Problem Set#1
Multiple Choice Test
Chapter 01.07 Taylors Series Revisited
COMPLETE SOLUTION SET
1. The coefficient of the term in the Maclaurin polynomial for is
(A)0
(B)0.0083333
(C)0.016667
(D)0.26667
Solution
The correct answer is (D).
The Maclaurin series for is
Hence, the coefficient of the term is 0.26667.
- Given , , , and all other higher order derivatives of are zero at , and assuming the function and all its derivatives exist and are continuous between and , the value of is
(A)38.000
(B)79.500
(C)126.00
(D)331.50
Solution
The correct answer is (C).
The Taylor series is given by
,
Since all the derivatives higher than second are zero,
- Given that is the solution to , the value of from a second order Taylor polynomial around x=0 is
(A)4.400
(B)8.800
(C)24.46
(D)29.00
Solution
The correct answer is (C).
The second order Taylor polynomial is
,
- The series is a Maclaurin series for the following function
(A)
(B)
(C)
(D)
Solution
The correct answer is (B).
5. The function is called the error function. It is used in the field of probability and cannot be calculated exactly. However, one can expand the integrand as a Taylor polynomial and conduct integration. The approximate value of using the first three terms of the Taylor series around is
(A)-0.75225
(B)0.99532
(C)1.5330
(D)2.8586
Solution
The correct answer is (A).
Rewrite the integral as
The first three terms of the Taylor series for around are
The first three terms of the Taylor series are
, or
Hence
Note: Compare with the exact value of
6. Using the remainder of Maclaurin polynomial of order for defined as
the order of the Maclaurin polynomial at least required to get an absolute true error of at most in the calculation of is (do not use the exact value of or to find the answer, but the knowledge that and ).
(A)3
(B)5
(C)7
(D)9
Solution
The correct answer is (B).
, ,
Since derivatives of are simply and , and
and
So when is
But since the Maclaurin series for only includes odd terms, .