Is the function $f(x)=frac{sin(sin{(x)}+x)}{2+cos{(lvert xrvert}+cos{(x)})}$ monotonic and/or periodic?
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Let $f: mathbb{R}rightarrow mathbb{R}$
$$f(x)=frac{sin(sin{(x)}+x)}{2+cos{(lvert xrvert}+cos{(x)})}$$
I am having trouble showing whether or not this function is monotonic and/or periodic.
Monotonic: The function is only monotonic if it is entirely non-increasing or entirely non-decreasing.
Monotonically increasing: $forall x,y$ such that $xle y$ it follows that $f(x) le f(y)$. Choose $x=pi,y=frac{3pi}{2}$: $$implies f(pi)=frac{sin{(sin{pi}+pi)}}{2+cos{(pi+cos{pi}})}=frac{cos{1}}{2}
\f(frac{3pi}{2})=frac{sin{(sin{frac{3 pi}{2}}+frac{3pi}{2})}}{2+cos(frac{3pi}{2}+cos{frac{3pi}{2}})}=frac{-cos{1}}{2} \ frac{cos{1}}{2} > -frac{cos{1}}{2} implies text{not monotonically increasing}$$Monotonically decreasing $forall x,y$ such that $x le y$ it follows that $f(x) ge f(y)$. Choose $x=0, y=frac{pi}{2}$:
$$implies f(0)=0 \ f(frac{pi}{2})=frac{sin{(sin{frac{pi}{2}}+frac{pi}{2})}}{2+cos{(frac{pi}{2}+cos{frac{pi}{2}}})}=frac{cos{1}}{2} \ 0 < frac{cos{1}}{2} implies text{not monotonically decreasing}$$
Is it possible to argue this way? Am I getting the definitions for
monotonicity right?
Periodic: This is where I am less sure and quite confused. I know that a function $f$ is periodic if $f(x+T)=f(x)$. I know that $sin{(x)}$ and $cos{(x)}$ are $2 pi$ periodic functions so I my attempt was to just plug in $f(x+2 pi)$ and see where it leads me.
$$implies f(x+2 pi)=frac{sin(sin{(x+2 pi)}+2 pi)}{2+cos{(lvert x+2 pirvert}+cos{(x+2 pi)})}=frac{sin{x}}{2+cos{(lvert x+2pirvert+cos{x})}}$$
If $x ge 0$ I get the desired solution and $f(x+2 pi)=f(x)$.
However, for negative $x$, $lvert x+2 pi rvert not=x+2 pi$. So I
cannot show periodicity for all negative $x$. How can I resolve this?
Can a function be periodic for positive $x$ and not for negative?
calculus periodic-functions monotone-functions
asked Nov 27 at 17:32
Nullspace
190110
add a comment |
up vote
3
down vote
favorite
Let $f: mathbb{R}rightarrow mathbb{R}$
$$f(x)=frac{sin(sin{(x)}+x)}{2+cos{(lvert xrvert}+cos{(x)})}$$
I am having trouble showing whether or not this function is monotonic and/or periodic.
Monotonic: The function is only monotonic if it is entirely non-increasing or entirely non-decreasing.
Monotonically increasing: $forall x,y$ such that $xle y$ it follows that $f(x) le f(y)$. Choose $x=pi,y=frac{3pi}{2}$: $$implies f(pi)=frac{sin{(sin{pi}+pi)}}{2+cos{(pi+cos{pi}})}=frac{cos{1}}{2}
\f(frac{3pi}{2})=frac{sin{(sin{frac{3 pi}{2}}+frac{3pi}{2})}}{2+cos(frac{3pi}{2}+cos{frac{3pi}{2}})}=frac{-cos{1}}{2} \ frac{cos{1}}{2} > -frac{cos{1}}{2} implies text{not monotonically increasing}$$Monotonically decreasing $forall x,y$ such that $x le y$ it follows that $f(x) ge f(y)$. Choose $x=0, y=frac{pi}{2}$:
$$implies f(0)=0 \ f(frac{pi}{2})=frac{sin{(sin{frac{pi}{2}}+frac{pi}{2})}}{2+cos{(frac{pi}{2}+cos{frac{pi}{2}}})}=frac{cos{1}}{2} \ 0 < frac{cos{1}}{2} implies text{not monotonically decreasing}$$
Is it possible to argue this way? Am I getting the definitions for
monotonicity right?
