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The intermediate value Theorem states that if $f:[a,b]\rightarrow\mathbb{C}$ is a continious function then $f$ also must take all values in between $f(a)$ and $f(b)$.

$\exp:\mathbb{R}\rightarrow\mathbb{C}$ is a continious function and derived from that the functions $\sin$ and $\cos$ are also continious functions.

(Note: So far we have defined $\cos\text{ and } \sin$ only for $\mathbb{R}$)

I have a question about an application of this Theorem now.

It was used in a proof that $\cos$ has at least one Zero Point.

It starts with the assumption that it has no Zero Point and then uses the Argument that $\cos$ is continious on $[0,+\infty]$ + the fact that $\cos(0)=1$ and then makes use of the intermediate value Theorem to conclude that $\cos$ has no negative values. [...]

But why can we use the Theorem here, i.e why are the conditions met that we can use it?

First of all $+\infty$ is not even in $\mathbb{R}$, I must have made an mistake when I wrote down from the blackboard, we could reformulate that $\cos$ is continious on $[0,+\infty)$. But what is the Definition of this interval again?

Why is it compact? - Because otherwise we could not use the Theorem

And why does it implicate that $\cos$ is continious in $\mathbb{R^+}$?

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closed as unclear what you're asking by Lord Shark the Unknown, max_zorn, metamorphy, José Carlos Santos, Cesareo Jan 25 at 10:38

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    $\begingroup$ Your title poses a question that appears to be absent from your text. $\endgroup$ – Lord Shark the Unknown Jan 24 at 22:15
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    $\begingroup$ It is not compact since it's not bounded $\endgroup$ – Heisenberg Jan 24 at 22:16
  • $\begingroup$ $[0,\infty)$ is closed, not bounded and hence not compact in $\mathbb{R}$. $\endgroup$ – LoveTooNap29 Jan 24 at 22:18
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    $\begingroup$ @LordSharktheUnknown Once a question of mine got closed because I did not put any context to it that's why I added the reason why I am asking this time. I have changed the title now. $\endgroup$ – RM777 Jan 24 at 22:19
  • $\begingroup$ @Heisenberg But why can I use the intermediate value Theorem? $\endgroup$ – RM777 Jan 24 at 22:21
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The cosine function is continuous on $[0,+\infty)$. This is not a compact interval, but it doesn't really matter. For every $b>0$, you can apply the intermediate value theorem on the interval $[0,b]$.

Suppose there exists $b$ such that $\cos b<0$, then the IVT applied to the interval $[0,b]$ tells us that $\cos c=0$, for some $c\in(0,b)$. Thus, if $\cos x\ne0$ for every $x\in[0,+\infty)$ we can conclude that $\cos x>0$, for every $x\in[0,+\infty)$.

The rest of the proof you are studying consists in deriving a contradiction from $\cos x>0$ for every $x\ge0$.

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