To prove: The Maclaurin series for the given function converges to that the function for all real
Explanation of Solution
Given information:
The function is
Calculation:
First let examine the derivative of
Hence,
Based on Example 4, the function is rising on the range
Since,
For each
And
Similarly, the function is rising on the range
Since,
For each
And
Given that
The convergence can be proved by using the Remainder Bounding Theorem with
Chapter 9 Solutions
AP CALCULUS TEST PREP-WORKBOOK
- Calculus: Early TranscendentalsCalculusISBN:9781285741550Author:James StewartPublisher:Cengage LearningThomas' Calculus (14th Edition)CalculusISBN:9780134438986Author:Joel R. Hass, Christopher E. Heil, Maurice D. WeirPublisher:PEARSONCalculus: Early Transcendentals (3rd Edition)CalculusISBN:9780134763644Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric SchulzPublisher:PEARSON
- Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. FreemanCalculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning