b.) P and q are propositions given by p: f is an odd function and q: ∫𝑓(𝑥)𝑑𝑥=0𝑘−𝑘
verify whether or not 𝑝↔𝑞
1
Expert's answer
2021-02-24T07:33:42-0500
A proposition, from a mathematical logic viewpoint, is a logical expression without free variables. From a viewpoint of "general" mathematics it is a statement meant to be proven.
First let's look at p→q. Suppose that f is an odd function. Strictly speaking, a proposition q has a meaning only if ∫−kkf(x)dx exists (i.e. f integrable), as we can have an odd function that is not Riemann-integrable/Lebesgue-integrable etc. Assuming that the expression ∫−kkf(x)dx makes sense, the implication p→q is true, as ∫−kkf(x)dx=∫0kf(x)dx+∫−k0f(x)dx=0 by changing the variables in the second integral. The other direction q→p depends on the class of f : if it is continuous, then it is true as F(k)=∫−kkf(x)dx=0,F′(x)=f(x)+f(−x)=0 and thus f(x)=−f(−x), it is odd. If we make no hypothesis on f, then it is obviously false (e.g. f(0)=1,f(x)=0,x=0) . Therefore for continuous functions this equivalence is true : ∀f∈C0(R),∀k∈R,f(−x)=−f(x)↔∫−kkf=0.
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