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| A Most Unusual Evaluation (Posted on 2006-01-04) |
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Let F be an increasing real function defined for all real X, where 0<=X<=1 such that:
(i) F (X/8) = F(X)/7 and
(ii) F(1-X) = 1 – F(X)
For all whole numbers M and N greater than zero, determine:
F ( 1/ ((8^M)* ( 8^N + 1)) ) in terms of M and N.
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Submitted by K Sengupta
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Rating: 3.0000 (2 votes)
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Solution:
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F ( 1/ ( (8^M)* ( 8^N + 1) ) ) = 1 / ( ( 7^M) * ( 7^N + 1) )
EXPLANATION:
One thing to note from rule (i) is since, F( X/ 8) = 1/7 * F(X) then F(X) = 7 * F ( X/8)
Another application of rule (i) to get an expression in terms for F(X). We observe that 0 <=X<=1 (given) so that:
0 <=X / (8^M) < 1 for all M>0 .
Now, F (X / (8^M))
= F (X / (8* (8^(M -1)))
= F (X / 8^(M -1))/ 7
= ...
= ((1/7)^(M-1)) * F(X/8)
= ((1/7)^ M) * F(X)..........(a)
Using rule (ii), F(1- X) = 1 – F(X) and substituting X = (8^N)/ (8^N +1 ) , we observe that for N>0 ;
F (1/ (8^N +1) ) = 1 – F( 8^N/ (8^N +1)) ). Since N>0, it follows that:
0 <= 8^N/ (8^N + 1) < 1 ........(b)
Accordingly, F( 8^N/ ( 8^N +1))
= 7 * F(8^(N-1)/ ( 8^N +1))
= ...
= (7^N) * F ( 1/( 8^N +1))
Hence, from (b):
F ( 1/( 8^N + 1)) = 1 – (7^N) * F ( 1/( 8^N + 1)), giving :
F ( 1/( 8^N + 1) ) = 1 / (7^N + 1 )
Consequently, substituting X = 1/( 8^N + 1) in (a) , we obtain:
F ( 1/ ( (8^M)* ( 8^N + 1) ) ) = 1 / ( ( 7^M) * ( 7^N + 1) ). |
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