Is $sqrt{sin x}$ periodic?
$begingroup$
$sin^2(x)$ has period $pi$ but it seems to me $sqrt{sin x}$ is not periodic since inside square root has to be positive and when it is negative, it is not defined.
Does it creates problem for periodicity? Can we say square root of the periodic function need to be periodic? Thanks for your help.
This is graph of the $sqrt{sin x}$ above.
calculus functions radicals periodic-functions
$endgroup$
add a comment |
$begingroup$
$sin^2(x)$ has period $pi$ but it seems to me $sqrt{sin x}$ is not periodic since inside square root has to be positive and when it is negative, it is not defined.
Does it creates problem for periodicity? Can we say square root of the periodic function need to be periodic? Thanks for your help.
This is graph of the $sqrt{sin x}$ above.
calculus functions radicals periodic-functions
$endgroup$
7
$begingroup$
Well, it's periodic where it is defined. That's something!
$endgroup$
– lulu
Mar 30 at 13:21
$begingroup$
You could ask the same question of $tan(x)$, because it's not defined at odd multiples of $pi/2$.
$endgroup$
– Barry Cipra
Mar 31 at 13:43
add a comment |
$begingroup$
$sin^2(x)$ has period $pi$ but it seems to me $sqrt{sin x}$ is not periodic since inside square root has to be positive and when it is negative, it is not defined.
Does it creates problem for periodicity? Can we say square root of the periodic function need to be periodic? Thanks for your help.
This is graph of the $sqrt{sin x}$ above.
calculus functions radicals periodic-functions
$endgroup$
$sin^2(x)$ has period $pi$ but it seems to me $sqrt{sin x}$ is not periodic since inside square root has to be positive and when it is negative, it is not defined.
Does it creates problem for periodicity? Can we say square root of the periodic function need to be periodic? Thanks for your help.
This is graph of the $sqrt{sin x}$ above.
calculus functions radicals periodic-functions
calculus functions radicals periodic-functions
edited Mar 30 at 17:18
user21820
40.1k544162
40.1k544162
asked Mar 30 at 13:20
izaagizaag
421210
421210
7
$begingroup$
Well, it's periodic where it is defined. That's something!
$endgroup$
– lulu
Mar 30 at 13:21
$begingroup$
You could ask the same question of $tan(x)$, because it's not defined at odd multiples of $pi/2$.
$endgroup$
– Barry Cipra
Mar 31 at 13:43
add a comment |
7
$begingroup$
Well, it's periodic where it is defined. That's something!
$endgroup$
– lulu
Mar 30 at 13:21
$begingroup$
You could ask the same question of $tan(x)$, because it's not defined at odd multiples of $pi/2$.
$endgroup$
– Barry Cipra
Mar 31 at 13:43
7
7
$begingroup$
Well, it's periodic where it is defined. That's something!
$endgroup$
– lulu
Mar 30 at 13:21
$begingroup$
Well, it's periodic where it is defined. That's something!
$endgroup$
– lulu
Mar 30 at 13:21
$begingroup$
You could ask the same question of $tan(x)$, because it's not defined at odd multiples of $pi/2$.
$endgroup$
– Barry Cipra
Mar 31 at 13:43
$begingroup$
You could ask the same question of $tan(x)$, because it's not defined at odd multiples of $pi/2$.
$endgroup$
– Barry Cipra
Mar 31 at 13:43
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
The function $sqrt{sin x}$ is periodic. If you want (but don't do it in public) think of "undefined" as a real number. Then if $sin x$ is negative, you have $$sqrt{sin x} = mbox{ undefined } =sqrt{sin(x+2pi)}.$$
Or you can extend to the complex numbers and you'll have periodicity everywhere.
$endgroup$
$begingroup$
Now we just need to choose how to define $sqrt{z}$ in general, but that's doable.
$endgroup$
– J.G.
Mar 31 at 13:21
add a comment |
$begingroup$
$f$ is periodic, if there is a positive real number $p $ (the period) such that for every $x$ from the domain of $f$ the value of $x$ is the same as value of $(x+p)$ - more formal and more exactly:
$$f text{ is periodic }iffexists p>0: forall x in D(f): x+p in D(f) land f(x) = f(x+p)$$
For the function $g = sqrt f$ of a periodic function $f$ obviously $$x in D(g) iff x + k in D(g)$$
(because $x in D(g) iff f(x) ge 0 iff f(x+k)=f(x) ge 0 $)
and
$$f(x) = f(x+p),$$
so the square root of a periodic function is a periodic one, too.
