{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) = tan x - 2x; x₀ = 1.2

f(x) = tan x - 2x; x₀ = 1.2

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. Using Newton's method to approximate a root of F of X is equal to tangent of X minus 2. With the initial approximation, X0 is equal to 1.0. Continue iterating until two successive approximations agree to 5 decimal places when rounded up. Present your work using a table. Awesome. So it appears for this particular prompt we're asked to take our function of F of X and we're asked to approximate a root of F of X, provided that we have an initial approximation of X0 equals 1.0, and then we're asked to continue iterating this specific function until we produce two successive approximations which agree to one another when it's rounded to 5 decimal places. And we're asked to take all of our data that we collect when we're iterating and put it into a table. So that is what we're ultimately trying to solve for. So, first off, in order to solve for this particular problem or problems similar to this, 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 N minus F of X N divided by F of X N. So we need to recall and note once again that F of X will be the derivative of F of X, and once again noting that this prime symbol specifically means we're taking the first derivative of F of X. So with that said, we're told that F of X, so once again we need to recall that we're told that F of X, our function is equal to tangent of X minus 2. So when we take the first derivative of F of X, we'll find that F of X is going to be equal to C2 X. Fantastic. So, Now, therefore, as a result, our iteration will therefore become X subscript N + 1 is going to be equal to X subscript N. Minus tangent of X minus 2 divided by c2 of X. Fantastic. So now what we're going to be doing is we're going to be taking our iteration that we determined. So we're taking our formula or our expression for the iteration, and we're now going to start performing our approximations. So we need to note that at X1, or X subscript 1, We need to note that at X or X subscript 0 is going to be equal to 1, or specifically 1.0. So now we need to use our iteration formula. So we need to note that X1 is going to be equal to, since we need to plug in a value of 1.0 in for XN or X subscript N. So let me do that, once again, We'll put a little note of red lines and so right now we're taking 1.0, which is X0 equal to 1.0, and we're plugging in 1.0 in for our value of X subscript N. So when we do that, we'll get 1.0 minus Tangent of 1.0 minus 2 divided by cant squared. Of 1.0. So when we plug that into our calculator and we round to five decimal places, we will find that it's equal to 1.12920. Fantastic. So now, continuing onwards, and so far we need to have more than 2 approximations because we're trying to find 2 successive approximations that agree to one another when rounded up to 5 decimal places. So now we need to solve for X2 or X subscript 2. So in order to do that, instead of plugging in a value of 1.0 in for X subscript N, we need to take our value of X1, which was 1.12920, and we need to plug that in for a value of X subscript N. Instead of rewriting out our entire expression again, I'm just going to give you the answer. So to quickly reiterate, so we're taking our value for X1, which is 1.12920, and we're plugging it in for a value of X subscript N. And when we do that, we will find that we'll get an answer of 1.10813. And so far we haven't had two successive approximations that agree to each other when rounded to five decimal places, so we need to take our third iteration, so X3. So now we need to take our value for X2, which is 1.10813, and plug it in our plug it in for our value of X subscript N into our expression, and when we do that, we will find that X. To the subscript of 3, so X3, or third iteration is going to be equal to 1.10715. Fantastic. And we have to take 1/4 iteration, because so far we don't have two successive approximations that agree to each other when rounded to five decimal places. So, when we take our fourth iteration, we need to take the value that we found in our third iteration, which is 1.10715, and plug it in for our value of X subscript N. And when we do that, we will find that our fourth iteration is going to be equal to 1.10715 once again when rounded the five decimal places, and that's it. We did it. We found our two successive approximations that agree to each other when rounded to five decimal places. So therefore, without a doubt, the two successive approximations when rounded to five decimal places being equal to each other, were from the 3rd and 4th iterations. So that means our final answer without a doubt has to be 1.10715. And when we collect all of our, Data that we found together and put it into our table, we should be able to produce a table that looks like so, that looks like this. So it's going to show that our values for n are going to be 0123, and 4, because those are different iterations since we did from 0 to 4, and that will mean there are values of X subscript N, when it's 0 to 4 it's going to be 1.000. because we have to make sure, even though we're told that X0 is equal to 1.0, we have to make sure it's rounded to 5 decimal places. And then for our first iteration, it was 1.12920, and then our second iteration, 1.10813, and then for our third iteration, 1.10715. And then for our fourth iteration. 1.10715 which they match to our 3rd and 4th iteration, and that's it though. That is 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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