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Show that $e^{frac{pi}{2}i }= i$ - Problem to understand a certain convergence
Prove that for $cos (alpha ) = frac{1}{3}$, $alpha < frac{pi}{2} - frac{1}{3}$Trigonometric identity from Fourier matrixHow to convert $3sin(x)cos(x)$ into expression involving only $sin (x)$Prove that $int_0^x frac {sin t}{t+1}dt geq 0 ~forall~xgeq 0$the double angle identities-sin2AUnderstanding this proof that $limlimits_{hto 0}frac{cos(h)-1}{h}=0$Prove that the sequence $left(frac{6+cos (n^2)}{n}right)_{nin mathbb{N}^*}$ approaches $0$Show that a certain parametric equation does not represent a cardioidProve that $1-frac{x^2}{2}<cos{x}<1-frac{x^2}{2}+frac{x^4}{24}$Show that $cos$ has at least one zeropoint
$begingroup$
We have defined $e^{ix}=cos x+ i sin x$ where $xinmathbb{R}$.
I also have proved earlier that $cos(frac{pi}{2})=0$ and know that $sin(0)=0$
Now I want to prove that
$e^{frac{pi}{2}i }= i$
I know that
$|e^{xi }|= 1,forall_{xinmathbb{R}}$
Therefore
$1=|e^{frac{pi}{2}i}|=|cos(frac{pi}{2})+isin(frac{pi}{2})|=|sin(frac{pi}{2})|$
And now the part that I don't understand
Because $sin'(x)=cos(x)>0,forall_{xin[0,frac{pi}{2})}$
$Longrightarrowsin(frac{pi}{2})=1$
$Longrightarrow e^{frac{pi}{2}i }= i$
Please explain me which Theorem we use here. And how this Situation can be generalised, so if I Encounter something similiar I directly know what to do
Edit:
Can it be shown constructively, i.e. with the mean-value-theorem for example?
real-analysis limits trigonometry
asked Jan 26 at 20:17
New2MathNew2Math
12312
$endgroup$
|
show 7 more comments
$begingroup$
We have defined $e^{ix}=cos x+ i sin x$ where $xinmathbb{R}$.
I also have proved earlier that $cos(frac{pi}{2})=0$ and know that $sin(0)=0$
Now I want to prove that
$e^{frac{pi}{2}i }= i$
I know that
$|e^{xi }|= 1,forall_{xinmathbb{R}}$
Therefore
$1=|e^{frac{pi}{2}i}|=|cos(frac{pi}{2})+isin(frac{pi}{2})|=|sin(frac{pi}{2})|$
And now the part that I don't understand
Because $sin'(x)=cos(x)>0,forall_{xin[0,frac{pi}{2})}$
$Longrightarrowsin(frac{pi}{2})=1$
$Longrightarrow e^{frac{pi}{2}i }= i$
Please explain me which Theorem we use here. And how this Situation can be generalised, so if I Encounter something similiar I directly know what to do
Edit:
Can it be shown constructively, i.e. with the mean-value-theorem for example?
real-analysis limits trigonometry
asked Jan 26 at 20:17
New2MathNew2Math
12312
$endgroup$
$begingroup$
"We have defined $e^{ix}=cos x+ i sin x$" - So just plug in $x = frac{pi}{2}$ and you get $i$..
$endgroup$
– user635162
Jan 26 at 20:38
$begingroup$
Is the point of your question to prove $cosfracpi2=0$ and $sinfracpi2=1$? What properties of sine and cosine are allowed to be known to this point?
$endgroup$
– user
Jan 26 at 20:40
$begingroup$
The Addition Theorem and the differentials. We don't know anything About the periodicity , the Intention was to Show $e^{ipi}=-1$
$endgroup$
– New2Math
Jan 26 at 20:43
$begingroup$
Tell us what definition of sine/cosine is used.
$endgroup$
– user
Jan 26 at 20:44
1
$begingroup$
There is simply no way we can answer this question without having all you axioms, definitions, and contexts explained explicitly.
