ENGR 391 - Numerical Methods in Engineering Multiple Choice Questions with correct selected answer s

ENGR 391 - Numerical Methods in Engineering Multiple Choice Questions with correct selected answer solution new version update Concordia University

Problem Set #1 Multiple Choice Test Chapter 01.05 Floating Point Representation COMPLETE SOLUTION SET 1. A hypothetical computer stores real numbers in floating point format in

8-bit words. The first bit is used for the sign of the number, the second bit for the sign of the exponent, the next two bits for the magnitude of the exponent, and the next four bits for the magnitude of the mantissa.

Represent e  2.718 in the 8-bit format.

(A) 00010101 (B) 00011010 (C) 00010011 (D) 00101010 Solution The correct answer is (A).

Finding 210  ?2

2/2 1/2

Quotient Remainde r 1 0 0 1

Hence 210  102 Finding 0.71810  ?2 Number 0.718 2 0.436  2 0.872  2 0.744  2

1.436 0.872 1.744 1.488

Number after decimal 0.436 0.872 0.744 0.488

Number before decimal 1 0 1 1

0.71810  0.1012 2.71810  10.1012 

 2110  2012

1.0101 2 

1.0101 2 Bits in mantissa  0101 Bits in exponent  01 Sign number bit  0 Sign exponent bit  0 The final floating point representation is 00010101.

2. A hypothetical computer stores real numbers in floating point format in

8-bit words. The first bit is used for the sign of the number, the second bit for the sign of the exponent, the next two bits for the magnitude of the exponent, and the next four bits for the magnitude of the mantissa. The base-10 number represents in the above (10100111)2 given 8-bit format is (A) -5.75 (B) -2.875 (C) -1.75 (D) -0.359375 Solution The correct answer is (A).

The number is 10100111 Sign number bit  1 Sign of number is negative Sign exponent bit  0 Sign of exponent is positive Bits in exponent  10

102  1 21  0  20  210

Bits in mantissa  0111

1.01112  1 20  0  21  1 22  1 23  1 24  (1.4375)10 Hence, the number in base-10 is =  1.4375  22  5.75

3. A hypothetical computer stores floating point numbers in 8-bit words.

The first bit is used for the sign of the number, the second bit for the sign of the exponent, the next two bits for the magnitude of the exponent, and the next four bits for the magnitude of the mantissa. The machine epsilon is most nearly 8 (A) 2 4 (B) 23 2 (C) 2 2 (D) Solution The correct answer is (B).

The machine epsilon is  mach  2 number of bits for mantissa   24

4. A machine stores floating point numbers in 7-bit word. The first bit is

used for the sign of the number, the next three for the biased exponent and the next three for the magnitude of the mantissa. The number (0010110)2 represented in base-10 is (A) 0.375 (B) 0.875 (C) 1.5 (D) 3.5 Solution The correct answer is (B).

Sign bit of number  0 Hence, the number is positive Bits in biased exponent  010 Thus, it is biased by 3 as the maximum value in the biased exponent is 1112  7 .

So the exponent is 0102  3  2  3  1 .

Bits in mantissa  110

1.110  1 20  1 21  1 22  0  23  1.7510 2 Hence, the number in base-10 is  1.75 21  0.87510

5. A machine stores floating point numbers in 7-bit word. The first bit is

stored for the sign of the number, the next three for the biased exponent and the next three for the magnitude of the mantissa. You are asked to represent 33.35 in the above word. The error you will get in this case would be (A) underflow (B) overflow (C) NaN (D) No error will be registered Solution The correct answer is (B). 3310  ?????2 Quotient Remaind er 33/2 16 1 16/2 8 0 8/2 4 0 4/2 2 0 2/2 1 0 1/2 0 1 Therefore, 3310  1000012 Number 0.35 2 0.70  2 0.40  2

0.70 1.40 0.80

Number after decimal 0.70 0.40 0.80

Number before decimal 0 1 0

0.3510  0.01…2 33.3510  100001.012  25 

1.00001012

 21012



1.0000101 2 The biased exponent has 3 bits, so the biased exponent varies from 0 to 1112  710 . So the exponent can vary from -3 to 4. Hence the number 33.3510 which has an exponent of 5 would overflow.

6. A hypothetical computer stores floating point numbers in 9-bit words.

The first bit is used for the sign of the number, the second bit for the sign of the exponent, the next three bits for the magnitude of the exponent, and the next four bits for the magnitude of the mantissa. Every second, the error between 0.1 and its binary representation in the 9-bit word is accumulated. The accumulated error after one day most nearly is (A) 0.002344 (B) 20.25 (C) 202.5 (D) 8640 Solution The correct answer is C. Number 0.1 2 0.2  2

0.2 0.4

Number after decimal 0.2 0.4

Number before decimal 0 0

0.4  2

0.8

0.8

0

0.8  2 0.6  2

1.6 1.2

0.6 0.2

1 1

0.2  2

0.4

0.4

0

0.4  2

0.8

0.8

0

0.8  2

1.6

0.6

1

0.6  2

1.2

0.2

1

0.110  0.000110011…2 

 24

1.10012  21002 

1.1001 2 Thus, 0.110 is represented approximately by

0.110  1 20  1 21  0  22  0  23  1 24  24  0.09765625

The difference is  0.1  0.09765625  0.00234375 This difference is accumulated every second for one day, giving the accumulated error as  0.00234375 60  60  24  202.5

Multiple-Choice Test Gaussian Elimination Simultaneous Linear Equations COMPLETE SOLUTION SET 1. The goal of forward elimination steps in the Naïve Gauss elimination method is

to reduce the coefficient matrix to a (an) (A) diagonal (B) identity (C) lower triangular (D) upper triangular

matrix.

Solution The correct answer is (D).

By reducing the coefficient matrix to an upper triangular matrix, starting from the last equation, each equation can be reduced to one equation-one unknown to be solved by back substitution.

2. Division by zero during forward elimination steps in Naïve Gaussian elimination

of the set of equations AX   C implies the coefficient matrix A (A) is invertible (B) is nonsingular (C) may be singular or nonsingular (D) is singular

Solution The correct answer is (C).

Division by zero during forward elimination does not relate to whether or not the coefficient matrix is singular or nonsingular. For example 3 0 5 x1  2 2    7  12 x 3   2    1 5 6x3  12 would give a division by zero error in the first step of forward elimination.

However, the coefficient matrix in this case is nonsingular.

In another example 3  x1   5  0 2 2 3 7  x   12   2    4 6 14x3  12 would also give a division by zero error in the first step of forward elimination. In this case the coefficient matrix is singular.

3. Using a computer with four significant digits with chopping, the Naïve

Gauss elimination solution to 0.0030x1  55.23x2  58.12 6.239x1  7.123x2  47.23

is (A) x1  (B) 26.66; (C) x1  8.769; x1  8.800;

x2  1.051 x2  1.051 x2  1.000

(D) x1  8.771; x2  1.052 Solution The correct answer is (A).

Writing all the entries with 4 significant digits 55.23  x1  58.12  0.003000 6.239  7.123x 47.23   2    Forward Elimination: Divide Row 1 by 0.003000 and multiply it by 6.239, giving the multiplier as 6.239  2079.

0.0030 00 [Row 1]  2079 gives Row 1 as

0.0030

55.23

58.12

00 |

6.239 1.148105 | 1.208105  Subtract the above result from Row 2 changes Row 2 to [0

1.148105 | 1.207 105 ]

and hence giving the set of equations at the end of the 1st step of forward elimination as 55.23  x1   58.12   0.0030 0 1.148 105  x  1.207 105    2   

This is also the end of all steps of forward elimination.

Back substitution: From the second equation 1.148105 2  1.207 105

 x x2 

1.207 105

1.148105  1.051 From the first equation, 0.003000x1  55.23x2  58.12

x1  58.12  55.23x2 0.003000 58.12  55.231.051  0.0030  58.12  58.04 0.003000  0.08000 0.003000  26.66

4. Using a computer with four significant digits with chopping, the Gaussian

elimination with partial pivoting solution to 0.0030x1  55.23x2  58.12 6.239x1  7.123x2  47.23 is (A) (B) (C) (D) Solutio n

x1  26.66; x1  8.769; x1  8.800; x1  8.771;

x2  1.051 x2  1.051 x2  1.000 x2  1.052

The correct answer is (B).

Writing all the entries with 4 significant digits 0.0030 55.23  x1  58.12  00  6.239  7.123x 47.23   2    Forward elimination: Now for the first step of forward elimination, the absolute value of first column elements is 0.003000 , 6.239 or 0.003000, 6.239 So we need to switch Row 1 with Row 2, to get  7.123 x1  47.12   6.239 0.003000 55.23  x  58.12   2    Divide Row 1 by 6.239 and multiply it by 0.00300, gives the multiplier as 0.003000  4.808104 6.239

[Row 1] 4.808104  gives Row 1 as [2.999 103

3.424 103 |

2.265102 ]

Subtract the above result from Row 2 changes Row 2 to [0

55.22 | 58.09]

and hence giving the set of equations at the end of the 1st step of forward elimination as  7.123   x1  47.23 6.239 0 55.22 x  58.09   2   

Back substitution: From the second equation 55.22x2  58.09 x2  58.09 55.22  1.051 Substituting the value of x2 in the first equation 6.239x1  7.123x2  47.23 x1  47.23  7.123x2 6.239  47.23  7.1231.051 6.239  47.23  7.486 6.239  54.71 6.239  8.769

5. At the end of the forward elimination steps of the Naïve Gauss elimination

method on the following equations 4.2857  9.2307 105 0 7 10 4.2857 107     6.5  

5  5.4619 10

 c1  7.887 103 

0 

7  4.2857 10

 c  

5.4619 105

  0.15384

6. 5

0 107

4.2857

0

2



 0.007 

 c3 

0.1538 4



0



   3.6057 105  c   4

 





the resulting equations in matrix form are given by 4.2857 107  9.2307 105  c  0

3.7688 105

0

0 

 4.2857 107 5.4619 105

 c 1

  7.887 103 











7.887 103

2

  

0

0

 26.9140



0

0

0

 c3   5.62500 105 c  0.57968 4

1.90336 104

  4  



The determinant of the original coefficient matrix is (A) 0.00 (B) 4.2857 107 19 (C) 5.486  10 20 (D)  2.445 10

Solutio n The correct answer is (D).



1.19530 102  

0





 

If a matrix is upper triangular, lower triangular or diagonal, then the determinant is n

a11  a22  .........  ann   aii i1

Thus, the determinant, D, of the matrix is D  4.2857 107  3.7688 105   26.9140 5.62500 105   2.445 1020

6. The following data is given for the velocity of the rocket as a function of time.

To find the velocity at t  21s , you are asked to use a quadratic polynomial, v(t)  at 2  bt  c to approximate the velocity profile. t v(t)

(s) (m / s)

0 14 15 20 30 35 0 227.04 362.78 517.35 602.97 901.67

The correct set of equations that will find a, b and c are 176 14 1 a 227.04 (A)

225 15 1 b  362.78  400

(B)

   20 1 c

225 15 1 a 362.78  400 20 1 b  517.35      900 30 1 c

(C)

 602.97

0 0 1 a 0   225 15 1b  362.78      400 20 1 c

(D)

517.35

 517.35

400 20 1 a 517.35    900 30 1 b  602.97  1225

   35 1 c 901.67

Solution The correct answer is (B).

