Difficulty: Intermediate | Prerequisites: Power series representations (see companion notes), basic derivative computations, factorial notation.
Taylor and Maclaurin series give you a systematic way to express any sufficiently smooth function as a power series, using its derivatives at a single point. Where the first set of notes built series by manipulating the geometric series, this approach works from the function itself: compute derivatives, evaluate them at the centre, and slot the values into a universal formula. This is the backbone of approximation theory, and it is how you derive series for exponentials, trigonometric functions, and binomial expressions. You should already be comfortable with power series representations before working through this material.
A Taylor series centred at a writes f(x) as Σ f⁽ⁿ⁾(a)/n! · (x - a)ⁿ. A Maclaurin series is the special case where a = 0. Taylor polynomials are truncated versions used for approximation, and the more terms you keep, the better the approximation near the centre.
Taylor series
The infinite series Σ f⁽ⁿ⁾(a)/n! · (x - a)ⁿ (n = 0 to ∞), representing f(x) near the point x = a. Think of it as building a function from its derivatives at a single point.
Maclaurin series
A Taylor series centred at a = 0, so the formula simplifies to Σ f⁽ⁿ⁾(0)/n! · xⁿ. In simple terms, it is the default Taylor series when no other centre is specified.
Taylor polynomial of degree n, Tₙ(x)
The partial sum of the Taylor series, keeping only terms up to (x - a)ⁿ. Think of it as the best polynomial approximation of degree n near x = a.
Binomial series
The Taylor series for (1 + x)ᵖ where p is any real number: Σ C(p, n) xⁿ, with C(p, n) = p(p-1)(p-2)...(p-n+1)/n!. In simple terms, this generalises the binomial theorem to non-integer exponents.
nth derivative at a point, f⁽ⁿ⁾(a)
The value of the nth derivative of f evaluated at x = a. This is the raw ingredient that determines each coefficient of the Taylor series.
The recipe: take derivatives of f, evaluate each at x = a, then form Tₙ(x) = f(a) + f'(a)(x - a) + f''(a)/2! · (x - a)² + ... + f⁽ⁿ⁾(a)/n! · (x - a)ⁿ.
For f(x) = 1/(4x + 1)^(1/2) centred at x = 0: this is a binomial-type function (1 + 4x)^(-1/2). Compute f(0) = 1, f'(0) = -2, f''(0) = 12, f'''(0) = -120. Then T₃(x) = 1 - 2x + 6x² - 20x³. (Note: each successive derivative picks up factors from the chain rule on the inner 4x.)
For f(x) = e^(4 - 3x): the nth derivative is f⁽ⁿ⁾(x) = (-3)ⁿ e^(4 - 3x). At x = 1: f⁽ⁿ⁾(1) = (-3)ⁿ · e. So f'''(1) = (-3)³ · e = -27e.
If T₄(x) = 3 - 6x + 2x² - 4x³ + 2x⁴ is the degree-4 Taylor polynomial for f at x = 0, then f(0.1) ≈ T₄(0.1). Just substitute: 3 - 0.6 + 0.02 - 0.004 + 0.0002 = 2.4162.
cos(u) = Σ (-1)ⁿ u²ⁿ/(2n)! for all u. For cos(4x), replace u with 4x: Σ (-1)ⁿ (4x)²ⁿ/(2n)! = Σ (-1)ⁿ 4²ⁿ/(2n)! · x²ⁿ.
eᵘ = Σ uⁿ/n! for all u. For 5ˣ, write 5ˣ = e^(x ln 5), so u = x ln 5: f(x) = Σ (ln 5)ⁿ/n! · xⁿ.
sin²(u) = (1 - cos(2u))/2. For sin²(3x): use cos(6x) = Σ (-1)ⁿ (6x)²ⁿ/(2n)!, then sin²(3x) = 1/2 - cos(6x)/2 = Σ (-1)ⁿ⁺¹ · 2²ⁿ⁻¹ · 3²ⁿ/(2n)! · x²ⁿ (n = 1 to ∞). The n = 0 constant terms cancel.
Some problems give you a series and ask for its sum. The trick is to recognise which function it represents.
Σ (-1)ⁿ 5²ⁿ/n! (n = 0 to ∞): this matches eᵘ = Σ uⁿ/n! with u = -25, since (-1)ⁿ · 5²ⁿ = (-25)ⁿ. The sum is e⁻²⁵.
When an integral has no closed-form antiderivative, expand the integrand as a series and integrate term by term.
For I = ∫₀³ 4e^(-x²) dx: write e^(-x²) = Σ (-1)ᵏ x²ᵏ/k!. Integrate term by term from 0 to 3: I = Σ (-1)ᵏ/(k!(2k + 1)) · 4 · 3²ᵏ⁺¹ (k = 0 to ∞).
For f(t) = ∫₀ᵗ s/(1 - s⁴) ds: write 1/(1 - s⁴) = Σ s⁴ⁿ, so the integrand is Σ s⁴ⁿ⁺¹. Integrate: f(t) = Σ t⁴ⁿ⁺²/(4n + 2) (n = 0 to ∞).
For f(x) = (5 + x)⁻³: factor out 5⁻³ to get (1/125)(1 + x/5)⁻³. Apply the generalised binomial series with p = -3 and u = x/5.
The general coefficient is C(-3, n) = (-3)(-4)(-5)...(-3-n+1)/n! = (-1)ⁿ (n+1)(n+2)/2.
