Unique fixed point on $[0, 2pi ]$
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In the following question I am looking for an explanation of the answer which is given. How did they find the critical point?
Prove that $f(x) = pi + frac{1}{2}sin left( frac{x}{2} right)$ has a unique fixed point on $[0,2 pi]$
Checking the range of the function $f$:
we have that $f(0) = pi $ and $f(2pi ) = pi$.
The derivative of $f(x)$ is:
$$f'(x) = frac{1}{4} cos left( frac{x}{2} right)$$.
Here is what I am having trouble understanding. The only critical point inside the interval is at $x = pi$. How did they find this critical point? Do I then take that critical point evaluate the function at it and conldude that $0 leq f(x) leq 2pi$ for all $x in [0, 2pi]$ and thus a fixed point exists?
calculus linear-algebra numerical-methods
asked Nov 15 at 18:20
fr14
36217
|
show 9 more comments
up vote
0
down vote
favorite
In the following question I am looking for an explanation of the answer which is given. How did they find the critical point?
Prove that $f(x) = pi + frac{1}{2}sin left( frac{x}{2} right)$ has a unique fixed point on $[0,2 pi]$
Checking the range of the function $f$:
we have that $f(0) = pi $ and $f(2pi ) = pi$.
The derivative of $f(x)$ is:
$$f'(x) = frac{1}{4} cos left( frac{x}{2} right)$$.
Here is what I am having trouble understanding. The only critical point inside the interval is at $x = pi$. How did they find this critical point? Do I then take that critical point evaluate the function at it and conldude that $0 leq f(x) leq 2pi$ for all $x in [0, 2pi]$ and thus a fixed point exists?
calculus linear-algebra numerical-methods
asked Nov 15 at 18:20
fr14
36217
To find the critical point see when $f'(x)=0$ for $xin[0,2pi]$
– Yadati Kiran
Nov 15 at 18:27
so then it is only 0 when x is $pi$? making $cos(pi /2) = 0$?
– fr14
Nov 15 at 18:29
1
Yes. Since here $xin[0,2pi]$. But in general when restriction on $x$ is not given then $f'(x)=0$ for $x=(2n+1)frac{pi}{2}$. You can use Banach fixed point theorem to prove it has a fixed point.
– Yadati Kiran
Nov 15 at 18:32
and then do I take this critical point evaluate it at the function and conclude that $0 leq f(x) leq 2pi $ for all x on the interval $[0, 2pi]$?
– fr14
Nov 15 at 18:34
Yes, the critical point is were function reaches maximum or minimum so find $f(x)$ at that point and see if the inequality holds. You've already checked both ends of the interval.
– Vasya
Nov 15 at 18:34
|
show 9 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In the following question I am looking for an explanation of the answer which is given. How did they find the critical point?
Prove that $f(x) = pi + frac{1}{2}sin left( frac{x}{2} right)$ has a unique fixed point on $[0,2 pi]$
Checking the range of the function $f$:
we have that $f(0) = pi $ and $f(2pi ) = pi$.
The derivative of $f(x)$ is:
$$f'(x) = frac{1}{4} cos left( frac{x}{2} right)$$.
Here is what I am having trouble understanding. The only critical point inside the interval is at $x = pi$. How did they find this critical point? Do I then take that critical point evaluate the function at it and conldude that $0 leq f(x) leq 2pi$ for all $x in [0, 2pi]$ and thus a fixed point exists?
calculus linear-algebra numerical-methods
asked Nov 15 at 18:20
fr14
36217
In the following question I am looking for an explanation of the answer which is given. How did they find the critical point?
Prove that $f(x) = pi + frac{1}{2}sin left( frac{x}{2} right)$ has a unique fixed point on $[0,2 pi]$
Checking the range of the function $f$:
we have that $f(0) = pi $ and $f(2pi ) = pi$.
The derivative of $f(x)$ is:
$$f'(x) = frac{1}{4} cos left( frac{x}{2} right)$$.
