Find the Taylor polynomials of order , and for centered at .
Table of contents
- 0. Functions4h 53m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation2h 18m
- 4. Derivatives of Exponential & Logarithmic Functions1h 16m
- 5. Applications of Derivatives2h 19m
- 6. Graphical Applications of Derivatives6h 0m
- 7. Antiderivatives & Indefinite Integrals48m
- 8. Definite Integrals4h 36m
- 9. Graphical Applications of Integrals1h 43m
- 10. Integrals of Inverse, Exponential, & Logarithmic Functions21m
- 11. Techniques of Integration2h 7m
- 12. Trigonometric Functions6h 54m
- Angles29m
- Trigonometric Functions on Right Triangles1h 8m
- Solving Right Triangles23m
- Trigonometric Functions on the Unit Circle1h 19m
- Graphs of Sine & Cosine46m
- Graphs of Other Trigonometric Functions32m
- Trigonometric Identities52m
- Derivatives of Trig Functions42m
- Integrals of Basic Trig Functions28m
- Integrals of Other Trig Functions10m
- 13: Intro to Differential Equations2h 23m
- 14. Sequences & Series2h 8m
- 15. Power Series2h 19m
- 16. Probability & Calculus45m
15. Power Series
Power Series & Taylor Series
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Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
Find the power series representation centered at x=0 of the following function. Give the interval of convergence for the resulting series.
f(x)=1−x31
A
∑n=0∞x3n; converges on (-∞,∞)
B
∑n=0∞x3n; converges on (-1,1)
C
∑n=0∞x3n1; converges on (-1,1)
D
∑n=0∞x3n1; diverges everywhere

1
Step 1: Recognize that the function f(x) = 1 / (1 - x^3) resembles the geometric series formula, which is 1 / (1 - r) = ∑_{n=0}^{∞} r^n, where |r| < 1 for convergence.
Step 2: Rewrite the function f(x) = 1 / (1 - x^3) in terms of the geometric series formula. Here, r = x^3, so the series representation becomes ∑_{n=0}^{∞} (x^3)^n.
Step 3: Simplify the series representation. Since (x^3)^n = x^(3n), the power series representation of f(x) becomes ∑_{n=0}^{∞} x^(3n).
Step 4: Determine the interval of convergence. For a geometric series to converge, |r| < 1. In this case, |x^3| < 1, which simplifies to |x| < 1. Therefore, the interval of convergence is (-1, 1).
Step 5: Verify the endpoints of the interval. At x = -1 and x = 1, the series diverges because the geometric series condition |r| < 1 is violated. Thus, the series converges only on (-1, 1).
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