Funky function-composed within itself umpteen thousand times
$begingroup$
$f(x)$ is a differentiable function satisfying the following conditions:
$$
0 < f(x) < 1 quad text{for all $x$ on the interval $0 le x le 1$.} \
0 < f'(x) < 1 quad text{for all $x$ on the interval $0 le x le 1$.}
$$
How many solutions does the equation
$$
underbrace{f(f(f( ldots f}_{2016~text{times}}(x) ldots) =x
$$
have on the interval $0leq xleq 1$?
This seems to be looking like chain rule
And since $f'$ is positive on the interval $[0,1]$ it seems to be increasing what does it do for $x>1$?
And what is the significance of 2016?
I dont think that matters. The function is composed within itself that many times but I think maybe it doesnt matter if its 2016 or 2019 !😀
functional-analysis functions contest-math
asked Jan 7 at 8:03
RandinRandin
339116
$endgroup$
add a comment |
$begingroup$
$f(x)$ is a differentiable function satisfying the following conditions:
$$
0 < f(x) < 1 quad text{for all $x$ on the interval $0 le x le 1$.} \
0 < f'(x) < 1 quad text{for all $x$ on the interval $0 le x le 1$.}
$$
How many solutions does the equation
$$
underbrace{f(f(f( ldots f}_{2016~text{times}}(x) ldots) =x
$$
have on the interval $0leq xleq 1$?
This seems to be looking like chain rule
And since $f'$ is positive on the interval $[0,1]$ it seems to be increasing what does it do for $x>1$?
And what is the significance of 2016?
I dont think that matters. The function is composed within itself that many times but I think maybe it doesnt matter if its 2016 or 2019 !😀
functional-analysis functions contest-math
asked Jan 7 at 8:03
RandinRandin
339116
$endgroup$
$begingroup$
Please rewrite the problem itself. What are the conditions? Please also use MathJax to typeset the equations. What have you tried?
$endgroup$
– Matti P.
Jan 7 at 8:05
$begingroup$
Ah ! Ok i will go back and do that thanks
$endgroup$
– Randin
Jan 7 at 13:08
$begingroup$
Just the green check mark below the upvotes ?
$endgroup$
– Randin
Jan 7 at 13:34
$begingroup$
Martin so only one accepted answer right ? And as for the question at hand - isnt this the same as the number of times the function crosses the y=x line ? The composing it 2016 times will mean it crosses multiple times right ?
$endgroup$
– Randin
Jan 7 at 14:20
add a comment |
$begingroup$
$f(x)$ is a differentiable function satisfying the following conditions:
$$
0 < f(x) < 1 quad text{for all $x$ on the interval $0 le x le 1$.} \
0 < f'(x) < 1 quad text{for all $x$ on the interval $0 le x le 1$.}
$$
How many solutions does the equation
$$
underbrace{f(f(f( ldots f}_{2016~text{times}}(x) ldots) =x
$$
have on the interval $0leq xleq 1$?
This seems to be looking like chain rule
And since $f'$ is positive on the interval $[0,1]$ it seems to be increasing what does it do for $x>1$?
And what is the significance of 2016?
I dont think that matters. The function is composed within itself that many times but I think maybe it doesnt matter if its 2016 or 2019 !😀
functional-analysis functions contest-math
asked Jan 7 at 8:03
RandinRandin
339116
$endgroup$
$f(x)$ is a differentiable function satisfying the following conditions:
$$
0 < f(x) < 1 quad text{for all $x$ on the interval $0 le x le 1$.} \
0 < f'(x) < 1 quad text{for all $x$ on the interval $0 le x le 1$.}
$$
How many solutions does the equation
$$
underbrace{f(f(f( ldots f}_{2016~text{times}}(x) ldots) =x
$$
have on the interval $0leq xleq 1$?
This seems to be looking like chain rule
And since $f'$ is positive on the interval $[0,1]$ it seems to be increasing what does it do for $x>1$?
And what is the significance of 2016?