First choose the three points closest to t  21s that also bracket it. t0  15 s, v(t0 )  362.78 m/s t1  20 s, v(t1 )  517.35 m/s t2  30 s, v(t2 )  602.97 m/s Such that v(15)  362.78  a(15)2  b(15)  c v(20)  517.35  a(20)2  b(20)  c v(30)  602.97  a(30)2  b(30)  c

This expands to

225a  15b  c  362.78 400a  20b  c  517.35 900a  30b  c  602.97 225a  15b  c  362.78 400a  20b  c  517.35       900a  30b  c 602.97 This can be rewritten as 15 1a 362.78  225 400 20 1b  517.35      900 30 1c 602.97

Multiple-Choice Test Gauss-Seidel Method of Solving Simultaneous Linear Equations COMPLETE SOLUTION SET 1. A square matrix

Ann

is diagonally dominant if n

(A) aii   aij , i  1,2,..., n j 1 i j n

aij , (B) aii   j 1 i j

n

i  1,2,..., n

and

aii

j 1 i j

n

(C) (D)

aij , ai   i  1,2,..., n j 1 i n and   aij , j 1 ai i  1,2,..., n

  aij , for any i  1,2,..., n

ai i

n

  aij , for any i  1,2,..., n j 1

i

Solutio n

The correct answer is (B). n

A n  n square matrix [A] is a diagonally dominant matrix if

ai   aij , for all i = 1, 2, …, n i

j 1 i j

n

an aii   aij for at least one i, that is, for each row, the absolute value of the j d diagonal 1 i j

element is greater than or equal to the sum of the absolute values of the rest of the elements of that row, and that the inequality is strictly greater than for at least one row. Diagonally dominant matrices are important in ensuring convergence in iterative schemes of solving simultaneous linear equations. 15 6 3 2  A  2  4

7  2 is a diagonally

dominant matrix as 6 

a11  15  15  a12   6  7  13 a13 a22   4  4  a21 a33  6  6  a31  a32

 a23

 2  2 4

 3 2 5

and for at least one row, that is Rows 1 and 3 in this case, the inequality is a strictly greater than inequality.

2. Using [x1, x2 , x3 ]  [1,3,5] as the initial guess, the values of [x1, x2 , x3 ] after three

iterations in the Gauss-Seidel method for 12 7 3  x1   2  1 5 1  x    5   2     2 7 11x3   6 are (A) [-1.4333 -1.9727] 2.8333 (B) [1.4959 0.90464 0.84914] (C) [0.9066 -1.0115 -1.0243] 6 (D) [1.2148 0.72060 0.82451 ] Solution The correct answer is (C).

Rewriting the equations gives 2  7x2  3x3 x1  12  5  x1  x3 x2  5 6  2x1  7x2 x3   11 Iteration #1 Given the initial guess of the solution vector as  x1  1 x   3  2   x3 

5

we get x1  2  73  35 12  2.8333 x2   5   2.8333  5 5  1.4333

x3  6  2 2.8333  7 1.4333  11  1.9727

At the end of the first iteration, the estimate of the solution vector is x  2.8333  x 1  1.4333  2   x3  1.9727 Iteration #2 Since the estimate of the solution vector at the end of Iteration #1 is x  2.8333  x 1  1.4333  2   x3  1.9727 we get x1  2  7 1.4333  3 1.9727 12  1.4960 x2   5  1.4960   1.9727 5  0.90465 x3  6  21.4960  7 0.90464  11  0.84915 At the end of the second iteration, the estimate of the solution vector is   xx1  1.4960 0.90465  2   x3   0.84915 Iteration #3 Since the estimate of the solution vector at the end of Iteration #2 is   xx1  1.4960 0.90465  2   x3   0.84915 we get

x1  2  7 0.90465  3 0.84915 12  0.90666 x2   5  0.90666   0.84914 5  1.0115 x3  6  20.90666  7 1.0115  11  1.0243 At the end of the third iteration, the estimate of the solution vector is   xx1  0.90666 1.0115  2   x3  1.0243

3. To ensure that the following system of equations,

2x1  7x2  11x3  6 x1  2x2  x3   5 7x1  5x2  2x3  17 converges using the Gauss-Seidel method, one can rewrite the above equations as 2 7 −11  x1   6  follows:     (A)

(B)

(C)

1 7  7 1

x = −5  2   2  x3  17  2   x1  17  5     2 1  x 2  = − 5 2 7 −11 x3   6  7 5 2   x1   6  1 2 1  x  = − 5  2   2 7 −11 x3  17  2 5

1

 (D)

The equations cannot be rewritten in a form to ensure convergence.

Solution The correct answer is (B).

A system of equations will converge using the Gauss-Seidel method if the coefficient matrix is diagonally dominant. Thus, rewriting the system of equations as 2   x1  17  7 5 1 2 1  x    5   2   2 7 11 x3   6  results in the coefficient matrix being diagonally dominant. a1  7  7  1 a12 a2  2  2  2 a21

 a13

a33  11  11  a31

 5 2 7 112

 a23  a32

 2  7 9

12 7 4.

For  1 5 initial 

3  x1 

 22 

1  x    7  and using x x  2  

1

2

x   1

2 1 as the

3

2 7 11 x3   2 guess, the values of x1 x2 x3  are found at the end of each iteration as Iteration # 1 2 3 4

x1 0.41667 0.93990 0.98908 0.99899

x2 1.1167 1.0184 1.0020 1.0003

x3 0.96818 1.0008 0.99931 1.0000

At what first iteration number would you trust at least 1 significant digit in your solution?

(A) 1 (B) 2 (C) 3 (D) 4 Solution The correct answer is (C).

The absolute relative approximate error at the end of the first iteration is 1.0000 a 1  0.41667 −3.2864% 0.41667

= 140.00% 1.1167 − 2.0000 a 2

1.1167

 = a 3 79.104% 0.96818 − 1.0000 0.96818

 =

100

1 00

100

The maximum absolute relative approximate error is 140%.

Since, 140%  0.5 10 2m , m  1. There are no significant digits correct. We are looking for 1 significant digit, so we must continue to conduct iterations.

The absolute relative approximate error at the end of the second iteration is

a 1  0.93990 − 0.41667 100 0.93990

= 55.669% 1.0184 − 1.1167 a 2

1.0184

100

 = a 3 9.6509% 1.0008 − 0.96818 100 1.0008

 = 3.2564% The maximum absolute relative approximate error is 55.669%.

Since, 55.669%  0.5 102m , m  1. There are no significant digits correct. We are looking for 1 significant digit so we must continue to conduct more iterations.

The absolute relative approximate error at the end of the third iteration is a 1  0.98908 - 0.9399  100 0.98908

= 4.9727% a 2 

1.0020 -1.0184 100 1.0020

= 1.6322% a 3  0.99930 -1.0007 100 0.99930

= 0.14660%

The maximum absolute approximate error is 4.9733%.

Since, 4.9733%  0.5 10 2m , m  1. There is at least one significant digit correct. Since we were looking for 1 correct significant digit, we need not conduct any more iterations.

5. The algorithm for the Gauss-Seidel method to solve

AX   C using n max iterations. The initial value of X  is stored in X .

(A) Sub Seidel (n, a, x, rhs, nmax)

For k  1 To nmax For i  1 To n For j  1 To n If ( i  j ) Then Sum = Sum + a(i, j) * x( j) endif Next j x(i)  (rhs(i)  Sum) / a(i,i)

Next i Next j End Sub (B) Sub Seidel (n, a, x, rhs, nmax)

For k  1 To nmax For i  1 To n Sum = 0 For j  1 To n If ( i  j ) Then Sum = Sum + a(i, j) * x( j) endif Next j x(i)  (rhs(i)  Sum) / a(i,i)

Next i Next k End Sub (C) Sub Seidel (n, a, x, rhs, nmax)

For k  1To

is given as follows when

nmax For i  1To n Sum = 0 For j  1 To n Sum = Sum + a(i, j) * x( j) Next j x(i)  (rhs(i)  Sum) / a(i,i)

Next i Next k End Sub

(D) Sub Seidel (n, a, x, rhs, nmax)

For k  1To nmax For i  1 To n Sum = 0 For j  1 To n If ( i  j ) Then Sum = Sum + a(i, j) * x( j) endif Next j x(i)  (rhs(i)  Sum) / a(i,i)

Next i Next k End Sub

Solution The correct answer is (B).

Sub Seidel(n, a, x, rhs, nmax) For k = 1 To nmax For i = 1 To n Sum = 0 For j = 1 To n If (i <> j) Then Sum = Sum + a(i, j) * x(j) endif Next j x(i) = (rhs(i) - Sum) / a(i, i) Next i Next k End Sub Choice (A) is incorrect because the value of the variable ‘Sum’ needs to be reset on each row.

Choice (C) is incorrect because it does not include an if statement that only adds the sum when the value of ‘i’ is not equal to the value of ‘j’. The value of ‘Sum’ would include the value of ‘a(i,i)*x(i)’, which would give an incorrect answer for the value of ‘x(i)’. Choice (D) is incorrect because it does not subtract the value of ‘Sum’ from ‘rhs(i)’, which again, would give the incorrect answer for ‘x(i)’.

6. Thermistors measure temperature, have a nonlinear output and are valued for

a limited range. So when a thermistor is manufactured, the manufacturer supplies a resistance vs. temperature curve. An accurate representation of the curve is generally given by 1

T0

 a  a ln(R)  a2 lnR2  a lnR3 1

3

where T is temperature in Kelvin, R is resistance in ohms, and a0 , a1 , a2 , a3 are constants of the calibration curve. Given the following for a thermistor R ohm 1101.0 911.3 636.0 451.1

T C 25.113 30.131 40.120 50.128

the value of temperature in C for a measured resistance of 900 ohms most nearly is (A) 30.002 (B) 30.473 (C) 31.272 (D) 31.445 Solution The correct answer is (B).

Give n

1  a  ln(R)  lnR2  a lnR3 T0 a1 a2 3

Then from the data in the table we can write four equations 1

2

3

3 a ln1101   ln(1101)  ln1101  25.11 a0 a1 a2 3

1

2

3

  ln(911.3)  ln911.3  a ln911.3 30.13 a0 a1 a2 1

1

2

3

  ln(636)  ln636  a ln636 40.12 a0 a1 a2 0 1

2

3

  ln(451.1)  ln451.1  a ln451.1 50.12 a0 a1 a2 8 3

which reduce to 0.039820  a0  7.0040a1  49.056a2  343.58a3 0.033188  a0  6.8149a1  46.442a2  316.50a3 0.024925  a0  6.4552a1  41.670a2  268.99a3 0.019949  a0  6.1117a1  37.353a2  228.29a3 In matrix form the equations can be rewritten as

3

3

343.58 49.056 7.0040 1a3  0.039820   316.5 46.442 6.8149 1  0.03318  a 2       268.99 41.670 6.4552 1  a1  0.024925  a 0.019949 228.29 37.353 6.1117 1 0    Using any method gives a3  0.011173 a2  0.20448 a1  1.2605 a0  2.5964 1  2.5964  1.2605ln(R)  0.20448lnR2  0.011173lnR3 T  2.5964  1.2605ln(900)  0.20448ln9002  0.011173ln9003  0.032816 1 T 0.032816  30.473C

Multiple-Choice Test Background Interpolation COMPLETE SOLUTION SET 1.

The number of different polynomials that can go through two fixed data points x1, y1  and x2 , y2  is A) 0 B) 1 C) 2 D) infinite

Solution The correct answer is (D).

The number of polynomials that can go through two fixed data points x1 , y1  and

x2 , y2  is infinite. For example the polynomials

y  5x  6 and

y  x 2 go through the two data points

(2,4) and (3,9). So does every polynomial of the form y  axm  bxn , where m and n are any positive integers of your choice.

2.

Given n+1 data pairs, a unique polynomial of degree passes through n  1 data points.

(A) n  1 (B) n  1 or less (C) n (D) n or less

Solution The correct answer is (D).

A unique polynomial of degree n or less passes through n  1 data points. If the polynomial is not unique, then at least two polynomials of order n or less pass through the n  1 data points.

Assume two polynomials Pn x and Qn x go through n  1 data points, x0 , y0 , x1 , y1 ,…, xn , yn  The n Sinc

Rn x  Pn x  Qn x Pn xand Qn x pass through all the n  1 data points, Pn xi   Qn xi , i  0, … , n Rn xi   Pn (xi )  Qn xi   0, i  0, … , n

e Henc e The nth order polynomial R x has n  1 zeros. A polynomial of order n can have n  1 zeros n only if it is identical to a zero polynomial, that is, Rn x  0 Hence Pn x  Qn x Extra Notes for the Student: How can one show that if a second order polynomial has three zeros, then it is zero everywhere?