$endgroup$
– fleablood
Jan 28 at 23:51
|
show 7 more comments
$begingroup$
We have defined $e^{ix}=cos x+ i sin x$ where $xinmathbb{R}$.
I also have proved earlier that $cos(frac{pi}{2})=0$ and know that $sin(0)=0$
Now I want to prove that
$e^{frac{pi}{2}i }= i$
I know that
$|e^{xi }|= 1,forall_{xinmathbb{R}}$
Therefore
$1=|e^{frac{pi}{2}i}|=|cos(frac{pi}{2})+isin(frac{pi}{2})|=|sin(frac{pi}{2})|$
And now the part that I don't understand
Because $sin'(x)=cos(x)>0,forall_{xin[0,frac{pi}{2})}$
$Longrightarrowsin(frac{pi}{2})=1$
$Longrightarrow e^{frac{pi}{2}i }= i$
Please explain me which Theorem we use here. And how this Situation can be generalised, so if I Encounter something similiar I directly know what to do
Edit:
Can it be shown constructively, i.e. with the mean-value-theorem for example?
real-analysis limits trigonometry
asked Jan 26 at 20:17
New2MathNew2Math
12312
$endgroup$
We have defined $e^{ix}=cos x+ i sin x$ where $xinmathbb{R}$.
I also have proved earlier that $cos(frac{pi}{2})=0$ and know that $sin(0)=0$
Now I want to prove that
$e^{frac{pi}{2}i }= i$
I know that
$|e^{xi }|= 1,forall_{xinmathbb{R}}$
Therefore
$1=|e^{frac{pi}{2}i}|=|cos(frac{pi}{2})+isin(frac{pi}{2})|=|sin(frac{pi}{2})|$
And now the part that I don't understand
Because $sin'(x)=cos(x)>0,forall_{xin[0,frac{pi}{2})}$
$Longrightarrowsin(frac{pi}{2})=1$
$Longrightarrow e^{frac{pi}{2}i }= i$
Please explain me which Theorem we use here. And how this Situation can be generalised, so if I Encounter something similiar I directly know what to do
Edit:
Can it be shown constructively, i.e. with the mean-value-theorem for example?
real-analysis limits trigonometry
real-analysis limits trigonometry
asked Jan 26 at 20:17
New2MathNew2Math
12312
asked Jan 26 at 20:17
New2MathNew2Math
12312
edited Jan 26 at 20:50
New2Math
asked Jan 26 at 20:17
New2MathNew2Math
12312
asked Jan 26 at 20:17
New2MathNew2Math
12312
asked Jan 26 at 20:17
New2MathNew2Math
12312
12312
$begingroup$
"We have defined $e^{ix}=cos x+ i sin x$" - So just plug in $x = frac{pi}{2}$ and you get $i$..
$endgroup$
– user635162
Jan 26 at 20:38
$begingroup$
Is the point of your question to prove $cosfracpi2=0$ and $sinfracpi2=1$? What properties of sine and cosine are allowed to be known to this point?
$endgroup$
– user
Jan 26 at 20:40
$begingroup$
The Addition Theorem and the differentials. We don't know anything About the periodicity , the Intention was to Show $e^{ipi}=-1$
$endgroup$
– New2Math
Jan 26 at 20:43
$begingroup$
Tell us what definition of sine/cosine is used.
$endgroup$
– user
Jan 26 at 20:44
1
$begingroup$
There is simply no way we can answer this question without having all you axioms, definitions, and contexts explained explicitly.
$endgroup$
– fleablood
Jan 28 at 23:51
|
show 7 more comments
$begingroup$
"We have defined $e^{ix}=cos x+ i sin x$" - So just plug in $x = frac{pi}{2}$ and you get $i$..
$endgroup$
– user635162
Jan 26 at 20:38
$begingroup$
Is the point of your question to prove $cosfracpi2=0$ and $sinfracpi2=1$? What properties of sine and cosine are allowed to be known to this point?
