Assignment Numec

Faculty of Mechanical Engineering

NOOR SYAHADAH BINTI YUSSOFF

2008407362

EMD5M1B

MEC 500: Numerical Method with Application

Assignment 01

Question 01

1a. Briefly describe round-off and truncation errors and measures to reduce these errors in

practice?

Answer: Round-off errors - Originate from the fact that computers retain only a fixed

number of significant figures during a calculation.

- Numbers such as π, e, or √ 7 cannot be expressed by a fixed

number of significant figures. Therefore, they cannot be

represented exactly by the computer.

Truncation errors - are those that result from using an approximation in place

of an exact mathematical procedure.

- was introduced into the numerical solution because the

difference equation only approximates the true value of the

derivatives.

1b. Briefly describe the terms true error, approximate error and stopping criterion.

Answer: i) True Error: True error,Et is the difference between the true value in a

calculation and the approximate value found using a numerical

method etc.

True Error, Et = True Value – Approximate Value (+/-)

ii) Approximate Error: Approximate error,Ea is defined as the difference between

the present approximation and the previous approximation.

Approximate Error (Ea) = Present Approximation – Previous

Approximation

iii) Stopping Criterion: Since an iterative method computes successive

approximations to the solution of a linear system, a practical

test is needed to determine when to stop the iteration. Ideally

this test would measure the distance of the last iterate to the

true solution, but this is not possible. Instead, various other

metrics are used, typically involving the residual.

A good stopping criterion should

1. identify when the error is small enough to stop,

2. stop if the error is no longer decreasing or decreasing too slowly

3. Limit the maximum amount of time spent iterating.

1c. Use zero –through third-order Taylor series expansions to predict f(6) for

f ( x )=x3+12 x2−100 x−6using a base point at x=1 . Compute the true percent relative

error, ε t, for each approximation. Comment on the results.

Answer:

f ( x )=x3+12 x2−100 x−6

x i=0 , h=1, x i+1=1

Zero order, f ( x ) : f ( x i+1 )≅ f ( x i )

f (1 )≅ f (0 )

≈−6 f (0 )=−6 , f (1 )=−93

True Error, Et = True Value – Approximation

= −6−¿(−93¿

= 87

True value, = Approximation + Error

= −93+ (−6 )

= −99

True percent Error, ε t = true errortrue value

× 100

=87

−99 ×100

= 87.7 %

First order, f ' ( x i ) = 3 x2+24 x−100

For n=1,

f ' (0 )=3¿

¿−100

f ( x i+1 )≅ f ( x i )+f ' ( x i ) h

≅−6−100 h

Substituting h=1 f ' (1 ) ≈−106


= −93−(−106)

= 13


= −106+13

= −93


× 100

=1393

×100

= 13.98 %

Second order, f ' ' ( x i ) = 6 x+24

For n=2 ,the second derivative is evaluated at x=0

f ' (0 )=6 (0 )+24

¿24

f ( x i+1 )≅ f ( x i )+f ' ( x i ) h+f ' ' ( x i ) h2

≅−6+ (−100 ) h+12 h2 Substituting

h=1 ≈−94


= −93−¿(−94 ¿

= 1


= −94+1

= −93

2


× 100

=1

−93 ×100

= −1.08 %

Third order, f ' ' ' ( x i ) = 6 x

For n=3 ,the second derivative is evaluated at x=0

f ' ' (0 )=6=6

f ( x i+1 )≅ f ( x i )+f ' ( x i ) h+f ' ' ( x i )

2!h2+

f ' ' ' ( x i )3 !

h3

≅−6+ (−100 )1+ 242 ×1

(1)2+ 63 ×2 ×1

(1 )3

Substituting h=1 ≈−93


= −93−¿(−93¿

= 0


= −93+0

= −93


× 100

=0

−93 ×100

= 0 %

The inclusion of the second and third derivatives results in exactly the same equation we started

with;

f ( x )=−6−100h+12h2 +h3

Where the remainder term;

R3 ¿f 4()4 !

h4=0 because the fourth derivative of a third-order polynomial is zero.

