Proving |sinx| =< |x| for all real x.
$begingroup$
Everything I did to solve this problem matches the book, except for the last bit.
Use the following to verify the statement $|sin x| leq |x|$.
a) Show that for all $x geq 0$, $f(x)=x-sin x$ is non-decreasing.
$f'(x) = 1-cos x$. Since $-1 leq cos x leq 1$, we have $f'(x) geq 0$ so that $f(x)$ is non-decreasing. Since $f(x)=0$ when $x=0$, we have $f(x)>0$ whenever $x>0$, and hence $sin x leq x$ when $xgeq 0$.
The book then says: And $|sin x|=sin xleq x=|x|$. How can we say that $|sin x|=sin x$?
proof-explanation
edited Jan 13 at 21:04
Namaste
1
asked Jan 13 at 20:52
mXdXmXdX
898
$endgroup$
|
show 4 more comments
$begingroup$
Everything I did to solve this problem matches the book, except for the last bit.
Use the following to verify the statement $|sin x| leq |x|$.
a) Show that for all $x geq 0$, $f(x)=x-sin x$ is non-decreasing.
$f'(x) = 1-cos x$. Since $-1 leq cos x leq 1$, we have $f'(x) geq 0$ so that $f(x)$ is non-decreasing. Since $f(x)=0$ when $x=0$, we have $f(x)>0$ whenever $x>0$, and hence $sin x leq x$ when $xgeq 0$.
The book then says: And $|sin x|=sin xleq x=|x|$. How can we say that $|sin x|=sin x$?
proof-explanation
edited Jan 13 at 21:04
Namaste
1
asked Jan 13 at 20:52
mXdXmXdX
898
$endgroup$
$begingroup$
You can’t because it isn’t true.
$endgroup$
– David G. Stork
Jan 13 at 20:55
$begingroup$
Do you have any idea why the book might say that in this context?
$endgroup$
– mXdX
Jan 13 at 20:55
2
$begingroup$
I can only imagine that the author is considering only $x$ between $0$ and $1$, since it is otherwise trivial. It should be explicitly stated though.
$endgroup$
– Git Gud
Jan 13 at 20:56
$begingroup$
The problem says for all x greater than or equal to zero, so I'm not really sure.
$endgroup$
– mXdX
Jan 13 at 20:58
1
$begingroup$
Observe it is enough to prove it for $0le xle 1$.
$endgroup$
– Bernard
Jan 13 at 21:14
|
show 4 more comments
$begingroup$
Everything I did to solve this problem matches the book, except for the last bit.
Use the following to verify the statement $|sin x| leq |x|$.
a) Show that for all $x geq 0$, $f(x)=x-sin x$ is non-decreasing.
$f'(x) = 1-cos x$. Since $-1 leq cos x leq 1$, we have $f'(x) geq 0$ so that $f(x)$ is non-decreasing. Since $f(x)=0$ when $x=0$, we have $f(x)>0$ whenever $x>0$, and hence $sin x leq x$ when $xgeq 0$.
The book then says: And $|sin x|=sin xleq x=|x|$. How can we say that $|sin x|=sin x$?
proof-explanation
edited Jan 13 at 21:04
Namaste
1
asked Jan 13 at 20:52
mXdXmXdX
898
$endgroup$
Everything I did to solve this problem matches the book, except for the last bit.
Use the following to verify the statement $|sin x| leq |x|$.
a) Show that for all $x geq 0$, $f(x)=x-sin x$ is non-decreasing.
$f'(x) = 1-cos x$. Since $-1 leq cos x leq 1$, we have $f'(x) geq 0$ so that $f(x)$ is non-decreasing. Since $f(x)=0$ when $x=0$, we have $f(x)>0$ whenever $x>0$, and hence $sin x leq x$ when $xgeq 0$.
The book then says: And $|sin x|=sin xleq x=|x|$. How can we say that $|sin x|=sin x$?
proof-explanation
proof-explanation
edited Jan 13 at 21:04
Namaste
1
asked Jan 13 at 20:52
mXdXmXdX
898
edited Jan 13 at 21:04
Namaste
1
asked Jan 13 at 20:52
mXdXmXdX
898
edited Jan 13 at 21:04
Namaste
1
edited Jan 13 at 21:04
Namaste
1
edited Jan 13 at 21:04
Namaste
1
1
asked Jan 13 at 20:52
mXdXmXdX
898
asked Jan 13 at 20:52
mXdXmXdX
898
asked Jan 13 at 20:52
mXdXmXdX
898
898
$begingroup$
You can’t because it isn’t true.
