Periodic function with period $T$; show that there exists a point in the interval with $f(x) = f(x + T /2).$
$begingroup$
Let $f$: $Rrightarrow R$ be a continuous periodic function with period $T > 0$.
Show: There is an $x in R$ such that $$f(x) = f(x + T /2).$$
Trivial case: If $f(x)$ is constant, this statement is always true, but this cannot happen as $T>0$, and for a constant function $T=0$, so we are not dealing with a constant function.
This really feels like an intermediate value theorem question, but I am very used to working with concrete functions and wouldn't immediately know how to apply this to an arbitrary function. What do I know about this function?
$f(x)$ is continuous, useful.
$f(x)=f(x+T)$ for every single $x$ in the domain.
My intuition tells me it would be useful to consider the values $x=0$ and $x=T$, but to which function would I apply this?
real-analysis
asked Dec 8 '18 at 14:27
Wesley StrikWesley Strik
1,741423
$endgroup$
add a comment |
$begingroup$
Let $f$: $Rrightarrow R$ be a continuous periodic function with period $T > 0$.
Show: There is an $x in R$ such that $$f(x) = f(x + T /2).$$
Trivial case: If $f(x)$ is constant, this statement is always true, but this cannot happen as $T>0$, and for a constant function $T=0$, so we are not dealing with a constant function.
This really feels like an intermediate value theorem question, but I am very used to working with concrete functions and wouldn't immediately know how to apply this to an arbitrary function. What do I know about this function?
$f(x)$ is continuous, useful.
$f(x)=f(x+T)$ for every single $x$ in the domain.
My intuition tells me it would be useful to consider the values $x=0$ and $x=T$, but to which function would I apply this?
real-analysis
asked Dec 8 '18 at 14:27
Wesley StrikWesley Strik
1,741423
$endgroup$
$begingroup$
You don't need to exclude $T=0$, because $f(x)=f(x)$ for all $x$.
$endgroup$
– TonyK
Dec 8 '18 at 14:31
$begingroup$
Yeah I know, it might as well be $T geq 0$
$endgroup$
– Wesley Strik
Dec 8 '18 at 14:32
add a comment |
$begingroup$
Let $f$: $Rrightarrow R$ be a continuous periodic function with period $T > 0$.
Show: There is an $x in R$ such that $$f(x) = f(x + T /2).$$
Trivial case: If $f(x)$ is constant, this statement is always true, but this cannot happen as $T>0$, and for a constant function $T=0$, so we are not dealing with a constant function.
This really feels like an intermediate value theorem question, but I am very used to working with concrete functions and wouldn't immediately know how to apply this to an arbitrary function. What do I know about this function?
$f(x)$ is continuous, useful.
$f(x)=f(x+T)$ for every single $x$ in the domain.
My intuition tells me it would be useful to consider the values $x=0$ and $x=T$, but to which function would I apply this?
real-analysis
asked Dec 8 '18 at 14:27
Wesley StrikWesley Strik
1,741423
$endgroup$
Let $f$: $Rrightarrow R$ be a continuous periodic function with period $T > 0$.
Show: There is an $x in R$ such that $$f(x) = f(x + T /2).$$
Trivial case: If $f(x)$ is constant, this statement is always true, but this cannot happen as $T>0$, and for a constant function $T=0$, so we are not dealing with a constant function.
This really feels like an intermediate value theorem question, but I am very used to working with concrete functions and wouldn't immediately know how to apply this to an arbitrary function. What do I know about this function?
$f(x)$ is continuous, useful.
$f(x)=f(x+T)$ for every single $x$ in the domain.
My intuition tells me it would be useful to consider the values $x=0$ and $x=T$, but to which function would I apply this?
real-analysis
real-analysis
asked Dec 8 '18 at 14:27
Wesley StrikWesley Strik
1,741423
asked Dec 8 '18 at 14:27
Wesley StrikWesley Strik
1,741423
asked Dec 8 '18 at 14:27
Wesley StrikWesley Strik
1,741423
asked Dec 8 '18 at 14:27
Wesley StrikWesley Strik
1,741423
asked Dec 8 '18 at 14:27
Wesley StrikWesley Strik
1,741423
1,741423
$begingroup$
You don't need to exclude $T=0$, because $f(x)=f(x)$ for all $x$.
