Source: MATH 308, Texas A&M University, Sections 11.10–11.11
Tags: Taylor series, Maclaurin series, Taylor polynomial, Taylor's inequality, remainder, Lagrange remainder, error bound, ASET, approximation, physics applications, small-angle approximation, relativity
The Taylor series expresses a "nice" function as an infinite power series built from its derivatives at a single point. Truncating after finitely many terms gives a Taylor polynomial, which approximates the function near that point. Taylor's inequality bounds the error. These tools let you sum tricky series in closed form, compute high-order derivatives without differentiating, evaluate integrals that have no elementary antiderivative, compute limits that would be painful with L'Hopital, and underpin the small-angle and low-speed approximations used throughout physics and engineering.
Taylor series (centred at a)
f(x) = ∑(n=0 to ∞) f⁽ⁿ⁾(a)/n! · (x − a)ⁿ = f(a) + f′(a)(x − a) + f″(a)(x − a)²/2! + f‴(a)(x − a)³/3! + …
Valid for |x − a| < R, where R is the radius of convergence.
Maclaurin series
The Taylor series centred at a = 0: f(x) = ∑ f⁽ⁿ⁾(0)/n! · xⁿ.
Taylor polynomial of degree n, Tₙ(x)
The partial sum of the Taylor series up to the (x − a)ⁿ term. It is a polynomial of degree at most n with n + 1 terms (beware: the n-th Taylor polynomial has n + 1 terms, not n).
Remainder / error, Rₙ(x)
f(x) = Tₙ(x) + Rₙ(x). The remainder Rₙ(x) is the error when you approximate f by its degree-n Taylor polynomial.
Taylor's inequality (Lagrange error bound)
If |f⁽ⁿ⁺¹⁾(t)| ≤ M for all t between a and x, then |Rₙ(x)| ≤ M/(n + 1)! · |x − a|^(n+1). This works for any Taylor series, alternating or not.
Lagrange form of the remainder
For some t between a and x: Rₙ(x) = f⁽ⁿ⁺¹⁾(t)/(n + 1)! · (x − a)^(n+1). Taylor's inequality follows immediately by bounding f⁽ⁿ⁺¹⁾(t) by M.
The recipe: compute f(a), f′(a), f″(a), f‴(a), … and slot them into the formula. The coefficient of (x − a)ⁿ is f⁽ⁿ⁾(a)/n!.
eˣ (Maclaurin, a = 0): Every derivative of eˣ is eˣ, so f⁽ⁿ⁾(0) = 1 for all n.
eˣ = ∑ xⁿ/n! = 1 + x + x²/2 + x³/6 + …, R = ∞
eˣ (Taylor, a = 3): f⁽ⁿ⁾(3) = e³ for all n, so eˣ = ∑ e³/n! · (x − 3)ⁿ.
cos x (Maclaurin): Derivatives cycle: cos, −sin, −cos, sin, cos, … At x = 0: 1, 0, −1, 0, 1, 0, −1, 0, … Only even-indexed terms survive.
cos x = ∑(n=0 to ∞) (−1)ⁿ x^(2n)/(2n)! = 1 − x²/2! + x⁴/4! − x⁶/6! + …, R = ∞
sin x (Maclaurin): Derivatives at 0: 0, 1, 0, −1, 0, 1, … Only odd-indexed terms survive.
sin x = ∑(n=0 to ∞) (−1)ⁿ x^(2n+1)/(2n+1)! = x − x³/3! + x⁵/5! − x⁷/7! + …, R = ∞
Connection to Euler's formula: cos x = (eⁱˣ + e⁻ⁱˣ)/2, which you can verify by substituting ix into the exponential series. The relation eᵖⁱ = −1 follows.
Rather than computing all derivatives from scratch, substitute into a known series.
x sin(x²): start from sin u = u − u³/3! + u⁵/5! − …, replace u with x², then multiply by x. Result: x³ − x⁷/3! + x¹¹/5! − …
cos(x⁵): replace x with x⁵ in the cosine series.
x²e^(−2x³): replace x with −2x³ in the eˣ series, then multiply by x².
This is faster and less error-prone than differentiating repeatedly.
Tₙ(x) is the partial sum: the first n + 1 terms of the Taylor series. It is a polynomial that approximates f near x = a.
