Prove $sin x$ is uniformly continuous on $mathbb R$
$begingroup$
How do I prove $sin x$ is uniformly continuous on $mathbb R$ with delta and epsilon?
I proved geometrically that $sin x<x$ and thus, $$|f(x_1)-f(x_2)|=|sin x_1 - sin x_2|le|sin x_1|+|sin x_2|<|x_1|+|x_2|$$
But this doesn't help me much finding a delta...
Thanks for any help!
P.S. I'm only at the beginning of calculus so I can't use many theorems and derivation (because they haven't been regorously proven).
calculus functions continuity
edited Jan 1 '13 at 19:48
Haskell Curry
15.2k3886
asked Jan 1 '13 at 19:27
HaroldHarold
2711315
$endgroup$
add a comment |
$begingroup$
How do I prove $sin x$ is uniformly continuous on $mathbb R$ with delta and epsilon?
I proved geometrically that $sin x<x$ and thus, $$|f(x_1)-f(x_2)|=|sin x_1 - sin x_2|le|sin x_1|+|sin x_2|<|x_1|+|x_2|$$
But this doesn't help me much finding a delta...
Thanks for any help!
P.S. I'm only at the beginning of calculus so I can't use many theorems and derivation (because they haven't been regorously proven).
calculus functions continuity
edited Jan 1 '13 at 19:48
Haskell Curry
15.2k3886
asked Jan 1 '13 at 19:27
HaroldHarold
2711315
$endgroup$
1
$begingroup$
I changed $sinx_1$, etc., to $sin x_1$. That is standard TeX usage.
$endgroup$
– Michael Hardy
Jan 1 '13 at 19:35
$begingroup$
This is a particular case of (at least) two more general results. First: a periodic continuous function on $mathbb{R}$ is uniformly continuous on $mathbb{R}$. Second: a Lipschitz function (mathworld.wolfram.com/LipschitzFunction.html) is uniformly continuous. To prove that $sin$ is Lipschitz, you can use a trigonometric identity like Nameless did in his answer, or you can claim that its derivative is bounded by $1$.
$endgroup$
– Julien
Jan 1 '13 at 19:41
add a comment |
$begingroup$
How do I prove $sin x$ is uniformly continuous on $mathbb R$ with delta and epsilon?
I proved geometrically that $sin x<x$ and thus, $$|f(x_1)-f(x_2)|=|sin x_1 - sin x_2|le|sin x_1|+|sin x_2|<|x_1|+|x_2|$$
But this doesn't help me much finding a delta...
Thanks for any help!
P.S. I'm only at the beginning of calculus so I can't use many theorems and derivation (because they haven't been regorously proven).
calculus functions continuity
edited Jan 1 '13 at 19:48
Haskell Curry
15.2k3886
asked Jan 1 '13 at 19:27
HaroldHarold
2711315
$endgroup$
How do I prove $sin x$ is uniformly continuous on $mathbb R$ with delta and epsilon?
I proved geometrically that $sin x<x$ and thus, $$|f(x_1)-f(x_2)|=|sin x_1 - sin x_2|le|sin x_1|+|sin x_2|<|x_1|+|x_2|$$
But this doesn't help me much finding a delta...
Thanks for any help!
P.S. I'm only at the beginning of calculus so I can't use many theorems and derivation (because they haven't been regorously proven).
calculus functions continuity
calculus functions continuity
edited Jan 1 '13 at 19:48
Haskell Curry
15.2k3886
asked Jan 1 '13 at 19:27
HaroldHarold
2711315
edited Jan 1 '13 at 19:48
Haskell Curry
15.2k3886
asked Jan 1 '13 at 19:27
HaroldHarold
2711315
edited Jan 1 '13 at 19:48
Haskell Curry
15.2k3886
edited Jan 1 '13 at 19:48
Haskell Curry
15.2k3886
edited Jan 1 '13 at 19:48
Haskell Curry
15.2k3886
15.2k3886
asked Jan 1 '13 at 19:27
HaroldHarold
2711315
asked Jan 1 '13 at 19:27
HaroldHarold
2711315
asked Jan 1 '13 at 19:27
HaroldHarold
2711315
2711315
1
$begingroup$
I changed $sinx_1$, etc., to $sin x_1$. That is standard TeX usage.