If R2 x  a0  a1 x  a2 x 2 , then if it has three zeros at x1, x2, and x3, then R x   a  a x  a x2  0 2 1 0 1 1 2 1 R x   a  a x  a x2  0 2 2 0 1 2 2 2 R x   a  a x  a x2  0 2

3

0

1 3

2 3

which in matrix 2form gives 1 x x a 0  1

 1 2  0    1 x x 2 a1    2 0 1 x x 2 a  0  3 3  2    The above set of equations has a trivial solution, that is, a1  a2  a3  0 . But is this the only solution? That is true if the coefficient matrix is invertible.

The determinant of the coefficient matrix can be found symbolically with the forward elimination steps of naïve Gauss elimination to give 1 x122  2 2 2 2 x xx   xx 2  x 2  x x x x x x x x 1 det 1 x 

2

2

3 2

1 x



2

2 3 3

1

1

3

1 2

1

2

x3 

3

 (x1  x2 )(x2  x3 )(x3  x1 ) Since x1  x2  x3 the determinant is non-zero. Hence the coefficient matrix is invertible. a1  a2  a3  0 is the only solution, that is, R2 x  0.

3.

The following function(s) can be used for interpolation: (A) polynomial (B) exponential (C) trigonometric (D) all of the above

Solution The correct answer is (D).

Polynomials are often used for interpolation because they are easy to evaluate, differentiate and integrate. However, other functions such as trigonometric and exponential can be used for interpolation. How is a polynomial easy to evaluate as compared to a trigonometric function?

Because terms such as xm in a polynomial involve multiplication of x to itself m 1 times. However, trigonometric and exponential functions include the use of computationally more involved calculations via a Taylor series.

4.

Polynomials are the most commonly used functions for interpolation because they are easy to (A) evaluate (B) differentiate (C) integrate (D) evaluate, differentiate and integrate

Solution The correct answer is (D).

Polynomials are often used for interpolation because they are easy to evaluate, differentiate and integrate. However, other functions such as trigonometric and exponential can be used for interpolation. How is a polynomial easy to evaluate as compared to a trigonometric function?

Because terms such as xm in a polynomial involve multiplication of x to itself m 1 times. However, trigonometric and exponential functions include the use of computationally more involved calculations via a Taylor series.

5.

Given n  1 data points x0 , y0 , x1 , y1 , ,xn1 , yn1 , xn , yn  , assume you pass a functio f through all the data points. If now the value of f (x) is n (x) the function required to be found outside the range of the given x-data, the procedure is called (A) extrapolation (B) interpolation (C) guessing (D) regression

Solution The correct answer is (A).

If x falls outside the range of x for which the data is given, it is no longer interpolation but instead is called extrapolation.

6.

Given three data points (1,6), (3,28), and (10, 231), it is found that the function y  2x2  3x  1 passes through the three data points. Your estimate of y at x  2 is most nearly (A) 6 (B) 15 (C) 17 (D) 28

Solution The correct answer is (B). y(x)  2x 2  3x 1 y(2)  2  22  3 2 1  8  6 1  15

Multiple-Choice Test Direct Method Interpolation COMPLETE SOLUTION SET 1. Given n+1 data pairs, a unique polynomial of degree

passes through n  1

data points.

(A) n  1 (B) n  1 or less (C) n (D) n or less Solution The correct answer is (D).

A unique polynomial of degree n or less passes through n  1 data points. If the polynomial is not unique, then at least two polynomials of order n or less pass through the n  1 data points.

Assume two polynomials Pn x and Qn x go through n  1 data points, x0 , y0 , x1 , y1 ,…, xn , yn  The n Sinc

Rn x  Pn x  Qn x Pn x and Qn x pass through all the n  1 data points, Pn xi   Qn xi , i  0,… , n Rn xi   Pn (xi )  Qn xi   0, i  0, … , n

e Henc e The nth order polynomial R x has n  1 zeros. A polynomial of order n can have n  1 zeros n only if it is identical to a zero polynomial, that is, Rn x  0 Hence

Pn x  Qn x Extra Notes for the Student: How can one show that if a second order polynomial has three zeros, then it is zero everywhere?

If R2 x  a0  a1 x  a2 x 2 , then if it has three zeros at x1 , x2 , and x3 , then R x   a  a x  a x2  0 2 1 0 1 1 2 1 R x   a  a x  a x2  0 2 2 0 1 2 2 2 R x   a  a x  a x2  0 2

3

0

1 3

2 3

which in matrix 2form gives 1 x x a 0  1

0

 1 2     1 x x 2 a1    2 0 1 x x 2 a  0  3 3  2    The above set of equations has a trivial solution, that is, a1  a2  a3  0 . But is this the only solution? That is true if the coefficient matrix is invertible.

The determinant of the coefficient matrix can be found symbolically with the forward elimination steps of naïve Gauss elimination to give 1 x122  2 2 2 2 x xx   xx 2  x 2  x x x x x x x x 1 det 1 x 

2

2

3 2

1 x



2

2 3 3

1

1

3

1 2

1

2

x3 

3

 (x1  x2 )(x2  x3 )(x3  x1 ) Since x1  x2  x3 the determinant is non-zero. Hence the coefficient matrix is invertible. a1  a2  a3  0 is the only solution, that is, R2 x  0.

2. The data of the velocity of a body as a function of time is given as follows.

0 15 18 22 24 Time (s) 22 24 37 25 123 Velocity (m/s) The velocity in m/s at 16 s using linear polynomial interpolation is most nearly (A) 27.867 (B) 28.333 (C) 30.429 (D) 43.000 Solution The correct answer is (B).

For the first order polynomial, we choose the velocity given by vt   a0  a1t Since we want to find the velocity at t  16 , we choose the two data points that are closest to t  16 and that also bracket t  16 . Those two points are t0  15 and t1  18 . Then t0  15, vt0   24 t1  18, vt1   37 gives v15  a0  a1 15  24 v18  a0  a1 18  37 Writing the equations in matrix form 15a0  24  1 1 18a   37   1    and solving the above two equations gives a0  41 a1  4.3333 Hence vt   a0  a1t

 41  4.3333t, 15  t  18

v16  41  4.333316  28.333 m/s

3. The following data of the velocity of a body as a function of time is given as follows.

0 15 18 22 24 Time (s) 22 24 37 25 123 Velocity (m/s) The velocity in m/s at 16 s using quadratic polynomial interpolation is most nearly (A) 27.867 (B) 28.333 (C) 30.429 (D) 43.000 Solution The correct answer is (C).

For second order polynomial interpolation (also called quadratic interpolation), we choose the velocity given by vt   a  a t  a t 2 0

1

2

Since we want to find the velocity at t  16 , we need to choose the three data points that are closest to t  16 and that also bracket t  16 . These three points are t0  15 , t1  18 , and t2  22 . t0  15, t1  18, give s

vt0   24 vt1   37 vt2   25

t2  22,

v15  a0  a1 15  a2 152  24 v18  a0  a1 18  a2 182  37 v22  a0  a1 22  a2 222  25 Writing the three equations in matrix form 1 15 225a0  24

1 18 324a   37   1    1 22 484a2 

25

and the solution of the above three equations gives

a0  323.86 a1  38.905 a2 Henc e

 1.0476 15  t  22 vt   323.86  38.905t 1.0476t 2 ,

At t  16 , v16  323.86  38.905161.0476162

 30.429 m/s

4. The following data of the velocity of a body is given as a function of time

0 15 18 22 24 Time (s) 22 24 37 25 123 Velocity (m/s) Using quadratic interpolation, the interpolant vt   8.667t 2  349.67t  3523,18  t  24 approximates the velocity of the body. From this information, one of the times in seconds at which the velocity of the body is 35 m/s during the above time interval of t  18 s to t  24 s is (A) 18.667 (B) 20.850 (C) 22.200 (D) 22.294 Solution The correct answer is (D).

Using the interpolant, set the velocity equal to 35 m/s and solve for time.

35  8.667t 2  349.67t  3523 0  8.667t 2  349.67t  3488 Using the quadratic equation solution t

− (−349.67)  (−349.67)2 − 4  8.667  3488

2  8.667  349.67  1347.1249 17.334 give s

t  22.294 s t  18.055 s

5. The following data of the velocity of a body is given as a function of time

0 15 18 22 24 Time (s) 22 24 37 25 123 Velocity (m/s) One of the interpolant approximations for the velocity from the above data is given as v(t)  8.6667t 2  349.67t  3523,18  t  24 Using the above interpolant, the distance in meters covered by the body between t  19s and t  22s is most nearly (A) 10.337 (B) 88.500 (C) 93.000 (D) 168.00

Solution The correct answer is (A).

Since v  integral

dx

, taking the integral of the velocity will give the location, x. By taking the

dt over the interval of t  19s to t  22s we can find the distance traveled, s , over that interval.

22

s   8.6667t 2  349.67t  3523dt 19

 2.8889t3 174.84t 2  3523t  19

22

 2.8889(22)3 174.84(22)2  3523(22) 2.8889(19)3 174.84(19)2  3523(19)  23647  23637  10.337 m

6. The following data of the velocity of a body is given as a function of time.

Time (s) Velocity (m/s)

0 15 18 22 24 22 24 37 25 123

If you were going to use quadratic interpolation to find the value of the velocity at t  14.9 seconds, what three data points of time would you choose for interpolation?

(A) 0, 15, 18 (B) 15, 18, 22 (C) 0, 15, 22 (D) 0, 18, 24 Solution The correct answer is (A).

We need to choose the three points closest to that also bracket t  14.9 s . Although t  14.9 s the data points in choice (B) are closest to 14.9, they do not bracket it. This would be performing extrapolation, not interpolation. Choices (C) and (D) both but they are not the bracket t  14.9 s closest three data points.

Time (s) Velocity (m/s) How far is t  14.9 s 0 22 14.9  0  14.9 15 24 14.9 15  0.1 18 37 14.9 18  3.1 22 25 14.9  22  7.1 24 123 14.9  24  9.1

Multiple-Choice Test Lagrange Method Interpolation COMPLETE SOLUTION SET 1.

Given n+1 data pairs, a unique polynomial of degree the n  1 data points.

(A) n  1 (B) n (C) n or less (D) n  1 or less

passes through

Solution The correct answer is (C).

A unique polynomial of degree n or less passes through n  1 data points. Assume two polynomials Pn x and Qn x go through n  1 data points, x0 , y0 , x1 , y1 ,…, xn , yn  The n

Rn x  Pn x  Qn x Pn x and Qn x pass through all the n  1 data points, Pn xi   Qn xi , i  0, … , n

Sinc

Rn xi   Pn (xi )  Qn xi   0, i  0, … , n

e Henc e The nth order polynomial R x has n  1 zeros. A polynomial of order n can have n  1 zeros n only if it is identical to a zero polynomial, that is, Rn x  0 Hence Pn x  Qn x How can one show that if a second order polynomial has three zeros, then it is zero everywhere?

If R2 x  a0  a1 x  a2 x 2 , then if it has three zeros at x1 , x2 , and x3 , then R x   a  a x  a x 2  0 2

1

0

1 1

2 1

R2 x2   a0  a1 x2  a2 2 x 2  0 R x   a  a x  a x 2  0 2

3

0

1 3

2 3

Which in matrix form gives

1 x 1

x 2 a  1

0

0

 2     1 x 2 x 2 a1    0 1 x x 2 a  0  3 3  2    The final solution a1  a2  a3  0 exists if the coefficient matrix is invertible. The determinant of the coefficient matrix can be found symbolically with the forward elimination steps of naïve Gauss elimination to give 1 x122  2 2 2 2 x xx   xx 2  x 2  xx x x x x x x 1 det 1 x 

2

2

3 2

1 x



2

2 3 3

1

1

3

1 2

1

2

x3 

3

 (x1  x2 )(x2  x3 )(x3  x1 ) Since x1  x2  x3 the determinant is non-zero. Hence, the coefficient matrix is invertible. Therefore, a1  a2  a3  0 is the only solution, that is, R2 x  0.

2.