Here is what I am having trouble understanding. The only critical point inside the interval is at $x = pi$. How did they find this critical point? Do I then take that critical point evaluate the function at it and conldude that $0 leq f(x) leq 2pi$ for all $x in [0, 2pi]$ and thus a fixed point exists?
calculus linear-algebra numerical-methods
calculus linear-algebra numerical-methods
asked Nov 15 at 18:20
fr14
36217
asked Nov 15 at 18:20
fr14
36217
asked Nov 15 at 18:20
fr14
36217
asked Nov 15 at 18:20
fr14
36217
asked Nov 15 at 18:20
fr14
36217
36217
To find the critical point see when $f'(x)=0$ for $xin[0,2pi]$
– Yadati Kiran
Nov 15 at 18:27
so then it is only 0 when x is $pi$? making $cos(pi /2) = 0$?
– fr14
Nov 15 at 18:29
1
Yes. Since here $xin[0,2pi]$. But in general when restriction on $x$ is not given then $f'(x)=0$ for $x=(2n+1)frac{pi}{2}$. You can use Banach fixed point theorem to prove it has a fixed point.
– Yadati Kiran
Nov 15 at 18:32
and then do I take this critical point evaluate it at the function and conclude that $0 leq f(x) leq 2pi $ for all x on the interval $[0, 2pi]$?
– fr14
Nov 15 at 18:34
Yes, the critical point is were function reaches maximum or minimum so find $f(x)$ at that point and see if the inequality holds. You've already checked both ends of the interval.
– Vasya
Nov 15 at 18:34
|
show 9 more comments
To find the critical point see when $f'(x)=0$ for $xin[0,2pi]$
– Yadati Kiran
Nov 15 at 18:27
so then it is only 0 when x is $pi$? making $cos(pi /2) = 0$?
– fr14
Nov 15 at 18:29
1
Yes. Since here $xin[0,2pi]$. But in general when restriction on $x$ is not given then $f'(x)=0$ for $x=(2n+1)frac{pi}{2}$. You can use Banach fixed point theorem to prove it has a fixed point.
– Yadati Kiran
Nov 15 at 18:32
and then do I take this critical point evaluate it at the function and conclude that $0 leq f(x) leq 2pi $ for all x on the interval $[0, 2pi]$?
– fr14
Nov 15 at 18:34
Yes, the critical point is were function reaches maximum or minimum so find $f(x)$ at that point and see if the inequality holds. You've already checked both ends of the interval.
– Vasya
Nov 15 at 18:34
To find the critical point see when $f'(x)=0$ for $xin[0,2pi]$
– Yadati Kiran
Nov 15 at 18:27
To find the critical point see when $f'(x)=0$ for $xin[0,2pi]$
– Yadati Kiran
Nov 15 at 18:27
so then it is only 0 when x is $pi$? making $cos(pi /2) = 0$?
– fr14
Nov 15 at 18:29
so then it is only 0 when x is $pi$? making $cos(pi /2) = 0$?
– fr14
Nov 15 at 18:29
1
1
Yes. Since here $xin[0,2pi]$. But in general when restriction on $x$ is not given then $f'(x)=0$ for $x=(2n+1)frac{pi}{2}$. You can use Banach fixed point theorem to prove it has a fixed point.
– Yadati Kiran
Nov 15 at 18:32
Yes. Since here $xin[0,2pi]$. But in general when restriction on $x$ is not given then $f'(x)=0$ for $x=(2n+1)frac{pi}{2}$. You can use Banach fixed point theorem to prove it has a fixed point.
– Yadati Kiran
Nov 15 at 18:32
and then do I take this critical point evaluate it at the function and conclude that $0 leq f(x) leq 2pi $ for all x on the interval $[0, 2pi]$?
– fr14
Nov 15 at 18:34
and then do I take this critical point evaluate it at the function and conclude that $0 leq f(x) leq 2pi $ for all x on the interval $[0, 2pi]$?
– fr14
Nov 15 at 18:34
Yes, the critical point is were function reaches maximum or minimum so find $f(x)$ at that point and see if the inequality holds. You've already checked both ends of the interval.
– Vasya
Nov 15 at 18:34
Yes, the critical point is were function reaches maximum or minimum so find $f(x)$ at that point and see if the inequality holds. You've already checked both ends of the interval.
