Hello, consider the well-known function f:(0,1)->R defined by
f(x) = 1/n if x = m/n, with m and n relatively prime,
f(x) = 0 if x is irrational.
Does there exist the right (left) limit of the difference quotient for x irrational? I think the answer is no, but a proof is not so trivial, beacuse it implies to study the set
S={ m - n*x | m<n are positive integers relatively prime}
Some idea?
Thank you veru much for your attention. My Best Regards, Maury Barbato
On Nov 10, 10:56 am, Maury Barbato <mauriziobarb...@aruba.it> wrote:
> Hello, > consider the well-known function f:(0,1)->R defined by
> f(x) = 1/n if x = m/n, with m and n relatively prime,
> f(x) = 0 if x is irrational.
> Does there exist the right (left) limit of the > difference quotient for x irrational? > I think the answer is no, but a proof is not so > trivial, beacuse it implies to study the set
> S={ m - n*x | m<n are positive integers relatively prime}
> Some idea?
It looks vaguely like the kind of considerations made in Hurwitz's theorem and variants thereof. For example,
Hurwitz's Theorem. Given any irrational number x, there exist infinitely many different rational numbers h/k (with h and k relatively prime) such that
| x - (h/k)| < 1/[sqrt(5)k^2].
Moreover, the sqrt(5) factor is best possible.
Not sure if that helps, but perhaps a way to start looking around.
In general, the subject of "Diophantine Approximation" deals with the problem of: given an irrational number x, to determine all solutions of |qx - p| < psi(q), where psi is a positive, decreasing function of real variable, and p and q are integers.
Don't know if that will help, but it looks similar to what you are looking at.
> On Nov 10, 10:56 am, Maury Barbato > <mauriziobarb...@aruba.it> wrote: > > Hello, > > consider the well-known function f:(0,1)->R defined > by
> > f(x) = 1/n if x = m/n, with m and n relatively > prime,
> > f(x) = 0 if x is irrational.
> > Does there exist the right (left) limit of the > > difference quotient for x irrational? > > I think the answer is no, but a proof is not so > > trivial, beacuse it implies to study the set
> > S={ m - n*x | m<n are positive integers relatively > prime}
> > Some idea?
> It looks vaguely like the kind of considerations made > in Hurwitz's > theorem and variants thereof. For example,
> Hurwitz's Theorem. Given any irrational number x, > there exist > infinitely many different rational numbers h/k (with > h and k > relatively prime) such that
> | x - (h/k)| < 1/[sqrt(5)k^2].
> Moreover, the sqrt(5) factor is best possible.
> Not sure if that helps, but perhaps a way to start > looking around.
> In general, the subject of "Diophantine > Approximation" deals with the > problem of: given an irrational number x, to > determine all solutions > of |qx - p| < psi(q), where psi is a positive, > decreasing function of > real variable, and p and q are integers.
> Don't know if that will help, but it looks similar to > what you are > looking at.
> -- > Arturo Magidin
For x irrational and rational r=h/k in lowest terms, the difference quotient is ** [f(r)-f(x)]/(r-x) = [1/k - 0]/(h/k - x). From Arturo's post, for infinitely many fractions h/k, | x - (h/k)| < 1/[sqrt(5)k^2]. From this, 1/|h/k - x| > sqrt(5)*k^2. So for these fractions h/k, the difference quotient is at least sqrt(5)*k in absolute value. So the difference quotient is unbounded (from either side).
On the other hand, if y is an irrational near the irrational x, the difference quotient is 0.
So I think this shows your function f(x) is not differentiable from either side at irrational x.
> Hello, > consider the well-known function f:(0,1)->R defined by
> f(x) = 1/n if x = m/n, with m and n relatively prime,
> f(x) = 0 if x is irrational.
> Does there exist the right (left) limit of the > difference quotient for x irrational? > I think the answer is no, but a proof is not so > trivial, beacuse it implies to study the set
> S={ m - n*x | m<n are positive integers relatively prime}
> Some idea?
> Thank you veru much for your attention. > My Best Regards, > Maury Barbato
Just to look at one specific irrational, consider [f(e/3) - f(S_n/3)]/[(e/3) - (S_n/3)], where S_n = sum(k=0,n) 1/k!. S_n/3 = integer/(3n!), so f(S_n/3) >= 1/(3n!). So in absolute value this difference quotient is >=
[1/(3n!)]/[e/3 - S_n/3]
= (1/n!)/(e - S_n) (1).
Now e - S_n = sum(k=n+1,oo) 1/k! < 1/(n+1)![1 + 1/(n+2) + 1/(n+2)^2 + ...]. In the brackets we have a geometric series that sums to (n+2)/(n+1), which shows (1) is on the order of n as n -> oo.
