b.) P and q are propositions given by p: f is an odd function and q: β«π(π₯)ππ₯=0πβπ
verify whether or not πβπ
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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 pβq. Suppose that f is an odd function. Strictly speaking, a proposition q has a meaning only if β«βkkβf(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 β«βkkβf(x)dx makes sense, the implication pβq is true, as β«βkkβf(x)dx=β«0kβf(x)dx+β«βk0βf(x)dx=0 by changing the variables in the second integral. The other direction qβp depends on the class of f : if it is continuous, then it is true as F(k)=β«βkkβf(x)dx=0,Fβ²(x)=f(x)+f(βx)=0 and thus f(x)=βf(βx), it is odd. If we make no hypothesis on f, then it is obviously false (e.g. f(0)=1,f(x)=0,xξ =0) . Therefore for continuous functions this equivalence is true : βfβC0(R),βkβR,f(βx)=βf(x)ββ«βkkβf=0.
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