Thoughts on the derivative of a function

  • #26
PeroK
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Hi,PF



I've been at the spanish forum, RM: we've came to the conclusion that some background on sequences and topology is needed to face the question, and I have none of them.
My decision is to give a last chance: the quote above. Is there an easier answer to the statement of the thread, based on an undergraduate's skills and lacks?.
I've also translated and wrote down the quote, but RM leads me again to topology.
My question: can you give me some more hints on Office_Shredder's quote?.
If I should provide more information, please tell me.

Greetings!
Sequences are usually the first thing to study in a course on real analysis. If you know the definition of a limit in terms of sequences, then this is fairly elementary. It's introductory pure mathematics and requires no topology.

Are you taking a course in real analysis?
 
  • #27
mcastillo356
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Sequences are usually the first thing to study in a course on real analysis. If you know the definition of a limit in terms of sequences, then this is fairly elementary. It's introductory pure mathematics and requires no topology.

Are you taking a course in real analysis?
I'm enrolled in real analysis, in the first college year of the Physics career. It's the only subject I've chosen (I had that chance, and I took this subject). At this moment, I've only taken a quick look at sequences. So I guess I need some more time.

Greetings. Thanks, PF!
 
  • #28
haruspex
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Hi,PF



I've been at the spanish forum, RM: we've came to the conclusion that some background on sequences and topology is needed to face the question, and I have none of them.
My decision is to give a last chance: the quote above. Is there an easier answer to the statement of the thread, based on an undergraduate's skills and lacks?.
I've also translated and wrote down the quote, but RM leads me again to topology.
My question: can you give me some more hints on Office_Shredder's quote?.
If I should provide more information, please tell me.

Greetings!
If you consider a sequence of rational numbers, xi converging to x, what does the sequence f(xi) look like? What does it converge to?
Then do the same for a sequence of irrationals converging to x.
 
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  • #29
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Hi, PF
It suffices to prove that every irrational number in ##\mathbb R## is the limit of a sequence of rational numbers, and every rational number in ##\mathbb R## is the limit of a sequence of irrational numbers? I've got some stuff on internet. It might be useful, but I need time to check it.
 
  • #30
PeroK
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Hi, PF
It suffices to prove that every irrational number in ##\mathbb R## is the limit of a sequence of rational numbers, and every rational number in ##\mathbb R## is the limit of a sequence of irrational numbers? I've got some stuff on internet. It might be useful, but I need time to check it.
I think you can assume that. This question doesn't expect you to prove that.
 
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  • #31
mcastillo356
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If you consider a sequence of rational numbers, xi converging to x, what does the sequence f(xi) look like? What does it converge to?
Then do the same for a sequence of irrationals converging to x.
##f(x_i)## is another sequence; it converges to ##f(x)##
The same for irrationals.
Is it that easy?. It's already proven: the polynomial functions converge to ##9## at ##x=3##:wideeyed:
 
  • #32
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##f(x_i)## is another sequence; it converges to ##f(x)##
The same for irrationals.
Is it that easy?. It's already proven: the polynomial functions converge to ##9## at ##x=3##:wideeyed:
If you have enrolled in a course on real analysis, you perhaps ought to take it seriously.
 
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  • #33
haruspex
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##f(x_i)## is another sequence; it converges to ##f(x)##
The same for irrationals.
Is it that easy?. It's already proven: the polynomial functions converge to ##9## at ##x=3##:wideeyed:
You do not seem to have understood what I asked you to do.
If you consider a sequence of rational numbers, xi converging to x, what does the sequence f(xi) look like?
I.e. write out the expression for that sequence, using what f looks like for rationals. Do the same for a sequence of irrationals. Post what you get.
 
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  • #34
Office_Shredder
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As a more concrete example, what is
$$\lim_{n\to \infty} f(\pi/n))$$
And what is
$$\lim_{n\to \infty} f(1/n)$$
 
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  • #35
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Well, here is my output. Sorry for the excessive time taken to post, and, forgive language mistakes.
 

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  • #36
PeroK
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Well, here is my output. Sorry for the excessive time taken to post, and, forgive language mistakes.
I think you need to talk to your tutor and sort out what you need to be able to do before you tackle a problem such as this.
 
