Difficulty: Advanced | Prerequisites: Derivatives, basic series from MATH 125, partial derivatives
Taylor polynomials approximate functions near a base point, and Taylor's inequality tells you how good that approximation is. Taylor series extend this idea to infinite sums, and you can build new series by manipulating known ones (1/(1-x), e^x, sin, cos). Tangent planes to surfaces z = f(x,y) use partial derivatives to generalise the tangent-line idea from single-variable calculus.
Taylor polynomial of degree n, Tn(x)
The polynomial that best approximates f(x) near a base point b, built from the function's derivatives at b. Tn(x) = f(b) + f'(b)(x-b) + f''(b)(x-b)^2/2! + ... + f^(n)(b)(x-b)^n/n!. Think of it as a polynomial that matches f and its first n derivatives at the point b.
Taylor series
The infinite version of a Taylor polynomial: the sum from k = 0 to infinity of f^(k)(b)(x-b)^k / k!. It equals the function f(x) wherever the series converges.
Interval of convergence
The set of x values for which a Taylor series converges. Found using the ratio test. For a series centred at b, the interval is symmetric about b (though endpoints must be checked separately).
Ratio test
For a series of terms a_k, compute L = lim |a_(k+1)/a_k|. If L < 1, the series converges absolutely. If L > 1, it diverges. If L = 1, the test is inconclusive.
Taylor's inequality (error bound)
The error |f(x) - Tn(x)| is at most M * |x - b|^(n+1) / (n+1)!, where M is an upper bound for |f^(n+1)(c)| on the interval between x and b.
Tangent plane
The plane that best approximates the surface z = f(x,y) at a point (x0, y0, f(x0,y0)). Its equation is z - f(x0,y0) = fx(x0,y0)(x - x0) + fy(x0,y0)(y - y0). Think of it as the 2D version of a tangent line.
Normal vector to a surface z = f(x,y)
At a point (x0, y0), the vector
To build Tn(x) at base point b, compute f(b), f'(b), f''(b), ..., f^(n)(b) and plug into the formula.
Worked example (from the exam): f(x) = (x-1)^(5/2) - cos(x-2), base point b = 2. Find T2(x).
First, let u = x - 2 (so x - 1 = u + 1). At u = 0 (i.e. x = 2):
f(2) = (1)^(5/2) - cos(0) = 1 - 1 = 0.
f'(x) = (5/2)(x-1)^(3/2) + sin(x-2). f'(2) = 5/2 + 0 = 5/2.
f''(x) = (15/4)(x-1)^(1/2) + cos(x-2). f''(2) = 15/4 + 1 = 19/4.
T2(x) = 0 + (5/2)(x-2) + (19/4)(x-2)^2 / 2 = (5/2)(x-2) + (19/8)(x-2)^2.
Taylor's inequality gives an upper bound on the approximation error:
|f(x) - Tn(x)| <= M * |x - b|^(n+1) / (n+1)!
where M is the maximum of |f^(n+1)(c)| for c between x and b.
The key steps:
Compute f^(n+1)(x). For T2, you need the third derivative.
Find the maximum of |f^(n+1)(x)| on the interval in question.
Find the maximum of |x - b| on that interval.
Plug into the formula.
Worked example (from the exam): f(x) = (x-1)^(5/2) - cos(x-2), bound |f(x) - T2(x)| on [1.8, 2.2].
f'''(x) = (15/8)(x-1)^(-1/2) + sin(x-2).
On [1.8, 2.2]: (x-1) ranges from 0.8 to 1.2, so (x-1)^(-1/2) ranges from 1/sqrt(1.2) to 1/sqrt(0.8). The maximum of (x-1)^(-1/2) is 1/sqrt(0.8).
|sin(x-2)| <= sin(0.2) (since |x-2| <= 0.2).
M = (15/8)(1/sqrt(0.8)) + sin(0.2). For a sharp bound, use calculator values or leave in exact form.
|x - 2| <= 0.2 on this interval.
Error bound = M * (0.2)^3 / 3! = M * 0.008 / 6 = M * 0.008/6.
