{Use of Tech} Finding roots with Newton’s method For the given...

Question

{Use of Tech} Finding roots with Newton’s method For the given function f and initial approximation x₀, use Newton’s method to approximate a root of f. Stop calculating approximations when two successive approximations agree to five digits to the right of the decimal point after rounding. Show your work by making a table similar to that in Example 1.

f(x) = sin x + x - 1; x₀ = 0.5

f(x) = sin x + x - 1; x₀ = 0.5

Accepted Answer

Hello there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Utilize Newton's method to approximate a root of F of X is equal to cosine of X minus X with the initial approximation X0 equals 0.5. Continue iterating until 2 successive approximations agree to 5 decimal places when rounded up. Present your work using a table. Awesome. So it appears for this particular problem we're asked to solve for two separate answers. So initially we're trying to figure out when we iterate, we're gonna need to continue iterating this specific function of F of X, given the initial approximation of X0 equals 0.5. And we're gonna be iterating this specific function until we reach two successive approximations which agree with one another when rounded to 5 decimal places. And we're going to be taking this information that we find while iterating, and we're going to put it into a table to keep our final answers organized and to make sure our work looks nice in the end. So essentially we're going to take all of our iterations or all the values we receive when we iterate, and we're going to put them into a table to present our data point. Our data points in an organized way. So with that said, now that we know that we're ultimately trying to solve for two successive approximations that agree within five decimal places of one another when rounded up, our first step in order to solve this type of problem is we need to recall and use the iterative formula for Newton's method, which, as we should recall, states that X subscript N + 1 is equal to X subscript N. Minus F of X N divided by F of X N, and we need to recall a note that F of X will be the derivative of F of X, where this prime symbol means specifically we're taking the first derivative of F of X. So we need to note that F of X is equal to. Cosine of X minus X, and when we take the first derivative of F of X, we will find that F of X is going to be equal to negatives of X minus 1. Fantastic. So, therefore, as a result, our iteration will become X N + 1 is equal to x N minus cosine of x N minus X N all divided by negative sign of X N minus 1. So, We now need to note that we're gonna solve for X1 or X subscript 1. So for X1, At X0 is equal to 0.5 because that's what we're told that the value of X0 is. So we need to note that since we're told that X0 is equal to 0.5, we need to plug in a value of 0.5 in for XN. So when we do that, we will find that XN. It's going, or rather, we need to note that it's X1 in this case. So X1 is going to be equal to XN, which once again we said is 0.5, so we need to say 0.5 minus cosine of 0.5 minus 0.5, because once again we have to plug in a value of 0.5 in for XN or X subscript N divided by negative signs, we need to take negative sign. Of 0.5 minus 1. So when we plug in that expression into our calculator and we round to 5 decimal places, we will find that it's equal to 0.75522. Fantastic. So now instead of rewriting our equation or expression that we found for our iteration every single time, I'm just going to give you the answer for X2. So we have to do this a few times. So we have to keep iterating essentially because once again we're trying to iterate until two successive approximations agree to 5 decimal places when rounded up. So we need to keep doing this until we can have two successive approximations agree with each other. So in order to do that now, instead of using 0.5 for a value for X subscript N. We now need to take 0.75522 and plug that in for our value of X subscript N. So when we plug in a value of 0.75522 N for 0.5, or instead of 0.5 into our expression, we will find that X2 is equal to when we round to 5 decimal places, 0.73914. And once again, that's when we plug in a value of 0.75522 in for our value of X subscript N. So now we need to go keep iterating because we still haven't had two successive approximations agree with one another. So now we need to take our answer that we found for X2, which is 0.73914, and we need to plug that value in for our value of XN. Or X subscript N. So when we do that, we will find that X3 is equal to 0.73909 when we round to five decimal places. But we haven't had our successive approximations like match or agree to each other to 5 deci places and rounded up, so we need to go and iterate another time. So, now we are at X subscript 4, so X4. And when we plug in a value of 0.73909, in for a value of X subscript N, we will find that it's going to be equal to 0.73909 when we round to five decimal places. And hooray, we did it. So, Our two successive approximations with five decimal places, or rather being rounded up to 5 decimal places being equal, were from the 3rd and 4th iterations, so that means our final answer without a doubt has to be 0.73909. And when we create our table, so with our iterations, we can create the following root approximations in our table. So as you can see in our table that we could create, I've gone ahead and done a digital version of our table. So, as you could see, looking at our table, Like we did for when we were solving for our iterations, we have our values for N, which is 0123, and 4, because we did 4 approximations. So we're told initially that our X 0 value is 0.5, so then making sure it rounds to 5 decimal places, we did 0.50,000. So then we had for our value for X1 is 0.75522. And then for X2, it was 0.73914, and then for X3, it was 0.73909 so 909, and then for our X4 or X subscript 4, it was 0.73909. Nice, and that's it. Those are our final answers. Hooray, we did it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Key Concepts

Newton's Method

Convergence Criteria

Function and Derivative Evaluation

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