– Vasya
Nov 15 at 18:34
|
show 9 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
You have that $f(0)=pi$, $f(pi)=pi+0.5$, $f(2pi)=pi$. According to intermediate value theorem, $pi le f(x) le pi+0.5$ when $0 le x le pi$. We cannot have fixed point there. On the other hand, $pi le f(x) le pi+0.5$ when $pi le x le 2pi$. Because there are no critical points between $pi$ and $2pi$, we know that $x=f(x)$ has only one solution on this interval.
answered Nov 15 at 19:08
Vasya
3,2291515
does your max and min give you your critical point for $f'(x)$ here our max and min is pi and that is what makes $f'(x) =0$
– fr14
Nov 15 at 19:13
1
@fr14: maximum is actually $pi+0.5$. To find critical points you find where $f'(x)=0$. But because you have the same value at both ends, you know that the function either constant or has a critical point.
– Vasya
Nov 15 at 19:19
add a comment |
up vote
1
down vote
It is not clear why you are checking for critical points.
Note that $sup_x |f'(x)| = {1 over 4} < 1 $, and $f([0,2 pi]) subset [0, 2 pi]$.
Hence $f$ is a contraction map and has a unique fixed point.
answered Nov 15 at 19:06
copper.hat
125k558158
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You have that $f(0)=pi$, $f(pi)=pi+0.5$, $f(2pi)=pi$. According to intermediate value theorem, $pi le f(x) le pi+0.5$ when $0 le x le pi$. We cannot have fixed point there. On the other hand, $pi le f(x) le pi+0.5$ when $pi le x le 2pi$. Because there are no critical points between $pi$ and $2pi$, we know that $x=f(x)$ has only one solution on this interval.
answered Nov 15 at 19:08
Vasya
3,2291515
does your max and min give you your critical point for $f'(x)$ here our max and min is pi and that is what makes $f'(x) =0$
– fr14
Nov 15 at 19:13
1
@fr14: maximum is actually $pi+0.5$. To find critical points you find where $f'(x)=0$. But because you have the same value at both ends, you know that the function either constant or has a critical point.
– Vasya
Nov 15 at 19:19
add a comment |
up vote
2
down vote
accepted
You have that $f(0)=pi$, $f(pi)=pi+0.5$, $f(2pi)=pi$. According to intermediate value theorem, $pi le f(x) le pi+0.5$ when $0 le x le pi$. We cannot have fixed point there. On the other hand, $pi le f(x) le pi+0.5$ when $pi le x le 2pi$. Because there are no critical points between $pi$ and $2pi$, we know that $x=f(x)$ has only one solution on this interval.
answered Nov 15 at 19:08
Vasya
3,2291515
does your max and min give you your critical point for $f'(x)$ here our max and min is pi and that is what makes $f'(x) =0$
– fr14
Nov 15 at 19:13
1
@fr14: maximum is actually $pi+0.5$. To find critical points you find where $f'(x)=0$. But because you have the same value at both ends, you know that the function either constant or has a critical point.
– Vasya
Nov 15 at 19:19
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You have that $f(0)=pi$, $f(pi)=pi+0.5$, $f(2pi)=pi$. According to intermediate value theorem, $pi le f(x) le pi+0.5$ when $0 le x le pi$. We cannot have fixed point there. On the other hand, $pi le f(x) le pi+0.5$ when $pi le x le 2pi$. Because there are no critical points between $pi$ and $2pi$, we know that $x=f(x)$ has only one solution on this interval.
answered Nov 15 at 19:08
Vasya
3,2291515
You have that $f(0)=pi$, $f(pi)=pi+0.5$, $f(2pi)=pi$. According to intermediate value theorem, $pi le f(x) le pi+0.5$ when $0 le x le pi$. We cannot have fixed point there. On the other hand, $pi le f(x) le pi+0.5$ when $pi le x le 2pi$. Because there are no critical points between $pi$ and $2pi$, we know that $x=f(x)$ has only one solution on this interval.
answered Nov 15 at 19:08
Vasya
3,2291515
answered Nov 15 at 19:08
Vasya
3,2291515
answered Nov 15 at 19:08
Vasya
3,2291515
answered Nov 15 at 19:08
Vasya
3,2291515
3,2291515
does your max and min give you your critical point for $f'(x)$ here our max and min is pi and that is what makes $f'(x) =0$
– fr14
Nov 15 at 19:13
1
@fr14: maximum is actually $pi+0.5$. To find critical points you find where $f'(x)=0$. But because you have the same value at both ends, you know that the function either constant or has a critical point.
