Existence of Derivative for Piecewise Function with Irrational Values?

[aid]

Homework Statement

Let's take function given by a condition:

[tex]f(x) = \begin{cases} \frac{1}{q^2} \ iff \ x = \frac{p}{q} \ $nieskracalny$,\\ 0 \ iff \ x \notin \mathbb{Q} \end{cases}[/tex]

Prove the existence of the derivative of [tex]f[/tex] in all points [tex]x \notin \mathbb{Q}[/tex].

The Attempt at a Solution

So, I am aware that if there was [tex]q[/tex] standing in the formula instead of [tex]q^2[/tex], the derivative wouldn't exist. The thing I couldn't figure out is, why would the replacement change anything?

 

[D H]

1/q² will be a lot closer to zero than will 1/q for a rational number p/q within some neighborhood of a given irrational number.

 

[micromass]

You'll need to take an irrational number a and evaluate

[tex]\lim_{x\rightarrow a}{\frac{f(x)-f(a)}{x-a}}=\lim_{x\rightarrow a}{\frac{f(x)}{x-a}}[/tex]

I would suggest using the definition of limit for this...

 

[D H]

He has to show that the limit exists. The limit, if it does exist, is obviously zero.

The same applies to Thomae's function, f(x)=0 if x is irrational, 1/q if x is a rational p/q with p,q relatively prime: The limit, if it exists is obviously zero. The limit does not exist in the case of Thomae's function. The square of Thomae's function however is differentiable at the irrationals.

 

[aid]

The square of Thomae's function however is differentiable at the irrationals.

I've tried to prove it straight from the definition of a derivative but it turns out not to be so easy. Any hints on how exactly should I do that?

 

[D H]

Given that the function is identically zero for all irrationals, it should be obvious that a derivative does not exist for rational x and that the only issue of concern for irrational x is the behavior of rationals in the vicinity of x.

What can you say about a rational p/q in the neighborhood of some irrational number x? Making this a bit more precise, define the rational neighborhood of some number x in terms of some number δ as N_ℚ(x;δ)=z∈ℚ:|x-z|<δ. What happens to the denominators of all members of N_ℚ(x;δ) as δ→0?

 

[aid]

What can you say about a rational p/q in the neighborhood of some irrational number x? Making this a bit more precise, define the rational neighborhood of some number x in terms of some number δ as N_ℚ(x;δ)=z∈ℚ:|x-z|<δ. What happens to the denominators of all members of N_ℚ(x;δ) as δ→0?

Well, if [tex]\delta \rightarrow 0[/tex], then the denominators [tex] \rightarrow \infty[/tex], is that right? Still don't see how this is going to suffice for the proof. :/

 

[D H]

I'm not going to give you a proof. We don't do that at this site. We instead help you solve your own homework problems.

 

[aid]

I'm not going to give you a proof.

I'm not asking for it either. Should the validity of the statement be seen out of the definition, i.e.

[tex]\lim_{x\rightarrow a}{\frac{f(x)-f(a)}{x-a}}= \lim_{\frac{p}{q} \rightarrow a} \frac{1}{q} \cdot \frac{\frac{1}{q}}{\frac{p}{q} - a}[/tex] ?

As you've pointed, [tex]\frac{1}{q} \rightarrow 0[/tex], but does anything guarantee that the second element of the product is limited?

 

[D H]

Consider the open interval (a-1/(2q),a+1/(2q)). This interval certainly contains a rational x of the form p/q, correct? (Prove this.) Suppose q is prime. What can you say about (f(x)-f(a))/(x-a)? What happens to this result as q goes to infinitity?

Now what happens when you remove the restriction on q being prime?

 

[aid]

Consider the open interval (a-1/(2q),a+1/(2q)). This interval certainly contains a rational x of the form p/q, correct? (Prove this.)

Well, given [tex]q \in \mathbb{N}[/tex], the statement of yours is equivalent to the existence of [tex]p \in \mathbb{Z}[/tex] such that:

[tex]aq - \frac{1}{2} < p < aq + \frac{1}{2}[/tex]

The latter is quite obvious since [tex]a \notin \mathbb{Q}[/tex].

The rest of your message, however, remains unclear to me. I understand that a sequence of the form:

[tex]\frac{p_n}{n}[/tex], where [tex]n \in \mathbb{N}[/tex] and [tex]p_n \in \mathbb{Z}[/tex] such that [tex]a - \frac{1}{2n} < \frac{p_n}{n} < a + \frac{1}{2n}[/tex] will ensure us that:

(T) [tex] \frac{f(\frac{p_n}{n} )}{\frac{p_n}{n} - a} \rightarrow 0[/tex]

But weren't we suppose to show that (T) holds if [tex]\frac{p_n}{n}[/tex] replaced with any sequence that convergs to given irrational [tex]a[/tex]?

Sorry if my questions sound very silly but I'm having serious trouble with this particular exercise.

 

[D H]

I understand that a sequence of the form:

[tex]\frac{p_n}{n}[/tex], where [tex]n \in \mathbb{N}[/tex] and [tex]p_n \in \mathbb{Z}[/tex] such that [tex]a - \frac{1}{2n} < \frac{p_n}{n} < a + \frac{1}{2n}[/tex] will ensure us that:

(T) [tex] \frac{f(\frac{p_n}{n} )}{\frac{p_n}{n} - a} \rightarrow 0[/tex]

Does it? How about the cases where p_n and n are not co-prime? For example, f(qⁿ/qⁿ⁺¹) is 1/q², not 1/q²ⁿ⁺².

But weren't we suppose to show that (T) holds if [tex]\frac{p_n}{n}[/tex] replaced with any sequence that convergs to given irrational [tex]a[/tex]?

No, you are supposed to show that the derivative exists at all irrational x. What is the definition of the derivative?

 

[aid]

No, you are supposed to show that the derivative exists at all irrational x. What is the definition of the derivative?

For a given [tex]f[/tex] and a number [tex]a[/tex] the derivative is the limit:

[tex] \lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h} [/tex]

So, in this case the derivative at a given irrational [tex]a[/tex] would be:

[tex]f'(a) = \lim_{x \rightarrow a} \frac{f(x)}{x - a}, x \in \mathbb{Q} [/tex]

But (from Heine's definition of a limit) [tex]b = f'(a)[/tex] iff:

[tex]\forall{(x_n)} \forall{n}(x_n \neq a \wedge (x_n) \rightarrow a) \rightarrow (f(x_n) \rightarrow b)[/tex]

So, in order to show that

[tex]\lim_{x \rightarrow a} \frac{f(x)}{x - a}[/tex] exists and is equal to some number [tex]b[/tex],

I have to show that for all [tex](x_n)[/tex] such that for all [tex]n \in \mathbb{N}, x_n \in \mathbb{Q}[/tex]:

[tex](x_n) \rightarrow a \Rightarrow \frac{f(x_n)}{x_n - a} \rightarrow b[/tex]

And that is what I was trying to say earlier.

 

[aid]

I'm confused: this article - http://kbeanland.files.wordpress.com/2010/01/beanlandrobstevensonmonthly.pdf (proposition 3.1 to be exact) - clearly implies that there are some [tex]x \notin \mathbb{Q}[/tex] such that [tex]f'(x)[/tex] doesn't exist. :/