Extrema and convergence of sequence

[mathmari]

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:

 

[I like Serena]

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

Is that correct? Or is it meant to calculate all extrema (local and global) in an other way?

Correct.
We've found a local minimum and a local maximum.
Are they also global extrema or not? (Wondering)

If $|x|\rightarrow \infty$ then $f_n$ does to $0$ since the denominator goes faster to infinity, right?

Yep. (Nod)

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?

I think we should check the zeros of $f_n''$ and see what its signs are between those zeros - just like we did for $f_n'$. (Thinking)

(b) For the pointwise convergence :
For the uniform convergence :
Is that correct and complete?

All correct and complete. (Sun)

 

[mathmari]

We've found a local minimum and a local maximum.
Are they also global extrema or not? (Wondering)

Do we have to calculate the values of the function at these points and take into consideration also the behaviour of the function when $|x|\rightarrow \infty$ ?:unsure:

I think we should check the zeros of $f_n''$ and see what its signs are between those zeros - just like we did for $f_n'$. (Thinking)

The second derivative is \begin{align*}f_n''(x)&=\frac{(-x^2-4nx+n)'\cdot (x^2+n)^2-(-x^2-4nx+n)\cdot ((x^2+n)^2)'}{((x^2+n)^2)^2}\\ & =\frac{(-2x-4n)\cdot (x^2+n)^2-(-x^2-4nx+n)\cdot (2(x^2+n)\cdot 2x)}{(x^2+n)^4}\\ & =\frac{(x^2+n)\cdot \left [(-2x-4n)\cdot (x^2+n)-(-x^2-4nx+n)\cdot (2\cdot 2x)\right ]}{(x^2+n)^4} \\ & =\frac{(x^2+n)\cdot \left [2x^3+12nx^2-6nx-4n^2\right ]}{(x^2+n)^4}\\ & =\frac{2x^3+12nx^2-6nx-4n^2}{(x^2+n)^3}\end{align*}
We set it equal to zero and we get \begin{align*}f_n''(x)=0 & \Rightarrow\frac{2x^3+12nx^2-6nx-4n^2}{(x^2+n)^3}=0 \\ & \Rightarrow 2x^3+12nx^2-6nx-4n^2=0\end{align*}
According to Wolfram the roots are these ones .At the exercise we have that " The $x_i$ depend on $n$, they don't have to be given explicitely. ", so I think they mean that we don't have to calculate the roots, or am I wrong?So it must be also an other way, or not? :unsure:

 

[I like Serena]

Do we have to calculate the values of the function at these points and take into consideration also the behaviour of the function when $|x|\rightarrow \infty$ ?

No need to calculate the function values.
Consider that at minus infinity the function approaches 0. Then it goes down to a local minumum, goes up to a local maximum, and goes down again until at plus infinity it approaches 0.
Can we deduce whether the local extrema are global extrema? (Wondering)

The second derivative is \begin{align*}f_n''(x)&=\frac{(-x^2-4nx+n)'\cdot (x^2+n)^2-(-x^2-4nx+n)\cdot ((x^2+n)^2)'}{((x^2+n)^2)^2}\\ & =\frac{(-2x-4n)\cdot (x^2+n)^2-(-x^2-4nx+n)\cdot (2(x^2+n)\cdot 2x)}{(x^2+n)^4}\\ & =\frac{(x^2+n)\cdot \left [(-2x-4n)\cdot (x^2+n)-(-x^2-4nx+n)\cdot (2\cdot 2x)\right ]}{(x^2+n)^4} \\ & =\frac{(x^2+n)\cdot \left [2x^3+12nx^2-6nx-4n^2\right ]}{(x^2+n)^4}\\ & =\frac{2x^3+12nx^2-6nx-4n^2}{(x^2+n)^3}\end{align*}
We set it equal to zero and we get \begin{align*}f_n''(x)=0 & \Rightarrow\frac{2x^3+12nx^2-6nx-4n^2}{(x^2+n)^3}=0 \\ & \Rightarrow 2x^3+12nx^2-6nx-4n^2=0\end{align*}
According to Wolfram the roots are these ones .At the exercise we have that " The $x_i$ depend on $n$, they don't have to be given explicitely. ", so I think they mean that we don't have to calculate the roots, or am I wrong?So it must be also an other way, or not? :unsure:

It suffices if we can deduce that there are 3 distinct zeros and if we can deduce it switches signs between those zeros.
Consider again the behavior of the function. Where must the inflection points be? (Wondering)

 

[mathmari]

No need to calculate the function values.
Consider that at minus infinity the function approaches 0. Then it goes down to a local minumum, goes up to a local maximum, and goes down again until at plus infinity it approaches 0.
Can we deduce whether the local extrema are global extrema? (Wondering)

So are the local extrema also global? :unsure:

It suffices if we can deduce that there are 3 distinct zeros and if we can deduce it switches signs between those zeros.
Consider again the behavior of the function. Where must the inflection points be? (Wondering)

I got stuck right now. Could you explain that further to me? :unsure:

 

[I like Serena]

So are the local extrema also global?

