Difficulty: Advanced | Prerequisites: Sequences and series, convergence tests, derivatives of exponential and polynomial functions, basic multivariable function notation.
Power series and Taylor polynomials let you represent complicated functions as infinite (or finite) sums of simple polynomial terms. This is how calculators compute sin, cos, e^x and ln. The interval of convergence tells you where the representation is valid. The multivariable calculus portion introduces partial derivatives and double integrals, extending everything you know about single-variable calculus into two dimensions. These are the first steps toward vector calculus and multivariable analysis.
A power series is a polynomial with infinitely many terms. Its interval of convergence, found via the ratio test and endpoint checks, tells you where it equals the function. Taylor polynomials approximate a function near a point using its derivatives. Partial derivatives and double integrals generalise differentiation and integration to functions of two variables.
Power Series
A series of the form sum of c_k (x - a)^k, where a is the centre and c_k are the coefficients. Think of it as an infinite polynomial centred at a.
Radius of Convergence
The value R such that the power series converges for |x - a| < R and diverges for |x - a| > R. Found using the ratio or root test on the coefficients.
Interval of Convergence
The set of all x for which the power series converges, including or excluding the endpoints. You must test endpoints separately, often with the alternating series test or p-series test.
Taylor Polynomial of Degree n
The polynomial T_n(x) = sum from k = 0 to n of f^(k)(a)/k! times (x - a)^k. It is the best polynomial approximation of f(x) near x = a, using the function's value and its first n derivatives at a.
Partial Derivative
The derivative of a multivariable function with respect to one variable, treating all others as constants. f_x means differentiate with respect to x, holding y fixed.
Double Integral
The integral of a function of two variables over a region in the plane. Computed as an iterated integral: integrate with respect to one variable first, then the other.
Reversing the Order of Integration
Rewriting a double integral so you integrate in the opposite order (dy dx instead of dx dy, or vice versa). Requires re-describing the region of integration. In simple terms, you are tracing the region with horizontal slices instead of vertical ones, or the reverse.
Start from the geometric series: 1/(1 - t) = sum of t^n for |t| < 1.
Substitute to match the target function.
Example: f(y) = 1/(4 - y^3). Rewrite as (1/4)(1/(1 - y^3/4)). Replace t with y^3/4: f(y) = sum from n = 0 to infinity of y^(3n)/4^(n+1).
Use the ratio test on the coefficients c_k to find the radius R.
Test endpoints separately (each may converge or diverge).
Example: sum of (2x - 1)^k/(k 2^k) = sum of c_k (x - 1/2)^k where c_k = 1/k.
Ratio test on c_k: lim |c_(k+1)/c_k| = lim k/(k+1) = 1, so R = 1. The series converges for |x - 1/2| < 1.
At x - 1/2 = -1 (x = -1/2): sum of (-1)^k/k converges by the alternating series test.
At x - 1/2 = 1 (x = 3/2): sum of 1/k diverges (harmonic series).
Interval of convergence: [-1/2, 3/2).
T_n(x) = f(a) + f'(a)(x-a) + f''(a)/2!(x-a)^2 + ... + f^(n)(a)/n!(x-a)^n.
The coefficient of (x-a)^k is f^(k)(a)/k!. This means you can read off derivatives from a given Taylor polynomial.
Example (reading derivatives): T_4(x) = 2 - 8(x-1) + 3(x-1)^2 - 8(x-1)^3 + 7(x-1)^4, centred at x = 1.
The coefficient of (x-1)^3 is f^(3)(1)/3! = -8, so f^(3)(1) = -8 times 3! = -48.
Example (building a Taylor polynomial): f(x) = xe^(-2x), centred at origin.
f(0) = 0, f'(0) = 1, f''(0)/2! = -2, f'''(0)/3! = 2.
T_3(x) = x - 2x^2 + 2x^3.
Differentiate with respect to one variable, treating the other as a constant.
Example: f(x, y) = 4x^2 + 2xy - 4y^2 + 2x - 3y.
f_x = 8x + 2y + 2 (treat y as constant).
f_y = 2x - 8y - 3 (treat x as constant).
f_x - f_y = 6x + 10y + 5.
For a rectangle A = {(x, y): 0 <= x <= 1, 0 <= y <= 2}, choose which variable to integrate first based on which is easier.
Example: double integral of xe^(3xy) over A.
Integrate with respect to y first (easier: the y-integral of xe^(3xy) is (1/3)e^(3xy)).
Inner integral from 0 to 2: (1/3)(e^(6x) - 1).
Outer integral from 0 to 1: (1/3)(e^(6x)/6 - x) evaluated from 0 to 1 = (1/18)(e^6 - 7).
Sketch the region. Re-describe it with the opposite variable as the outer integral.
Example: integral from 0 to 4 of (integral from y/2 to 2 of f(x,y) dx) dy.
Region: y/2 <= x <= 2, 0 <= y <= 4. Equivalently: 0 <= y <= 2x, 0 <= x <= 2.