$endgroup$
– David G. Stork
Jan 13 at 20:55
$begingroup$
Do you have any idea why the book might say that in this context?
$endgroup$
– mXdX
Jan 13 at 20:55
2
$begingroup$
I can only imagine that the author is considering only $x$ between $0$ and $1$, since it is otherwise trivial. It should be explicitly stated though.
$endgroup$
– Git Gud
Jan 13 at 20:56
$begingroup$
The problem says for all x greater than or equal to zero, so I'm not really sure.
$endgroup$
– mXdX
Jan 13 at 20:58
1
$begingroup$
Observe it is enough to prove it for $0le xle 1$.
$endgroup$
– Bernard
Jan 13 at 21:14
|
show 4 more comments
$begingroup$
You can’t because it isn’t true.
$endgroup$
– David G. Stork
Jan 13 at 20:55
$begingroup$
Do you have any idea why the book might say that in this context?
$endgroup$
– mXdX
Jan 13 at 20:55
2
$begingroup$
I can only imagine that the author is considering only $x$ between $0$ and $1$, since it is otherwise trivial. It should be explicitly stated though.
$endgroup$
– Git Gud
Jan 13 at 20:56
$begingroup$
The problem says for all x greater than or equal to zero, so I'm not really sure.
$endgroup$
– mXdX
Jan 13 at 20:58
1
$begingroup$
Observe it is enough to prove it for $0le xle 1$.
$endgroup$
– Bernard
Jan 13 at 21:14
$begingroup$
You can’t because it isn’t true.
$endgroup$
– David G. Stork
Jan 13 at 20:55
$begingroup$
You can’t because it isn’t true.
$endgroup$
– David G. Stork
Jan 13 at 20:55
$begingroup$
Do you have any idea why the book might say that in this context?
$endgroup$
– mXdX
Jan 13 at 20:55
$begingroup$
Do you have any idea why the book might say that in this context?
$endgroup$
– mXdX
Jan 13 at 20:55
2
2
$begingroup$
I can only imagine that the author is considering only $x$ between $0$ and $1$, since it is otherwise trivial. It should be explicitly stated though.
$endgroup$
– Git Gud
Jan 13 at 20:56
$begingroup$
I can only imagine that the author is considering only $x$ between $0$ and $1$, since it is otherwise trivial. It should be explicitly stated though.
$endgroup$
– Git Gud
Jan 13 at 20:56
$begingroup$
The problem says for all x greater than or equal to zero, so I'm not really sure.
$endgroup$
– mXdX
Jan 13 at 20:58
$begingroup$
The problem says for all x greater than or equal to zero, so I'm not really sure.
$endgroup$
– mXdX
Jan 13 at 20:58
1
1
$begingroup$
Observe it is enough to prove it for $0le xle 1$.
$endgroup$
– Bernard
Jan 13 at 21:14
$begingroup$
Observe it is enough to prove it for $0le xle 1$.
$endgroup$
– Bernard
Jan 13 at 21:14
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
$|sin x|=sin xleq x=|x|$ is of course wrong. So what is the correct argument?
For $lvert x rvert ge 1$ we trivially have $|sin x| le|x|$.
Moreover we know that $sin xleq x = lvert x rvert$ for $x ge 0$. It remains to consider $-1 le x le 0$. In this range
$$lvert sin x rvert = -sin(x) =sin(-x) le -x = lvert x rvert.$$
answered Jan 13 at 22:09
Paul FrostPaul Frost
11.6k3935
$endgroup$
add a comment |
$begingroup$
Since both $|sin x|$ and $|x|$ are even functions, you only need to show the statement when $xge0$. Also for $x>1$ the inequality holds true for obvious reasons and in the interval $[0,1]$ we have $$|sin x|=sin x\|x|=x$$
answered Jan 13 at 22:31
Mostafa AyazMostafa Ayaz
17k31039
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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oldest
votes
$begingroup$
$|sin x|=sin xleq x=|x|$ is of course wrong. So what is the correct argument?
For $lvert x rvert ge 1$ we trivially have $|sin x| le|x|$.
Moreover we know that $sin xleq x = lvert x rvert$ for $x ge 0$. It remains to consider $-1 le x le 0$. In this range
$$lvert sin x rvert = -sin(x) =sin(-x) le -x = lvert x rvert.$$
answered Jan 13 at 22:09
Paul FrostPaul Frost
11.6k3935
$endgroup$
add a comment |
$begingroup$
$|sin x|=sin xleq x=|x|$ is of course wrong. So what is the correct argument?
