A problem by Arpita Karkera

Level pending

If f ( x ) = 4 x 4 x + 2 f\left( x \right) =\frac { { 4 }^{ x } }{ { 4 }^{ x }+2 } , then ⌊ f ( 1 2008 ) + f ( 2 2008 ) + f ( 3 2008 ) + . . . + f ( 2007 2008 ) ⌋ − 1000 \left\lfloor f\left( \frac { 1 }{ 2008 } \right) +f\left( \frac { 2 }{ 2008 } \right) +f\left( \frac { 3 }{ 2008 } \right) +...+f\left( \frac { 2007 }{ 2008 } \right) \right\rfloor -1000 is equal to... (where, ⌊ . ⌋ \left\lfloor . \right\rfloor denotes the greatest integer function)


The answer is 3.

Arpita Karkera by Arpita Karkera , 24, India

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1 solution

Otto Bretscher
May 12, 2015

f ( x ) f(x) is a logistic functon whose graph is symmetric with respect to the point ( 1 2 , 1 2 ) , \left(\frac{1}{2},\frac{1}{2}\right), its point of inflection. Thus f ( x ) + f ( 1 − x ) = 1 f(x)+f(1-x)=1 for all x x . Now f ( k 2008 ) + f ( 2008 − k 2008 ) = 1 f\left(\frac{k}{2008}\right)+f\left(\frac{2008-k}{2008}\right) = 1 for k = 1...1003 k=1...1003 and f ( 1004 2008 ) = f ( 1 2 ) = 1 2 f\left(\frac{1004}{2008}\right)=f\left(\frac{1}{2}\right)=\frac{1}{2} , so ⌊ 1003.5 ⌋ − 1000 = 3 . \lfloor{1003.5}\rfloor-1000=\boxed{3}.

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