Function with Periodicity

Calculus Level 4

Let $f : \mathbb R \rightarrow \mathbb R$ be a continuous function and $f(x) = f(2x)$ is true $\forall x \in \mathbb R$ . If $f(1) = 3$ , then find the value of

$\large\ \int _{ -1 }^{ 1 }{ f\left( f\left( f\left( x \right) \right) \right) }\, dx$ .

The answer is 6.

2 solutions

Aaghaz Mahajan
May 16, 2018

@Priyanshu Mishra Here is my solution, please check whether it is right.....

After iterating the given equation by putting x = 1/2 , 1/4 , 1/8 , 1/16 , and so on........we get f(0) = 3......
Then, proceeding similarly, after iterating the function by putting x = x/2 , x/4 , x/8 , x/16 , and so on......we get f(x) = 3......... Now it is simple integration....!!

@Aaghaz Mahajan , no i am not able to understand how you are trying to show that f is a constant. Merely checking few values of f cannot not guarantee that f is indeed constant.

Priyanshu Mishra - 3 years ago

@Priyanshu Mishra No no.!! You don't get my point......I am not checking a few values....... infact I am checking all values of the form 1/(2^n)......

The first equation results in the following set of equation.....
f(1) = f(1/2)
f(1/2) = f(1/4)
f(1/4) = f(1/8)
f(1/8) = f(1/16)
...........
f(1/[2^{n-1}]) = f(1/[2^n])

Now, letting n tend to infinity and adding up all the equations, we get
f(0) = f(1) = 3
Similarly we can proceed and find that f(x) = 3.......
Do you get it now??

Aaghaz Mahajan - 3 years ago

@Priyanshu Mishra Do you get it now???

Aaghaz Mahajan - 3 years ago

Priyanshu Mishra
May 15, 2018

@Ayush Mishra , please post your solution.

I got $f(x)$ as constant function

A Former Brilliant Member - 3 years ago

Function with Periodicity

The answer is 6.

2 solutions

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