How to calculate compounding frequency for an loan (Using Lambert-W Function)












4












$begingroup$



A bond will become worth 500 dollars when it becomes due in 5 years. If the bond was purchased today for 450 dollars at 2.13% per year, determine how frequently the interest was compounded.




I tried to solve the question with trial and error, but is there a better way to solve it?



The function looks like this:
$$500 = 450(1 +{0.0213over x})^{5x}$$






exponential-function lambert-w







share|cite|improve this question











$endgroup$

















    4












    $begingroup$



    A bond will become worth 500 dollars when it becomes due in 5 years. If the bond was purchased today for 450 dollars at 2.13% per year, determine how frequently the interest was compounded.




    I tried to solve the question with trial and error, but is there a better way to solve it?



    The function looks like this:
    $$500 = 450(1 +{0.0213over x})^{5x}$$






    exponential-function lambert-w







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$



      A bond will become worth 500 dollars when it becomes due in 5 years. If the bond was purchased today for 450 dollars at 2.13% per year, determine how frequently the interest was compounded.




      I tried to solve the question with trial and error, but is there a better way to solve it?



      The function looks like this:
      $$500 = 450(1 +{0.0213over x})^{5x}$$






      exponential-function lambert-w







      share|cite|improve this question











      $endgroup$





      A bond will become worth 500 dollars when it becomes due in 5 years. If the bond was purchased today for 450 dollars at 2.13% per year, determine how frequently the interest was compounded.




      I tried to solve the question with trial and error, but is there a better way to solve it?



      The function looks like this:
      $$500 = 450(1 +{0.0213over x})^{5x}$$






      exponential-function lambert-w




      exponential-function lambert-w






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 15 at 11:58







      Kevin

















      asked Jan 14 at 1:59









      KevinKevin

      496




      496






















          1 Answer
          1






          active

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          1












          $begingroup$

          Consider the more general case of the equation
          $$a=left(1+frac{b}{x}right)^{c x}$$ The solution of it is given in terms of Lambert function. I shall skip all intermediate steps and just provide the result; it write
          $$x=-frac{b log (a)}{log (a)+b, c, W_{-1}(-t)}qquad text{where} qquad t=frac{ a^{-frac{1}{b, c}}}{b ,c}log (a)$$
          Just remember than any equation which can write or rewrite as
          $$A+B x+C log(D+Ex)=0$$ has solution(s) i terms of Lambert function.



          For your specific problem, since graphing, you see that the solution is close to $x=1$, you could have done the following : take logarithms of both sides and use Taylor expansion around $x=1$. Working with whole numbers you would get
          $$left(5 log left(frac{10213}{10000}right)-log
          left(frac{10}{9}right)right)+(x-1) left(5 log
          left(frac{10213}{10000}right)-frac{1065}{10213}right)+Oleft((x-1)^2right)$$           and ignoring the higher order terms, this would give, as an approximation,
          $$x=1-frac{10213 left(5 log left(frac{10213}{10000}right)-log
          left(frac{10}{9}right)right)}{5 left(10213 log
          left(frac{10213}{10000}right)-213right)}approx 0.98086$$           while the exact solution is $0.98121$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            The explanation is really good, wish I could give you an upvote if I am reputation 15
            $endgroup$
            – Kevin
            Jan 14 at 20:29










          • $begingroup$
            @KevinXu. You are very welcome ! I upvoted you question because it reveals an intelligent attitude (curiosity). Cheers :-)
            $endgroup$
            – Claude Leibovici
            Jan 15 at 3:36











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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Consider the more general case of the equation
          $$a=left(1+frac{b}{x}right)^{c x}$$ The solution of it is given in terms of Lambert function. I shall skip all intermediate steps and just provide the result; it write
          $$x=-frac{b log (a)}{log (a)+b, c, W_{-1}(-t)}qquad text{where} qquad t=frac{ a^{-frac{1}{b, c}}}{b ,c}log (a)$$
          Just remember than any equation which can write or rewrite as
          $$A+B x+C log(D+Ex)=0$$ has solution(s) i terms of Lambert function.



          For your specific problem, since graphing, you see that the solution is close to $x=1$, you could have done the following : take logarithms of both sides and use Taylor expansion around $x=1$. Working with whole numbers you would get
          $$left(5 log left(frac{10213}{10000}right)-log
          left(frac{10}{9}right)right)+(x-1) left(5 log
          left(frac{10213}{10000}right)-frac{1065}{10213}right)+Oleft((x-1)^2right)$$           and ignoring the higher order terms, this would give, as an approximation,
          $$x=1-frac{10213 left(5 log left(frac{10213}{10000}right)-log
          left(frac{10}{9}right)right)}{5 left(10213 log
          left(frac{10213}{10000}right)-213right)}approx 0.98086$$           while the exact solution is $0.98121$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            The explanation is really good, wish I could give you an upvote if I am reputation 15
            $endgroup$
            – Kevin
            Jan 14 at 20:29










          • $begingroup$
            @KevinXu. You are very welcome ! I upvoted you question because it reveals an intelligent attitude (curiosity). Cheers :-)
            $endgroup$
            – Claude Leibovici
            Jan 15 at 3:36
















          1












          $begingroup$

          Consider the more general case of the equation
          $$a=left(1+frac{b}{x}right)^{c x}$$ The solution of it is given in terms of Lambert function. I shall skip all intermediate steps and just provide the result; it write
          $$x=-frac{b log (a)}{log (a)+b, c, W_{-1}(-t)}qquad text{where} qquad t=frac{ a^{-frac{1}{b, c}}}{b ,c}log (a)$$
          Just remember than any equation which can write or rewrite as
          $$A+B x+C log(D+Ex)=0$$ has solution(s) i terms of Lambert function.