I dont think that matters. The function is composed within itself that many times but I think maybe it doesnt matter if its 2016 or 2019 !😀
functional-analysis functions contest-math
functional-analysis functions contest-math
asked Jan 7 at 8:03
RandinRandin
339116
asked Jan 7 at 8:03
RandinRandin
339116
edited Jan 7 at 13:36
Randin
asked Jan 7 at 8:03
RandinRandin
339116
asked Jan 7 at 8:03
RandinRandin
339116
asked Jan 7 at 8:03
RandinRandin
339116
339116
$begingroup$
Please rewrite the problem itself. What are the conditions? Please also use MathJax to typeset the equations. What have you tried?
$endgroup$
– Matti P.
Jan 7 at 8:05
$begingroup$
Ah ! Ok i will go back and do that thanks
$endgroup$
– Randin
Jan 7 at 13:08
$begingroup$
Just the green check mark below the upvotes ?
$endgroup$
– Randin
Jan 7 at 13:34
$begingroup$
Martin so only one accepted answer right ? And as for the question at hand - isnt this the same as the number of times the function crosses the y=x line ? The composing it 2016 times will mean it crosses multiple times right ?
$endgroup$
– Randin
Jan 7 at 14:20
add a comment |
$begingroup$
Please rewrite the problem itself. What are the conditions? Please also use MathJax to typeset the equations. What have you tried?
$endgroup$
– Matti P.
Jan 7 at 8:05
$begingroup$
Ah ! Ok i will go back and do that thanks
$endgroup$
– Randin
Jan 7 at 13:08
$begingroup$
Just the green check mark below the upvotes ?
$endgroup$
– Randin
Jan 7 at 13:34
$begingroup$
Martin so only one accepted answer right ? And as for the question at hand - isnt this the same as the number of times the function crosses the y=x line ? The composing it 2016 times will mean it crosses multiple times right ?
$endgroup$
– Randin
Jan 7 at 14:20
$begingroup$
Please rewrite the problem itself. What are the conditions? Please also use MathJax to typeset the equations. What have you tried?
$endgroup$
– Matti P.
Jan 7 at 8:05
$begingroup$
Please rewrite the problem itself. What are the conditions? Please also use MathJax to typeset the equations. What have you tried?
$endgroup$
– Matti P.
Jan 7 at 8:05
$begingroup$
Ah ! Ok i will go back and do that thanks
$endgroup$
– Randin
Jan 7 at 13:08
$begingroup$
Ah ! Ok i will go back and do that thanks
$endgroup$
– Randin
Jan 7 at 13:08
$begingroup$
Just the green check mark below the upvotes ?
$endgroup$
– Randin
Jan 7 at 13:34
$begingroup$
Just the green check mark below the upvotes ?
$endgroup$
– Randin
Jan 7 at 13:34
$begingroup$
Martin so only one accepted answer right ? And as for the question at hand - isnt this the same as the number of times the function crosses the y=x line ? The composing it 2016 times will mean it crosses multiple times right ?
$endgroup$
– Randin
Jan 7 at 14:20
$begingroup$
Martin so only one accepted answer right ? And as for the question at hand - isnt this the same as the number of times the function crosses the y=x line ? The composing it 2016 times will mean it crosses multiple times right ?
$endgroup$
– Randin
Jan 7 at 14:20
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since $f'(x)>0$ for all $x$ in the concerned interval, $f(x)$ is strictly increasing.
So:
$x<y implies f(x)<f(y)$
(assuming $x,y in [0,1]$)
Now, suppose for some $a in [0,1],f(a) ne a$.
Then, either $f(a)<a$ or $f(a)>a$.
Looking at the first case and using the fact that $f$ is strictly increasing and, crucially, that $f$ maps from $[0,1]$ back into $[0,1]$, $f(f(a))<f(a)$. Of course, by the initial assumption, $f(a)<a$, so now we get $f(f(a))<a$. We can apply this reasoning again $-f(f(f(a)))<f(a)$, but also, $f(a)<a$, so $f(f(f(a)))<a$. But if we repeat this line of reasoning an extra $2013$ times, we get that $f^{circ 2016}(a)<a$.
This contradicts our requirement that $f^{circ 2016}$ fix all of $[0,1]$ so $f(a) nless a$ for all $a in [0,1]$.
We can reason analogously for the case where $f(a)$ is assumed to be $>a$, ruling it as an impossibility.