Instead of computing derivatives from scratch, build new Taylor series by substituting into known ones. The four standard series to memorise:
1/(1-x) = sum of x^k for k = 0 to infinity, valid for |x| < 1.
e^x = sum of x^k / k! for k = 0 to infinity, valid for all x.
sin(x) = sum of (-1)^k * x^(2k+1) / (2k+1)! for k = 0 to infinity.
cos(x) = sum of (-1)^k * x^(2k) / (2k)! for k = 0 to infinity.
Worked example (from the exam): Find the Taylor series for f(x) = x / (5 + x^2) at b = 0.
Rewrite: x / (5 + x^2) = x * 1/(5(1 + x^2/5)) = (x/5) * 1/(1 - (-x^2/5)).
Substitute u = -x^2/5 into the geometric series: 1/(1-u) = sum of u^k = sum of (-x^2/5)^k = sum of (-1)^k * x^(2k) / 5^k.
Multiply by x/5: f(x) = sum from k = 0 to infinity of (-1)^k * x^(2k+1) / 5^(k+1).
Finding g(x) = ln(5 + x^2) from f(x): Notice that f(x) = x/(5+x^2) and the derivative of ln(5 + x^2) is 2x/(5+x^2) = 2f(x). So ln(5+x^2) = integral of 2f(x) dx. Integrate the series term by term:
ln(5 + x^2) = ln(5) + sum from k = 0 to infinity of (-1)^k * 2x^(2k+2) / ((2k+2) * 5^(k+1)).
The constant ln(5) comes from evaluating at x = 0.
Use the ratio test on the general term of the series. For the series sum of (-1)^k * x^(2k+1) / 5^(k+1):
|a_(k+1) / a_k| = |x^(2k+3) / 5^(k+2)| * |5^(k+1) / x^(2k+1)| = |x|^2 / 5.
This is less than 1 when |x|^2 < 5, i.e. |x| < sqrt(5).
The open interval of convergence is (-sqrt(5), sqrt(5)).
Endpoints must be checked separately (substituting x = sqrt(5) and x = -sqrt(5) into the series and testing for convergence using the alternating series test or other criteria). For the exam, if only the open interval is asked for, you can stop at the ratio test.
For a surface z = f(x, y), the tangent plane at (x0, y0, f(x0, y0)) has equation:
z - f(x0, y0) = fx(x0, y0)(x - x0) + fy(x0, y0)(y - y0)
A normal vector to this plane is n =
Worked example (from the exam, Problem 9): f(x, y) = 100 - x^2 - y^2 + xy.
fx = -2x + y, fy = -2y + x.
Normal vector at (x0, y0): n = <-2x0 + y0, -2y0 + x0, -1>.
Finding the steepest tangent plane (part b): The angle between the tangent plane and the xy-plane depends on how tilted the tangent plane is. The tilt is determined by the magnitude of
The angle theta between the tangent plane and the xy-plane satisfies tan(theta) = sqrt(fx^2 + fy^2). To maximise this angle, maximise fx^2 + fy^2 = (-2x + y)^2 + (-2y + x)^2.
Expanding: 4x^2 - 4xy + y^2 + 4y^2 - 4xy + x^2 = 5x^2 - 8xy + 5y^2.
On the region -2 <= x <= 2, -2 <= y <= 2, this is maximised at a corner. Check (2, -2): 20 + 16 + 20 = 56. Check (2, 2): 20 - 32 + 20 = 8. Check (-2, 2): 20 + 16 + 20 = 56.
Maximum is 56, achieved at (2, -2) and (-2, 2). At (2, -2): f(2, -2) = 100 - 4 - 4 - 4 = 88. So P = (2, -2, 88) or (-2, 2, 88).
Taylor polynomial Tn(x) at base b: Tn(x) = sum from k = 0 to n of f^(k)(b) * (x - b)^k / k!
Taylor's inequality: |f(x) - Tn(x)| <= M |x - b|^(n+1) / (n+1)!, where M = max|f^(n+1)(c)| on the interval.
Four standard Taylor series (at b = 0): 1/(1-x) = sum x^k, |x| < 1. e^x = sum x^k/k!, all x. sin(x) = sum (-1)^k x^(2k+1)/(2k+1)!, all x. cos(x) = sum (-1)^k x^(2k)/(2k)!, all x.
Ratio test for convergence: L = lim |a_(k+1)/a_k|. Converges if L < 1.
