How Can I calculate the Taylor series of $ln(cos(x))$?











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I found it in an exercise about limits and i don't know how to solve it.
If it Is possible I would like to understand the steps to calculate It.
Many thanks in advance.






limits trigonometry logarithms







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    The same way you calculate the Taylor series of any function. Do you know how to do that?
    – Gerry Myerson
    23 hours ago










  • This have the same example you're asking for en.wikipedia.org/wiki/Taylor_series
    – hamza boulahia
    23 hours ago










  • Yes, but i don't undetstand how to calculate this in particular because, probably, I don't understand deeply the argument. what I want is simply a clarification on how to calculate this particularly if it is possible, thanks
    – Aldo
    23 hours ago















up vote
0
down vote

favorite












I found it in an exercise about limits and i don't know how to solve it.
If it Is possible I would like to understand the steps to calculate It.
Many thanks in advance.






limits trigonometry logarithms







share|cite|improve this question




















  • 1




    The same way you calculate the Taylor series of any function. Do you know how to do that?
    – Gerry Myerson
    23 hours ago










  • This have the same example you're asking for en.wikipedia.org/wiki/Taylor_series
    – hamza boulahia
    23 hours ago










  • Yes, but i don't undetstand how to calculate this in particular because, probably, I don't understand deeply the argument. what I want is simply a clarification on how to calculate this particularly if it is possible, thanks
    – Aldo
    23 hours ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I found it in an exercise about limits and i don't know how to solve it.
If it Is possible I would like to understand the steps to calculate It.
Many thanks in advance.






limits trigonometry logarithms







share|cite|improve this question















I found it in an exercise about limits and i don't know how to solve it.
If it Is possible I would like to understand the steps to calculate It.
Many thanks in advance.






limits trigonometry logarithms




limits trigonometry logarithms






share|cite|improve this question















share|cite|improve this question













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edited 23 hours ago









Tianlalu

2,559632




2,559632










asked 23 hours ago









Aldo

32




32








  • 1




    The same way you calculate the Taylor series of any function. Do you know how to do that?
    – Gerry Myerson
    23 hours ago










  • This have the same example you're asking for en.wikipedia.org/wiki/Taylor_series
    – hamza boulahia
    23 hours ago










  • Yes, but i don't undetstand how to calculate this in particular because, probably, I don't understand deeply the argument. what I want is simply a clarification on how to calculate this particularly if it is possible, thanks
    – Aldo
    23 hours ago














  • 1




    The same way you calculate the Taylor series of any function. Do you know how to do that?
    – Gerry Myerson
    23 hours ago










  • This have the same example you're asking for en.wikipedia.org/wiki/Taylor_series
    – hamza boulahia
    23 hours ago










  • Yes, but i don't undetstand how to calculate this in particular because, probably, I don't understand deeply the argument. what I want is simply a clarification on how to calculate this particularly if it is possible, thanks
    – Aldo
    23 hours ago








1




1




The same way you calculate the Taylor series of any function. Do you know how to do that?
– Gerry Myerson
23 hours ago




The same way you calculate the Taylor series of any function. Do you know how to do that?
– Gerry Myerson
23 hours ago












This have the same example you're asking for en.wikipedia.org/wiki/Taylor_series
– hamza boulahia
23 hours ago




This have the same example you're asking for en.wikipedia.org/wiki/Taylor_series
– hamza boulahia
23 hours ago












Yes, but i don't undetstand how to calculate this in particular because, probably, I don't understand deeply the argument. what I want is simply a clarification on how to calculate this particularly if it is possible, thanks
– Aldo
23 hours ago




Yes, but i don't undetstand how to calculate this in particular because, probably, I don't understand deeply the argument. what I want is simply a clarification on how to calculate this particularly if it is possible, thanks
– Aldo
23 hours ago










5 Answers
5






active

oldest

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up vote
1
down vote



accepted










A quick hack is often to partially express some function in terms of a taylor approximation, since higher terms go to zero if we are considering limits for $x rightarrow 0$. To really answer your question we need to know what the original question was. But with the current information I would say that:



$$cos(x)=1 - frac{1}{2}x^2 + mathcal{O} (x^4)$$
You can add more terms if you need to. Now we write:
$$ ln(1+(- frac{1}{2}x^2) ) = dots$$
Do you know the standard Taylor series for this function?