$endgroup$
add a comment |
$begingroup$
If a function $f$ is periodic, then there exists $kne0$ such that $f(x)=f(x+k)$ for all $xinmathbb{R}$. So if $g$ is its square root, we have
$$g(x)=sqrt{f(x)}=sqrt{f(x+k)}=g(x+k).$$
And, hence, $g$ is also periodic. BUT note that the square root is only defined in the intervals in which $f$ is nonnegative. Out of them, $g$ is not periodic because it is not even defined.
$endgroup$
add a comment |
$begingroup$
The periodicity holds if we can show $f(x+T) = f(x)$. We can see if the periodicity holds if we do
$$
begin{align}
f(x+T) &= sqrt{sin(x+T)} \
&= sqrt{sin(x)cos(T)+cos(t)sin(T)} tag{1}
end{align}
$$
We need to search for the smallest value of $T$ in Eq.(1) that leads to $f(x)$. Indeed, if $T=2pi$ sec, then Eq.(1) becomes
$$
begin{align}
f(x+T)
&= sqrt{sin(x)cos(T)+cos(t)sin(T)} \
&= sqrt{sin(x)(1)+cos(t)(0)} \
&= sqrt{sin(x)} \
&= f(x) tag{2}
end{align}
$$
Eq.(2) proves the periodicity of the function with a fundamental period $T_o=2pi$ sec. We can plot the function and check if it indeed repeats itself after $T=2pi=6.28$ sec.
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The function $sqrt{sin x}$ is periodic. If you want (but don't do it in public) think of "undefined" as a real number. Then if $sin x$ is negative, you have $$sqrt{sin x} = mbox{ undefined } =sqrt{sin(x+2pi)}.$$
Or you can extend to the complex numbers and you'll have periodicity everywhere.
$endgroup$
$begingroup$
Now we just need to choose how to define $sqrt{z}$ in general, but that's doable.
$endgroup$
– J.G.
Mar 31 at 13:21
add a comment |
$begingroup$
The function $sqrt{sin x}$ is periodic. If you want (but don't do it in public) think of "undefined" as a real number. Then if $sin x$ is negative, you have $$sqrt{sin x} = mbox{ undefined } =sqrt{sin(x+2pi)}.$$
Or you can extend to the complex numbers and you'll have periodicity everywhere.
$endgroup$
$begingroup$
Now we just need to choose how to define $sqrt{z}$ in general, but that's doable.
$endgroup$
– J.G.
Mar 31 at 13:21
add a comment |
$begingroup$
The function $sqrt{sin x}$ is periodic. If you want (but don't do it in public) think of "undefined" as a real number. Then if $sin x$ is negative, you have $$sqrt{sin x} = mbox{ undefined } =sqrt{sin(x+2pi)}.$$
Or you can extend to the complex numbers and you'll have periodicity everywhere.
$endgroup$
The function $sqrt{sin x}$ is periodic. If you want (but don't do it in public) think of "undefined" as a real number. Then if $sin x$ is negative, you have $$sqrt{sin x} = mbox{ undefined } =sqrt{sin(x+2pi)}.$$
Or you can extend to the complex numbers and you'll have periodicity everywhere.
answered Mar 30 at 13:46
B. GoddardB. Goddard
20.2k21543
20.2k21543
$begingroup$
Now we just need to choose how to define $sqrt{z}$ in general, but that's doable.
$endgroup$
– J.G.
Mar 31 at 13:21
add a comment |
$begingroup$
Now we just need to choose how to define $sqrt{z}$ in general, but that's doable.
$endgroup$
– J.G.
Mar 31 at 13:21
$begingroup$
Now we just need to choose how to define $sqrt{z}$ in general, but that's doable.
$endgroup$
– J.G.
Mar 31 at 13:21
$begingroup$
Now we just need to choose how to define $sqrt{z}$ in general, but that's doable.
$endgroup$
– J.G.