Tangent plane at (x0, y0): z = f(x0,y0) + fx(x0,y0)(x-x0) + fy(x0,y0)(y-y0).
Normal vector to z = f(x,y): n =
Angle between tangent plane and xy-plane: tan(theta) = sqrt(fx^2 + fy^2). Steeper gradient means steeper tangent plane.
Students often confuse the base point b with the point where they evaluate the polynomial. The base point is where you compute derivatives; you then plug in other x values to approximate f there.
When building series from the geometric series 1/(1-x), students forget that x must be replaced by the entire expression (like -x^2/5), not just part of it.
A common error with Taylor's inequality is using the wrong derivative order. For Tn, you need the (n+1)th derivative, not the nth.
Students sometimes write the normal vector to z = f(x,y) as
Taylor polynomial problems (Problem 7) are worth 14 points on this exam. Compute derivatives carefully and keep track of factorial denominators.
Taylor series from known series (Problem 8) is another 14 points. The substitution trick with the geometric series is the core technique.
The interval-of-convergence question always uses the ratio test. Practise it until the algebra is automatic.
The connection between f(x) and its antiderivative/derivative via term-by-term integration or differentiation of the series (Problem 8c) is a frequent exam question.
Tangent-plane problems (Problem 9) tie together partial derivatives and geometry. The "steepest tangent plane" variant tests whether you understand that the gradient magnitude controls the tilt.
True or False: T2(x) for f(x) at b = 0 is f(0) + f'(0)x + f''(0)x^2.
Fill in the blank: The open interval of convergence of the geometric series sum x^k is ______.
True or False: The normal vector to z = x^2 + y^2 at (1, 1, 2) is <2, 2, -1>.
Fill in the blank: In Taylor's inequality for T2, you need an upper bound on the ______ derivative.
True or False: To find the Taylor series for ln(1 + x), you can integrate the series for 1/(1 + x) term by term.
Answers: 1. False (the last term should be f''(0)x^2 / 2). 2. (-1, 1). 3. True (fx = 2x = 2, fy = 2y = 2). 4. Third. 5. True.
Q: Find T1(x) for f(x) = e^(2x) at b = 0.
A: f(0) = 1, f'(x) = 2e^(2x), f'(0) = 2. T1(x) = 1 + 2x.
Q: Use the known series for 1/(1-x) to find the Taylor series for 1/(1+x^3) at b = 0.
A: Replace x with -x^3: 1/(1+x^3) = sum from k = 0 to infinity of (-1)^k * x^(3k). Converges for |x| < 1.
Q: Find a normal vector to the surface z = xy at the point (3, 2, 6).
A: fx = y = 2, fy = x = 3. Normal vector: <2, 3, -1>.
Q: Use Taylor's inequality to bound |sin(x) - x| for |x| <= 0.1.
A: T1(x) for sin(x) at b = 0 is just x. The second derivative of sin is -sin(x), and the third derivative is -cos(x). For Taylor's inequality with n = 1, M = max|f''(c)| = max|sin(c)| <= sin(0.1) on [-0.1, 0.1]. Bound: sin(0.1) * (0.1)^2 / 2 <= 0.005. (A tighter bound uses the third-order remainder, giving (0.1)^3/6 approx 0.000167.)
Taylor polynomials are the single-variable version of the multivariable linear and quadratic approximations you use for tangent planes and the second derivative test. The tangent plane z = f(a,b) + fx(a,b)(x-a) + fy(a,b)(y-b) is the 2D analogue of T1(x) = f(b) + f'(b)(x-b).
The geometric series 1/(1-x) = sum x^k is the foundation for partial fractions, generating functions, and many applied-maths techniques beyond this course.
Normal vectors to surfaces connect to gradient vectors, which you will use extensively in vector calculus courses (MATH 324) for line integrals, surface integrals, and the divergence and curl theorems.
Taylor polynomial, Taylor series, Maclaurin series, base point, centre, Taylor's inequality, Taylor remainder, error bound, ratio test, interval of convergence, radius of convergence, geometric series, power series, term-by-term integration, term-by-term differentiation, tangent plane, normal vector, gradient, surface, partial derivatives, steepest tangent plane, angle between planes, MATH 126, Calculus III, University of Washington, multivariable calculus, series convergence