inequality2

Status
Not open for further replies.
C

chrislav

Junior Member
Joined
Jun 22, 2017
Messages
124
Given [math]A>0[/math]Find a [math]B>0[/math]such that : [math]|\frac{x}{x-\lfloor x^2\rfloor}|<B[/math]Without using the concept of the limit
 
chrislav said:
Given [imath]A>0[/imath]
Find a [imath]B>0[/imath]
such that : [imath]|\frac{x}{x-\lfloor x^2\rfloor}|<B[/imath]
Without using the concept of the limit
Click to expand...
By now, you should know that we expect you to show some of your own thinking, rather than just ask us to provide an answer. Please follow the rules!

And you even made the same mistake again, defining A but not using it.
 
yes you are right it should be:
Given: [math]A>0[/math]find a [math]B>0[/math]such that :


[math]\forall x[x>B\implies|\frac{x}{x-\lfloor x^2\rfloor}|<A][/math]
The point here is that i cannot initiate a start
What should the B be
Sorry for the inconviniace
 
chrislav said:
yes you are right it should be:
Given: [math]A>0[/math]find a [math]B>0[/math]such that :


[math]\forall x[x>B\implies|\frac{x}{x-\lfloor x^2\rfloor}|<A][/math]
The point here is that i cannot initiate a start
What should the B be
Sorry for the inconviniace
Click to expand...
I think this can be solved in a way similar to the other one I helped with. Please show us some sort of thinking; just saying you can't start at all is lazy. We can't steer you until you start moving.

First, what sign will the denominator have for large x? Use that to eliminate the absolute value.

Then, you might consider using reciprocals, and/or comparing to something simpler. There are lots of possibilities; and even showing an idea that didn't work can give us something to work with.
 
nope no ideas
Dr.Peterson said:
I think this can be solved in a way similar to the other one I helped with. Please show us some sort of thinking; just saying you can't start at all is lazy. We can't steer you until you start moving.

First, what sign will the denominator have for large x? Use that to eliminate the absolute value.

Then, you might consider using reciprocals, and/or comparing to something simpler. There are lots of possibilities; and even showing an idea that didn't work can give us something to work with.
Click to expand...
 
chrislav said:
He is suggesting reciprocals or something simpler
Click to expand...
Yes, so please try something! I'm waiting for any kind of effort on your part so I can make more suggestions!

Try doing something generally like what you did here:

inequality

Given : A>0 find k\in N such that: \forall n[n\geq k\implies\frac{1}{(n+1)!}
www.freemathhelp.com
 
I cannot find anything bigger than [math]|\frac{x}{x-\lfloor x^2\rfloor}|[/math]
 
Please try at least answering my specific question:
Dr.Peterson said:
First, what sign will the denominator have for large x? Use that to eliminate the absolute value.
Click to expand...
Part of what I am suggesting as a possible way forward after that is to turn [math]\left|\frac{x}{x-\lfloor x^2\rfloor}\right|<A[/math] into [math]\left|\frac{x-\lfloor x^2\rfloor}{x}\right|>\frac{1}{A}[/math]
And what can you write, without the floor function, to replace [imath]\lfloor x^2\rfloor[/imath] in an inequality?
 
Dr.Peterson said:
That looks a lot like what I found (which I wrote as [imath]\frac{1}{A}+1[/imath]); but what do you mean by "for x>1"?

You may show your work, if you like.
Click to expand...
please show your work because i checked my work and it is wrong
 
chrislav said:
please show your work because i checked my work and it is wrong
Click to expand...
You said you got an answer, and I said it looked at least partly right. Please show your work, so we can see if there is just a little error somewhere. And show your check, because that could be wrong too.

Why are you so resistant to taking responsibility for your own learning?
 
Sporry for the delay but my P.C broke down
Yes you right
[math]|\frac{x}{\lfloor x^2\rfloor-x}|<[/math]...........................................[math]\frac{1}{x-1}[/math]And that gives [math]B=1+\frac{1}{A}[/math]
 
That's rather abbreviated, but those are a couple central steps in the correct work.
 
Dr.Peterson said:
That's rather abbreviated, but those are a couple central steps in the correct work.
Click to expand...
you want me to state all the steps and all the theorems, definitions taking place in the proof?
I can do that if you wish so
Most of mathematical proofs are abbreviated to a certain degree
What is a central step?
 
chrislav said:
you want me to state all the steps and all the theorems, definitions taking place in the proof?
I can do that if you wish so
Most of mathematical proofs are abbreviated to a certain degree
What is a central step?
Click to expand...
I was just a little surprised that you didn't show a little more, as you did in the answer I referred to in post #8. Nothing more than that.

I'm not demanding anything of you; and I said that you showed enough to indicate that you had completed the problem and no longer felt that your work was wrong. There's no need to overreact.
 
No,No
please unswer my questions
1)Most of mathematical proofs are abbreviated to a certain degree:true or false
2)What is a central step in a mathematical proof
That is if you wish to do so
Please free to demand anything you like when i write a mathematical proof
For example in the said proof do you also want me to mention explicitly the laws of logic taking place in the proof?
 
Status
Not open for further replies.
Top