Periodic: This is where I am less sure and quite confused. I know that a function $f$ is periodic if $f(x+T)=f(x)$. I know that $sin{(x)}$ and $cos{(x)}$ are $2 pi$ periodic functions so I my attempt was to just plug in $f(x+2 pi)$ and see where it leads me.
$$implies f(x+2 pi)=frac{sin(sin{(x+2 pi)}+2 pi)}{2+cos{(lvert x+2 pirvert}+cos{(x+2 pi)})}=frac{sin{x}}{2+cos{(lvert x+2pirvert+cos{x})}}$$
If $x ge 0$ I get the desired solution and $f(x+2 pi)=f(x)$.
However, for negative $x$, $lvert x+2 pi rvert not=x+2 pi$. So I
cannot show periodicity for all negative $x$. How can I resolve this?
Can a function be periodic for positive $x$ and not for negative?
calculus periodic-functions monotone-functions
asked Nov 27 at 17:32
Nullspace
190110
First of all, this is not Real Analysis. That tag is getting abused way too much. Secondly, the exact same question was asked some days ago. I updated the tags to fit the post.
– Rebellos
Nov 27 at 17:45
2
@Rebellos The same questions was NOT asked a few days ago because a few days ago I asked if the infimum and supremum could be obtained analytically to which the answer in a comment was "no" so I decided to delete it. This question refers to the same function but I am asking about monotonicity and periodicity. I would appreciate it if you could change that part of the comment. Otherwise, people are just going to look at the comment and assume this is a duplicate. Sorry about the tags. I never know which ones to use.
– Nullspace
Nov 27 at 17:50
1
Note that $f(x) = -f(-x)$, so periodicity on the interval $[0, infty)$ implies periodicity on the interval $(-infty, 0]$. Of course, a function could be periodic for positive and negative $x$ separately but not periodic overall.
– Connor Harris
Nov 27 at 20:51
@ConnorHarris Thanks for your comment. Building on your answer, can we now say that the function is periodic even though it has a different periodicity on $[0, infty)$ and $(-infty.0]?$
– Nullspace
Nov 27 at 21:17
1
@Nullspace that's a matter of the precise definition of periodicity that you were handed (I assume this is for some sort of school assignment).
– Connor Harris
Nov 28 at 14:37
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $f: mathbb{R}rightarrow mathbb{R}$
$$f(x)=frac{sin(sin{(x)}+x)}{2+cos{(lvert xrvert}+cos{(x)})}$$
I am having trouble showing whether or not this function is monotonic and/or periodic.
Monotonic: The function is only monotonic if it is entirely non-increasing or entirely non-decreasing.
Monotonically increasing: $forall x,y$ such that $xle y$ it follows that $f(x) le f(y)$. Choose $x=pi,y=frac{3pi}{2}$: $$implies f(pi)=frac{sin{(sin{pi}+pi)}}{2+cos{(pi+cos{pi}})}=frac{cos{1}}{2}
\f(frac{3pi}{2})=frac{sin{(sin{frac{3 pi}{2}}+frac{3pi}{2})}}{2+cos(frac{3pi}{2}+cos{frac{3pi}{2}})}=frac{-cos{1}}{2} \ frac{cos{1}}{2} > -frac{cos{1}}{2} implies text{not monotonically increasing}$$Monotonically decreasing $forall x,y$ such that $x le y$ it follows that $f(x) ge f(y)$. Choose $x=0, y=frac{pi}{2}$:
$$implies f(0)=0 \ f(frac{pi}{2})=frac{sin{(sin{frac{pi}{2}}+frac{pi}{2})}}{2+cos{(frac{pi}{2}+cos{frac{pi}{2}}})}=frac{cos{1}}{2} \ 0 < frac{cos{1}}{2} implies text{not monotonically decreasing}$$
Is it possible to argue this way? Am I getting the definitions for
monotonicity right?
Periodic: This is where I am less sure and quite confused. I know that a function $f$ is periodic if $f(x+T)=f(x)$. I know that $sin{(x)}$ and $cos{(x)}$ are $2 pi$ periodic functions so I my attempt was to just plug in $f(x+2 pi)$ and see where it leads me.
$$implies f(x+2 pi)=frac{sin(sin{(x+2 pi)}+2 pi)}{2+cos{(lvert x+2 pirvert}+cos{(x+2 pi)})}=frac{sin{x}}{2+cos{(lvert x+2pirvert+cos{x})}}$$
If $x ge 0$ I get the desired solution and $f(x+2 pi)=f(x)$.
However, for negative $x$, $lvert x+2 pi rvert not=x+2 pi$. So I
cannot show periodicity for all negative $x$. How can I resolve this?
Can a function be periodic for positive $x$ and not for negative?
calculus periodic-functions monotone-functions
asked Nov 27 at 17:32
Nullspace
190110
Let $f: mathbb{R}rightarrow mathbb{R}$
$$f(x)=frac{sin(sin{(x)}+x)}{2+cos{(lvert xrvert}+cos{(x)})}$$
I am having trouble showing whether or not this function is monotonic and/or periodic.