$endgroup$
– Michael Hardy
Jan 1 '13 at 19:35
$begingroup$
This is a particular case of (at least) two more general results. First: a periodic continuous function on $mathbb{R}$ is uniformly continuous on $mathbb{R}$. Second: a Lipschitz function (mathworld.wolfram.com/LipschitzFunction.html) is uniformly continuous. To prove that $sin$ is Lipschitz, you can use a trigonometric identity like Nameless did in his answer, or you can claim that its derivative is bounded by $1$.
$endgroup$
– Julien
Jan 1 '13 at 19:41
add a comment |
1
$begingroup$
I changed $sinx_1$, etc., to $sin x_1$. That is standard TeX usage.
$endgroup$
– Michael Hardy
Jan 1 '13 at 19:35
$begingroup$
This is a particular case of (at least) two more general results. First: a periodic continuous function on $mathbb{R}$ is uniformly continuous on $mathbb{R}$. Second: a Lipschitz function (mathworld.wolfram.com/LipschitzFunction.html) is uniformly continuous. To prove that $sin$ is Lipschitz, you can use a trigonometric identity like Nameless did in his answer, or you can claim that its derivative is bounded by $1$.
$endgroup$
– Julien
Jan 1 '13 at 19:41
1
1
$begingroup$
I changed $sinx_1$, etc., to $sin x_1$. That is standard TeX usage.
$endgroup$
– Michael Hardy
Jan 1 '13 at 19:35
$begingroup$
I changed $sinx_1$, etc., to $sin x_1$. That is standard TeX usage.
$endgroup$
– Michael Hardy
Jan 1 '13 at 19:35
$begingroup$
This is a particular case of (at least) two more general results. First: a periodic continuous function on $mathbb{R}$ is uniformly continuous on $mathbb{R}$. Second: a Lipschitz function (mathworld.wolfram.com/LipschitzFunction.html) is uniformly continuous. To prove that $sin$ is Lipschitz, you can use a trigonometric identity like Nameless did in his answer, or you can claim that its derivative is bounded by $1$.
$endgroup$
– Julien
Jan 1 '13 at 19:41
$begingroup$
This is a particular case of (at least) two more general results. First: a periodic continuous function on $mathbb{R}$ is uniformly continuous on $mathbb{R}$. Second: a Lipschitz function (mathworld.wolfram.com/LipschitzFunction.html) is uniformly continuous. To prove that $sin$ is Lipschitz, you can use a trigonometric identity like Nameless did in his answer, or you can claim that its derivative is bounded by $1$.
$endgroup$
– Julien
Jan 1 '13 at 19:41
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Let $epsilon>0$ and $x,yin mathbb{R}$. We want
$$left|f(x)-f(y)right|<epsilonimplies left|sin x-sin yright|<epsilonimplies left|2cosfrac{x+y}2sinfrac{x-y}2right|$$
Because
$$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$
it suffices
$$2left|sinfrac{x-y}2right|<epsilon$$
when
$$left|x-yright|<deltaimplies left|frac{x-y}2right|<delta$$
SInce $left|sin xright|le left|xright|$,
$$2left|sinfrac{x-y}2right|le 2left|frac{x-y}2right|<2delta$$
Choosing $delta=frac{epsilon}{2}>0$ will do the trick. Because $delta$ doesn't depend on $x,y$, the continuity is uniform
answered Jan 1 '13 at 19:29
NamelessNameless
10.4k12155
$endgroup$
$begingroup$
Hi, thanks for the fast response! Why $$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$ is correct? Also, I know how to prove $sinx<x$, but how do I show $|sinx|<|x|$? Thanks for the reply again!
$endgroup$
– Harold
Jan 1 '13 at 19:32
1
$begingroup$
because $cos alpha leqslant 1 $, $forall alpha in mathbb{R}$.
$endgroup$
– Amihai Zivan
Jan 1 '13 at 19:34
2
$begingroup$
Because $left|cos(bullet)right|le 1$.
$endgroup$
– Michael Hardy
Jan 1 '13 at 19:34
1
$begingroup$
@Harold $sin x$ is odd
$endgroup$
– Nameless
Jan 1 '13 at 19:48
2
$begingroup$
@Harold You want $left|sin yright|le left|yright|iff -sin yle -yiff sin yge y$ for $y<0$ near $0$. But $sin xle x$ for $x>0$. Multipliying by $-1$ gives $-sin xge -xiff sin (-x)ge -xiff sin yge y$
$endgroup$
– Nameless
Jan 1 '13 at 20:10
|
show 3 more comments
$begingroup$
By Mean Value Theorem,
$$ |sin{x}- sin{y}| leq |x-y| |cos{xi}| leq |x-y|, quad xleqxi leq y.$$
Hence, you may choose $epsilon=delta$.
answered Jan 1 '13 at 19:35
BerkheimerBerkheimer
1,4371024
$endgroup$
4
$begingroup$
This is a circular reasoning. For MVT you need that sine is differentiable. Since each differentiable function is continuous, you have used what you are proving.
