{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) = x² - 10; x₀ = 3

f(x) = x² - 10; x₀ = 3

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. Use Newton's method to approximate a root of F of X is equal to X2 minus 17 with the initial approximation X0 equals 4. Continue iterating until 2 successive approximations agree to 5 decimal places when rounding up. Show your work in a table. Awesome. So it appears for this particular problem, we're asked to create a table to show all of our work, but what we're ultimately trying to do is we're trying to iterate until two successive approximations agree with one another to 5 decimal places when it's rounded up. So that's what we're essentially trying to do. So we're trying to use Newton's method to approximate a root of our function of F of X, provided an initial approximation of X0 equals 4, or specifically X subscript 0 equals 4, and we're going to be iterating until we find two successive approximations that agree with one another to 5 decimal places when they're rounded up. So now that we know what we're ultimately trying to solve for, our first step that we need to take is we need to recall and use the iterative formula for Newton's method, which states that X subscript N + 1 is equal to X subscript N. Minus F of X N divided by F of X N. Fantastic. And we need to note that F of X is going to be the derivative of F of X, where this prime symbol means specifically the first derivative of F of X. So therefore, we need to note that F of X is going to be equal to X2 minus 17, but when we take the first derivative of F of X, we will find that F of X is going to be simply equal to 2 X. Thus, as a result, our iteration will become, when we simplify our expression, it'll be X subscript. N + 1 is equal to X subscript N minus. X to the power of 2, so it's gonna be X N to the power of 2 minus 17 divided by 2 multiplied by X N. So we need to note, and let's write this in red, we need to note that at X subscript 1, so at X1, at specifically X0 equals 4, we can go ahead and plug in our known variables, so we could say that X1 is equal to, so we're going to say, since we know that X0 is equal to 4, we're going to take this value of 4 and we're going to plug it in for a value of XN. So with that said, we need to take and plug in our value of 4, N for X subscript N, and then we need to subtract 4, and then we need to take it to the power of 2, since once again we're taking our value of XN and we're plugging in a value of 4 in for XN. So 42 minus 17, divided by 2 multiplied by 4, fantastic. So then we need to take 2 multiplied by 4, fantastic. So when we plug that into our calculator and we round to 5 decimal places, we will find that it's equal to 4.12500. So once again, it's 4.12500. So now in order to find X2, we need to take our value that we found for X1, so we need to take this 4.12,500 and plug it in for our value of XN. So instead of rewriting our expression all over again, I'm just going to give you the answer, but essentially instead of plugging in a value of 4, you're going to be plugging in a value of 4.12500 into our expression. So when you do that and you round to, once again you round to 5 decimal places, you will find that X2 is going to be equal to 4.12311. So once again, 4.1, 23, 11. Fantastic. And similarly for X. So subscript 3, so X subscript 3, we're going to do the same thing except instead of plugging in a value for 4 or plugging in a value of 4.12500, you're going to plug in a value of 4.12, 3, 11 into our expression, and when you do, when you round to 5 decial places, you will find that it also is equal to 4.12311. When rounded to 5 decimal points. Which is the same, so that means we have iterated until 2 successive approximations agreed to 5 decimal places. So that means we have iterated down to as many times we kept iterating until we didn't have to iterate anymore. So that means we found the approximation that agrees to 5 decimal places. So that means without a doubt, our final answer has to be 4.12311. So now we need to create a table to collect our data points that we found together. So as we could see, when we look at our table that We've created. We know that our values for N, which is going to be from 0 going to 3, because we did 012, and 3, we will find that our values for XN is going to be 4, but in order for it to match 5 decial places, you have to put 50s after 4. So then we found our first one for XN. For our for when it's when X is 1, so X subscript 1, it's 4.12500, and then for 2, when the value of XN and the value of N is 2, then it's 4.12311, and then for X3, it was 4.2 123 11. And that's it. We've officially solved for this answer. Or we solve for this problem, we found the final answer. Hooray, we did it. So this problem is very easy to do, as long as you're able to recall and use the iterative formula for Newton's method. And once you're able to recall that and perform your derivatives and make sure you plug in your values properly into your calculator, this is a very simplistic, quick problem to solve. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

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