Finding the Period of a Trig Function without Graphing

[sutupidmath]

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]

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.

 

[phlegmy]

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 derivative], 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 isn't 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

 

[HallsofIvy]

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).

 

[kamerling]

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]

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

 

[sutupidmath]

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!

 

[kamerling]

??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]

??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.

 

[sutupidmath]

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?

 

[Eidos]

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.

 

[sutupidmath]

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, let's 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]<br /> =sin(ax+2r_1\pi+b)+cos(cx+2r_2\pi+d)=sin[(ax+b)+2r_1\pi]+cos[(cx+d)+2r_2\pi]<br /> <br /> Now since 2r_1\pi, 2r_2\pi will always be even, it means that the above expretion is equal to <br /> <br /> [texf(x)=sin(ax+b)+cos(cx+d) 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??

 

[Eidos]

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).