So $f(x)=x$ for all $x in [0,1]$.
But this contradicts the requirement that $f'(x)<1$ for all $0<x<1$, so no solution exists.
answered Jan 7 at 20:28
Cardioid_Ass_22Cardioid_Ass_22
27413
$endgroup$
$begingroup$
It seems that we interpret the question differently. What I thought is that – for a given function $f$ with those properties, the question is about the number of solutions $x in [0, 1]$ of the equation $f(f(f( ldots f)(x) ldots) =x$, and not that $f^{circ 2016}$ fixes all $x in [0, 1]$. – But I may be wrong of course.
$endgroup$
– Martin R
Jan 8 at 14:18
$begingroup$
Oh, that's certainly fair. Yeah, I was a bit confused when I first saw your answer. Upon rereading the question, it seems your interpretation is more likely to be the right one. It's just generally questions of this sort of format that I've seen ask for solutions in functions.
$endgroup$
– Cardioid_Ass_22
Jan 8 at 14:31
$begingroup$
So does the 2016 have any significance ?
$endgroup$
– Randin
Jan 8 at 17:26
$begingroup$
@Randin For this question, no.
$endgroup$
– Cardioid_Ass_22
Jan 9 at 5:17
add a comment |
$begingroup$
First use induction to show that each iterate $f_n(x) = underbrace{f(f(f( ldots f}_{n~text{times}}(x) ldots)$ satisfies:
$$
begin{align}
0 < f_n(x) < 1 quad &text{for all $x$ on the interval $0 le x le 1$,} tag 1\
0 < f_n'(x) < 1 quad &text{for all $x$ on the interval $0 le x le 1$.} tag 2
end{align}
$$
Then use $(1)$ and the intermediate value theorem to show that $f_n(x) - x$ has at least one zero in $[0, 1]$.
Finally use $(2)$ and the mean-value theorem to show that $f_n(x) - x$ has at most one zero in $[0, 1]$.
(Remark: Instead of $0 < f'(x) < 1$ it would be sufficient to require that $|f'(x)| < 1$ on the interval.)
answered Jan 7 at 9:10
Martin RMartin R
27.8k33255
$endgroup$
$begingroup$
I still dont get it
$endgroup$
– Randin
Jan 8 at 17:25
$begingroup$
@Randin: What exactly is unclear?
$endgroup$
– Martin R
Jan 8 at 17:54
$begingroup$
The use of Ivt and mvt
$endgroup$
– Randin
Jan 10 at 0:57
$begingroup$
@Randin: Consider $g(x) = f_n(x) - x$. Then $g(0) > 0 > g(1)$, and the IVT states that $g$ has a zero, which means that $f_n$ has a fixed point. If $f_n$ has two distinct fixed points $a, b$ then from the MVT $1 = frac{f_n(a)-f_n(b)}{a-b} = f_n'(c)$ for some $c$.
$endgroup$
– Martin R
Jan 10 at 5:59
$begingroup$
Martin R ,I get the ivt part now that u wrote that thanx .can u write the second part ..the mvt part in math jax?
$endgroup$
– Randin
Jan 11 at 0:39
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $f'(x)>0$ for all $x$ in the concerned interval, $f(x)$ is strictly increasing.
So:
$x<y implies f(x)<f(y)$
(assuming $x,y in [0,1]$)
Now, suppose for some $a in [0,1],f(a) ne a$.
Then, either $f(a)<a$ or $f(a)>a$.
Looking at the first case and using the fact that $f$ is strictly increasing and, crucially, that $f$ maps from $[0,1]$ back into $[0,1]$, $f(f(a))<f(a)$. Of course, by the initial assumption, $f(a)<a$, so now we get $f(f(a))<a$. We can apply this reasoning again $-f(f(f(a)))<f(a)$, but also, $f(a)<a$, so $f(f(f(a)))<a$. But if we repeat this line of reasoning an extra $2013$ times, we get that $f^{circ 2016}(a)<a$.
This contradicts our requirement that $f^{circ 2016}$ fix all of $[0,1]$ so $f(a) nless a$ for all $a in [0,1]$.
We can reason analogously for the case where $f(a)$ is assumed to be $>a$, ruling it as an impossibility.
