Parity Party

Algebra Level 2

Suppose there is an even function f ( x ) f(x) . Is it possible that f ( x ) f(x) was also an odd function?

Yes No

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2 solutions

For any n N n \in \N , let S S be a set of real numbers such that: S = i = 1 n S i S = \bigcup\limits_{i=1}^{n} S_i , where S i = [ l i , r i ) S_i = [l_i, r_i) , l 1 0 l_1 \geq 0 and l i < r i l i + 1 l_i < r_i \leq l_{i+1} . Let: S = i = 1 n S i \overline{S} = \bigcup\limits_{i=1}^{n} \overline{S_i} , where S i = ( r i , l i ] \overline{S_i} = (-r_i, -l_i] . Now, let: X = S S X = S \cup \overline{S} . Any function f ( x ) f(x) defined on X X such that f : X { 0 } f : X \to \{0\} is both odd and even, since f ( x ) = f ( x ) = f ( x ) = 0 f(x) = f(-x) = -f(-x) = 0 , x X \forall x \in X .

Your subset does not need to be in this form. You could just let be S R + S\subset \R_+ be any subset of positive reals, and f : S S { 0 } f:S\cup-S\mapsto \{0\} works, without all the complication.

Noe Blassel - 1 year, 8 months ago

I know, I just wanted to learn more about union of sets and how to use these neat symbols in LaTeX :D

Novak Radivojević - 1 year, 8 months ago

If we set the function to be f ( x ) = 0 f(x)=0 , it is as even as odd function

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