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  • #37
haruspex
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Well, here is my output. Sorry for the excessive time taken to post, and, forgive language mistakes.
Any algebra which may have been in that image seems to have evaporated. That makes your thoughts hard to interpret.

There is not really any such thing as a "piecewise function". You can have a piecewise definition of a function, but that is a property of the way you choose to define it, not of the function.
Perhaps you mean piecewise continuous. Such a function is one which can be described in a piecewise manner such that each sub function is continuous in its own interval.
Either way, the pieces need to be intervals, not isolated points.

Please try to do what I asked you to do in post #33. The answers must be in the form of algebraic expressions, not just verbiage.
 
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  • #38
PeroK
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Here's an example of some real analysis. Suppose we have:
$$f(x) =\bigg\{ \begin{matrix} g(x) & x < a \\ h(x) & x \ge a \end{matrix}$$ Where ##g## and ##h## are continuous functions on ##\mathbb R## with ##g(a) = h(a)##. We want to show that ##f## is continuous at ##a##.

Proof: Let ##\epsilon > 0##. As ##g## is continuous at ##a##, there exists ##\delta_1## such that: $$|x - a| < \delta_1 \ \Rightarrow \ |g(x) - g(a)| < \epsilon$$ And, as ##h## is continuous at ##a##, there exists ##\delta_2 > 0## such that $$|x - a| < \delta_2 \ \Rightarrow \ |h(x) - h(a)| < \epsilon$$ Letting ##\delta = min\{\delta_1, \delta_2 \}## we have for ##x \ge a##:
$$|x - a| < \delta \ \Rightarrow \ |f(x) - f(a)| = |h(x) - h(a)| < \epsilon$$ And, for ##x < a##:$$|x - a| < \delta \ \Rightarrow \ |f(x) - f(a)| = |g(x) - g(a)| < \epsilon$$ And, it follows that ##f## is continuous at ##a##.

Now, suppose that ##g(a) \ne h(a)##. We show that ##f## is not continuous at ##a##.

Proof: let ##\epsilon = |g(a) - h(a)|/2##. As ##g## is continuous at ##a## we can find ##\delta_0## such that:
$$|x - a| < \delta_0 \ \Rightarrow \ |g(x) - g(a)| < |g(a) - h(a)|/2$$ Now, by the triangle inequality we have for ##a -\delta_0 < x < a##:
$$|f(x) - f(a)| = |g(x) - h(a)| \ge |g(a) - h(a)| - |g(x) - g(a)| > |g(a) - h(a)| - |g(a) - h(a)|/2$$ Hence:
$$|f(x) - f(a)| > |g(a) - h(a)|/2 = \epsilon$$
In summary, we have found ##\epsilon > 0## and ##\delta_0 > 0## such that:$$a -\delta_0 < x < a \ \Rightarrow \ |f(x) - f(a)| > \epsilon$$ And that is enough to show that ##f## is not continuous at ##a##, because:

For any ##\delta > 0## we can find ##x## with ##|x- a| < \delta## and ##|f(x) - f(a)| > \epsilon##.

And you can see that real analysis is hard work. It's taken a lot of effort to prove something that is rather obvious. The idea is that you use the same techniques to prove things that are not so obvious.

In any case, this is the sort of work I would expect you to produce for your question.
 
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  • #39
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Teines que aprender la definición de Continuity y Differentiation. Cuando digo que aprendas la definición, arriéndela usando epsilon/delta y también la forma usando sequences.

Estas asciendo las cosas muy complicadas. Cuando aprendas la definición, usa las (las dos formas) para probar problemas fáciles de matemáticas. Cuando hagas esto por un tiempo, attempta el problema que escribiste aquí.

Tables mira el libro Abbot: Understanding Analysis. También puedes mirar Bartle: Elements Of Real Analysis. No mi recuerdo si Abbot tiene la definición de Continuity usando sequences. Si Abbot no lo teine, mira la definición de Bartle usando sequences. También ley cuando dice como enseñar que una función no es continuous.


