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anonymous

  • one year ago

help here

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  1. anonymous
    • one year ago
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    @Michele_Laino

  2. anonymous
    • one year ago
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  3. anonymous
    • one year ago
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    please help with question number 5b

  4. Michele_Laino
    • one year ago
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    here 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

  5. Michele_Laino
    • one year ago
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    so, 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

  6. Michele_Laino
    • one year ago
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    in other words those three statements above allow us to say that max(x,y) and min(x,y) are continuous functions

  7. ganeshie8
    • one year ago
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    can w use epsilon-delta definition to prove the statements ?

  8. Michele_Laino
    • one year ago
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    I don't know, since f and g are generic functions

  9. Michele_Laino
    • one year ago
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    yes! we can prove those last three statements using epsilon-delta

  10. Michele_Laino
    • one year ago
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    nevertheless , 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

  11. ganeshie8
    • one year ago
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    correct 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 \(\delta_k=\max{\{\delta_f,~\delta_g\}}\)

  12. Michele_Laino
    • one year ago
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    ok!

  13. ganeshie8
    • one year ago
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    we want to show below : \(|x-a|\lt \delta_h \implies |h(x)-h(a)| \lt \epsilon\) \(|x-a|\lt \delta_k \implies |k(x)-h(a)| \lt \epsilon\)

  14. Michele_Laino
    • one year ago
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    ok!

  15. Michele_Laino
    • one year ago
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    yes! I think that we can prove the statement using the epsilon-delta, 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

  16. Michele_Laino
    • one year ago
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    greater*

  17. anonymous
    • one year ago
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    is the proving still going on?

  18. ganeshie8
    • one year ago
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    I 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 : \(|x-a|\lt \delta_h \implies\) \(\begin{align}|h(x)-h(a)| &=|\max\{f(x),g(x)\}-\max\{f(a),g(a)\}|\\~\\ &=|\max\{f(x),g(x)\}-f(a)\}|\end{align}\) 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

  19. Michele_Laino
    • one year ago
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    referring 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

  20. ganeshie8
    • one year ago
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    Please double check.. it might contain errors as im really bad with real analysis...

  21. Michele_Laino
    • one year ago
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    that's right! @ganeshie8

  22. Michele_Laino
    • one year ago
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    I was thinking to your method :)

  23. ganeshie8
    • one year ago
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    xD

  24. Michele_Laino
    • one year ago
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    now you have two different solutions @GIL.ojei

  25. anonymous
    • one year ago
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    thank you sirs but is that all from you @ganeshie8

  26. ganeshie8
    • one year ago
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    I 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...

  27. anonymous
    • one year ago
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    I 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

  28. anonymous
    • one year ago
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    @ganeshie8

  29. ganeshie8
    • one year ago
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    (... continued) \(|x-a|\lt \delta_k \implies\) \(\begin{align}|k(x)-k(a)| &=|\min\{f(x),g(x)\}-\min\{f(a),g(a)\}|\\~\\ &=|\min\{f(x),g(x)\}-g(a)\}|\end{align}\) 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 \((\color{red}{\diamond})\) \(\blacksquare\)

  30. anonymous
    • one year ago
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    @ganeshie8

  31. anonymous
    • one year ago
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