Mixing
Is $[0,+infty)$ closed and can the intermediate value theorem can be applied to functions of this domain?...
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The intermediate value Theorem states that if $f:[a,b]rightarrowmathbb{C}$ is a continious function then $f$ also must take all values in between $f(a)$ and $f(b)$.
$exp:mathbb{R}rightarrowmathbb{C}$ is a continious function and derived from that the functions $sin$ and $cos$ are also continious functions.
(Note: So far we have defined $costext{ and } sin$ 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 $mathbb{R^+}$?
real-analysis
asked Jan 24 at 22:13
RM777RM777
38312
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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
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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show 1 more comment
$begingroup$
The intermediate value Theorem states that if $f:[a,b]rightarrowmathbb{C}$ is a continious function then $f$ also must take all values in between $f(a)$ and $f(b)$.
$exp:mathbb{R}rightarrowmathbb{C}$ is a continious function and derived from that the functions $sin$ and $cos$ are also continious functions.
(Note: So far we have defined $costext{ and } sin$ 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 $mathbb{R^+}$?
real-analysis
asked Jan 24 at 22:13
RM777RM777
38312
$endgroup$
closed as unclear what you're asking by Lord Shark the Unknown, max_zorn, metamorphy, José Carlos Santos, Cesareo Jan 25 at 10:38
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
Your title poses a question that appears to be absent from your text.
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– Lord Shark the Unknown
Jan 24 at 22:15
1
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It is not compact since it's not bounded
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– 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
1
$begingroup$
@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
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@Heisenberg But why can I use the intermediate value Theorem?
$endgroup$
– RM777
Jan 24 at 22:21
|
show 1 more comment
$begingroup$
The intermediate value Theorem states that if $f:[a,b]rightarrowmathbb{C}$ is a continious function then $f$ also must take all values in between $f(a)$ and $f(b)$.
$exp:mathbb{R}rightarrowmathbb{C}$ is a continious function and derived from that the functions $sin$ and $cos$ are also continious functions.
(Note: So far we have defined $costext{ and } sin$ 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 $mathbb{R^+}$?
real-analysis
asked Jan 24 at 22:13
RM777RM777
38312
$endgroup$
The intermediate value Theorem states that if $f:[a,b]rightarrowmathbb{C}$ is a continious function then $f$ also must take all values in between $f(a)$ and $f(b)$.
$exp:mathbb{R}rightarrowmathbb{C}$ is a continious function and derived from that the functions $sin$ and $cos$ are also continious functions.
(Note: So far we have defined $costext{ and } sin$ 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 $mathbb{R^+}$?
real-analysis
real-analysis
asked Jan 24 at 22:13
RM777RM777
38312
asked Jan 24 at 22:13
RM777RM777
38312
edited Jan 24 at 22:23
RM777
asked Jan 24 at 22:13
RM777RM777
38312
asked Jan 24 at 22:13
RM777RM777
38312
asked Jan 24 at 22:13
RM777RM777
38312
38312
closed as unclear what you're asking by Lord Shark the Unknown, max_zorn, metamorphy, José Carlos Santos, Cesareo Jan 25 at 10:38
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Lord Shark the Unknown, max_zorn, metamorphy, José Carlos Santos, Cesareo Jan 25 at 10:38
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
Your title poses a question that appears to be absent from your text.
$endgroup$
– Lord Shark the Unknown
Jan 24 at 22:15
1
$begingroup$
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
1
$begingroup$
@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
|
show 1 more comment
4
$begingroup$
Your title poses a question that appears to be absent from your text.
$endgroup$
– Lord Shark the Unknown
Jan 24 at 22:15
1
$begingroup$
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
1
$begingroup$
@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
4
4
$begingroup$
Your title poses a question that appears to be absent from your text.
$endgroup$
– Lord Shark the Unknown
Jan 24 at 22:15
$begingroup$
Your title poses a question that appears to be absent from your text.
$endgroup$
– Lord Shark the Unknown
Jan 24 at 22:15
1
1
$begingroup$
It is not compact since it's not bounded
$endgroup$
– Heisenberg
Jan 24 at 22:16
$begingroup$
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
$begingroup$
$[0,infty)$ is closed, not bounded and hence not compact in $mathbb{R}$.
$endgroup$
– LoveTooNap29
Jan 24 at 22:18
1
1
$begingroup$
@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
$begingroup$
@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
$begingroup$
@Heisenberg But why can I use the intermediate value Theorem?
$endgroup$
– RM777
Jan 24 at 22:21
|
show 1 more comment
1 Answer
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oldest
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$begingroup$
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 $cin(0,b)$. Thus, if $cos xne0$ for every $xin[0,+infty)$ we can conclude that $cos x>0$, for every $xin[0,+infty)$.
The rest of the proof you are studying consists in deriving a contradiction from $cos x>0$ for every $xge0$.
answered Jan 24 at 22:34
egregegreg
184k1486205
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $cin(0,b)$. Thus, if $cos xne0$ for every $xin[0,+infty)$ we can conclude that $cos x>0$, for every $xin[0,+infty)$.
The rest of the proof you are studying consists in deriving a contradiction from $cos x>0$ for every $xge0$.
answered Jan 24 at 22:34
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
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 $cin(0,b)$. Thus, if $cos xne0$ for every $xin[0,+infty)$ we can conclude that $cos x>0$, for every $xin[0,+infty)$.
The rest of the proof you are studying consists in deriving a contradiction from $cos x>0$ for every $xge0$.
answered Jan 24 at 22:34
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
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 $cin(0,b)$. Thus, if $cos xne0$ for every $xin[0,+infty)$ we can conclude that $cos x>0$, for every $xin[0,+infty)$.
The rest of the proof you are studying consists in deriving a contradiction from $cos x>0$ for every $xge0$.
answered Jan 24 at 22:34
egregegreg
184k1486205
$endgroup$
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 $cin(0,b)$. Thus, if $cos xne0$ for every $xin[0,+infty)$ we can conclude that $cos x>0$, for every $xin[0,+infty)$.
The rest of the proof you are studying consists in deriving a contradiction from $cos x>0$ for every $xge0$.
answered Jan 24 at 22:34
egregegreg
184k1486205
answered Jan 24 at 22:34
egregegreg
184k1486205
answered Jan 24 at 22:34
egregegreg
184k1486205
answered Jan 24 at 22:34
egregegreg
184k1486205
184k1486205
add a comment |
add a comment |
4
$begingroup$
Your title poses a question that appears to be absent from your text.
$endgroup$
– Lord Shark the Unknown
Jan 24 at 22:15
1
$begingroup$
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
1
$begingroup$
@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