Sorry for writing in Spanish. It is clear that English is not the OP native language. I suggested that he first learn the definition of continuity (both epsilon/delta and sequence), practice on more simple problems, then attempt the problem he posted. Since it is a waste of his time attempting this problem when he does not know how to do a basic epsilon/delta proof. I also recommend that he read Abbot (a gentle and well motivated book), and maybe look at Bartle for sequence characterization of continuity (I do not remember if Abbot contains it).

At this point. It seems he is being hardheaded...
 
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  • #40
mcastillo356
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Another approach. First of all, my tutor has given me a non worthposting answer. So I'll keep on trying: a theorem: $$\displaystyle{ \lim_{x\to c}g(x)=L\iff \lim_{n\to\infty}g(x_n)=L\text{ for all sequence }\{x_n\}_{n\in \mathbb N}\text{ so that }\lim_{n\to\infty }x_n=c }$$

Now,
Hi, PF
(...) every irrational number in ##\mathbb R## is the limit of a sequence of rational numbers, and every rational number in ##\mathbb R## is the limit of a sequence of irrational numbers (...)

Be ##\{a_n\}_{n=1}^{+\infty}## a sequence s.t.
1.) ##a_n## is irrational for all natural.
2.) ##a_n \neq 3## for all natural
3.) ##\displaystyle \lim_{n \to +\infty} a_n = 3 ##.

Be ##\{b_n\}_{n=1}^{+\infty}## a sequence s.t.
1.) ##b_n## is rational for all natural.
2.) ##b_n \neq 3## for all natural
3.) ##\displaystyle \lim_{n \to +\infty} b_n = 3##.


##\displaystyle \lim_{n \to +\infty} \dfrac{f(a_n) - f(3)}{a_n-3} =\lim_{n \to +\infty} \dfrac{6\cdot(a_n-3) + 9 - 9}{a_n-3} = 6##

##\displaystyle \lim_{n \to +\infty} \dfrac{f(b_n)-f(3)}{b_n-3} = \lim_{n \to +\infty} \dfrac{b_n^2 -9}{b_n-3} = \lim_{n \to +\infty} \dfrac{(b_n -3) \cdot (b_n +3)}{b_n-3} = \lim_{n \to +\infty} b_n + 3 = 6##

Comments:
This theorem has got a proof in terms of topological issues
If I'm right, I've proven that ##f(x)## is differentiable if and only if ##x=3##

Greetings!
 
  • #41
PeroK
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Another approach. First of all, my tutor has given me a non worthposting answer. So I'll keep on trying: a theorem: $$\displaystyle{ \lim_{x\to c}g(x)=L\iff \lim_{n\to\infty}g(x_n)=L\text{ for all sequence }\{x_n\}_{n\in \mathbb N}\text{ so that }\lim_{n\to\infty }x_n=c }$$

Now,


Be ##\{a_n\}_{n=1}^{+\infty}## a sequence s.t.
1.) ##a_n## is irrational for all natural.
2.) ##a_n \neq 3## for all natural
3.) ##\displaystyle \lim_{n \to +\infty} a_n = 3 ##.

Be ##\{b_n\}_{n=1}^{+\infty}## a sequence s.t.
1.) ##b_n## is rational for all natural.
2.) ##b_n \neq 3## for all natural
3.) ##\displaystyle \lim_{n \to +\infty} b_n = 3##.


##\displaystyle \lim_{n \to +\infty} \dfrac{f(a_n) - f(3)}{a_n-3} =\lim_{n \to +\infty} \dfrac{6\cdot(a_n-3) + 9 - 9}{a_n-3} = 6##

##\displaystyle \lim_{n \to +\infty} \dfrac{f(b_n)-f(3)}{b_n-3} = \lim_{n \to +\infty} \dfrac{b_n^2 -9}{b_n-3} = \lim_{n \to +\infty} \dfrac{(b_n -3) \cdot (b_n +3)}{b_n-3} = \lim_{n \to +\infty} b_n + 3 = 6##

Comments:
This theorem has got a proof in terms of topological issues
If I'm right, I've proven that ##f(x)## is differentiable if and only if ##x=3##

Greetings!
This is not a proof either in content or format. It's a useful start.

The definition of continuity says for all sequences. What about all the sequences that are a mixture of rational and irrational numbers?
 
