perplexus.info :: Calculus : A Power Problem

A Power Problem (Posted on 2006-09-03)

Given f(x)= 19^x²/(19^x²+19^x)

determine if f(1/200000)+f(2/200000)+...+f(99999/200000) equals
f(100001/200000)+f(100002/200000)+...+f(199999/200000)

See The Solution Submitted by K Sengupta

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Method

| Comment 1 of 3

f(x)=1/(1+19^(x-x^2))=1/(1+19^(x(1-x)) and x(1-x) is symmetrical, so f(1/200000)=f(199999/200000), f(2/200000)=f(199998/200000), and so on.

Posted by e.g. on 2006-09-03 11:45:37

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