Consequently, the Taylor series expansion to the third derivative yields an exact estimate at

x i+1=1 ;

f (1 )=−6−100 (1 )+12 (1 )2 +(1 )3=−93

1d. Use forward, backward and centered difference approximations to estimate the first

derivative of f ( x )=x3+12 x2−100−6. Evaluate the derivative at x=2 using step size of

h=0.5. Compare the result with the true value of the derivative. Interpret the result on

the basis of the remainder term of the Taylor series expansion.

True Value,f (x) = x3+12 x2−100 x−6

f '( x) = 3 x2+24 x−100

f '(2) =3¿

= −40

Solution:

h=0.5

x i−1=1.5

x i=2

x i+1=2.5

f ( x i−1 )=x3+12 x2−100 x−6

=(1.5)3+12¿

= −125.625

f ( x i )=x3+12 x2−100 x−6

=(2)3+12¿

¿ −150

f ( x i+1 )=x3+12 x2−100 x−6

=(2.5)3+12¿

= −165.375

Forward, f ' ( x i )=f ( x i+1)−f (x i)

h+0(h)

¿(−165.375 )−(−150)

0.5+0 (0.5)

¿ −30.75

True percent Error, |εt| = true errortrue value

× 100

=−40−(−30.75)

40 ×100

= 23.125 %

Backward, f ' ( x i )=f ( x i )−f (x i−1)

h+0 (h)

¿(−150 )−(−125.625)

0.5+0 (0.5)

¿ −48.75

True percent Error,|εt| = true errortrue value

× 100

=−40−(−48.75)

−40 ×100

= 21.875 %

Centered, f ' ( x i )=f ( x i+1)− f (x i−1)

2 h+0(h2)

¿(−165.375 )−(−125.625)

2(0.5)+0¿

¿ −39.75

True percent Error, |εt| = true errortrue value

× 100

=−40−(−39.75)

−40 ×100

= 0.625 %

To estimate the first derivative of f ( x )=x3+12 x2−100−6 , the centered difference

approximation will used to get the value more accurately.

Question 02

2a. In Finding roots of equations, briefly describe, how do we determine all the possible

roots of an equation?

One method to obtain an approximate solution is to plot the function and determine where it

crosses the x axis. This point, which represents the x value for which f(x) = 0, is the roots. Although

the graphical methods are useful for obtaining rough estimates of roots, they are limited because of

their lack of precision. Numerical Method such as fixed point iteration, The Newton-Raphson method,

bisection method and so on, is an alternative approach that be used trial and error.

2b. Roots finding using open methods such as Newton-Raphson and Secant methods may

Sometimes NOT converge to the true values. Describe the procedure for checking of the

possibility of convergence of open methods.

The procedure for checking of the possibility of convergence of open methods can be

depicted graphically. We graphed a function to visualize its structure and behavior. An alternative

graphical approach is to separate the equation into two component parts, can be plotted separately.

The x values corresponding to the intersections of these functions represent the roots of f(x) = 0.

2c. Newton-Raphson formula given below is based on first order Taylor expansion of the

function f(x) about xi . Derive the Newton-Raphson formula using the second order

Taylor expansion of the function f(x) about xi.

Answer:

Let F ( x )=0

For the derivation of the formula used for solving a one-dimensional problem, we simply make a first-

order Taylor series expansion of the function F(x)

F (x)=F (x )+hF ’ (x) eq.1

Use the following notation for the x-values:

xi= x

x i+1=x+h eq.2

Substitute eq.1 in eq.2

F (x)=F (x i)+( x i+1−x i) F ’ (x i)

When x=0 ,

F (0)=F (x i)+( x i+1−x i)F ’( xi)

Then,

x i+1=x i−f (x i)f '(x i)

