Trig Functions Periodicity: Which Function is Not Periodic?

[PeterO]

Lets get back to the question, the (a) option is |sin(3x)|+sin²x.
The period of |sin(3x)| is pi/3 and the period of sin²x is pi.
Is this function periodic?

Yes it is. The Period is pi. in the 0 ti pi range, the first function goes through 3 cycles, while the second goes through 1.

If a function consists of the sum or difference of two periodic functions, it will itself be periodic.

 

[Saitama]

Yes it is. The Period is pi. in the 0 ti pi range, the first function goes through 3 cycles, while the second goes through 1.

If a function consists of the sum or difference of two periodic functions, it will itself be periodic.

Are there some tricks to remember whether a given function is periodic or not?
Like you said "If a function consists of the sum or difference of two periodic functions, it will itself be periodic." If it was like this, one of the function is periodic and other is not, then what would be the result?

 

[PeterO]

Are there some tricks to remember whether a given function is periodic or not?
Like you said "If a function consists of the sum or difference of two periodic functions, it will itself be periodic." If it was like this, one of the function is periodic and other is not, then what would be the result?

As soon as one is not periodic, the whole thing isn't.

The easiest ones to recognise as non-periodic are those where the argument is an indice.

In sin (x) , sin is the function, x is the argument

sin (x^2) is not periodic [remember x^2 is an indice]

What ever you do, don't confuse (sin x ) ^2 and sin (x^2)

If you used a calculator to evaluate those two , then for the first you would take the sign of the angle [getting an answer between -1 and 1] then square the answer. For the second you would first square x, then plug that answer into the sine function.

 

[Saitama]

As soon as one is not periodic, the whole thing isn't.

The easiest ones to recognise as non-periodic are those where the argument is an indice.

In sin (x) , sin is the function, x is the argument

sin (x^2) is not periodic [remember x^2 is an indice]

What ever you do, don't confuse (sin x ) ^2 and sin (x^2)

If you used a calculator to evaluate those two , then for the first you would take the sign of the angle [getting an answer between -1 and 1] then square the answer. For the second you would first square x, then plug that answer into the sine function.

What is an indice?

 

[PeterO]

What is an indice?

There is a whole chapter on them, including the laws to use with indices, in every maths book.
The laws are sometimes called index laws.
An indice has a base and an index.
Common indices can have x as the base and a number as the index, often written here as x^2 or x^3

 

[Saitama]

May i know what's the period of tan²x because i am getting stuck in the (b) and (c) option.

 

[PeterO]

May i know what's the period of tan²x because i am getting stuck in the (b) and (c) option.

Probably pi.

 

[SammyS]

What is an indice?

The plural of index is indices, so by "indice", Peter O probably means index.

However, I think he really is talking about an exponent, that is to say a power of x, not a power of the trig function.

 

[Saitama]

Probably pi.

How? I Like Serena gave me an equation to solve it and deduce the period of sin²x in https://www.physicsforums.com/showpost.php?p=3448043&postcount=16". How would i solve that equation?

The plural of index is indices, so by "indice", Peter O probably means index.

However, I think he really is talking about an exponent, that is to say a power of x, not a power of the trig function.

Thanks for solving the confusion,

 

[I like Serena]

How? I Like Serena gave me an equation to solve it and deduce the period of sin²x in https://www.physicsforums.com/showpost.php?p=3448043&postcount=16". How would i solve that equation?

Try tan²x = sin²x / cos²x and apply the cos 2x formulas.

 

[Saitama]

Try tan²x = sin²x / cos²x and apply the cos 2x formulas.

Applying cos(2x) formulas, i get tan^2x=\frac{1-cos(2x)}{cos(2x)+1}.
What next?

 

[I like Serena]

Applying cos(2x) formulas, i get tan^2x=\frac{1-cos(2x)}{cos(2x)+1}.
What next?

What is the period of cos(2x)?

 

[Saitama]

What is the period of cos(2x)?

Is it pi?

 

[I like Serena]

Is it pi?

Perhaps. Why would it be pi?

 

[Saitama]

Perhaps. Why would it be pi?

Since cos(x)=cos(x+2pi)=cos(x+4pi)...
therefore cos(2x)=cos(2x+2pi)=cos(2x+4pi)...
or cos2(x)=cos 2(x+pi)=cos 2(x+2pi)...
So, the period is pi.

Btw, is this relation correct? If a function is given, cos n(x), where n is an integer then its period would be \frac{2\pi}{n}.

 

[I like Serena]

Since cos(x)=cos(x+2pi)=cos(x+4pi)...
therefore cos(2x)=cos(2x+2pi)=cos(2x+4pi)...
or cos2(x)=cos 2(x+pi)=cos 2(x+2pi)...
So, the period is pi.

Btw, is this relation correct? If a function is given, cos n(x), where n is an integer then its period would be \frac{2\pi}{n}.

Yep!

So the period of tan²x...?

 

[Saitama]

Yep!

So the period of tan²x...?

Is it pi? If it is so, then is it becuase the period of cos(2x) is pi?

 

[I like Serena]

Is it pi? If it is so, then is it becuase the period of cos(2x) is pi?

