# Period of trig functions

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 $$\frac{2\pi}{w}$$.

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

$$f(x)=asin(wx+c)+kcos(f \pi x+d)-rtan(zx)$$

Is there any way to prove it analiytically?

letters in front of x are constants.

tiny-tim
Homework Helper
Hi sutupidmath! 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. 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

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HallsofIvy
Homework Helper
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).

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
Homework Helper
??Any finite collection of numbers has a common multiple: multiply them together!

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!!!

??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
Homework Helper
??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.

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?

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 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.

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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 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 :

$$f(x)=sin(ax+b)+cos(cx+d)$$ where a,c are integers.

Let:

$$f_1(x)=sin(ax+b),f_2(x)=cos(cx+d)$$

Then from here i can show that $$T_1=\frac{2\pi}{a},T_2=\frac{2\pi}{c}$$

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

$$n\pi$$ such that :

$$r_1 *\frac{2\pi}{a}=n\pi,and,r_2 * \frac{2\pi}{c}=n\pi=>2r_1\pi=an\pi,2r_2\pi=cn\pi$$

Where r_1,r_2 are also integers.

Now, let's try to show that

$$f(x+n\pi)=f(x)...????$$

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

Now since $$2r_1\pi, 2r_2\pi$$ 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??

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Now, let's try to show that

$$f(x+n\pi)=f(x)...????$$

$$f(x+n\pi)=\sin(ax+an\pi+b)+\cos(cx+cn\pi+d)=$$

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

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

$$f(x)=sin(ax+b)+cos(cx+d)$$ 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. {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).

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