How to use Newton's Method to choose a Xo guess that bounces back and forth infinitely

Bluewolf1986

New member
Joined
Sep 15, 2019
Messages
17
Here's my question from my Calculus 1: Differential Calculus Class on the Topic of Newton's Method:

Find a positive initial guess x0 for the zero of x−x^3=0 for which Newton's method bounces back and forth infinitely. (Use symmetry.)

(Type ∗ for multiplication; type / for division; type ∧ for exponentiation. You may type sqrt for √. You may also enter answer as a decimal correct to 3 decimal places.)

My initial method was to use symmetry to guess the initial Xo value. Since the function x-x^3 only has symmetry around (0,0), I thought that zero may be an answer, since it does technically go in a loop when using the formula for Newton's Method. However, this was wrong probably because it is the an actual root and not an approximation. I am struggling how to further proceed. Any hints about the graph points or how to arrange Newton's formula would greatly help.

Thank you!
 
Here's my question from my Calculus 1: Differential Calculus Class on the Topic of Newton's Method:

Find a positive initial guess x0 for the zero of x−x^3=0 for which Newton's method bounces back and forth infinitely. (Use symmetry.)

(Type ∗ for multiplication; type / for division; type ∧ for exponentiation. You may type sqrt for √. You may also enter answer as a decimal correct to 3 decimal places.)

My initial method was to use symmetry to guess the initial Xo value. Since the function x-x^3 only has symmetry around (0,0), I thought that zero may be an answer, since it does technically go in a loop when using the formula for Newton's Method. However, this was wrong probably because it is the an actual root and not an approximation. I am struggling how to further proceed. Any hints about the graph points or how to arrange Newton's formula would greatly help.

Thank you!
Where does f'(x) become 0?
 
Given a function, f(x), Newton's method for solving f(x)= 0, is to iterate xn+1=f(xn)f(xn)f(xn)\displaystyle x_{n+1}= f(x_n)- \frac{f'(x_n)}{f(x_n)}. If that converges then it converges to a root of f(x)= 0. It will "bounce back and forth" between two numbers, a and b, if b=af(a)f(a)\displaystyle b= a- \frac{f'(a)}{f(a)} and a=bf(b)f(b)\displaystyle a= b- \frac{f'(b)}{f(b)}. That second equation can be written as b=af(b)f(b)\displaystyle -b= -a- \frac{f'(b)}{f(b)} and, adding that to the first equation. (f(a)f(a)+f(b)f(b))=0\displaystyle -\left(\frac{f'(a)}{f(a)}+ \frac{f'(b)}{f(b)}\right)= 0. We must have f(a)f(a)=f(b)f(b)\displaystyle \frac{f'(a)}{f(a)}= -\frac{f'(b)}{f(b)}. Or, if you don't like fractions, f(a)f(b)=f(b)f(a)\displaystyle f'(a)f(b)= -f'(b)f(a).
 
Find a positive initial guess x0 for the zero of x−x^3=0 for which Newton's method bounces back and forth infinitely. (Use symmetry.)

I thought that zero may be an answer
You were asked to guess a positive value for x0 and you chose x0=0. The main reason your guess is wrong is because 0 is not a positive number.
 
Top