Hint: it is of the form $ln(1+u)$






share|cite|improve this answer






























    up vote
    3
    down vote













    You are given $f(x)=log(cos x)$. Then, by the chain rule, $f'(x)=-{sin xovercos x}=-tan x$. Since you know that $tan' x=1+tan^2 x$ you now can "mechanically" compute all desired derivatives $f^{(j)}(x)$, $>jgeq0$. For the Taylor series of the given $f$ computed at $x=0$ you have to evaluate these derivatives at $x=0$. The resulting numbers do not follow a simple law that you might recognize.






    share|cite|improve this answer





















    • Or use the taylor series for the tangent, should he happen to know that
      – vrugtehagel
      21 hours ago


















    up vote
    1
    down vote













    Since $lncos x=-int_0^xtan t dt$, this problem reduces to knowing how to express the coefficients in the Taylor series of $tan x$ in terms of the up/down numbers.






    share|cite|improve this answer




























      up vote
      0
      down vote













      If $$ell(1+u)=u-frac{u^2}{2}+frac{u^3}{3}-cdots$$ and $$c(x)=1-frac{x^2}{2}+frac{x^4}{4}-cdots$$ try $$ellbigl( c(x)-1 bigr)$$






      share|cite|improve this answer




























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        Since
        $$cos(x)=prod_{ngeq 0}left(1-frac{4x^2}{pi^2(2n+1)^2}right)$$
        we have
        $$ -logcos x = sum_{ngeq 0}sum_{mgeq 1}frac{4^m x^{2m}}{mpi^{2m}(2n+1)^{2m}}=sum_{mgeq 1}frac{4^m x^{2m}}{mpi^{2m}}left[zeta(2m)-frac{zeta(2m)}{4^m}right] $$
        and
        $$ logcos x = sum_{mgeq 1}(1-4^m)frac{zeta(2m)}{mpi^{2m}} x^{2m}. $$






        share|cite|improve this answer





















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          5 Answers
          5






          active

          oldest

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          5 Answers
          5






          active

          oldest

          votes









          active

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          active

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          up vote
          1
          down vote



          accepted










          A quick hack is often to partially express some function in terms of a taylor approximation, since higher terms go to zero if we are considering limits for $x rightarrow 0$. To really answer your question we need to know what the original question was. But with the current information I would say that:



          $$cos(x)=1 - frac{1}{2}x^2 + mathcal{O} (x^4)$$
          You can add more terms if you need to. Now we write:
          $$ ln(1+(- frac{1}{2}x^2) ) = dots$$
          Do you know the standard Taylor series for this function?



          Hint: it is of the form $ln(1+u)$






          share|cite|improve this answer



























            up vote
            1
            down vote



            accepted










            A quick hack is often to partially express some function in terms of a taylor approximation, since higher terms go to zero if we are considering limits for $x rightarrow 0$. To really answer your question we need to know what the original question was. But with the current information I would say that:



            $$cos(x)=1 - frac{1}{2}x^2 + mathcal{O} (x^4)$$
            You can add more terms if you need to. Now we write:
            $$ ln(1+(- frac{1}{2}x^2) ) = dots$$
            Do you know the standard Taylor series for this function?



            Hint: it is of the form $ln(1+u)$






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted







              up vote
              1
              down vote



              accepted






              A quick hack is often to partially express some function in terms of a taylor approximation, since higher terms go to zero if we are considering limits for $x rightarrow 0$. To really answer your question we need to know what the original question was. But with the current information I would say that:



              $$cos(x)=1 - frac{1}{2}x^2 + mathcal{O} (x^4)$$
              You can add more terms if you need to. Now we write:
              $$ ln(1+(- frac{1}{2}x^2) ) = dots$$
              Do you know the standard Taylor series for this function?



              Hint: it is of the form $ln(1+u)$






              share|cite|improve this answer














              A quick hack is often to partially express some function in terms of a taylor approximation, since higher terms go to zero if we are considering limits for $x rightarrow 0$. To really answer your question we need to know what the original question was. But with the current information I would say that:



              $$cos(x)=1 - frac{1}{2}x^2 + mathcal{O} (x^4)$$
              You can add more terms if you need to. Now we write:
              $$ ln(1+(- frac{1}{2}x^2) ) = dots$$
              Do you know the standard Taylor series for this function?



