answer

f⁽⁵⁾(0) = 20. The product rule applied five times gets the same number with far more writing. The page's first row lists all six derivatives at 0: 0, 0, 2, 6, 12, 20.
How many terms for six decimal places 3 steps
How many terms of the Maclaurin series of eˣ are needed to compute e with an error below 0.0000005?

Open this problem in the tool
Show solution
1. At x = 1 Lagrange's bound for degree n is M/(n + 1)!, with M the largest eᵗ on [0, 1], which is e < 3:
  
  $∣ R_{n} (1) ∣ \leq \frac{e}{( n + 1 )!} < \frac{3}{( n + 1 )!}$
2. Find the first n where that drops below 0.0000005:
  
  $\frac{3}{10 !} = 8.3 \times 1 0^{- 7} > 5 \times 1 0^{- 7}, \frac{3}{11 !} = 7.5 \times 1 0^{- 8} < 5 \times 1 0^{- 7}$
3. So n = 10, which is eleven terms counting the constant. The page at degree 10 gives:
  
  $P_{10} (1) = 2.718282, \frac{e}{11 !} = 6.81 \times 1 0^{- 8}, ∣ e - P_{10} (1) ∣ = 2.73 \times 1 0^{- 8}$
Answer

Degree 10: eleven terms. By the bound, degree 9 is not enough (7.4909 × 10⁻⁷), although its actual error, 3.0289 × 10⁻⁷, would have done. The bound settles what can be promised in advance; that degree 9 is close enough nobody could have said before computing e.

Learning path

Why everything looks straight up close

References (1)

What turns “local” from a caution into a quantity — the remainder term: U. Abel, "On the Lagrange Remainder of the Taylor Formula." The American Mathematical Monthly 110(7), 627–633, 2003.