$endgroup$
– wdacda
Feb 1 '16 at 19:55
2
$begingroup$
@wdacda How is that circular reasoning?
$endgroup$
– 3x89g2
Sep 5 '16 at 5:27
add a comment |
$begingroup$
Since $sin x$ is a periodic continuous function with a period $2pi$, it suffices to prove that it is uniformly continuous on $[0, 2pi]$. Since $[0, 2pi]$ is compact, this follows from the well-known theorem.
answered Jan 1 '13 at 19:59
Makoto KatoMakoto Kato
23k560162
$endgroup$
add a comment |
$begingroup$
There is an elementary geometric way to do this. Let $x$ and $y$ be real numbers; for now, assume $x, yin(-pi, pi]$. Start off at $(1,0)$ and march off signed distance $x$ to get to point $a$ and $y$ to get to point $b$ on the unit circle. Then $|x - y|$ is the distance from $a$ to $b$ along the unit circle. $|sin(x) - sin(y)|$ is the distance between the $y$-coordiates of $a$ and $b$. Hence, in this case
$$|sin(x) - sin(y) | le |x - y|.$$
This gives us uniform continuity on $(-pi, pi]$, so by periodciity the sine function is uniformly continuous on the entire line.
answered Jan 1 '13 at 21:03
ncmathsadistncmathsadist
43k260103
$endgroup$
$begingroup$
Another way to prove this identity: math.stackexchange.com/questions/620305/…
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– GinKin
Dec 28 '13 at 10:08
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This answer deliberately does not appeal to the MVT; it uses purely geometric properties of the sine and cosine functions.
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– ncmathsadist
Dec 29 '13 at 19:56
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Yeah I know. But for a test I'm not sure if your approach would be rigorous enough. No offense.
$endgroup$
– GinKin
Dec 29 '13 at 20:03
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What is not rigorous about the fact that, given a line and a point, the shortest distance between them is achieved along the perpendicular dropped from the point to the line? In fact, this solution is unique.
$endgroup$
– ncmathsadist
Dec 30 '13 at 0:46
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $epsilon>0$ and $x,yin mathbb{R}$. We want
$$left|f(x)-f(y)right|<epsilonimplies left|sin x-sin yright|<epsilonimplies left|2cosfrac{x+y}2sinfrac{x-y}2right|$$
Because
$$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$
it suffices
$$2left|sinfrac{x-y}2right|<epsilon$$
when
$$left|x-yright|<deltaimplies left|frac{x-y}2right|<delta$$
SInce $left|sin xright|le left|xright|$,
$$2left|sinfrac{x-y}2right|le 2left|frac{x-y}2right|<2delta$$
Choosing $delta=frac{epsilon}{2}>0$ will do the trick. Because $delta$ doesn't depend on $x,y$, the continuity is uniform
answered Jan 1 '13 at 19:29
NamelessNameless
10.4k12155
$endgroup$
$begingroup$
Hi, thanks for the fast response! Why $$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$ is correct? Also, I know how to prove $sinx<x$, but how do I show $|sinx|<|x|$? Thanks for the reply again!
$endgroup$
– Harold
Jan 1 '13 at 19:32
1
$begingroup$
because $cos alpha leqslant 1 $, $forall alpha in mathbb{R}$.
$endgroup$
– Amihai Zivan
Jan 1 '13 at 19:34
2
$begingroup$
Because $left|cos(bullet)right|le 1$.
