Uniform Convergence of f{_n} in Complex Analysis on S=[0,infinity)

[Poirot1]

Let S=[0,infinity) and let f{_n}(z)=n^2ze^-(nz) Show that f{_n} -> 0. Is the function uniformly convergent? Sorry about it being unclear but TEX tags don't see to work. f{_n} means f subscript n. Thanks

 

[chisigma]

Let S=[0,infinity) and let f{_n}(z)=n^2ze^-(nz) Show that f{_n} -> 0. Is the function uniformly convergent? Sorry about it being unclear but TEX tags don't see to work. f{_n} means f subscript n. Thanks

If f(*) is defined in $[0,\infty)$ that means that it is a function of real variable x, so that is...

$\displaystyle f_{n}(x)= n^{2}\ x\ e^{-n x}$ (1)

If we consider an $x_{0} \in [0;\infty)$ and a $\varepsilon>0$ it exists an $n_{0}$ for which $\forall n>n_{0}$ is $f_{n}(x_{0})<\varepsilon$ so that $f_{n} \rightarrow 0$. Each $f_{n}(x)$ however has a maximum in $x=\frac{1}{n}$ and here is $f_{n}(\frac{1}{n})= \frac{n}{e}$ that increases without limit with n so that the function doesn't converges uniformly...

Kind regards

$\chi$ $\sigma$

 

[Poirot1]

Thanks, how would you show

\displaystyle f_{n}(x)= n^{2}\ x\ e^{-n x} tends to 0?

 

[chisigma]

Thanks, how would you show

\displaystyle f_{n}(x)= n^{2}\ x\ e^{-n x} tends to 0?

For $x=x_{0} \in [0,\infty)$ is $\displaystyle f_{n}(x_{0})= n^{2}\ x_{0}\ e^{- n x_{0}}$ so that for an $\varepsilon >0$ it will be $\displaystyle f_{n}(x_{0})<\varepsilon\ \forall n>\frac{2 \ln n -\ln \varepsilon}{\ln x_{0}}$...

Kind regards

$\chi$ $\sigma$

 

[blue_raver22]

I can confirm that the function f{_n}(z)=n^2ze^-(nz) converges to 0 as n approaches infinity. This can be seen by taking the limit as n approaches infinity, which results in the function approaching 0 regardless of the value of z. However, in order for the function to be uniformly convergent, it must converge to 0 at a uniform rate for all values of z in the interval S=[0,infinity). In this case, the function does not converge uniformly as the rate of convergence depends on the value of z. Therefore, we can conclude that the function is not uniformly convergent on the interval S=[0,infinity).