Is there a real function $f$ / algebraic expression, such that
- $f(-3) = 1$
- $f(3) = 1$
- $f(x) = - f(x)$ for all $-3 < x < 3$
I was able to construct some $f$ with Fourier analysis and complex arguments but this wasn't what I was looking for. :(
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something is missing -- why not take $f(x) = 0$ when $|x| \ne 3$ and $f(-3) = f(3) = 1$? Did you mean a continuous function perhaps? That'd likely be hard since $f(3) \ne - f(-3)$... – gt6989b Oct 25 '22 at 03:35
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What does your 'algebraic expression' mean? If dividing cases are possible, @gt6989b 's function is easily possible, too, and there are lots of functions... (Even if it's not possible, $f(x)=[|x/3|]$ is possible, too.) – RDK Oct 25 '22 at 06:56