So f(x) = Σ (-1)ⁿ (n+1)(n+2)/(2 · 5ⁿ⁺³) · xⁿ (n = 0 to ∞).
f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}\,(x-a)^ne^u = \sum_{n=0}^{\infty} \frac{u^n}{n!}\cos(u) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\,u^{2n}\sin^2(u) = \frac{1 - \cos(2u)}{2}(1+u)^p = \sum_{n=0}^{\infty} \binom{p}{n} u^n, \quad \binom{p}{n} = \frac{p(p-1)\cdots(p-n+1)}{n!}5^x = e^{x\ln 5} = \sum_{n=0}^{\infty} \frac{(\ln 5)^n}{n!}\,x^nTaylor polynomials underpin numerical methods across engineering and physics. When a computer evaluates sin(x) or eˣ, it uses a polynomial approximation derived from the Maclaurin series, truncated to enough terms for the required precision. The binomial series for fractional exponents appears in relativistic mechanics (the Lorentz factor expansion) and in electrical engineering for small-signal analysis of nonlinear circuits.
Students often confuse the Taylor series (infinite) with the Taylor polynomial (finite, truncated). The polynomial is an approximation; the series is the exact representation (within the radius of convergence).
When computing f'''(1) for f(x) = e^(4 - 3x), students frequently forget the chain rule factor. Each derivative brings down a factor of -3, so the nth derivative is (-3)ⁿ e^(4 - 3x), not just e^(4 - 3x).
For sin²(3x), students sometimes try to square the sine series term by term. This does not work. Use the identity sin²(u) = (1 - cos(2u))/2 instead.
When finding the sum of Σ (-1)ⁿ 5²ⁿ/n!, students often write the answer as e⁻⁵ instead of e⁻²⁵. Rewrite (-1)ⁿ 5²ⁿ as (-25)ⁿ first, then recognise the exponential series.
⚠️ You will almost certainly be asked to compute a low-degree Taylor polynomial (T₃ or T₄) by hand. Practise until the derivative-evaluate-divide-by-factorial pipeline is automatic.
⚠️ Recognising a given series as a known function (e.g. spotting that Σ (-1)ⁿ 5²ⁿ/n! = e⁻²⁵) is a very common exam question.
⚠️ Evaluating definite integrals via series expansion (like ∫₀³ 4e^(-x²) dx) appears regularly. Know the steps: expand, integrate term by term, apply bounds.
⚠️ The identity sin²(u) = (1 - cos(2u))/2 is essential for the Maclaurin series of sin². This comes up almost every time trig-squared series are tested.
Fill in the blank: The Maclaurin series for eᵘ is Σ ______. (uⁿ/n!, n = 0 to ∞.)
True or False: To find the Maclaurin series for sin²(3x), you square the series for sin(3x) term by term. (False. Use the identity sin²(u) = (1 - cos(2u))/2.)
Fill in the blank: f'''(1) for f(x) = e^(4 - 3x) equals ______. (-27e.)
True or False: A Taylor polynomial of degree 4 includes a term with x⁵. (False. The highest power is x⁴.)
Fill in the blank: 5ˣ can be rewritten as e^(______) before expanding as a Maclaurin series. (x ln 5.)
Q: Find T₃(x) for f(x) = 1/(4x + 1)^(1/2) centred at x = 0.
A: Compute f(0) = 1, f'(0) = -2, f''(0) = 12, f'''(0) = -120. Then T₃(x) = 1 - 2x + (12/2)x² + (-120/6)x³ = 1 - 2x + 6x² - 20x³.
Q: Find f'''(1) when f(x) = e^(4 - 3x).
A: f⁽ⁿ⁾(x) = (-3)ⁿ e^(4 - 3x). At x = 1: f'''(1) = (-3)³ · e^(4 - 3) = -27e.
Q: Find the Maclaurin series for cos(4x).
A: Replace u with 4x in the cosine series: cos(4x) = Σ (-1)ⁿ (4x)²ⁿ/(2n)! = Σ (-1)ⁿ 4²ⁿ/(2n)! · x²ⁿ (n = 0 to ∞).
Q: Find the sum of Σ (-1)ⁿ 5²ⁿ/n! (n = 0 to ∞).
A: Rewrite as Σ (-25)ⁿ/n! = e⁻²⁵.
Q: Use T₄(x) = 3 - 6x + 2x² - 4x³ + 2x⁴ to estimate f(0.1).
A: f(0.1) ≈ 3 - 0.6 + 0.02 - 0.004 + 0.0002 = 2.4162.
Q: Find the power series for f(x) = (5 + x)⁻³.
A: Write as (1/125)(1 + x/5)⁻³. Apply binomial series with p = -3: f(x) = Σ (-1)ⁿ (n+1)(n+2)/(2 · 5ⁿ⁺³) · xⁿ (n = 0 to ∞).
Q: Evaluate I = ∫₀³ 4e^(-x²) dx as a power series.
A: Expand e^(-x²) = Σ (-1)ᵏ x²ᵏ/k!. Multiply by 4 and integrate: I = Σ (-1)ᵏ · 4 · 3²ᵏ⁺¹ / (k!(2k + 1)) (k = 0 to ∞).
Taylor series connect back to the power series representation techniques in the companion notes: the geometric series approach is often faster when it works, but the Taylor formula handles functions (like eˣ, cos, and binomial expressions) where no simple geometric manipulation is available. This material feeds forward into error estimation (Taylor remainder theorem), which tells you how good your polynomial approximation actually is. In later courses, Taylor expansions appear in multivariable calculus (linearisation of surfaces) and in differential equations (series solutions near regular singular points).
Taylor polynomial, Maclaurin series, Taylor series centred at zero, degree n approximation, binomial series, generalised binomial coefficient, cos(4x) series, sin squared Maclaurin series, 5^x power series, e^(-x^2) integral, Taylor remainder, polynomial approximation, integral calculus UT Austin, HW 13