2d. Using incremental search, prove that function f ( x )=6 x3−26 x2+32 x−10 has

three roots between x=0 andx=3

Initial guess for this function is x i=0 and xu=3

First iteration;

xr 1=x i+xu

2

¿0+3

2

¿1.5

f ( x i 1 )=f (0 )=6 (0 )3−26 (0 )2+32 (0 )−10=−10

f ( xr 1)=f (1.5 )=¿ 6 (1.5 )3−26 (1.5 )2+32 (1.5 )−10=−0.25

f ( x i ) ∙ f ( xr 1 )=2.5 (¿0 ) Root is in upper interval

Second iteration;

xr 2=0+1.5

2

¿0.75

f ( x i 2 )=f (0 )=6 (0 )3−26 (0 )2+32 (0 )−10=−10

f ( xr 2)=f (0.75 )=¿ 6 (0.75 )3−26 (0.75 )2+32 (0.75 )−10=1.90625

f ( x i 2 ) ∙ f ( xr 2 )=−19.0625 (¿0 ) Root is in lower interval

ε a=0.75−1.5

0.75 ×100 %

¿−100 %

Third iteration;

xr 3=0.75+1.5

2

¿1.125

f ( x i 3 )=f (0.75 )=6 (0.75 )3−26 (0.75 )2+32 (0.75 )−10=1.90625

f ( xr 3)=f (1.125 )=¿ 6 (1.125 )3−26 (1.125 )2+32 (1.125 )−10=1.637

f ( x i 3 ) ∙ f ( xr 3 )=3.121 (¿0 ) Root is in upper interval

ε a=1.125−0.75

1.125 ×100 %

¿33.33 %

Fourth iteration;

xr 4=1.125+1.5

2

¿1.3125

f ( x i 4 )=f (1.125 )=¿ 6 (1.125 )3−26 (1.125 )2+32 (1.125 )−10=1.637

f ( xr 4 )=f (1.3125 )=6 (1.3125 )3−26 (1.3125 )2+32 (1.3125 )−10=0.777

f ( x i 4 )∙ f ( xr 4 )=1.2712 (¿0 ) Root is in upper interval

ε a=1.3125−1.125

1.3125 ×100%

¿14.28 %

sign change in calculation, so there is the root between x=0 andx=3.

2e. Use MATLAB to graphically locate all roots of function f ( x )=6 x3−26 x2+32 x−10.

Clearly indicate all roots on the plot of the graph.

2f. Use Bisection method, conduct four iterations with initial guesses of x i=2.0 and xu=3.0

to locate the highest root of the function and compute the percentage approximation relative

error.

Problem statement. Use bisection method with four iterations with initial guesses of x i=2.0

and xu=3.0 to locate the highest root of the function.

First iteration;

xr 1=x i+xu

2

¿2.0+3.0

2

¿2.5

f ( x i 1 )=f (2.0 )=6 (2.0 )3−26 (2.0 )2+32 (2.0 )−10=−2

f ( xr 1)=f (2.5 )=¿ 6 (2.5 )3−26 (2.5 )2+32 (2.5 )−10=1.25

f ( x i ) ∙ f ( xr 1 )=−2.5 (¿0 ) root is in lower interval

Second iteration;

xr 2=2.0+2.5

2

¿2.25

f ( x i 2 )=f (2.0 )=6 (2.0 )3−26 (2.0 )2+32 (2.0 )−10=−2

f ( xr 2)=f (2.25 )=¿ 6 (2.25 )3−26 (2.25 )2+32 (2.25 )−10=−1.28125

f ( x i 2 ) ∙ f ( xr 2 )=2.563 (¿0 ) root is in upper interval

ε a=2.25−2.5

2.25 ×100%

¿−11.1%

Third iteration;

xr 3=2.25+2.5

2

¿2.375

f ( x i 3 )=f (2.25 )=6 (2.25 )3−26 (2.25 )2+32 (2.25 )−10=−1.281

f ( xr 3)=f (2.375 )=¿ 6 (2.375 )3−26 (2.375 )2+32 (2.375 )−10=−0.2773

f ( x i 3 ) ∙ f ( xr 3 )=0.3552 (¿0 ) root is in upper interval

ε a=2.375−2.25

2.375 ×100%

¿5.26 %

Fourth iteration;

xr 4=2.375+2.5

2

¿2.4375

f ( x i 4 )=f (2.375 )=¿ 6 (2.375 )3−26 (2.375 )2+32 (2.375 )−10=−0.2773

f ( xr 4 )=f (2.4375 )=6 (2.4375 )3−26 (2.4375 )2+32 (2.4375 )−10=0.4165

f ( x i 4 )∙ f ( xr 4 )=−0.1155 (¿0 ) root is in lower interval

ε a=2.4375−2.375

2.4375 ×100%

¿2.56%

So, the highest root of the function is x=2.563

2g. Use the Newton-Raphson’s method and with an initial guess ofx0=2.0, try estimate the

highest root f ( x )=6 x3−26 x2+32 x−10. Give your comment on what happens and propose

solution to this problem?