$endgroup$
– user
Jan 26 at 20:40
$begingroup$
The Addition Theorem and the differentials. We don't know anything About the periodicity , the Intention was to Show $e^{ipi}=-1$
$endgroup$
– New2Math
Jan 26 at 20:43
$begingroup$
Tell us what definition of sine/cosine is used.
$endgroup$
– user
Jan 26 at 20:44
1
$begingroup$
There is simply no way we can answer this question without having all you axioms, definitions, and contexts explained explicitly.
$endgroup$
– fleablood
Jan 28 at 23:51
$begingroup$
"We have defined $e^{ix}=cos x+ i sin x$" - So just plug in $x = frac{pi}{2}$ and you get $i$..
$endgroup$
– user635162
Jan 26 at 20:38
$begingroup$
"We have defined $e^{ix}=cos x+ i sin x$" - So just plug in $x = frac{pi}{2}$ and you get $i$..
$endgroup$
– user635162
Jan 26 at 20:38
$begingroup$
Is the point of your question to prove $cosfracpi2=0$ and $sinfracpi2=1$? What properties of sine and cosine are allowed to be known to this point?
$endgroup$
– user
Jan 26 at 20:40
$begingroup$
Is the point of your question to prove $cosfracpi2=0$ and $sinfracpi2=1$? What properties of sine and cosine are allowed to be known to this point?
$endgroup$
– user
Jan 26 at 20:40
$begingroup$
The Addition Theorem and the differentials. We don't know anything About the periodicity , the Intention was to Show $e^{ipi}=-1$
$endgroup$
– New2Math
Jan 26 at 20:43
$begingroup$
The Addition Theorem and the differentials. We don't know anything About the periodicity , the Intention was to Show $e^{ipi}=-1$
$endgroup$
– New2Math
Jan 26 at 20:43
$begingroup$
Tell us what definition of sine/cosine is used.
$endgroup$
– user
Jan 26 at 20:44
$begingroup$
Tell us what definition of sine/cosine is used.
$endgroup$
– user
Jan 26 at 20:44
1
1
$begingroup$
There is simply no way we can answer this question without having all you axioms, definitions, and contexts explained explicitly.
$endgroup$
– fleablood
Jan 28 at 23:51
$begingroup$
There is simply no way we can answer this question without having all you axioms, definitions, and contexts explained explicitly.
$endgroup$
– fleablood
Jan 28 at 23:51
|
show 7 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Essentially we need only to prove $sinfracpi2>0$. The answer will use the mean value theorem (MVT) as asked in question.
The definition of $e^z,sin z, cos z$ implies that the functions are continuous and differentiable. Particularly:
$$
forall zinmathbb C: sin'z=cos z.tag1
$$
As $fracpi2$ is by definition (see discussion in comments) the least positive root of $cos x$ and $cos(0)=1>0$ we have from the continuity of the function:
$$
forall xinleft(0,fracpi2right): cos x>0.
$$
Together with (1) this implies:
$$
forall xinleft(0,fracpi2right): sin' x>0.tag2
$$
Now by MVT there exists such a point $0<c<fracpi2$ that
$$
sin'(c)=frac{sinfracpi2-sin0}{fracpi2-0}
implies sinfracpi2=fracpi2sin'(c).