$endgroup$
– Michael Hardy
Jan 1 '13 at 19:34
1
$begingroup$
@Harold $sin x$ is odd
$endgroup$
– Nameless
Jan 1 '13 at 19:48
2
$begingroup$
@Harold You want $left|sin yright|le left|yright|iff -sin yle -yiff sin yge y$ for $y<0$ near $0$. But $sin xle x$ for $x>0$. Multipliying by $-1$ gives $-sin xge -xiff sin (-x)ge -xiff sin yge y$
$endgroup$
– Nameless
Jan 1 '13 at 20:10
|
show 3 more comments
$begingroup$
Let $epsilon>0$ and $x,yin mathbb{R}$. We want
$$left|f(x)-f(y)right|<epsilonimplies left|sin x-sin yright|<epsilonimplies left|2cosfrac{x+y}2sinfrac{x-y}2right|$$
Because
$$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$
it suffices
$$2left|sinfrac{x-y}2right|<epsilon$$
when
$$left|x-yright|<deltaimplies left|frac{x-y}2right|<delta$$
SInce $left|sin xright|le left|xright|$,
$$2left|sinfrac{x-y}2right|le 2left|frac{x-y}2right|<2delta$$
Choosing $delta=frac{epsilon}{2}>0$ will do the trick. Because $delta$ doesn't depend on $x,y$, the continuity is uniform
answered Jan 1 '13 at 19:29
NamelessNameless
10.4k12155
$endgroup$
$begingroup$
Hi, thanks for the fast response! Why $$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$ is correct? Also, I know how to prove $sinx<x$, but how do I show $|sinx|<|x|$? Thanks for the reply again!
$endgroup$
– Harold
Jan 1 '13 at 19:32
1
$begingroup$
because $cos alpha leqslant 1 $, $forall alpha in mathbb{R}$.
$endgroup$
– Amihai Zivan
Jan 1 '13 at 19:34
2
$begingroup$
Because $left|cos(bullet)right|le 1$.
$endgroup$
– Michael Hardy
Jan 1 '13 at 19:34
1
$begingroup$
@Harold $sin x$ is odd
$endgroup$
– Nameless
Jan 1 '13 at 19:48
2
$begingroup$
@Harold You want $left|sin yright|le left|yright|iff -sin yle -yiff sin yge y$ for $y<0$ near $0$. But $sin xle x$ for $x>0$. Multipliying by $-1$ gives $-sin xge -xiff sin (-x)ge -xiff sin yge y$
$endgroup$
– Nameless
Jan 1 '13 at 20:10
|
show 3 more comments
$begingroup$
Let $epsilon>0$ and $x,yin mathbb{R}$. We want
$$left|f(x)-f(y)right|<epsilonimplies left|sin x-sin yright|<epsilonimplies left|2cosfrac{x+y}2sinfrac{x-y}2right|$$
Because
$$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$
it suffices
$$2left|sinfrac{x-y}2right|<epsilon$$
when
$$left|x-yright|<deltaimplies left|frac{x-y}2right|<delta$$
SInce $left|sin xright|le left|xright|$,
$$2left|sinfrac{x-y}2right|le 2left|frac{x-y}2right|<2delta$$
Choosing $delta=frac{epsilon}{2}>0$ will do the trick. Because $delta$ doesn't depend on $x,y$, the continuity is uniform
answered Jan 1 '13 at 19:29
NamelessNameless
10.4k12155
$endgroup$
Let $epsilon>0$ and $x,yin mathbb{R}$. We want
$$left|f(x)-f(y)right|<epsilonimplies left|sin x-sin yright|<epsilonimplies left|2cosfrac{x+y}2sinfrac{x-y}2right|$$
Because
$$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$
it suffices
$$2left|sinfrac{x-y}2right|<epsilon$$
when
$$left|x-yright|<deltaimplies left|frac{x-y}2right|<delta$$
SInce $left|sin xright|le left|xright|$,
$$2left|sinfrac{x-y}2right|le 2left|frac{x-y}2right|<2delta$$
Choosing $delta=frac{epsilon}{2}>0$ will do the trick. Because $delta$ doesn't depend on $x,y$, the continuity is uniform
answered Jan 1 '13 at 19:29
NamelessNameless
10.4k12155
answered Jan 1 '13 at 19:29
NamelessNameless
10.4k12155
answered Jan 1 '13 at 19:29
NamelessNameless
10.4k12155
answered Jan 1 '13 at 19:29
NamelessNameless
10.4k12155
10.4k12155
$begingroup$
Hi, thanks for the fast response! Why $$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$ is correct? Also, I know how to prove $sinx<x$, but how do I show $|sinx|<|x|$? Thanks for the reply again!
$endgroup$
– Harold
Jan 1 '13 at 19:32
1
$begingroup$
because $cos alpha leqslant 1 $, $forall alpha in mathbb{R}$.
$endgroup$
– Amihai Zivan
Jan 1 '13 at 19:34
2
$begingroup$
Because $left|cos(bullet)right|le 1$.
