Periodic functions

Calculus Level 5

Let f : R → R f: \mathbb {R \to R} be a continuous function satisfying f ( x + 1 ) = f ( x ) − 5 f ( x ) − 3 f(x+1)= \dfrac {f(x)-5}{f(x)-3} , where x ∈ R x \in \mathbb R . Find the minimum period of the function.


The answer is 4.

Yash Shroff by Yash Shroff , 22, India

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2 solutions

Chew-Seong Cheong
Feb 23, 2019

f ( x + 1 ) = f ( x ) − 5 f ( x ) − 3 f ( x + 2 ) = f ( x + 1 ) − 5 f ( x + 1 ) − 3 = f ( x ) − 5 f ( x ) − 3 − 5 f ( x ) − 5 f ( x ) − 3 − 3 = 10 − 4 f ( x ) 4 − 2 f ( x ) = 5 − 2 f ( x ) 2 − f ( x ) f ( x + 3 ) = 5 − 2 f ( x ) 2 − f ( x ) − 5 5 − 2 f ( x ) 2 − f ( x ) − 3 = 3 f ( x ) − 5 f ( x ) − 1 f ( x + 4 ) = 3 f ( x ) − 5 f ( x ) − 1 − 5 3 f ( x ) − 5 f ( x ) − 1 − 3 = − 2 f ( x ) − 2 = f ( x ) \begin{aligned} f(x+1) & = \frac {f(x)-5}{f(x)-3} \\ f(x+2) & = \frac {f(x+1)-5}{f(x+1)-3} = \frac {\frac {f(x)-5}{f(x)-3}-5}{\frac {f(x)-5}{f(x)-3}-3} = \frac {10-4f(x)}{4-2f(x)} = \frac {5-2f(x)}{2-f(x)} \\ f(x+3) & = \frac {\frac {5-2f(x)}{2-f(x)}-5}{\frac {5-2f(x)}{2-f(x)}-3} = \frac {3f(x)-5}{f(x)-1} \\ f(x+4) & = \frac {\frac {3f(x)-5}{f(x)-1}-5}{\frac {3f(x)-5}{f(x)-1}-3} = \frac {-2f(x)}{-2} = f(x) \end{aligned}

Therefore, the principal period of f ( x ) f(x) is 4 \boxed{4} .

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@Yash Shroff , I think you meant "principal period" instead of "minimum period".

Chew-Seong Cheong - 2 years, 3 months ago
Joe Mansley
Jul 18, 2018

Applying the definition on f(x+1) gives you f(x+2)=(2f(x)-5)/(f(x)-2).

Similarly f(x+3)=(3f(x)-5)/(f(x)-1)

And f(x+4)=f(x).

Thus the period must be at least 4.

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