Prove that if $forall_xinBbbR f(x)=f(x+1)$ and $f$ is continuous then there are infinitely many $c$ such that $f(c+pi)=f(c)$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)A Topology such that the continuous functions are exactly the polynomialsLet $f,g:[0,1]toBbbR$ such that $f(1)<0<f(0)$, $g$ is continuous on $[0,1]$ and $f+g$ is nondecreasing, Prove that $exists x_0in [0,1],f(x_0)=0$Determining whether a given set can be an image of continuous functionLet $fcolonBbb Rto Bbb R-3$ be a function such that there exist $T>0$ with $f(x+T)=fracf(x)-5f(x)-3$ for every $xinBbb R$.Function problem involving limits and periodicityProblem with periodic, continuous functionProve that there exists a c such that f(c-1/8)=f(c+1/8)Explain precisely why a continuous surjection $f: [0,1] mapsto [1, infty)$ does / does not existthe existence of sequences that approach infimum and supremum without knowing monotonicityProve there doesn't exist a such continuous function $f$ on $BbbR$
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Prove that if $forall_xinBbbR f(x)=f(x+1)$ and $f$ is continuous then there are infinitely many $c$ such that $f(c+pi)=f(c)$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)A Topology such that the continuous functions are exactly the polynomialsLet $f,g:[0,1]toBbbR$ such that $f(1)<0<f(0)$, $g$ is continuous on $[0,1]$ and $f+g$ is nondecreasing, Prove that $exists x_0in [0,1],f(x_0)=0$Determining whether a given set can be an image of continuous functionLet $fcolonBbb Rto Bbb R-3$ be a function such that there exist $T>0$ with $f(x+T)=fracf(x)-5f(x)-3$ for every $xinBbb R$.Function problem involving limits and periodicityProblem with periodic, continuous functionProve that there exists a c such that f(c-1/8)=f(c+1/8)Explain precisely why a continuous surjection $f: [0,1] mapsto [1, infty)$ does / does not existthe existence of sequences that approach infimum and supremum without knowing monotonicityProve there doesn't exist a such continuous function $f$ on $BbbR$
$begingroup$
We have 3 assumptions about $f$:
- $f: BbbR rightarrow BbbR$
- $f$ is continuous
- $forall_xinBbbR f(x)=f(x+1)$
The problem asks us to prove 2 things: That $f$ reaches its supremum and infimum and also there exist infinitely many $cinBbbR$ such that $f(pi+c)=f(c)$.
So the first thing can be proven (I think) discovering that $f([0,1]) = f(BbbR)$ (because the function is periodic) so supremum and infinimum is in $f([0,1])$ and by the extreme value theorem it can be reached.
As for the second problem I can't see why it should be true. That would mean that there exist $xinBbbR$ such that $f(pi +x+1) = f(x+1) = f(x)$. Wouldn't that mean that function must have 2 periods - $1$ and $pi$? But this $pi$ makes no sense then. Am I understanding this correctly?
calculus limits functions continuity
asked Feb 3 '15 at 12:01
qiubitqiubit
1,1171923
$endgroup$
add a comment |
$begingroup$
We have 3 assumptions about $f$:
- $f: BbbR rightarrow BbbR$
- $f$ is continuous
- $forall_xinBbbR f(x)=f(x+1)$
The problem asks us to prove 2 things: That $f$ reaches its supremum and infimum and also there exist infinitely many $cinBbbR$ such that $f(pi+c)=f(c)$.
So the first thing can be proven (I think) discovering that $f([0,1]) = f(BbbR)$ (because the function is periodic) so supremum and infinimum is in $f([0,1])$ and by the extreme value theorem it can be reached.