$endgroup$
– Michael Hardy
Jan 1 '13 at 19:34
1
$begingroup$
@Harold $sin x$ is odd
$endgroup$
– Nameless
Jan 1 '13 at 19:48
2
$begingroup$
@Harold You want $left|sin yright|le left|yright|iff -sin yle -yiff sin yge y$ for $y<0$ near $0$. But $sin xle x$ for $x>0$. Multipliying by $-1$ gives $-sin xge -xiff sin (-x)ge -xiff sin yge y$
$endgroup$
– Nameless
Jan 1 '13 at 20:10
|
show 3 more comments
$begingroup$
Hi, thanks for the fast response! Why $$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$ is correct? Also, I know how to prove $sinx<x$, but how do I show $|sinx|<|x|$? Thanks for the reply again!
$endgroup$
– Harold
Jan 1 '13 at 19:32
1
$begingroup$
because $cos alpha leqslant 1 $, $forall alpha in mathbb{R}$.
$endgroup$
– Amihai Zivan
Jan 1 '13 at 19:34
2
$begingroup$
Because $left|cos(bullet)right|le 1$.
$endgroup$
– Michael Hardy
Jan 1 '13 at 19:34
1
$begingroup$
@Harold $sin x$ is odd
$endgroup$
– Nameless
Jan 1 '13 at 19:48
2
$begingroup$
@Harold You want $left|sin yright|le left|yright|iff -sin yle -yiff sin yge y$ for $y<0$ near $0$. But $sin xle x$ for $x>0$. Multipliying by $-1$ gives $-sin xge -xiff sin (-x)ge -xiff sin yge y$
$endgroup$
– Nameless
Jan 1 '13 at 20:10
$begingroup$
Hi, thanks for the fast response! Why $$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$ is correct? Also, I know how to prove $sinx<x$, but how do I show $|sinx|<|x|$? Thanks for the reply again!
$endgroup$
– Harold
Jan 1 '13 at 19:32
$begingroup$
Hi, thanks for the fast response! Why $$left|2cosfrac{x+y}2sinfrac{x-y}2right|le 2left|sinfrac{x-y}2right|$$ is correct? Also, I know how to prove $sinx<x$, but how do I show $|sinx|<|x|$? Thanks for the reply again!
$endgroup$
– Harold
Jan 1 '13 at 19:32
1
1
$begingroup$
because $cos alpha leqslant 1 $, $forall alpha in mathbb{R}$.
$endgroup$
– Amihai Zivan
Jan 1 '13 at 19:34
$begingroup$
because $cos alpha leqslant 1 $, $forall alpha in mathbb{R}$.
$endgroup$
– Amihai Zivan
Jan 1 '13 at 19:34
2
2
$begingroup$
Because $left|cos(bullet)right|le 1$.
$endgroup$
– Michael Hardy
Jan 1 '13 at 19:34
$begingroup$
Because $left|cos(bullet)right|le 1$.
$endgroup$
– Michael Hardy
Jan 1 '13 at 19:34
1
1
$begingroup$
@Harold $sin x$ is odd
$endgroup$
– Nameless
Jan 1 '13 at 19:48
$begingroup$
@Harold $sin x$ is odd
$endgroup$
– Nameless
Jan 1 '13 at 19:48
2
2
$begingroup$
@Harold You want $left|sin yright|le left|yright|iff -sin yle -yiff sin yge y$ for $y<0$ near $0$. But $sin xle x$ for $x>0$. Multipliying by $-1$ gives $-sin xge -xiff sin (-x)ge -xiff sin yge y$
$endgroup$
– Nameless
Jan 1 '13 at 20:10
$begingroup$
@Harold You want $left|sin yright|le left|yright|iff -sin yle -yiff sin yge y$ for $y<0$ near $0$. But $sin xle x$ for $x>0$. Multipliying by $-1$ gives $-sin xge -xiff sin (-x)ge -xiff sin yge y$
$endgroup$
– Nameless
Jan 1 '13 at 20:10
|
show 3 more comments
$begingroup$
By Mean Value Theorem,
$$ |sin{x}- sin{y}| leq |x-y| |cos{xi}| leq |x-y|, quad xleqxi leq y.$$
Hence, you may choose $epsilon=delta$.
answered Jan 1 '13 at 19:35
BerkheimerBerkheimer
1,4371024
$endgroup$
4
$begingroup$
This is a circular reasoning. For MVT you need that sine is differentiable. Since each differentiable function is continuous, you have used what you are proving.
$endgroup$
– wdacda
Feb 1 '16 at 19:55
2
$begingroup$
@wdacda How is that circular reasoning?
