Change of sign method Essay
Change of sign method
I use the same method to keep doing decimal research. to work out a more accurate answer. Take increments in size 0. 1 within the interval [1. 5,1. 6] From the above we can see that must lie between [1. 52,1. 53]. It can be very clear if I use graph. I will do a better research this time by using the same method to prove my result is correct. Now I can find a more accurate result from the research which lies between [1. 521,. 1522]. Here is a graph to prove the interval is right. However, I will take 4 decimal places to improve the accuracy of the interval.
Also I will use the same method again. As we can see the root is between [1. 5213,1. 5214] Same again. Autograph is used to prove my solution is right. Error bounds This is the process which check how the accuracy of the roots are. From those 4 decimal search I have done so far, I can say that the answer is between 1. 5213 and 1. 5214. These can be improved the accuracy. Assume X=1. 5213 f(x)=(1. 5213)^3-1. 5213-2=-0. 00047 X=1. 5214 f(x)=(1. 5214)^3-1. 5214-2=0. 000121 Because the answer is -0. 00047<0<0. 000121. So the answer must between 1. 5213 and 1.5214.
However , these are not the exact answer so I have to estimate them. In this case, X=1. 5213. 5, so the error bound is . Because this is the middle point between the interval. Fail example by using Exel It is not guaranteed to use this method, because there still has some problems in it. See the graph below: As we can see the curve touches the x axis. The root lies between 0 and 1. I am going to use Exel program to prove it. There is no change of sign of this equation. So we can say that the change of sign method is failed. Newton-Raphson method
This is another fixed point estimation method, and as for the previous method it is necessary to use an estimate of the root as a starting point. The process can be repeated to give a sequence of points x2, x3… I am going to use the following equation. As we can see there are 2 roots in this function. The first root lies close to +1. But I will estimate the first root is x1 = +2. I will show it in graphical as +2 is a starting point. There is a technical way to do Newton-Raphson method by using Autograph. I will do it step by step with showing the graph.
I click the curve then right click it and chook the “Newton Raphson Iteration” option. I have entered the value that I estimated, then press the right side button. The solutions appear automatically. The answer that I got is 1. 27202. Error bound Because my solution is 5. bp. So the answer will be x=1. 27202 The numbers that I squared shows how close to the real answer. So we can say there are some error in it. I am going to try another root of the equation. I have estimated the x1 = -2. As I can see from the graph, -1. 27202 is the best answer I can get. Then I will check whether the solutions are correct.
Subject: Critical method,
University/College: University of Chicago
Type of paper: Thesis/Dissertation Chapter
Date: 10 July 2017
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