Global Extrema of a Trigonometric Function

[mathmari]

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:

 

[I like Serena]

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.

 

[mathmari]

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:

 

[I like Serena]

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]

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:

 

[I like Serena]

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]

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:

 

[I like Serena]

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]

Yep. (Nod)

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:

 

[I like Serena]

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]

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:

 

[I like Serena]

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]

Yep. (Nod)

Ah the minimum in this case is $0$ because of the square, right? :unsure:

 

[I like Serena]

Ah the minimum in this case is $0$ because of the square, right?

Indeed. (Nod)

 

[mathmari]

Indeed. (Nod)

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:

 

[I like Serena]

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]

Correct and complete. :geek:

Great! Thank you very much!