How to find the $pi$? [duplicate]

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We know as we increase the number of sides in regular polygon, after infinite repetition it will give us a circle. So, is there any way to find a function which approaches to value of $pi$ when we consider $limlimits_nto infty$?




proof-explanation circle pi




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marked as duplicate by Did, Jyrki Lahtonen, Claude Leibovici, N. F. Taussig, mfl Aug 23 at 8:55


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    There are a lot of such functions. Do you seek one in general or one that uses your polygon approach?
    – James
    Aug 23 at 7:59














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This question already has an answer here:



  • Calculating pi manually

    12 answers



We know as we increase the number of sides in regular polygon, after infinite repetition it will give us a circle. So, is there any way to find a function which approaches to value of $pi$ when we consider $limlimits_nto infty$?




proof-explanation circle pi




share|cite|improve this question














marked as duplicate by Did, Jyrki Lahtonen, Claude Leibovici, N. F. Taussig, mfl Aug 23 at 8:55


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    There are a lot of such functions. Do you seek one in general or one that uses your polygon approach?
    – James
    Aug 23 at 7:59












up vote
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favorite









up vote
1
down vote

favorite












This question already has an answer here:



  • Calculating pi manually

    12 answers



We know as we increase the number of sides in regular polygon, after infinite repetition it will give us a circle. So, is there any way to find a function which approaches to value of $pi$ when we consider $limlimits_nto infty$?




proof-explanation circle pi




share|cite|improve this question















This question already has an answer here:



  • Calculating pi manually

    12 answers



We know as we increase the number of sides in regular polygon, after infinite repetition it will give us a circle. So, is there any way to find a function which approaches to value of $pi$ when we consider $limlimits_nto infty$?





This question already has an answer here:



  • Calculating pi manually

    12 answers





proof-explanation circle pi





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 23 at 8:29









miracle173

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7,17922247










asked Aug 23 at 7:39









user422780

164




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marked as duplicate by Did, Jyrki Lahtonen, Claude Leibovici, N. F. Taussig, mfl Aug 23 at 8:55


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Did, Jyrki Lahtonen, Claude Leibovici, N. F. Taussig, mfl Aug 23 at 8:55


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









  • 1




    There are a lot of such functions. Do you seek one in general or one that uses your polygon approach?
    – James
    Aug 23 at 7:59












  • 1




    There are a lot of such functions. Do you seek one in general or one that uses your polygon approach?
    – James
    Aug 23 at 7:59







1




1




There are a lot of such functions. Do you seek one in general or one that uses your polygon approach?
– James
Aug 23 at 7:59




There are a lot of such functions. Do you seek one in general or one that uses your polygon approach?
– James
Aug 23 at 7:59










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The area of a regular $n$-gon circumscribed by the circle of radius $r$ is
$$A_n = frac12nr^2sinleft(frac2pinright).$$ You can get the proof here. Taking limit, we will get circle when $ntoinfty$. We can use this formula for $pi$, $$pi=fractextArea of circle with radius rtextradius^2=frac1r^2cdot lim_ntoinftyA_n=lim_nto inftyfracn2sinleft(frac2pinright).$$



Using this result you can tend to $pi$ putting bigger and bigger values of $n$ in $fracn2sinleft(frac2pinright)$. Here I have plotted $pi -fracx2sinleft(frac2pixright)$, for first few values of $x$ very fast, then slowly approaches $0$, means $fracn2sinleft(frac2pinright)topi$ very slowly but, the methods explained in the given link in comments tend $pi$ much faster than this.






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    1 Answer
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    up vote
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    The area of a regular $n$-gon circumscribed by the circle of radius $r$ is
    $$A_n = frac12nr^2sinleft(frac2pinright).$$ You can get the proof here. Taking limit, we will get circle when $ntoinfty$. We can use this formula for $pi$, $$pi=fractextArea of circle with radius rtextradius^2=frac1r^2cdot lim_ntoinftyA_n=lim_nto inftyfracn2sinleft(frac2pinright).$$



    Using this result you can tend to $pi$ putting bigger and bigger values of $n$ in $fracn2sinleft(frac2pinright)$. Here I have plotted $pi -fracx2sinleft(frac2pixright)$, for first few values of $x$ very fast, then slowly approaches $0$, means $fracn2sinleft(frac2pinright)topi$ very slowly but, the methods explained in the given link in comments tend $pi$ much faster than this.






    share|cite|improve this answer


























      up vote
      4
      down vote













      The area of a regular $n$-gon circumscribed by the circle of radius $r$ is
      $$A_n = frac12nr^2sinleft(frac2pinright).$$ You can get the proof here. Taking limit, we will get circle when $ntoinfty$. We can use this formula for $pi$, $$pi=fractextArea of circle with radius rtextradius^2=frac1r^2cdot lim_ntoinftyA_n=lim_nto inftyfracn2sinleft(frac2pinright).$$



      Using this result you can tend to $pi$ putting bigger and bigger values of $n$ in $fracn2sinleft(frac2pinright)$. Here I have plotted $pi -fracx2sinleft(frac2pixright)$, for first few values of $x$ very fast, then slowly approaches $0$, means $fracn2sinleft(frac2pinright)topi$ very slowly but, the methods explained in the given link in comments tend $pi$ much faster than this.






      share|cite|improve this answer
























        up vote
        4
        down vote










        up vote
        4
        down vote









        The area of a regular $n$-gon circumscribed by the circle of radius $r$ is
        $$A_n = frac12nr^2sinleft(frac2pinright).$$ You can get the proof here. Taking limit, we will get circle when $ntoinfty$. We can use this formula for $pi$, $$pi=fractextArea of circle with radius rtextradius^2=frac1r^2cdot lim_ntoinftyA_n=lim_nto inftyfracn2sinleft(frac2pinright).$$



        Using this result you can tend to $pi$ putting bigger and bigger values of $n$ in $fracn2sinleft(frac2pinright)$. Here I have plotted $pi -fracx2sinleft(frac2pixright)$, for first few values of $x$ very fast, then slowly approaches $0$, means $fracn2sinleft(frac2pinright)topi$ very slowly but, the methods explained in the given link in comments tend $pi$ much faster than this.






        share|cite|improve this answer














        The area of a regular $n$-gon circumscribed by the circle of radius $r$ is
        $$A_n = frac12nr^2sinleft(frac2pinright).$$ You can get the proof here. Taking limit, we will get circle when $ntoinfty$. We can use this formula for $pi$, $$pi=fractextArea of circle with radius rtextradius^2=frac1r^2cdot lim_ntoinftyA_n=lim_nto inftyfracn2sinleft(frac2pinright).$$



        Using this result you can tend to $pi$ putting bigger and bigger values of $n$ in $fracn2sinleft(frac2pinright)$. Here I have plotted $pi -fracx2sinleft(frac2pixright)$, for first few values of $x$ very fast, then slowly approaches $0$, means $fracn2sinleft(frac2pinright)topi$ very slowly but, the methods explained in the given link in comments tend $pi$ much faster than this.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 23 at 8:20









        user1729

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        16.8k64082










        answered Aug 23 at 7:59









        tarit goswami

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