For $lvert x rvert ge 1$ we trivially have $|sin x| le|x|$.
Moreover we know that $sin xleq x = lvert x rvert$ for $x ge 0$. It remains to consider $-1 le x le 0$. In this range
$$lvert sin x rvert = -sin(x) =sin(-x) le -x = lvert x rvert.$$
answered Jan 13 at 22:09
Paul FrostPaul Frost
11.6k3935
$endgroup$
add a comment |
$begingroup$
$|sin x|=sin xleq x=|x|$ is of course wrong. So what is the correct argument?
For $lvert x rvert ge 1$ we trivially have $|sin x| le|x|$.
Moreover we know that $sin xleq x = lvert x rvert$ for $x ge 0$. It remains to consider $-1 le x le 0$. In this range
$$lvert sin x rvert = -sin(x) =sin(-x) le -x = lvert x rvert.$$
answered Jan 13 at 22:09
Paul FrostPaul Frost
11.6k3935
$endgroup$
$|sin x|=sin xleq x=|x|$ is of course wrong. So what is the correct argument?
For $lvert x rvert ge 1$ we trivially have $|sin x| le|x|$.
Moreover we know that $sin xleq x = lvert x rvert$ for $x ge 0$. It remains to consider $-1 le x le 0$. In this range
$$lvert sin x rvert = -sin(x) =sin(-x) le -x = lvert x rvert.$$
answered Jan 13 at 22:09
Paul FrostPaul Frost
11.6k3935
answered Jan 13 at 22:09
Paul FrostPaul Frost
11.6k3935
answered Jan 13 at 22:09
Paul FrostPaul Frost
11.6k3935
answered Jan 13 at 22:09
Paul FrostPaul Frost
11.6k3935
11.6k3935
add a comment |
add a comment |
$begingroup$
Since both $|sin x|$ and $|x|$ are even functions, you only need to show the statement when $xge0$. Also for $x>1$ the inequality holds true for obvious reasons and in the interval $[0,1]$ we have $$|sin x|=sin x\|x|=x$$
answered Jan 13 at 22:31
Mostafa AyazMostafa Ayaz
17k31039
$endgroup$
add a comment |
$begingroup$
Since both $|sin x|$ and $|x|$ are even functions, you only need to show the statement when $xge0$. Also for $x>1$ the inequality holds true for obvious reasons and in the interval $[0,1]$ we have $$|sin x|=sin x\|x|=x$$
answered Jan 13 at 22:31
Mostafa AyazMostafa Ayaz
17k31039
$endgroup$
add a comment |
$begingroup$
Since both $|sin x|$ and $|x|$ are even functions, you only need to show the statement when $xge0$. Also for $x>1$ the inequality holds true for obvious reasons and in the interval $[0,1]$ we have $$|sin x|=sin x\|x|=x$$
answered Jan 13 at 22:31
Mostafa AyazMostafa Ayaz
17k31039
$endgroup$
Since both $|sin x|$ and $|x|$ are even functions, you only need to show the statement when $xge0$. Also for $x>1$ the inequality holds true for obvious reasons and in the interval $[0,1]$ we have $$|sin x|=sin x\|x|=x$$
answered Jan 13 at 22:31
Mostafa AyazMostafa Ayaz
17k31039
answered Jan 13 at 22:31
Mostafa AyazMostafa Ayaz
17k31039
answered Jan 13 at 22:31
Mostafa AyazMostafa Ayaz
17k31039
answered Jan 13 at 22:31
Mostafa AyazMostafa Ayaz
17k31039
17k31039
add a comment |
add a comment |
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$begingroup$
You can’t because it isn’t true.
$endgroup$
– David G. Stork
Jan 13 at 20:55
$begingroup$
Do you have any idea why the book might say that in this context?
$endgroup$
– mXdX
Jan 13 at 20:55
2
$begingroup$
I can only imagine that the author is considering only $x$ between $0$ and $1$, since it is otherwise trivial. It should be explicitly stated though.
$endgroup$
– Git Gud
Jan 13 at 20:56
$begingroup$
The problem says for all x greater than or equal to zero, so I'm not really sure.
$endgroup$
– mXdX
Jan 13 at 20:58
1
$begingroup$
Observe it is enough to prove it for $0le xle 1$.
$endgroup$
– Bernard
Jan 13 at 21:14