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.
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.