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Real Functions Determination (Posted on 2015-05-02) Difficulty: 3 of 5
Find all real valued functions G(x) on integers is such that:

(i) G(1) = 5/2, and:
(ii) G(0) is not zero, and:
(iii) G(A)*G(B) = G(A+B)+G(A-B), for all A and B

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Solution Determined Function (spoiler) | Comment 2 of 4 |
From (iii), G(A) * G(0) = 2*G(A), so G(0) = 2. 
Not clear why the problem needs to specify that G(0) is not 0

G(A)*G(A) = G(2A)+G(0), so G(2A) = G(A)^2 - 2
G(2) = 17/4 = 4 + 1/4
G(4) = 257/16 = 16 + 1/16

It looks like G(A) = 2^A + 2^(-A)

Yes, that works!

G(A)*G(B) = (2^A + 2^(-A))*(2^B + 2^(-B))
    = 2^(A+B) + 2^(A-B) + 2^(B-A) + 2^(-B-A)
    = G(A+B) + G(A-B)
    
In fact, the property that G(A)*G(B) = G(A+B) + G(A-B) 
is true if G(A) = x^A + x^(-A)

So it is just a matter of solving for the initial condition.
G(1) = x + 1/x = 2 + 1/2  means that x = 2 or 1/2

And the function is symmetric around 0, which we actually noticed initially because G(A)*G(A) = G(A)*G(-A)

Also, it looks like the range can be bigger than just integers 

  Posted by Steve Herman on 2015-05-02 11:19:33
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