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Is $sqrtsin x$ periodic?

7

$begingroup$

$sin^2(x)$ has period $pi$ but it seems to me $sqrtsin x$ 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 $sqrtsin x$ above.

calculus functions radicals periodic-functions

share|cite|improve this question

edited 5 hours ago

user21820

39.8k544158

asked 9 hours ago

izaagizaag

416210

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Well, it's periodic where it is defined. That's something!
  $endgroup$
  – lulu
  9 hours ago
  

  
  
  
  

add a comment | 

7

$begingroup$

$sin^2(x)$ has period $pi$ but it seems to me $sqrtsin x$ 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 $sqrtsin x$ above.

calculus functions radicals periodic-functions

share|cite|improve this question

edited 5 hours ago

user21820

39.8k544158

asked 9 hours ago

izaagizaag

416210

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Well, it's periodic where it is defined. That's something!
  $endgroup$
  – lulu
  9 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

$sin^2(x)$ has period $pi$ but it seems to me $sqrtsin x$ 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 $sqrtsin x$ above.

calculus functions radicals periodic-functions

share|cite|improve this question

edited 5 hours ago

user21820

39.8k544158

asked 9 hours ago

izaagizaag

416210

$endgroup$

share|cite|improve this question

edited 5 hours ago

user21820

39.8k544158

asked 9 hours ago

izaagizaag

416210

share|cite|improve this question

edited 5 hours ago

user21820

39.8k544158

asked 9 hours ago

izaagizaag

416210

edited 5 hours ago

user21820

39.8k544158

edited 5 hours ago

user21820

39.8k544158

asked 9 hours ago

izaagizaag

416210

asked 9 hours ago

izaagizaag

416210

3 Answers
3

active

oldest

votes

5

$begingroup$

The function $sqrtsin x$ 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 $$sqrtsin x = mbox undefined =sqrtsin(x+2pi).$$

Or you can extend to the complex numbers and you'll have periodicity everywhere.

share|cite|improve this answer

answered 9 hours ago

B. GoddardB. Goddard

19.9k21442

$endgroup$

add a comment | 

5

$begingroup$

$f$ 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.

share|cite|improve this answer

edited 6 hours ago

answered 9 hours ago

MarianDMarianD

1,8661617

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Matt, thank you! For this proof itisn't so important if $p>0$ or $p<0$, but definitely it must be nonzero, and the definition of the period as a positive number is in agreement with you. (I corrected my answer.)
  $endgroup$
  – MarianD
  5 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

If a function $f$ is periodic, then there exists $kinmathbbR$ such that $f(x)=f(x+k)$ for all $xinmathbbR$. So if $g$ is its square root, we have
$$g(x)=sqrtf(x)=sqrtf(x+k)=g(x+k).$$
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.

share|cite|improve this answer

answered 9 hours ago

AugSBAugSB

3,39921734

$endgroup$

add a comment | 

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5

$begingroup$

The function $sqrtsin x$ 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 $$sqrtsin x = mbox undefined =sqrtsin(x+2pi).$$

Or you can extend to the complex numbers and you'll have periodicity everywhere.

share|cite|improve this answer

answered 9 hours ago

B. GoddardB. Goddard

19.9k21442

$endgroup$

add a comment | 

5

$begingroup$

The function $sqrtsin x$ 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 $$sqrtsin x = mbox undefined =sqrtsin(x+2pi).$$

Or you can extend to the complex numbers and you'll have periodicity everywhere.

share|cite|improve this answer

answered 9 hours ago

B. GoddardB. Goddard

19.9k21442

$endgroup$

add a comment | 

$begingroup$

The function $sqrtsin x$ 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 $$sqrtsin x = mbox undefined =sqrtsin(x+2pi).$$

Or you can extend to the complex numbers and you'll have periodicity everywhere.

share|cite|improve this answer

answered 9 hours ago

B. GoddardB. Goddard

19.9k21442

$endgroup$

share|cite|improve this answer

answered 9 hours ago

B. GoddardB. Goddard

19.9k21442

answered 9 hours ago

B. GoddardB. Goddard

19.9k21442

answered 9 hours ago

B. GoddardB. Goddard

19.9k21442

5

$begingroup$

$f$ 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.

share|cite|improve this answer

edited 6 hours ago

answered 9 hours ago

MarianDMarianD

1,8661617

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Matt, thank you! For this proof itisn't so important if $p>0$ or $p<0$, but definitely it must be nonzero, and the definition of the period as a positive number is in agreement with you. (I corrected my answer.)
  $endgroup$
  – MarianD
  5 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

$f$ 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.

share|cite|improve this answer

edited 6 hours ago

answered 9 hours ago

MarianDMarianD

1,8661617

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Matt, thank you! For this proof itisn't so important if $p>0$ or $p<0$, but definitely it must be nonzero, and the definition of the period as a positive number is in agreement with you. (I corrected my answer.)
  $endgroup$
  – MarianD
  5 hours ago
  

  
  
  
  

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.

share|cite|improve this answer

edited 6 hours ago

answered 9 hours ago

MarianDMarianD

1,8661617

$endgroup$

share|cite|improve this answer

edited 6 hours ago

answered 9 hours ago

MarianDMarianD

1,8661617

answered 9 hours ago

MarianDMarianD

1,8661617

answered 9 hours ago

MarianDMarianD

1,8661617

0

$begingroup$

If a function $f$ is periodic, then there exists $kinmathbbR$ such that $f(x)=f(x+k)$ for all $xinmathbbR$. So if $g$ is its square root, we have
$$g(x)=sqrtf(x)=sqrtf(x+k)=g(x+k).$$
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.

share|cite|improve this answer

answered 9 hours ago

AugSBAugSB

3,39921734

$endgroup$

add a comment | 

0

$begingroup$

If a function $f$ is periodic, then there exists $kinmathbbR$ such that $f(x)=f(x+k)$ for all $xinmathbbR$. So if $g$ is its square root, we have
$$g(x)=sqrtf(x)=sqrtf(x+k)=g(x+k).$$
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.

share|cite|improve this answer

answered 9 hours ago

AugSBAugSB

3,39921734

$endgroup$

add a comment | 

$begingroup$

If a function $f$ is periodic, then there exists $kinmathbbR$ such that $f(x)=f(x+k)$ for all $xinmathbbR$. So if $g$ is its square root, we have
$$g(x)=sqrtf(x)=sqrtf(x+k)=g(x+k).$$
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.

share|cite|improve this answer

answered 9 hours ago

AugSBAugSB

3,39921734

$endgroup$

share|cite|improve this answer

answered 9 hours ago

AugSBAugSB

3,39921734

answered 9 hours ago

AugSBAugSB

3,39921734

answered 9 hours ago

AugSBAugSB

3,39921734