$$
As both $sin'(c)$ and $fracpi2$ are positive, we obtain $$sinfracpi2>0.$$
answered Jan 28 at 23:45
useruser
5,41411030
$endgroup$
add a comment |
$begingroup$
You know that $ sin(x) in mathbb{R}$, so that
$$leftvert sin left( frac{pi}{2} right) rightvert=1 Rightarrow sin left( frac{pi}{2} right)= pm1$$
If $sin'(x)=cos(x) >0$ for all $x in [0,frac{pi}{2})$, then this means that the sinus function is strictly growing on $x in [0,frac{pi}{2})$. With $sin(0)=0$, with $sin$ being an continuous function, then the only possibility is $$sin left( frac{pi}{2} right)=+1$$
Hence $$e^{ifrac{pi}{2}} = i$$
Edit If you want to proove that $sin'(x)=cos(x)$, so note that $$frac{d}{dx}left( e^{ix} right) = ie^{ix} = icos(x) - sin(x)$$
Implies that
$$ cos'(x) + isin'(x) = -sin(x) + icos(x)$$
Prooving that $sin'(x)=cos(x)$
answered Jan 26 at 20:38
Thomas LesgourguesThomas Lesgourgues
1,128119
$endgroup$
$begingroup$
But why is it the only possibility?
$endgroup$
– New2Math
Jan 26 at 20:40
$begingroup$
it can be only +1 or -1, because it is in $mathbb{R}$. If the function is growing, starting at 0, it cannot be -1
$endgroup$
– Thomas Lesgourgues
Jan 26 at 20:43
$begingroup$
Because otherwise it would contradict continuity, Right?
$endgroup$
– New2Math
Jan 26 at 20:45
$begingroup$
yes, exactly...
$endgroup$
– Thomas Lesgourgues
Jan 26 at 20:46
$begingroup$
I was wondering if one could Show the property with a constructive proof and not by contradiction because I feel like I should know some Theorems by now which would imply the result directly
$endgroup$
– New2Math
Jan 26 at 20:48
|
show 1 more comment
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Essentially we need only to prove $sinfracpi2>0$. The answer will use the mean value theorem (MVT) as asked in question.
The definition of $e^z,sin z, cos z$ implies that the functions are continuous and differentiable. Particularly:
$$
forall zinmathbb C: sin'z=cos z.tag1
$$
As $fracpi2$ is by definition (see discussion in comments) the least positive root of $cos x$ and $cos(0)=1>0$ we have from the continuity of the function:
$$
forall xinleft(0,fracpi2right): cos x>0.
$$
Together with (1) this implies:
$$
forall xinleft(0,fracpi2right): sin' x>0.tag2
$$
Now by MVT there exists such a point $0<c<fracpi2$ that
$$
sin'(c)=frac{sinfracpi2-sin0}{fracpi2-0}
implies sinfracpi2=fracpi2sin'(c).
$$
As both $sin'(c)$ and $fracpi2$ are positive, we obtain $$sinfracpi2>0.$$
answered Jan 28 at 23:45
useruser
5,41411030
$endgroup$
add a comment |
$begingroup$
Essentially we need only to prove $sinfracpi2>0$. The answer will use the mean value theorem (MVT) as asked in question.
The definition of $e^z,sin z, cos z$ implies that the functions are continuous and differentiable. Particularly:
$$
forall zinmathbb C: sin'z=cos z.tag1
$$
As $fracpi2$ is by definition (see discussion in comments) the least positive root of $cos x$ and $cos(0)=1>0$ we have from the continuity of the function:
$$
forall xinleft(0,fracpi2right): cos x>0.
$$
Together with (1) this implies:
$$
forall xinleft(0,fracpi2right): sin' x>0.tag2
$$
Now by MVT there exists such a point $0<c<fracpi2$ that
$$
sin'(c)=frac{sinfracpi2-sin0}{fracpi2-0}
implies sinfracpi2=fracpi2sin'(c).
$$
As both $sin'(c)$ and $fracpi2$ are positive, we obtain $$sinfracpi2>0.$$
answered Jan 28 at 23:45
useruser
5,41411030
$endgroup$
add a comment |
$begingroup$
Essentially we need only to prove $sinfracpi2>0$. The answer will use the mean value theorem (MVT) as asked in question.