Given the two points a, f a, b, f b, the linear Lagrange polynomial through these two points is given by x f a  f b a a a b

x  (A) f  b 1

b

x

x f x f b (B) f1 x b  a b   a a (C) f1 x f a 

f b f a b  a b  a

xb

f a  f b

a b

ba

(D) f x  xa 1

Solution The correct answer is (D). n

f n (x)   Li (x) f (xi ) i0 1

f1 (x)   Li (x) f (xi ) i0

 L0 x f x0   L1 x f x1   L0 x f a  L1 x f b n

Li (x)  

j0 ji 1

L0 (x)  

x  xj x i x j x  xj

j0 x

 xj

0 j0

 x  x1 x0  x1  xb

f1 x that passes

a1  b x  xj L1 (x)  

j0 x 1 j1

 xj

 x x0 x1  x0  xa ba f1 (x)  L0 x f a  L1 x f b x f b  x  b f a  ab a b a

3.

The Lagrange polynomial that passes through the 3 data points is given by x y

15 18 22 24 37 25

f 2 x  L0 x24 L1 x37 L2 x25 The value of L1 x at x  16 is (A) –0.071430 (B) 0.50000 (C) 0.57143 (D) 4.3333 Solution The correct answer is (B). n

xxj

Li (x)  

j xi 0 j i 2

L1 (x)  

xj

xxj

j 0 x

xj

1 j 1

 x  x2    x  x0  x x x x  1 0  1 2    16 1516  22  18   15  18  22   1   6  =  3   4     

 0.50000

4.

The following data of the velocity of a body is given as a function of time.

Time (s) 10 15 18 22 24 Velocity (m/s) 22 24 37 25 123

A quadratic Lagrange interpolant is found using three data points, t  15 , 18 and 22. From this information, at what of the times given in seconds is the velocity of the body 26 m/s during the time interval of t  15 to seconds. t  22 (A) 20.173 (B) 21.858 (C) 21.667 (D) 22.020 Solution The correct answer is (B).

n

vn (t)   Li (t)v(ti ) i0

wher e t0  15, t1  18, give s

vt0   24 vt1   37 vt2   25

t2  22,

n

t tj

Li (t)  

j0 t i ji 2

L0 (t)  

tj

t tj

j 0 t 0 j 0

tj

  t  t1  t  t2  t t t t  0 1  0 2  2 t t j L1 (t)   j 0 t  t j 1 j 1

 t  t0  t  t2      t t  t  1 0  2  t1 2

L2 (t)  

t tj

j 0 t 2 j 2

tj

  t  t0  t  t1  t t t t  2 0  2 1   t  t1  t  t2  t   t  t2  t   t  t1    t0 t0 v2 (t)  v(t )  v(t0 )   t   1 v(t2 )  t t  t t  t t t t t  0 1  0 2  1 0  1 2  2 0  2 1      t  18  t  22 v(15)   t 15  t  22 v(18)   t 15  t  18 v(22)            15  18 15  22 

18 15 18  22 

 22 15  22  18 

26   t  18  t  22   24   t 15  t  22   37   t 15  t  18   25            15 18 15  22 

18  15  18  22 

 22 15  22 18 

 t  22   24   t  15  t  22   37   t  15  t  18   25 26   t 18         3  4   7  4   − 3   7       t 2  40t  396  26  

21

 t 2  37t  330    24  

1 2

 t 2  33t  270    37  



28

  25

     26  1.1429t 2  45.714t  452.57  3.0833t 2  114.08t 1017.5  0.89286t 2  29.464t  241.07 26  1.0476t 2  38.905t  323.86 0  1.0476t 2  38.905t  349.86 t

− 38.905 

(38.905)2 − 41.858  −1.0476  −349.86  2

2   1 .

0 4 7 6

and

15.

27 8

5. The path that a robot is following on a x-y plane is found by interpolating four data

points as

x 2 y 7.5

4.5 5.5 7.5 6

7 5

yx  0.15238x3  2.2571x2  9.6048x  3.9000 The length of the path x  2 x  7 is to− 2)2 from (7.5 − 7.5)2 + (4.5 (5 − 6)2 + (7 − 5.5)2  6  7.52  5.5  4.52 

(A) 7

1  (0.15238x3  2.2571x2  9.6048x  3.9000)2 dx

(B)

 2 7

1  (0.45714x2  4.5142x  9.6048)2 dx

(C)

 2 7

(D)  (0.15238x3  2.2571x2  9.6048x  3.9000)dx 2

Solution The correct answer is (C). yx from a to b is given by

The length S of the curve b

S

2

 dy  1+    dx 

dx

a

wher a2 e b7 7

2

givin S  1 +  dy  dx 2  dx  g

yx  0.15238x3  2.2571x2  9.6048x  3.9000 Thus dy  0.45714x2  4.5142x  9.6048 , dx 7

S   1  0.45714x  4.5142x  9.6048 dx 2

2

2

6. The following data of the velocity of a body is given as a function of time.

0 15 18 22 24 Time (s) 22 24 37 25 123 Velocity (m/s) If you were going to use quadratic interpolation to find the value of the velocity at t  14.9 seconds, what three data points of time would you choose for interpolation?

(A) 0, 15, 18 (B) 15, 18, 22 (C) 0, 15, 22 (D) 0, 18, 24 Solution The correct answer is (A).

We need to choose the three points closest to that also bracket t  14.9 s . Although t  14.9 s the data points in choice (B) are closest to 14.9, they do not bracket it. This would be performing extrapolation, not interpolation. Choices (C) and (D) both but they are not the bracket t  14.9 s closest three data points.

Time (s) Velocity (m/s) How far is t  14.9 s 0 22 14.9  0  14.9 15 24 14.9 15  0.1 18 37 14.9 18  3.1 22 25 14.9  22  7.1 24 123 14.9  24  9.1

Multiple-Choice Test Newton’s Divided Difference Polynomial Method Interpolation COMPLETE SOLUTION SET 1. If a polynomial of degree n has more than n zeros, then the polynomial is (A) oscillatory (B) zero everywhere (C) quadratic (D) not defined

Solution The correct answer is (B).

A unique polynomial of degree n or less passes through n  1 data points. Assume two polynomials Pn xand Qn x go through n  1 data points, x0 , y0 , x1 , y1 ,…, xn , yn  The n

Rn x  Pn x  Qn x Pn x and Qn x pass through all the n  1 data points, Pn xi   Qn xi , i  0, … , n

Sinc

Rn xi   Pn (xi )  Qn xi   0, i  0, … , n

e Henc e The nth order polynomial R x has n  1 zeros. A polynomial of order n can have n  1 zeros n only if it is identical to a zero polynomial, that is, Rn x  0 Hence Pn x  Qn x How can one show that if a second order polynomial has three zeros, then it is zero everywhere?

If R2 x  a0  a1 x  a2 x 2 , then if it has three zeros at x1 , x2 , and x3 , then

R x   a  a x  a x2  0 2 1 0 1 1 2 1 R x   a  a x  a x2  0 2 2 0 1 2 2 2 R x   a  a x  a x2  0 2

3

0

1 3

2 3

Which in matrix2form gives 1 x x a 0  1

1

0

 2     1 x 2 x 2 a1    0 1 x x 2 a  0  3 3  2   

The final solution a1  a2  a3  0 exists if the coefficient matrix is invertible. The determinant of the coefficient matrix can be found symbolically with the forward elimination steps of naïve Gauss elimination to give 1 x2 2 2 x x2    x x2  x 2 x x 12  x2xx x  x xx 1 det 1 x 

2

2

3 2

1 x



2

2 3 3

1

1 3

1 2

1 2

x3 

3

 x1  x2 x2  x3 x3  x1  Sinc e

x1  x2  x3

the determinant is non-zero. Hence, the coefficient matrix is invertible. Therefore, a1  a2  a3  0 is the only solution, that is, R2 x  0.

2. The following x - y data is given.

x 15 18 22 y 24 37 25 The Newton’s divided difference second order polynomial for the above data is given by f2 (x)  b0  b1 x  15  b2 x  15x  18 The value of b1 is most nearly (A) –1.0480 (B) 0.14333 (C) 4.3333 (D) 24.000 Solution The correct answer is (C).

Give x0  15 n x1  18 f2 x0   24 f2 x1   37 we have f2 (x)  b0  b1x  x0   b2 x  x0 x  x1  The f2 (x0 )  b0  b1 x0  x0  b2 x0  x0 x0  x1  n  b0 f2 (x1 )  b0  b1 x1  x0  b2 x1  x0 x1  x1   b0  b1 x1  x0   f2 ( x0 )  b1 x1  x0  Thus , b1  f2 (x1 )  f2 (x0 ) x1  x0  37  24 18  15  4.3333

3. The polynomial that passes through the following x - y data

x 18 22 24 y ? 25 123 is given by 8.125x2  324.75x  3237,18  x  24 The corresponding polynomial using Newton’s divided difference polynomial is given by f2 (x)  b0  b1 x  18  b2 x  18x  22 The value of b2 is (A) 0.25000 (B) 8.1250 (C) 24.000 (D) not obtainable with the information given

Solution The correct answer is (B).

Expanding, f2 (x)  b0  b1x  18  b2 x  18x  22  b  b x  18b  b x2  40x  396 0 1 1 2  b  18b  396b   b  40b x  b x2 0

1

2

This needs to be the same as 8.125x2  324.75x  3237 Hence b2  8.125

1

2

2

4. Velocity vs. time data for a body is approximated by a second order

Newton’s divided difference polynomial as vt   b0  39.622t  20  0.5540t  20t  15, 10  t  20 The acceleration in m/s2 at t  15 is (A) 0.55400 (B) 39.622 (C) 36.852 (D) not obtainable with the given information

Solution The correct answer is (C). vt   b0  39.622t  20  0.5540t  20t  15,

10  t  20

d at  vt  dt   39.6221  0.5540t  20  0.5540t  15  39.622  0.5540t  20  0.5540t  15, 10  t  20 a15  39.622  0.554015  20  0.554015  15  39.622  2.7700  36.852 m/s2

5. The path that a robot is following on a x-y plane is found by interpolating four

data points as

x 2 y 7.5

4.5 5.5 7.5 6

7 5

yx  0.15238x3  2.2571x2  9.6048x  3.9000 The length of the path x  2 x  7 is to− 2)2 from (7.5 − 7.5)2 + (4.5  6  7.52  5.5  4.52 

(A) 7

1  (0.15238x3  2.2571x2  9.6048x  3.9000)2 dx

(B)



2 7

1  (0.45714x2  4.5142x  9.6048)2 dx

(C)



2 7

(D)  (0.15238x3  2.2571x2  9.6048x  3.9000)dx 2

Solution The correct answer is (C). yx from a to b is given by

The length S of the curve b

S

2

 dy  1+    dx 

dx

a

wher a  2 e b7

7

2

 dy  givin S   1 +  dx  dx 2  

g

yx  0.15238x3  2.2571x2  9.6048x  3.9000 Thus dy  0.45714x2  4.5142x  9.6048 , dx 7

S   1  0.45714x  4.5142x  9.6048 dx 2

2

2

6. The following data of the velocity of a body is given as a function of time.

0 15 18 22 24 Time (s) 22 24 37 25 123 Velocity (m/s) If you were going to use quadratic interpolation to find the value of the velocity at t  14.9

seconds, the three data points of time you would choose for interpolation are (A) 0, 15, 18 (B) 15, 18, 22 (C) 0, 15, 22 (D) 0, 18, 24 Solution The correct answer is (A).

We need to choose the three points closest to

that also bracket t  14.9 s . Although the data points in choice (B) are closest to 14.9, they do not bracket it. This would be performing extrapolation, not interpolation. Choices (C) and (D) both but they are not the bracket t  14.9 s closest three data points. t  14.9 s

Time (s) Velocity (m/s) How far is t  14.9 s 0 22 14.9  0  14.9 15 24 14.9 15  0.1 18 37 14.9 18  3.1 22 25 14.9  22  7.1 24 123 14.9  24  9.1

Multiple-Choice Test Spline Method Interpolation COMPLETE SOLUTION SET 1.

The following n data points, x1 , y1 , x2 , y2 , xn , yn , are given. For conducting quadratic spline interpolation the x-data needs to be (A) equally spaced (B) placed in ascending or descending order of x -values (C) integers (D) positive

Solution The correct answer is (B).