          For your specific problem, since graphing, you see that the solution is close to $x=1$, you could have done the following : take logarithms of both sides and use Taylor expansion around $x=1$. Working with whole numbers you would get
          $$left(5 log left(frac{10213}{10000}right)-log
          left(frac{10}{9}right)right)+(x-1) left(5 log
          left(frac{10213}{10000}right)-frac{1065}{10213}right)+Oleft((x-1)^2right)$$           and ignoring the higher order terms, this would give, as an approximation,
          $$x=1-frac{10213 left(5 log left(frac{10213}{10000}right)-log
          left(frac{10}{9}right)right)}{5 left(10213 log
          left(frac{10213}{10000}right)-213right)}approx 0.98086$$           while the exact solution is $0.98121$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            The explanation is really good, wish I could give you an upvote if I am reputation 15
            $endgroup$
            – Kevin
            Jan 14 at 20:29










          • $begingroup$
            @KevinXu. You are very welcome ! I upvoted you question because it reveals an intelligent attitude (curiosity). Cheers :-)
            $endgroup$
            – Claude Leibovici
            Jan 15 at 3:36














          1












          1








          1





          $begingroup$

          Consider the more general case of the equation
          $$a=left(1+frac{b}{x}right)^{c x}$$ The solution of it is given in terms of Lambert function. I shall skip all intermediate steps and just provide the result; it write
          $$x=-frac{b log (a)}{log (a)+b, c, W_{-1}(-t)}qquad text{where} qquad t=frac{ a^{-frac{1}{b, c}}}{b ,c}log (a)$$
          Just remember than any equation which can write or rewrite as
          $$A+B x+C log(D+Ex)=0$$ has solution(s) i terms of Lambert function.



          For your specific problem, since graphing, you see that the solution is close to $x=1$, you could have done the following : take logarithms of both sides and use Taylor expansion around $x=1$. Working with whole numbers you would get
          $$left(5 log left(frac{10213}{10000}right)-log
          left(frac{10}{9}right)right)+(x-1) left(5 log
          left(frac{10213}{10000}right)-frac{1065}{10213}right)+Oleft((x-1)^2right)$$           and ignoring the higher order terms, this would give, as an approximation,
          $$x=1-frac{10213 left(5 log left(frac{10213}{10000}right)-log
          left(frac{10}{9}right)right)}{5 left(10213 log
          left(frac{10213}{10000}right)-213right)}approx 0.98086$$           while the exact solution is $0.98121$.






          share|cite|improve this answer











          $endgroup$



          Consider the more general case of the equation
          $$a=left(1+frac{b}{x}right)^{c x}$$ The solution of it is given in terms of Lambert function. I shall skip all intermediate steps and just provide the result; it write
          $$x=-frac{b log (a)}{log (a)+b, c, W_{-1}(-t)}qquad text{where} qquad t=frac{ a^{-frac{1}{b, c}}}{b ,c}log (a)$$
          Just remember than any equation which can write or rewrite as
          $$A+B x+C log(D+Ex)=0$$ has solution(s) i terms of Lambert function.



          For your specific problem, since graphing, you see that the solution is close to $x=1$, you could have done the following : take logarithms of both sides and use Taylor expansion around $x=1$. Working with whole numbers you would get
          $$left(5 log left(frac{10213}{10000}right)-log
          left(frac{10}{9}right)right)+(x-1) left(5 log
          left(frac{10213}{10000}right)-frac{1065}{10213}right)+Oleft((x-1)^2right)$$           and ignoring the higher order terms, this would give, as an approximation,
          $$x=1-frac{10213 left(5 log left(frac{10213}{10000}right)-log
          left(frac{10}{9}right)right)}{5 left(10213 log
          left(frac{10213}{10000}right)-213right)}approx 0.98086$$           while the exact solution is $0.98121$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 14 at 6:32

























          answered Jan 14 at 6:19









          Claude LeiboviciClaude Leibovici

          124k1158135




          124k1158135












          • $begingroup$
            The explanation is really good, wish I could give you an upvote if I am reputation 15
            $endgroup$
            – Kevin
            Jan 14 at 20:29










          • $begingroup$
            @KevinXu. You are very welcome ! I upvoted you question because it reveals an intelligent attitude (curiosity). Cheers :-)
            $endgroup$
            – Claude Leibovici
            Jan 15 at 3:36


















          • $begingroup$
            The explanation is really good, wish I could give you an upvote if I am reputation 15
            $endgroup$
            – Kevin
            Jan 14 at 20:29










          • $begingroup$
            @KevinXu. You are very welcome ! I upvoted you question because it reveals an intelligent attitude (curiosity). Cheers :-)
            $endgroup$
            – Claude Leibovici
            Jan 15 at 3:36
















          $begingroup$
          The explanation is really good, wish I could give you an upvote if I am reputation 15
          $endgroup$
          – Kevin
          Jan 14 at 20:29




          $begingroup$
          The explanation is really good, wish I could give you an upvote if I am reputation 15
          $endgroup$
          – Kevin
          Jan 14 at 20:29












          $begingroup$
          @KevinXu. You are very welcome ! I upvoted you question because it reveals an intelligent attitude (curiosity). Cheers :-)
          $endgroup$
          – Claude Leibovici
          Jan 15 at 3:36




          $begingroup$
          @KevinXu. You are very welcome ! I upvoted you question because it reveals an intelligent attitude (curiosity). Cheers :-)
          $endgroup$
          – Claude Leibovici
          Jan 15 at 3:36


















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