The definition of $e^z,sin z, cos z$ implies that the functions are continuous and differentiable. Particularly:
$$
forall zinmathbb C: sin'z=cos z.tag1
$$
As $fracpi2$ is by definition (see discussion in comments) the least positive root of $cos x$ and $cos(0)=1>0$ we have from the continuity of the function:
$$
forall xinleft(0,fracpi2right): cos x>0.
$$
Together with (1) this implies:
$$
forall xinleft(0,fracpi2right): sin' x>0.tag2
$$
Now by MVT there exists such a point $0<c<fracpi2$ that
$$
sin'(c)=frac{sinfracpi2-sin0}{fracpi2-0}
implies sinfracpi2=fracpi2sin'(c).
$$
As both $sin'(c)$ and $fracpi2$ are positive, we obtain $$sinfracpi2>0.$$
answered Jan 28 at 23:45
useruser
5,41411030
$endgroup$
Essentially we need only to prove $sinfracpi2>0$. The answer will use the mean value theorem (MVT) as asked in question.
The definition of $e^z,sin z, cos z$ implies that the functions are continuous and differentiable. Particularly:
$$
forall zinmathbb C: sin'z=cos z.tag1
$$
As $fracpi2$ is by definition (see discussion in comments) the least positive root of $cos x$ and $cos(0)=1>0$ we have from the continuity of the function:
$$
forall xinleft(0,fracpi2right): cos x>0.
$$
Together with (1) this implies:
$$
forall xinleft(0,fracpi2right): sin' x>0.tag2
$$
Now by MVT there exists such a point $0<c<fracpi2$ that
$$
sin'(c)=frac{sinfracpi2-sin0}{fracpi2-0}
implies sinfracpi2=fracpi2sin'(c).
$$
As both $sin'(c)$ and $fracpi2$ are positive, we obtain $$sinfracpi2>0.$$
answered Jan 28 at 23:45
useruser
5,41411030
edited Mar 11 at 10:03
answered Jan 28 at 23:45
useruser
5,41411030
answered Jan 28 at 23:45
useruser
5,41411030
answered Jan 28 at 23:45
useruser
5,41411030
5,41411030
add a comment |
add a comment |
$begingroup$
You know that $ sin(x) in mathbb{R}$, so that
$$leftvert sin left( frac{pi}{2} right) rightvert=1 Rightarrow sin left( frac{pi}{2} right)= pm1$$
If $sin'(x)=cos(x) >0$ for all $x in [0,frac{pi}{2})$, then this means that the sinus function is strictly growing on $x in [0,frac{pi}{2})$. With $sin(0)=0$, with $sin$ being an continuous function, then the only possibility is $$sin left( frac{pi}{2} right)=+1$$
Hence $$e^{ifrac{pi}{2}} = i$$
Edit If you want to proove that $sin'(x)=cos(x)$, so note that $$frac{d}{dx}left( e^{ix} right) = ie^{ix} = icos(x) - sin(x)$$
Implies that
$$ cos'(x) + isin'(x) = -sin(x) + icos(x)$$
Prooving that $sin'(x)=cos(x)$
answered Jan 26 at 20:38
Thomas LesgourguesThomas Lesgourgues
1,128119
$endgroup$
$begingroup$
But why is it the only possibility?
$endgroup$
– New2Math
Jan 26 at 20:40
$begingroup$
it can be only +1 or -1, because it is in $mathbb{R}$. If the function is growing, starting at 0, it cannot be -1
$endgroup$
– Thomas Lesgourgues
Jan 26 at 20:43
$begingroup$
Because otherwise it would contradict continuity, Right?
$endgroup$
– New2Math
Jan 26 at 20:45
$begingroup$
yes, exactly...
