

Re: MTH603 Assignment 2 Solution and Discussion
Assignment NO. 2 MTH603 (Fall 2020)
Maximum Marks: 20
Due Date: February 15, 2021DON’T MISS THESE: Important instructions before attempting the solution of this assignment:
• To solve this assignment, you should have good command over 23  35 lectures.
Try to get the concepts, consolidate your concepts and ideas from these questions which you learn in the 2335lectures.
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Also remember that you are supposed to submit your assignment in Word format any other like scan images etc. will not be accepted and we will give zero mark corresponding to these assignments.Question 1:
Obtain the value of using Simpson’s 1/3 rule correct to 3 decimal places.
MARKS 10
Question 2:Using Newton’s divided difference formula, find the quadratic equation for the following data:
X 1 2 5
Y 8 14 44Hence find y(3). MARKS 10



Re: MTH603 Assignment 1 Solution and Discussion
Question #1: Find the root of the equation x^3+x^2+x1 =0 correct to two decimal places by using bisection method.
Question #2: Solve the system of linear equations with the help of Gaussian elimination method.
2x + y + z = 9;3x −2y + 4z = 9;x +y2z = 3Solution File

Question 1:
Convert the decimal number 80 into its binary equivalent.
Question 2:
Convert the binary number 2 (11001100) to its decimal equivalent.
Question 3:
Find the relative error when 17 is considered upto four decimal places.
Question 4:
Find the interval in which atleast one root of the equation 3 2 xx x 2 10 lies.
Question 5:
Find the real root of the equation 4 x x 10 0 in the interval [1, 2] by bisection method upto
two iterations. 
MTH603_Final_Term_(GIGA_FILE_by_Ishfaq_V11.02.02).pdf
MTH603 Spring_2010_FinalTerm_OPKST_.doc
MTH603 solvedbyAtifAli…doc
MTH603 finalterm paper 2.doc
MTH603 finalterm paper 1.doc
MTH603 final.pdf MTH603 Final.doc MTH603  Final Term Papers.pdf
MTH603  Final Term Papers 02.pdf
MTH603 Final term papers in one file.pdf 
Assignment NO. 2 MTH603 (Spring 2020)
Maximum Marks: 20 Due Date: August 13, 2020
DON’T MISS THESE: Important instructions before attempting the solution of this assignment:
• To solve this assignment, you should have good command over 23  30 lectures.
Try to get the concepts, consolidate your concepts and ideas from these questions which you learn in the 2330 lectures.
• Upload assignments properly through LMS, No Assignment will be accepted through email.
• Write your ID on the top of your solution file.
Don’t use colourful back grounds in your solution files.
Use Math Type or Equation Editor Etc. for mathematical symbols.
You should remember that if we found the solution files of some students are same then we will reward zero marks to all those students.
Try to make solution by yourself and protect your work from other students, otherwise you and the student who send same solution file as you will be given zero mark.
Also remember that you are supposed to submit your assignment in Word format any other like scan images etc. will not be accepted and we will give zero mark corresponding to these assignments.Question :
Using difference operator formulas (Δ and ∇) and the values given in the table below,
x 0.3 0.5 0.7 0.9 1.1 1.3
y 3.9118 3.8234 3.6773 3.4807 3.2408 2.9648estimate the value of
y^' (0.3) Marks 10 y''(1.3) Marks 10 
Grand Quiz Total Questions : 30
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Re: MTH603 Assignment 1 Solution and Discussion
Question 1: Find the root of on equation f(x) =2coshx sinx1 taking initial value x0 = 0.4, using Newton Raphson Method. Convert Up to four decimal places.
Question 2: Evaluate √167 by Newton Raphson Method correct up to 4 decimal places.

Assignment NO. 1 MTH603 (Fall 2019)
Maximum Marks: 20 Due Date: 24 112019
DON’T MISS THESE: Important instructions before attempting the solution of this assignment:
• To solve this assignment, you should have good command over 01  12 lectures.
• Try to get the concepts, consolidate your concepts and ideas from these questions which you learn in the 01 to 12 lectures.
• Upload assignments properly through LMS, No Assignment will be accepted through email.
• Write your ID on the top of your solution file.
• Don’t use colourful back grounds in your solution files.
• Use Math Type or Equation Editor Etc. for mathematical symbols.
• You should remember that if we found the solution files of some students are same then we will reward zero marks to all those students.
• Try to make solution by yourself and protect your work from other students, otherwise you and the student who send same solution file as you will be given zero mark.
• Also remember that you are supposed to submit your assignment in Word format any other like scan images etc. will not be accepted and we will give zero mark corresponding to these assignments.Question #1: Find the root of the equation, Perform three iteration of the equation,
ln (x −1) + sinx = 0 by using Newton Raphson method.
Question #2: Solve the system of linear equations with the help of Gaussian elimination method.
x + y + z = 6; 2x − y + z = 3; x + z = 4
MTH603 Assignment 1 Solution and Discussion