Dan Cass wrote: > > On Nov 10, 10:56 am, Maury Barbato > > <mauriziobarb...@aruba.it> wrote: > > > Hello, > > > consider the well-known function f:(0,1)->R > defined > > by
> > > f(x) = 1/n if x = m/n, with m and n relatively > > prime,
> > > f(x) = 0 if x is irrational.
> > > Does there exist the right (left) limit of the > > > difference quotient for x irrational? > > > I think the answer is no, but a proof is not so > > > trivial, beacuse it implies to study the set
> > > S={ m - n*x | m<n are positive integers > relatively > > prime}
> > > Some idea?
> > It looks vaguely like the kind of considerations > made > > in Hurwitz's > > theorem and variants thereof. For example,
> > Hurwitz's Theorem. Given any irrational number x, > > there exist > > infinitely many different rational numbers h/k > (with > > h and k > > relatively prime) such that
> > | x - (h/k)| < 1/[sqrt(5)k^2].
> > Moreover, the sqrt(5) factor is best possible.
> > Not sure if that helps, but perhaps a way to start > > looking around.
> > In general, the subject of "Diophantine > > Approximation" deals with the > > problem of: given an irrational number x, to > > determine all solutions > > of |qx - p| < psi(q), where psi is a positive, > > decreasing function of > > real variable, and p and q are integers.
> > Don't know if that will help, but it looks similar > to > > what you are > > looking at.
> > -- > > Arturo Magidin
> For x irrational and rational r=h/k in lowest terms,
> the difference quotient is > ** [f(r)-f(x)]/(r-x) = [1/k - 0]/(h/k - x). > From Arturo's post, for infinitely many fractions > h/k, > | x - (h/k)| < 1/[sqrt(5)k^2]. > From this, 1/|h/k - x| > sqrt(5)*k^2. > So for these fractions h/k, the difference quotient > is at least sqrt(5)*k in absolute value. > So the difference quotient is unbounded (from either > side).
> On the other hand, if y is an irrational near the > irrational x, the difference quotient is 0.
> So I think this shows your function f(x) is not > differentiable from either side at irrational x.
Good, this surely works! Thank you very very much, Dan. My Best Regards, Maury Barbato
Arturo Magdin wrote: > On Nov 10, 10:56 am, Maury Barbato > <mauriziobarb...@aruba.it> wrote: > > Hello, > > consider the well-known function f:(0,1)->R defined > by
> > f(x) = 1/n if x = m/n, with m and n relatively > prime,
> > f(x) = 0 if x is irrational.
> > Does there exist the right (left) limit of the > > difference quotient for x irrational? > > I think the answer is no, but a proof is not so > > trivial, beacuse it implies to study the set
> > S={ m - n*x | m<n are positive integers relatively > prime}
> > Some idea?
> It looks vaguely like the kind of considerations made > in Hurwitz's > theorem and variants thereof. For example,
> Hurwitz's Theorem. Given any irrational number x, > there exist > infinitely many different rational numbers h/k (with > h and k > relatively prime) such that
> | x - (h/k)| < 1/[sqrt(5)k^2].
> Moreover, the sqrt(5) factor is best possible.
> Not sure if that helps, but perhaps a way to start > looking around.
> In general, the subject of "Diophantine > Approximation" deals with the > problem of: given an irrational number x, to > determine all solutions > of |qx - p| < psi(q), where psi is a positive, > decreasing function of > real variable, and p and q are integers.
> Don't know if that will help, but it looks similar to > what you are > looking at.
> -- > Arturo Magidin
Numbers are a wonderful wood, where I often get lost, but when I always get out of it, taking some hidden treasure!
Thank you, very much, Arturo (this name sounds quite familiar to me: I'm Italian).
> Hello, > consider the well-known function f:(0,1)->R defined by
> f(x) = 1/n if x = m/n, with m and n relatively prime,
> f(x) = 0 if x is irrational.
> Does there exist the right (left) limit of the > difference quotient for x irrational? > I think the answer is no, but a proof is not so > trivial, beacuse it implies to study the set
> S={ m - n*x | m<n are positive integers relatively prime}
> Some idea?
> Thank you veru much for your attention. > My Best Regards, > Maury Barbato
Avoiding Hurwitz's Theorem. All we need is | x - (h/k)| < 1/k which is immediate.
Let y_1, y_2, y_3 be a sequence of irrationals with limit x. Then the difference quotient is 0.
For each n, choose an integer k(n) so that k(n)/n is smaller that x and x-k(n)/n is a minimum. Clearly k(n)/n has limit x. Now consider (f(k(n)/n) - f(x))/ (k(n)/n - x). The denominator has is absolute value less than 1/n. The numerator is greater than or equal to 1/n. Thus the difference quotient has absolute value greater than 1.