Problem statement. Use the Newton-Raphson method to estimate the highest root of

f ( x )=6 x3−26 x2+32 x−10, employing an initial guess of x0=2.0. The first derivative of the

function can be evaluated as f ' ( x )=18x2−52 x+32 which can be substituted along with the original

function into

equation:

x i+1=x i−f (x i)f ' (x i)

to give:

x i+1=x i−6 x i

3−26 x i2+32 x i−10

18 x i2−52 x i+32

starting with an initial guess of x0=2.0, this iterative equation can be applied to compute:

i=0 , x0=2.0 ;

x1=x0−6 x0

3−26 x02+32 x0−10

18 x02−52 x0+32

x1=2.0−6 (2.0 )3−26 (2.0 )2+32 (2.0 )−10

18 (2.0 )2−52 (2.0 )+32

x1=2.0

i=1 , x1=2.0 ;

x2=x1−6 x1

3−26 x12+32 x1−10

18 x12−52 x1+32

x2=2.0−6 (2.0 )3−26 (2.0 )2+32 (2.0 )−10

18 (2.0 )2−52 (2.0 )+32

x2=2.0

Based on the calculation above, the value of x1and x2 was the same which is 2.0 if we start with initial

guess of 2.0. If we wrongly choose the initial guess, the problem cannot be settled. So, the best

solution to this problem is we start the estimating highest root with initial guess of x0=0.

i=0 , x0=0 ;

x1=x0−6 x0

3−26 x02+32 x0−10

18 x02−52 x0+32

x1=0−6 ( 0 )3−26 (0 )2+32 (0 )−10

18 (0 )2−52 (0 )+32

x1=0.3125

ε t=x1−x0

x1

×100 %

ε t=0.3125−0

0.3125 ×100 %

¿100 %

i=1 , x1=0.3125 ;

x2=x1−6 x1

3−26 x12+32 x1−10

18 x12−52 x1+32

x2=0.3125−6 (0.3125 )3−26 (0.3125 )2+32 (0.3125 )−10

18 (0.3125 )2−52 (0.3125 )+32

x2=0.4471

ε t=x2−x1

x2

×100%

ε t=0.4471−0.3125

0.4471 ×100%

¿30.11%

i=2 , x2=0.4471 ;

x3=x2−6 x2

3−26 x22+32 x2−10

18 x22−52 x2+32

x3=0.4471−6 (0.4471 )3−26 (0.4471 )2+32 ( 0.4471 )−10

18 (0.4471 )2−52 (0.4471 )+32

x3=0.4758

ε t=x3−x2

x3

×100 %

ε t=0.4758−0.4471

0.4758 ×100%

¿6.03%

i=3 , x3=0.4758 ;

x4=x3−6 x3

3−26 x32+32 x3−10

18 x32−52 x3+32

x4=0.4758−6 (0.4758 )3−26 (0.4758 )2+32 (0.4758 )−10

18 (0.4758 )2−52 (0.4758 )+32

x4=0.4770

ε t=x4−x3

x4

×100 %

ε t=0.4770−0.4758

0.4770 ×100%

¿0.25 %

i=4 , x4=0.4770 ;

x5=x4−6 x4

3−26 x42+32 x4−10

18 x 42−52 x 4+32

x5=0.4770−6 (0.4770 )3−26 (0.4770 )2+32 (0.4770 )−10

18 (0.4770 )2−52 (0.4770 )+32

x5=0.4771

ε t=x5−x4

x5

×100 %

ε t=0.4771−0.4770

0.4771 ×100%

¿0.02 %

2h. Use MATLAB’s fzero function to determine all roots of f ( x )=6 x3−26 x2+32 x−10.

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Assignment Numec