$endgroup$
– TonyK
Dec 8 '18 at 14:31
$begingroup$
Yeah I know, it might as well be $T geq 0$
$endgroup$
– Wesley Strik
Dec 8 '18 at 14:32
add a comment |
$begingroup$
You don't need to exclude $T=0$, because $f(x)=f(x)$ for all $x$.
$endgroup$
– TonyK
Dec 8 '18 at 14:31
$begingroup$
Yeah I know, it might as well be $T geq 0$
$endgroup$
– Wesley Strik
Dec 8 '18 at 14:32
$begingroup$
You don't need to exclude $T=0$, because $f(x)=f(x)$ for all $x$.
$endgroup$
– TonyK
Dec 8 '18 at 14:31
$begingroup$
You don't need to exclude $T=0$, because $f(x)=f(x)$ for all $x$.
$endgroup$
– TonyK
Dec 8 '18 at 14:31
$begingroup$
Yeah I know, it might as well be $T geq 0$
$endgroup$
– Wesley Strik
Dec 8 '18 at 14:32
$begingroup$
Yeah I know, it might as well be $T geq 0$
$endgroup$
– Wesley Strik
Dec 8 '18 at 14:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$g(x)=f(x)-f(x+T/2)$, $g(0)=f(0)-f(T/2)$, $g(T/2)=f(T/2)-f(T)=f(T/2)-f(0)$ apply intermediate value to $[0,T/2]$ since $g(0)$ and $g(T/2)$ have opposite signs.
answered Dec 8 '18 at 14:31
Tsemo AristideTsemo Aristide
57.5k11444
$endgroup$
$begingroup$
I see, this argument makes sense, thank you.
$endgroup$
– Wesley Strik
Dec 8 '18 at 14:52
$begingroup$
I can argue that for some value $c$, we have that $g(c)=0$ and Bob's your uncle, thanks!
$endgroup$
– Wesley Strik
Dec 8 '18 at 14:53
$begingroup$
A similar approach is to argue that if $f(x) neq f(x+T/2)$, then we must have $f(x) > f(x+T/2)$ for all $x$, or $f(x) < f(x+T/2)$ for all $x$. Then $f(x+T) > f(x+T/2) > f(x) = f(x+T)$, contradiction.
$endgroup$
– sdcvvc
Dec 8 '18 at 14:59
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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$begingroup$
$g(x)=f(x)-f(x+T/2)$, $g(0)=f(0)-f(T/2)$, $g(T/2)=f(T/2)-f(T)=f(T/2)-f(0)$ apply intermediate value to $[0,T/2]$ since $g(0)$ and $g(T/2)$ have opposite signs.
answered Dec 8 '18 at 14:31
Tsemo AristideTsemo Aristide
57.5k11444
$endgroup$
$begingroup$
I see, this argument makes sense, thank you.
$endgroup$
– Wesley Strik
Dec 8 '18 at 14:52
$begingroup$
I can argue that for some value $c$, we have that $g(c)=0$ and Bob's your uncle, thanks!
$endgroup$
– Wesley Strik
Dec 8 '18 at 14:53
$begingroup$
A similar approach is to argue that if $f(x) neq f(x+T/2)$, then we must have $f(x) > f(x+T/2)$ for all $x$, or $f(x) < f(x+T/2)$ for all $x$. Then $f(x+T) > f(x+T/2) > f(x) = f(x+T)$, contradiction.
$endgroup$
– sdcvvc
Dec 8 '18 at 14:59
add a comment |
$begingroup$
$g(x)=f(x)-f(x+T/2)$, $g(0)=f(0)-f(T/2)$, $g(T/2)=f(T/2)-f(T)=f(T/2)-f(0)$ apply intermediate value to $[0,T/2]$ since $g(0)$ and $g(T/2)$ have opposite signs.
answered Dec 8 '18 at 14:31
Tsemo AristideTsemo Aristide
57.5k11444
$endgroup$
$begingroup$
I see, this argument makes sense, thank you.
$endgroup$
– Wesley Strik
Dec 8 '18 at 14:52
$begingroup$
I can argue that for some value $c$, we have that $g(c)=0$ and Bob's your uncle, thanks!
$endgroup$
– Wesley Strik
Dec 8 '18 at 14:53
$begingroup$
A similar approach is to argue that if $f(x) neq f(x+T/2)$, then we must have $f(x) > f(x+T/2)$ for all $x$, or $f(x) < f(x+T/2)$ for all $x$. Then $f(x+T) > f(x+T/2) > f(x) = f(x+T)$, contradiction.