So $f(x)=x$ for all $x in [0,1]$.
But this contradicts the requirement that $f'(x)<1$ for all $0<x<1$, so no solution exists.
answered Jan 7 at 20:28
Cardioid_Ass_22Cardioid_Ass_22
27413
$endgroup$
$begingroup$
It seems that we interpret the question differently. What I thought is that – for a given function $f$ with those properties, the question is about the number of solutions $x in [0, 1]$ of the equation $f(f(f( ldots f)(x) ldots) =x$, and not that $f^{circ 2016}$ fixes all $x in [0, 1]$. – But I may be wrong of course.
$endgroup$
– Martin R
Jan 8 at 14:18
$begingroup$
Oh, that's certainly fair. Yeah, I was a bit confused when I first saw your answer. Upon rereading the question, it seems your interpretation is more likely to be the right one. It's just generally questions of this sort of format that I've seen ask for solutions in functions.
$endgroup$
– Cardioid_Ass_22
Jan 8 at 14:31
$begingroup$
So does the 2016 have any significance ?
$endgroup$
– Randin
Jan 8 at 17:26
$begingroup$
@Randin For this question, no.
$endgroup$
– Cardioid_Ass_22
Jan 9 at 5:17
add a comment |
$begingroup$
Since $f'(x)>0$ for all $x$ in the concerned interval, $f(x)$ is strictly increasing.
So:
$x<y implies f(x)<f(y)$
(assuming $x,y in [0,1]$)
Now, suppose for some $a in [0,1],f(a) ne a$.
Then, either $f(a)<a$ or $f(a)>a$.
Looking at the first case and using the fact that $f$ is strictly increasing and, crucially, that $f$ maps from $[0,1]$ back into $[0,1]$, $f(f(a))<f(a)$. Of course, by the initial assumption, $f(a)<a$, so now we get $f(f(a))<a$. We can apply this reasoning again $-f(f(f(a)))<f(a)$, but also, $f(a)<a$, so $f(f(f(a)))<a$. But if we repeat this line of reasoning an extra $2013$ times, we get that $f^{circ 2016}(a)<a$.
This contradicts our requirement that $f^{circ 2016}$ fix all of $[0,1]$ so $f(a) nless a$ for all $a in [0,1]$.
We can reason analogously for the case where $f(a)$ is assumed to be $>a$, ruling it as an impossibility.
So $f(x)=x$ for all $x in [0,1]$.
But this contradicts the requirement that $f'(x)<1$ for all $0<x<1$, so no solution exists.
answered Jan 7 at 20:28
Cardioid_Ass_22Cardioid_Ass_22
27413
$endgroup$
$begingroup$
It seems that we interpret the question differently. What I thought is that – for a given function $f$ with those properties, the question is about the number of solutions $x in [0, 1]$ of the equation $f(f(f( ldots f)(x) ldots) =x$, and not that $f^{circ 2016}$ fixes all $x in [0, 1]$. – But I may be wrong of course.
$endgroup$
– Martin R
Jan 8 at 14:18
$begingroup$
Oh, that's certainly fair. Yeah, I was a bit confused when I first saw your answer. Upon rereading the question, it seems your interpretation is more likely to be the right one. It's just generally questions of this sort of format that I've seen ask for solutions in functions.
$endgroup$
– Cardioid_Ass_22
Jan 8 at 14:31
$begingroup$
So does the 2016 have any significance ?
$endgroup$
– Randin
Jan 8 at 17:26
$begingroup$
@Randin For this question, no.
$endgroup$
– Cardioid_Ass_22
Jan 9 at 5:17
add a comment |
$begingroup$
Since $f'(x)>0$ for all $x$ in the concerned interval, $f(x)$ is strictly increasing.
So:
$x<y implies f(x)<f(y)$
(assuming $x,y in [0,1]$)
Now, suppose for some $a in [0,1],f(a) ne a$.
Then, either $f(a)<a$ or $f(a)>a$.
