At 5:16, the last equation you see, subtract 1 from both sides. Put both sides back in rational form. Then solve for x(i+1). The expressions you are getting are right; the expressions I am showing are to reconcile it with the approximation to Newton Raphson Method.
@boganix11 жыл бұрын
i wish this man was my maths teacher...i learnt so much in 6:59 minutes than i ever did through 4 years of geometry in school! My maths teacher was lousy to say the least!
@numericalmethodsguy14 жыл бұрын
@divinenuker Thanks for your comments, The mistake pointed out is not a mistake. x(i+1) is used as the notation because it is the next guess after x(i-1) and x(i). Whether it turns out to the left or right of the initial guesses does not affect what the next subscript should be. The subscripts are used to keep track of the new found guess and what the next two guesses should be for the next iteration.
@benjaminjohn575 жыл бұрын
This video made me achieve my dream. Thank you sir for that be blessed
@numericalmethodsguy14 жыл бұрын
@divinenuker Try a problem like x raised to power 2-4-0 and use x(0)=9 and x(1)=6 as the initial guesses. That will not affect how the algorithm works even when I am choosing x(1) less than x(0).
@numericalmethodsguy14 жыл бұрын
@numericalmethodsguy divinenuker - Sorry your comment got deleted which was as follows: There is a trivial but I believe there was a tiny mistake in identifying the zero of f(x) with x(i+1), when an increasing x value is labeled as x(i-1). I know what you meant but for people who perhaps were confused. Brilliant derivation btw... can't believe how much you actually understand the method if you learn the derivation! Thanks!
@hasanalihasanali91585 жыл бұрын
I need just a help. The question can be asked by given negative intervals?
@numericalmethodsguy5 жыл бұрын
Yes, you can any numbers (negative or positive or both) for the two initial guesses. Convergence is not guaranteed for secant method though!
@Malik_MKhan6 жыл бұрын
Sir it was really helpful, but I have a doubt, According to graph when you started derivation with initial guess, the root was not lying in [(x_i-1),(x_i)], so how can you start with such an interval in which root not lies. Please correct me if I am wrong , in examples when we find approximate root then our initial guesses will be in such a way (graphically as well as mathematically) that the function value is positive at one initial guess and is negative at other initial guess,but in this derivation the root is not graphically seem to lies between initial guesses. I am just a beginner in this subject so if I asked some thing wrong then I am sorry for that and thank you very much for this informative lecture.
@numericalmethodsguy6 жыл бұрын
Secant method is an open method and hence the initial guesses or future estimates do not have to bracket the root. On the other hand, another numerical method of solving nonlinear equations called the false position method, which is a modified bisection method, also uses a secant line to find roots. The false position method is, in fact, a bracketed method - the upper and lower limits of the bracket do have to bracket the root. For more details, see mathforcollege.com/nm/topics/false_position.html
@Malik_MKhan6 жыл бұрын
Thank you very much sir , I was really confused in "Regula Falsi and Secant methods ". Because both have same formula for next iterate, but now i understand it.
@numericalmethodsguy12 жыл бұрын
There is no requirement for x(i+1) to be more than x(i). The argument (i+1) just stands for the (i+1)th guess after the ith guess. There is no requirement for x(i-1) to be less than x(i) either.
@bashirahmad85592 жыл бұрын
Sir please also explain householder method
@Fatih-hy7jj4 жыл бұрын
Is it okay to draw the second secant from x_(i+1) to x_(i-1) instead of x_i? Thanks.
@profautarkaw4 жыл бұрын
Why not, but wait? Numerical methods need to be programmed and have to follow some iterative logic here. When you have x(i-1) and x(i), you get x(i+1), then you increment i by 1, and you use x(i-1) and x(i) to get x(i+1), you see how it goes. Start with x1 and x2 to give you x3, then x2 and x3 give you x4, then x3 and x4 give you x5, and so on!
