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In Newton’s method, the stopping criterion is set as |pn+1 − pn|< T OL. Explain why this stopping criterion is a goodapproximation when pn is close ...

Question

In Newton’s method, the stopping criterion is set as |pn+1 − pn|< T OL. Explain why this stopping criterion is a goodapproximation when pn is close to the true solution p.

In Newton’s method, the stopping criterion is set as |pn+1 − pn| < T OL. Explain why this stopping criterion is a good approximation when pn is close to the true solution p.



Answers

Use the improved Euler's method with tolerance to approximate the solution to
$$ \frac{d x}{d t}=1+t \sin (t x), \quad x(0)=0 $$
at $ t=1 $ For a tolerance of $ \varepsilon=0.01 $ use a stopping procedure based on the absolute error.

Question 33. We're looking at our function as extra 1/3 but a key root of X and a prime of X and, of course, is 1/3 x to the negative 2/3. So x to the n plus one is an equal x minus x to the 1/3 over 1/3 X to the negative 2/3. So simplify that I would have, um, one there x to the to certain so back to the I'm just simply crying this part right here, x and we're when they're x 2/3 I'm going to invert and multiply, So I'm gonna have X to the 1/3 times three x to the 2/3 over one. So I'm just gonna multiply them when I must play and the exponents. So I have three x. So now I have X minus three X, which gives me and negative to X and negative X will not converge

Uh, the answer to that question is yes, news Mother does fail when our initial guesses at a relative maximum. I want you to look at this example are saying that this point here is, uh, our initial guess for noon. Demented new method graphically. What's going on? If you draw the tangent line one join at this point and then we take the X in effect of the mind and our new brother next death of their ex, too. However, at Annapolis at a relative maximum, uh, smoke with this minus zero and equal zero for the fine. And so it's a horizontal line. There is no exit except oh, there will not be any next guess. Um, and algebraic Leawood happening is that when we divide by the derivative, ah, of our initial guys in new method, we have dividing my hero, and that is undefined. So this is why new method fails it. We take our initial guess to be a relative maximum. It will also be able for the same reason to be take a point at a relative minimum

Question 34. I am Effort X, the groom's ex to the 1/2 when X is greater than zero or positive in f of X equals negative Negative X to the 1/2 1 X is negative or less than zero. Now you do my crime of eggs. You have 1/2 X to the negative 1/2 in my F crime here. Good. I'm gonna have negative 1/2 over negative X to the negative 1/2 So now I'm gonna in purple Just worry about my next to the end Plus one equals x minus x to the 1/2 over 1/2 Invert and multiply. I get to X so I have X minus two X, which gives me a negative X And that's the end, of course. And that's when X is positive. And now I do the same thing when X is negative. So I have negative square root of negative X except then minus. And then I have negative 1/2 negative. Yeah, invert and multiply. And I'm gonna get next to the end minus two x, which is gonna give me a negative X to the end as well. When exes Lutz, then zero so negative exit the end will not converge

So to understand why you failed when a crime of passion equals zero, we could do with a couple different things. First, we can look at the formula afternoons method the federation. So the method said that extent plus one equal with the previous number accent minus thy function That accent divided by the value of the derivative it at FedEx and from of eccentric with zero. Then we have divided my view here that's going to under fine. So performing live, not going to work. And we can also understand why that true? Graphically. So you get your function like this when the derivative of hero that means the smoke of attention. I have zero. I must say that, for example, that we have, um I say we have here their ex. Why is that? If you can hear that you hear there, why our fee? I might say that accented right here. So the derivative you're here, that means the tangent line of zero here. And remember that new movement that this accent quest one is really the X intercept of this tangible on. But when the smoke of the tangent line is zero, it's a foregone tangent line and or is on a tangent line do not have any extra steps so graphically we do not approximate. They want to find the second step. Sure, we won't get any closer to that point. There's no other points we can approximate with. Whereas if the smoke of non zero maybe we'll get it. Can you write this that dish extend? Then we could approximate this accident, check with this other reconnect up here and then continue doing the approximation. But smoking zero You admitted it not work.


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