$endgroup$
– 3x89g2
Sep 5 '16 at 5:27
add a comment |
$begingroup$
By Mean Value Theorem,
$$ |sin{x}- sin{y}| leq |x-y| |cos{xi}| leq |x-y|, quad xleqxi leq y.$$
Hence, you may choose $epsilon=delta$.
answered Jan 1 '13 at 19:35
BerkheimerBerkheimer
1,4371024
$endgroup$
4
$begingroup$
This is a circular reasoning. For MVT you need that sine is differentiable. Since each differentiable function is continuous, you have used what you are proving.
$endgroup$
– wdacda
Feb 1 '16 at 19:55
2
$begingroup$
@wdacda How is that circular reasoning?
$endgroup$
– 3x89g2
Sep 5 '16 at 5:27
add a comment |
$begingroup$
By Mean Value Theorem,
$$ |sin{x}- sin{y}| leq |x-y| |cos{xi}| leq |x-y|, quad xleqxi leq y.$$
Hence, you may choose $epsilon=delta$.
answered Jan 1 '13 at 19:35
BerkheimerBerkheimer
1,4371024
$endgroup$
By Mean Value Theorem,
$$ |sin{x}- sin{y}| leq |x-y| |cos{xi}| leq |x-y|, quad xleqxi leq y.$$
Hence, you may choose $epsilon=delta$.
answered Jan 1 '13 at 19:35
BerkheimerBerkheimer
1,4371024
answered Jan 1 '13 at 19:35
BerkheimerBerkheimer
1,4371024
answered Jan 1 '13 at 19:35
BerkheimerBerkheimer
1,4371024
answered Jan 1 '13 at 19:35
BerkheimerBerkheimer
1,4371024
1,4371024
4
$begingroup$
This is a circular reasoning. For MVT you need that sine is differentiable. Since each differentiable function is continuous, you have used what you are proving.
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– wdacda
Feb 1 '16 at 19:55
2
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@wdacda How is that circular reasoning?
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– 3x89g2
Sep 5 '16 at 5:27
add a comment |
4
$begingroup$
This is a circular reasoning. For MVT you need that sine is differentiable. Since each differentiable function is continuous, you have used what you are proving.
$endgroup$
– wdacda
Feb 1 '16 at 19:55
2
$begingroup$
@wdacda How is that circular reasoning?
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– 3x89g2
Sep 5 '16 at 5:27
4
4
$begingroup$
This is a circular reasoning. For MVT you need that sine is differentiable. Since each differentiable function is continuous, you have used what you are proving.
$endgroup$
– wdacda
Feb 1 '16 at 19:55
$begingroup$
This is a circular reasoning. For MVT you need that sine is differentiable. Since each differentiable function is continuous, you have used what you are proving.
$endgroup$
– wdacda
Feb 1 '16 at 19:55
2
2
$begingroup$
@wdacda How is that circular reasoning?
$endgroup$
– 3x89g2
Sep 5 '16 at 5:27
$begingroup$
@wdacda How is that circular reasoning?
$endgroup$
– 3x89g2
Sep 5 '16 at 5:27
add a comment |
$begingroup$
Since $sin x$ is a periodic continuous function with a period $2pi$, it suffices to prove that it is uniformly continuous on $[0, 2pi]$. Since $[0, 2pi]$ is compact, this follows from the well-known theorem.
answered Jan 1 '13 at 19:59
Makoto KatoMakoto Kato
23k560162
$endgroup$
add a comment |
$begingroup$
Since $sin x$ is a periodic continuous function with a period $2pi$, it suffices to prove that it is uniformly continuous on $[0, 2pi]$. Since $[0, 2pi]$ is compact, this follows from the well-known theorem.
answered Jan 1 '13 at 19:59
Makoto KatoMakoto Kato
23k560162
$endgroup$
add a comment |
$begingroup$
Since $sin x$ is a periodic continuous function with a period $2pi$, it suffices to prove that it is uniformly continuous on $[0, 2pi]$. Since $[0, 2pi]$ is compact, this follows from the well-known theorem.
answered Jan 1 '13 at 19:59
Makoto KatoMakoto Kato
23k560162
$endgroup$
Since $sin x$ is a periodic continuous function with a period $2pi$, it suffices to prove that it is uniformly continuous on $[0, 2pi]$. Since $[0, 2pi]$ is compact, this follows from the well-known theorem.
