Hi Manuj,

The period of f(x) will be given by the LCM of T1 and T2 -------- (ie if the LCM exists).

In case T1 and T2 do not have an LCM, then f(x) will not be periodic.

Hope this helps.

Wish you all the best.

Regards,

Ashwin (IIT Madras).

Sir,

Im not getting it . . . Is the condition always true ? ? ?

Though it seems valid for most cases but not for all ...

How can be the period , a common multiple T₁  and T₂

Contradiction:  tan(x) = sin(x)/cos(x)

Both sin(x) and cos(x) have periods 2pi so tan (x) must have a perion 2pi(a common multiple) but on the other hand the period of tan is pi.

Hi Swapnil,

You can just address by the name yaar !!!. (No need to use any Sir).

What you have stated is very true.

This is valid only (and in almost all cases if T1 and T2 are different) when there exists no smaller number T (>0) less than the LCM of T1 and T2, which satisfies f(x+T) = f(x).

If there exists a smaller T, then T will be the period, as in your example of tanx.

Regards,

Ashwin (IIT MadraS).

Dear,

One more classical example is the function

f(x) = |sinx| + |cosx|.

Individually |sinx| and |cosx| has period pi.

But the addition will have period pi/2 ------ (and not pi, which is the LCM).

As we can find a smaller number pi/2, which leads to the interchanging of the functions, we have pi/2 as the period of f(x) in this case (Another exception).

So one needs to be careful while using the LCM concept.

Hope it helps.

Best Regards,

Ashwin (IIT MadraS).

Hi,

In that case , how is the period of 2sin(x)/sec(x)  is pi not 2pi (least common multiple).(As sec does not satisfies sec(x+T)=sec(x) fr pi/2 not pi )

Swapnil,

I dont understand your Question here exactly.

Yes the period of f(x) = sin2x = 2sinx/secx is pi.

If you go by LCM Concept, ther period is 2pi.

But pi is smallest number for which f(x+pi) = f(x).

Thats You and I have stated accordingly.

So now I think we agree on the common thing.

 Best Regards,

Ashwin (IIT Madras).

Im extremely sorry for my typing mistake.

Actually the question is sin(x)/sec(2x) not 2sin(x)/sec(x). Here f(x+t)=f(X) for sec(2x) is satisfied for pi/2 (sec2(x+pi/2))=sec(x). but the answer is not p/2 The answer is 2pi .

Dude,

It is like this.

Say you have found period using LCM of T1, T2.... say it is T.

Now say we have g(x) = f(x) + h(x) -------- (T1, T2 are periods of f and h respectively).

And in case suppose you used 1/2*LCM, which is T/2.

And say supose f(x+T/2) = h(x)

and h(x+T/2) = f(x).

In this case, T/2 will be the period of g, since f and h complement each other by an interval of T/2.

Next in case g(x) = f(x)/h(x), or g(x) = f(x)*h(x).

And say f(x+T/2) = -f(x)

and h(x+T/2) = -h(x).

Even in this case T/2 will be the period of g(x), as f and h are complementary even (though they are not actually even)

Eg, would be g(x) = sinx*cosx ------ each has period 2pi. But for 1/2(2pi) = pi, they are complementary even.

Hence pi is the period of sinx*cosx.

Hope this will makes it clear.

With Best Regards,

Ashwin (IIT Madras).

Thanks  ASHWIN SIR.It was vry helpful sir 

@ manuj mittal :Please mention his name as Ashwin Sir,He is an iitian......