– Vasya
Nov 15 at 19:19
add a comment |
does your max and min give you your critical point for $f'(x)$ here our max and min is pi and that is what makes $f'(x) =0$
– fr14
Nov 15 at 19:13
1
@fr14: maximum is actually $pi+0.5$. To find critical points you find where $f'(x)=0$. But because you have the same value at both ends, you know that the function either constant or has a critical point.
– Vasya
Nov 15 at 19:19
does your max and min give you your critical point for $f'(x)$ here our max and min is pi and that is what makes $f'(x) =0$
– fr14
Nov 15 at 19:13
does your max and min give you your critical point for $f'(x)$ here our max and min is pi and that is what makes $f'(x) =0$
– fr14
Nov 15 at 19:13
1
1
@fr14: maximum is actually $pi+0.5$. To find critical points you find where $f'(x)=0$. But because you have the same value at both ends, you know that the function either constant or has a critical point.
– Vasya
Nov 15 at 19:19
@fr14: maximum is actually $pi+0.5$. To find critical points you find where $f'(x)=0$. But because you have the same value at both ends, you know that the function either constant or has a critical point.
– Vasya
Nov 15 at 19:19
add a comment |
up vote
1
down vote
It is not clear why you are checking for critical points.
Note that $sup_x |f'(x)| = {1 over 4} < 1 $, and $f([0,2 pi]) subset [0, 2 pi]$.
Hence $f$ is a contraction map and has a unique fixed point.
answered Nov 15 at 19:06
copper.hat
125k558158
add a comment |
up vote
1
down vote
It is not clear why you are checking for critical points.
Note that $sup_x |f'(x)| = {1 over 4} < 1 $, and $f([0,2 pi]) subset [0, 2 pi]$.
Hence $f$ is a contraction map and has a unique fixed point.
answered Nov 15 at 19:06
copper.hat
125k558158
add a comment |
up vote
1
down vote
up vote
1
down vote
It is not clear why you are checking for critical points.
Note that $sup_x |f'(x)| = {1 over 4} < 1 $, and $f([0,2 pi]) subset [0, 2 pi]$.
Hence $f$ is a contraction map and has a unique fixed point.
answered Nov 15 at 19:06
copper.hat
125k558158
It is not clear why you are checking for critical points.
Note that $sup_x |f'(x)| = {1 over 4} < 1 $, and $f([0,2 pi]) subset [0, 2 pi]$.
Hence $f$ is a contraction map and has a unique fixed point.
answered Nov 15 at 19:06
copper.hat
125k558158
answered Nov 15 at 19:06
copper.hat
125k558158
answered Nov 15 at 19:06
copper.hat
125k558158
answered Nov 15 at 19:06
copper.hat
125k558158
125k558158
add a comment |
add a comment |
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To find the critical point see when $f'(x)=0$ for $xin[0,2pi]$
– Yadati Kiran
Nov 15 at 18:27
so then it is only 0 when x is $pi$? making $cos(pi /2) = 0$?
– fr14
Nov 15 at 18:29
1
Yes. Since here $xin[0,2pi]$. But in general when restriction on $x$ is not given then $f'(x)=0$ for $x=(2n+1)frac{pi}{2}$. You can use Banach fixed point theorem to prove it has a fixed point.
– Yadati Kiran
Nov 15 at 18:32
and then do I take this critical point evaluate it at the function and conclude that $0 leq f(x) leq 2pi $ for all x on the interval $[0, 2pi]$?
– fr14
Nov 15 at 18:34
Yes, the critical point is were function reaches maximum or minimum so find $f(x)$ at that point and see if the inequality holds. You've already checked both ends of the interval.
– Vasya
Nov 15 at 18:34