As for the second problem I can't see why it should be true. That would mean that there exist $xinBbbR$ such that $f(pi +x+1) = f(x+1) = f(x)$. Wouldn't that mean that function must have 2 periods - $1$ and $pi$? But this $pi$ makes no sense then. Am I understanding this correctly?
calculus limits functions continuity
asked Feb 3 '15 at 12:01
qiubitqiubit
1,1171923
$endgroup$
$begingroup$
Look at the function $g:x mapsto f(x+pi) - f(x)$. You know that $g$ is $1$-periodic. Your claim is equivalent to showing that $g$ has infinitely many zeros. By the intermediate value theorem, it suffices that ??? I will let you take it from here.
$endgroup$
– PhoemueX
Feb 3 '15 at 12:13
add a comment |
$begingroup$
We have 3 assumptions about $f$:
- $f: BbbR rightarrow BbbR$
- $f$ is continuous
- $forall_xinBbbR f(x)=f(x+1)$
The problem asks us to prove 2 things: That $f$ reaches its supremum and infimum and also there exist infinitely many $cinBbbR$ such that $f(pi+c)=f(c)$.
So the first thing can be proven (I think) discovering that $f([0,1]) = f(BbbR)$ (because the function is periodic) so supremum and infinimum is in $f([0,1])$ and by the extreme value theorem it can be reached.
As for the second problem I can't see why it should be true. That would mean that there exist $xinBbbR$ such that $f(pi +x+1) = f(x+1) = f(x)$. Wouldn't that mean that function must have 2 periods - $1$ and $pi$? But this $pi$ makes no sense then. Am I understanding this correctly?
calculus limits functions continuity
asked Feb 3 '15 at 12:01
qiubitqiubit
1,1171923
$endgroup$
We have 3 assumptions about $f$:
- $f: BbbR rightarrow BbbR$
- $f$ is continuous
- $forall_xinBbbR f(x)=f(x+1)$
The problem asks us to prove 2 things: That $f$ reaches its supremum and infimum and also there exist infinitely many $cinBbbR$ such that $f(pi+c)=f(c)$.
So the first thing can be proven (I think) discovering that $f([0,1]) = f(BbbR)$ (because the function is periodic) so supremum and infinimum is in $f([0,1])$ and by the extreme value theorem it can be reached.
As for the second problem I can't see why it should be true. That would mean that there exist $xinBbbR$ such that $f(pi +x+1) = f(x+1) = f(x)$. Wouldn't that mean that function must have 2 periods - $1$ and $pi$? But this $pi$ makes no sense then. Am I understanding this correctly?
calculus limits functions continuity
calculus limits functions continuity
asked Feb 3 '15 at 12:01
qiubitqiubit
1,1171923
asked Feb 3 '15 at 12:01
qiubitqiubit
1,1171923
asked Feb 3 '15 at 12:01
qiubitqiubit
1,1171923
asked Feb 3 '15 at 12:01
qiubitqiubit
1,1171923
asked Feb 3 '15 at 12:01
qiubitqiubit
1,1171923
1,1171923
$begingroup$
Look at the function $g:x mapsto f(x+pi) - f(x)$. You know that $g$ is $1$-periodic. Your claim is equivalent to showing that $g$ has infinitely many zeros. By the intermediate value theorem, it suffices that ??? I will let you take it from here.
$endgroup$
– PhoemueX
Feb 3 '15 at 12:13
add a comment |
$begingroup$
Look at the function $g:x mapsto f(x+pi) - f(x)$. You know that $g$ is $1$-periodic. Your claim is equivalent to showing that $g$ has infinitely many zeros. By the intermediate value theorem, it suffices that ??? I will let you take it from here.
$endgroup$
– PhoemueX
Feb 3 '15 at 12:13
$begingroup$
Look at the function $g:x mapsto f(x+pi) - f(x)$. You know that $g$ is $1$-periodic. Your claim is equivalent to showing that $g$ has infinitely many zeros. By the intermediate value theorem, it suffices that ??? I will let you take it from here.
$endgroup$
– PhoemueX
Feb 3 '15 at 12:13
$begingroup$
Look at the function $g:x mapsto f(x+pi) - f(x)$. You know that $g$ is $1$-periodic. Your claim is equivalent to showing that $g$ has infinitely many zeros. By the intermediate value theorem, it suffices that ??? I will let you take it from here.
