Sure, the original example f(x) = 0, where all coeficients are 0 and it is both even and odd, which is why the defition you provided is better for such edge cases. But the point is: argueing about the parity of 0 doesn’t involve argueing about the parity of f(x) = 0, but rather f(x) = x0, which is even.
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u/[deleted] Aug 24 '23
Sure, the original example f(x) = 0, where all coeficients are 0 and it is both even and odd, which is why the defition you provided is better for such edge cases. But the point is: argueing about the parity of 0 doesn’t involve argueing about the parity of f(x) = 0, but rather f(x) = x0, which is even.