Difficulty: Intermediate | Prerequisites: Derivatives through the chain rule, factorial notation
Taylor polynomials let you approximate complicated functions with simple polynomials built from derivatives at a single point. You compute successive derivatives of the function, evaluate each at your chosen centre, divide by the matching factorial, and multiply by the appropriate power of (x minus centre). The result is a polynomial that hugs the original function near that centre, and the more terms you include, the tighter the fit.
Taylor polynomial of degree n
The polynomial P_n(x) = f(a) + f'(a)(x - a) + f''(a)(x - a)^2 / 2! + ... + f^(n)(a)(x - a)^n / n!, which matches the function f and its first n derivatives at the point x = a.
In simple terms, it is the best possible degree-n polynomial approximation of f near x = a.
Maclaurin polynomial
A Taylor polynomial centred at a = 0. Every Maclaurin polynomial is a Taylor polynomial; the name just tells you the centre is zero.
Think of it as the special case where you expand around the origin.
Centre of expansion (a)
The x-value at which you evaluate all the derivatives. The approximation is most accurate near this point and generally worsens as you move away from it.
Degree of the polynomial (n)
The highest power of (x - a) that appears with a non-zero coefficient. A higher degree means more derivative information is baked in and, usually, a better approximation over a wider interval.
Remainder / error
The difference between the true function value and the polynomial's output: R_n(x) = f(x) - P_n(x). On exams you may be asked for the absolute error |P_n(x) - f(x)|.
Choose the function f(x) and the centre a.
Compute f(a), f'(a), f''(a), ..., f^(n)(a).
Plug into the formula: each term is f^(k)(a) / k! multiplied by (x - a)^k.
Collect terms to write the polynomial.
This is Exam 4, Question 1. The goal is to find P_0, P_1, and P_2.
f(x) = (1 - x)^(-1/2), so f(0) = 1.
f'(x) = (1/2)(1 - x)^(-3/2), so f'(0) = 1/2.
f''(x) = (3/4)(1 - x)^(-5/2), so f''(0) = 3/4.
Assembling the polynomials:
P_0(x) = 1
P_1(x) = 1 + (1/2)x
P_2(x) = 1 + (1/2)x + (3/8)x^2
Note: the coefficient of x^2 is f''(0) / 2! = (3/4) / 2 = 3/8.
Exam 4, Question 2. Use f(x) = e^(-x) and its second-degree Maclaurin polynomial.
f(x) = e^(-x), f(0) = 1.
f'(x) = -e^(-x), f'(0) = -1.
f''(x) = e^(-x), f''(0) = 1.
So P_2(x) = 1 - x + x^2/2.
Substitute x = 0.13:
P_2(0.13) = 1 - 0.13 + (0.13)^2 / 2 = 1 - 0.13 + 0.0085 = 0.8785.
The true value is e^(-0.13) approximately equals 0.87810.
Absolute error: |P_2(0.13) - e^(-0.13)| is roughly 3.55 x 10^(-4), which is very small.
Exam 4, Question 8. Build P_3(x) at a = 0, then evaluate at x = 0.05.
n | f^(n)(x) | f^(n)(0) | Coefficient f^(n)(0)/n! |
|---|---|---|---|
0 | (1 + x)^(-2) | 1 | 1 |
1 | -2(1 + x)^(-3) | -2 | -2 |
2 | 6(1 + x)^(-4) | 6 | 3 |
3 | -24(1 + x)^(-5) | -24 | -4 |
So P_3(x) = 1 - 2x + 3x^2 - 4x^3.
Evaluating at x = 0.05:
P_3(0.05) = 1 - 0.1 + 0.0075 - 0.0005 = 0.907.
This approximates 1/(1.05)^2, which equals roughly 0.90703.
P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x - a)^kWhen a = 0 (Maclaurin case):
P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!} x^kUseful shortcut: if you already know the Maclaurin series for a standard function (e^x, sin x, cos x, 1/(1 - x)), you can substitute directly into it instead of re-deriving from scratch.
Students often forget to divide by k! when writing each term. The coefficient of (x - a)^k is f^(k)(a) / k!, not just f^(k)(a).
Confusing the degree of the polynomial with the number of non-zero terms. P_2 means the highest power is x^2, regardless of whether some coefficients happen to be zero.
Evaluating derivatives at x instead of at the centre a. Every derivative in the formula is evaluated at a single fixed point.
Mixing up Taylor and Maclaurin. A Maclaurin polynomial is simply a Taylor polynomial with a = 0. If the problem says "centred at 0," it is Maclaurin.
Building P_n from scratch (computing derivatives, evaluating at a, assembling the polynomial) is a near-guaranteed exam question.
Approximation questions pair naturally with Taylor polynomials: "Use P_3 to approximate f(0.05)" is a standard format.
You may be asked for the error |P_n(c) - f(c)| as well as the approximation itself.
Sketching the original function alongside P_0, P_1, P_2 to show how the fit improves may appear as a graphing question.
True or false: A Maclaurin polynomial is a Taylor polynomial centred at a = 1. (False, a = 0.)
Fill in the blank: The coefficient of (x - a)^3 in a Taylor polynomial is ___ / 3!. (f'''(a).)
True or false: P_2(x) for e^(-x) at a = 0 is 1 + x + x^2/2. (False, it is 1 - x + x^2/2.)
Fill in the blank: If f(0) = 1 and f'(0) = -2 and f''(0) = 6, then P_2(x) = ___. (1 - 2x + 3x^2.)
True or false: Adding more terms to a Taylor polynomial always improves the approximation at points far from the centre. (False, it improves near the centre; far away, the polynomial may diverge.)
Q: Let f(x) = (1 - x)^(-1/2). Find P_2(x) centred at a = 0.
A: Compute f(0) = 1, f'(0) = 1/2, f''(0) = 3/4. Then P_2(x) = 1 + (1/2)x + (3/8)x^2.
Q: Use the degree-2 Maclaurin polynomial for e^(-x) to approximate e^(-0.13). What is the approximate value?
A: P_2(x) = 1 - x + x^2/2. So P_2(0.13) = 1 - 0.13 + 0.0085 = 0.8785.
Q: For f(x) = (1 + x)^(-2), find P_3(x) and use it to approximate 1/(1.05)^2.
A: Derivatives give coefficients 1, -2, 3, -4. So P_3(x) = 1 - 2x + 3x^2 - 4x^3. Evaluating at x = 0.05 gives P_3(0.05) = 0.907.
Q: What is the general formula for the k-th term of a Taylor polynomial centred at a?
A: The k-th term is f^(k)(a) / k! times (x - a)^k.
Taylor polynomials are finite truncations of Taylor series (covered in the power series notes). The error term R_n connects to the Lagrange remainder and integral remainder, which bound how far off your approximation can be. The technique of substituting into known Maclaurin series (e.g. replacing x with -x in the series for e^x) reappears heavily in the series-applications notes.
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Taylor polynomial, Maclaurin polynomial, Taylor approximation, polynomial approximation, degree-n approximation, Taylor expansion, local approximation, linearisation, quadratic approximation, Taylor's theorem, remainder, error bound, M 2414, integral calculus, UT Austin