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

  • Thread starter GangSines
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  • #1
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Homework Statement


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.


Homework Equations


None.


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?
 

Answers and Replies

  • #2
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You're on the right track. Can you show that [itex]\lim_{x\rightarrow x_0}\frac{f(x)}{x-x_0}[/itex] exists?
 
  • #3
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After using L'hospital's rule, you find that the limit is f'(x0) ≠ g(x0). So g is not continuous at x0.
 
  • #4
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After using L'hospital's rule, you find that the limit is f'(x0) ≠ g(x0). So g is not continuous at x0.
How is g(x0) defined?
 
  • #5
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My idea was to have g(x) = f(x)/(x-x0), that satisfies f(x) = g(x)(x - x0), but not the continuity part
 
  • #6
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My idea was to have g(x) = f(x)/(x-x0), that satisfies f(x) = g(x)(x - x0), but not the continuity part
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)?
 
  • #7
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I would just have to choose that number such that it makes g(x0) continuous?
 
  • #8
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I would just have to choose that number such that it makes g(x0) continuous?
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.
 

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