Real analysis. Infimum of a sequence.

[abhishekkgp]

Find the infimum of the sequence \(\displaystyle x_n= \text{frac} \{ n \sqrt{3} \}\) where \(\displaystyle \text{frac} \{ x \}\) is the fractional part of \(\displaystyle x\).

I think that the infimum should be zero. I was able to produce a monotonically decreasing subsequence of \(\displaystyle (x_n)\) but that does not help. Any ideas?

 

[pka]

Find the infimum of the sequence \(\displaystyle x_n= \text{frac} \{ n \sqrt{3} \}\) where \(\displaystyle \text{frac} \{ x \}\) is the fractional part of \(\displaystyle x\).
I think that the infimum should be zero. I was able to produce a monotonically decreasing subsequence of \(\displaystyle (x_n)\) but that does not help. Any ideas?

I think that this is a strange question to be in an analysis course. I am sure that I do not know enough number theory to answer it completely.

That said, in order for zero to be the answer, you would have to show that infinitely many of those fractional numbers have infinitely leading zeros after the decimal. I have no idea how to do that.

 

[abhishekkgp]

I think that this is a strange question to be in an analysis course. I am sure that I do not know enough number theory to answer it completely.

That said, in order for zero to be the answer, you would have to show that infinitely many of those fractional numbers have infinitely leading zeros after the decimal. I have no idea how to do that.

Actually this can be done using Pigeon hole principle.

Let \(\displaystyle n \in \mathbb{Z}^{+}\). Partition the interval \(\displaystyle (0,1)\) into \(\displaystyle n\) parts, viz, \(\displaystyle (0,\frac{1}{n}),(\frac{1}{n},\frac{2}{n}), \ldots, (\frac{n-1}{n},1)\).

Consider \(\displaystyle n+1\) numbers, \(\displaystyle x_1, x_2, \ldots, x_{n+1}\).

By PHP there exist \(\displaystyle i,j, i \neq j\) such that \(\displaystyle x_i,x_j \in (\frac{k}{n},\frac{k+1}{n})\).

This implies \(\displaystyle |i-j|\sqrt{3} < \frac{1}{n}\). Thus \(\displaystyle x_{|i-j|} < \frac{1}{n}\).

This will do it.

But I couldn't think of this solution. I would not even think of PHP to solve this problem. I posted it here so that I may get a solution which has more to do with real analysis. Anyways, if you find anything then please post it here. Thanks for your reply.

 

[pka]

Actually this can be done using Pigeon hole principle.
Let \(\displaystyle n \in \mathbb{Z}^{+}\). Partition the interval \(\displaystyle (0,1)\) into \(\displaystyle n\) parts, viz, \(\displaystyle (0,\frac{1}{n}),(\frac{1}{n},\frac{2}{n}), \ldots, (\frac{n-1}{n},1)\).
Consider \(\displaystyle n+1\) numbers, \(\displaystyle x_1, x_2, \ldots, x_{n+1}\).
By PHP there exist \(\displaystyle i,j, i \neq j\) such that \(\displaystyle x_i,x_j \in (\frac{k}{n},\frac{k+1}{n})\).
This will do it. ...has more to do with real analysis.

I too was blinded by the analysis part. It is important that each of those endpoint is rational and \(\displaystyle x_j\) is irrational.
There is a similar idea in an analysis book by Serge Lang. It has to do with the density of such a set. But I did not connect those dots, until I sat through a dull committee meeting from which I am just back. I thought of the Lang example during the meeting.

 

[daon2]

Actually this can be done using Pigeon hole principle.

Let \(\displaystyle n \in \mathbb{Z}^{+}\). Partition the interval \(\displaystyle (0,1)\) into \(\displaystyle n\) parts, viz, \(\displaystyle (0,\frac{1}{n}),(\frac{1}{n},\frac{2}{n}), \ldots, (\frac{n-1}{n},1)\).

Consider \(\displaystyle n+1\) numbers, \(\displaystyle x_1, x_2, \ldots, x_{n+1}\).

By PHP there exist \(\displaystyle i,j, i \neq j\) such that \(\displaystyle x_i,x_j \in (\frac{k}{n},\frac{k+1}{n})\).

This implies \(\displaystyle |i-j|\sqrt{3} < \frac{1}{n}\). Thus \(\displaystyle x_{|i-j|} < \frac{1}{n}\).

This will do it.

But I couldn't think of this solution. I would not even think of PHP to solve this problem. I posted it here so that I may get a solution which has more to do with real analysis. Anyways, if you find anything then please post it here. Thanks for your reply.

How could it possibly be true that \(\displaystyle |i-j|\sqrt{3} < \frac{1}{n}\)? The left hand side is at least \(\displaystyle \sqrt{3}\) for distinct i,j.

 

[abhishekkgp]

How could it possibly be true that \(\displaystyle |i-j|\sqrt{3} < \frac{1}{n}\)? The left hand side is at least \(\displaystyle \sqrt{3}\) for distinct i,j.

I am so sorry.. actually it should be \(\displaystyle \text{frac} \{ |i-j| \sqrt{3} \} < \frac{1}{n}\)

 

[abhishekkgp]

I too was blinded by the analysis part. It is important that each of those endpoint is rational and \(\displaystyle x_j\) is irrational.
There is a similar idea in an analysis book by Serge Lang. It has to do with the density of such a set. But I did not connect those dots, until I sat through a dull committee meeting from which I am just back. I thought of the Lang example during the meeting.

Thank you for your reply pka.
Can you, if its not too much trouble, tell me which example it is? or if you can please post the details here.
The fact that this question can't be done without PHP just doesn't feel right.

 

[pka]

Thank you for your reply pka.
Can you, if its not too much trouble, tell me which example it is? or if you can please post the details here.
The fact that this question can't be done without PHP just doesn't feel right.

It is not the exact problem. Look at Lang's Undergraduate Analysis section on real numbers. There is a sequence of problems in part 4.
But I hate to tell you that I think he uses some ideas closely related to PHP. I cannot find my copy at the moment to post exact details.
I hope that my memory is not faulty here.
Also the Wikipedia entry on the floor function may be useful.

P.S.
I found that \(\displaystyle \{nx\}\) (the fractional part) where \(\displaystyle n\in\mathbb{Z}^+\) and \(\displaystyle x\) is irrational is dense in \(\displaystyle (0,1)\). That proves the problem