The following n data points, x1 , y1 , x2 , y2 , xn , yn , are given. For conducting quadratic spline interpolation the x-data needs to be arranged in ascending or descending order.

2.

In cubic spline interpolation, (A) the first derivatives of the splines are continuous at the interior data points (B) the second derivatives of the splines are continuous at the interior data points (C) the first and the second derivatives of the splines are continuous at the interior data points (D) the third derivatives of the splines are continuous at the interior data points

Solution The correct answer is (C).

In cubic spline interpolation, the first and the second derivatives of the splines are continuous at the interior data points. In quadratic spline interpolation, only the first derivatives of the splines are continuous at the interior data points.

3.

The following incomplete y vs. x data is given.

x 1 2 4 6 7 y 5 11 ???? ???? 32 The data is fit by quadratic spline interpolants given by f x  ax  1, 1  x  2 f x  2x 2  14x  9, 2  x  4 f x  bx 2  cx  d , 4  x  6 f x  25x 2  303x  928, 6  x  7 where a, b, c, and d are constants. The value of c is most nearly (A)  303.00 (B) 144.50 (C) 0.0000 (D) 14.000 Solution The correct answer is (C).

Solution Method 1: Since the first derivatives of two quadratic splines are continuous at the interior points, at x = 4 d d bx 2  cx    2x 2  14x d  dx x4  9 x4 dx 2bx  cx =4   4x  14

x=4

2(4)b  c  4(4)  14 8b  c  2

(1)

and at x = 6 d d bx2  cx   25x2  303x  d  dx x 928 6 dx 2bx  cx =6  50x  303

x=6

x6

2(6)b  c  50(6)  303 12b  c  3

(2) Equations (1) and (2) in matrix form  8 1b  2 12 1c   3      Solving these equations gives b  0.25 c0

Solution Method 2: The third spline bx 2  cx  d goes through x  4 . However, so does the second spline.

We can use this latter knowledge to find the value of y at x  4. f (x)  2x 2  14x  9, 2  x  4 f (4)  2(4)2  14(4)  9  15 The third spline bx 2  cx  d goes through x  4 . Hence b(4)2  c(4)  d  15 (1) 16b  4c  d  15 The third spline bx 2  cx  d goes through x  6 . However, so does the fourth spline. We can use this latter knowledge to find the value of y at x  6 . f (x)  25x 2  303x  928, 6  x  7 f (6)  25(6)2  303(6)  928  10 The third spline bx 2  cx  d goes through x  6 . Hence b(6)2  c(6)  d  10 36b  6c  d  10 Since the first derivatives of second and third quadratic splines are continuous at the interior points, at x = 4

(2)

d bx 2  cx    2x 2  14x d  dx x4  9 x4 d

dx

2bx  cx =4   4x  14

x=4

2(4)b  c  4(4)  14 8b  c  2 Equations (1), (2) and (3) are then 16b  4c  d  15 36b  6c  d  10 8b  c  0d

(3)

 2 Putting these equations in matrix form gives 4 1  b  15  16   36 6 1 c  10       2   8 1 0 d Solving the above equations gives b  0.25 c0 d  19

4.

The following incomplete y vs. x data is given.

x 1 2 4 6 7 y 5 11 ???? ???? 32 The data is fit by quadratic spline interpolants given by f x  ax  1, 1  x  2 , f x  2x 2  14x  9, 2  x  4 f x  bx 2  cx  d , 4  x  6 f x  ex 2  fx  g, 6  x  7 df where a, b, c, d f , and are constants. The at x  dx 2.6 value of , e, g (A) 144.5 (B) 0  4.0000 (C) 3.6000 (D) 12.200

Solution The correct answer is (C).

Since the spline f x  2x 2  14x  9 2  x  4 , the derivative at x = 2.6 is is valid in the interval df x  4x 14 dx df 2.6  4  2.6 14 dx  10.4 14  3.6000

most nearly is

5.

The following incomplete y vs. x data is given.

x 1 2 4 6 7 y 5 11 ???? ???? 32 The data is fit by quadratic spline interpolants given by f x  ax  1, 1  x  2 , f x  2x 2  14x  9, 2  x  4 f x  bx 2  cx  d , 4  x  6 f x  25x 2  303x  928, 6  x  7 3.5

Where a, b, c, and d

are constants. What is the value of

 f xdx ?

1.5

(A) 23.500 (B) 25.667 (C) 25.750 (D) 28.000

Solution The correct answer is (C).

3.5

2

3.5

To find  f xdx we must take  ax  1dx    2x 2  14x  9dx but first we have to find the 1.5

1.5

2

value of the constant a. Since at x = 2, y = 11 a  2  1  11 a  11  1 2 6 Thus, 3.5 2 3.5

f xdx  6x 1dx    2x2 14x  9dx  1.5 1.5 2  6

2





 2 14

2 x3   x  x1.  2 5  3

3.5

 

2

x2  9x 2

 3 22  2 31.52 1.5

 2 2    3.53  7  3.52  9  3.5     23  7  22  9  2      3   3   12  2 6.75 1.5  28.583  85.75  31.5  5.3333  28 18  4.75 21  25.75

6.

A robot needs to follow a path that passes consecutively through six points as shown in the figure. To find the shortest path that is also smooth you would recommend which of the following?

(A) Pass a fifth order polynomial through the data (B) Pass linear splines through the data (C) Pass quadratic splines through the data (D) Regress the data to a second order polynomial

y

Path of a Robot

8 7 6 5 4 3 2 1 0 0

5

10

15

x

Solution The correct answer is (C).

Using linear splines (Choice B) would create a straight-line path between consecutive points. Although this will be the shortest path it will not be smooth.

Regressing the data to a second order polynomial (Choice D) will result in a smooth path but it will not pass through all the points. As demonstrated in the following figure, using polynomial interpolation such as choice (A) is a bad idea and will result in a long path. By using quadratic spline interpolation (choice C), the path will be short as well as smooth.

10 8

y

6 4 2 0 0

2

4

6

8

10

12

x Cubic Spline

Fifth Order Polynomial

Multiple-Choice Test LU Decomposition Method Simultaneous Linear Equations COMPLETE SOLUTION SET 1. The LU  decomposition method is computationally more efficient than

Naïve Gauss elimination for solving (A) a single set of simultaneous linear equations.

(B) multiple sets of simultaneous linear equations with different coefficient matrices and the same right hand side vectors.

(C) multiple sets of simultaneous linear equations with the same coefficient matrix and different right hand side vectors.

(D) less than ten simultaneous linear equations.

Solution The correct answer is (C).

An example can be demonstrated by finding the inverse of the matrix A. The problem of finding the inverse reduces to solving n sets of equations with the coefficient matrix staying the same and n different RHS vectors. These n RHS vectors are the n columns of the identity matrix.

For calculations of each column of the inverse of the A matrix, the coefficient matrix A in the set of equations AX   C does not change. So if we use the LU Decomposition method, the A  LU  decomposition needs to be done only once and forward substitution LZ   C and the back substitution U X   Z  still needs to be done n times.

The total computational time CT |inverse LU required to find the inverse of a matrix using LU decomposition is CT |inverse LU = 1  CT |DE + n  CT |FS + n  CT |BS  3 20n   + n  T(  4n+ n  T  12n  8n  2 4n2 = 1 T  4n 4n2

wher e

3   3  32n3 2 20n  = T  12n   3   3

the computational time1 DECT | given by

1

to decompose the [ A] matrix to [L][U ] form is

The time is calculated by first separately calculating the number of additions, subtractions, multiplications, and divisions in a procedure such as back substitution, etc. We then assume 4

CT |DE

 8n3 2 20n  = T   4n   , 3   3

the computational time CT |FS to solve LZ   C by forward substitution is given by CT |FS = T 4n2  4n, and the computational time CT |BS to solve U X   Z  by back substitution is given by CT |BS = T 4n2  12n In comparison, if Gaussian elimination method were used to find the inverse of matrix [A], the forward elimination as well as the back substitution will have to be done n times. The total computational time CT |inverse GE required to find the inverse of a matrix by using Gaussian elimination then is CT |inverse GE = n  CT |FE + n  CT |BS  3 32n   + n  T(  12n  8n  2 4n2 8n = n T

wher e

3   3 2  8n4 3 4n  = T   12n   3   3

the computational time CT |FE for the forward elimination part is given by  8n3 2 32n   , and CT |FE = T   8n 3 3    the computational time CT |BS CT |BS = T 4n2  12n

for the back substitution part is given by

Clearly for large n , CT | /3

>> CT |

inverse GE

inverse LU

as CT | inverse GE

has the dominating terms of 8n4

and CT inverse | LU has the dominating terms of 32n3 / 3 . For large values of n , Gaussian elimination method would take more computational time (approximately n / 4 times – prove it) than the LU decomposition method. Typical values of the ratio of the computational time for different values of n are given in Table 1.

Table 1 Comparing computational times of finding inverse of a matrix using LU decomposition and Gaussian elimination. n CT |inverse GE / CT |inverse

10

100

1000 1000 0 3.28 25.83 250.8 2501

LU

clock cycles each for an add, subtract, or multiply operation, and 16 clock cycles for a divide operation as is the case for a typical AMD®-K7 chip. http://www.isi.edu/~draper/papers/mwscas07_kwon.pdf

2. The lower triangular matrix L in the LU  decomposition of the matrix given

below 25

4  1

5

0



u13 

2

0 0 u

1

 

21

 8 12

22

  1

25

22

23

ℓ31 ℓ 32 1 0 0 0 

0

u33 

4  14.400  

(B) 5 0  6

 0 0  4.2400  1 0 0 10 1 0   8 0 1 2

 1

u 



 0 (A) 0.400 1 00 1.733 1  0.320 3 00

(C)

u1

11

10 8 16  ℓ

is

0u

0

0



(D)

 0.400 0 1 00 1.500 1  0.320 0 00

Solution The correct answer is (A).

We must first complete the first step of forward elimination. 25 5 4  10 8 16    8 22 1 2 10  0.4 , and subtract the results from Row 2 First step: Multiply Row 25 1 by 25 5 4  Row 2 Row1 (0.4)   0 6 14.4  8 12 22  8  0.32 , and subtract the results from Row 3 Multiply Row 1 25 by

4  25 5 Row 3 Row1 (0.32)   0 6 14.4   0 10.4 20.72 To find ℓ 21 and ℓ 31 , what multiplier was used to a 21 make the and

a31 elements zero in the first

step of forward elimination using the Naïve Gauss elimination method? They are ℓ 21  0.4

ℓ 31  0.32 To find ℓ 32 , what multiplier would be used to make the a32 element zero?

Remember the a32 element is made zero in the second step of forward elimination. So ℓ  32

10.4

 1.7333

6

Henc e 0 0  1 L  0.40000 1 0 0.32000 1.733 1 3

3. The upper triangular matrix U  in the LU  decomposition of the matrix given

below 25

4  1

5

0



1

u13 

2

0 0 u

 

21

 0 12

ℓ ℓ 22

32

31



(A)

u1

11

 0 8 16  ℓ

is

0u

 1 0.400

0

25

4  14.400 



22

23

1 0

0

u33 

0  0 1 00 1.733 1  0.320 3 00



(B) 5 0  6

 0 0  4.2400 25 (C) 5 0  8  0 0

4 16    2

u  

(D) Solution

 0.200 0.16000 1 2.4000  0 0   4.240  1  0 0

The correct answer is (C).

The U  matrix is the same as the coefficient matrix that is found at the end of the forward elimination steps of the Naïve Gauss elimination method, that is 25 5 4  U    0 8 16  0 22 1 2 First step: The first step of forward elimination does not need to be conducted as already zero.

Second step: Multiply Row 2 by  1.5 8

12

a21 and

and subtract the results from Row 3

a3 are 1

25

Thus ,

5 Row 3 Row 2 (1.5)   0 8  0 0 25 5 4  U    0 8 16  0 0  2

4 16    2

4. For a given 2000  2000 matrix A, assume that it takes about 15 seconds to

find the inverse of A by the use of the LU  decomposition method, that is, finding the LU  once, and then doing forward substitution and back substitution 2000 times using the 2000 columns of the identity matrix as the right hand side vector. The approximate time, in seconds, that it will take to find the inverse if found by repeated use of the Naive Gauss elimination method, that is, doing forward elimination and back substitution 2000 times by using the 2000 columns of the identity matrix as the right hand side vector is most nearly (A) 300 (B) 1500 (C) 7500 (D) 30000 Solution The correct solution is (C).

