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

[Bluewolf1986]

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!

 

[Deleted member 4993]

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?

 

[HallsofIvy]

Given a function, f(x), Newton's method for solving f(x)= 0, is to iterate \(\displaystyle x_{n+1}= f(x_n)- \frac{f'(x_n)}{f(x_n)}\). 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 \(\displaystyle b= a- \frac{f'(a)}{f(a)}\) and \(\displaystyle a= b- \frac{f'(b)}{f(b)}\). That second equation can be written as \(\displaystyle -b= -a- \frac{f'(b)}{f(b)}\) and, adding that to the first equation. \(\displaystyle -\left(\frac{f'(a)}{f(a)}+ \frac{f'(b)}{f(b)}\right)= 0\). We must have \(\displaystyle \frac{f'(a)}{f(a)}= -\frac{f'(b)}{f(b)}\). Or, if you don't like fractions, \(\displaystyle f'(a)f(b)= -f'(b)f(a)\).

 

[HallsofIvy]

Subhotosh Khan has pointed out to me that my f'/f should be f/f'.

 

[Steven G]

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.