We are here with you hands in hands to facilitate your learning & don't appreciate the idea of copying or replicating solutions. Read More>>
+ Link For Assignments, GDBs & Online Quizzes Solution |
+ Link For Past Papers, Solved MCQs, Short Notes & More |
Maximum Marks: 30
Due Date: 12 May, 2014
INSTRUCTIONS
Please read the following instructions before attempting the solution of this assignment:
• Upload assignments properly through LMS, No Assignment will be accepted through email.
• Write your ID on the top of your solution file.
Question: 1 Marks: 10
The roots of the following equation are and. Using Muller method approximates the root upto four decimal places.
Question: 2 Marks: 10
Using Newton-Raphson method approximates the root of the following equation in the interval upto the three decimal places.
Note: All the calculations should be in radian.
Question: 3 Marks: 10
Using Regula Falsi method approximates the root of the following equation upto four decimal places.
Note: The following graph of above equation can be very helpful in finding the interval of the root. So, I am not mentioning the interval there as mentioned above. So, try to find the appropriate interval then apply the method. Otherwise, the convergence would be very slow.
Tags:
+ How to Follow the New Added Discussions at Your Mail Address?
+ How to Join Subject Study Groups & Get Helping Material? + How to become Top Reputation, Angels, Intellectual, Featured Members & Moderators? + VU Students Reserves The Right to Delete Your Profile, If?.
+ http://bit.ly/vucodes (Link for Assignments, GDBs & Online Quizzes Solution)+ http://bit.ly/papersvu (Link for Past Papers, Solved MCQs, Short Notes & More)
+ Click Here to Search (Looking For something at vustudents.ning.com?) + Click Here To Join (Our facebook study Group)muller method konsa ha
Muller's method is a generalization of the secant method. Instead of starting with two initial values and then joining them with a straight line in secant method, Mullers method starts with three initial approximations to the root and then join them with a second degree polynomial (a parabola), then the quadratic formula is used to find a root of the quadratic for the next approximation. That is if x_{0}, x_{1} and x_{ 2} are the initial approximations then x_{3} is obtained by solving the quadratic which is obtained by means of x_{0}, x_{ 1} and x_{2}. Then two values among x_{0}, x_{1} and x_{2} which are close to x_{3}are chosen for the next iteration
x_{3} = x_{2} + z ( * )
where z = | -2c |
b±Ö( b^{2}-4ac) |
a = D_{1}/D_{2} , b = D_{ 2 }/D and c = f(x_{2})
D = h_{0}h_{1}(h_{0}-h_{1}) , D_{1} = (f_{0}-c)h_{1}-(f_{1}-c)h_{0 }, D_{2} = (f_{1} -c)h_{0}^{2} - (f_{0}-c)h_{1}^{2}
h_{0} = x_{0}-x_{2 }, h_{1 }= x_{1}-x_{2}
Given an equation f(x) = 0
Let the initial guesses be x_{0}, x_{1 }and x_{2}
Let x_{i} = x_{2}, x_{i-1} = x_{1 }and_{ }x_{i-2} = x_{0}
Compute f(x_{i-2}), f(x_{i-1}) and f(x_{i})
_{Do}
Compute
h = x - x_{i}, h_{i} = x_{i} - x_{i-1} and h_{i-1} = x_{i-1} - x_{i-2}
l_{i} = h_{i} / h_{i-1}, d_{i} = 1 + l_{i}
g_{i} = l_{i}^{2} f_{i-2} - d_{i}^{2} f_{i-1} + ( l_{i }+ d_{i} ) f_{i}
c_{i} = l_{i }(l_{i} f_{i-2} - d_{i} f_{i-1} + f_{i} )
l_{i+1} = min ( -2d_{i} f_{i} / (g_{i }± Ö(g_{i}^{2}- 4d_{i} f_{i}c_{i}) ) )
(take minimum in absolute sense)
Then x_{i+1} = x_{i} + ( x_{i } - _{ }x_{i-1 }) l_{i+1} , i = 2, 3, 4, . . .
Please Discuss here about this assignment.Thanks
Our main purpose here discussion not just Solution
We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.
koi es asgnmnt ka sol nhiiiiiiiiiiiiiiiiiiiiii day ga kya
Numerical methods for solving either algebraic or transcendental equation are classified
into two groups
Direct Methods:
Those methods which do not require any information about the initial approximation of
root to start the solution are known as direct methods.
(All these methods do not require any type of initial approximation.)
Iteratative Methods:
These methods require an initial approximation to start.
neton raphson method ka solution k ly b guid kr dain
What is root?
x + 7= 0 is a linear equation
x = -7 -7 is root of equation x+7= 0
x-3=2 is a linear equation
x=5 5 is root of equation x-3=2
Regula-Falsi method (Method of false position)
Here we choose two points xn and n 1 x − such that 1 ( ) ( ) n n f x and f x− have opposite signs.
Intermediate value property suggests that the graph of the y=f(x) crosses the x-axis
between these two points and therefore, a root lies between these two points.
© 2019 Created by + M.Tariq Malik. Powered by
Promote Us | Report an Issue | Privacy Policy | Terms of Service
VU Students reserves the right to delete any profile, which does not show any Activity at site nor has not any activity more than 01 month.
We are user-generated contents site. All product, videos, pictures & others contents on vustudents.ning.com don't seem to be beneath our Copyrights & belong to their respected owners & freely available on public domains. We believe in Our Policy & do according to them. If Any content is offensive in your Copyrights then please email at m.tariqmalik@gmail.com or Contact us at contact Page with copyright detail & We will happy to remove it immediately.
Management: Admins ::: Moderators
Awards Badges List | Moderators Group
All Members | Featured Members | Top Reputation Members | Angels Members | Intellectual Members | Criteria for Selection
Become a Team Member | Safety Guidelines for New | Site FAQ & Rules | Safety Matters | Online Safety | Rules For Blog Post