Taylor & Maclaurin Series in Depth: Convergence, Radius, and Error Bounds in Applications

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| Attempts: 11 | Questions: 15 | Updated: Feb 6, 2026

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1) The Taylor polynomial of degree 3 for f(x) = tan(x) centered at a = π/4 is

1 + 2(x-π/4) + 2(x-π/4)² + (8/3)(x-π/4)³

1 + 2(x-π/4) + 2(x-π/4)² + (4/3)(x-π/4)³

1 + 2(x-π/4) + (x-π/4)² + (8/3)(x-π/4)³

1 + 2(x-π/4) + 2(x-π/4)² + (⅔)(x-π/4)³

We start with f(x) = tan(x), so f(π/4) = 1. Then we find f'(x) = sec²(x), so f'(π/4) = 2. Next we find f''(x) = 2 sec²(x) tan(x), so f''(π/4) = 221 = 4. Then we find f'''(x) = 2 sec⁴(x) + 4 sec²(x) tan²(x), so f'''(π/4) = 24 + 42*1 = 8 + 8 = 16. The Taylor polynomial is 1 + 2(x-π/4) + 4(x-π/4)² /2 + 16(x-π/4)³ /6. Simplifying 4/2 = 2, 16/6 = 8/3 gives 1 + 2(x-π/4) + 2(x-π/4)² + (8/3)(x-π/4)³.

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About This Quiz

Taylor & Maclaurin Series In Depth: Convergence, Radius, And Error Bounds In Applications - Quiz

Taylor polynomials give us a powerful way to approximate functions by expanding them around a chosen center and capturing the behavior of their derivatives. In this quiz, you’ll work with Taylor expansions of trigonometric, exponential, logarithmic, and rational functions—not only identifying coefficients and constructing polynomials, but also applying them to... see moreevaluate limits, estimate integrals, and approximate function values near specific points. You will practice using both the Lagrange error bound and the alternating-series error bound to measure how accurate an approximation is, and explore how the radius of convergence determines where a Taylor series behaves well. These problems highlight the interplay between derivatives, series structure, and real-world estimation—essential skills for mastering Taylor series in calculus.
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2) Using the 3rd-degree Taylor polynomial for tan(x) centered at π/4, approximate tan(π/4 + 0.1).

1.2227

1.2238

1.2200

1.2240

The 3rd-degree Taylor polynomial for tan(x) centered at π/4 is 1 + 2(x-π/4) + 2(x-π/4)² + (8/3)(x-π/4)³. At x = π/4 + 0.1, (x-π/4) = 0.1, so 1 + 2(0.1) + 2(0.01) + (8/3)(0.001). Compute 20.1 = 0.2, 20.01 = 0.02, (8/3)0.001 ≈ 2.6660.001 = 0.002667. So 1 + 0.2 + 0.02 + 0.002667 = 1.222667, which rounds to 1.2227.

Explanation

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3) The Lagrange error bound for approximating e^{1.2} using the 4th-degree Taylor polynomial centered at 1 is at most

E^{1.2} (0.2)⁵ /120

E^{1.2} (0.2)⁴ /24

E^{1.2} (0.2)⁵ /5

E^{1.2} (0.2)³ /6

The Lagrange remainder for degree 4 is |R4(x)| ≤ max |f^(5)(ξ)| |x-1|⁵ /5! for ξ between 1 and 1.2. For f(x) = eˣ, f^(5)(ξ) = e^ξ, the max is at ξ=1.2, e^{1.2}. So the bound is e^{1.2} (0.2)⁵ /120.

Explanation

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4) As the degree of the Taylor polynomial for ln(x) centered at 1 increases, the approximation improves over what interval?

0 < x < 2

X > 0

|x-1| < 1

The entire real line

The Taylor series for ln(x) centered at 1 has radius of convergence 1, so converges for |x-1| <1, but since it's the principal log, and the series converges at the endpoint x=2 (alternating harmonic), but diverges at x=0, so the approximation improves as degree increases for 0 < x < 2.

