Show that there is a z such that f(z)=0

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In summary, the conversation discusses a given exercise involving a continuous function and the need to show the existence of a point where the function equals zero. Various approaches are suggested, including the use of the Bolzano-Weierstrass theorem and the Extreme Value Theorem. Ultimately, it is determined that the function must have a minimal value and a contradiction can be reached, leading to the conclusion that there is a point where the function equals zero.
  • #1
evinda
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Hey! :)
I am given the following exercise:
$f:[a,b] \to \mathbb{R}$ continuous and $\forall x$ there is a $y$ such that $|f(y)| < \frac{|f(x)|}{2}$ .Show that there is a $z$ such that $f(z)=0$.
That's what I have tried:
Suppose that there is not a $z$ such that $f(z)=0$.Then $f(x)>0 , \forall x$ or $f(x)<0 , \forall x$.
If $f(x)>0 , \forall x$ then from the relation $|f(y)| <\frac{|f(x)|}{2}$ for $x=y$ we find $\frac{-f(x)}{2}>0$,that can't be true,as we have supposed that $f$ is positive $\forall$ x.
Now,suppose that $f(x)<0 , \forall x$,from the relation $|f(y)| < \frac{|f(x)|}{2}$ for $x=y$ we find $|f(x)|<0$,that also can't be true.

So,we conclude that it can't be true that $f$ doesn't change sign,so there has to be a $z$ such that $f(z)=0$.

Could you tell me if it is right?
 
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  • #2
evinda said:
Hey! :)
I am given the following exercise:
$f:[a,b] \to \mathbb{R}$ continuous and $\forall x$ there is a $y$ such that $|f(y)| < \frac{|f(x)|}{2}$ .Show that there is a $z$ such that $f(z)=0$.
That's what I have tried:
Suppose that there is not a $z$ such that $f(z)=0$.Then $f(x)>0 , \forall x$ or $f(x)<0 , \forall x$.
If $f(x)>0 , \forall x$ then from the relation $|f(y)| <\frac{|f(x)|}{2}$ for $x=y$ we find $\frac{-f(x)}{2}>0$,that can't be true,as we have supposed that $f$ is positive $\forall$ x.
Now,suppose that $f(x)<0 , \forall x$,from the relation $|f(y)| < \frac{|f(x)|}{2}$ for $x=y$ we find $|f(x)|<0$,that also can't be true.

So,we conclude that it can't be true that $f$ doesn't change sign,so there has to be a $z$ such that $f(z)=0$.

Could you tell me if it is right?
No, that argument does not work: you are not permitted to take $y=x$. You are given that for each $x$ in $[a,b]$ there exists a $y$ in $[a,b]$ such that $|f(y)| < |f(x)|/2$, but there is no reason to think that $y=x$.

Start with an arbitrary point $x_1$ in $[a,b]$, and use the given condition to construct inductively a sequence $(x_n)$ such that $|f(x_{n+1})| < |f(x_n)|/2$ for each $n$. Then apply a theorem which says that this sequence must have a convergent subsequence.
 
  • #3
Opalg said:
No, that argument does not work: you are not permitted to take $y=x$. You are given that for each $x$ in $[a,b]$ there exists a $y$ in $[a,b]$ such that $|f(y)| < |f(x)|/2$, but there is no reason to think that $y=x$.

Start with an arbitrary point $x_1$ in $[a,b]$, and use the given condition to construct inductively a sequence $(x_n)$ such that $|f(x_{n+1})| < |f(x_n)|/2$ for each $n$. Then apply a theorem which says that this sequence must have a convergent subsequence.

Which theorem could I use for example? :confused:
 
  • #4
  • #5
Here is another idea.

Consider, $|f|:[a,b]\to \mathbb{R}$ this function is continous.
By EVT it has a minimal value $m$.

If $m=0$ the proof of your claim is complete.

If $m>0$ choose $x$ such that $|f(x)| = m$. By hypothesis of your problem there is a $y$ such that $|f(y)| < \tfrac{1}{2}|f(x)|$. This leads to a contradiction ...
 
  • #6
Opalg said:

Ok!Thanks a lot! :)

- - - Updated - - -

ThePerfectHacker said:
Here is another idea.

Consider, $|f|:[a,b]\to \mathbb{R}$ this function is continous.
By EVT it has a minimal value $m$.

If $m=0$ the proof of your claim is complete.

If $m>0$ choose $x$ such that $|f(x)| = m$. By hypothesis of your problem there is a $y$ such that $|f(y)| < \tfrac{1}{2}|f(x)|$. This leads to a contradiction ...

Ok..Thank you! :eek:
 

FAQ: Show that there is a z such that f(z)=0

How do you show that there is a z such that f(z)=0?

To show that there is a z such that f(z)=0, you can use the Intermediate Value Theorem. This theorem states that if a continuous function f(x) has different signs at two points a and b, then there exists at least one point c between a and b where f(c)=0. Therefore, if you can find two points where f(x) has different signs, you can conclude that there must be a z such that f(z)=0.

What is the significance of finding a z such that f(z)=0?

Finding a z such that f(z)=0 is significant because it means that the function f(x) has a root or a solution. This can help us solve equations or find the zeros of a function, which can have many practical applications in fields like engineering, physics, and economics.

Can you use any method to show that there is a z such that f(z)=0?

No, you cannot use any method to show that there is a z such that f(z)=0. The method you use will depend on the specific function and the information you have about it. Some common methods include the Intermediate Value Theorem, the Root Finding Algorithm, and the Bisection Method.

Can you use a computer program to show that there is a z such that f(z)=0?

Yes, you can use a computer program to show that there is a z such that f(z)=0. In fact, many numerical methods for finding roots of functions rely on computer programs to perform calculations and iterate towards a solution. However, it is important to verify the results and understand the limitations of the method being used.

Are there any other ways to prove that there is a z such that f(z)=0?

Yes, there are other ways to prove that there is a z such that f(z)=0. Some common techniques include using the Mean Value Theorem, the Cauchy-Riemann equations, and the Bolzano-Weierstrass Theorem. However, these methods may only apply to specific types of functions or may require additional conditions to be met.

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