Warning: The n-th Taylor polynomial Tₙ has degree n but n + 1 terms (the constant term counts).
Example: f(x) = xeˣ centred at a = 2.
f(2) = 2e², f′(x) = (x+1)eˣ so f′(2) = 3e², f″(x) = (x+2)eˣ so f″(2) = 4e², f‴(x) = (x+3)eˣ so f‴(2) = 5e².
T₁(x) = 2e² + 3e²(x − 2)
T₂(x) = 2e² + 3e²(x − 2) + 2e²(x − 2)²
T₃(x) = 2e² + 3e²(x − 2) + 2e²(x − 2)² + (5e²/6)(x − 2)³
Example: T₆ for x²e^(−2x³) at a = 0. From eˣ = 1 + x + x²/2 + …, substituting −2x³ gives e^(−2x³) = 1 − 2x³ + 2x⁶ − … Multiplying by x² yields x² − 2x⁵ + 2x⁸ − … So T₆(x) = x² − 2x⁵ (the x⁸ term is beyond degree 6).
Example: sin x has T₅(x) = x − x³/6 + x⁵/120. Because the x⁶ coefficient is zero, T₅ = T₆ for sin x.
Method 1: ASET (alternating series estimation theorem)
If the Taylor series is alternating and satisfies the AST conditions, the error is bounded by the first omitted term: |Rₙ| ≤ |first dropped term|.
Method 2: Taylor's inequality (Lagrange error bound)
Works for all Taylor series. Find M = max|f⁽ⁿ⁺¹⁾(t)| for t between a and x. Then:
|Rₙ(x)| ≤ M/(n + 1)! · |x − a|^(n+1)
This is essentially the magnitude of the "next term," with the derivative evaluated at its worst-case value.
Example: Approximate e^(0.1) using T₃(x) = 1 + x + x²/2 + x³/6. For 0 < x < 0.1, the fourth derivative of eˣ is eˣ, which is largest at x = 0.1. So M = e^(0.1) < 2 (rough bound). Error < 2 · (0.1)⁴/4! = 2 · 0.0001/24 < 0.00001. So e^(0.1) ≈ 1.1052 is accurate to about five decimal places.
Example: For sin x ≈ T₅(x) = x − x³/6 + x⁵/120, for which x is the error below 1/1000? This series is alternating, so ASET gives |R₅| ≤ |x|⁷/7! = |x|⁷/5040 < 0.001. Solve: |x|⁷ < 5.04, so |x| < 5.04^(1/7) ≈ 1.26.
Recognise a numerical series as a known Taylor series evaluated at a specific x.
∑ 1/n! = e (set x = 1 in eˣ = ∑xⁿ/n!)
∑ (−1)ⁿ/n! = e⁻¹ = 1/e (set x = −1)
∑ (−1)ⁿ 2ⁿ x^(3n)/n! = e^(−2x³) (recognise as eˣ with x replaced by −2x³)
∑ (−1)ⁿ π^(2n) / (3^(2n)(2n)!) = cos(π/3) = 1/2
∑ 1/((n+1)·3^(n+1)) = ln 3 − ln 2 (from the ln(1 − x) series with an appropriate x)
∑ n/2^(n−1) = 4 (from differentiating the geometric series)
If you know a power series expansion, you can read off any derivative at the centre without differentiating.
The coefficient of (x − a)ⁿ in a Taylor series is f⁽ⁿ⁾(a)/n!, so f⁽ⁿ⁾(a) = n! · [coefficient of (x − a)ⁿ].
Example: f has expansion ∑ 3ⁿ(n+4)/(n+1)! · (x − 7)ⁿ. Then f⁽¹⁰⁾(7) = 10! · 3¹⁰ · 14/11! = 3¹⁰ · 14/11.
Example: Find (cos(x⁵))⁽¹⁷⁾ at 0. The Maclaurin series of cos(x⁵) replaces x with x⁵ in ∑(−1)ⁿ x^(2n)/(2n)!, giving terms x⁰, x¹⁰, x²⁰, … There is no x¹⁷ term, so the 17th derivative at 0 is zero.
