mean value theorem: $ | cos x-1 | leq | x | $
I've been studying mean value theorem, and I was told that I should use MTV for solving problems with $leq$. So I've been trying to show that $ | cos x - 1 | leq | x | $ for all x values, using MTV, but I just don't get how MTV can be used...
I've been stuck on this for days now and I would be really grateful for any help.
calculus real-analysis trigonometry inequality
edited Jan 11 '15 at 14:36
Martin Sleziak
44.7k7115270
asked Mar 31 '14 at 15:28
user125342
346415
add a comment |
I've been studying mean value theorem, and I was told that I should use MTV for solving problems with $leq$. So I've been trying to show that $ | cos x - 1 | leq | x | $ for all x values, using MTV, but I just don't get how MTV can be used...
I've been stuck on this for days now and I would be really grateful for any help.
calculus real-analysis trigonometry inequality
edited Jan 11 '15 at 14:36
Martin Sleziak
44.7k7115270
asked Mar 31 '14 at 15:28
user125342
346415
add a comment |
I've been studying mean value theorem, and I was told that I should use MTV for solving problems with $leq$. So I've been trying to show that $ | cos x - 1 | leq | x | $ for all x values, using MTV, but I just don't get how MTV can be used...
I've been stuck on this for days now and I would be really grateful for any help.
calculus real-analysis trigonometry inequality
edited Jan 11 '15 at 14:36
Martin Sleziak
44.7k7115270
asked Mar 31 '14 at 15:28
user125342
346415
I've been studying mean value theorem, and I was told that I should use MTV for solving problems with $leq$. So I've been trying to show that $ | cos x - 1 | leq | x | $ for all x values, using MTV, but I just don't get how MTV can be used...
I've been stuck on this for days now and I would be really grateful for any help.
calculus real-analysis trigonometry inequality
calculus real-analysis trigonometry inequality
edited Jan 11 '15 at 14:36
Martin Sleziak
44.7k7115270
asked Mar 31 '14 at 15:28
user125342
346415
edited Jan 11 '15 at 14:36
Martin Sleziak
44.7k7115270
asked Mar 31 '14 at 15:28
user125342
346415
edited Jan 11 '15 at 14:36
Martin Sleziak
44.7k7115270
edited Jan 11 '15 at 14:36
Martin Sleziak
44.7k7115270
edited Jan 11 '15 at 14:36
Martin Sleziak
44.7k7115270
44.7k7115270
asked Mar 31 '14 at 15:28
user125342
346415
asked Mar 31 '14 at 15:28
user125342
346415
asked Mar 31 '14 at 15:28
user125342
346415
346415
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
For $xne0$ you should know that MVT says
$$frac{cos x-cos0}{x-0}=cos'xi$$
for some $xiin(0,x)$ or $(x,0)$ depending on the sign of $x$.
Therefore $|cos x-1|=|cos x-cos 0|=|x-0||cos'xi|=|x||!!-!sinxi|le|x|cdot1=|x|.$
Verification for $x=0$ is trivial.
answered Mar 31 '14 at 15:33
user2345215
14.3k11952
Thank you!! I think I'm finally understanding this...
– user125342
Mar 31 '14 at 16:09
add a comment |
$$
|cos x-1|=|cos x-cos0|=|cos'(xi)(x-0)|le ?
$$
$xi$ is an intermediate point between $0$ and $x$.
answered Mar 31 '14 at 15:33
Julián Aguirre
67.6k24094
add a comment |
Put $f(x):= cos(x)$. Then $|f'(x)| = |-sin(x)| leq 1 quad forall x in mathbb{R}$.
For any $x > 0$ there exists (by the MVT) a $theta in (0,x)$ such that
$$f(x) - f(0) = f'(theta)(x-0)$$
or in other words
$$cos(x) - 1 = f'(theta)x.$$
Taking the modulus gives
$$|cos(x) - 1| = |f'(theta)| cdot |x| leq 1 cdot |x| = |x|.$$
You can deal with the case $x<0$ similarly. The case $x=0$ is trivial.
answered Mar 31 '14 at 15:33
Frank
2,429920
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
For $xne0$ you should know that MVT says
$$frac{cos x-cos0}{x-0}=cos'xi$$
for some $xiin(0,x)$ or $(x,0)$ depending on the sign of $x$.
Therefore $|cos x-1|=|cos x-cos 0|=|x-0||cos'xi|=|x||!!-!sinxi|le|x|cdot1=|x|.$
Verification for $x=0$ is trivial.
answered Mar 31 '14 at 15:33
user2345215
14.3k11952
Thank you!! I think I'm finally understanding this...
– user125342
Mar 31 '14 at 16:09
add a comment |
For $xne0$ you should know that MVT says
$$frac{cos x-cos0}{x-0}=cos'xi$$
for some $xiin(0,x)$ or $(x,0)$ depending on the sign of $x$.
Therefore $|cos x-1|=|cos x-cos 0|=|x-0||cos'xi|=|x||!!-!sinxi|le|x|cdot1=|x|.$
Verification for $x=0$ is trivial.
answered Mar 31 '14 at 15:33
user2345215
14.3k11952
Thank you!! I think I'm finally understanding this...
– user125342
Mar 31 '14 at 16:09
add a comment |
For $xne0$ you should know that MVT says
$$frac{cos x-cos0}{x-0}=cos'xi$$
for some $xiin(0,x)$ or $(x,0)$ depending on the sign of $x$.
