What Is a Taylor Series?
A Taylor series represents a function as an infinite polynomial — a sum of powers of (x − a) weighted by the function's derivatives at a point a.
The key idea is remarkable: if you know the value of a function and all its derivatives at a single point, you can reconstruct the entire function (at least within a certain radius). This turns complicated functions like sin(x), eˣ, or ln(1+x) into polynomials, which are vastly easier to compute, differentiate, integrate, and analyze.
The Taylor series formula for a function f(x) centered at a is: f(x) = f(a) + f'(a)(x − a) + f''(a)(x − a)²/2! + f'''(a)(x − a)³/3! + ... = Σ f⁽ⁿ⁾(a)(x − a)ⁿ/n! for n = 0 to ∞.
A Maclaurin series is just a Taylor series centered at a = 0: f(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + ....
Most common Maclaurin series are worth memorizing because they appear constantly in calculus, differential equations, physics, and numerical analysis.
The key caveat is convergence: the series only equals the original function within a specific interval around the center point.
The Taylor Series Formula and Taylor Polynomials
The full Taylor series is an infinite sum, but in practice we often use a finite truncation called a Taylor polynomial.
The nth-degree Taylor polynomial of f centered at a is T_n(x) = Σ (k=0 to n) f⁽ᵏ⁾(a)(x − a)ᵏ/k!. This is the best polynomial approximation of degree n near the point a: T_n matches f and its first n derivatives at a.
Example: Find the 3rd-degree Taylor polynomial of f(x) = √x centered at a = 4.
Compute derivatives: f(x) = x^(1/2), f(4) = 2.
f'(x) = (1/2)x^(−1/2), f'(4) = 1/4.
f''(x) = −(1/4)x^(−3/2), f''(4) = −1/32.
f'''(x) = (3/8)x^(−5/2), f'''(4) = 3/256.
Assemble: T_3(x) = 2 + (1/4)(x − 4) − (1/32)(x − 4)²/2 + (3/256)(x − 4)³/6 = 2 + (x − 4)/4 − (x − 4)²/64 + (x − 4)³/512.
Use this to approximate √4.2: plug in x = 4.2 so x − 4 = 0.2. T_3(4.2) ≈ 2 + 0.05 − 0.000625 + 0.0000156 ≈ 2.04939.
√4.2 ≈ 2.04939Actual: √4.2 ≈ 2.04939. Matched to 5 decimal places with a cubic!
Maclaurin Series of Common Functions
Several Maclaurin series appear so often that they're worth committing to memory:
| Function | Maclaurin Series | Interval of Convergence | Notes |
|---|---|---|---|
| Exponential | eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + ... = Σ xⁿ/n! | All real x | Converges everywhere. |
| Sine | sin(x) = x − x³/3! + x⁵/5! − x⁷/7! + ... = Σ (−1)ⁿ x^(2n+1)/(2n+1)! | All real x | Only odd powers (because sin is odd). |
| Cosine | cos(x) = 1 − x²/2! + x⁴/4! − x⁶/6! + ... = Σ (−1)ⁿ x^(2n)/(2n)! | All real x | Only even powers (because cos is even). |
| Geometric / 1/(1 − x) | 1/(1 − x) = 1 + x + x² + x³ + ... = Σ xⁿ | |x| < 1 | Diverges at endpoints. |
| Natural log (shifted) | ln(1 + x) = x − x²/2 + x³/3 − x⁴/4 + ... = Σ (−1)^(n+1) xⁿ/n | −1 < x ≤ 1 | Converges at right endpoint. |
| Arctangent | arctan(x) = x − x³/3 + x⁵/5 − x⁷/7 + ... = Σ (−1)ⁿ x^(2n+1)/(2n+1) | |x| ≤ 1 | See Leibniz formula below. |
| Binomial | (1 + x)^α = 1 + αx + α(α − 1)x²/2! + α(α − 1)(α − 2)x³/3! + ... | |x| < 1 | When α is a positive integer, this terminates as the ordinary binomial theorem. |
Setting x = 1 gives the Leibniz formula for π: π/4 = 1 − 1/3 + 1/5 − 1/7 + ....
Deriving a Maclaurin Series
Deriving a Maclaurin series directly from the definition is methodical: compute derivatives, evaluate at 0, plug into the formula.
Example — derive the Maclaurin series for eˣ:
All derivatives of eˣ are eˣ. At x = 0, eˣ = 1.