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  • #42
Office_Shredder
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I don't think the goal is to prove that theorem, the goal is to use the theorem to solve the problem. If that's the case,I think you correctly have shown f is continuous at 3 in your post. The other parts can be done similarly.
 
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  • #43
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I don't think the goal is to prove that theorem, the goal is to use the theorem to solve the problem. If that's the case,I think you correctly have shown f is continuous at 3 in your post. The other parts can be done similarly.
Are you sure that constitutes a valid proof that ##f## is differentiable at ##x = 3##? Note that it's not showing that ##f## is continuous.
 
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  • #44
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I think there is something missing... I must think it over.
I will take some time. :headbang:
 
  • #45
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I think you need to talk to your tutor and sort out what you need to be able to do before you tackle a problem such as this.

Yes. My initial reaction is that you use the theorem that iff a function is continuous at a point, every subsequence must converge to the point.

Thanks
Bill
 
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  • #46
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I think you need to talk to your tutor and sort out what you need to be able to do before you tackle a problem such as this.
Well, my decission is to write a last post, in a week. If I still show no progress, let me know. The alternative would be to return to the textbook; leave the exercise.
 
  • #47
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Hi, PF

Why ##f(x)=\begin{cases}{g(x)=x^2}&\text{if}& x\in \mathbb Q\\h(x)=6(x-3)+9 & \text{if}& x\in \mathbb R \setminus \mathbb Q\end{cases}## is differentiable only at ##x=3##?

(1)- Why the function has derivative at ##x=3##?

##\displaystyle\lim_{x \to{3}}{\dfrac{f(x)-f(3)}{x-3}}##

This derivate turns into two:

(a)-##\displaystyle\lim_{x \in \mathbb Q, x \to{3}}{\dfrac{x^2-9}{x-3}}=6##

(b)-##\displaystyle\lim_{x \in \mathbb I, x \to{3}}{\dfrac{(6(x-3)+9)-9}{x-3}}=6##

Proof for (a)

For ##x\in \mathbb Q\;\forall{\epsilon_1>0}\;\exists\;{\delta_1}## such that if ##0<|x-3|<\delta_1\Rightarrow{|g(x)-9|=|x^2-9|<\epsilon_1}##:

##\delta_1=\mbox{mín}\left ({1,\dfrac{\epsilon_1}{3}}\right )##

Proof of (b)

For ##x\in \mathbb I\;\forall{\epsilon_2>0}\;\exists\;{\delta_2}## such that if ##0<|x-3|<\delta_2\Rightarrow{|h(x)-9|=|(6(x-3)+9)-9|<\epsilon_2}##:

##\delta_2=\dfrac{\epsilon_2}{6}##

##\delta=\mbox{mín}\left ({1,\dfrac{\epsilon_1}{3},\dfrac{\epsilon_2}{6}}\right )##


If we separate the domain of some function into two subsets, and approaching to a point ##a## by each one, they reach the same value, the function approaches to this value as it comes near ##a## by all the domain.

Proof:

##f:\mathbb R\rightarrow{\mathbb R}##

##h:\mathbb Q\rightarrow{\mathbb R}##

##g:\mathbb R\setminus{\mathbb Q}\rightarrow{\mathbb R}##

(2)- Why is not differentiable if ##x\neq 3##?

Now, suppose that ##g(a)\neq h(a)## (this happens iff ##x\neq 3##). We show that ##f## is not continous at ##a##.

Proof

Let ##\epsilon=|g(a)-h(a)|/2##. As ##g## is continous at ##a## we can find ##\delta_{\mathbb Q}## such that

##|x-a|<\delta_{\mathbb Q}\Rightarrow{|g(x)-g(a)|<|g(a)-h(a)|/2}##

Now, by the triangle inequality we have for ##|x-3|<\delta_{\mathbb Q}##

##|f(x)-f(a)|=|g(x)-h(a)|\geq |g(a)-h(a)|-|g(x)-g(a)|>|g(a)-h(a)|-|g(a)-h(a)|/2##

Hence

##|f(x)-f(a)|>|g(a)-h(a)|/2=\epsilon##

In summary, we have found ##\epsilon>0## and ##\delta_{\mathbb Q}>0## such that

##|x-a|<\delta_{\mathbb Q}\Rightarrow{|f(x)-f(a)|>\epsilon}##

And that is enough to show that ##f## is not continous at ##a## (if ##a\neq 3##), because for any ##\delta_{\mathbb Q}>0## we can find ##\epsilon>0## such that ##|x-a|<\delta_{\mathbb Q}## and ##|f(x)-f(a)|>\epsilon##

Proven it is not continous, it is not differentiable at that points
 
  • #48
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If we separate the domain of some function into two subsets, and approaching to a point a by each one, they reach the same value, the function approaches to this value as it comes near a by all the domain.

