Uniform convergence with exponential function

K

Kolmogorovv

New member
Joined
Jan 31, 2021
Messages
3
Im having problem with this exercise:

study the pointwise limit and the uniform convergence of this function for [MATH]x>0[/MATH]
[MATH]f_n(x)=\frac {ne^x+xe^{-x}}{n+x}[/MATH]
The pointwise limit is [MATH]e^x[/MATH], for the uniform convergence i study:

[MATH]g_n(x)=|f_n(x)-f(x)|=|\frac {ne^x+xe^{-x}}{n+x}-e^x|=|\frac{xe^{-x}-xe^x}{n+x}|=\frac{x(e^x-e^{-x})}{n+x}[/MATH]. Now I don't understand the thing my book says:

"since [MATH]g_n(n)[/MATH] goes to infinity as n goes to infinity, there's no uniform convergence". I don't understand this last sentence, could you help me?

I thought of doing so:

[MATH]\frac{x(e^x-e^{-x})}{n+x} \leq \frac{x(e^x-e^{-x})}{n}[/MATH] which goes to 0 as n goes to infinity so there is uniform convergence, but it's not correct, I don't know why.
 
Kolmogorovv said:
Im having problem with this exercise:

study the pointwise limit and the uniform convergence of this function for [MATH]x>0[/MATH]
[MATH]f_n(x)=\frac {ne^x+xe^{-x}}{n+x}[/MATH]
The pointwise limit is [MATH]e^x[/MATH], for the uniform convergence i study:

[MATH]g_n(x)=|f_n(x)-f(x)|=|\frac {ne^x+xe^{-x}}{n+x}-e^x|=|\frac{xe^{-x}-xe^x}{n+x}|=\frac{x(e^x-e^{-x})}{n+x}[/MATH]. Now I don't understand the thing my book says:

"since [MATH]g_n(n)[/MATH] goes to infinity as n goes to infinity, there's no uniform convergence". I don't understand this last sentence, could you help me?

I thought of doing so:

[MATH]\frac{x(e^x-e^{-x})}{n+x} \leq \frac{x(e^x-e^{-x})}{n}[/MATH] which goes to 0 as n goes to infinity so there is uniform convergence, but it's not correct, I don't know why.
Click to expand...
Could you post the relevant excerpt from the book?
 
  • Like
Reactions: pka
since gn(n) goes to infinity as n goes to infinity, there's no uniform convergence
Who said that is true. 2nd request for the entire excerpt.
 
Kolmogorovv said:
Im having problem with this exercise:
study the pointwise limit and the uniform convergence of this function for [MATH]x>0[/MATH][MATH]f_n(x)=\frac {ne^x+xe^{-x}}{n+x}[/MATH]"since [MATH]g_n(n)[/MATH] goes to infinity as n goes to infinity, there's no uniform convergence".
Click to expand...
Does your text really say:
\(\mathop {\lim }\limits_{n \to \infty } \dfrac{{x\left( {{e^{ - x}} - x{e^x}} \right)}}{{n + x}} = \infty~~? \)
 
Let [MATH]g_n(x)=\frac{|x(e^{-x}-e^x)|}{|n+x|}[/MATH] since [MATH]g_n(n)[/MATH] goes to infinity as n goes to infinity, there's no uniform convergence in [0, +infinity).


Anyway beyond the book, could you show me how you would have done?
 
I asked you a simple question about what you textbook actually said.
Please look at this link.
You see that \(\mathop {\lim }\limits_{n \to \infty } g_n(x)=\dfrac{{\left| {x\left( {{e^{ - x}} - {e^x}} \right)} \right|}}{{(n - x)}} = 0\)
So once again, is that different from your tectbook?
 
If you really want help from this forum you need to answer our questions/requests.
 
You must log in or register to reply here.
Top