$endgroup$
– Thomas Lesgourgues
Jan 26 at 20:46
$begingroup$
I was wondering if one could Show the property with a constructive proof and not by contradiction because I feel like I should know some Theorems by now which would imply the result directly
$endgroup$
– New2Math
Jan 26 at 20:48
|
show 1 more comment
$begingroup$
You know that $ sin(x) in mathbb{R}$, so that
$$leftvert sin left( frac{pi}{2} right) rightvert=1 Rightarrow sin left( frac{pi}{2} right)= pm1$$
If $sin'(x)=cos(x) >0$ for all $x in [0,frac{pi}{2})$, then this means that the sinus function is strictly growing on $x in [0,frac{pi}{2})$. With $sin(0)=0$, with $sin$ being an continuous function, then the only possibility is $$sin left( frac{pi}{2} right)=+1$$
Hence $$e^{ifrac{pi}{2}} = i$$
Edit If you want to proove that $sin'(x)=cos(x)$, so note that $$frac{d}{dx}left( e^{ix} right) = ie^{ix} = icos(x) - sin(x)$$
Implies that
$$ cos'(x) + isin'(x) = -sin(x) + icos(x)$$
Prooving that $sin'(x)=cos(x)$
answered Jan 26 at 20:38
Thomas LesgourguesThomas Lesgourgues
1,128119
$endgroup$
$begingroup$
But why is it the only possibility?
$endgroup$
– New2Math
Jan 26 at 20:40
$begingroup$
it can be only +1 or -1, because it is in $mathbb{R}$. If the function is growing, starting at 0, it cannot be -1
$endgroup$
– Thomas Lesgourgues
Jan 26 at 20:43
$begingroup$
Because otherwise it would contradict continuity, Right?
$endgroup$
– New2Math
Jan 26 at 20:45
$begingroup$
yes, exactly...
$endgroup$
– Thomas Lesgourgues
Jan 26 at 20:46
$begingroup$
I was wondering if one could Show the property with a constructive proof and not by contradiction because I feel like I should know some Theorems by now which would imply the result directly
$endgroup$
– New2Math
Jan 26 at 20:48
|
show 1 more comment
$begingroup$
You know that $ sin(x) in mathbb{R}$, so that
$$leftvert sin left( frac{pi}{2} right) rightvert=1 Rightarrow sin left( frac{pi}{2} right)= pm1$$
If $sin'(x)=cos(x) >0$ for all $x in [0,frac{pi}{2})$, then this means that the sinus function is strictly growing on $x in [0,frac{pi}{2})$. With $sin(0)=0$, with $sin$ being an continuous function, then the only possibility is $$sin left( frac{pi}{2} right)=+1$$
Hence $$e^{ifrac{pi}{2}} = i$$
Edit If you want to proove that $sin'(x)=cos(x)$, so note that $$frac{d}{dx}left( e^{ix} right) = ie^{ix} = icos(x) - sin(x)$$
Implies that
$$ cos'(x) + isin'(x) = -sin(x) + icos(x)$$
Prooving that $sin'(x)=cos(x)$
answered Jan 26 at 20:38
Thomas LesgourguesThomas Lesgourgues
1,128119
$endgroup$
You know that $ sin(x) in mathbb{R}$, so that
$$leftvert sin left( frac{pi}{2} right) rightvert=1 Rightarrow sin left( frac{pi}{2} right)= pm1$$
If $sin'(x)=cos(x) >0$ for all $x in [0,frac{pi}{2})$, then this means that the sinus function is strictly growing on $x in [0,frac{pi}{2})$. With $sin(0)=0$, with $sin$ being an continuous function, then the only possibility is $$sin left( frac{pi}{2} right)=+1$$
Hence $$e^{ifrac{pi}{2}} = i$$
Edit If you want to proove that $sin'(x)=cos(x)$, so note that $$frac{d}{dx}left( e^{ix} right) = ie^{ix} = icos(x) - sin(x)$$
Implies that
$$ cos'(x) + isin'(x) = -sin(x) + icos(x)$$
Prooving that $sin'(x)=cos(x)$
answered Jan 26 at 20:38
Thomas LesgourguesThomas Lesgourgues
1,128119
edited Jan 26 at 20:46
answered Jan 26 at 20:38
Thomas LesgourguesThomas Lesgourgues
1,128119
answered Jan 26 at 20:38
Thomas LesgourguesThomas Lesgourgues
1,128119
answered Jan 26 at 20:38
Thomas LesgourguesThomas Lesgourgues
1,128119
1,128119
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But why is it the only possibility?