              Hint: it is of the form $ln(1+u)$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 23 hours ago

























              answered 23 hours ago









              WesleyGroupshaveFeelingsToo

              816217




              816217






















                  up vote
                  3
                  down vote













                  You are given $f(x)=log(cos x)$. Then, by the chain rule, $f'(x)=-{sin xovercos x}=-tan x$. Since you know that $tan' x=1+tan^2 x$ you now can "mechanically" compute all desired derivatives $f^{(j)}(x)$, $>jgeq0$. For the Taylor series of the given $f$ computed at $x=0$ you have to evaluate these derivatives at $x=0$. The resulting numbers do not follow a simple law that you might recognize.






                  share|cite|improve this answer





















                  • Or use the taylor series for the tangent, should he happen to know that
                    – vrugtehagel
                    21 hours ago















                  up vote
                  3
                  down vote













                  You are given $f(x)=log(cos x)$. Then, by the chain rule, $f'(x)=-{sin xovercos x}=-tan x$. Since you know that $tan' x=1+tan^2 x$ you now can "mechanically" compute all desired derivatives $f^{(j)}(x)$, $>jgeq0$. For the Taylor series of the given $f$ computed at $x=0$ you have to evaluate these derivatives at $x=0$. The resulting numbers do not follow a simple law that you might recognize.






                  share|cite|improve this answer





















                  • Or use the taylor series for the tangent, should he happen to know that
                    – vrugtehagel
                    21 hours ago













                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  You are given $f(x)=log(cos x)$. Then, by the chain rule, $f'(x)=-{sin xovercos x}=-tan x$. Since you know that $tan' x=1+tan^2 x$ you now can "mechanically" compute all desired derivatives $f^{(j)}(x)$, $>jgeq0$. For the Taylor series of the given $f$ computed at $x=0$ you have to evaluate these derivatives at $x=0$. The resulting numbers do not follow a simple law that you might recognize.






                  share|cite|improve this answer












                  You are given $f(x)=log(cos x)$. Then, by the chain rule, $f'(x)=-{sin xovercos x}=-tan x$. Since you know that $tan' x=1+tan^2 x$ you now can "mechanically" compute all desired derivatives $f^{(j)}(x)$, $>jgeq0$. For the Taylor series of the given $f$ computed at $x=0$ you have to evaluate these derivatives at $x=0$. The resulting numbers do not follow a simple law that you might recognize.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 23 hours ago









                  Christian Blatter

                  170k7111324




                  170k7111324












                  • Or use the taylor series for the tangent, should he happen to know that
                    – vrugtehagel
                    21 hours ago


















                  • Or use the taylor series for the tangent, should he happen to know that
                    – vrugtehagel
                    21 hours ago
















                  Or use the taylor series for the tangent, should he happen to know that
                  – vrugtehagel
                  21 hours ago




                  Or use the taylor series for the tangent, should he happen to know that
                  – vrugtehagel
                  21 hours ago










                  up vote
                  1
                  down vote













                  Since $lncos x=-int_0^xtan t dt$, this problem reduces to knowing how to express the coefficients in the Taylor series of $tan x$ in terms of the up/down numbers.






                  share|cite|improve this answer

























                    up vote
                    1
                    down vote













                    Since $lncos x=-int_0^xtan t dt$, this problem reduces to knowing how to express the coefficients in the Taylor series of $tan x$ in terms of the up/down numbers.






                    share|cite|improve this answer























                      up vote
                      1
                      down vote










                      up vote
                      1
                      down vote









                      Since $lncos x=-int_0^xtan t dt$, this problem reduces to knowing how to express the coefficients in the Taylor series of $tan x$ in terms of the up/down numbers.






                      share|cite|improve this answer












                      Since $lncos x=-int_0^xtan t dt$, this problem reduces to knowing how to express the coefficients in the Taylor series of $tan x$ in terms of the up/down numbers.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 21 hours ago









                      J.G.