Looking at the first case and using the fact that $f$ is strictly increasing and, crucially, that $f$ maps from $[0,1]$ back into $[0,1]$, $f(f(a))<f(a)$. Of course, by the initial assumption, $f(a)<a$, so now we get $f(f(a))<a$. We can apply this reasoning again $-f(f(f(a)))<f(a)$, but also, $f(a)<a$, so $f(f(f(a)))<a$. But if we repeat this line of reasoning an extra $2013$ times, we get that $f^{circ 2016}(a)<a$.
This contradicts our requirement that $f^{circ 2016}$ fix all of $[0,1]$ so $f(a) nless a$ for all $a in [0,1]$.
We can reason analogously for the case where $f(a)$ is assumed to be $>a$, ruling it as an impossibility.
So $f(x)=x$ for all $x in [0,1]$.
But this contradicts the requirement that $f'(x)<1$ for all $0<x<1$, so no solution exists.
answered Jan 7 at 20:28
Cardioid_Ass_22Cardioid_Ass_22
27413
$endgroup$
Since $f'(x)>0$ for all $x$ in the concerned interval, $f(x)$ is strictly increasing.
So:
$x<y implies f(x)<f(y)$
(assuming $x,y in [0,1]$)
Now, suppose for some $a in [0,1],f(a) ne a$.
Then, either $f(a)<a$ or $f(a)>a$.
Looking at the first case and using the fact that $f$ is strictly increasing and, crucially, that $f$ maps from $[0,1]$ back into $[0,1]$, $f(f(a))<f(a)$. Of course, by the initial assumption, $f(a)<a$, so now we get $f(f(a))<a$. We can apply this reasoning again $-f(f(f(a)))<f(a)$, but also, $f(a)<a$, so $f(f(f(a)))<a$. But if we repeat this line of reasoning an extra $2013$ times, we get that $f^{circ 2016}(a)<a$.
This contradicts our requirement that $f^{circ 2016}$ fix all of $[0,1]$ so $f(a) nless a$ for all $a in [0,1]$.
We can reason analogously for the case where $f(a)$ is assumed to be $>a$, ruling it as an impossibility.
So $f(x)=x$ for all $x in [0,1]$.
But this contradicts the requirement that $f'(x)<1$ for all $0<x<1$, so no solution exists.
answered Jan 7 at 20:28
Cardioid_Ass_22Cardioid_Ass_22
27413
edited Jan 8 at 10:27
answered Jan 7 at 20:28
Cardioid_Ass_22Cardioid_Ass_22
27413
answered Jan 7 at 20:28
Cardioid_Ass_22Cardioid_Ass_22
27413
answered Jan 7 at 20:28
Cardioid_Ass_22Cardioid_Ass_22
27413
27413
$begingroup$
It seems that we interpret the question differently. What I thought is that – for a given function $f$ with those properties, the question is about the number of solutions $x in [0, 1]$ of the equation $f(f(f( ldots f)(x) ldots) =x$, and not that $f^{circ 2016}$ fixes all $x in [0, 1]$. – But I may be wrong of course.
$endgroup$
– Martin R
Jan 8 at 14:18
$begingroup$
Oh, that's certainly fair. Yeah, I was a bit confused when I first saw your answer. Upon rereading the question, it seems your interpretation is more likely to be the right one. It's just generally questions of this sort of format that I've seen ask for solutions in functions.
$endgroup$
– Cardioid_Ass_22
Jan 8 at 14:31
$begingroup$
So does the 2016 have any significance ?
$endgroup$
– Randin
Jan 8 at 17:26
$begingroup$
@Randin For this question, no.
$endgroup$
– Cardioid_Ass_22
Jan 9 at 5:17
add a comment |
$begingroup$
It seems that we interpret the question differently. What I thought is that – for a given function $f$ with those properties, the question is about the number of solutions $x in [0, 1]$ of the equation $f(f(f( ldots f)(x) ldots) =x$, and not that $f^{circ 2016}$ fixes all $x in [0, 1]$. – But I may be wrong of course.
$endgroup$
– Martin R
Jan 8 at 14:18
$begingroup$
Oh, that's certainly fair. Yeah, I was a bit confused when I first saw your answer. Upon rereading the question, it seems your interpretation is more likely to be the right one. It's just generally questions of this sort of format that I've seen ask for solutions in functions.