answered Jan 1 '13 at 19:59
Makoto KatoMakoto Kato
23k560162
answered Jan 1 '13 at 19:59
Makoto KatoMakoto Kato
23k560162
answered Jan 1 '13 at 19:59
Makoto KatoMakoto Kato
23k560162
answered Jan 1 '13 at 19:59
Makoto KatoMakoto Kato
23k560162
23k560162
add a comment |
add a comment |
$begingroup$
There is an elementary geometric way to do this. Let $x$ and $y$ be real numbers; for now, assume $x, yin(-pi, pi]$. Start off at $(1,0)$ and march off signed distance $x$ to get to point $a$ and $y$ to get to point $b$ on the unit circle. Then $|x - y|$ is the distance from $a$ to $b$ along the unit circle. $|sin(x) - sin(y)|$ is the distance between the $y$-coordiates of $a$ and $b$. Hence, in this case
$$|sin(x) - sin(y) | le |x - y|.$$
This gives us uniform continuity on $(-pi, pi]$, so by periodciity the sine function is uniformly continuous on the entire line.
answered Jan 1 '13 at 21:03
ncmathsadistncmathsadist
43k260103
$endgroup$
$begingroup$
Another way to prove this identity: math.stackexchange.com/questions/620305/…
$endgroup$
– GinKin
Dec 28 '13 at 10:08
$begingroup$
This answer deliberately does not appeal to the MVT; it uses purely geometric properties of the sine and cosine functions.
$endgroup$
– ncmathsadist
Dec 29 '13 at 19:56
$begingroup$
Yeah I know. But for a test I'm not sure if your approach would be rigorous enough. No offense.
$endgroup$
– GinKin
Dec 29 '13 at 20:03
$begingroup$
What is not rigorous about the fact that, given a line and a point, the shortest distance between them is achieved along the perpendicular dropped from the point to the line? In fact, this solution is unique.
$endgroup$
– ncmathsadist
Dec 30 '13 at 0:46
add a comment |
$begingroup$
There is an elementary geometric way to do this. Let $x$ and $y$ be real numbers; for now, assume $x, yin(-pi, pi]$. Start off at $(1,0)$ and march off signed distance $x$ to get to point $a$ and $y$ to get to point $b$ on the unit circle. Then $|x - y|$ is the distance from $a$ to $b$ along the unit circle. $|sin(x) - sin(y)|$ is the distance between the $y$-coordiates of $a$ and $b$. Hence, in this case
$$|sin(x) - sin(y) | le |x - y|.$$
This gives us uniform continuity on $(-pi, pi]$, so by periodciity the sine function is uniformly continuous on the entire line.
answered Jan 1 '13 at 21:03
ncmathsadistncmathsadist
43k260103
$endgroup$
$begingroup$
Another way to prove this identity: math.stackexchange.com/questions/620305/…
$endgroup$
– GinKin
Dec 28 '13 at 10:08
$begingroup$
This answer deliberately does not appeal to the MVT; it uses purely geometric properties of the sine and cosine functions.
$endgroup$
– ncmathsadist
Dec 29 '13 at 19:56
$begingroup$
Yeah I know. But for a test I'm not sure if your approach would be rigorous enough. No offense.
$endgroup$
– GinKin
Dec 29 '13 at 20:03
$begingroup$
What is not rigorous about the fact that, given a line and a point, the shortest distance between them is achieved along the perpendicular dropped from the point to the line? In fact, this solution is unique.
$endgroup$
– ncmathsadist
Dec 30 '13 at 0:46
add a comment |
$begingroup$
There is an elementary geometric way to do this. Let $x$ and $y$ be real numbers; for now, assume $x, yin(-pi, pi]$. Start off at $(1,0)$ and march off signed distance $x$ to get to point $a$ and $y$ to get to point $b$ on the unit circle. Then $|x - y|$ is the distance from $a$ to $b$ along the unit circle. $|sin(x) - sin(y)|$ is the distance between the $y$-coordiates of $a$ and $b$. Hence, in this case
$$|sin(x) - sin(y) | le |x - y|.$$
This gives us uniform continuity on $(-pi, pi]$, so by periodciity the sine function is uniformly continuous on the entire line.
answered Jan 1 '13 at 21:03
ncmathsadistncmathsadist
43k260103
$endgroup$
There is an elementary geometric way to do this. Let $x$ and $y$ be real numbers; for now, assume $x, yin(-pi, pi]$. Start off at $(1,0)$ and march off signed distance $x$ to get to point $a$ and $y$ to get to point $b$ on the unit circle. Then $|x - y|$ is the distance from $a$ to $b$ along the unit circle. $|sin(x) - sin(y)|$ is the distance between the $y$-coordiates of $a$ and $b$. Hence, in this case
$$|sin(x) - sin(y) | le |x - y|.$$
This gives us uniform continuity on $(-pi, pi]$, so by periodciity the sine function is uniformly continuous on the entire line.
