Assignment 12 Solutions

Assignment 12 Solutions - MATH 138 Calculus 2, Solutions to...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 138 Calculus 2, Solutions to Assignment 12 1: Estimate the following values so the error is 1 1000 . (a) ln(5 / 4) Solution: ln(1 + x ) = x- 1 2 x 2 + 1 3 x 3- 1 4 x 4 + , so we have ln ( 5 4 ) = 1 4- 1 2 4 2 + 1 3 4 3- 1 4 4 4 + = 1 4- 1 2 4 2 + 1 3 4 3 = 1 4- 1 32 + 1 192 = 43 192 with error E 1 4 4 4 = 1 1024 < 1 1000 by the A.S.T. (b) 5 e Solution: e x = 1 + x + 1 2! x 2 + 1 3! x 3 + 1 4! x 4 + , so we have 5 e = e 1 5 = 1 + 1 5 + 1 5 2 2! + 1 5 3 3! + 1 5 4 4! + = 1 + 1 5 + 1 5 2 2! + 1 5 3 3! = 1 + 1 5 + 1 50 + 1 750 = 916 750 with error E = 1 5 4 4! + 1 5 5 5! + 1 5 6 6! + = 1 5 4 4! 1 + 1 5 5 + 1 5 2 5 6 + 1 5 3 5 6 7 + 1 5 4 4! 1 + 1 5 2 + 1 5 4 + 1 5 6 + = 1 5 4 4! 1 1- 1 25 = 1 5 4 4! 25 24 = 1 13200 where we used the C.T. and the formula for the sum of a geometric series. 2: Approximate the value of the definite integral Z 1 p 4 + x 3 dx so the error is at most 1 1000 ....
View Full Document

This note was uploaded on 09/18/2008 for the course MATH MAth 138 taught by Professor Marshman during the Spring '08 term at Waterloo.

Page1 / 4

Assignment 12 Solutions - MATH 138 Calculus 2, Solutions to...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online