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π° Question: A philosopher of science considers a function $ k(x) $ modeling the "distance" from theory to observation, satisfying $ k(x + y) = k(x) + k(y) - 2k(xy) $. If $ k(1) = 1 $, find $ k(2) $.
π° Solution: Set $ x = y = 1 $: $ k(2) = k(1) + k(1) - 2k(1 \cdot 1) = 1 + 1 - 2(1) = 0 $. Verify consistency: Try $ x = 2, y = 1 $: $ k(3) = k(2) + k(1) - 2k(2) = 0 + 1 - 0 = 1 $. Try $ x = y = 2 $: $ k(4) = k(2) + k(2) - 2k(4) $ Γ’ΒΒ $ k(4) + 2k(4) = 0 + 0 $ Γ’ΒΒ $ 3k(4) = 0 $ Γ’ΒΒ $ k(4) = 0 $. Assume $ k(x) = 0 $ for all $ x $, but $ k(1) = 1
π° eq 0 $. Contradiction? Wait, from $ k(2) = 0 $, check $ x = 1, y = -1 $: $ k(0) = k(1) + k(-1) - 2k(-1) = 1 + k(-1) - 2k(-1) = 1 - k(-1) $. Also $ k(0) = k(0 + 0) = 2k(0) - 2k(0) = 0 $? No: $ k(0) = k(0 + 0) = 2k(0) - 2k(0) = 0 $. So $ k(0) = 0 $. Then $ 0 = 1 - k(-1) $ Γ’ΒΒ $ k(-1) = 1 $. Then $ x = -1, y = -1 $: $ k(-2) = 2k(-1) - 2k(1) = 2(1) - 2(1) = 0 $. $ x = 1, y = -1 $: $ k(0) = k(1) + k(-1) - 2k(-1) = 1 + 1 - 2(1) = 0 $, consistent. Now $ x = 2, y = -1 $: $ k(1) = k(2) + k(-1) - 2k(-2) = 0 + 1 - 0 = 1 $, matches. No contradiction. Thus $ k(2) = 0 $. Final answer: $ oxed{0} $.
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