uniform convergrence

[triclino]

Let : [math]f_{n}(x) =\frac{x}{x^2+n}[/math] be a sequence of functions in the Real Nos.

 

I can prove point wise convergence to the zero function as n goes to infinity .

 

But is there a uniform convergence ??

 

If there is, can anyone prove it ,please??

[D H]

[math]N=\bigl{\lceil}1/(4\epsilon^2)\bigr{\rceil}[/math]

[triclino]

[math]N=\bigl{\lceil}1/(4\epsilon^2)\bigr{\rceil}[/math]

 

Do you know any natural No N to be equal to [math] \frac{1}{4\epsilon^2}[/math]

[D H]

Learn your symbols, triclino. [math]\lceil x\rceil[/math] is the the ceiling function. For example, [math]\lceil \pi \rceil = 4[/math].

 

Edit

I should have said [math]N=\lceil 1/(4\epsilon^2)\rceil+1[/math] to handle the rare case where [math]1/(4\epsilon^2)[/math] is an integer.

[triclino]

Learn your symbols, triclino. [math]\lceil x\rceil[/math] is the the ceiling function. For example, [math]\lceil \pi \rceil = 4[/math].

 

Edit

I should have said [math]N=\lceil 1/(4\epsilon^2)\rceil+1[/math] to handle the rare case where [math]1/(4\epsilon^2)[/math] is an integer.

 

You mixing up ceiling function with the floor function.

 

But according to what axiom or theorem you came to the conclusion :

 

[math]N=\lceil 1/(4\epsilon^2)\rceil+1[/math] or [math]N=\lceil 1/(4\epsilon^2)\rceil[/math]

[Dave]

You mixing up ceiling function with the floor function.

 

No, this is standard notation. The floor function is [math]\lfloor x\rfloor[/math].

 

But according to what axiom or theorem you came to the conclusion :

 

[math]N=\lceil 1/(4\epsilon^2)\rceil+1[/math] or [math]N=\lceil 1/(4\epsilon^2)\rceil[/math]

 

Plug it into the definition of uniform continuity and check that the definition is satisfied.

[D H]

For crying out loud, triclino. Do look things up first. Here: http://en.wikipedia.org/wiki/Floor_and_ceiling_functions.

 

What is the definition of uniform convergence? What is the maximum of the absolute value of your function less the limit of the sequence?

[triclino]

No, this is standard notation. The floor function is [math]\lfloor x\rfloor[/math].

 

 

 

Plug it into the definition of uniform continuity and check that the definition is satisfied.

 

You mixing up uniform continuity with uniform convergence .

 

But in my very 1st post i ask if any body could prove uniform convergence that i could not prove.

 

Now by producing a No and plugging it into the definition of uniform convergence to find out if the definition is satisfied or not ,i am sorry is not much of a help

[D H]

Triclino, instead of arguing (and arguing with an administrator is not very bright), try the hint. Heck, it wasn't a hint. I gave you the answer.