answered Jan 1 '13 at 21:03
ncmathsadistncmathsadist
43k260103
answered Jan 1 '13 at 21:03
ncmathsadistncmathsadist
43k260103
answered Jan 1 '13 at 21:03
ncmathsadistncmathsadist
43k260103
answered Jan 1 '13 at 21:03
ncmathsadistncmathsadist
43k260103
43k260103
$begingroup$
Another way to prove this identity: math.stackexchange.com/questions/620305/…
$endgroup$
– GinKin
Dec 28 '13 at 10:08
$begingroup$
This answer deliberately does not appeal to the MVT; it uses purely geometric properties of the sine and cosine functions.
$endgroup$
– ncmathsadist
Dec 29 '13 at 19:56
$begingroup$
Yeah I know. But for a test I'm not sure if your approach would be rigorous enough. No offense.
$endgroup$
– GinKin
Dec 29 '13 at 20:03
$begingroup$
What is not rigorous about the fact that, given a line and a point, the shortest distance between them is achieved along the perpendicular dropped from the point to the line? In fact, this solution is unique.
$endgroup$
– ncmathsadist
Dec 30 '13 at 0:46
add a comment |
$begingroup$
Another way to prove this identity: math.stackexchange.com/questions/620305/…
$endgroup$
– GinKin
Dec 28 '13 at 10:08
$begingroup$
This answer deliberately does not appeal to the MVT; it uses purely geometric properties of the sine and cosine functions.
$endgroup$
– ncmathsadist
Dec 29 '13 at 19:56
$begingroup$
Yeah I know. But for a test I'm not sure if your approach would be rigorous enough. No offense.
$endgroup$
– GinKin
Dec 29 '13 at 20:03
$begingroup$
What is not rigorous about the fact that, given a line and a point, the shortest distance between them is achieved along the perpendicular dropped from the point to the line? In fact, this solution is unique.
$endgroup$
– ncmathsadist
Dec 30 '13 at 0:46
$begingroup$
Another way to prove this identity: math.stackexchange.com/questions/620305/…
$endgroup$
– GinKin
Dec 28 '13 at 10:08
$begingroup$
Another way to prove this identity: math.stackexchange.com/questions/620305/…
$endgroup$
– GinKin
Dec 28 '13 at 10:08
$begingroup$
This answer deliberately does not appeal to the MVT; it uses purely geometric properties of the sine and cosine functions.
$endgroup$
– ncmathsadist
Dec 29 '13 at 19:56
$begingroup$
This answer deliberately does not appeal to the MVT; it uses purely geometric properties of the sine and cosine functions.
$endgroup$
– ncmathsadist
Dec 29 '13 at 19:56
$begingroup$
Yeah I know. But for a test I'm not sure if your approach would be rigorous enough. No offense.
$endgroup$
– GinKin
Dec 29 '13 at 20:03
$begingroup$
Yeah I know. But for a test I'm not sure if your approach would be rigorous enough. No offense.
$endgroup$
– GinKin
Dec 29 '13 at 20:03
$begingroup$
What is not rigorous about the fact that, given a line and a point, the shortest distance between them is achieved along the perpendicular dropped from the point to the line? In fact, this solution is unique.
$endgroup$
– ncmathsadist
Dec 30 '13 at 0:46
$begingroup$
What is not rigorous about the fact that, given a line and a point, the shortest distance between them is achieved along the perpendicular dropped from the point to the line? In fact, this solution is unique.
$endgroup$
– ncmathsadist
Dec 30 '13 at 0:46
add a comment |
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I changed $sinx_1$, etc., to $sin x_1$. That is standard TeX usage.
$endgroup$
– Michael Hardy
Jan 1 '13 at 19:35
$begingroup$
This is a particular case of (at least) two more general results. First: a periodic continuous function on $mathbb{R}$ is uniformly continuous on $mathbb{R}$. Second: a Lipschitz function (mathworld.wolfram.com/LipschitzFunction.html) is uniformly continuous. To prove that $sin$ is Lipschitz, you can use a trigonometric identity like Nameless did in his answer, or you can claim that its derivative is bounded by $1$.
$endgroup$
– Julien
Jan 1 '13 at 19:41