I When does an inequality indicate a maximum or minimum value?

[Mr Davis 97]

If I write $-1 \le\cos x \le 1$, we all clearly know what this means, that not only is $\cos x$ contained in this interval but that its max is 1 and its min is -1. However, what if I write $-1 \le \cos n \le 1$, where $n \in \mathbb{N}$. What does the inequality mean in this case? This inequality is still true, but it doesn't say anything about the max or the min. I guess my question is this, in what cases can we suppose that an inequality is saying something about a max or a min, and in what cases is the inequality simply stating that something is bounded above or below?

 

[fresh_42]

If I write $-1 \le\cos x \le 1$, we all clearly know what this means, that not only is $\cos x$ contained in this interval but that its max is 1 and its min is -1.

But only because we know it from some other sources.

However, what if I write $-1 \le \cos n \le 1$, where $n \in \mathbb{N}$. What does the inequality mean in this case?

The same, that the values of $\cos n \in [-1,1]$.

This inequality is still true, but it doesn't say anything about the max or the min.

Neither did the first.

I guess my question is this, in what cases can we suppose that an inequality is saying something about a max or a min,...

Never, unless we add information. I mean, to know that $\cos \mathbb{N}$ doesn't reach the boundaries requires the knowledge, that $\pi$ is transcendental. Not a trivial result.

... and in what cases is the inequality simply stating that something is bounded above or below?

Always.

There are of course situations where we can conclude something about the boundaries, e.g. in linear optimization circumstances, or if the function has monotony properties. Even the knowledge, that the maximum is inside a set requires a compact set and a continuous function.

 

[Mr Davis 97]

But only because we know it from some other sources.

The same, that the values of $\cos n \in [-1,1]$.

Neither did the first.

Never, unless we add information. I mean, to know that $\cos \mathbb{N}$ doesn't reach the boundaries requires the knowledge, that $\pi$ is transcendental. Not a trivial result.

Always.

There are of course situations where we can conclude something about the boundaries, e.g. in linear optimization circumstances, or if the function has monotony properties. Even the knowledge, that the maximum is inside a set requires a compact set and a continuous function.

Alright, I think I understand now. One more thing though, as an example. By AM-GM, we have that $x+1/x \ge 2$. In trying to find the minimum value of $f(x) = x+1/x$ with $x \ge 0$, how can how can we conclude that 2 is the minimum, if technically all it is is a lower bound?

 

[fresh_42]

Alright, I think I understand now. One more thing though, as an example. By AM-GM, we have that $x+1/x \ge 2$. In trying to find the minimum value of $f(x) = x+1/x$ with $x \ge 0$, how can how can we conclude that 2 is the minimum, if technically all it is is a lower bound?

Better to define $x > 0$. You can either use the knowledge that $f(1)=2$ as you did, when you said $\cos x$ takes the boundary values $\pm 1$ because you already knew $\cos 0 = 1$ and $\cos \pi = -1$, or by solving $f\,'(x)=0$ and look for the minima and maxima of this differential function. Or you prove that $f$ is strictly monotone decreasing on $0<x\le 2$ and strictly monotone increasing for $x \geq 2$, which appears to be more work to do.

 

[member 587159]

If I write $-1 \le\cos x \le 1$, we all clearly know what this means, that not only is $\cos x$ contained in this interval but that its max is 1 and its min is -1. However, what if I write $-1 \le \cos n \le 1$, where $n \in \mathbb{N}$. What does the inequality mean in this case? This inequality is still true, but it doesn't say anything about the max or the min. I guess my question is this, in what cases can we suppose that an inequality is saying something about a max or a min, and in what cases is the inequality simply stating that something is bounded above or below?

We can also write $-100 \leq \cos x \leq 100$ and this is true as well. Really no problem here, but your statement contained more information.

 

[nuuskur]

In a closed bounded set, a continuous function always attains its extremal values (the general result is much more..well..general). This is fine to talk about for subsets of $\mathbb R$, but not for $\mathbb N$. Continuity is a topological property, which subsets do you call open sets in $\mathbb N$?

Naturally, $\cos 0$ is the maximum value, but you won't find any other multiple of $\pi$ in $\mathbb N_0$. You can ask whether you get a minimum value too for some $\mathbb N$, but that is not obvious.

 

[nuuskur]

for some $\mathbb N$

bah.. for some $n\in\mathbb N$ is what I meant.

 

[FactChecker]

If I write $-1 \le\cos x \le 1$, we all clearly know what this means, that not only is $\cos x$ contained in this interval but that its max is 1 and its min is -1.

Not true. Just to focus on an important part of other comments, you are reading too much into $-1 \le\cos x \le 1$. It says nothing about the min and max except that they must be somewhere in the closed interval [-1,1].

 

[jbriggs444]

Not true. Just to focus on an important part of other comments, you are reading too much into $-1 \le\cos x \le 1$. It says nothing about the min and max except that they must be somewhere in the closed interval [-1,1].

To be pickier, by itself it does not guarantee that either min or max even exist. Although, as @nuuskur points out, the continutity of cos x does allow one to deduce existence.