Period of trig functions

1,631
4

Main Question or Discussion Point

Period of trig functions!!

This is a rather easy-silly quiestion but i just don't know how to show it.

I know how to find the period of trig functions of the form f(x)=Asin(wx+c) etc. i mean i know how to show that the period of this is [tex]\frac{2\pi}{w}[/tex].

However, how would one find the period of let's say the following function, is there any analitycal method of showing it:

[tex]f(x)=asin(wx+c)+kcos(f \pi x+d)-rtan(zx)[/tex]

Is there any way to prove it analiytically?

letters in front of x are constants.
 

Answers and Replies

tiny-tim
Science Advisor
Homework Helper
25,789
249
Hi sutupidmath! :smile:

There almost certainly isn't a period.

The three individual parts have their own periods. Unless those periods have rational-number ratios, there won't be a common period!

For example, the orbit of the moon is a combination of three periodic functions (one is called synodic … I forget the names of the other two), and they nearly coincide every 18 years, but not quite, and so eclipses aren't regular, but come in groups which only last a few hundred years. :cry:
 
119
0
THIS is probably a more difficult question than i'm cabable of answering but here are my thoughts for what its worth

i guess if you got the roots of the equation that would give you a starting point
i could call the first root the start of the first period

then i'm guessing at the start of each subsequent period the values of
d^nf(x)/dx^n [thats the nth derivitive], should be the same

for sin(x) for eg the roots all differ by pi
then the first deriv is cos(x)
evaluated at the roots gives you 1,-1, 1,-1, 1,-1
so obviously every second root has the same value so the period is 2*pi

i'm pretty darn sure this method is 99% unreliable
and god knows how many derivites you'd have to go to
but as tiny tim says there almost certainly isnt a period
[i actually think there almost certainly should be if all terms are trig?]

anyway i doubt this is of any use apart from food for thought
 
Last edited:
HallsofIvy
Science Advisor
Homework Helper
41,738
897
I thought it was well known that the period a sum of periodic functions was the least common multiple of their periods. If f has period p1 and g has period p2, P= np1, P= mp2, then (f+g)(P)= f(np1)+ g(np2)= f(0)+ g(0)= (f+g)(0).
 
454
0
I thought it was well known that the period a sum of periodic functions was the least common multiple of their periods. If f has period p1 and g has period p2, P= np1, P= mp2, then (f+g)(P)= f(np1)+ g(np2)= f(0)+ g(0)= (f+g)(0).
that works fine if there IS a common multiple.
 
HallsofIvy
Science Advisor
Homework Helper
41,738
897
??Any finite collection of numbers has a common multiple: multiply them together!
 
1,631
4
I thought it was well known that the period a sum of periodic functions was the least common multiple of their periods. If f has period p1 and g has period p2, P= np1, P= mp2, then (f+g)(P)= f(np1)+ g(np2)= f(0)+ g(0)= (f+g)(0).
Halls, pardone my ignorance, but can you just elaborate a little bit more this. MOreover, my question should have read this way: How do we show that the least common multiple of a sum of periodic functions is their period?

P.S. I was so ashamed to post this question here, i thought it was too obvious, but i am glad it is not that obvious by the way!!!
 
454
0
??Any finite collection of numbers has a common multiple: multiply them together!
Ok, I should have said if there is least common multiple. There is no such thing in the real numbers, since everything is a multiple of everything else.
 
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
12,038
128
??Any finite collection of numbers has a common multiple: multiply them together!
We require that the LCM be an integer multiple of all the numbers involved. So we really want the Least Common INTEGER Multiple.
 
1,631
4
Ok let's suppose that the constants before "x" are all integers, then how do we show that the LCM is the period of the sum of trig functions?
 
108
1
I would think that you could evaluate it at a point, say x=0. Then evaluate it at x = common integer multiple (CIM) + 0 and show that the function has the same value in this case. You could then use induction to show that it is periodic with n*(CIM), using trig. identities and what not :smile:

So:
Find CIM.
Show f(x+CIM)=f(x) for some arbitrary x.
Assume, f(x+n*CIM)=f(x).
Show that f(x+(n+1)*CIM)=f(x+n*CIM)

That should work, please correct me if I'm wrong.
 
Last edited:
1,631
4
I would think that you could evaluate it at a point, say x=0. Then evaluate it at x = common integer multiple (CIM) + 0 and show that the function has the same value in this case. You could then use induction to show that it is periodic with n*(CIM), using trig. identities and what not :smile:

So:
Find CIM.
Show f(x+CIM)=f(x) for some arbitrary x.
Assume, f(x+n*CIM)=f(x).
Show that f(x+(n+1)*CIM)=f(x+n*CIM)

That should work, please correct me if I'm wrong.

I think this suggestion works, lets see how i went about it, correct me if i am missing sth.

Let :

[tex]f(x)=sin(ax+b)+cos(cx+d)[/tex] where a,c are integers.

Let:

[tex]f_1(x)=sin(ax+b),f_2(x)=cos(cx+d)[/tex]

Then from here i can show that [tex] T_1=\frac{2\pi}{a},T_2=\frac{2\pi}{c}[/tex]

Now, let the Leas Common Multiple of T_1,T_2 be :

[tex] n\pi[/tex] such that :

[tex]r_1 *\frac{2\pi}{a}=n\pi,and,r_2 * \frac{2\pi}{c}=n\pi=>2r_1\pi=an\pi,2r_2\pi=cn\pi[/tex]

Where r_1,r_2 are also integers.

Now, let's try to show that

[tex]f(x+n\pi)=f(x)...????[/tex]

[tex]f(x+n\pi)=sin(ax+an\pi+b)+cos(cx+cn\pi+d)=[\tex]
[tex]=sin(ax+2r_1\pi+b)+cos(cx+2r_2\pi+d)=sin[(ax+b)+2r_1\pi]+cos[(cx+d)+2r_2\pi][/tex]

Now since [tex] 2r_1\pi, 2r_2\pi[/tex] will always be even, it means that the above expretion is equal to

[texf(x)=sin(ax+b)+cos(cx+d)[/tex] What we needed to show.

Now i think i could extend it to n+1 also, by just supposing first that its period is npi, and then trying to show that also its period is (n+1)pi.

Would this be approximately correct??
 
Last edited:
108
1
Now, let's try to show that

[tex]f(x+n\pi)=f(x)...????[/tex]

[tex]f(x+n\pi)=\sin(ax+an\pi+b)+\cos(cx+cn\pi+d)=[/tex]

[tex]\sin(ax+2r_1\pi+b)+\cos(cx+2r_2\pi+d)=\sin[(ax+b)+2r_1\pi]+\cos[(cx+d)+2r_2\pi][/tex]

Now since [tex] 2r_1\pi, 2r_2\pi[/tex] will always be even, it means that the above expression is equal to

[tex]f(x)=sin(ax+b)+cos(cx+d)[/tex] What we needed to show.
In performing the above step you have shown that the inductive hypothesis
f(x)=f(x+n*LCM) is true, so you're done. :smile: {Assuming its correct if course, I could only skim through the working}
The normal approach is to assume this statement and show that f(x+(n+1)*LCM)=f(x+n*LCM).
 
Last edited:

Related Threads for: Period of trig functions

Replies
8
Views
13K
Replies
4
Views
21K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
5
Views
2K
Replies
3
Views
2K
  • Last Post
Replies
2
Views
433
  • Last Post
Replies
4
Views
884
  • Last Post
Replies
9
Views
6K
Top