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Local & global extrema

mathmari

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Apr 14, 2013
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Hey !! :giggle:

Let $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$, $f(x,y)=\sin^2(x)\cdot \cos^2(y)$.

- Show that $f$ has at $\left (\frac{\pi}{2}, 0\right )$ a strictly local maximum and that is also a global maximum.

- Determine all points at which $f$ gets its global minimum.



I have sone the following :

It holds that $|\sin (x)|\leq 1$ so $\sin^2(x)\leq 1$ and also that $|\cos (x)|\leq 1$ so $\cos^2(x)\leq 1$.

From that we get that $f(x,y)\leq 1$, that means that $1$ is a maximum of $f$.

At the point $\left (\frac{\pi}{2}, 0\right )$ the function gets the value $1$, that means that $f$ has at $\left (\frac{\pi}{2}, 0\right )$ a strictly local maximum.

Is that correct so far? :unsure:
 

Klaas van Aarsen

MHB Seeker
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Mar 5, 2012
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It holds that $|\sin (x)|\leq 1$ so $\sin^2(x)\leq 1$ and also that $|\cos (x)|\leq 1$ so $\cos^2(x)\leq 1$.

From that we get that $f(x,y)\leq 1$, that means that $1$ is a maximum of $f$.
Hey mathmari !!

At this stage we don't know yet if $1$ is an actual maximum of the function.
We only know that $1$ is an upper bound of the function. 🤔

At the point $\left (\frac{\pi}{2}, 0\right )$ the function gets the value $1$, that means that $f$ has at $\left (\frac{\pi}{2}, 0\right )$ a strictly local maximum.
We now have a point where we achieve the upper bound, which implies that $1$ is indeed a maximum of the function.
However, at this stage we don't know yet, if it is a strictly local maximum. 🤔
 
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mathmari

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Apr 14, 2013
4,713
However, at this stage we don't know yet, if it is a strictly local maximum. 🤔
Do we have to show that in the neighbourhood of the point the function achieves only at this point the value $1$ ? Or how do we show that? :unsure:
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
9,591
Do we have to show that in the neighbourhood of the point the function achieves only at this point the value $1$ ? Or how do we show that?
Yes. That is what it means that a local maximum is strict. (Nod)

The function $\sin x$ has a strict local maximum at $x=\frac\pi 2$, doesn't it? 🤔
 

mathmari

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MHB Site Helper
Apr 14, 2013
4,713
The function $\sin x$ has a strict local maximum at $x=\frac\pi 2$, doesn't it? 🤔
So we have that for each other $x$ it holds that $\sin(x)<1$ and so we get that $\sin^2 (x)\cdot \cos^2 (y)<1$ for $x\neq \frac{\pi}{2}$, right?

So can we say that this means that at $\left (\frac{\pi}{2}, 0\right )$ the function has a strictly local maximum?

:unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
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So we have that for each other $x$ it holds that $\sin(x)<1$ and so we get that $\sin^2 (x)\cdot \cos^2 (y)<1$ for $x\neq \frac{\pi}{2}$, right?
Only in a small enough neighborhood of $(\frac\pi 2,0)$. 🧐

So can we say that this means that at $\left (\frac{\pi}{2}, 0\right )$ the function has a strictly local maximum?
We still have the case that we are in a small enough neighborhood with $x= \frac{\pi}{2}$.
Can we deduce in that case as well that the function value is strictly less than $1$? (Wondering)
 

mathmari

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MHB Site Helper
Apr 14, 2013
4,713
Only in a small enough neighborhood of $(\frac\pi 2,0)$. 🧐



We still have the case that we are in a small enough neighborhood with $x= \frac{\pi}{2}$.
Can we deduce in that case as well that the function value is strictly less than $1$? (Wondering)
In a small neighborhood of $0$ the function $\cos^2(y)$ is only $1$ at $y=0$ and in a small neighborhood of $\frac{\pi}{2}$ the function $\sin^2(x)$ is only $1$ at $x=\frac{\pi}{2}$, that means that in a small neighborhood of $(\frac\pi 2,0)$ the function $f$ is less than $1$.

Ia that correct so far? :unsure:
 

Klaas van Aarsen

MHB Seeker
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Mar 5, 2012
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In a small neighborhood of $0$ the function $\cos^2(y)$ is only $1$ at $y=0$ and in a small neighborhood of $\frac{\pi}{2}$ the function $\sin^2(x)$ is only $1$ at $x=\frac{\pi}{2}$, that means that in a small neighborhood of $(\frac\pi 2,0)$ the function $f$ is less than $1$.

Is that correct so far?
Yep. (Nod)
 

mathmari

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MHB Site Helper
Apr 14, 2013
4,713
So we get that $f$ has at this point a strictly local maximum , right?

To show that this is also a global maximum do we say that it cannot be greater than $1$ ?

Also can we just mention the parts with the small neighborhoods at #7 or do we have to prove that?

:unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
So we get that $f$ has at this point a strictly local maximum , right?

To show that this is also a global maximum do we say that it cannot be greater than $1$ ?
Indeed.
Suppose it was not a global maximum. Then there would have to be another point where $f$ is greater than $1$.
But we already established that $1$ is an upper bound, leading to a contradiction.
So it is indeed also a global maximum. (Nod)

Also can we just mention the parts with the small neighborhoods at #7 or do we have to prove that?
The functions $\sin$ and $\cos$ are well known to have a strict local maximum at $\frac\pi 2$ respectively $0$.
So they are less than $1$ in a sufficiently small neighborhood around those points.
Consequently their squares and product are also less than $1$ in a sufficiently small neighborhood.
I believe that is sufficient to serve as proof. 🤔
 

mathmari

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Apr 14, 2013
4,713
Indeed.
Suppose it was not a global maximum. Then there would have to be another point where $f$ is greater than $1$.
But we already established that $1$ is an upper bound, leading to a contradiction.
So it is indeed also a global maximum. (Nod)


The functions $\sin$ and $\cos$ are well known to have a strict local maximum at $\frac\pi 2$ respectively $0$.
So they are less than $1$ in a sufficiently small neighborhood around those points.
Consequently their squares and product are also less than $1$ in a sufficiently small neighborhood.
I believe that is sufficient to serve as proof. 🤔
Ok! I see!! :geek:

At the second part,where we have to determine all points at which $f$ gets its global minimum, do we have to do the same as in the first part just for minimum instead of maximum ? :unsure:
 

Klaas van Aarsen

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Mar 5, 2012
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At the second part,where we have to determine all points at which $f$ gets its global minimum, do we have to do the same as in the first part just for minimum instead of maximum ?
Yep. (Nod)
 

mathmari

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Apr 14, 2013
4,713

Klaas van Aarsen

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Mar 5, 2012
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mathmari

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MHB Site Helper
Apr 14, 2013
4,713
So we have that $\sin^2(x)\cdot \cos^2(y)\geq0$ and at all points with $x=n\pi$ and all points with $ y=(2k-1)\frac{\pi}{2}$ the function achieves that minimum, which is a global minimum because there is no other lower value of the function.

Is that correct and complete? Or do we have to prove something more?

:unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
So we have that $\sin^2(x)\cdot \cos^2(y)\geq0$ and at all points with $x=n\pi$ and all points with $ y=(2k-1)\frac{\pi}{2}$ the function achieves that minimum, which is a global minimum because there is no other lower value of the function.

Is that correct and complete? Or do we have to prove something more?
Correct and complete. :geek:
 

mathmari

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MHB Site Helper
Apr 14, 2013
4,713