Monotonic: The function is only monotonic if it is entirely non-increasing or entirely non-decreasing.
Monotonically increasing: $forall x,y$ such that $xle y$ it follows that $f(x) le f(y)$. Choose $x=pi,y=frac{3pi}{2}$: $$implies f(pi)=frac{sin{(sin{pi}+pi)}}{2+cos{(pi+cos{pi}})}=frac{cos{1}}{2}
\f(frac{3pi}{2})=frac{sin{(sin{frac{3 pi}{2}}+frac{3pi}{2})}}{2+cos(frac{3pi}{2}+cos{frac{3pi}{2}})}=frac{-cos{1}}{2} \ frac{cos{1}}{2} > -frac{cos{1}}{2} implies text{not monotonically increasing}$$Monotonically decreasing $forall x,y$ such that $x le y$ it follows that $f(x) ge f(y)$. Choose $x=0, y=frac{pi}{2}$:
$$implies f(0)=0 \ f(frac{pi}{2})=frac{sin{(sin{frac{pi}{2}}+frac{pi}{2})}}{2+cos{(frac{pi}{2}+cos{frac{pi}{2}}})}=frac{cos{1}}{2} \ 0 < frac{cos{1}}{2} implies text{not monotonically decreasing}$$
Is it possible to argue this way? Am I getting the definitions for
monotonicity right?
Periodic: This is where I am less sure and quite confused. I know that a function $f$ is periodic if $f(x+T)=f(x)$. I know that $sin{(x)}$ and $cos{(x)}$ are $2 pi$ periodic functions so I my attempt was to just plug in $f(x+2 pi)$ and see where it leads me.
$$implies f(x+2 pi)=frac{sin(sin{(x+2 pi)}+2 pi)}{2+cos{(lvert x+2 pirvert}+cos{(x+2 pi)})}=frac{sin{x}}{2+cos{(lvert x+2pirvert+cos{x})}}$$
If $x ge 0$ I get the desired solution and $f(x+2 pi)=f(x)$.
However, for negative $x$, $lvert x+2 pi rvert not=x+2 pi$. So I
cannot show periodicity for all negative $x$. How can I resolve this?
Can a function be periodic for positive $x$ and not for negative?
calculus periodic-functions monotone-functions
calculus periodic-functions monotone-functions
asked Nov 27 at 17:32
Nullspace
190110
asked Nov 27 at 17:32
Nullspace
190110
edited Nov 27 at 20:24
asked Nov 27 at 17:32
Nullspace
190110
asked Nov 27 at 17:32
Nullspace
190110
asked Nov 27 at 17:32
Nullspace
190110
190110
First of all, this is not Real Analysis. That tag is getting abused way too much. Secondly, the exact same question was asked some days ago. I updated the tags to fit the post.
– Rebellos
Nov 27 at 17:45
2
@Rebellos The same questions was NOT asked a few days ago because a few days ago I asked if the infimum and supremum could be obtained analytically to which the answer in a comment was "no" so I decided to delete it. This question refers to the same function but I am asking about monotonicity and periodicity. I would appreciate it if you could change that part of the comment. Otherwise, people are just going to look at the comment and assume this is a duplicate. Sorry about the tags. I never know which ones to use.
– Nullspace
Nov 27 at 17:50
1
Note that $f(x) = -f(-x)$, so periodicity on the interval $[0, infty)$ implies periodicity on the interval $(-infty, 0]$. Of course, a function could be periodic for positive and negative $x$ separately but not periodic overall.
– Connor Harris
Nov 27 at 20:51
@ConnorHarris Thanks for your comment. Building on your answer, can we now say that the function is periodic even though it has a different periodicity on $[0, infty)$ and $(-infty.0]?$
– Nullspace
Nov 27 at 21:17
1
@Nullspace that's a matter of the precise definition of periodicity that you were handed (I assume this is for some sort of school assignment).
– Connor Harris
Nov 28 at 14:37
add a comment |
First of all, this is not Real Analysis. That tag is getting abused way too much. Secondly, the exact same question was asked some days ago. I updated the tags to fit the post.
– Rebellos
Nov 27 at 17:45
2
@Rebellos The same questions was NOT asked a few days ago because a few days ago I asked if the infimum and supremum could be obtained analytically to which the answer in a comment was "no" so I decided to delete it. This question refers to the same function but I am asking about monotonicity and periodicity. I would appreciate it if you could change that part of the comment. Otherwise, people are just going to look at the comment and assume this is a duplicate. Sorry about the tags. I never know which ones to use.
