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Given: $$f(x) = f(x+2)$$ $$f(3)=20$$ prove or disprove $f(4)=20$

In my opinion $f(4)$ can have any value according to the question...since the only data is about the odd range values...I thought of a function like $f(x)=(K-20)(-1)^x + K$

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    $\begingroup$ If f(x) = f(x+2) then the function is periodic (compare to sin(x) = sin(x +2pi)). But knowing the value at x=3 doesn't say anything about values other than x = 3+2, 3+4, .... $\endgroup$ Feb 20, 2014 at 9:51

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The value is not unique take any function whose period is 2 such that f(3)=20. So f(4) could be any value.

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this function is not homomorphic and i would start looking at the range and domain of the function. is it natural numbers? reals? .

assume natural numbers. then f(3)=f(5)=...f(any odd) = 20 . then consider x to be even then f(2)=f(4)=...f(any even) = ?

well clearly this functions is not a homomorphism and you can not use an argument that it is an equivalence relation. if you can prove that it forms an equivalence relation, then these two sets will be disjoint, and thus f(4) cant be equal to an even output.

so just check if the odd functions above are reflexvive transitive and symmetric. if so, the you have the proof.

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  • $\begingroup$ The range/domain is f: R -> R by the way. And it doesn't say anything else. Thank you for your reply :) $\endgroup$
    – new one
    Feb 20, 2014 at 10:57

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