                      18.1k11830




                      18.1k11830






















                          up vote
                          0
                          down vote













                          If $$ell(1+u)=u-frac{u^2}{2}+frac{u^3}{3}-cdots$$ and $$c(x)=1-frac{x^2}{2}+frac{x^4}{4}-cdots$$ try $$ellbigl( c(x)-1 bigr)$$






                          share|cite|improve this answer

























                            up vote
                            0
                            down vote













                            If $$ell(1+u)=u-frac{u^2}{2}+frac{u^3}{3}-cdots$$ and $$c(x)=1-frac{x^2}{2}+frac{x^4}{4}-cdots$$ try $$ellbigl( c(x)-1 bigr)$$






                            share|cite|improve this answer























                              up vote
                              0
                              down vote










                              up vote
                              0
                              down vote









                              If $$ell(1+u)=u-frac{u^2}{2}+frac{u^3}{3}-cdots$$ and $$c(x)=1-frac{x^2}{2}+frac{x^4}{4}-cdots$$ try $$ellbigl( c(x)-1 bigr)$$






                              share|cite|improve this answer












                              If $$ell(1+u)=u-frac{u^2}{2}+frac{u^3}{3}-cdots$$ and $$c(x)=1-frac{x^2}{2}+frac{x^4}{4}-cdots$$ try $$ellbigl( c(x)-1 bigr)$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 21 hours ago









                              Chase Ryan Taylor

                              4,35521530




                              4,35521530






















                                  up vote
                                  0
                                  down vote













                                  Since
                                  $$cos(x)=prod_{ngeq 0}left(1-frac{4x^2}{pi^2(2n+1)^2}right)$$
                                  we have
                                  $$ -logcos x = sum_{ngeq 0}sum_{mgeq 1}frac{4^m x^{2m}}{mpi^{2m}(2n+1)^{2m}}=sum_{mgeq 1}frac{4^m x^{2m}}{mpi^{2m}}left[zeta(2m)-frac{zeta(2m)}{4^m}right] $$
                                  and
                                  $$ logcos x = sum_{mgeq 1}(1-4^m)frac{zeta(2m)}{mpi^{2m}} x^{2m}. $$






                                  share|cite|improve this answer

























                                    up vote
                                    0
                                    down vote













                                    Since
                                    $$cos(x)=prod_{ngeq 0}left(1-frac{4x^2}{pi^2(2n+1)^2}right)$$
                                    we have
                                    $$ -logcos x = sum_{ngeq 0}sum_{mgeq 1}frac{4^m x^{2m}}{mpi^{2m}(2n+1)^{2m}}=sum_{mgeq 1}frac{4^m x^{2m}}{mpi^{2m}}left[zeta(2m)-frac{zeta(2m)}{4^m}right] $$
                                    and
                                    $$ logcos x = sum_{mgeq 1}(1-4^m)frac{zeta(2m)}{mpi^{2m}} x^{2m}. $$






                                    share|cite|improve this answer























                                      up vote
                                      0
                                      down vote










                                      up vote
                                      0
                                      down vote









                                      Since
                                      $$cos(x)=prod_{ngeq 0}left(1-frac{4x^2}{pi^2(2n+1)^2}right)$$
                                      we have
                                      $$ -logcos x = sum_{ngeq 0}sum_{mgeq 1}frac{4^m x^{2m}}{mpi^{2m}(2n+1)^{2m}}=sum_{mgeq 1}frac{4^m x^{2m}}{mpi^{2m}}left[zeta(2m)-frac{zeta(2m)}{4^m}right] $$
                                      and
                                      $$ logcos x = sum_{mgeq 1}(1-4^m)frac{zeta(2m)}{mpi^{2m}} x^{2m}. $$






                                      share|cite|improve this answer












                                      Since
                                      $$cos(x)=prod_{ngeq 0}left(1-frac{4x^2}{pi^2(2n+1)^2}right)$$
                                      we have
                                      $$ -logcos x = sum_{ngeq 0}sum_{mgeq 1}frac{4^m x^{2m}}{mpi^{2m}(2n+1)^{2m}}=sum_{mgeq 1}frac{4^m x^{2m}}{mpi^{2m}}left[zeta(2m)-frac{zeta(2m)}{4^m}right] $$
                                      and
                                      $$ logcos x = sum_{mgeq 1}(1-4^m)frac{zeta(2m)}{mpi^{2m}} x^{2m}. $$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 20 hours ago









                                      Jack D'Aurizio

                                      282k33274653




                                      282k33274653






























                                           

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