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| A Power Problem (Posted on 2006-09-03) |
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Given f(x)= 19x²/(19x²+19x)
determine if f(1/200000)+f(2/200000)+...+f(99999/200000)
equals
f(100001/200000)+f(100002/200000)+...+f(199999/200000)
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Submitted by K Sengupta
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Solution:
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We observe that:
f(x)
=19^(x^2)/( 19^(x^2)) + 19^x )
= 1/ [1 + 19^(x-x^2)}
= 1/ [1 + 19^{x(1-x)}]
= 1/ [1 + 19^{(1-x)*x}]
= f(1-x)
Accordingly, f(j/(2m)) = f(1 – (j/2m)); whenever j= 1, 2,...., 2m
Consequently:
f(1/2m) + f(2/2m) + ....+ f((m-1)/2m)
= f((m+1)/m) + f((m+2)/2m) + .... + f((2m-1)/2m)
Substituting, m = 100000, we observe that the two expressions given in the problem are indeed equal to each other.
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