Proof:
What proof?

Now, suppose that g(a)≠h(a)
There is no point a at which g and h are both defined.
 
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  • #49
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It's clear I must study further. I thoght I had clear the concepts of function, continuity, derivative, proof... I must return to the textbook studied and the subject I passed a year ago. I feel very thankful for your support and advice at this thread. Thank you, PeroK, haruspex, ...Loving support. Loving indeed.
 
  • #50
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Hi, PF, sorry, I am a little bit boring:

¿Why ##f(x)=\begin{cases}{x^2}&\text{if}& x\in \mathbb Q\\6(x-3)+9 & \text{if}& x\in \mathbb R \setminus \mathbb Q\end{cases}## is differentiable only at ##x=3##?

1- ##\displaystyle\lim_{x \to{3}}{\dfrac{f(x)-f(3)}{x-3}}## (*)

##\displaystyle\lim_{x\in \Bbb Q,\,x \to 3}{}\dfrac{x^2-9}{x-3}##(**)

##\displaystyle\lim_{x\in \Bbb I,\,x \to 3}{}\dfrac{6(x-3)+9-9}{x-3}##(***)

For this limit (*), it exists iff (**) and (***) are equal.

(**)##\displaystyle\lim_{x\in \Bbb Q,\,x \to 3}{}\dfrac{x^2-9}{x-3}=6##

Proof:

For ##x \in\Bbb Q##, ##\forall{\varepsilon}## ##\exists\;{\delta_1}## such that if ##0<|x-3|<\delta_1\Rightarrow{\left |{\dfrac{x^2-9}{x-3}-6}\right |=|x+3-6|=|x-3|<\varepsilon}##. Hence, ##\delta_1=\varepsilon##


(***)##\displaystyle\lim_{x\in \Bbb I,\,x \to 3}{}\dfrac{6(x-3)+9-9}{x-3}=6##

Proof:

For ##x \in\Bbb R\setminus{\Bbb Q}##, ##\forall{\varepsilon}## ##\exists\;{\delta_2}## such that if ##0<|x-3|<\delta_2\Rightarrow{\left |{\dfrac{(6(x-3)+9)-9}{x-3}-6}\right |=|6-6|=0<\varepsilon}##. Hence, ##\delta_2## can take any value, and the implication will be true.

Choose ##\delta_1=\varepsilon##, and ##f(x)## is differentiable at ##x=3##

2- ¿Why is it not continous at ##x \neq 3##?

##\displaystyle\lim_{x \to x_0,x\in \Bbb Q}{}x^2=x_0^2##

Proof :

##|x^2-x_0^2|=|x-x_0||x+x_0|##

First restriction: ##|x-x_0|<1\Leftrightarrow{x_0-1<x<x_0+1}\rightarrow{|x+x_0|<2|x_0|+1}##

Hence, for ##\delta=\varepsilon/(2|x_0|+1)## this limit exists.

##\displaystyle\lim_{x \to x_0,x\in \Bbb I}{}(6(x-3)+9)=6x_0-9##

Proof:

If ##0<|x-x_0|<\delta\Rightarrow{|(6x-9)-(6x_0-9)|<\varepsilon}##

##|6x-9-6x_0+9|<\varepsilon\Rightarrow{6|x-x_0|<\varepsilon}##

For ##\delta=\varepsilon/6##

and ##x_0^2=6x_0-9## iff ##x_0=3##

Well, this was not really homework; I left aside my textbook when I met this exercise, which captivated me. I didn't know I was facing my absolute ignorance about concepts I had already passed on september. I actually passed with a 5,6 over 10.
 

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