Mar 31 at 13:21
add a comment |
$begingroup$
$f$ is periodic, if there is a positive real number $p $ (the period) such that for every $x$ from the domain of $f$ the value of $x$ is the same as value of $(x+p)$ - more formal and more exactly:
$$f text{ is periodic }iffexists p>0: forall x in D(f): x+p in D(f) land f(x) = f(x+p)$$
For the function $g = sqrt f$ of a periodic function $f$ obviously $$x in D(g) iff x + k in D(g)$$
(because $x in D(g) iff f(x) ge 0 iff f(x+k)=f(x) ge 0 $)
and
$$f(x) = f(x+p),$$
so the square root of a periodic function is a periodic one, too.
$endgroup$
add a comment |
$begingroup$
$f$ is periodic, if there is a positive real number $p $ (the period) such that for every $x$ from the domain of $f$ the value of $x$ is the same as value of $(x+p)$ - more formal and more exactly:
$$f text{ is periodic }iffexists p>0: forall x in D(f): x+p in D(f) land f(x) = f(x+p)$$
For the function $g = sqrt f$ of a periodic function $f$ obviously $$x in D(g) iff x + k in D(g)$$
(because $x in D(g) iff f(x) ge 0 iff f(x+k)=f(x) ge 0 $)
and
$$f(x) = f(x+p),$$
so the square root of a periodic function is a periodic one, too.
$endgroup$
add a comment |
$begingroup$
$f$ is periodic, if there is a positive real number $p $ (the period) such that for every $x$ from the domain of $f$ the value of $x$ is the same as value of $(x+p)$ - more formal and more exactly:
$$f text{ is periodic }iffexists p>0: forall x in D(f): x+p in D(f) land f(x) = f(x+p)$$
For the function $g = sqrt f$ of a periodic function $f$ obviously $$x in D(g) iff x + k in D(g)$$
(because $x in D(g) iff f(x) ge 0 iff f(x+k)=f(x) ge 0 $)
and
$$f(x) = f(x+p),$$
so the square root of a periodic function is a periodic one, too.
$endgroup$
$f$ is periodic, if there is a positive real number $p $ (the period) such that for every $x$ from the domain of $f$ the value of $x$ is the same as value of $(x+p)$ - more formal and more exactly:
$$f text{ is periodic }iffexists p>0: forall x in D(f): x+p in D(f) land f(x) = f(x+p)$$
For the function $g = sqrt f$ of a periodic function $f$ obviously $$x in D(g) iff x + k in D(g)$$
(because $x in D(g) iff f(x) ge 0 iff f(x+k)=f(x) ge 0 $)
and
$$f(x) = f(x+p),$$
so the square root of a periodic function is a periodic one, too.
edited Mar 30 at 16:55
answered Mar 30 at 13:54
MarianDMarianD
2,2611618
2,2611618
add a comment |
add a comment |
$begingroup$
If a function $f$ is periodic, then there exists $kne0$ such that $f(x)=f(x+k)$ for all $xinmathbb{R}$. So if $g$ is its square root, we have
$$g(x)=sqrt{f(x)}=sqrt{f(x+k)}=g(x+k).$$
And, hence, $g$ is also periodic. BUT note that the square root is only defined in the intervals in which $f$ is nonnegative. Out of them, $g$ is not periodic because it is not even defined.
$endgroup$
add a comment |
$begingroup$
If a function $f$ is periodic, then there exists $kne0$ such that $f(x)=f(x+k)$ for all $xinmathbb{R}$. So if $g$ is its square root, we have
$$g(x)=sqrt{f(x)}=sqrt{f(x+k)}=g(x+k).$$
And, hence, $g$ is also periodic. BUT note that the square root is only defined in the intervals in which $f$ is nonnegative. Out of them, $g$ is not periodic because it is not even defined.
$endgroup$
add a comment |
$begingroup$
If a function $f$ is periodic, then there exists $kne0$ such that $f(x)=f(x+k)$ for all $xinmathbb{R}$. So if $g$ is its square root, we have
$$g(x)=sqrt{f(x)}=sqrt{f(x+k)}=g(x+k).$$
And, hence, $g$ is also periodic. BUT note that the square root is only defined in the intervals in which $f$ is nonnegative. Out of them, $g$ is not periodic because it is not even defined.