Yes.

I got stuck right now. Could you explain that further to me?

At minus infinity $f_n$ has a horizontal slope.
Then it decreases to a local minimum.
The slope first became more and more negative, and then became less and less negative until the slope was horizontal in the local minimum.
It means that there is an inflection point between minus infinity and the local minimum.
An inflection point means that $f_n''$ is zero.

Similarly we have an inflection point between the minimum and the maximum.
And we must also have an inflection point between the maximum and plus infinity where we approach a horizontal slope again.

Pick $x_1$ left of the first inflection point where $f_n''(x_1)<0$.
And pick $x_2$ between the first and second inflection point where $f_n''(x_2)>0$.
Then $f_n''(x_1)\cdot f_n''(x_2)<0$.

 

[I like Serena]

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?

We can indeed use the mean value theorem. (Nod)

You have found that there are $a$ and $b$ with $a<b$, $f_n'(a)=0$, and $f_n'(b)=0$.
Since $f_n'$ is differentiable, the mean value theorem tells us that there must be a $c$ with $a<c<b$ for which we have $f_n''(c)=0$.

Can we find if $f_n''$ is zero somewhere for $x<a$? (Wondering)

 

[I like Serena]

We can do a function analysis and draw the following chart of zeroes and signs.
\begin{tikzpicture}
\begin{scope}
\draw[latex-latex] (-6,0) node[ left ] {$f_n$} -- (6,0);
\draw foreach \i/\x/\y in {-6/{-\infty}/0, -2.5/{a_n}/{}, 0/{-2n}/0, 2.5/{b_n}/{}, 6/\infty/0} { (\i,0.1) node[above] {$\y$} -- (\i,-0.1) node[below] {$\x$} };
\draw foreach \i/\y in {-4/{descending}, -1.25/{ascending}, 1.25/{ascending}, 4/{descending}} { (\i,0.1) node[above] {$\y$} };
\end{scope}
\begin{scope}[yshift=-2cm]
\draw[latex-latex] (-6,0) node[ left ] {$f_n'$} -- (6,0);
\draw foreach \i/\x/\y in {-6/{-\infty}/0, -2.5/{a_n}/0, 2.5/{b_n}/0, 6/\infty/0} { (\i,0.1) node[above] {$\y$} -- (\i,-0.1) node[below] {$\x$} };
\draw foreach \i/\y in {-4/-, 0/+, 4/-} { (\i,0.1) node[above] {$\y$} };
\end{scope}
\begin{scope}[yshift=-4cm]
\draw[latex-latex] (-6,0) node[ left ] {$f_n''$} -- (6,0);
\draw foreach \i/\x/\y in {-6/{-\infty}/0, -4//0, 0/c/0, 4//0, 6/\infty/0} { (\i,0.1) node[above] {$\y$} -- (\i,-0.1) node[below] {$\x$} };
\draw foreach \i/\x/\y in {-5/{x_1}/-, -2/{x_2}/+, 2/{x_3}/-, 5/{x_4}/+} { (\i,0.1) node[above] {$\y$} (\i,-0.1) node[below] {$\x$} };
\end{scope}
\end{tikzpicture}
It shows where $f_n$, $f_n'$, and $f_n''$ are zero, what their signs are, and how they behave at the infinities.
We can pick $x_i$ as in the diagram can't we?

 

[mathmari]

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 that we check for $\c\rightarrow \infty$ ?
In that way we show that the supremum is greater or equal to $2$ and so the limit when $n\rightarrow \infty$ goes to $2$ or greater, and not to $0$, that's why we conclude that the convergence is not uniform.
Is that correct?

:unsure:

 

[I like Serena]

Is that correct that we check for $\c\rightarrow \infty$ ?
In that way we show that the supremum is greater or equal to $2$ and so the limit when $n\rightarrow \infty$ goes to $2$ or greater, and not to $0$, that's why we conclude that the convergence is not uniform.
Is that correct?

Yep. Correct assuming you meant $x\to\infty$ instead of $\c\rightarrow \infty$. (Nod)