Questions:Taylor Series - Mathematical Methods of Physics, UGC - NET Physics | Physics for IIT JAM, UGC - NET, CSIR NET PDF Download

Problem Set # 1 Multiple Choice Test 

COMPLETE SOLUTION SET

1. The coefficient of the x⁵ term in the Maclaurin polynomial for sin (2 x ) is

(A) 0
(B) 0.0083333
(C) 0.016667
(D) 0.26667

Solution The correct answer is (D).
The Maclaurin series for sin(2 x ) is

Hence, the coefficient of the x⁵ term is 0.26667.

2. Given f (3) = 6 , f ′(3) = 8 , f ′′(3) = 11 , and all other higher order derivatives of f (x ) are zero at x = 3 , and assuming the function and all its derivatives exist and are continuous between x = 3 and x = 7 , the value of f (7 ) 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,

3. Given that y(x ) is the solution to y (0) = 3 the value of y(0.2) 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

4. The series is a Maclaurin series for the following function

(A) cos(x)
(B) cos(2 x)
(C) sin (x)
(D) sin (2 x)

Solution: The correct answer is (B).

5.   The function dt 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 erf (2.0) using the first three terms of the Taylor series around t = 0 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 t = 0 are

The first three terms of the Taylor series are

Hence

Note:  Compare with the exact value of erf (2)

6.  Using the remainder of Maclaurin polynomial of n^th order for f (x ) defined as

the order of the Maclaurin polynomial at least required to get an absolute true error of at most 10 ⁻⁶ in the calculation of sin (0.1) is (do not use the exact value of sin (0.1) or cos(0.1) to find the answer, but the knowledge that |sin( x)| ≤ 1 and | cos( x) |≤ 1 ).
(A) 3
(B) 5
(C) 7
(D) 9

Solution : The correct answer is (B).

So when is 

But since the Maclaurin series for sin (x ) only includes odd terms, n ≥ 5 .