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 5x ( )x2sin is

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

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

( ) ( ) "++=!5

2!3

22)2sin(53 xxxx

"++=12032

682

53 xxx

"++= 53 26667.03333.12 xxxHence, the coefficient of the term is 0.26667. 5x

2. Given ( ) 63 =f , ( ) 83 =f , ( ) 113 =f , and all other higher order derivatives of ( )xf are zero at 3=x , and assuming the function and all its derivatives exist and are continuous between 3=x and 7=x , the value of ( )7f 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

( ) ( ) ( ) ( ) ( ) "++++=+ 32!3!2

hxfhxfhxfxfhxf

, 3=x 437 ==h( ) ( ) ( ) ( ) ( ) "++++=+ 32 4

!334

!2343343 fffff

( ) ++++= 32 4!3

)3(4!2

)3(4)3()3(7 fffff

Since all the derivatives higher than second are zero,

( ) ( ) ( ) ( ) 24!234337 ffff

++=

24!2

11486 ++= 126=

3. Given that ( )xy is the solution to 23 += ydxdy , 3)0( =y the value of ( )2.0y from a

second order Taylor polynomial 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

( ) ( ) ( ) ( ) 2!2

hxyhxyxyhxy++=+

, 0=x 2.002.0 ==h ( ) ( ) ( ) ( ) 22.0

!202.0002.00 yyyy

++=+ ( ) ( ) ( ) ( ) 02.002.0002.0 ++= yyyy ( ) 30 =y ( ) 23 += yxy ( ) 230 3 +=y 29= ( )

dxdyyxy 23=

( )23 32 += yy ( ) ( ) ( )23330 32 +=y 783= ( ) 02.07832.02932.0 ++=y 46.24=

4. The series ( )=

0

2

4)!2(

1n

nn

n

nx is a Maclaurin series for the following function

(A) ( )xcos (B) ( )x2cos (C) ( )xsin(D) ( )x2sin

Solution The correct answer is (B).

( ) ( ) ( )

==

0

2

!21cos

n

nn

nxx

( ) ( ) ( )( )

==

0

2

!2212cos

n

nn

nxx

( ) ( )

==

0

22

!221

n

nnn

nx

( ) ( )

==

0

2

!241

n

nnn

nx

5. The function ( ) = x t dtexerf0

22 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

( 0.2erf )0=t is

(A) -0.75225 (B) 0.99532 (C) 1.5330 (D) 2.8586

Solution The correct answer is (A).

The first three terms of the Taylor series for ( ) 22 tetf = around 0=t are ( ) 22 tetf = ( ) 2020 = ef

2=

( ) ( )tetf t 22 2 = ( ) ( )( )0220 2 = tef 0= ( ) ( )( ) ( )22222 22 += tt ettetf

( ) ( )( ) ( )( ) ( )22020220 22 00 += eef

4= The first three terms of the Taylor series are

2!2

)()()()( hxfhxfxfhxf++=+

2!2

)0()0()0()0( hfhffhf++=+

2!2

)0()0()0()( hfhffhf++=

!2

4)(022hh +=

!2

42 2h =

222 h = Hence

( )

=x

dttxerf0

222

x

tt0

3

322

=

3

22 3xx =

( ) ( )322222

3

=erf 75225.0=Note: Compare with the exact value of ( )2erf

6. Using the remainder of Maclaurin polynomial of order for thn ( )xf defined as ( ) ( ) ( ) ( ) xcncfn

xxR nn

n +=+

+0 ,0 ,

!11

1

the order of the Maclaurin polynomial at least required to get an absolute true error of at most in the calculation of 610 ( )1.0sin is (do not use the exact value of or

to find the answer, but the knowledge that (0.1sin )

)(0.1cos 1)sin( |x| and ). 1|)cos(| x(A) 3 (B) 5 (C) 7 (D) 9

Solution The correct answer is (B).

( ) ( ) ( )cfnxxR n

n

n1

1

)!1(+

+

+= , , 0n xc 0

( ) ( ) ( )( ) 1.00,0,)!1(

1.01.0 11

+=+

+cncf

nR n

n

n

Since derivatives of are simply ( )xf ( )xsin and ( )xcos , and ( ) 1sin x and ( ) 1cos x ( ) ( ) 11 + cf n ( ) ( ) ( )1

)!1(1.01.0

1

++

nR

n

n

( ))!1(

1.0 1

+=+

n

n

So when is ( ) 6101.0