So every coefficient f⁽ⁿ⁾(0)/n! = 1/n!.
eˣ = 1 + x + x²/2! + x³/3! + ...Example — derive the Maclaurin series for sin(x):
f(x) = sin(x), f(0) = 0.
f'(x) = cos(x), f'(0) = 1.
f''(x) = −sin(x), f''(0) = 0.
f'''(x) = −cos(x), f'''(0) = −1.
f⁽⁴⁾(x) = sin(x), f⁽⁴⁾(0) = 0.
Pattern repeats every 4 terms: 0, 1, 0, −1. Series: sin(x) = 0 + 1·x + 0 − x³/3! + 0 + x⁵/5! − ...
sin(x) = x − x³/6 + x⁵/120 − ...Shortcut tricks
Shortcut tricks: You can often derive a series without computing all derivatives.
- Substitution
To find the series of
e^(x²), substituteu = x²into theeˣseries:e^(x²) = 1 + x² + x⁴/2! + x⁶/3! + .... - Differentiation/integration
To find
cos(x)series, differentiate thesin(x)series term by term: derivative ofx − x³/6 + x⁵/120 − ...is1 − x²/2 + x⁴/24 − .... - Multiplication
The series for
x · sin(x)=x · (x − x³/6 + ...) = x² − x⁴/6 + x⁶/120 − ....
These algebraic manipulations are usually faster than direct differentiation.
Radius and Interval of Convergence
A Taylor series may not equal the original function everywhere — only within a specific interval of convergence. The radius of convergence R is half the width of this interval: the series converges for |x − a| < R and diverges for |x − a| > R. Behavior at |x − a| = R (the endpoints) must be checked separately.
To find R, the Ratio Test is standard. For a series Σ aₙ, compute lim |aₙ₊₁/aₙ|. If the limit is L, series converges when L < 1.
Example: Find the radius of convergence of the Maclaurin series for ln(1 + x) = Σ (−1)^(n+1) xⁿ/n.
Ratio: |a_(n+1)/a_n| = |x^(n+1)/(n+1)| · |n/xⁿ| = |x| · n/(n+1) → |x| as n → ∞.
Series converges when |x| < 1, so R = 1.
Check endpoints: x = 1 gives the alternating harmonic series (converges). x = −1 gives the negative harmonic series (diverges).
Interval of convergence: (−1, 1]Functions with R = ∞ vs finite R
| Case | Functions | Radius | Why |
|---|---|---|---|
| R = ∞ | eˣ, sin(x), cos(x), sinh(x), cosh(x) | Infinite | Their Maclaurin series converge for every real x. |
| Finite R | 1/(1 − x) | R = 1 | The function has a vertical asymptote at x = 1. |
| Finite R | ln(1 + x) | R = 1 | A singularity at x = −1. |
A Taylor series cannot cross a singularity, so R equals the distance from the center point to the nearest singularity in the complex plane.
Applications and Remainder Estimation
Taylor series have enormous practical applications in calculus, numerical analysis, and physics.
Approximating functions
Approximating functions: Calculators don't compute sin(x) by some magic — they use a truncated Maclaurin series.
Computing sin(0.1): T_3(0.1) = 0.1 − 0.001/6 ≈ 0.09983.
sin(0.1) ≈ 0.09983341Evaluating limits
Evaluating limits: L'Hôpital's rule is great, but Taylor series often give cleaner answers. Example: lim(x→0) (sin(x) − x)/x³.
Substitute: sin(x) − x = (x − x³/6 + x⁵/120 − ...) − x = −x³/6 + x⁵/120 − ....
Divide: (sin(x) − x)/x³ = −1/6 + x²/120 − ....
As x → 0, limit = −1/6.
−1/6Other applications
Solving differential equations: Power series methods solve ODEs when closed-form solutions don't exist. Physics: Small-angle approximations (sin(θ) ≈ θ, cos(θ) ≈ 1 − θ²/2) are just low-order Taylor truncations. Special relativity's γ = 1/√(1 − v²/c²) ≈ 1 + v²/(2c²) for small v/c is a binomial series.
Remainder / error estimation
Remainder / error estimation: The error from truncating a Taylor series after the nth term is bounded by Taylor's Remainder Theorem: |R_n(x)| ≤ M · |x − a|^(n+1)/(n+1)!, where M is an upper bound on |f^(n+1)| in the interval.
Example: For sin(x) approximated by T_5(x) on |x| ≤ 1, |f⁽⁶⁾| ≤ 1, so |R_5(x)| ≤ 1/720 ≈ 0.00139.
|R_5(x)| ≤ 0.00139This gives a rigorous bound on approximation accuracy.
If you're working through Taylor series problems — derivations, convergence tests, or error estimates — scan them with Solver AI for full step-by-step solutions.