$endgroup$
– PhoemueX
Feb 3 '15 at 12:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since $g(x) := f(x+pi)-f(x)$ is also periodic, we only have to show, that $g$ has a zero.
If not, we can assume that $g$ is strictly positive by the intermediate value theorem. This gives us an increasing sequence $f(npi)$, which converges, since $f$ is bounded.
But the sequence equals the sequence $f(npi-lfloor npi rfloor)$. Since $npi-lfloor npi rfloor$ has all points in $[0,1]$ as accumulation points, by continuity of $f$ we get that $f(npi-lfloor npi rfloor)$ has all points of $f([0,1])$ as accumulation points. But $f(npi-lfloor npi rfloor)$ converges, hence admits a unique accumulation point. We deduce that $f([0,1])$ is a singleton, hence $f$ is constant and $g=0$. Contradiction!
answered Feb 3 '15 at 12:18
MooSMooS
27.1k11334
$endgroup$
$begingroup$
Why is $g(x)$ periodic? Is this because a difference of 2 $1$-periodic functions is periodic?
$endgroup$
– qiubit
Feb 3 '15 at 13:03
$begingroup$
Yes. Just check $f(x+1+pi)-f(x+1)=f(x+pi)-f(x)$.
$endgroup$
– MooS
Feb 3 '15 at 18:33
add a comment |
Your Answer
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1 Answer
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$begingroup$
Since $g(x) := f(x+pi)-f(x)$ is also periodic, we only have to show, that $g$ has a zero.
If not, we can assume that $g$ is strictly positive by the intermediate value theorem. This gives us an increasing sequence $f(npi)$, which converges, since $f$ is bounded.
But the sequence equals the sequence $f(npi-lfloor npi rfloor)$. Since $npi-lfloor npi rfloor$ has all points in $[0,1]$ as accumulation points, by continuity of $f$ we get that $f(npi-lfloor npi rfloor)$ has all points of $f([0,1])$ as accumulation points. But $f(npi-lfloor npi rfloor)$ converges, hence admits a unique accumulation point. We deduce that $f([0,1])$ is a singleton, hence $f$ is constant and $g=0$. Contradiction!
answered Feb 3 '15 at 12:18
MooSMooS
27.1k11334
$endgroup$
$begingroup$
Why is $g(x)$ periodic? Is this because a difference of 2 $1$-periodic functions is periodic?
$endgroup$
– qiubit
Feb 3 '15 at 13:03
$begingroup$
Yes. Just check $f(x+1+pi)-f(x+1)=f(x+pi)-f(x)$.
$endgroup$
– MooS
Feb 3 '15 at 18:33
add a comment |
$begingroup$
Since $g(x) := f(x+pi)-f(x)$ is also periodic, we only have to show, that $g$ has a zero.
If not, we can assume that $g$ is strictly positive by the intermediate value theorem. This gives us an increasing sequence $f(npi)$, which converges, since $f$ is bounded.
But the sequence equals the sequence $f(npi-lfloor npi rfloor)$. Since $npi-lfloor npi rfloor$ has all points in $[0,1]$ as accumulation points, by continuity of $f$ we get that $f(npi-lfloor npi rfloor)$ has all points of $f([0,1])$ as accumulation points. But $f(npi-lfloor npi rfloor)$ converges, hence admits a unique accumulation point. We deduce that $f([0,1])$ is a singleton, hence $f$ is constant and $g=0$. Contradiction!
answered Feb 3 '15 at 12:18
MooSMooS
27.1k11334
$endgroup$
$begingroup$
Why is $g(x)$ periodic? Is this because a difference of 2 $1$-periodic functions is periodic?
$endgroup$
– qiubit
Feb 3 '15 at 13:03
$begingroup$
Yes. Just check $f(x+1+pi)-f(x+1)=f(x+pi)-f(x)$.
