- #1

- 5,049

- 7

Hey! :giggle:

For $n\in \mathbb{N}$ let $f_n:\mathbb{R}\rightarrow \mathbb{R}$ given by $f_n(x)=\frac{x+2n}{x^2+n}$.

(a) Determine all (local and global) extrema of $f_n$ and the behaviour for $|x|\rightarrow \infty$. Make a sketch for $f_n$ and $f_n'$. Show that there exists $x_1<x_2<x_3<x_4$ with $f_n''(x_i)\cdot f_n''(x_{i+1})<0$ for $i=1,2,3$. The $x_i$ depend on $n$, they don't have to be given explicitely.

(b) Is the sequence $(f_n)_{n\in \mathbb{N}}$ pointwise or uniformly convergent to a function $f$ ?I have done the following :

(a) The derivative as for $x$ is \begin{align*}f_n'(x)&=\frac{(x+2n)'\cdot (x^2+n)-(x+2n)\cdot (x^2+n)'}{(x^2+n)^2}=\frac{(x^2+n)-(x+2n)\cdot 2x}{(x^2+n)^2}=\frac{x^2+n-2x^2-4nx}{(x^2+n)^2}\\ & =\frac{-x^2-4nx+n}{(x^2+n)^2}\end{align*}

We set that equal to $0$ and we get : \begin{align*}f_n'(x)=0&\Rightarrow \frac{-x^2-4nx+n}{(x^2+n)^2}=0\\ &\Rightarrow -x^2-4nx+n=0\\ &\Rightarrow x_{1,2}=\frac{-(-4n)\pm \sqrt{(-4n)^2-4\cdot (-1)\cdot n}}{2\cdot (-1)}=\frac{4n\pm \sqrt{16n^2+4 n}}{-2}\end{align*}

Outside the roots we have that $f_n'<0$ and so $f_n$ is there decreasing and inside the roots $f_n'>0$ and so $f_n$ is there increasing.

That means that $f_n$ has a minimum at $\frac{4n- \sqrt{16n^2+4 n}}{-2}$ and amaximum at $\frac{4n+ \sqrt{16n^2+4 n}}{-2}$.

Is that correct? Or is it meant tocalculate all extrema (local and global) in an other way? :unsure:If $|x|\rightarrow \infty$ then $f_n$ does to $0$ since the denominator goes faster to infinity, right? :unsure:To show the existence of $x_1<x_2<x_3<x_4$ with $f_n''(x_i)\cdot f_n''(x_{i+1})<0$ do we have to use the mean value theorem? :unsure:

(b) For the pointwise convergence :

\begin{equation*}\lim_{n\rightarrow \infty}f_n(x)=\lim_{n\rightarrow \infty}\frac{x+2n}{x^2+n}=2\end{equation*}

So the sequence converges pointwise to $f(x)=2$.

For the uniform convergence :

\begin{equation*}f_n(x)-f(x)=\frac{x+2n}{x^2+n}-2=\frac{x+2n-2x^2-2n}{x^2+n}=\frac{x-2x^2}{x^2+n}\rightarrow |f_n(x)-f(x)|=\left |\frac{x-2x^2}{x^2+n}\right |\end{equation*} The supremum of that expression doesn't go to zero, since when $x\rightarrow \infty$ then that expression goes to $2$. Therefore $f_n$ doesn't converge uniformly to $2$.

Is that correct and complete? :unsure:

For $n\in \mathbb{N}$ let $f_n:\mathbb{R}\rightarrow \mathbb{R}$ given by $f_n(x)=\frac{x+2n}{x^2+n}$.

(a) Determine all (local and global) extrema of $f_n$ and the behaviour for $|x|\rightarrow \infty$. Make a sketch for $f_n$ and $f_n'$. Show that there exists $x_1<x_2<x_3<x_4$ with $f_n''(x_i)\cdot f_n''(x_{i+1})<0$ for $i=1,2,3$. The $x_i$ depend on $n$, they don't have to be given explicitely.

(b) Is the sequence $(f_n)_{n\in \mathbb{N}}$ pointwise or uniformly convergent to a function $f$ ?I have done the following :

(a) The derivative as for $x$ is \begin{align*}f_n'(x)&=\frac{(x+2n)'\cdot (x^2+n)-(x+2n)\cdot (x^2+n)'}{(x^2+n)^2}=\frac{(x^2+n)-(x+2n)\cdot 2x}{(x^2+n)^2}=\frac{x^2+n-2x^2-4nx}{(x^2+n)^2}\\ & =\frac{-x^2-4nx+n}{(x^2+n)^2}\end{align*}

We set that equal to $0$ and we get : \begin{align*}f_n'(x)=0&\Rightarrow \frac{-x^2-4nx+n}{(x^2+n)^2}=0\\ &\Rightarrow -x^2-4nx+n=0\\ &\Rightarrow x_{1,2}=\frac{-(-4n)\pm \sqrt{(-4n)^2-4\cdot (-1)\cdot n}}{2\cdot (-1)}=\frac{4n\pm \sqrt{16n^2+4 n}}{-2}\end{align*}

Outside the roots we have that $f_n'<0$ and so $f_n$ is there decreasing and inside the roots $f_n'>0$ and so $f_n$ is there increasing.

That means that $f_n$ has a minimum at $\frac{4n- \sqrt{16n^2+4 n}}{-2}$ and amaximum at $\frac{4n+ \sqrt{16n^2+4 n}}{-2}$.

Is that correct? Or is it meant tocalculate all extrema (local and global) in an other way? :unsure:If $|x|\rightarrow \infty$ then $f_n$ does to $0$ since the denominator goes faster to infinity, right? :unsure:To show the existence of $x_1<x_2<x_3<x_4$ with $f_n''(x_i)\cdot f_n''(x_{i+1})<0$ do we have to use the mean value theorem? :unsure:

(b) For the pointwise convergence :

\begin{equation*}\lim_{n\rightarrow \infty}f_n(x)=\lim_{n\rightarrow \infty}\frac{x+2n}{x^2+n}=2\end{equation*}

So the sequence converges pointwise to $f(x)=2$.

For the uniform convergence :

\begin{equation*}f_n(x)-f(x)=\frac{x+2n}{x^2+n}-2=\frac{x+2n-2x^2-2n}{x^2+n}=\frac{x-2x^2}{x^2+n}\rightarrow |f_n(x)-f(x)|=\left |\frac{x-2x^2}{x^2+n}\right |\end{equation*} The supremum of that expression doesn't go to zero, since when $x\rightarrow \infty$ then that expression goes to $2$. Therefore $f_n$ doesn't converge uniformly to $2$.

Is that correct and complete? :unsure:

Last edited by a moderator: