Let f(x)=x e^-x. (a) Show that a sequence obtained by...

Let $f(x)=x e^{-x}$. (a) Show that a sequence obtained by applying Newton's Method to $f$ satisfies $x_{n+1}=\frac{x_{n}^{2}}{x_{n}-1}$. (b) Compute $x_{1}, x_{2}, \ldots, x_{10}$ separately with $x_{0}=0.8$ and $x_{0}=5$. Discuss what appears to be happening in each case. (Note: the only root of $f$ is at $x=0 .$

Let $f(x)=x e^{-x}$.
(a) Show that a sequence obtained by applying Newton's Method to $f$ satisfies $x_{n+1}=\frac{x_{n}^{2}}{x_{n}-1}$.
(b) Compute $x_{1}, x_{2}, \ldots, x_{10}$ separately with $x_{0}=0.8$ and $x_{0}=5$. Discuss what appears to be happening in each case. (Note: the only root of $f$ is at $x=0 .$

Key Concepts

Sensitivity to Initial Conditions

The performance of iterative root-finding methods like Newton's Method is highly sensitive to the initial guess. An appropriate initial value can lead to rapid convergence to a nearby root, whereas a poorly chosen starting point may result in divergence or convergence to an unintended root. This sensitivity underscores the importance of understanding the function's behavior before applying the method.

Convergence and Divergence in Iterative Methods

Convergence refers to the process where the sequence of approximations gradually and consistently moves closer to a true root, eventually stabilizing at it. Divergence, on the other hand, occurs when the iterative approximations fail to settle and instead move away from the root, often due to poor initial guesses, the presence of local extrema, or instabilities in the iterative formula.

Iterative Sequence Formulation

The derivation of an iterative sequence from Newton's Method involves reformulating the update step x??? = x? - f(x?)/f'(x?) into an recursive relationship that explicitly expresses the next term as a function of the current term. This formulation is essential for analyzing the behavior of the sequence, either analytically or through numerical computation.

Newton's Method

Newton's Method is an iterative procedure used to approximate the roots of a real-valued function. By starting from an initial guess and sequentially applying an update rule based on the function’s value and its derivative, the method refines the approximation to approach a true root. This process leverages the idea of linearizing the function around an estimated point to predict the next point closer to the root.

Transcript

00:01 Hello very much we are going to solve problem number 31 for newton method which is using to solve nonlinear equation so we have to remember our steps okay move it okay our steps okay so step number one we have to get the derivative f dash of x number two we have to use initial solution for x that will be given in the question or we have to assume it okay the third step is substituting n neutron formula x n plus 1 equal x n minus f of x n over f dash of x n okay the question we have the main function f of x equal x multiply e to the power negative x okay we have at number a we have to we have to to get the approved okay and number p we have to get or so for number a okay we can get the derivative f dash of x it will be equal e to the power negative x minus e sorry minus x multiply e to the power negative x okay so when we substitute in newton formula x n plus 1 equal x n minus f x n over f dash of x n okay so we can sorry so we can that equal will be equal x n x n minus x n minus n minus n minus x n e to the power negative x n over e to the power negative x n over e to the power negative x n minus x n multiply e to the power negative x n this will be x n minus x n e to the power negative x n and and over we can get e to the 0.

02:42 Negative x n add common factor, multiply 1 minus xn.

02:49 Okay, so after simplification, okay, we can get finally it will be equal x n squared over x n minus 1.

03:04 Okay, that's the first proof for number 8.

03:08 Okay.

03:08 For number b we have the given initial solution for that main function will be 0 .8 and 5 okay so when we get the graph of that main function it will be like that okay so we have only solution for that main function as 0 and infinity 0s after that because there are infinity intersected point between the main graph and the x -axis okay so when we form our table like that, for the first initial solution for 0 .8, we can get 3 as xm plus 1.

03:54 Okay, by using the formula we get what we got...

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