$endgroup$
– MooS
Feb 3 '15 at 18:33
add a comment |
$begingroup$
Since $g(x) := f(x+pi)-f(x)$ is also periodic, we only have to show, that $g$ has a zero.
If not, we can assume that $g$ is strictly positive by the intermediate value theorem. This gives us an increasing sequence $f(npi)$, which converges, since $f$ is bounded.
But the sequence equals the sequence $f(npi-lfloor npi rfloor)$. Since $npi-lfloor npi rfloor$ has all points in $[0,1]$ as accumulation points, by continuity of $f$ we get that $f(npi-lfloor npi rfloor)$ has all points of $f([0,1])$ as accumulation points. But $f(npi-lfloor npi rfloor)$ converges, hence admits a unique accumulation point. We deduce that $f([0,1])$ is a singleton, hence $f$ is constant and $g=0$. Contradiction!
answered Feb 3 '15 at 12:18
MooSMooS
27.1k11334
$endgroup$
Since $g(x) := f(x+pi)-f(x)$ is also periodic, we only have to show, that $g$ has a zero.
If not, we can assume that $g$ is strictly positive by the intermediate value theorem. This gives us an increasing sequence $f(npi)$, which converges, since $f$ is bounded.
But the sequence equals the sequence $f(npi-lfloor npi rfloor)$. Since $npi-lfloor npi rfloor$ has all points in $[0,1]$ as accumulation points, by continuity of $f$ we get that $f(npi-lfloor npi rfloor)$ has all points of $f([0,1])$ as accumulation points. But $f(npi-lfloor npi rfloor)$ converges, hence admits a unique accumulation point. We deduce that $f([0,1])$ is a singleton, hence $f$ is constant and $g=0$. Contradiction!
answered Feb 3 '15 at 12:18
MooSMooS
27.1k11334
answered Feb 3 '15 at 12:18
MooSMooS
27.1k11334
answered Feb 3 '15 at 12:18
MooSMooS
27.1k11334
answered Feb 3 '15 at 12:18
MooSMooS
27.1k11334
27.1k11334
$begingroup$
Why is $g(x)$ periodic? Is this because a difference of 2 $1$-periodic functions is periodic?
$endgroup$
– qiubit
Feb 3 '15 at 13:03
$begingroup$
Yes. Just check $f(x+1+pi)-f(x+1)=f(x+pi)-f(x)$.
$endgroup$
– MooS
Feb 3 '15 at 18:33
add a comment |
$begingroup$
Why is $g(x)$ periodic? Is this because a difference of 2 $1$-periodic functions is periodic?
$endgroup$
– qiubit
Feb 3 '15 at 13:03
$begingroup$
Yes. Just check $f(x+1+pi)-f(x+1)=f(x+pi)-f(x)$.
$endgroup$
– MooS
Feb 3 '15 at 18:33
$begingroup$
Why is $g(x)$ periodic? Is this because a difference of 2 $1$-periodic functions is periodic?
$endgroup$
– qiubit
Feb 3 '15 at 13:03
$begingroup$
Why is $g(x)$ periodic? Is this because a difference of 2 $1$-periodic functions is periodic?
$endgroup$
– qiubit
Feb 3 '15 at 13:03
$begingroup$
Yes. Just check $f(x+1+pi)-f(x+1)=f(x+pi)-f(x)$.
$endgroup$
– MooS
Feb 3 '15 at 18:33
$begingroup$
Yes. Just check $f(x+1+pi)-f(x+1)=f(x+pi)-f(x)$.
$endgroup$
– MooS
Feb 3 '15 at 18:33
add a comment |
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$begingroup$
Look at the function $g:x mapsto f(x+pi) - f(x)$. You know that $g$ is $1$-periodic. Your claim is equivalent to showing that $g$ has infinitely many zeros. By the intermediate value theorem, it suffices that ??? I will let you take it from here.
$endgroup$
– PhoemueX
Feb 3 '15 at 12:13