Proving $lnleft(cos frac{1}{2^n}right) = Oleft(frac{1}{4^n}right)$












0












$begingroup$


I would like to show that :



$$lnleft(cos frac{1}{2^n}right) = Oleft(frac{1}{4^n}right)$$



Attempt :



let's show that the limit :



$limlimits_{n to infty} 4^nlnleft(cos frac{1}{2^n}right)$ is bounded. Now the problem is that I need to control at which speed $ln(...)$ go to $0$. Yet I don't se how to do so.



Thank you !






real-analysis calculus asymptotics







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  • 1




    $begingroup$
    $cos x = 1- frac{x^2}{2} + O(x^3)$ and $ln(1+x) simeq x$.
    $endgroup$
    – mathcounterexamples.net
    Jan 7 at 21:48












  • $begingroup$
    $cos x$ goes to 1 for $xto0$, so you need to know $ln$ close to 1.
    $endgroup$
    – Fabian
    Jan 7 at 21:48












  • $begingroup$
    @mathcounterexamples.net we even have : $cos x = 1 -frac{x^2}{2} + O(x^2)$ which is better to use here I guess
    $endgroup$
    – Thinking
    Jan 7 at 21:51










  • $begingroup$
    @Thinking with a little o, else the $-x^2/2$ is absorbed by $O(x^2)$.
    $endgroup$
    – zwim
    Jan 7 at 21:54










  • $begingroup$
    For technical reasons, posting titles that are all MathJax is discouraged - on default settings, it overrides the right-click menu that would be used to open the question in a new tab. I've edited a word into this one. In the future, please keep this in mind when posting questions.
    $endgroup$
    – jmerry
    Jan 7 at 21:57
















0












$begingroup$


I would like to show that :



$$lnleft(cos frac{1}{2^n}right) = Oleft(frac{1}{4^n}right)$$



Attempt :



let's show that the limit :



$limlimits_{n to infty} 4^nlnleft(cos frac{1}{2^n}right)$ is bounded. Now the problem is that I need to control at which speed $ln(...)$ go to $0$. Yet I don't se how to do so.



Thank you !






real-analysis calculus asymptotics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $cos x = 1- frac{x^2}{2} + O(x^3)$ and $ln(1+x) simeq x$.
    $endgroup$
    – mathcounterexamples.net
    Jan 7 at 21:48












  • $begingroup$
    $cos x$ goes to 1 for $xto0$, so you need to know $ln$ close to 1.
    $endgroup$
    – Fabian
    Jan 7 at 21:48












  • $begingroup$
    @mathcounterexamples.net we even have : $cos x = 1 -frac{x^2}{2} + O(x^2)$ which is better to use here I guess
    $endgroup$
    – Thinking
    Jan 7 at 21:51










  • $begingroup$
    @Thinking with a little o, else the $-x^2/2$ is absorbed by $O(x^2)$.
    $endgroup$
    – zwim
    Jan 7 at 21:54










  • $begingroup$
    For technical reasons, posting titles that are all MathJax is discouraged - on default settings, it overrides the right-click menu that would be used to open the question in a new tab. I've edited a word into this one. In the future, please keep this in mind when posting questions.
    $endgroup$
    – jmerry
    Jan 7 at 21:57














0












0








0





$begingroup$


I would like to show that :



$$lnleft(cos frac{1}{2^n}right) = Oleft(frac{1}{4^n}right)$$



Attempt :



let's show that the limit :



$limlimits_{n to infty} 4^nlnleft(cos frac{1}{2^n}right)$ is bounded. Now the problem is that I need to control at which speed $ln(...)$ go to $0$. Yet I don't se how to do so.



Thank you !






real-analysis calculus asymptotics







share|cite|improve this question











$endgroup$




I would like to show that :



$$lnleft(cos frac{1}{2^n}right) = Oleft(frac{1}{4^n}right)$$



Attempt :



let's show that the limit :



$limlimits_{n to infty} 4^nlnleft(cos frac{1}{2^n}right)$ is bounded. Now the problem is that I need to control at which speed $ln(...)$ go to $0$. Yet I don't se how to do so.



