Factor the polynomials in exercises 37 and 39 . Find a relationship between the factored polynom

Factor the polynomials in exercises 37 and 39 . Find a...

Key Concepts

Non-Factorable Functions

For functions that do not factor neatly, as seen in the exercise where the function does not factor, traditional insights such as visible multiplicity from factorization are not readily available. Analyzing convergence in these cases relies on studying the function's derivative near the root to understand how the convergence rate might be affected, emphasizing the importance of derivative behavior in predicting the effectiveness of Newton's method.

Rate of Convergence

The rate at which an iterative method like Newton's method converges to a zero depends on the derivative's behavior at that zero. The factored form of a polynomial can expose repeated factors that cause a reduction in effective derivative magnitude, thereby decreasing the convergence rate from quadratic to a lower order.

Newton's Method

Newton's method is an iterative technique used to approximate the roots of a function by successively applying a formula that incorporates both the function and its derivative. Its convergence is highly dependent on the behavior of the function near the root and is typically quadratic for simple roots.

Polynomial Factorization

Polynomial factorization involves rewriting a polynomial as a product of simpler factors, which can reveal the locations and multiplicities of its zeros. This process is instrumental in understanding the structure of the polynomial, especially in relation to how its roots are approached by numerical methods.

Multiplicity of Roots

The multiplicity of a root indicates how many times a particular solution occurs within a polynomial. In the context of Newton’s method, roots with higher multiplicities tend to slow the convergence rate, often changing the method’s usual quadratic convergence for simple (multiplicity one) roots into a linear convergence behavior.

Transcript

00:01 In problem 50 we want to get back to problems 47 48 and 49 to find a relationship between the factorized polynomials and the conversion rate.

00:17 We have seen that in problem 47 we have have f of x equals x cubed minus 5 x squared plus 8 x minus 4 it has 2 root the first root is x equals 1 this means we can rewrite f of x equals x minus 1 and the second root was 2 this means we have another factor which is x minus 2 and to find the third factor we can see that we have here x cubed with a coefficient of 1 which means we have here x to make x multiplied by x multiplied by x equals 1 and we can estimate the number here by seeing the number here which is minus 4 and to multiply minus 1 by minus 2 gives 2 and to get minus 4 by multiplying it here we have must here minus 1 or minus 2 yes 2 yes we must have here minus 1 or minus 2 yeah we must have here minus 2 to make the multiplication of minus 1 and minus 2 and minus 2 to get minus 4 this is a factorization of problem 47 and from problem 47 we've seen that the conversion rate to reach x equals 1 is much faster than the conversion rate and the conversion to get the root x minus 2 then 1 is much faster here.

02:23 Let's revise problem 49.

02:25 Problem 49 has a function f of x equals x cubed minus 3x minus 2 and we have two factors from problem 49 we have we have two roots the first root is minus 1 which means we have x plus 1 as a factor and the second root was 2 meaning that we have x minus 2 as a factor the third factor comes from estimating it because we have here the coefficient of x cubed equals 1 then we must have here x to make x multiplied by x multiplied by x multiplied by x equals x cubed and we can get the number here by multiplying 1 multiplied by minus 2 gives minus 2 and we must multiply here by positive 1 to get the number here minus 2 plus 1 multiplied by negative 2 multiplied by positive 1 equals negative 2 and from solving problem 49 we have concluded that the conversion to get the root x equals 2 is much faster than to get the root of x equals negative 1 we have to get 2 is much faster here then from problem 47 and problem 49 we can get a relation if there is a repeated route like here and like here x plus 1 here x plus minus 2 the conversion rate comes very slow and comes very low in a standard or in contrary to if we have only single root for the function as we have here 1 and 2 1 and 2 to fit the problem of 48 into this relationship let's revise problem 48 problem 48 we have f of x equals x sine x there is no need for factorization because we have it factorized...

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