Explanation

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5) Which expression approximates the integral from 1 to 1.1 of sin(x-1)/(x-1) dx using the first two non-zero terms of the Taylor series centered at x=1?

Integral from 1 to 1.1 of (1 - (x-1)²/6) dx

Integral from 1 to 1.1 of ((x-1) - (x-1)³/6) dx

Integral from 1 to 1.1 of (1 - (x-1)/2) dx

Integral from 1 to 1.1 of (1 + (x-1)) dx

We know sin(u) = u - u³/3! + ... Let u = x-1. Then sin(x-1) = (x-1) - (x-1)³/6 + ... Dividing by (x-1) gives 1 - (x-1)²/6. These are the first two non-zero terms to be integrated.

Explanation

We know sin(u) = u - u³/3! + ... Let u = x-1. Then sin(x-1) = (x-1) - (x-1)³/6 + ... Dividing by (x-1) gives 1 - (x-1)²/6. These are the first two non-zero terms to be integrated.

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6) The error in approximating e^(x-1) at x=1.1 using a Taylor polynomial centered at 1 is bounded by 3 * (0.1)^(n+1) / (n+1)!. What is the smallest degree n required to ensure the error is less than 0.0001?

N=1

N=2

N=3

N=10

We can test the values given in the choices. For n=2, we get 3*(0.001)/6 = 0.0005, which is too big. For n=3, we have 3*(0.0001)/24 = 0.0003/24 = 0.0000125. Since 0.0000125 < 0.0001, n=3 is sufficient.

Explanation

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7) The Lagrange error bound for approximating tan(π/4 + 0.1) using the 3rd-degree Taylor polynomial centered at π/4 is at most

|f^(4)(ξ)|(0.1)^4 /24

|f^(4)(ξ)|(0.1)^3 /6

|f^(4)(π/4)|(0.1)^4 /24

|f^(4)(π/4+0.1)|(0.1)^4 /24

The Lagrange remainder for degree 3 is R3(x) = f^(4)(ξ) (x-a)⁴ /4! for some ξ between a and x. For f(x) = tan(x), a = π/4, x = π/4 + 0.1, (x-a) = 0.1, 4! =24, so |f^(4)(ξ)| (0.1)⁴ /24 for some ξ between π/4 and π/4 + 0.1.

Explanation

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8) Using the alternating series error bound for the Taylor series of ln(x) centered at 1, the error in approximating ln(1.3) with the first 3 terms is less than

(0.3)^4/4

(0.3)^3/3

(0.3)^5/5

(0.3)^4/8

The series is (x-1) - (x-1)²/2 + (x-1)³/3 - (x-1)⁴/4 + ..., alternating. For x=1.3, h=0.3, the first 3 terms are up to k=3, the next term is - (0.3)⁴ /4, so error < (0.3)⁴ /4.

Explanation

The series is (x-1) - (x-1)²/2 + (x-1)³/3 - (x-1)⁴/4 + ..., alternating. For x=1.3, h=0.3, the first 3 terms are up to k=3, the next term is - (0.3)⁴ /4, so error < (0.3)⁴ /4.

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9) As the degree of the Taylor polynomial for 1/(1-x) centered at a=2 is increased, the approximation improves over what interval?

|x-2|<1

Entire real line

X>1

|x-2|<2

The Taylor series for 1/(1-x) centered at 2 is the geometric series in terms of (x-2), but first, 1/(1-x) = -1/(x-1) = -1 / ((x-2) +1) = -1 * 1/(1 + (x-2)) = - Σ (-1)^k (x-2)^k for |x-2| <1. The radius of convergence is 1, so the approximation improves as degree increases for |x-2| <1.

Explanation

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10) Calculate the remainder term using Lagrange form for approximating sin(π/3 + 0.2) using the 3rd-degree Taylor polynomial centered at π/3.