Therefore $|cos x-1|=|cos x-cos 0|=|x-0||cos'xi|=|x||!!-!sinxi|le|x|cdot1=|x|.$
Verification for $x=0$ is trivial.
answered Mar 31 '14 at 15:33
user2345215
14.3k11952
For $xne0$ you should know that MVT says
$$frac{cos x-cos0}{x-0}=cos'xi$$
for some $xiin(0,x)$ or $(x,0)$ depending on the sign of $x$.
Therefore $|cos x-1|=|cos x-cos 0|=|x-0||cos'xi|=|x||!!-!sinxi|le|x|cdot1=|x|.$
Verification for $x=0$ is trivial.
answered Mar 31 '14 at 15:33
user2345215
14.3k11952
edited Mar 31 '14 at 15:40
answered Mar 31 '14 at 15:33
user2345215
14.3k11952
answered Mar 31 '14 at 15:33
user2345215
14.3k11952
answered Mar 31 '14 at 15:33
user2345215
14.3k11952
14.3k11952
Thank you!! I think I'm finally understanding this...
– user125342
Mar 31 '14 at 16:09
add a comment |
Thank you!! I think I'm finally understanding this...
– user125342
Mar 31 '14 at 16:09
Thank you!! I think I'm finally understanding this...
– user125342
Mar 31 '14 at 16:09
Thank you!! I think I'm finally understanding this...
– user125342
Mar 31 '14 at 16:09
add a comment |
$$
|cos x-1|=|cos x-cos0|=|cos'(xi)(x-0)|le ?
$$
$xi$ is an intermediate point between $0$ and $x$.
answered Mar 31 '14 at 15:33
Julián Aguirre
67.6k24094
add a comment |
$$
|cos x-1|=|cos x-cos0|=|cos'(xi)(x-0)|le ?
$$
$xi$ is an intermediate point between $0$ and $x$.
answered Mar 31 '14 at 15:33
Julián Aguirre
67.6k24094
add a comment |
$$
|cos x-1|=|cos x-cos0|=|cos'(xi)(x-0)|le ?
$$
$xi$ is an intermediate point between $0$ and $x$.
answered Mar 31 '14 at 15:33
Julián Aguirre
67.6k24094
$$
|cos x-1|=|cos x-cos0|=|cos'(xi)(x-0)|le ?
$$
$xi$ is an intermediate point between $0$ and $x$.
answered Mar 31 '14 at 15:33
Julián Aguirre
67.6k24094
answered Mar 31 '14 at 15:33
Julián Aguirre
67.6k24094
answered Mar 31 '14 at 15:33
Julián Aguirre
67.6k24094
answered Mar 31 '14 at 15:33
Julián Aguirre
67.6k24094
67.6k24094
add a comment |
add a comment |
Put $f(x):= cos(x)$. Then $|f'(x)| = |-sin(x)| leq 1 quad forall x in mathbb{R}$.
For any $x > 0$ there exists (by the MVT) a $theta in (0,x)$ such that
$$f(x) - f(0) = f'(theta)(x-0)$$
or in other words
$$cos(x) - 1 = f'(theta)x.$$
Taking the modulus gives
$$|cos(x) - 1| = |f'(theta)| cdot |x| leq 1 cdot |x| = |x|.$$
You can deal with the case $x<0$ similarly. The case $x=0$ is trivial.
answered Mar 31 '14 at 15:33
Frank
2,429920
add a comment |
Put $f(x):= cos(x)$. Then $|f'(x)| = |-sin(x)| leq 1 quad forall x in mathbb{R}$.
For any $x > 0$ there exists (by the MVT) a $theta in (0,x)$ such that
$$f(x) - f(0) = f'(theta)(x-0)$$
or in other words
$$cos(x) - 1 = f'(theta)x.$$
Taking the modulus gives
$$|cos(x) - 1| = |f'(theta)| cdot |x| leq 1 cdot |x| = |x|.$$
You can deal with the case $x<0$ similarly. The case $x=0$ is trivial.
answered Mar 31 '14 at 15:33
Frank
2,429920
add a comment |
Put $f(x):= cos(x)$. Then $|f'(x)| = |-sin(x)| leq 1 quad forall x in mathbb{R}$.
For any $x > 0$ there exists (by the MVT) a $theta in (0,x)$ such that
$$f(x) - f(0) = f'(theta)(x-0)$$
or in other words
$$cos(x) - 1 = f'(theta)x.$$
Taking the modulus gives
$$|cos(x) - 1| = |f'(theta)| cdot |x| leq 1 cdot |x| = |x|.$$
You can deal with the case $x<0$ similarly. The case $x=0$ is trivial.
answered Mar 31 '14 at 15:33
Frank
2,429920
Put $f(x):= cos(x)$. Then $|f'(x)| = |-sin(x)| leq 1 quad forall x in mathbb{R}$.
For any $x > 0$ there exists (by the MVT) a $theta in (0,x)$ such that
$$f(x) - f(0) = f'(theta)(x-0)$$
or in other words
$$cos(x) - 1 = f'(theta)x.$$
Taking the modulus gives
$$|cos(x) - 1| = |f'(theta)| cdot |x| leq 1 cdot |x| = |x|.$$
You can deal with the case $x<0$ similarly. The case $x=0$ is trivial.
answered Mar 31 '14 at 15:33
Frank
2,429920
answered Mar 31 '14 at 15:33
Frank
2,429920
answered Mar 31 '14 at 15:33
Frank
2,429920
answered Mar 31 '14 at 15:33
Frank
2,429920
2,429920
add a comment |
add a comment |
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