$endgroup$
– Cardioid_Ass_22
Jan 8 at 14:31
$begingroup$
So does the 2016 have any significance ?
$endgroup$
– Randin
Jan 8 at 17:26
$begingroup$
@Randin For this question, no.
$endgroup$
– Cardioid_Ass_22
Jan 9 at 5:17
$begingroup$
It seems that we interpret the question differently. What I thought is that – for a given function $f$ with those properties, the question is about the number of solutions $x in [0, 1]$ of the equation $f(f(f( ldots f)(x) ldots) =x$, and not that $f^{circ 2016}$ fixes all $x in [0, 1]$. – But I may be wrong of course.
$endgroup$
– Martin R
Jan 8 at 14:18
$begingroup$
It seems that we interpret the question differently. What I thought is that – for a given function $f$ with those properties, the question is about the number of solutions $x in [0, 1]$ of the equation $f(f(f( ldots f)(x) ldots) =x$, and not that $f^{circ 2016}$ fixes all $x in [0, 1]$. – But I may be wrong of course.
$endgroup$
– Martin R
Jan 8 at 14:18
$begingroup$
Oh, that's certainly fair. Yeah, I was a bit confused when I first saw your answer. Upon rereading the question, it seems your interpretation is more likely to be the right one. It's just generally questions of this sort of format that I've seen ask for solutions in functions.
$endgroup$
– Cardioid_Ass_22
Jan 8 at 14:31
$begingroup$
Oh, that's certainly fair. Yeah, I was a bit confused when I first saw your answer. Upon rereading the question, it seems your interpretation is more likely to be the right one. It's just generally questions of this sort of format that I've seen ask for solutions in functions.
$endgroup$
– Cardioid_Ass_22
Jan 8 at 14:31
$begingroup$
So does the 2016 have any significance ?
$endgroup$
– Randin
Jan 8 at 17:26
$begingroup$
So does the 2016 have any significance ?
$endgroup$
– Randin
Jan 8 at 17:26
$begingroup$
@Randin For this question, no.
$endgroup$
– Cardioid_Ass_22
Jan 9 at 5:17
$begingroup$
@Randin For this question, no.
$endgroup$
– Cardioid_Ass_22
Jan 9 at 5:17
add a comment |
$begingroup$
First use induction to show that each iterate $f_n(x) = underbrace{f(f(f( ldots f}_{n~text{times}}(x) ldots)$ satisfies:
$$
begin{align}
0 < f_n(x) < 1 quad &text{for all $x$ on the interval $0 le x le 1$,} tag 1\
0 < f_n'(x) < 1 quad &text{for all $x$ on the interval $0 le x le 1$.} tag 2
end{align}
$$
Then use $(1)$ and the intermediate value theorem to show that $f_n(x) - x$ has at least one zero in $[0, 1]$.
Finally use $(2)$ and the mean-value theorem to show that $f_n(x) - x$ has at most one zero in $[0, 1]$.
(Remark: Instead of $0 < f'(x) < 1$ it would be sufficient to require that $|f'(x)| < 1$ on the interval.)
answered Jan 7 at 9:10
Martin RMartin R
27.8k33255
$endgroup$
$begingroup$
I still dont get it
$endgroup$
– Randin
Jan 8 at 17:25
$begingroup$
@Randin: What exactly is unclear?
$endgroup$
– Martin R
Jan 8 at 17:54
$begingroup$
The use of Ivt and mvt
$endgroup$
– Randin
Jan 10 at 0:57
$begingroup$
@Randin: Consider $g(x) = f_n(x) - x$. Then $g(0) > 0 > g(1)$, and the IVT states that $g$ has a zero, which means that $f_n$ has a fixed point. If $f_n$ has two distinct fixed points $a, b$ then from the MVT $1 = frac{f_n(a)-f_n(b)}{a-b} = f_n'(c)$ for some $c$.
$endgroup$
– Martin R
Jan 10 at 5:59
$begingroup$
Martin R ,I get the ivt part now that u wrote that thanx .can u write the second part ..the mvt part in math jax?