Thank you !






real-analysis calculus asymptotics




real-analysis calculus asymptotics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 22:52









rtybase

10.7k21533




10.7k21533










asked Jan 7 at 21:46









hfrjeaklhfuzlhfrjeaklhfuzl

132




132








  • 1




    $begingroup$
    $cos x = 1- frac{x^2}{2} + O(x^3)$ and $ln(1+x) simeq x$.
    $endgroup$
    – mathcounterexamples.net
    Jan 7 at 21:48












  • $begingroup$
    $cos x$ goes to 1 for $xto0$, so you need to know $ln$ close to 1.
    $endgroup$
    – Fabian
    Jan 7 at 21:48












  • $begingroup$
    @mathcounterexamples.net we even have : $cos x = 1 -frac{x^2}{2} + O(x^2)$ which is better to use here I guess
    $endgroup$
    – Thinking
    Jan 7 at 21:51










  • $begingroup$
    @Thinking with a little o, else the $-x^2/2$ is absorbed by $O(x^2)$.
    $endgroup$
    – zwim
    Jan 7 at 21:54










  • $begingroup$
    For technical reasons, posting titles that are all MathJax is discouraged - on default settings, it overrides the right-click menu that would be used to open the question in a new tab. I've edited a word into this one. In the future, please keep this in mind when posting questions.
    $endgroup$
    – jmerry
    Jan 7 at 21:57














  • 1




    $begingroup$
    $cos x = 1- frac{x^2}{2} + O(x^3)$ and $ln(1+x) simeq x$.
    $endgroup$
    – mathcounterexamples.net
    Jan 7 at 21:48












  • $begingroup$
    $cos x$ goes to 1 for $xto0$, so you need to know $ln$ close to 1.
    $endgroup$
    – Fabian
    Jan 7 at 21:48












  • $begingroup$
    @mathcounterexamples.net we even have : $cos x = 1 -frac{x^2}{2} + O(x^2)$ which is better to use here I guess
    $endgroup$
    – Thinking
    Jan 7 at 21:51










  • $begingroup$
    @Thinking with a little o, else the $-x^2/2$ is absorbed by $O(x^2)$.
    $endgroup$
    – zwim
    Jan 7 at 21:54










  • $begingroup$
    For technical reasons, posting titles that are all MathJax is discouraged - on default settings, it overrides the right-click menu that would be used to open the question in a new tab. I've edited a word into this one. In the future, please keep this in mind when posting questions.
    $endgroup$
    – jmerry
    Jan 7 at 21:57








1




1




$begingroup$
$cos x = 1- frac{x^2}{2} + O(x^3)$ and $ln(1+x) simeq x$.
$endgroup$
– mathcounterexamples.net
Jan 7 at 21:48






$begingroup$
$cos x = 1- frac{x^2}{2} + O(x^3)$ and $ln(1+x) simeq x$.
$endgroup$
– mathcounterexamples.net
Jan 7 at 21:48














$begingroup$
$cos x$ goes to 1 for $xto0$, so you need to know $ln$ close to 1.
$endgroup$
– Fabian
Jan 7 at 21:48






$begingroup$
$cos x$ goes to 1 for $xto0$, so you need to know $ln$ close to 1.
$endgroup$
– Fabian
Jan 7 at 21:48














$begingroup$
@mathcounterexamples.net we even have : $cos x = 1 -frac{x^2}{2} + O(x^2)$ which is better to use here I guess
$endgroup$
– Thinking
Jan 7 at 21:51




$begingroup$
@mathcounterexamples.net we even have : $cos x = 1 -frac{x^2}{2} + O(x^2)$ which is better to use here I guess
$endgroup$
– Thinking
Jan 7 at 21:51












$begingroup$
@Thinking with a little o, else the $-x^2/2$ is absorbed by $O(x^2)$.
$endgroup$
– zwim
Jan 7 at 21:54




$begingroup$
@Thinking with a little o, else the $-x^2/2$ is absorbed by $O(x^2)$.
$endgroup$
– zwim
Jan 7 at 21:54












$begingroup$
For technical reasons, posting titles that are all MathJax is discouraged - on default settings, it overrides the right-click menu that would be used to open the question in a new tab. I've edited a word into this one. In the future, please keep this in mind when posting questions.
$endgroup$
– jmerry
Jan 7 at 21:57




$begingroup$
For technical reasons, posting titles that are all MathJax is discouraged - on default settings, it overrides the right-click menu that would be used to open the question in a new tab. I've edited a word into this one. In the future, please keep this in mind when posting questions.
$endgroup$
– jmerry
Jan 7 at 21:57










2 Answers
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We have $$cos x=1-frac{x^2}2+o(x^2)\cosfrac1{2^n}=?\ln (1-x)=x+o(x)\lnleft(cosfrac1{2^n}right)=?$$



Can you fill in the blanks?






