Guides / Series

Taylor series and linear approximation: divide by k!, expand in powers of x − a, and stay inside the radius

A Taylor polynomial replaces a function near a point a with the polynomial that has the same value and the same first few derivatives at a. The tangent line is the degree-1 case, and it is already a useful approximation; each extra term makes it better near a. The mistakes come from leaving out the k! under each derivative, from writing powers of x where powers of x − a belong, and from a remainder bound that takes the derivative's size at the wrong end of the interval.

Why it works

Write a polynomial in powers of x − a: p(x) = c₀ + c₁(x − a) + c₂(x − a)² + c₃(x − a)³ + ⋯. At x = a every term but the first is 0, so p(a) = c₀. Differentiate once and set x = a: p′(a) = c₁. Twice: p″(a) = 2c₂, since (x − a)² differentiates to 2 · 1. Three times: p‴(a) = 3 · 2 · 1 · c₃. In general p^(k)(a) = k! c_k, so matching f's derivatives at a needs c_k = f^(k)(a)/k!. That is where the factorial comes from, and why the powers have to be powers of x − a: they are what vanish at a.

The error after degree n behaves like the next term, with the derivative taken at some unknown point c between a and x instead of at a. You don't know c, so a guaranteed bound uses the worst case.

Example 1: a tangent-line estimate and the sign of its error

Estimate ∛8.3 with the linear approximation of f(x) = x^1/3 at a = 8.

1. f(8) = 2, f′(x) = 13x^−2/3, f′(8) = 13 · 4 = 1128 is the nearest point where the cube root is easy: 8^2/3 = 4.
2. ∛x ≈ 2 + 112(x − 8)The tangent line at 8, in powers of x − 8.
3. ∛8.3 ≈ 2 + 0.312 = 2.025The true value is 2.02469…, so the error is about 0.0003.
4. f″(x) = −29x^−5/3 < 0, so 2.025 is too highThe cube root is concave down for x > 0, so it bends below its tangent lines, and every tangent-line estimate overshoots.

Example 2: a Taylor polynomial about a ≠ 0

Find the degree-3 Taylor polynomial of f(x) = 1/x about a = 2, and use it to estimate 1/2.2.

1. f(2) = 12, f′(2) = −14, f″(2) = 28 = 14, f‴(2) = −616 = −38The derivatives are −x⁻², 2x⁻³ and −6x⁻⁴, evaluated at 2.
2. c₂ = 1/42! = 18, c₃ = −3/83! = −116Divide each derivative by k!.
3. T₃(x) = 12 − (x − 2)/4 + (x − 2)²8 − (x − 2)³16The pattern is (−1)^k(x − 2)^k/2^k+1: a geometric series with ratio −(x − 2)/2, which converges for |x − 2| < 2.
4. T₃(2.2) = 0.5 − 0.05 + 0.005 − 0.0005 = 0.4545With x − 2 = 0.2. The true value is 0.454545…, and each term shrinks by a factor of 10 here.

Example 3: a series from a series you already know

Find the Maclaurin series of arctan x, and the fifth derivative of arctan at 0.

1. 11 + t² = 1 − t² + t⁴ − t⁶ + ⋯, |t| < 1The geometric series 1/(1 − r) with r = −t².
2. arctan x = ∫₀^x dt1 + t² = x − x³3 + x⁵5 − x⁷7 + ⋯Integrate term by term, which is allowed inside the radius. arctan 0 = 0, so there is no constant.
3. f⁽⁵⁾(0)5! = 15 ⇒ f⁽⁵⁾(0) = 1205 = 24A power series is the Taylor series of its sum, so the coefficient of x⁵ is f⁽⁵⁾(0)/5!. Differentiating arctan five times by hand would take much longer.

Example 4: bounding the error

Estimate e^0.5 with the degree-3 Maclaurin polynomial of e^x, and bound the error.

1. T₃(0.5) = 1 + 0.5 + 0.252 + 0.1256 ≈ 1.645833Every derivative of e^x at 0 is 1, so the coefficients are 1/k!.
2. |error| = e^c4!(0.5)⁴, 0 < c < 0.5The Lagrange remainder with n = 3: the fourth derivative at an unknown c between 0 and 0.5.
3. e^c < e^0.5 < 2e^c is increasing, so its largest value is at the right end. Bound it by a number you know without the calculator: e < 4, so √e < 2.
4. |error| < (2 × 0.0625)/24 ≈ 0.0052The actual error is e^0.5 − 1.645833 ≈ 0.0029, inside the bound, as it must be.

Common mistakes, and the exact line they happen on

1. Leaving out the k!

The polynomial from Example 2, stopped at degree 2.

1. f(2) = 12, f′(2) = −14, f″(2) = 14Right so far.
2. T₂(x) = 12 − (x − 2)/4 + 14(x − 2)²This is the mistake. The coefficient of (x − 2)² is f″(2)/2! = 1/8. In the linear term the divisor 1! = 1 is invisible, which is why it's easy to forget from the second term on. This version gives 0.46 at x = 2.2, off by 0.0055, which is worse than the tangent line's 0.45 (off by 0.0045). The correct T₂(2.2) = 0.455 is off by 0.0005.

2. Powers of x instead of powers of x − a

The tangent line from Example 1.

1. f(8) = 2, f′(8) = 112Right so far.
2. ∛x ≈ 2 + x12, so ∛8.3 ≈ 2.692This is the mistake. The derivative was taken at 8, so the step has to be measured from 8: x − 8, not x. This line doesn't even pass through (8, 2): at x = 8 it gives 2.667. With x − 8 the estimate is 2.025.

3. Bounding the remainder at the wrong end

The bound from Example 4.

1. |error| = e^c24(0.5)⁴, 0 < c < 0.5Right so far.
2. c = 0 ⇒ |error| ≤ 0.062524 ≈ 0.0026This is the mistake. c is unknown, and a bound has to cover every c in the interval. e^c is smallest at c = 0, so this is the smallest the error could be, not the largest. It fails: the actual error, 0.0029, is bigger than this "bound." Take the largest value of the derivative, at c = 0.5 here.

Practice

1. Use the linear approximation of tan x at π/4 to estimate tan 0.8.

   Show answer

   tan(π/4) = 1 and sec²(π/4) = 2, so tan 0.8 ≈ 1 + 2(0.8 − π/4) ≈ 1.0292. The true value is 1.0296.

2. Find the Maclaurin series of x e^x through the x⁴ term.

   Show answer

   Multiply the series of e^x by x: x + x² + x³2 + x⁴6.

3. Find the Maclaurin series of 11 − 2x and its radius of convergence.

   Show answer

   The geometric series with ratio 2x: 1 + 2x + 4x² + 8x³ + ⋯ = Σ 2ⁿxⁿ, valid for |2x| < 1, so the radius is ½.

4. Use series to find lim_x→0 (sin x − x)/x³.

   Show answer

   sin x − x = −x³6 + x⁵120 − ⋯, so the quotient is −16 + x²120 − ⋯ and the limit is −16.

5. Bound the error of cos x ≈ 1 − x²/2 for |x| ≤ 0.2.

   Show answer

   The x³ term of cos x is 0, so 1 − x²/2 is also T₃, and the remainder uses the fourth derivative, cos c, with |cos c| ≤ 1: |error| ≤ (0.2)⁴24 ≈ 6.7 × 10⁻⁵.