Newton's method fails. Explain why the method fails and, if possible, find a root by correcting

Newton's method fails. Explain why the method fails and, if...

Key Concepts

Algebraic Simplification of Rational Equations

For equations involving rational functions set to zero, it is often more effective to simplify the problem by focusing on the numerator, provided that the denominator does not vanish at the solution. This algebraic simplification reduces the complexity of the problem and avoids numerical instabilities that can occur when applying iterative methods directly to the original form.

Convergence Issues and Failure Conditions

Newton's method can fail or converge slowly if the derivative at an iterate is zero or close to zero, since this causes the update step to be undefined or excessively large. In the context of rational functions, singularities or points where the derivative behaves poorly due to the structure of the function can impair convergence.

Newton's Method

Newton's method is an iterative procedure for approximating the roots of a real-valued function. It uses the formula x??? = x? - f(x?)/f'(x?), updating an initial guess by taking into account both the function value and its derivative. Under suitable conditions, particularly when the derivative is nonzero near the root, the method converges quadratically.

Derivative and the Quotient Rule

The computation of a function's derivative is crucial in Newton’s method because the update step depends on f'(x). When the function is given as a ratio of two polynomials (a rational function), the quotient rule must be used to correctly differentiate it. Errors in differentiation or issues like near-zero derivatives can lead to failure or divergence of the method.

Transcript

00:01 In problem 44 when we use newton's method to get an approximate route for this equation using this initial guess we have a problem we want to find the cause of this problem let's find recall what newton's method is newton's method is based on iterations to get an approximate route for any equation or a function using an initial guess the next root x n plus 1 equals the brevest root x n minus f of x n divided by f dash of x n this means we need to define a function for this the newton's method the function is the left -hand side of the equation only if if the right -hand side equals 0 this means f of x equals the left -hand side of the equation we have for x squared minus 8x plus 1 divided by 4x squared minus 3x minus 7 now we can define or get f dash of x by differentiating f of x we have here a fraction we square the denominator 4x squared minus 3x minus 7 all squared and in denominator we have denominator we have denominator for x squared minus 3x minus 7 multiplied by the differential of the denominator multiplied by 8x minus 8 minus we have the nominator for x squared minus 8x plus 1 multiplied by the differential of the denominator 8x minus 3 this is the differential of the function above x and by starting by x node equals minus 1 we can use newton's method this newton's method equation to get x1 x1 equals x node minus f of x node divided by f dash of x node we have x node given in the equation equals minus 1 minus f of x node at this means we get we substitute by x equals minus minus 1 in this equation we get the value of 1 divided by 0 or in number divided by 0 if we we substitute by minus 1 we have 4 minus 8 4 plus 8 which is 12 plus 1 which is 13 divided by 0 13 divided by 0 and we don't know we don't need to find what is f -dash of x because we have a division by 0 in the nominator here which makes this term infinity and x1 is and find and we can't continue iterations to find x2 and x3 and so on this means that there is a problem during iterations of newton's method and this is the problem and this is the cause of the problem to solve this problem this problem we can start using any other value any other initial value of x node we can start using for example x node equal 0 and to make our calculations clear we can make a table for all our calculations and start by the iteration the initial value x equals 0 f of x equals when we substitute here by x equals 0 we get the value of minus 0 .148286 then we get f dash of x in substitute in here by x equal 0 we get the value of 1 .2048 and by applying newton's method we can get x1 which equals 0 minus f of x divided by f dash of x equals 0 .11684 as long as x n plus 1 is not equal to x n when continue our iterations then we get f of x1 by substituting here by x equals x1 and it equals minus 0 .01 4668 and we get f dash of x1 by substituting here by x equals x1 and it gives 0 .9704 using newton's method equation we get x2 which equals 0 .13378 again x2 is not equal to x1 we continue iterations we get f of x2 f of x2 equals minus 0 .0 .019 we get f dash of x2 equals 0 .94548...

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