Example: Find (x²eˣ)⁽⁷⁾ at 0. Multiply the eˣ series by x² to get ∑ x^(n+2)/n!. The x⁷ term has n + 2 = 7, so n = 5, coefficient = 1/5! = 1/120. Then f⁽⁷⁾(0) = 7!/5! = 42.
Some antiderivatives have no elementary formula but can be expressed as a convergent power series.
∫ e^(−x²) dx: Replace x with −x² in eˣ = ∑xⁿ/n! to get e^(−x²) = ∑(−1)ⁿ x^(2n)/n!. Integrate term by term: ∫ e^(−x²) dx = C + ∑(−1)ⁿ x^(2n+1)/((2n+1)·n!). This is essential in probability (the Gaussian integral).
∫ sin(x²)/x dx: Write sin(x²) = ∑(−1)ⁿ x^(4n+2)/(2n+1)!, divide by x, integrate.
Taylor expansions can replace repeated applications of L'Hopital's rule.
Example: lim(x→0) [(sin x³)² − x⁶] / x¹². Using sin u = u − u³/6 + …, sin(x³) = x³ − x⁹/6 + …, so (sin x³)² = x⁶ − x¹²/3 + … Then (sin x³)² − x⁶ = −x¹²/3 + higher order terms. Divide by x¹² and take the limit: result is −1/3. Far cleaner than differentiating 12 times.
By the generalised mean value theorem (Lagrange remainder): Rₙ(x) = f⁽ⁿ⁺¹⁾(t)/(n+1)! · (x − a)^(n+1) for some t between a and x. Bounding |f⁽ⁿ⁺¹⁾(t)| by M gives the inequality.
The function f(x) = e^(−1/x²) for x ≠ 0 and f(0) = 0 has f⁽ⁿ⁾(0) = 0 for every n (because e^(−1/x²) decays faster than any power of x). Its Maclaurin series is identically zero, yet f itself is not zero. The Taylor series converges (to 0), but not to f. This is why the remainder estimates matter: they confirm the series converges to the right thing.
For small x, these are the standard linearisations and quadratic approximations:
1/(1 + x) ≈ 1 − x + x²
1/(1 − x) ≈ 1 + x + x²
eˣ ≈ 1 + x + x²/2
sin x ≈ x
cos x ≈ 1 − x²/2
tan x ≈ x
ln(1 − x) ≈ −x − x²/2
ln(1 + x) ≈ x − x²/2
1/(1 + x²) ≈ 1 − x²
√(1 + x) ≈ 1 + x/2 − x²/8
1/√(1 + x) ≈ 1 − x/2 + 3x²/8
The equation of motion for a pendulum is ℓθ″(t) = −g sin θ(t), which is nonlinear and hard to solve. The small-angle approximation sin θ ≈ θ turns it into ℓθ″ ≈ −gθ, a linear ODE with solution θ(t) = A sin(α + ηt). For larger angles, using sin θ ≈ θ − θ³/6 gives a better (but still approximate) nonlinear equation.
Einstein's relativistic mass is m = m₀(1 − v²/c²)^(−1/2). Using the approximation (1 − x²)^(−1/2) ≈ 1 + x²/2 for small x = v/c gives m ≈ m₀(1 + v²/(2c²)). The kinetic energy K = mc² − m₀c² then simplifies to approximately (1/2)m₀v², which is Newton's formula. Newtonian mechanics is the low-speed Taylor approximation of relativistic mechanics.
Convert to radians: 69° = π/3 + π/20. Expand cos x in a Taylor series around a = π/3 (where cos and sin are known exactly). The small quantity is h = π/20 ≈ 0.157. Four terms of the expansion give cos 69° ≈ 0.35837, with error bounded by Taylor's inequality.
Taylor series: f(x) = ∑ f⁽ⁿ⁾(a)/n! · (x − a)ⁿ
Key Maclaurin series:
eˣ = ∑ xⁿ/n!, R = ∞
sin x = ∑ (−1)ⁿ x^(2n+1)/(2n+1)!, R = ∞
cos x = ∑ (−1)ⁿ x^(2n)/(2n)!, R = ∞
1/(1 − x) = ∑ xⁿ, R = 1
ln(1 − x) = −∑ xⁿ/n (n ≥ 1), R = 1
tan⁻¹(x) = ∑ (−1)ⁿ x^(2n+1)/(2n+1), R = 1
Taylor's inequality: |Rₙ(x)| ≤ M/(n+1)! · |x − a|^(n+1), where M = max|f⁽ⁿ⁺¹⁾(t)| on the interval.
