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

• GangSines
In summary, we aim to prove that a function f has a simple zero at x0 if and only if it can be represented as f(x) = g(x)(x - x0), where g is continuous at x0 and differentiable in a deleted neighborhood of x0, and g(x0) ≠ 0. To do so, we must find a function g(x) that satisfies these properties. One potential function is g(x) = f(x)/(x - x0), but this is not continuous at x0. To fix this, we can choose a value for g(x0) such that g is continuous at x0. By making this choice, we can prove that f has a simple zero at x0.
GangSines

## 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.

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?

You're on the right track. Can you show that $\lim_{x\rightarrow x_0}\frac{f(x)}{x-x_0}$ exists?

After using L'hospital's rule, you find that the limit is f'(x0) ≠ g(x0). So g is not continuous at x0.

GangSines said:
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?

My idea was to have g(x) = f(x)/(x-x0), that satisfies f(x) = g(x)(x - x0), but not the continuity part

GangSines said:
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)?

I would just have to choose that number such that it makes g(x0) continuous?

GangSines said:
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.

## 1. What is Real Analysis?

Real Analysis is a branch of mathematics that deals with the study of real numbers and their properties. It involves the use of rigorous mathematical proofs to establish the truth of various mathematical statements.

## 2. What does multiplicity = 1 mean in regards to simple zeros?

Multiplicity = 1 in regards to simple zeros means that a function has a single root at a certain point, and the function crosses the x-axis at that point without touching or crossing it again.

## 3. How do you prove that a zero has multiplicity = 1?

To prove that a zero has multiplicity = 1, you must show that the function has a single root at a certain point and that it does not touch or cross the x-axis again at that point. This is typically done using the definition of a simple zero and the properties of continuous functions.

## 4. Can a function have multiple simple zeros with multiplicity = 1?

No, a function can only have one simple zero at a given point. This is because a simple zero means that the function crosses the x-axis at that point without touching or crossing it again. If a function had multiple simple zeros at the same point, it would have to cross the x-axis multiple times, which is not possible for a continuous function.

## 5. Why is it important to prove the multiplicity of zeros in Real Analysis?

It is important to prove the multiplicity of zeros in Real Analysis because it helps us understand the behavior of a function at a specific point. The multiplicity of a zero can affect the slope and curvature of a function, which can have significant implications in various mathematical applications. Additionally, proving the multiplicity of zeros strengthens the validity of our mathematical proofs and allows us to make more accurate conclusions about the behavior of a function.

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