(Piecewise-defined Functions) If I have a piecewise function where both intervals are open and right next to each other, is the overlapping point defined? (Example below)

Question

mendicant_bias · Answer

E.g. if I have a function f(x) equal to
$$1, \ \ \ -\pi/2<x<\pi/2,$$$$0, \ \ \ \pi/2<x<3\pi/2.$$

Is my function defined at pi/2? It isn't right?

welshfella · Answer

it isn;t  correctft

welshfella · Answer

* correct

phi · Answer

If your function were defined at pi/2 , then you could say its value.
But we can't tell the function's value at pi/2 ... so it's not defined

mendicant_bias · Answer

Could you elaborate? What you said meant literally nothing to me. "It isn't correct". What? What isn't correct? What I said, the problem itself?

Thanks, Phi.

mendicant_bias · Answer

My question was stemming from something dealing w/Fourier Series, are you at all good with Mathematica?

phi · Answer

not really, but I can look at your question