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Consider that, for small $epsilon$
    $$cos(epsilon)=1-frac{epsilon ^2}{2}+frac{epsilon ^4}{24}+Oleft(epsilon ^6right)$$
    Continue with Taylor series to get
    $$log (cos (epsilon ))=-frac{epsilon ^2}{2}-frac{epsilon ^4}{12}+Oleft(epsilon ^5right)$$ Make $epsilon=frac 1{2^n}$ and ... continue






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

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      active

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      0












      $begingroup$

      We have $$cos x=1-frac{x^2}2+o(x^2)\cosfrac1{2^n}=?\ln (1-x)=x+o(x)\lnleft(cosfrac1{2^n}right)=?$$



      Can you fill in the blanks?






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        We have $$cos x=1-frac{x^2}2+o(x^2)\cosfrac1{2^n}=?\ln (1-x)=x+o(x)\lnleft(cosfrac1{2^n}right)=?$$



        Can you fill in the blanks?






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          We have $$cos x=1-frac{x^2}2+o(x^2)\cosfrac1{2^n}=?\ln (1-x)=x+o(x)\lnleft(cosfrac1{2^n}right)=?$$



          Can you fill in the blanks?






          share|cite|improve this answer









          $endgroup$



          We have $$cos x=1-frac{x^2}2+o(x^2)\cosfrac1{2^n}=?\ln (1-x)=x+o(x)\lnleft(cosfrac1{2^n}right)=?$$



          Can you fill in the blanks?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 21:57









          John DoeJohn Doe

          11.1k11238




          11.1k11238























              0












              $begingroup$

              Consider that, for small $epsilon$
              $$cos(epsilon)=1-frac{epsilon ^2}{2}+frac{epsilon ^4}{24}+Oleft(epsilon ^6right)$$
              Continue with Taylor series to get
              $$log (cos (epsilon ))=-frac{epsilon ^2}{2}-frac{epsilon ^4}{12}+Oleft(epsilon ^5right)$$ Make $epsilon=frac 1{2^n}$ and ... continue






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Consider that, for small $epsilon$
                $$cos(epsilon)=1-frac{epsilon ^2}{2}+frac{epsilon ^4}{24}+Oleft(epsilon ^6right)$$
                Continue with Taylor series to get
                $$log (cos (epsilon ))=-frac{epsilon ^2}{2}-frac{epsilon ^4}{12}+Oleft(epsilon ^5right)$$ Make $epsilon=frac 1{2^n}$ and ... continue






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Consider that, for small $epsilon$
                  $$cos(epsilon)=1-frac{epsilon ^2}{2}+frac{epsilon ^4}{24}+Oleft(epsilon ^6right)$$
                  Continue with Taylor series to get
                  $$log (cos (epsilon ))=-frac{epsilon ^2}{2}-frac{epsilon ^4}{12}+Oleft(epsilon ^5right)$$ Make $epsilon=frac 1{2^n}$ and ... continue






                  share|cite|improve this answer









                  $endgroup$



                  Consider that, for small $epsilon$
                  $$cos(epsilon)=1-frac{epsilon ^2}{2}+frac{epsilon ^4}{24}+Oleft(epsilon ^6right)$$
                  Continue with Taylor series to get
                  $$log (cos (epsilon ))=-frac{epsilon ^2}{2}-frac{epsilon ^4}{12}+Oleft(epsilon ^5right)$$ Make $epsilon=frac 1{2^n}$ and ... continue







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 8 at 6:23









                  Claude LeiboviciClaude Leibovici

                  120k1157132




                  120k1157132






























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