- Thread starter
- Banned
- #1

- Thread starter Poirot
- Start date

- Thread starter
- Banned
- #1

- Feb 13, 2012

- 1,704

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

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

Kind regards

$\chi$ $\sigma$

- Thread starter
- Banned
- #3

- Feb 13, 2012

- 1,704

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 $\displaystyle f_{n}(x_{0})<\varepsilon\ \forall n>\frac{2 \ln n -\ln \varepsilon}{\ln x_{0}}$...Thanks, how would you show

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

Kind regards

$\chi$ $\sigma$