Use Newton's method to find an approximate root (accurate to six decimal places). Sketch the gra

Use Newton's method to find an approximate root (accurate to...

Key Concepts

Initial Guess Strategy

The strategy for determining an initial guess is crucial in iterative methods like Newton's method. A good starting point is typically chosen based on insights from the function’s graph, which reveals approximate locations of the roots. Selecting an appropriate initial guess improves the likelihood of convergence to the correct root and minimizes the number of iterations required for achieving the desired accuracy.

Newton's Method

Newton's method is an iterative numerical technique used to approximate the roots of a real-valued function. By starting from an initial guess, the method uses the function's derivative to predict where the function crosses zero, refining the approximation with each iteration. This technique is especially powerful for functions that are well-behaved near the root, but its success is contingent on a good initial estimate and the function’s differentiability in the region of interest.

Graphical Analysis

Graphical analysis involves plotting the function to understand its behavior over a range of values. This visualization helps in identifying where the function may cross the x-axis, thereby indicating the possible location of roots. A well-chosen initial guess for iterative methods can often be obtained by inspecting the graph, making the process of numerical approximation more efficient and reliable.

Root Approximation

Root approximation involves finding numerical solutions to equations where analytic solutions are either difficult or impossible to obtain. Ensuring a high degree of accuracy, such as six decimal places, often requires iterative methods like Newton's method, combined with a good understanding of the function's behavior. This concept underscores the importance of precision and the need for convergence criteria in numerical analysis.

Transcript

00:01 In problem 26 we want to use newton's method to get an approximate root for this function let's recall newton's method neuton's method depends on iterations we can get the root x n plus 1 by calculating the previous root x n and subtracting from it the value of the function at xn divided by the value of the differential of this function at x -end this means we first need to get f of x define f of x equals the left hand side of this equation as long as the right -hand side equal 0 x to the bar of 4 minus 4 x to the power of 3 plus x squared minus 1 and now we can give the differential of this function if dash of x equals 4 x cubed 12 x squared plus 2x and finally we need to define an initial guess we can start by x node equals minus 1 and using this equation we can get x1 x1 equals x node minus f of x node divided by f dash of x node and so on x2 equals x1 minus f of x node and so on x2 equals x1 minus f of x1 divided by f dash of x1 we can get the value of the function by substituting here and we can get the value of the differential by substituting by x equals x note from here and for easier calculations we can use a table where we have here x n we have here of x n and we have here f of x of x of x n we have here f of x of x n we have started by x node equals minus 1 the value of f of minus 1 equals 5 by substituting here as we've said and the value of the differential at x n or x node equals minus 1 equals minus 18 by substituting here as we've said now using the newton's method we can get x1 by substituting here x1 equals 1 equals x node minus 1 minus 5 divided by minus 18 which equals minus 0 .7 2222 and so on we can to write the root to 3 to 6 decimal place we have here 4 2 2 and 2 the 7th decimal place now we repeat the process to get f of x1 and f dash of x1 using f of x and f dash of x we can get it 1 .3 0 .3035 and the value of the differential equals minus 8 9 by 0 .2156 and using newton's method we can get xx2 by substituting here x2 equals x1 minus f of x1 divided by f dash of x1 we have this value minus this value divided by this value it equals minus 0 .5 -8 -1 -0218 the value of x node x22 it's not equal to x1 to the 6 decimal place which means we need to get another iteration we get the value of f of x2 which is 0 .231 and the value of f -dash of x2 minus 5 .99766 using newton's method we can get x3 x3 equals minus 0 .5416513 again x3 is not equal to x2 to 6 decimal places we need another iteration we get f of x3 0 .015113 and f dash of x3 minus 5 .2 399599 using newton's method we can get x4 which is equals minus 0 .5387 387 668 it's it's not equal to x3 and this means we need another iteration we can get f of x4 which equals 7 .69 to multiply by 10 to the bar of negative 5 and the value of the differential equals minus 5 .18632 we can get x x5 u.

06:34 Using nettance method which equals minus 0 .5 387 -387 520 x5.

06:50 X5 is approximately equals x4 but it's not for the 6 decimal places.

06:58 We need another iteration we get f of x node equals 2 .02 multiplied to 10 to the 1 of negative 9 and the value of the differential equals minus 5 .1 8605.

07:17 Using newton's method, we can get x6, which equals minus 0 .538750...

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