Show that $ln 2 approx frac{pi^2}{16}Bigl(1+frac{pi^2}{96}Bigr)$












0














Show the Maclaurin Series for $ln(cos x)=-frac{1}{2}x^2-frac{1}{12}x^4 + ...$ (1st part of the question already worked out )



Show that $ln 2 approx frac{pi^2}{16}Bigl(1+frac{pi^2}{96}Bigr)$



Given $x=frac{pi}{4}$






sequences-and-series taylor-expansion







share|cite|improve this question
























  • well $ln 2approx x$, to any $xinBbb C$
    – Masacroso
    Nov 20 at 12:11












  • Hint: $2=frac1{1/2}$
    – Arthur
    Nov 20 at 12:11
















0














Show the Maclaurin Series for $ln(cos x)=-frac{1}{2}x^2-frac{1}{12}x^4 + ...$ (1st part of the question already worked out )



Show that $ln 2 approx frac{pi^2}{16}Bigl(1+frac{pi^2}{96}Bigr)$



Given $x=frac{pi}{4}$






sequences-and-series taylor-expansion







share|cite|improve this question
























  • well $ln 2approx x$, to any $xinBbb C$
    – Masacroso
    Nov 20 at 12:11












  • Hint: $2=frac1{1/2}$
    – Arthur
    Nov 20 at 12:11














0












0








0







Show the Maclaurin Series for $ln(cos x)=-frac{1}{2}x^2-frac{1}{12}x^4 + ...$ (1st part of the question already worked out )



Show that $ln 2 approx frac{pi^2}{16}Bigl(1+frac{pi^2}{96}Bigr)$



Given $x=frac{pi}{4}$






sequences-and-series taylor-expansion







share|cite|improve this question















Show the Maclaurin Series for $ln(cos x)=-frac{1}{2}x^2-frac{1}{12}x^4 + ...$ (1st part of the question already worked out )



Show that $ln 2 approx frac{pi^2}{16}Bigl(1+frac{pi^2}{96}Bigr)$



Given $x=frac{pi}{4}$






sequences-and-series taylor-expansion




sequences-and-series taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 20 at 12:09









Bernard

118k639112




118k639112










asked Nov 20 at 12:08









Henias

615




615












  • well $ln 2approx x$, to any $xinBbb C$
    – Masacroso
    Nov 20 at 12:11












  • Hint: $2=frac1{1/2}$
    – Arthur
    Nov 20 at 12:11


















  • well $ln 2approx x$, to any $xinBbb C$
    – Masacroso
    Nov 20 at 12:11












  • Hint: $2=frac1{1/2}$
    – Arthur
    Nov 20 at 12:11
















well $ln 2approx x$, to any $xinBbb C$
– Masacroso
Nov 20 at 12:11






well $ln 2approx x$, to any $xinBbb C$
– Masacroso
Nov 20 at 12:11














Hint: $2=frac1{1/2}$
– Arthur
Nov 20 at 12:11




Hint: $2=frac1{1/2}$
– Arthur
Nov 20 at 12:11










1 Answer
1






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4














Hint:



Since $cos(frac{pi}{4})=frac{1}{sqrt 2}$, you can write $$lnleft(frac{1}{sqrt 2}right)=ln(2^{-1/2})=-frac{1}{2}ln(2)$$




Edit for completion:



For small values of $cos(x)$ we have $$ln(cos x)approx-frac{1}{2}x^2-frac{1}{12}x^4tag{1}$$ When $x=frac{pi}{4}$ we get $$-frac{1}{2}ln(2)=ln(cos frac{pi}{4})approx-frac{1}{2}left(frac{pi}{4}right)^2-frac{1}{12}left(frac{pi}{4}right)^4\ln(2)approxfrac{pi^2}{16}left(1-frac{pi^2}{96}right)$$






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    1 Answer
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    active

    oldest

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    1 Answer
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    active

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    active

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    active

    oldest

    votes









    4














    Hint:



    Since $cos(frac{pi}{4})=frac{1}{sqrt 2}$, you can write $$lnleft(frac{1}{sqrt 2}right)=ln(2^{-1/2})=-frac{1}{2}ln(2)$$




    Edit for completion:



    For small values of $cos(x)$ we have $$ln(cos x)approx-frac{1}{2}x^2-frac{1}{12}x^4tag{1}$$ When $x=frac{pi}{4}$ we get $$-frac{1}{2}ln(2)=ln(cos frac{pi}{4})approx-frac{1}{2}left(frac{pi}{4}right)^2-frac{1}{12}left(frac{pi}{4}right)^4\ln(2)approxfrac{pi^2}{16}left(1-frac{pi^2}{96}right)$$






    share|cite|improve this answer




























      4














      Hint:



      Since $cos(frac{pi}{4})=frac{1}{sqrt 2}$, you can write $$lnleft(frac{1}{sqrt 2}right)=ln(2^{-1/2})=-frac{1}{2}ln(2)$$




      Edit for completion:



      For small values of $cos(x)$ we have $$ln(cos x)approx-frac{1}{2}x^2-frac{1}{12}x^4tag{1}$$ When $x=frac{pi}{4}$ we get $$-frac{1}{2}ln(2)=ln(cos frac{pi}{4})approx-frac{1}{2}left(frac{pi}{4}right)^2-frac{1}{12}left(frac{pi}{4}right)^4\ln(2)approxfrac{pi^2}{16}left(1-frac{pi^2}{96}right)$$






      share|cite|improve this answer


























        4












        4








        4






        Hint:



        Since $cos(frac{pi}{4})=frac{1}{sqrt 2}$, you can write $$lnleft(frac{1}{sqrt 2}right)=ln(2^{-1/2})=-frac{1}{2}ln(2)$$




        Edit for completion:



        For small values of $cos(x)$ we have $$ln(cos x)approx-frac{1}{2}x^2-frac{1}{12}x^4tag{1}$$ When $x=frac{pi}{4}$ we get $$-frac{1}{2}ln(2)=ln(cos frac{pi}{4})approx-frac{1}{2}left(frac{pi}{4}right)^2-frac{1}{12}left(frac{pi}{4}right)^4\ln(2)approxfrac{pi^2}{16}left(1-frac{pi^2}{96}right)$$






        share|cite|improve this answer














        Hint:



        Since $cos(frac{pi}{4})=frac{1}{sqrt 2}$, you can write $$lnleft(frac{1}{sqrt 2}right)=ln(2^{-1/2})=-frac{1}{2}ln(2)$$




        Edit for completion:



        For small values of $cos(x)$ we have $$ln(cos x)approx-frac{1}{2}x^2-frac{1}{12}x^4tag{1}$$ When $x=frac{pi}{4}$ we get $$-frac{1}{2}ln(2)=ln(cos frac{pi}{4})approx-frac{1}{2}left(frac{pi}{4}right)^2-frac{1}{12}left(frac{pi}{4}right)^4\ln(2)approxfrac{pi^2}{16}left(1-frac{pi^2}{96}right)$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 21 at 16:58

























        answered Nov 20 at 12:13









        cansomeonehelpmeout

        6,6433834




        6,6433834






























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