{Use of Tech} Modified Newton’s method The function f has a ro...

Question

{Use of Tech} Modified Newton’s method The function f has a root of multiplicity 2 at r if f(r) = f'(r) = 0 and f"(r) ≠ 0. In this case, a slight modification of Newton’s method, known as the modified (or accelerated) Newton’s method, is given by the formula xₙ + 1 = xₙ - (2f(xₙ)/(f'(xₙ), for n = 0, 1, 2, . . . . This modified form generally increases the rate of convergence.
b. Apply Newton’s method and the modified Newton’s method using x₀ = 0.1 to find the value of x₃ in each case. Compare the accuracy of these values of x₃.

b. Apply Newton’s method and the modified Newton’s method using x₀ = 0.1 to find the value of x₃ in each case. Compare the accuracy of these values of x₃.

Accepted Answer

Welcome back, everyone. In this problem, the function F of X equals X minus 9 X cub plus 20X2 as a root of multiplicity 2 at X equals 0. Apply both the traditional Newton's method and the modified Newton's method starting from X0 equals 0.5 to find X3. Which method provides a closer approximation to the root? We're told that the modified Newton's method for a multiplicity of 2 is XN + 1 equals XN minus 2 F X F XN divided by the derivative of F of XN. A says the modified Newton's method gives a closer approximation to the root at X equals 0 after 3 iterations. B says the traditional Newton's method gives a closer approximation. C says the traditional and modified Newton's methods result in equal root approximations after 3 iterations, and D says neither the traditional nor the modified Newton's method gives an accurate root approximation after 3 iteration methods. Now, in this problem we're basically comparing the modified method to the traditional method. What do we know about the traditional Newton's method? Well, recall that it basically says that. XN + 1 is going to be equal to XN minus F of XN. Divided by the derivative of F of XN. So it's like Newton's modified method. The only difference is that our quotient is not being multiplied by 2. So notice that in both of these we'll need the derivative of FF X and for our function FF X if we differentiate it with respect to X, the derivative is going to be equal to 4 X cubed, OK, X, the derivative of X to the 4th is 4 X cubed. Minus 27 X2 + 40 X. So now that we have the derivative, we know what the function is and we know that we can start from X0 for 0.5. Let's try to compute the first two iterations, that is X1, X2, and then X3 for both of Newton's methods and compare to see which one has a closer approximation, if any. Now, let's start by using uh Newton's method, OK? So now, buy a newt method. OK. Starting by finding X1. X1 is going to be equal to X0 minus F of X0 divided by the derivative of F of X0. We know that X is 0.5, so this is going to really be 0.5 minus F of 0.5, which is going to be 0.5 to the fourth, minus 9 multiplied by 0.5 to, let me fix that here, 0.5 to the third. Plus 20 multiplied by 0.5 squared, OK. Divided by the derivative of F of X not. So that's no wherever we see X in the derivative, we're going to put 0.5. So that's 4 multiplied by 0.5 cubed. Minus 27 multiplied by 0.5 squared, OK. Plus 40 multiplied by 0.5. No, when we work that out for X1, then we should get it approximately equal to 0.2136363636. OK? So this is going to be the value of X1. Now we're going to repeat, we're going to continue our iterations for X2 and X3. Gained by newer method, this must mean then that X2 is going to be equal to X1 minus F of X1. Divided by the derivative of F of X1. Now, in the interest of time, I'll save us writing out uh this value 0.2136363636 uh into our formula for X2. But what we know it, we were basically doing the same thing. We're replacing X wherever we see it with our value for X1. And now when you do that, then you should find that you get X2. To be approximately equal to 0.1011335883. Now that we have that value for X2, OK. Then X3 is gonna be equal to X2 minus F of X2. Divided by the derivative of F of X2. And then we substitute X2 into this formula, then we should get X3 to be approximately equal to 0.04936091773. So this is the value of X3 using Newton's traditional method of approximation. Now, let's try to repeat the process with the modified Neoans method. Let me put that in red here. So now by the modified Newton's method. Where we know that. XN + 1. Is equal to XN. Minus 2 F of XN divided by the derivative of F of XN in this case for a multiplicity of 2, then we can pretty much do the same thing, OK? So that means, for example, X1. Is going to be equal to x knot. Minus 2 multiplied by F of X0. Divided by the derivative of F of X knot. We know that X0 is 0.5, so this is going to be 0.5 minus 2 multiplied by F of X0. 0.5 to the fourth minus 9 multiplied by 0.5 cubed. Plus 20 multiplied by 0.5 squared. All divided by. The derivative of F of X not, so that's gonna be 4 multiplied by 0.5 cubed minus 27 multiplied by 0.5 squared. Plus 40 multiplied by 0.5. Pretty much kind of similar to what we did before. I know when we do that by Newton's modified method, you should find that we get X1 to be equal to -0.07272727273. OK? And again, just like we did with Newton's traditional method, we're going to repeat the process with our new value X1. So this means then in iteration 2, X2 will be X1 minus 2 multiplied by F of X1 divided by the derivative of F of X1. And when we plug in our values, then we should get that to be approximately equal to negative 0.001152146962. Now if we move on to iteration 3. X3 equals X2 minus 2, F of X1, or F of X2, sorry, divided by the derivative of F of X2. And when we calculate here, then we should get it to be approximately 2.985, 188. 6 to 8 multiplied by 10 to the negative 7. And again, we know that we're just basically using the same formula but replacing our values of X with uh whether it be X2, X3, and so on. So now we have our values based on the modified Newton's method and the traditional Newton's method, which is closer. Well, by observation, you should be able to see that the root approximated by the modified Newton's method is closer to the root of X equals 0. If we look back at our answer choices, that tells us then that A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Newton's Method

Modified Newton's Method

Convergence Rate

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