#31 Measure Your Calibre

Geometry Level 2

Find the fundamental period of the function f ( x ) = e 3 ( x x ) \large f(x) = e^{3(x - \lfloor x \rfloor )} .

Notation: \lfloor \cdot \rfloor denotes the floor function .

The answer is 1.

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1 solution

Oliver Papillo
Apr 12, 2017

Since x + 1 x + 1 = x x x+1-\lfloor x+1 \rfloor = x-\lfloor x \rfloor

But x + 1 2 x + 1 2 x x x+\frac{1}{2}-\lfloor x+\frac{1}{2} \rfloor \neq x-\lfloor x \rfloor

The period of the function is 1 1 .

Can you please provide some details , I couldn't understand.

And also tell me whether I am correct with this logic , that all the values of x x x-\lfloor{x}\rfloor can be attained in varying the values between any two consecutive integers and hence the period becomes 1 1 .

@Oliver Papillo @Md Zuhair

Ankit Kumar Jain - 4 years, 1 month ago

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