Yes.
Note that if you evaluate the function for x+pi instead of for x, you'll get the same result.

 

[I like Serena]

Btw, there is a catch.

Consider for instance |sin x| and |1 + sin x|.
What are their respective periods?

 

[Saitama]

Btw, there is a catch.

Consider for instance |sin x| and |1 + sin x|.
What are their respective periods?

|sin x| period is pi. First i thought that the period for |1+sin x| is also pi but when i checked it on wolfram alpha it is 2pi. Why is it so?

 

[I like Serena]

|sin x| period is pi. First i thought that the period for |1+sin x| is also pi but when i checked it on wolfram alpha it is 2pi. Why is it so?

What did you see on WolframAlpha?

 

[Saitama]

What did you see on WolframAlpha?

This is the link:-http://www.wolframalpha.com/input/?i=|1+sin+x|"

[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP153119ggif1081a6cdhb00005f313i3bi0hec616?MSPStoreType=image/gif&s=40&w=185&h=18

 

[I like Serena]

This is the link:-http://www.wolframalpha.com/input/?i=|1+sin+x|"

[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP153119ggif1081a6cdhb00005f313i3bi0hec616?MSPStoreType=image/gif&s=40&w=185&h=18[/QUOTE]

All right, here's my counter:
http://www.wolframalpha.com/input/?i=|sin+x|,+|1+sin+x|"

Can you interpret what you see in the graph?

 

[Saitama]

All right, here's my counter:
http://www.wolframalpha.com/input/?i=|sin+x|,+|1+sin+x|"

Can you interpret what you see in the graph?

I can interpret that |sin(x)| is periodic with pi and |1+sin(x)| is periodic with 2pi. Also both the graphs intersect at two points between (0,2pi).

 

[I like Serena]

I can interpret that |sin(x)| is periodic with pi and |1+sin(x)| is periodic with 2pi. Also both the graphs intersect at two points between (0,2pi).

Yes but why?

Actually, since sin x has a period of 2pi, one might expect that |sin x| also has a period of 2pi.
How come it has a shorter period (other than that Wolfram says so )?

 

[Saitama]

Yes but why?

Actually, since sin x has a period of 2pi, one might expect that |sin x| also has a period of 2pi.
How come it has a shorter period (other than that Wolfram says so )?

|sin x| period is pi since sin(x) is negative in third and fourth quadrant. Applying the modulus to sin(x), the negative values becomes positive and therefore the period become pi and it results in a graph like this:-
[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP46719gh00f8c6ai9f2100003i3ic66d1763h0b2?MSPStoreType=image/gif&s=35&w=299&h=142&cdf=Coordinates&cdf=Tooltips

 

[I like Serena]

|sin x| period is pi since sin(x) is negative in third and fourth quadrant. Applying the modulus to sin(x), the negative values becomes positive and therefore the period become pi

Yep!

More generally, we can say that if you know that a function repeats itself after for instance 2pi (like |sin x|), that it is still possible that the actual (shortest) period of the function is shorter, but it will have to be a divider of the period you found.

 

[Saitama]

... that it is still possible that the actual (shortest) period of the function is shorter, but it will have to be a divider of the period you found.

Sorry i didn't get you.

Btw, i got why the period of |1+sin(x)| is 2pi. The graph of sin(x) is[PLAIN]http://www3.wolframalpha.com/Calculate/MSP/MSP133419ggii6778e7h11g0000594571daf619f270?MSPStoreType=image/gif&s=2&w=299&h=131&cdf=Coordinates&cdf=Tooltips

If we add one to sin(x), that means we are adding one to all the outputs of sin(x) which makes the graph to flow over zero. Now that means if we apply the modulus function, it doesn't affect the graph since all the values of 1+sin(x) are positive. Therefore the period of |1+sin(x)| is 2pi.

 

[I like Serena]

Btw, i got why the period of |1+sin(x)| is 2pi.

If we add one to sin(x), that means we are adding one to all the outputs of sin(x) which makes the graph to flow over zero. Now that means if we apply the modulus function, it doesn't affect the graph since all the values of 1+sin(x) are positive. Therefore the period of |1+sin(x)| is 2pi.

Yes. I think you're getting the hang of it how to interpret graphs!

Sorry i didn't get you.

I just meant, that as far as you can tell without looking at the graph, at first you would assume that |sin x| has period 2pi.
When you look at the graph, or if you otherwise think about it some more, you'd see that in this case the actual (shortest) period is pi, which is half of 2pi.

This is also why it is so important to look at the graph and interpret it.

 

[Saitama]

I just meant, that as far as you can tell without looking at the graph, at first you would assume that |sin x| has period 2pi.
When you look at the graph, or if you otherwise think about it some more, you'd see that in this case the actual (shortest) period is pi, which is half of 2pi.

This is also why it is so important to look at the graph and interpret it.

Yeah, at first i thought that the period of |sin(x)| is 2pi but my teacher corrected me that i didn't take care of modulus function which is the absolute value function.

(btw, in a few days (maybe 27th august), a test is going to be conducted in my classes, would you be willing to tell me some short tricks for solving trigonometry questions and other topics. (i will ask other topics to my teacher which are included in the syllabus tomorrow) )[/size]