$endgroup$
– Randin
Jan 11 at 0:39
add a comment |
$begingroup$
First use induction to show that each iterate $f_n(x) = underbrace{f(f(f( ldots f}_{n~text{times}}(x) ldots)$ satisfies:
$$
begin{align}
0 < f_n(x) < 1 quad &text{for all $x$ on the interval $0 le x le 1$,} tag 1\
0 < f_n'(x) < 1 quad &text{for all $x$ on the interval $0 le x le 1$.} tag 2
end{align}
$$
Then use $(1)$ and the intermediate value theorem to show that $f_n(x) - x$ has at least one zero in $[0, 1]$.
Finally use $(2)$ and the mean-value theorem to show that $f_n(x) - x$ has at most one zero in $[0, 1]$.
(Remark: Instead of $0 < f'(x) < 1$ it would be sufficient to require that $|f'(x)| < 1$ on the interval.)
answered Jan 7 at 9:10
Martin RMartin R
27.8k33255
$endgroup$
$begingroup$
I still dont get it
$endgroup$
– Randin
Jan 8 at 17:25
$begingroup$
@Randin: What exactly is unclear?
$endgroup$
– Martin R
Jan 8 at 17:54
$begingroup$
The use of Ivt and mvt
$endgroup$
– Randin
Jan 10 at 0:57
$begingroup$
@Randin: Consider $g(x) = f_n(x) - x$. Then $g(0) > 0 > g(1)$, and the IVT states that $g$ has a zero, which means that $f_n$ has a fixed point. If $f_n$ has two distinct fixed points $a, b$ then from the MVT $1 = frac{f_n(a)-f_n(b)}{a-b} = f_n'(c)$ for some $c$.
$endgroup$
– Martin R
Jan 10 at 5:59
$begingroup$
Martin R ,I get the ivt part now that u wrote that thanx .can u write the second part ..the mvt part in math jax?
$endgroup$
– Randin
Jan 11 at 0:39
add a comment |
$begingroup$
First use induction to show that each iterate $f_n(x) = underbrace{f(f(f( ldots f}_{n~text{times}}(x) ldots)$ satisfies:
$$
begin{align}
0 < f_n(x) < 1 quad &text{for all $x$ on the interval $0 le x le 1$,} tag 1\
0 < f_n'(x) < 1 quad &text{for all $x$ on the interval $0 le x le 1$.} tag 2
end{align}
$$
Then use $(1)$ and the intermediate value theorem to show that $f_n(x) - x$ has at least one zero in $[0, 1]$.
Finally use $(2)$ and the mean-value theorem to show that $f_n(x) - x$ has at most one zero in $[0, 1]$.
(Remark: Instead of $0 < f'(x) < 1$ it would be sufficient to require that $|f'(x)| < 1$ on the interval.)
answered Jan 7 at 9:10
Martin RMartin R
27.8k33255
$endgroup$
First use induction to show that each iterate $f_n(x) = underbrace{f(f(f( ldots f}_{n~text{times}}(x) ldots)$ satisfies:
$$
begin{align}
0 < f_n(x) < 1 quad &text{for all $x$ on the interval $0 le x le 1$,} tag 1\
0 < f_n'(x) < 1 quad &text{for all $x$ on the interval $0 le x le 1$.} tag 2
end{align}
$$
Then use $(1)$ and the intermediate value theorem to show that $f_n(x) - x$ has at least one zero in $[0, 1]$.
Finally use $(2)$ and the mean-value theorem to show that $f_n(x) - x$ has at most one zero in $[0, 1]$.
(Remark: Instead of $0 < f'(x) < 1$ it would be sufficient to require that $|f'(x)| < 1$ on the interval.)
answered Jan 7 at 9:10
Martin RMartin R
27.8k33255
edited Jan 7 at 9:37
answered Jan 7 at 9:10
Martin RMartin R
27.8k33255
answered Jan 7 at 9:10
Martin RMartin R
27.8k33255
answered Jan 7 at 9:10
Martin RMartin R
27.8k33255
27.8k33255
$begingroup$
I still dont get it
$endgroup$
– Randin
Jan 8 at 17:25
$begingroup$
@Randin: What exactly is unclear?