From the solution of Problem 1 of this MC question test, the computational times for finding the inverse of A by Naïve Gaussian elimination and LU decomposition is given by 2  4  3 3 4n 2 20n   8n 32n T   12n   and T   12n   , respectively, where 3  3   3  3 T = clock cycle time n = number of rows or columns in square matrix [A].

The ratio of computational time, R between the two methods then is 2  8n4 3 4n  T  3  12n  3   R  32n3 2 20n  T  12n   3 3   For a 2000  2000 matrix 2  8  20004 3 4  2000  T 3  12  2000  3   R  32  20003 2 20  2000   T  12  2000  3 3    500.84

Since it takes 15 seconds to find the inverse by LU decomposition, it would take about 500.84 15  7512.6 seconds to find the inverse by using Naïve Gaussian elimination.

5. The algorithm for solving a set of n equations AX   C, where A 

LU  involves solving LZ   C by forward substitution. The algorithm to solve LZ   C is given by (A) z1  c1 / l11

for i from 2 to n do sum = 0 for j from 1 to i do sum = sum + lij * z j end do zi  (ci  sum) / lii end do (B) z1  c1 / l11 for i from 2 to n do sum = 0 for j from 1 to (i  1) do sum = sum + lij * z j end do zi  (ci  sum) / lii end do (C) z1  c1 / l11 for i from 2 to n do for j from 1 to (i  1) do sum = sum + lij * z j end do zi  (ci  sum) / lii end do (D) for i from 2 to n do sum = 0 for j from 1 to (i  1) do sum = sum + lij * z j end do zi  (ci  sum) / lii end do

Solution The correct answer is (B).

Since LZ   C is written as

… 0 z1 c1  

1 

0



ℓ 21

  c 2 

 

⁝ z 2

1 ℓ 31

⋱ 0 z3   c3 

⁝  ℓ n1 ℓ n 2 

 ⁝   ⁝      cn  1zn 

We know that the first step of forward substitution is z1  c1 / l11 The next equations would be c  ℓ 21  z1 z2  2 ℓ22 c  (ℓ 31  z1  ℓ 32  z2 ) z3  3 ℓ33 ⁝ i1

ci   (ℓ ij  z j ) zi 

j 1

ℓ ii

Thus, the algorithm is z1  c1 / l11 for i from 2 to n do sum = 0 loop)

(the variable sum must be cleared each time through the i1

for j from 1 to (i-1) do

(this loop finds (ℓ ij  z j ) ) j 1

sum = sum + lij * z j end do zi = (ci – sum) / lii end do

6. To solve boundary value problems, a numerical method based on finite

difference method is used. This results in simultaneous linear equations with tridiagonal coefficient matrices. These are solved using a specialized LU  decomposition method.

Choose the set of equations that approximately solves the boundary value problem d2y 2  2 6x 0.5x y0  y12 0 , 0 x  12 dx  , 0,   The second derivative in the above equation is approximated by the secondorder accurate central divided difference approximation as learned in the differentiation module (Chapter 02.02). A step size of h  4 is used, and hence the value of y can be found approximately at equidistantly placed 4 nodes between x=0 and x=12.

i =1

i=2

i=3

i=4

x=0

x=4

x=8

x = 12

 1 0  y1   0  0 0   0.0625 0.125 0.0625  0  y 2  16.0    = (A)  0 0.0625 0.125 0.0625 y3  16.0      0 0 1 y 4   0   0  1 0.0625

 y1   0   y  16.0 − 0.125 0.0625  2  =   (B)   0 0.0625 − 0.125 0.0625 y3  16.0      1  y 4   0  0 0 0  0  y1   0   1 0 0  0 1  y 2   0  0 0  = (C)  0  y 0.0625 − 0.125 0.0625 3   16.0   0. y 6.0  0625 1 0 0.0625 −  4    0.125 0  y1   0  0 0  1     0 1  0 0  y 2  =  0  (D) 0.0625 0.125 0.0625 0   y3  16.0    0. y 6.0 0625 4  1  0 0.0625 0.125



0

0

0 0

Solution The correct answer is (B). d2y 2  2 6x 0.5x y0  y12 0 , 0 x  dx  , 12 0,   The first equation is from the boundary condition y0  0

(1)

and that is the value of y at node 1, giving y1  0 d2y

(a)

The approximation with a second order accurate central divided difference dx method is for 2

d2y dx2

y  2y  y 

i1

i

h2

i1

Hence the ordinary differential equation (1) can be approximated at node i as yi1  2 yi  yi1   (2) 0.5x2 6x where At

h2

i

i

xi 1  xi  h

node i  2, x2  x1  h 04 4 Then at node i=2, equation (2) can be written as y1  2 y2   6  4  0.5  42 y3 42 0.0625 y1  0.125 y2  0.0625 y3  16 At node i  3, x3  x2  h 44 8 Then at node i=3, equation (2) can be written as

(b)

y2  2 y3   6  8  0.5  82 y4 42 0.0625 y2  0.125 y3  0.0625 y4  16 The final equation is from the boundary condition y12  0

(c)

or y4  0

(d)

Writing equations (a), (b), (c) and (d) in matrix form, we get 0 0  y1  0   1 0  0.06   y 16.0 0.062 0 5 0.125 25    0   0

 2     0.0625  0.125 0.0625 y3  16.0     0 0 1 y 4   0 

Multiple-Choice Test Newton-Raphson Method Nonlinear Equations COMPLETE SOLUTION SET 1. The Newton-Raphson method of finding roots of nonlinear equations falls under the category of

methods.

(A) bracketing (B) open (C) random (D) graphical

Solution The correct answer is (B).

The Newton-Raphson method is an open method since the guess of the root that is needed to get the iterative method started is a single point. Other open methods such as the secant method use two initial guesses of the root but they do not have to bracket the root.

2. The Newton-Raphson method formula for finding the square root of a real number R from the equatio x2  R  0 is, n (A) x



xi



2 3xi

i1

(B) x i1

2 1 R (C) xi1   xi   2 x i  1 R (D) xi1  3xi   2 x i  Solution The correct answer is (C).

The Newton-Raphson method formula for solving f(x) = 0 is

wher e

Thus,

xi  xi 1 

f (xi ) f (xi )

f (x)  x 2  R f (x)  2x x 2 i R xi1  xi  2xi R   xi x + 2x i i 2 1 xi + R 2xi 2 1 R   xi   2 ix



3. The next iterative value of the root x2  4  0 using the Newton-Raphson method, if the of initial guess is 3, is (A) 1.5 (B) 2.067 (C) 2.167 (D) 3.000

Solution The correct answer is (C).

The estimate of the root is

xi1  xi 

f (xi ) f (xi )

Chose i = 0,

x0  3 f (x0 )  x0 2  4  32  4 5 f (x0 )  2x0  23 6 Thus,

x  x 1



f (x0 ) 0

0

f (x ) 5  3 6  2.166

4. The root of the equation f (x)  0 is found by using the Newton-Raphson method. The initial estimate of the root is x0  3,

f 3  5 . The angle the line tangent to the

f (x) makes

function

at x  3 is 57 with respect to the x-axis. The next estimate of the x1 most nearly is root, (A) –3.2470 (B) −0.24704 (C) 3.2470 (D) 6.2470

Solution The correct answer is (B).



x

Since,

  57, x0  3, and f (x0 )  5 rise tan( )  run tan(57)



tan(57) f (x0 )  f (x1 )  x x 0

50 3  x1

1

x 15  3(tan(57))  tan(57)  5  3  1.5399  1.5399  0.24704

5. The root of x3  4 is found by using the Newton-Raphson method. The successive iterative values of the root are given in the table below.

Iteration Number

Value of Root

0

2.0000

1

1.6667

2

1.5911

3

1.5874

4

1.5874

The iteration number at which I would first trust at least two significant digits in the answer is (A) 1 (B) 2 (C) 3 (D) 4 Solution The correct answer is (C).

The absolute relative approximate error for the first iteration is

 a  1.6667  2 100 1.6667   0.19999 100  19.999% Since, 19.99%  0.5 102m , m  0 . There are no significant digits correct.

The absolute relative approximate error for the second iteration is

 a  1.5911  1.6667 100 1.5911   0.047514 100  4.7514% Since, 4.7514%  0.5 102m , m  1. There is at least one significant digit correct.

The absolute relative approximate error for the third iteration is

 a  1.5874  1.5911 100 1.5874   0.0023308 100  0.23308%

Since, 0.23308%  0.5 102m , m  2 . There are at least two significant digits correct.

Thus, the third iteration is the first iteration in which at least two significant digits are correct.

6. The ideal gas law is given by pv  RT where p is the pressure, v is the specific volume, R is the universal gas constant, and T is the absolute temperature. This equation is only accurate for a limited range of pressure and temperature. Vander Waals came up with an equation that was accurate for larger range of pressure and temperature given by

 p a v  b  RT   v2   

where a and b are empirical constants dependent on a particular gas. Given the value of R  0.08 , a  3.592 , b  0.04267 , p  10 and T  300 (assume all units are consistent), one is going to find the specific volume, v , for the above values. Without finding the solution from the Vander Waals equation, what would be a good initial guess for v ?

(A) 0 (B) 1.2 (C) 2.4 (D) 3.6

Solution The correct answer is (C).

From the physics of the problem, the initial guess can be found from the linear relationship of the ideal gas law

pv  RT v  RT p  0.08  300 10  2.4

Problem Set#1 Multiple Choice Test Chapter 01.06 Propagation Errors COMPLETE SOLUTION SET 1. If A  3.56  0.05

and (A) (B) (C) (D)

B  3.25  0.04 , the values of

6.81  A  B  6.90 6.72  A  B  6.90 6.81  A  B  6.81 6.71  A  B  6.91

Solution The correct answer is (B).

A  3.50  0.05 Henc 3.56  0.05  A  3.56  0.05 e 3.51  A  3.61 B  3.25  0.04 3.25  0.04  B  3.25  0.04 3.21  B  3.29 Henc e

Henc e

3.51 3.21  A  B  3.61 3.29 6.72  A  B  6.90

A  B are

2. A number A is correctly rounded to 3.18 from a given

A B  C,

number B. Then where C is (A) 0.005 (B) 0.01 (C) 0.18 (D) 0.09999 Solution The correct answer is (A).

Since A is rounded off to 3.18, the number can be 3.17XYZ … where X is a number between 5 and 9 or wher X is a number between 0 and 4. Hence, 3.18XYZ e A  B  C makes C  0.005

3. Two numbers A and B are approximated as C and D, respectively. The relative

error in C  is given by D

(A)  A  C    B  D       A   B  (B)  A  C    B  D    A  C    B  D         A   A   B     B  (C)  A  C    B  D    A  C    B  D   

 

A 

  B 

    A B    

(D)  A  C    B  D       A   B  Solution The correct answer is (C).

Rel C  D A  B  C  D A B 

True Error = True Value – Approximate Value Approximate Value = True Value – True Error C  A  D  B Where and  are the true errors in the representation of A and B, respectively.

Rel CD  AB  ( A   )(B   ) AB  AB  AB  B  A   AB AB cancels which yields,    Rel (CD)    A B AB Rel(A) + Rel(B) + Rel(A)Rel(B)   AC    B  D   AC    B  D

 

 

A 

  B 

     A   B 

4.

The formula for normal strain in a longitudinal bar is given where F by  AE F = normal force applied A = cross-sectional area of the bar E = Young’s modulus

If F  50  0.5 N , A  0.2  0.002 m2 , E  210 109  1109 Pa , the maximum and error in the measurement of strain is (A) 10 12 11 (B) 2.95 10 1.22 10 9 (C) 1.19 10 9 (D) Solution The correct answer is (B).

