WebWhen we say that a function is “well-defined”, we generally mean that for every (valid) input value there is one, and only one, output value. WebApr 21, 2010 · Apr 20, 2010. #1. Hi. I am trying to show that for f belonging to L^2 (-pi;pi) the integral that defines the complex Fourier Coefficients is well defined. In other words what I need to show is that. int_from -pi to pi ( f (x)*exp (-i*k*x) dx) < infinity (limited) I was thinking that since f belongs to L^2 (-pi;pi) then the integral of this will ...
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WebTherefore to show that some function is well-defined you need to prove that no element of the domain maps to more than one element of the codomain and that every element of the domain gets mapped to something in the codomain. For example, take the inverse of the squared function (square root): WebMar 24, 2014 · $\begingroup$ Many people have pointed out that students need to encounter cases where something is not well defined. A related issue is deciding whether a certain definition is useful. For example, every semester I have my students do a group discussion question that asks, is it a good idea to define positive and negative vectors -- … how to write to a csv
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WebTo show that a function is injective, you show that if f (x) = f (y), then x = y. If you think back to college algebra, this is the "horizontal line test". To show that it's well-defined, you show that if x = y, then f (x) = f (y). This is the "vertical line test". WebJun 23, 2024 · Your argument is sufficient to show the dot operator is well-defined. My guess is that they meant to ask a different question. The hint is that they use G for the set, which suggests they may be thinking of groups. I think they meant to specify that where those two elements are considered as Abelian groups under addition. Webthe \function" we’ve de ned above has the property that f([2]) 6= f([7]). This is a clear violation of the de nition of a function, and so f is not really a function at all. Mathematicians would say that the \function" fis not well-de ned, which really just means that fis not a function. The issue is that the \rule" for fmaps orkin sherman