Re: MTH603 Assignment 1 Solution and Discussion
Question #1: Find the root of the equation x^3+x^2+x1 =0 correct to two decimal places by using bisection method.
Question #2: Solve the system of linear equations with the help of Gaussian elimination method.
2x + y + z = 9;3x −2y + 4z = 9;x +y2z = 3Solution File

@zareen said in MTH603 Assignment 1 Solution and Discussion:
Question #2: Solve the system of linear equations with the help of Gaussian elimination method.
2x + y + z = 9;3x −2y + 4z = 9;x +y2z = 3System of Linear Equations entered :
[1] 2x + y + z = 9
[2] 3x  2y + 4z = 9
[3] x + y  2z = 3Solve by Substitution :
// Solve equation [3] for the variable y
[3] y = x + 2z + 3
// Plug this in for variable y in equation [1]
[1] 2x + (x +2z+3) + z = 9
[1] x + 3z = 6
// Plug this in for variable y in equation [2][2] 3x  2•(x +2z+3) + 4z = 9
[2] 5x = 15
// Solve equation [2] for the variable x[2] 5x = 15
[2] x = 3
// Plug this in for variable x in equation [1][1] (3) + 3z = 6
[1] 3z = 3
// Solve equation [1] for the variable z[1] 3z = 3
[1] z = 1
// By now we know this much :x = 3
y = x+2z+3
z = 1
// Use the x and z values to solve for y
y = (3)+2(1)+3 = 2Solution :
{x,y,z} = {3,2,1} 
@zareen said in MTH603 Assignment 1 Solution and Discussion:
Question #1: Find the root of the equation x^3+x^2+x1 =0 correct to two decimal places by using bisection method.
We now use the Bisection Method to approximate one of the solutions. The Bisection Method is an iterative procedure to approximate a root (Root is another name for a solution of an equation).
The function is F(x) = x3 + x2 + x  1
At x= 0.00 F(x) is equal to 1.00
At x= 1.00 F(x) is equal to 2.00Intuitively we feel, and justly so, that since F(x) is negative on one side of the interval, and positive on the other side then, somewhere inside this interval, F(x) is zero
Procedure :
(1) Find a point “Left” where F(Left) < 0(2) Find a point ‘Right’ where F(Right) > 0
(3) Compute ‘Middle’ the middle point of the interval [Left,Right]
(4) Calculate Value = F(Middle)
(5) If Value is close enough to zero goto Step (7)
Else :
If Value < 0 then : Left < Middle
If Value > 0 then : Right < Middle(6) Loop back to Step (3)
(7) Done!! The approximation found is Middle
Follow Middle movements to understand how it works :
Left Value(Left) Right Value(Right) 0.000000000 1.000000000 1.000000000 2.000000000 0.000000000 1.000000000 1.000000000 2.000000000 0.500000000 0.125000000 1.000000000 2.000000000 0.500000000 0.125000000 0.750000000 0.734375000 0.500000000 0.125000000 0.625000000 0.259765625 0.500000000 0.125000000 0.562500000 0.056884766 0.531250000 0.036590576 0.562500000 0.056884766 0.531250000 0.036590576 0.546875000 0.009502411 0.539062500 0.013703823 0.546875000 0.009502411 0.542968750 0.002140820 0.546875000 0.009502411 0.542968750 0.002140820 0.544921875 0.003670745 0.542968750 0.002140820 0.543945312 0.000762452 0.543457031 0.000689811 0.543945312 0.000762452 0.543457031 0.000689811 0.543701172 0.000036164 0.543579102 0.000326863 0.543701172 0.000036164 0.543640137 0.000145359 0.543701172 0.000036164 0.543670654 0.000054600 0.543701172 0.000036164 0.543685913 0.000009219 0.543701172 0.000036164 0.543685913 0.000009219 0.543693542 0.000013472 0.543685913 0.000009219 0.543689728 0.000002127 Next Middle will get us close enough to zero:
F( 0.543688774 ) is 0.000000710
The desired approximation of the solution is:
x ≓ 0.543688774
Note, ≓ is the approximation symbol
One solution was found :
x ≓ 0.543688774