$endgroup$
– sdcvvc
Dec 8 '18 at 14:59
add a comment |
$begingroup$
$g(x)=f(x)-f(x+T/2)$, $g(0)=f(0)-f(T/2)$, $g(T/2)=f(T/2)-f(T)=f(T/2)-f(0)$ apply intermediate value to $[0,T/2]$ since $g(0)$ and $g(T/2)$ have opposite signs.
answered Dec 8 '18 at 14:31
Tsemo AristideTsemo Aristide
57.5k11444
$endgroup$
$g(x)=f(x)-f(x+T/2)$, $g(0)=f(0)-f(T/2)$, $g(T/2)=f(T/2)-f(T)=f(T/2)-f(0)$ apply intermediate value to $[0,T/2]$ since $g(0)$ and $g(T/2)$ have opposite signs.
answered Dec 8 '18 at 14:31
Tsemo AristideTsemo Aristide
57.5k11444
answered Dec 8 '18 at 14:31
Tsemo AristideTsemo Aristide
57.5k11444
answered Dec 8 '18 at 14:31
Tsemo AristideTsemo Aristide
57.5k11444
answered Dec 8 '18 at 14:31
Tsemo AristideTsemo Aristide
57.5k11444
57.5k11444
$begingroup$
I see, this argument makes sense, thank you.
$endgroup$
– Wesley Strik
Dec 8 '18 at 14:52
$begingroup$
I can argue that for some value $c$, we have that $g(c)=0$ and Bob's your uncle, thanks!
$endgroup$
– Wesley Strik
Dec 8 '18 at 14:53
$begingroup$
A similar approach is to argue that if $f(x) neq f(x+T/2)$, then we must have $f(x) > f(x+T/2)$ for all $x$, or $f(x) < f(x+T/2)$ for all $x$. Then $f(x+T) > f(x+T/2) > f(x) = f(x+T)$, contradiction.
$endgroup$
– sdcvvc
Dec 8 '18 at 14:59
add a comment |
$begingroup$
I see, this argument makes sense, thank you.
$endgroup$
– Wesley Strik
Dec 8 '18 at 14:52
$begingroup$
I can argue that for some value $c$, we have that $g(c)=0$ and Bob's your uncle, thanks!
$endgroup$
– Wesley Strik
Dec 8 '18 at 14:53
$begingroup$
A similar approach is to argue that if $f(x) neq f(x+T/2)$, then we must have $f(x) > f(x+T/2)$ for all $x$, or $f(x) < f(x+T/2)$ for all $x$. Then $f(x+T) > f(x+T/2) > f(x) = f(x+T)$, contradiction.
$endgroup$
– sdcvvc
Dec 8 '18 at 14:59
$begingroup$
I see, this argument makes sense, thank you.
$endgroup$
– Wesley Strik
Dec 8 '18 at 14:52
$begingroup$
I see, this argument makes sense, thank you.
$endgroup$
– Wesley Strik
Dec 8 '18 at 14:52
$begingroup$
I can argue that for some value $c$, we have that $g(c)=0$ and Bob's your uncle, thanks!
$endgroup$
– Wesley Strik
Dec 8 '18 at 14:53
$begingroup$
I can argue that for some value $c$, we have that $g(c)=0$ and Bob's your uncle, thanks!
$endgroup$
– Wesley Strik
Dec 8 '18 at 14:53
$begingroup$
A similar approach is to argue that if $f(x) neq f(x+T/2)$, then we must have $f(x) > f(x+T/2)$ for all $x$, or $f(x) < f(x+T/2)$ for all $x$. Then $f(x+T) > f(x+T/2) > f(x) = f(x+T)$, contradiction.
$endgroup$
– sdcvvc
Dec 8 '18 at 14:59
$begingroup$
A similar approach is to argue that if $f(x) neq f(x+T/2)$, then we must have $f(x) > f(x+T/2)$ for all $x$, or $f(x) < f(x+T/2)$ for all $x$. Then $f(x+T) > f(x+T/2) > f(x) = f(x+T)$, contradiction.
$endgroup$
– sdcvvc
Dec 8 '18 at 14:59
add a comment |
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$begingroup$
You don't need to exclude $T=0$, because $f(x)=f(x)$ for all $x$.
$endgroup$
– TonyK
Dec 8 '18 at 14:31
$begingroup$
Yeah I know, it might as well be $T geq 0$
$endgroup$
– Wesley Strik
Dec 8 '18 at 14:32