$endgroup$
– Martin R
Jan 8 at 17:54
$begingroup$
The use of Ivt and mvt
$endgroup$
– Randin
Jan 10 at 0:57
$begingroup$
@Randin: Consider $g(x) = f_n(x) - x$. Then $g(0) > 0 > g(1)$, and the IVT states that $g$ has a zero, which means that $f_n$ has a fixed point. If $f_n$ has two distinct fixed points $a, b$ then from the MVT $1 = frac{f_n(a)-f_n(b)}{a-b} = f_n'(c)$ for some $c$.
$endgroup$
– Martin R
Jan 10 at 5:59
$begingroup$
Martin R ,I get the ivt part now that u wrote that thanx .can u write the second part ..the mvt part in math jax?
$endgroup$
– Randin
Jan 11 at 0:39
add a comment |
$begingroup$
I still dont get it
$endgroup$
– Randin
Jan 8 at 17:25
$begingroup$
@Randin: What exactly is unclear?
$endgroup$
– Martin R
Jan 8 at 17:54
$begingroup$
The use of Ivt and mvt
$endgroup$
– Randin
Jan 10 at 0:57
$begingroup$
@Randin: Consider $g(x) = f_n(x) - x$. Then $g(0) > 0 > g(1)$, and the IVT states that $g$ has a zero, which means that $f_n$ has a fixed point. If $f_n$ has two distinct fixed points $a, b$ then from the MVT $1 = frac{f_n(a)-f_n(b)}{a-b} = f_n'(c)$ for some $c$.
$endgroup$
– Martin R
Jan 10 at 5:59
$begingroup$
Martin R ,I get the ivt part now that u wrote that thanx .can u write the second part ..the mvt part in math jax?
$endgroup$
– Randin
Jan 11 at 0:39
$begingroup$
I still dont get it
$endgroup$
– Randin
Jan 8 at 17:25
$begingroup$
I still dont get it
$endgroup$
– Randin
Jan 8 at 17:25
$begingroup$
@Randin: What exactly is unclear?
$endgroup$
– Martin R
Jan 8 at 17:54
$begingroup$
@Randin: What exactly is unclear?
$endgroup$
– Martin R
Jan 8 at 17:54
$begingroup$
The use of Ivt and mvt
$endgroup$
– Randin
Jan 10 at 0:57
$begingroup$
The use of Ivt and mvt
$endgroup$
– Randin
Jan 10 at 0:57
$begingroup$
@Randin: Consider $g(x) = f_n(x) - x$. Then $g(0) > 0 > g(1)$, and the IVT states that $g$ has a zero, which means that $f_n$ has a fixed point. If $f_n$ has two distinct fixed points $a, b$ then from the MVT $1 = frac{f_n(a)-f_n(b)}{a-b} = f_n'(c)$ for some $c$.
$endgroup$
– Martin R
Jan 10 at 5:59
$begingroup$
@Randin: Consider $g(x) = f_n(x) - x$. Then $g(0) > 0 > g(1)$, and the IVT states that $g$ has a zero, which means that $f_n$ has a fixed point. If $f_n$ has two distinct fixed points $a, b$ then from the MVT $1 = frac{f_n(a)-f_n(b)}{a-b} = f_n'(c)$ for some $c$.
$endgroup$
– Martin R
Jan 10 at 5:59
$begingroup$
Martin R ,I get the ivt part now that u wrote that thanx .can u write the second part ..the mvt part in math jax?
$endgroup$
– Randin
Jan 11 at 0:39
$begingroup$
Martin R ,I get the ivt part now that u wrote that thanx .can u write the second part ..the mvt part in math jax?
$endgroup$
– Randin
Jan 11 at 0:39
add a comment |
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Please rewrite the problem itself. What are the conditions? Please also use MathJax to typeset the equations. What have you tried?
$endgroup$
– Matti P.
Jan 7 at 8:05
$begingroup$
Ah ! Ok i will go back and do that thanks
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– Randin
Jan 7 at 13:08
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Just the green check mark below the upvotes ?
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– Randin
Jan 7 at 13:34
$begingroup$
Martin so only one accepted answer right ? And as for the question at hand - isnt this the same as the number of times the function crosses the y=x line ? The composing it 2016 times will mean it crosses multiple times right ?
$endgroup$
– Randin
Jan 7 at 14:20