F^(4)(ξ)(0.2)^4/24

F^(4)(ξ)(0.2)^3/6

F^(4)(π/3)(0.2)^4/24

F^(4)(π/3+0.2)(0.2)^4/24

The Lagrange remainder for degree 3 is R3(x) = f^(4)(ξ) (x-a)⁴ /4! for some ξ between a and x. For f(x) = sin(x), a = π/3, x = π/3 + 0.2, (x-a) = 0.2, 4! =24, so f^(4)(ξ) (0.2)⁴ /24 for some ξ between π/3 and π/3 + 0.2.

Explanation

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11) The alternating series error bound can be used for the Taylor series of ln(x) centered at 1 when the terms alternate and decrease, which occurs for

1 < x < 2

0 < x < 1

X > 2

All x > 0

For x >1, the terms (-1)^{k+1} (x-1)^k /k are alternating because (x-1)^k is positive, and the signs are +, -, +, -, ..., and for 1<x<2, the magnitudes decrease and approach 0, so the alternating series error bound applies for 1 < x < 2.

Explanation

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12) A function g has derivatives of all orders and |g^{(6)}(x)| ≤ 15 on the interval from x = 1 to x = 4. Let P₅(x) be the fifth-degree Taylor polynomial for g centered at x = 1. The Lagrange error bound guarantees that |g(4) − P₅(4)| ≤ k. What is the smallest such k among the choices?

(3^6/6!)·15

(3^5/5!)·15

(3^6/6!)·8

(4^6/6!)·15

For a fifth-degree polynomial (n = 5), the remainder involves the sixth derivative. The distance from the center x = 1 to x = 4 is |4−1| = 3. The Lagrange bound is |R₅(4)| ≤ (max |g^{(6)}(z)|) · |4−1|⁶ / 6! ≤ 15 · 3⁶ / 720. This is exactly choice A.

Explanation

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13) The function f has all derivatives and |f'''(x)| ≤ 8 on the interval from x = 1 to x = 0.5. Let P₂(x) be the second-degree Taylor polynomial centered at x = 1. Using the Lagrange error bound, what is the maximum possible error when approximating f(0.5) with P₂(0.5)?

(0.5^3/3!)·8

(0.5^2/2!)·8

(0.5^3/3!)·6

(1^3/3!)·8

Second-degree polynomial means n = 2, so the remainder uses the third derivative. Distance from center x = 1 to x = 0.5 is |0.5−1| = 0.5. The bound is |R₂(0.5)| ≤ (max |f’’’(z)|) · (0.5)³ / 3! ≤ 8 · (0.125)/6 = 8 · (0.5³/3!), which is choice A.

Explanation

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14) When using the Lagrange error bound for a Taylor polynomial of degree n centered at a, the exponent on the distance |x−a| in the error formula is always

N+1

N−1

The Lagrange remainder is |Rₙ(x)| = |f^{(n+1)}(c)| · |x−a|ⁿ⁺¹ / (n+1)! for some c between a and x. The power of the distance is always one more than the degree of the polynomial.

Explanation

The Lagrange remainder is |Rₙ(x)| = |f^{(n+1)}(c)| · |x−a|ⁿ⁺¹ / (n+1)! for some c between a and x. The power of the distance is always one more than the degree of the polynomial.

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15) To get the smallest Lagrange error bound when approximating f(b) with Pₙ(b) centered at a, you should use

Largest bound on |f^{(n+1)}| & exact |b−a|^{n+1}/(n+1)!

Smaller derivative bound

Lower degree polynomial

Use alternating series bound

The Lagrange bound is an upper bound, so using the actual maximum of |f^{(n+1)}| on the interval and the exact distance gives the smallest k that is guaranteed to work.

Explanation

The Lagrange bound is an upper bound, so using the actual maximum of |f^{(n+1)}| on the interval and the exact distance gives the smallest k that is guaranteed to work.

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