– Nullspace
Nov 27 at 17:50
1
Note that $f(x) = -f(-x)$, so periodicity on the interval $[0, infty)$ implies periodicity on the interval $(-infty, 0]$. Of course, a function could be periodic for positive and negative $x$ separately but not periodic overall.
– Connor Harris
Nov 27 at 20:51
@ConnorHarris Thanks for your comment. Building on your answer, can we now say that the function is periodic even though it has a different periodicity on $[0, infty)$ and $(-infty.0]?$
– Nullspace
Nov 27 at 21:17
1
@Nullspace that's a matter of the precise definition of periodicity that you were handed (I assume this is for some sort of school assignment).
– Connor Harris
Nov 28 at 14:37
First of all, this is not Real Analysis. That tag is getting abused way too much. Secondly, the exact same question was asked some days ago. I updated the tags to fit the post.
– Rebellos
Nov 27 at 17:45
First of all, this is not Real Analysis. That tag is getting abused way too much. Secondly, the exact same question was asked some days ago. I updated the tags to fit the post.
– Rebellos
Nov 27 at 17:45
2
2
@Rebellos The same questions was NOT asked a few days ago because a few days ago I asked if the infimum and supremum could be obtained analytically to which the answer in a comment was "no" so I decided to delete it. This question refers to the same function but I am asking about monotonicity and periodicity. I would appreciate it if you could change that part of the comment. Otherwise, people are just going to look at the comment and assume this is a duplicate. Sorry about the tags. I never know which ones to use.
– Nullspace
Nov 27 at 17:50
@Rebellos The same questions was NOT asked a few days ago because a few days ago I asked if the infimum and supremum could be obtained analytically to which the answer in a comment was "no" so I decided to delete it. This question refers to the same function but I am asking about monotonicity and periodicity. I would appreciate it if you could change that part of the comment. Otherwise, people are just going to look at the comment and assume this is a duplicate. Sorry about the tags. I never know which ones to use.
– Nullspace
Nov 27 at 17:50
1
1
Note that $f(x) = -f(-x)$, so periodicity on the interval $[0, infty)$ implies periodicity on the interval $(-infty, 0]$. Of course, a function could be periodic for positive and negative $x$ separately but not periodic overall.
– Connor Harris
Nov 27 at 20:51
Note that $f(x) = -f(-x)$, so periodicity on the interval $[0, infty)$ implies periodicity on the interval $(-infty, 0]$. Of course, a function could be periodic for positive and negative $x$ separately but not periodic overall.
– Connor Harris
Nov 27 at 20:51
@ConnorHarris Thanks for your comment. Building on your answer, can we now say that the function is periodic even though it has a different periodicity on $[0, infty)$ and $(-infty.0]?$
– Nullspace
Nov 27 at 21:17
@ConnorHarris Thanks for your comment. Building on your answer, can we now say that the function is periodic even though it has a different periodicity on $[0, infty)$ and $(-infty.0]?$
– Nullspace
Nov 27 at 21:17
1
1
@Nullspace that's a matter of the precise definition of periodicity that you were handed (I assume this is for some sort of school assignment).
– Connor Harris
Nov 28 at 14:37
@Nullspace that's a matter of the precise definition of periodicity that you were handed (I assume this is for some sort of school assignment).
– Connor Harris
Nov 28 at 14:37
add a comment |
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First of all, this is not Real Analysis. That tag is getting abused way too much. Secondly, the exact same question was asked some days ago. I updated the tags to fit the post.
– Rebellos
Nov 27 at 17:45
2
@Rebellos The same questions was NOT asked a few days ago because a few days ago I asked if the infimum and supremum could be obtained analytically to which the answer in a comment was "no" so I decided to delete it. This question refers to the same function but I am asking about monotonicity and periodicity. I would appreciate it if you could change that part of the comment. Otherwise, people are just going to look at the comment and assume this is a duplicate. Sorry about the tags. I never know which ones to use.
– Nullspace
Nov 27 at 17:50
1
Note that $f(x) = -f(-x)$, so periodicity on the interval $[0, infty)$ implies periodicity on the interval $(-infty, 0]$. Of course, a function could be periodic for positive and negative $x$ separately but not periodic overall.
– Connor Harris
Nov 27 at 20:51
@ConnorHarris Thanks for your comment. Building on your answer, can we now say that the function is periodic even though it has a different periodicity on $[0, infty)$ and $(-infty.0]?$
– Nullspace
Nov 27 at 21:17
1
@Nullspace that's a matter of the precise definition of periodicity that you were handed (I assume this is for some sort of school assignment).
– Connor Harris
Nov 28 at 14:37