Reading derivatives from series: f⁽ⁿ⁾(a) = n! · (coefficient of (x − a)ⁿ in the Taylor series).
⚠️ The n-th degree Taylor polynomial Tₙ has n + 1 terms (constant through degree n). Don't confuse "degree n" with "n terms."
⚠️ When building a Maclaurin series by substitution (e.g. sin(x²) or e^(−x³)), remember to adjust the powers. The degree of Tₙ refers to the final variable, not the intermediate substitution.
⚠️ For error estimation: use ASET if the series is alternating, Taylor's inequality otherwise. Know which to apply.
⚠️ To bound |Rₙ| with Taylor's inequality, you need to find M, the maximum of |f⁽ⁿ⁺¹⁾| on the relevant interval. For eˣ on [0, 0.1], that is e^(0.1). A rough upper bound (like 2 or 3) is fine as long as it really is an upper bound.
⚠️ "Find f⁽¹⁷⁾(0)" questions are solved by reading the coefficient of x¹⁷ from the Maclaurin series and multiplying by 17!. Do not differentiate 17 times.
⚠️ A Taylor series can converge without converging to f. The function e^(−1/x²) is the classic counterexample. Taylor's inequality (showing Rₙ → 0) is what proves the series converges to f.
⚠️ Exam problems on approximating values like cos 69° typically involve choosing a centre where trig values are known exactly (like π/3) and expanding in the small deviation.
Q: Write the Maclaurin series for eˣ and state its radius of convergence.
A: eˣ = ∑(n=0 to ∞) xⁿ/n! = 1 + x + x²/2! + x³/3! + … with R = ∞.
Q: What is the Maclaurin series for sin x?
A: sin x = ∑(n=0 to ∞) (−1)ⁿ x^(2n+1)/(2n+1)! = x − x³/3! + x⁵/5! − … with R = ∞.
Q: Find ∑(n=0 to ∞) (−1)ⁿ π^(2n) / (3^(2n)(2n)!).
A: Recognise this as cos(π/3) = 1/2, since cos u = ∑(−1)ⁿ u^(2n)/(2n)! with u = π/3.
Q: Find f⁽⁷⁾(0) for f(x) = x²eˣ.
A: x²eˣ = ∑ x^(n+2)/n!. The x⁷ term has n = 5, coefficient 1/5!. So f⁽⁷⁾(0) = 7! · (1/5!) = 7·6 = 42.
Q: Use Taylor's inequality to bound the error of e^(0.1) ≈ T₃(0.1).
A: The 4th derivative of eˣ is eˣ ≤ e^(0.1) < 2 on [0, 0.1]. Error ≤ 2 · (0.1)⁴/4! = 0.0002/24 < 0.00001.
Q: Express ∫ e^(−x²) dx as a power series.
A: ∫ e^(−x²) dx = C + ∑(n=0 to ∞) (−1)ⁿ x^(2n+1) / ((2n+1)·n!).
Q: Why can't the small-angle approximation sin θ ≈ θ be used for large θ?
A: It drops the cubic and higher terms (−θ³/6 + …). For small θ those are negligible, but for larger θ they contribute significantly, making the linear approximation inaccurate.
Q: Compute lim(x→0) [(sin x³)² − x⁶] / x¹² using Taylor series.
A: sin(x³) = x³ − x⁹/6 + …, so (sin x³)² = x⁶ − 2·x³·x⁹/6 + … = x⁶ − x¹²/3 + … Then the numerator is −x¹²/3 + higher order terms, and dividing by x¹² gives the limit −1/3.
Taylor series, Maclaurin series, Taylor polynomial, Taylor's inequality, Lagrange remainder, Lagrange error bound, ASET, alternating series estimation, approximation, error bound, eˣ series, sin x series, cos x series, small-angle approximation, relativistic kinetic energy, computing limits with series, integrals as power series, extracting derivatives, MATH 308, calculus II, Texas A&M