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anonymous
 one year ago
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anonymous
 one year ago
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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0please help with question number 5b

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here we have to keep in mind that: 1) the addition is a continuous function on R, namely R x R > R 2) the opposite value function is continuous in R, namely R>R 3) the absolute value function is continuous in R, namely R> R+ of course all those properties can be proved

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so, using the theorem which states that the composition of two continuous functions is again a continuous function and the fact that the previous functions, namely max(x,y) and min(x,y) are continuous functions, then we get your thesis

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1in other words those three statements above allow us to say that max(x,y) and min(x,y) are continuous functions

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2can w use epsilondelta definition to prove the statements ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I don't know, since f and g are generic functions

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! we can prove those last three statements using epsilondelta

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1nevertheless , since we can view this function: f(x)+g(x) as the composition of the functions: x> (f(x),g(x)) and the addition function, then we can apply the theorem of composition of continuous function, in order to say that the function: f(x)+g(x) is a continuous function when also f and are continuous

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2correct me if im wrong michele Since \(f\) and \(g\) are continuous at \(a\), we have : \[\forall \varepsilon > 0\ \exists \delta_f > 0\ \text{s.t. } 0 < x a < \delta_f \implies f(x)  f(a) < \varepsilon\] \[\forall \varepsilon > 0\ \exists \delta_g > 0\ \text{s.t. } 0 < x a < \delta_g \implies g(x)  g(a) < \varepsilon\] let \(\delta_h=\min{\{\delta_f,~\delta_g\}}\) and \(\delta_k=\max{\{\delta_f,~\delta_g\}}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2we want to show below : \(xa\lt \delta_h \implies h(x)h(a) \lt \epsilon\) \(xa\lt \delta_k \implies k(x)h(a) \lt \epsilon\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! I think that we can prove the statement using the epsilondelta, if we consider these 2 cases, namely when f greter or equal to g, and when f less or equal to g. In those cases, we have: h(x)= f(x) and k(x) = g(x), and vice versa for second case, and by hypothesis both f and g are continuous

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is the proving still going on?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I think the same \(\delta\) works for both \(h(x)\) and \(k(x)\). Here is the complete proof : Since \(f\) and \(g\) are continuous at \(a\), we have : \[\forall \varepsilon > 0\ \exists \delta_f > 0\ \text{s.t. } 0 < x a < \delta_f \implies f(x)  f(a) < \varepsilon ~~(\color{red}{\star}) \] \[\forall \varepsilon > 0\ \exists \delta_g > 0\ \text{s.t. } 0 < x a < \delta_g \implies g(x)  g(a) < \varepsilon ~~(\color{red}{\diamond}) \] Let \(\delta_h=\delta_k=\min{\{\delta_f,~\delta_g\}}\) and suppose that \(f(a)\ge g(a)\), then : \(xa\lt \delta_h \implies\) \(\begin{align}h(x)h(a) &=\max\{f(x),g(x)\}\max\{f(a),g(a)\}\\~\\ &=\max\{f(x),g(x)\}f(a)\}\end{align}\) Case1 : \(f(x)\ge g(x)\) \(\max\{f(x),g(x)\}f(a)\}=f(x)f(a)\lt \epsilon\) by \((\color{red}{\star})\) Case2 : \(g(x)\ge f(x)\) \(\max\{f(x),g(x)\}f(a)\}=g(x)f(a) \lt g(x)g(a)\lt \epsilon\) by \((\color{red}{\diamond})\) similarly we can prove the other statement by considering two cases

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1referring to my proof we have finished. So reassuming, in order to prove your statement, we need of the subsequent properties: 1) the addition is a continuous function on R, namely R x R > R 2) the opposite value function is continuous in R, namely R>R 3) the absolute value function is continuous in R, namely R> R+ furthermore, using the theorem of composition of continuous functions, and the fact that this function: x>(f(x),g(x)) is continuous, then we get your thesis

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Please double check.. it might contain errors as im really bad with real analysis...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1that's right! @ganeshie8

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I was thinking to your method :)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now you have two different solutions @GIL.ojei

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you sirs but is that all from you @ganeshie8

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I think what I had there proves that \(h(x)\) is continuous at \(a\). I'm hoping you can mimic the same easily for proving continuity of \(k(x)\) too...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I WILL BUT IT WOULD BE MORE EASIER FOR ME IF YOU COMPLETE IT IN OTHER NOT TO HAVE ERRORS, COS DON'T KNOW HOW TO USE THE SYMBOLS IN THIS SITE

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2(... continued) \(xa\lt \delta_k \implies\) \(\begin{align}k(x)k(a) &=\min\{f(x),g(x)\}\min\{f(a),g(a)\}\\~\\ &=\min\{f(x),g(x)\}g(a)\}\end{align}\) Case1 : \(f(x)\ge g(x)\) \(\min\{f(x),g(x)\}g(a)\}=g(x)g(a)\lt \epsilon\) by \((\color{red}{\star})\) Case2 : \(g(x)\ge f(x)\) \(\min\{f(x),g(x)\}g(a)\}=f(x)g(a) \lt g(x)g(a)\lt \epsilon\) by \((\color{red}{\diamond})\) \(\blacksquare\)
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