@Fatih-hy7jj4 жыл бұрын
@@profautarkaw thanks. I would like to ask about false position method. What is the difference between it and the secant method? By the way, can we find complex roots using secant method or FPM?
@Fatih-hy7jj4 жыл бұрын
Is the difference only about the signs of first two approximations? If so, do we go with keeping signs opposite for the other f(x) approximations? And lastly, what do we do when the root is a double root?
@numericalmethodsguy4 жыл бұрын
@@Fatih-hy7jj False position method is a closed method (root estimates have to bracket the root) - secant method is an open method (root estimates do not have to bracket the root). False position method also uses a secant "line" but to improve bisection method. The two estimates of the root in that case always bracket the root. See nm.mathforcollege.com/topics/false_position.html An extension of secant method called the Muller's method may help in finding complex roots, but not secant method.
@numericalmethodsguy4 жыл бұрын
@@Fatih-hy7jj The sign of the functions or the sign of the estimates of the roots has no bearing on the method. They can be same or of opposite signs. When the root is a double root, you follow the secant method as usual. You can come up with improvements though as given here math.oregonstate.edu/~gibsonn/Teaching/MTH351-001W12/Supplements/King.pdf
@ironman-bz7wx3 жыл бұрын
sir how that point A can be x+1 ..it should be lesser than x-1 right
@numericalmethodsguy3 жыл бұрын
it is not x+1 and x-1. It is x(i+1) and x(i-1). The elements of the vector of root estimates, x will end up less or more than the previous estimate. x(i+1) is used as the notation because it is the next estimate after x(i-1) and x(i). Whether it turns out to the left or right of the initial estimate on the x-axis does not affect what the next subscript should be. The subscripts are used to keep track of the newfound estimate and what the next two estimates should be for the next iteration.
@ironman-bz7wx3 жыл бұрын
@@numericalmethodsguy thank u so much sir..i got it..😃
@computersciencebasis60514 жыл бұрын
Stay Blessed. Hitting the bulls eye!!!.. I am not good with math background, still, i would ask how x of i+1 is formed. It was supposed to be x of i-2 right?. i mean i, i -1, i -2 would be a right sequence. But we use i, i-1, i+1. How is that possible?
@erikramirez26334 жыл бұрын
could someone explain how to go from triangle equations to the secant method equation the algebra steps
@profautarkaw4 жыл бұрын
Here is the blog that reconciles it. autarkaw.org/2013/10/01/reconciling-secant-method-formulas/
@puneettripathi7406 жыл бұрын
Sir video was really helpful thank you very much.
@muwongeevanspaul91662 жыл бұрын
Explain your thing well
@numericalmethodsguy2 жыл бұрын
Is this a compliment or a complaint? Look at this blog.autarkaw.com/2013/10/01/reconciling-secant-method-formulas/ and go here for more details and examples: nm.mathforcollege.com/chapter-03.05-secant-method/
@Taiseerghulam201111 жыл бұрын
im an instructor in Yemen ur videos helps me a lot thank u question while algebric simplifications i did not get the same rule ??? i got xi+1 = (f(xi-1)xi - f(xi)xi-1) ) / ( f(xi-1)- f(xi) ) ????
Newton's method and secant method are closely related, aren't they?
@numericalmethodsguy10 жыл бұрын
Yes, they are closely related. The derivative of the function in Newton Raphson method is replaced by the backward divided difference approximation of the derivative of the function in secant method. It is less convergent than Newton method but does not require symbolic derivation of the derivative of the function. Go to the About section of the video and you will see a link for more resources explaining all this!
I put a blog about it. autarkaw(dot)wordpress(dot)com/2013/10/01/reconciling-secant-method-formulas/
@numericalmethodsguy11 жыл бұрын
At 5:16, the last equation you see, subtract 1 from both sides. Put both sides back in rational form. Then solve for x(i+1). The expressions you are getting are right; the expressions I am showing are to reconcile it with the approximation to Newton Raphson Method.