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– New2Math
Jan 26 at 20:40
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it can be only +1 or -1, because it is in $mathbb{R}$. If the function is growing, starting at 0, it cannot be -1
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– Thomas Lesgourgues
Jan 26 at 20:43
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Because otherwise it would contradict continuity, Right?
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– New2Math
Jan 26 at 20:45
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yes, exactly...
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– Thomas Lesgourgues
Jan 26 at 20:46
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I was wondering if one could Show the property with a constructive proof and not by contradiction because I feel like I should know some Theorems by now which would imply the result directly
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– New2Math
Jan 26 at 20:48
|
show 1 more comment
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But why is it the only possibility?
$endgroup$
– New2Math
Jan 26 at 20:40
$begingroup$
it can be only +1 or -1, because it is in $mathbb{R}$. If the function is growing, starting at 0, it cannot be -1
$endgroup$
– Thomas Lesgourgues
Jan 26 at 20:43
$begingroup$
Because otherwise it would contradict continuity, Right?
$endgroup$
– New2Math
Jan 26 at 20:45
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yes, exactly...
$endgroup$
– Thomas Lesgourgues
Jan 26 at 20:46
$begingroup$
I was wondering if one could Show the property with a constructive proof and not by contradiction because I feel like I should know some Theorems by now which would imply the result directly
$endgroup$
– New2Math
Jan 26 at 20:48
$begingroup$
But why is it the only possibility?
$endgroup$
– New2Math
Jan 26 at 20:40
$begingroup$
But why is it the only possibility?
$endgroup$
– New2Math
Jan 26 at 20:40
$begingroup$
it can be only +1 or -1, because it is in $mathbb{R}$. If the function is growing, starting at 0, it cannot be -1
$endgroup$
– Thomas Lesgourgues
Jan 26 at 20:43
$begingroup$
it can be only +1 or -1, because it is in $mathbb{R}$. If the function is growing, starting at 0, it cannot be -1
$endgroup$
– Thomas Lesgourgues
Jan 26 at 20:43
$begingroup$
Because otherwise it would contradict continuity, Right?
$endgroup$
– New2Math
Jan 26 at 20:45
$begingroup$
Because otherwise it would contradict continuity, Right?
$endgroup$
– New2Math
Jan 26 at 20:45
$begingroup$
yes, exactly...
$endgroup$
– Thomas Lesgourgues
Jan 26 at 20:46
$begingroup$
yes, exactly...
$endgroup$
– Thomas Lesgourgues
Jan 26 at 20:46
$begingroup$
I was wondering if one could Show the property with a constructive proof and not by contradiction because I feel like I should know some Theorems by now which would imply the result directly
$endgroup$
– New2Math
Jan 26 at 20:48
$begingroup$
I was wondering if one could Show the property with a constructive proof and not by contradiction because I feel like I should know some Theorems by now which would imply the result directly
$endgroup$
– New2Math
Jan 26 at 20:48
|
show 1 more comment
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$begingroup$
"We have defined $e^{ix}=cos x+ i sin x$" - So just plug in $x = frac{pi}{2}$ and you get $i$..
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– user635162
Jan 26 at 20:38
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Is the point of your question to prove $cosfracpi2=0$ and $sinfracpi2=1$? What properties of sine and cosine are allowed to be known to this point?
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– user
Jan 26 at 20:40
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The Addition Theorem and the differentials. We don't know anything About the periodicity , the Intention was to Show $e^{ipi}=-1$
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– New2Math
Jan 26 at 20:43
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Tell us what definition of sine/cosine is used.
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– user
Jan 26 at 20:44
1
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There is simply no way we can answer this question without having all you axioms, definitions, and contexts explained explicitly.
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– fleablood
Jan 28 at 23:51