Mersenne this before?

Calculus Level 3

Let f ( x ) = x q − ⌊ x ⌋ f(x)=x^{q}-\left\lfloor{x}\right\rfloor and g ( x ) = ∫ − 1 1 f ( x ) d x \displaystyle g(x)= \int_{-1}^{1} f(x) dx . If g ( x ) = a b g(x)=\dfrac{a}{b} , where a a and b b are relatively prime and q q is 1 less than the smallest number in a twin prime pair, find a − b a-b .


The answer is 2.

David Holcer by David Holcer , 20, Canada

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1 solution

Tom Engelsman
Dec 9, 2017

Calculating g ( x ) g(x) yields:

g ( x ) = ∫ − 1 1 x q − ⌊ x ⌋ d x = x q + 1 q + 1 ∣ − 1 1 + 1 = 1 − ( − 1 ) q + 1 q + 1 + 1 g(x) = \int_{-1}^{1} x^q - \lfloor x \rfloor dx = \frac{x^{q+1}}{q+1} |_{-1}^{1} + 1 = \frac{1 - (-1)^{q+1}}{q+1} + 1 .

The smallest possible twin prime pair occurs at 3 , 5 {3, 5} , which yields q = 3 − 1 ⇒ q + 1 = 3 q = 3 - 1 \Rightarrow q + 1 = 3 . Substituting this value into g ( x ) g(x) gives:

g ( x ) = 1 − ( − 1 ) 3 3 + 1 = 2 3 + 1 = 5 3 . g(x) = \frac{1 - (-1)^3}{3} + 1 = \frac{2}{3} + 1 = \boxed{\frac{5}{3}}.

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