$endgroup$
If a function $f$ is periodic, then there exists $kne0$ such that $f(x)=f(x+k)$ for all $xinmathbb{R}$. So if $g$ is its square root, we have
$$g(x)=sqrt{f(x)}=sqrt{f(x+k)}=g(x+k).$$
And, hence, $g$ is also periodic. BUT note that the square root is only defined in the intervals in which $f$ is nonnegative. Out of them, $g$ is not periodic because it is not even defined.
edited Mar 31 at 13:09
answered Mar 30 at 13:54
AugSBAugSB
3,46421734
3,46421734
add a comment |
add a comment |
$begingroup$
The periodicity holds if we can show $f(x+T) = f(x)$. We can see if the periodicity holds if we do
$$
begin{align}
f(x+T) &= sqrt{sin(x+T)} \
&= sqrt{sin(x)cos(T)+cos(t)sin(T)} tag{1}
end{align}
$$
We need to search for the smallest value of $T$ in Eq.(1) that leads to $f(x)$. Indeed, if $T=2pi$ sec, then Eq.(1) becomes
$$
begin{align}
f(x+T)
&= sqrt{sin(x)cos(T)+cos(t)sin(T)} \
&= sqrt{sin(x)(1)+cos(t)(0)} \
&= sqrt{sin(x)} \
&= f(x) tag{2}
end{align}
$$
Eq.(2) proves the periodicity of the function with a fundamental period $T_o=2pi$ sec. We can plot the function and check if it indeed repeats itself after $T=2pi=6.28$ sec.
$endgroup$
add a comment |
$begingroup$
The periodicity holds if we can show $f(x+T) = f(x)$. We can see if the periodicity holds if we do
$$
begin{align}
f(x+T) &= sqrt{sin(x+T)} \
&= sqrt{sin(x)cos(T)+cos(t)sin(T)} tag{1}
end{align}
$$
We need to search for the smallest value of $T$ in Eq.(1) that leads to $f(x)$. Indeed, if $T=2pi$ sec, then Eq.(1) becomes
$$
begin{align}
f(x+T)
&= sqrt{sin(x)cos(T)+cos(t)sin(T)} \
&= sqrt{sin(x)(1)+cos(t)(0)} \
&= sqrt{sin(x)} \
&= f(x) tag{2}
end{align}
$$
Eq.(2) proves the periodicity of the function with a fundamental period $T_o=2pi$ sec. We can plot the function and check if it indeed repeats itself after $T=2pi=6.28$ sec.
$endgroup$
add a comment |
$begingroup$
The periodicity holds if we can show $f(x+T) = f(x)$. We can see if the periodicity holds if we do
$$
begin{align}
f(x+T) &= sqrt{sin(x+T)} \
&= sqrt{sin(x)cos(T)+cos(t)sin(T)} tag{1}
end{align}
$$
We need to search for the smallest value of $T$ in Eq.(1) that leads to $f(x)$. Indeed, if $T=2pi$ sec, then Eq.(1) becomes
$$
begin{align}
f(x+T)
&= sqrt{sin(x)cos(T)+cos(t)sin(T)} \
&= sqrt{sin(x)(1)+cos(t)(0)} \
&= sqrt{sin(x)} \
&= f(x) tag{2}
end{align}
$$
Eq.(2) proves the periodicity of the function with a fundamental period $T_o=2pi$ sec. We can plot the function and check if it indeed repeats itself after $T=2pi=6.28$ sec.
$endgroup$
The periodicity holds if we can show $f(x+T) = f(x)$. We can see if the periodicity holds if we do
$$
begin{align}
f(x+T) &= sqrt{sin(x+T)} \
&= sqrt{sin(x)cos(T)+cos(t)sin(T)} tag{1}
end{align}
$$
We need to search for the smallest value of $T$ in Eq.(1) that leads to $f(x)$. Indeed, if $T=2pi$ sec, then Eq.(1) becomes
$$
begin{align}
f(x+T)
&= sqrt{sin(x)cos(T)+cos(t)sin(T)} \
&= sqrt{sin(x)(1)+cos(t)(0)} \
&= sqrt{sin(x)} \
&= f(x) tag{2}
end{align}
$$
Eq.(2) proves the periodicity of the function with a fundamental period $T_o=2pi$ sec. We can plot the function and check if it indeed repeats itself after $T=2pi=6.28$ sec.
answered Apr 4 at 5:15
CroCoCroCo
283221
283221
add a comment |
add a comment |
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7
$begingroup$
Well, it's periodic where it is defined. That's something!
$endgroup$
– lulu
Mar 30 at 13:21
$begingroup$
You could ask the same question of $tan(x)$, because it's not defined at odd multiples of $pi/2$.
$endgroup$
– Barry Cipra
Mar 31 at 13:43