The total error for strain is given by 





|  | F F  A A  E E The partial derivatives are then  1  ,    F ,    F F

AE

A2 E

A

E

AE 2

The total error for 1 strain is then |  |     F A    E F F   AE   |  | 



1

 2  AE    (0.5) 

 2 AE    50 

 (0.002) 

 9 (0.2)(210 10 )

 2 9 (0.2) (210 10 )













 50

 (1 109 )

(0.2)(210 109 )

|  |  1.19 1011  1.19 1011  5.67 1012  2.95 1011





5. A wooden block is measured to be 60mm by a ruler and the

measurements are considered to be good to 1/4th of a millimeter. Then in the measurement 60mm, we have . . significant digits (A) 0 (B) 1 (C) 2 (D) 3 Solution The correct answer is (C).

We are given the uncertainty is within 1/4th of a millimeter so at least 2 significant digits are accurate.

6. In the calculation of the volume of a cube of nominal size 5" , the

uncertainty in the measurement of each side is 10%. The uncertainty in the measurement of the volume would be (A) 5.477% (B) 10.00% (C) 17.32% (D) 30.00% Solution The correct answer is (D).

For this problem, V  a 3 where a is the length of the side of the cube. V 

dV a da

 3a 2a

V  3

a3

 aa V V  3 a a V  3 a V a Plugging in numbers yields V V  3 0.1

 0.3  30%

Multiple-Choice Test Linear Regression Regression COMPLETE SOLUTION SET

1. Given x1 , y1 ,x2 , y2 , .. ,xn , yn best fitting

,

y  f x by least squares

data to

requires minimization of n

(A)   y i  f xi  i1 n

(B)  yi  f xi  i1

2

n

(C)

i1yi  f xi  n

(D)

2

 yi n

yi  y , y  ni1 i1

Solution The correct answer is (C).

A measure of goodness of fit, that is, how f x predicts the response variable y is the magnitude of the residual Ei at each of the n data points.

Ei  yi  f xi , i  1,2,. , n 1, n Ideally, if all the residuals Ei are zero, one may have found an equation in which all the points lie on the model. But, this is practically impossible. Thus, minimization of the residuals is an objective of obtaining regression coefficients.

The most popular method to minimize the residual is the least squares method, where the estimates of the constants of the models are chosen such that the sum n of the squared residuals is

minimized, that is minimize  E 2 . i1

y  f x by least squares requires minimization of

Thus, best fitting data to n

i

2

i1yi  f xi 

2. The following data

x 1 20 30 40 y 1 400 800 1300 is regressed with least squares regression to A) 27.480 B) 28.956 C) 32.625 D) 40.000

y  a0  a1 x . The value of a1 most nearly is

Solution The correct answer is (C). n

a1 

n

n

i1

i1

n xi yi  xi  yi i1 n

2

 n  n x    x  i i i1  i1  2

n n

n

n

i1x 2 i1 y  i1x i1x y 2 a  n n  2  n x    x  i i

i

i

i

i

0

i

i1

Sinc e

 i1 

n 4 4

i1 xi yi  11 20  400  30  800  40 1300  84001 4

i1 xi  1  20  30  40  91 4

i1 yi  1  400  800 1300  2501 4

 x2 i112  202  302  402  2901 i

the

n

a1 4  840 01 91 250 1 4  2 9 0 1 

 9 1

 2

 108 413 3 3 2 3  3 2 .

6 2 5

3. The following data

x 1 20 30 40 y 1 400 800 1300 is regressed with least squares regression to A) 27.480 B) 28.956 C) 32.625

y  a1 x . The value of a1 most nearly is

D) 40.000

Solution The correct answer is (B).

Using the least squares criterion, we minimize 2 n n S   E 2  y  a x  r

i

i

i1

1 i

i1

To find a1 we minimize Sr

with respect a1 . to

dS r  2   y  a x  x   0 n

i

da givin g

1 i

i

i1

1 n

n

  y x   a x2  0 i i

1 i

i1 n

i1

n

  y x  a  x2  0 i i

1

i1

i

i1

Solving the above equation for a1 gives n

 yi xi a1  i1n

i1 x 2 i

Can you show that this corresponds to a local minimum, and then the

absolute minimum. Since n4 4

i1 xi yi  11 20  400  30  800  40 1300  84001 4

xi = 1 + 20 + 30 + 40 = 2901  i1

then

2

2

2

2

2

a1  84001 2901  28.956

4. An instructor gives the same y vs. x data as given below to four students and

asks them to regress the data with least squares regression to

y  a0  a1 x .

x 1 10 20 30 40 y 1 100 400 600 1200 They each come up with four different answers for the straight-line regression model. Only one is correct. The correct model is A) y  60x  1200 B) y  30x  200 C) y  139.43  29.684x D) y  1  22.782x

Solution The correct answer is (C).

We know for the straight line regression model n

a1 

n

n

i1

i1

y  a0  a1 x the values of a0

and a1 are

n xi yi  xi  yi i1 n

2

 n  n x    x  i i i1  i1  2

a0  y  a1 x Sinc e

n5 5

i1 xi yi  1110 100  20  400  30  600  40 1200  75001 5

 xi  1 10  20  30  40  101 i1 5

i1 yi  1 100  400  600 1200  2301 5

i1 x2  12 102  202  302  402  3001 i

the

n

a  5 750 01 101  230 1 1

5 30 01  1 01

2

4 8 0 4  2 9 .

6 8 4

 1 4 2 6 0 4

101  29.684  0 5 5  139.42 Hence, the linear regression model is y  29.684x  139.42 Can you think of a different way to do the problem? a 

2301

5. A torsion spring of a mousetrap is twisted through an angle of 180°. The

torque vs. angle data is given below.

Torsion, T (N-m) 0.110 0.189 0.230 0.250 Angle,  (rad) 0.10 0.50 1.1 1.5 The relationship between the torque and the angle is T  a0  a1 .

The amount of strain energy stored in the mousetrap spring in Joules is A) 0.29872 B) 0.41740 C) 0.84208 D) 1561.8 Solution The correct answer is (C).

The linear regression curve for the torque is T  a0  a1 . The values of a0

a1 

n

n

n

i1 n

i1

i1

and a1 are

niTi i Ti 2

 n  n i2     i  i1  i1  a0  y  a1 x Sinc e

n4 4

i1i Ti  0.10  0.110  0.50  0.189 1.1 0.230 1.5 0.250  0.7335 4

i1i  0.10  0.50 1.11.5  3.2 4

i1Ti  0.110  0.189  0.230  0.250  0.779 4

i1 2  0.102  0.502 1.12 1.52  3.72 i

the n

a1  4  0.7335  3.2 2 0.779 4  3.72  3.2  0.4412

4 .

6 4  0 .

0 9 5 0 8 6 N m / r a d i a n

3.2  0.095086  0 4 4  0.11868 N - m Hence, the linear regression model is T  0.095086  0.11868 N  m a 

0.779

The potential energy U stored in the spring from   is be given by 0 to   



U  Td 0 

 (0.095086θ  0.11868)d 0

  2    0.11868 = 0.09508  6 



2

0  0.047543  0.11868  0 2

 0.047543 2  0.11868  0.047543 02  0.11868 0  0.84208 J

6. A scientist finds that regressing the y vs. x data given y  a0  a1x results in

below to the coefficient of determination for the straight-line model, r 2 to being zero. x 1 3 11 17 y 2 6 22 ?

The missing value for x  17 y at A)  2.4444 B) 2.0000 C) 6.8889 D) 34.000

Solutio n The correct answer is (A).

Let y17  b Sinc r2  0 e

this means that St  Sr wher n e  y 2 St  yi

i1

and

n

S    y   a x 2 a r

i i1

0

1 i

most nearly is

The two being equal is possible if and a0  y . a1  0 , then a1  0 If n

n

n

n xi yi   xi  yi  0 i1

i1

i1

41 2  3 6 11 22 17  b 1 3 11172  6  22  b  0 b  2.4444 Extra notes for the student What is a0 ? n

n

n

n

i1x i1y  i1x i1x y 2 a0  n  n  n x2    x  i 2 i

i

i

i

i

i1

 i1 

 6.8889

Is a0

the average value of y?

i

n

 yi

y  i1 n 4

 yi  i14  2  6  22  2.4444 4  6.8889 Question: Can you prove that if r 2 for a straight line regression 0 model

y a0

 then the regression a1x

model reduces to y  y ?

Hint: Solve the three equations Sr 0 a0 Sr 0 a1 St  Sr This will give a1  0 and a0  y.

Check the blog entry also: http://numericalmethods.eng.usf.edu/blog_entries.html

Multiple-Choice Test Nonlinear Regression Regression COMPLETE SOLUTION SET 1. When using the transformed data model to find the constants of the regression

model y  aebx to best fit x1 , y1 , x2 , y2 , xn , yn , the sum of the square of the residuals that is minimized is n

(A) y  ae

bxi

i 1 n

2

i

(B) ln( y )  lna  bx 2 i

i

i 1 n

(C)   y lna  bx 2 i

i

i 1 n

(D) ln( y )  lna b ln(x )2 i

i

i 1

Solutio n The correct answer is (B).

Taking the natural log of both sides of the regression model y  aebx give lny  lna  bx s The residual at each data point xi is Ei  lnyi   lna  bxi The sum of the square of the residuals for the transformed data is n

Sr   E 2i i 1

 n lny   lna  bx 2 i

i1

i

2. It is suspected from theoretical considerations that the rate of water flow from

a firehouse is proportional to some power of the nozzle pressure. Assume pressure data is more accurate. You are transforming the data.

Flow rate, F 96 129 135 145 168 235 (gallons/min) 11 17 20 25 40 55 Pressure, p (psi) The exponent of the nozzle pressure in the regression model (A) 0.49721 (B) 0.55625 (C) 0.57821 (D) 0.67876 Solution The correct answer is (A).

The transforming of the above data is done as follows.

F  ap b ln(F )  ln(a)  b ln( p) z  a0  bx wher z  lnF  e x  ln( p) a0  lna implying a  ea0 There is a linear relationship between z and x. Linear regression constants are given by n

b

n

n

n xi zi  xi  zi i 1 n

i 1

n

n

n

i 1x2 i 1z  i 1x i 1x2z i

a0 

i 1 2

 n  n x    x  i i i 1  i 1  n 2

i

n

i

 n 

i i

F  apb most nearly is

n x2    x  i i i 1  i 1  Sinc e

n6 6

i1 xi zi  ln(11)  ln(96)  ln(17)  ln(129)  ln(20)  ln(135)  ln(25)  ln(145)  ln(40)  ln(168)  ln(55)  ln(235)  96.208 6

i1 xi  ln(11)  ln(17)  ln(20)  ln(25)  ln(40)  ln(55)  19.142 6

i1 zi  ln(96)  ln(129)  ln(135)  ln(145)  ln(168)  ln(235)  29.890 6

i1 x2  (ln(11))2  (ln(17))2  (ln(20))2  (ln(25))2  (ln(40))2  (ln(55))2  62.779 i

the n

b  6  96.208  19.142  29.890 6  62.779  19.1422  577.25  572.15 376.67  366.41  0.49721 Can you now find what a is?

3. The transformed data model for the stress-strain curve   k ek2 for concrete in 1

compression, where  is the stress and  is the strain, is (A) ln lnk1   ln   k2 (B) ln  lnk   k    1 2   (C) ln  lnk   k    1 2   (D) ln   ln(k1 )  k2 Solutio n The correct answer is (B)

  k1 ek2 The model can be rewritten as   k1 ek2  To transform  the data, we take the natural log of both sides ln   lnk ek2    1     lnk1   lnek2   lnk1   k2

4. In nonlinear regression, finding the constants of the model requires solving

simultaneous nonlinear equations. However in the exponential model y  aebx that is best fit to x1, y1 ,x2 , y2 ,........,xn , yn , the value of b can be found as a solution of a single nonlinear equation. That nonlinear equation is given by n n n (A)  y x ebxi   y ebxi  x  0 i i

i

i1

i

i1 n

i1

bxi n i1y ie (B) n y x ebxi    x e2bxi  0 i i

i

n

i1

i1

i1e2bx

i

n

 yie

n

bx i

(C)  y x ebxi  i1 i i

ebx  0 i

n

i1

n

i1

i1e2bxi n

i n i1 y e (D)  y e   x e2bx  0  i i n

bxi

bxi

i

n

i1

i1 2bxi

i1e Solutio n The correct answer is (B).