Reversed: integral from 0 to 2 of (integral from 0 to 2x of f(x,y) dy) dx.
\frac{1}{1-t} = \sum_{n=0}^{\infty} t^n, \quad |t| < 1T_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k\iint_A f(x,y)\,dA = \int_a^b \int_c^d f(x,y)\,dy\,dxTaylor polynomials are how computers approximate transcendental functions: your calculator computes sin(x) using a Taylor expansion. Power series representations arise in solving differential equations that have no closed-form solution. Partial derivatives are fundamental to optimisation in economics, machine learning and physics. Double integrals compute mass, centre of mass and probability over two-dimensional regions.
Students forget to test the endpoints of a power series separately. The ratio test tells you the radius, not what happens at the boundary. Each endpoint must be checked with a different test (alternating series, p-series, comparison).
When reading off f^(k)(a) from a Taylor polynomial, students often forget to multiply by k!. The coefficient of (x - a)^k is f^(k)(a)/k!, not f^(k)(a) itself.
With partial derivatives, students sometimes differentiate with respect to both variables at once. f_x means only x varies; y is treated as a constant throughout.
When reversing the order of integration, students sometimes swap limits mechanically without sketching the region. Always draw the region first to identify the correct new limits.
Power series questions often ask you to manipulate 1/(1 - t) with a substitution. Practise recognising which substitution transforms your function into that form.
Taylor polynomial questions come in two flavours: building one from derivatives and reading derivatives from a given polynomial. Both are common exam formats.
Partial derivative questions are typically straightforward computation but require care with signs and terms.
Reversing the order of integration requires a correct sketch of the region. The exam tests whether you can re-describe a region bounded by lines or curves.
Fill in the blank: 1/(1 - t) = sum of t^n, valid for |t| ____ 1. (less than)
True or False: The coefficient of (x - a)^3 in a Taylor polynomial equals f'''(a). _____ (False: it equals f'''(a)/3! = f'''(a)/6.)
Fill in the blank: To compute f_x for f(x, y), treat ____ as a constant. (y)
True or False: When reversing the order of integration, the integrand f(x, y) changes. _____ (False: only the limits and the order of dx/dy change.)
True or False: The interval of convergence of a power series always includes both endpoints. _____ (False: each endpoint must be tested separately. It may include neither, one, or both.)
Q: Find a power series representation for f(y) = 1/(4 - y^3).
A: Rewrite as (1/4)(1/(1 - y^3/4)). Substitute t = y^3/4 into the geometric series: f(y) = sum from n = 0 to infinity of y^(3n)/4^(n+1).
Q: Find the interval of convergence of sum of (2x - 1)^k / (k 2^k).
A: Rewrite as sum of (1/k)(x - 1/2)^k. Ratio test gives R = 1. At x = -1/2: alternating harmonic series, converges. At x = 3/2: harmonic series, diverges. Interval: [-1/2, 3/2).
Q: Given T_4(x) = 2 - 8(x-1) + 3(x-1)^2 - 8(x-1)^3 + 7(x-1)^4, find f^(3)(1).
A: The coefficient of (x-1)^3 is f^(3)(1)/3! = -8. So f^(3)(1) = -8 times 6 = -48.
Q: Find T_3(x) for f(x) = xe^(-2x) centred at x = 0.
A: Compute f(0) = 0, f'(0) = 1, f''(0) = -4 so f''(0)/2! = -2, f'''(0) = 12 so f'''(0)/3! = 2. Thus T_3(x) = x - 2x^2 + 2x^3.
Q: Find f_x - f_y for f(x, y) = 4x^2 + 2xy - 4y^2 + 2x - 3y.
A: f_x = 8x + 2y + 2, f_y = 2x - 8y - 3. Difference: 6x + 10y + 5.
Q: Evaluate the double integral of xe^(3xy) over the rectangle 0 <= x <= 1, 0 <= y <= 2.
A: Integrate y first: inner integral gives (1/3)(e^(6x) - 1). Outer integral: (1/3)(e^(6x)/6 - x) from 0 to 1 = (1/18)(e^6 - 7).
Q: Reverse the order of integration for integral from 0 to 4 of (integral from y/2 to 2 of f(x,y) dx) dy.
A: The region is bounded by x = y/2, x = 2, y = 0, y = 4. Equivalently: y from 0 to 2x, x from 0 to 2. Reversed: integral from 0 to 2 of (integral from 0 to 2x of f(x,y) dy) dx.
Power series connect to convergence tests (you use the ratio test to find the radius of convergence) and to integration techniques (term-by-term integration of power series). Taylor polynomials are the foundation for numerical methods and error analysis. Partial derivatives lead into gradient, divergence and curl in vector calculus. Double integrals extend to triple integrals, surface integrals and the theorems of Green, Stokes and Gauss.
Power series, Taylor polynomial, Taylor series, Maclaurin series, radius of convergence, interval of convergence, geometric series representation, partial derivative, multivariable calculus, double integral, iterated integral, reversing order of integration, Fubini's theorem, MATH 101, integral calculus, UT Austin