1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Real Analysis: Proof about simple zeros (multiplicity = 1)

  1. Sep 16, 2013 #1
    1. The problem statement, all variables and given/known data
    A function f has a simple zero (or zero of multiplicity 1) at x0 if f is differentiable in a neighborhood of x0 and f(x0) = 0 while f(x0) ≠ 0.

    Prove that f has a simple zero at x0 iff f(x) = g(x)(x - x0), where g is continuous at x0 and differentiable in a deleted neighborhood of x0, and g(x0) ≠ 0.


    2. Relevant equations
    None.


    3. The attempt at a solution
    You need to find a function g(x) such that it fufills the properties in the problem. I thought of using the function g(x) = f(x)/(x-x0), but that is not continuous at x0. Am I close in trying to find this function g or should the proof go in a different direction?
     
  2. jcsd
  3. Sep 16, 2013 #2
    You're on the right track. Can you show that [itex]\lim_{x\rightarrow x_0}\frac{f(x)}{x-x_0}[/itex] exists?
     
  4. Sep 16, 2013 #3
    After using L'hospital's rule, you find that the limit is f'(x0) ≠ g(x0). So g is not continuous at x0.
     
  5. Sep 16, 2013 #4
    How is g(x0) defined?
     
  6. Sep 16, 2013 #5
    My idea was to have g(x) = f(x)/(x-x0), that satisfies f(x) = g(x)(x - x0), but not the continuity part
     
  7. Sep 16, 2013 #6
    That formula, f(x)/(x-x0), defines g(x) for x≠x0 and is undefined at x0. How are you defining g(x0)? What is preventing you from making g(x0)=f'(x0) (or any other number, for that matter)?
     
  8. Sep 16, 2013 #7
    I would just have to choose that number such that it makes g(x0) continuous?
     
  9. Sep 16, 2013 #8
    Right. And that number would be ...

    Also (and I apologize for the pedantry), you should say (and write) "g continuous at x0" rather than "g(x0) continuous". g(x0) is a number, and continuity is not a property of numbers.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Real Analysis: Proof about simple zeros (multiplicity = 1)
Loading...