Given x , y ,x , y ,........,x , y , best fit y  aebx to the data. The variables a and b are the 1

1

2

2

n

n

constants of the exponential model. The residual at each data point xi is

(1)

Ei  yi  aebxi The sum of the square of the residuals is n

Sr   E 2i i 1 n



i 1y

 aebxi 

2

i

(2)

To find the constants a and b of the exponential model, we find where Sr is a local minimum or maximum by differentiating with respect to a and b and equating the resulting equations to zero. Sr n 2 − aebxi  ebxi   0  i  i1 n y a − aebxi i ax ebxi  Sr i1 2  0  yi b





(3a,b)

or

n

n

  y ie n

 ae2bxi  0

bxi

i 1

i 1 n

y x ebx  a x e2bx   0 i

i

i i

(4a,b)

i

i 1

i 1

Equations (4a) and (4b) are simultaneous nonlinear equations with constants a and b . This is unlike linear regression where the equations to find the constants of the model are simultaneous but linear. In general, iterative methods (such as the Gauss-Newton iteration method, Method of Steepest Descent, Marquardt's Method, Direct search, etc) must be used to find values of a and b .

However, in this case, from Equation (4a), a can be written explicitly in terms of b as n

yi e  i 1

bxi

a n

(5)

i 1e2bx

i

Substituting Equation (5) in (4b) gives n

bxi bxi  i 1y ie y x e   n x e2bxi  0 n

i i

i 1

n

i

i 1e

2bxi

i 1

This equation is still a nonlinear equation in terms of b , and can be solved best by numerical methods such as the bisection method or the secant method.

You can now show that these values of of a and b, correspond to a local minimum, and since the above nonlinear equation has only one real solution, it corresponds to an absolute minimum.

5. There is a functional relationship between the mass density p of air and the

altitude h above the sea level.

0.32 0.64 1.28 1.60 Altitude above sea level, h (km) 1.15 1.10 1.05 0.95 Mass Density,  (kg/m3) k 2 h In the regression model 1  k e , the constant k is found as k  0.1315 .

2 Assuming the 2 mass density of air at the top of the atmosphere is 1/1000th of the mass density of air at sea level.

The altitude in kilometers of the top of the atmosphere most nearly is (A) 46.2 (B) 46.6 (C) 49.7 (D) 52.5 Solution The correct answer is (D).

Note to the student: See the alternative answer given later as that is quite a bit shorter.

Since k 2  0.1315 is given, the sum of the square of the residual is 2

n

Sr  



i

− k1e

0.1315hi

)

i 1

First we need to find the value of the constant k1 . Sr n 2  k e0.1315hi  e0.1315hi  0  1 k  i i  1

1

n

n

   ie0.1315hi  k1  e  20.1315hi  0 i 1

i 1

Thus ,

n

 ie h

0.1315 i

k1  i 1n

i 1e0.263h

i

Sinc e

n4 n

  ie

0.1315hi

 1.15e0.13150.32  1.10e0.13150.64  1.05e0.13151.28  0.95e0.13151.60

i 1

 1.15  0.95879  1.10  0.91928  1.05  0.84508  0.95  0.81026  3.7709

n

i 1e0.263hi  e

0.2630.32

 e0.2630.64  e0.2631.28  e0.2631.60

 0.91928  0.84508  0.71417  0.65652  3.1351 the value of the k1 is constant k  3.7709 3.1351  1.2028

1

Henc e   k1 ek2 h  kg m3 1.2028e0.1315 h

sea level =1.2028e 0.13150

 1.2028 kg m3 1 top   100 sea level 0  1 1.2028 1000  0.0012028 kg m3 top  k e0.1315htop 1

e0.1315ht  op

0.0012028 1.2028

ln0.001 htop   0.1315  52.530 km

Alternative Answer: Note to the student: Do we really need to find k1 for this problem?   k 1e0.1315h  sea level = k1e 0.13150

 k1  top = k1e 0.1315htop



sea level

 top



k

1 0.1315htop

k1 e

sealevel  1 0.1315htop 1 sealev e 100 el 0 1  ln1000  h   top  0.1315  52.530 km

6. A steel cylinder at 80° F of length 12" is placed in a commercially available

liquid nitrogen bath (315 F) . If the thermal expansion coefficient of steel behaves as a second order polynomial function of temperature and the polynomial is found by regressing the data below, Temperature, T (°F)  320  240 160  80 0 80

Thermal expansion Coefficient,  (  in/in/°F) 2.76 3.83 4.72 5.43 6.00 6.47

the reduction in the length of the cylinder in inches most nearly is (A) 0.0219 (B) 0.0231 (C) 0.0235 (D) 0.0307 Solution The correct answer is (C).

We are fitting the above data to the following polynomial.   a aT a T2 0

1

2

Sr  i  a0  a1 Ti  a2 iT 2 

2

There is a quadratic relationship between the thermal expansion coefficient and the temperature, and the coefficients a0 , a1 , and a2 are found as follows Sr a0 Sr a1 Sr a2

 n 2  a  a T  a T 2  1  0 i

0

1 i

2 i

i1

 n 2  a  a T  a T 2  T   0 i

0

1 i

2 i

i

i1

 n 2  a  a T  a T 2  T 2   0 i

i1

0

1 i

2 i

i

which gives

 

n

  

 nT

n   T2

 i

 i     i    i1  a0   i1

 i1 

 n



n     n T  T  3 2  T T     n n   i   i     ia  1  i i   a  i1     i1  i1 i1  n n     2   n 2   n  

2 3    Ti    T i 

  Ti 4     i1

 i1  

 Ti i    i1



i1

Table 1 Summations for calculating constants of model.  (in/in/oF) T 2 i T (oF) T3 6.4700  10–6 6.4000  5.1200  105 1 80 103 2 0 6.0000  10–6 0.0000 0.0000  80 5.4300  10–6 6.4000  –5.1200  3 3 10 105 160 4 4.7200  10–6 2.5600  –4.0960  4 10 106 3.8300  10–6 5.7600  –1.3824  5  240 104 107  320 2.7600  10–6 1.0240  –3.2768  6 105 107 6

 i 1

 7.2000  102 2.9210  10–5 1.9840  105

Table 1 (cont) i T4 4.0960  1 107 2 0.0000 4.0960  3 107 4 6.5536 

T  T 2  5.1760  10– 4.1408  10– 4

2

0.0000 –4.3440  10–4 –7.5520 

0.0000 3.4752  10– 2

1.2083  10–

 5.0688  107

5 6 6



i 1

108

10–4

1

3.3178  109 1.0486  1010 1.4541  1010

–9.1920  10–4 –8.8320  10–4 –2.4744  10–3

2.2061  10–

We have 

1

2.8262  10– 1

7.0022  10– 1

 2.9210 105 

 7.2000 102

1.9840 105 a 

 7.2000  102

1.9840 105

 5.0688107 a   3   1    2.4744 10  

 1.9840 105

 5.0688107

1.45411010 a

6

0

2



 7.0022 101 

Solving the above system of simultaneous linear equations, we get

6   6.0238 10  

a0   

9

6.3319 1011 a1   1.1965 a 10  2   The polynomial regression model is α  a aT a T2 0

1

2

 6.0237 106  6.3375109T 1.1942 1011T 2 Sinc e

Tfluid

L  L0   dT Troom

315

 12  ( 6.0237 106  6.3375109T 1.1942 10) 11T 2 dT 80

  12  6.0237 106T  

6.337510 9

2

T2 

1.1942 1011

315

 T  80 3

3

12 3  12  6.0237 106T  3.1687 109T 2  3.9807 10 T  80  12  6.0237 106  315 3.1687 109  3152  3.9807 1012  3153  12  6.0237 106 80 3.1687 109 802  3.9807 1012 803  315

 12  1.4586 103  5.0014 104   0.023505"

Problem Set#1 Multiple Choice Test Chapter 01.07 Taylors Series Revisited COMPLETE SOLUTION SET 1. The coefficient of x5 term in the Maclaurin polynomial for sin2x is

the (A) 0 (B) 0.008333

3 (C) 0.016667 (D) 0.26667

Solution The correct answer is (D).

The Maclaurin series for sin(2x) is 5 2x +2x +  sin(2x)  2x 3 5!   3!

32x5   2x  8x 3   120 6  2x  1.3333x3  0.26667x5  Hence, the coefficient of the x 5 term is 0.26667.

2. Give f 3  f 3

n

6,

 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) 38.00 0 (B) 79.50 0 (C) 126.0 0 (D) 331.5 0 Solution The correct answer is (C).

The Taylor series is given by f  x 2 f x 3 f x  h  f x  f xh  h  h  2!

3! x  3, h  7  3  4 f  3 2 f 3 3 f 3  4  f 3  f 34  4  4  2!

3! f  (3) 42 f (3) 3 4 … f 7  f (3)  f 3!  (3)4  2!

Since all the derivatives higher than second are zero, f  3 2 4 f 7  f 3  f 34  2!

11  6  8  4  42 2!  126

3. Given

that

yx is the solution to

dy 2, dx

 y3 

y(0)  3 the value of

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 y x 2 yx  h  yx  yxh  h 2! x  0 , h  0.2  0  0.2 y 0 2 y0  0.2  y0  y0 0.2  0.2 2! y0.2  y0  y0 0.2  y0 0.02 y 0  3 yx  y 3  2 y0  33  2  29 dy y x  3y 2 dx 2  3y y 3  2 y0  332 33  2  783 y0.2  3  29  0.2  783  0.02  24.46

y0.2 from a



4. The

series

2n

x n 4 is a Maclaurin series for the following function n0  (2n)!

1n

(A) cosx (B) cos2x (C) sinx (D) sin2x Solution The correct answer is (B).  x 2n   cosx   2n! n0 1n 

    n 2x2n cos 2x  1



2n!

n0

   n0

   n0

2n

2n

n

2n

n 2 x  1 2n!

n 4 x 1 2n!

x

5. The function erf x  

2

2

 et dt is called the error function. It is used in the

field of 0

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) 0.9953 2 (C) 1.533 0 (D) 2.858 6 Solution The correct answer is (A).

Rewrite the integral as x 2 t 2 erf x   e dt 0



The first three terms of the Taylor series for

f t  

2



f t   2 et

2



f 0 2 0 e  

2

2   f t 2 et  2t  2



2

  f 2 et  20 

0 

2 e 2 (− 2t )(− 2t ) + 2 e −t   −t

2

2 e− 2(− 2(0))(− 2(0)) + 2   0

2

e t around t  0 are

0 f  t  

 2

f  0 

e0  2



2

4



The first three terms of the Taylor series are f  (x) h 2 f (x  h) f (x)  f 2! (x)h  

f  (0) h 2 2!

f (0  h)  f (0)  f (0)h 

f  (0) h 2 2!

f (h)  f (0)  f (0)h 

4 h2!2

 2  0(h)   2

 2  4 h 

 2!

 2 

2 h2



2

2  2

2

eh   2

 

2

x   2 er x

f



h 2 , or

2  2 x2

e x  

Henc e



0 



2

 t 2 dt  

3 x  2  t 2 = t−     3 0  2 x  2 x3   3

erf 2 

2 2 2 23  3   0.75225

Note: Compare with the exact value of erf 2

6. Using the remainder of Maclaurin polynomial of nth order for f x defined as

R n x  

xn1 n  f 1!

n1

c,

n 0,

0c x

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). xn1 n1 R n x   c, n  0 , 0  c  x (n  1)!  f n1

 = 0.1 f n 1   

Rn 0.1

(n  1)!

c , n  0, 0  c  0.1

Since derivatives of f x are simply sinx and cosx, and sinx  1 and cosx  1 f n1 c  1 n1   0.1 1  Rn 0.1 (n  1)! 0.1n1  (n  1)!

So when is Rn0.1  106

0.1 n1

(n  1)! n4

 106

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