Do you know its property ? -9

Algebra Level 4

Which of the following conditions make the function $f(x)=\sin(ax) +\cos(bx)$ periodic?

A. $a=\frac{3\pi}{2},b=\pi$

B. $a=\sqrt{3},b=5\sqrt{3}$

C. $a=3\sqrt{2},b=2\sqrt{3}$

D. $a,b \in\mathbb{R}$

E. $f(x)$ is not periodic if $a,b \notin\mathbb{Q}$

Note: $\mathbb{R}$ is the set of Real Numbers and $\mathbb{Q}$ is the set of Rational Numbers.

You can try more such problems of the set Do you know its property ?

A, B and E A, B, C and D A, B A, B and C

2 solutions

Sandeep Bhardwaj
Oct 19, 2014

Given $f(x)=sin(ax) + cos(bx)$

So Fundamental Period of f(x)= $L.C.M. (T_1,T_2)$ if $T_1$ is the L.C.M. of $sin(ax)$ and $T_2$ is the $L.C.M. of cos(bx)$ .

And if period of f(x) is T, then period of f(ax+b) will be $\large \frac{T}{|a|}$ .

$\boxed{A}$ . $L.C.M. \large (\frac{2\pi}{\frac{3\pi}{2}},\frac{2\pi}{\pi})=L.C.M.(\frac{4}{3},2)=4$ . So in this case f(x) is periodic with Fundamental Period $4$

$\boxed{B}$ . $L.C.M. \large (\frac{2\pi}{\sqrt{3}},\frac{2\pi}{5.\sqrt{3}})=\frac{2\pi}{\sqrt{3}}$ . So in this case $f(x)$ is periodic with Fundamental Period $\large \frac{2\pi}{\sqrt{3}}$

$\boxed{C}$ . $L.C.M. \large (\frac{2\pi}{3.\sqrt{2}},\frac{2\pi}{2.\sqrt{3}})= does \space not \space exist$

$\boxed{D}$ . Not Periodic for $\boxed{C}$ . So $\boxed{D}$ is incorrect.

$\boxed{E}$ . Periodic for $\boxed{B}$ which violates $\boxed{E}$ . So $\boxed{E}$ is not correct.

So only A,B are correct !

enjoy!

Do you know how to prove the fact that

If the fundamental period of $f$ is $\alpha$ and the fundamental period of $g$ is $\beta$ , then the fundamental period of $f + g$ is the LCM of $\alpha, \beta$ ?

Note that you have to be careful of the following:
1) If $\frac{ \alpha } { \beta }$ is rational, then their LCM exists. It is clearly a period, but we also need to show that this is the minimum) .

2) If $\frac{ \alpha } { \beta }$ is irrational, then their LCM does not exist, and we have to show that the function cannot ever be periodic (with any period). This is much tricker.

Note: At least one of the above statement is not true. Why?

I think this is an interesting discussion and have moved it to Fundamental Period Claim

Calvin Lin Staff - 6 years, 7 months ago

Yeah.. And one example is : Let $f(x)=|sinx|+|cosx|$ . The fundamental period of $f(x)$ is $\frac{\pi}{2}$ . But L.C.M. of fundamental periods of $|sinx|$ and $|cosx|$ is $\pi$ .

Sandeep Bhardwaj - 6 years, 7 months ago

$|\sin x|$ & $|\cos x|$ interchanged by adding $\frac{π}{2}$ & the result being $|\cos x| +|\sin x|=|\sin x|+|\cos x|$ hence the period is $\frac{π}{2}$ instead of $π$ .

Sanjeet Raria - 6 years, 7 months ago

Let LCM of $T_1$ & $T_2$ be $T$ then $T$ will be the period of $(f+g)$ , provided there does not exist a positive number $K(<T)$ for which $f(x+K)+g(x+K)=f(x)+g(x)$ else $K$ will be the period. Generally the situation arises due to the interchanging of $f(x)$ and $g(x)$ by addition of the positive number $K$ .

Sanjeet Raria - 6 years, 7 months ago

Good observation for 1.

All we know is that LCM $( \alpha , \beta )$ will be a period of $f + g$ . We do not know that it must be the fundamental period. Another example that I was thinking of was $f = \sin x$ and $g = - \sin x$ .

Calvin Lin Staff - 6 years, 7 months ago

@Calvin Lin – Wow!! Didn't think bout that..... $\sin x-\sin x$ . :)

Sanjeet Raria - 6 years, 7 months ago

I don't know how to prove these. And why it is not necessary that at least one of the above statements is true ? Would you please explain the proofs ?

Sandeep Bhardwaj - 6 years, 7 months ago

Sir , I have been taught that for complementary even functions the resultant period is halved. Maybe that's the reason for the period being pi/2 and not pi.

Keshav Tiwari - 6 years, 7 months ago

@Keshav Tiwari – The reason behind this can be justified just by looking at the graph of the given function !

Sandeep Bhardwaj - 6 years, 7 months ago

Sanjeet Raria
Oct 19, 2014

In periodicity of the functions, we know that if $f$ and $g$ are periodic functions from $R \to R$ having periods $T_1$ and $T_2$ respectively then $(f+g)$ is periodic with period as LCM $(T_1,T_2)$ . Now LCM of two numbers $T_1$ & $T_2$ makes sense $\textbf {only when}$ $\frac{T_1}{T_2}$ is rational.

So obviously a,b both is the answer.

P.S Let LCM of $T_1$ & $T_2$ be $T$ then $T$ will be the period of $(f+g)$ , provided there does not exist a positive number $K(<T)$ for which $f(x+K)+g(x+K)=f(x)+g(x)$ else $K$ will be the period. Generally the situation arises due to the interchanging of $f(x)$ and $g(x)$ by addition of the positive number $K$ .

How do you establish that fact?

Calvin Lin Staff - 6 years, 7 months ago

Note: I believe that there exists functions such that $\frac{ T_1} { T_2}$ is irrational, and $f+g$ is periodic.

See A fundamental period claim for the discussion.

Calvin Lin Staff - 6 years, 7 months ago

Do you know its property ? -9

2 solutions

0 pending reports