Existence of Derivative for Piecewise Function with Irrational Values?

[aid]

Homework Statement

Let's take function given by a condition:

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

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

The Attempt at a Solution

So, I am aware that if there was q standing in the formula instead of q^2, 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

\lim_{x\rightarrow a}{\frac{f(x)-f(a)}{x-a}}=\lim_{x\rightarrow a}{\frac{f(x)}{x-a}}

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 \delta \rightarrow 0, then the denominators \rightarrow \infty, 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.

\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} ?

As you've pointed, \frac{1}{q} \rightarrow 0, 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 q \in \mathbb{N}, the statement of yours is equivalent to the existence of p \in \mathbb{Z} such that:

aq - \frac{1}{2} &lt; p &lt; aq + \frac{1}{2}

The latter is quite obvious since a \notin \mathbb{Q}.

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

\frac{p_n}{n}, where n \in \mathbb{N} and p_n \in \mathbb{Z} such that a - \frac{1}{2n} &lt; \frac{p_n}{n} &lt; a + \frac{1}{2n} will ensure us that:

(T) \frac{f(\frac{p_n}{n} )}{\frac{p_n}{n} - a} \rightarrow 0

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

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:

\frac{p_n}{n}, where n \in \mathbb{N} and p_n \in \mathbb{Z} such that a - \frac{1}{2n} &lt; \frac{p_n}{n} &lt; a + \frac{1}{2n} will ensure us that:

(T) \frac{f(\frac{p_n}{n} )}{\frac{p_n}{n} - a} \rightarrow 0

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 \frac{p_n}{n} replaced with any sequence that convergs to given irrational a?

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 f and a number a the derivative is the limit:

\lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h}

So, in this case the derivative at a given irrational a would be:

f&#039;(a) = \lim_{x \rightarrow a} \frac{f(x)}{x - a}, x \in \mathbb{Q}

But (from Heine's definition of a limit) b = f&#039;(a) iff:

\forall{(x_n)} \forall{n}(x_n \neq a \wedge (x_n) \rightarrow a) \rightarrow (f(x_n) \rightarrow b)

So, in order to show that

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

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

(x_n) \rightarrow a \Rightarrow \frac{f(x_n)}{x_n - a} \rightarrow b

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 x \notin \mathbb{Q} such that f&#039;(x) doesn't exist. :/