Where is the error in my attempted derivation of the result $i^{(p)} = p[(1+i)^frac{1}{p}-1]$?











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Suppose $i^{(p)}$ denotes the nominal interest rate, payable as $frac{i^{(p)}}{p}$, at a rate of $p$ times per period, and assume that $i$ is the effective interest rate (payable once per period) that is equivalent.



Then, by considering the accumulated value of the interest payments made under the nominal rate, we deduce that



$$
frac{i^{(p)}}{p} + frac{i^{(p)}}{p}(1+i) + frac{i^{(p)}}{p}(1+i)^2 + dots + frac{i^{(p)}}{p}(1+i)^frac{p-1}{p} = i hspace{10mm} (*)
$$



Then, by using the rule
$$
sum_{k=0}^{n-1}ar^k = a cdot frac{(1-r^n)}{1-r}
$$

and taking
$$
a = frac{i^{(p)}}{p}, hspace{10mm} r = (1+i)^frac{1}{p}, hspace{10mm} n = p
$$

we can express $(*)$ as
$$
frac{i^{(p)}}{p} cdot frac{1 - left( (1+i)^frac{1}{p} right)^p}{1 - (1+i)^frac{1}{p}} = frac{i^{(p)}}{p} cdot frac{i}{1 - (1+i)^frac{1}{p}} = i
$$

Thus, by dividing through by $i$ we have
$$
frac{i^{(p)}}{p[1 - (1+i)^frac{1}{p}]} = 1
$$

Giving the result
$$
i^{(p)} = p[1 - (1+i)^frac{1}{p}]
$$



This result is incorrect, since the correct result should be
$$
i^{(p)} = p[(1+i)^frac{1}{p}-1]
$$



Where in my attempted derivation did I go wrong?










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    up vote
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    down vote

    favorite












    Suppose $i^{(p)}$ denotes the nominal interest rate, payable as $frac{i^{(p)}}{p}$, at a rate of $p$ times per period, and assume that $i$ is the effective interest rate (payable once per period) that is equivalent.



    Then, by considering the accumulated value of the interest payments made under the nominal rate, we deduce that



    $$
    frac{i^{(p)}}{p} + frac{i^{(p)}}{p}(1+i) + frac{i^{(p)}}{p}(1+i)^2 + dots + frac{i^{(p)}}{p}(1+i)^frac{p-1}{p} = i hspace{10mm} (*)
    $$



    Then, by using the rule
    $$
    sum_{k=0}^{n-1}ar^k = a cdot frac{(1-r^n)}{1-r}
    $$

    and taking
    $$
    a = frac{i^{(p)}}{p}, hspace{10mm} r = (1+i)^frac{1}{p}, hspace{10mm} n = p
    $$

    we can express $(*)$ as
    $$
    frac{i^{(p)}}{p} cdot frac{1 - left( (1+i)^frac{1}{p} right)^p}{1 - (1+i)^frac{1}{p}} = frac{i^{(p)}}{p} cdot frac{i}{1 - (1+i)^frac{1}{p}} = i
    $$

    Thus, by dividing through by $i$ we have
    $$
    frac{i^{(p)}}{p[1 - (1+i)^frac{1}{p}]} = 1
    $$

    Giving the result
    $$
    i^{(p)} = p[1 - (1+i)^frac{1}{p}]
    $$



    This result is incorrect, since the correct result should be
    $$
    i^{(p)} = p[(1+i)^frac{1}{p}-1]
    $$



    Where in my attempted derivation did I go wrong?










    share|cite|improve this question
























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Suppose $i^{(p)}$ denotes the nominal interest rate, payable as $frac{i^{(p)}}{p}$, at a rate of $p$ times per period, and assume that $i$ is the effective interest rate (payable once per period) that is equivalent.



      Then, by considering the accumulated value of the interest payments made under the nominal rate, we deduce that



      $$
      frac{i^{(p)}}{p} + frac{i^{(p)}}{p}(1+i) + frac{i^{(p)}}{p}(1+i)^2 + dots + frac{i^{(p)}}{p}(1+i)^frac{p-1}{p} = i hspace{10mm} (*)
      $$



      Then, by using the rule
      $$
      sum_{k=0}^{n-1}ar^k = a cdot frac{(1-r^n)}{1-r}
      $$

      and taking
      $$
      a = frac{i^{(p)}}{p}, hspace{10mm} r = (1+i)^frac{1}{p}, hspace{10mm} n = p
      $$

      we can express $(*)$ as
      $$
      frac{i^{(p)}}{p} cdot frac{1 - left( (1+i)^frac{1}{p} right)^p}{1 - (1+i)^frac{1}{p}} = frac{i^{(p)}}{p} cdot frac{i}{1 - (1+i)^frac{1}{p}} = i
      $$

      Thus, by dividing through by $i$ we have
      $$
      frac{i^{(p)}}{p[1 - (1+i)^frac{1}{p}]} = 1
      $$

      Giving the result
      $$
      i^{(p)} = p[1 - (1+i)^frac{1}{p}]
      $$



      This result is incorrect, since the correct result should be
      $$
      i^{(p)} = p[(1+i)^frac{1}{p}-1]
      $$



      Where in my attempted derivation did I go wrong?










      share|cite|improve this question













      Suppose $i^{(p)}$ denotes the nominal interest rate, payable as $frac{i^{(p)}}{p}$, at a rate of $p$ times per period, and assume that $i$ is the effective interest rate (payable once per period) that is equivalent.



      Then, by considering the accumulated value of the interest payments made under the nominal rate, we deduce that



      $$
      frac{i^{(p)}}{p} + frac{i^{(p)}}{p}(1+i) + frac{i^{(p)}}{p}(1+i)^2 + dots + frac{i^{(p)}}{p}(1+i)^frac{p-1}{p} = i hspace{10mm} (*)
      $$



      Then, by using the rule
      $$
      sum_{k=0}^{n-1}ar^k = a cdot frac{(1-r^n)}{1-r}
      $$

      and taking
      $$
      a = frac{i^{(p)}}{p}, hspace{10mm} r = (1+i)^frac{1}{p}, hspace{10mm} n = p
      $$

      we can express $(*)$ as
      $$
      frac{i^{(p)}}{p} cdot frac{1 - left( (1+i)^frac{1}{p} right)^p}{1 - (1+i)^frac{1}{p}} = frac{i^{(p)}}{p} cdot frac{i}{1 - (1+i)^frac{1}{p}} = i
      $$

      Thus, by dividing through by $i$ we have
      $$
      frac{i^{(p)}}{p[1 - (1+i)^frac{1}{p}]} = 1
      $$

      Giving the result
      $$
      i^{(p)} = p[1 - (1+i)^frac{1}{p}]
      $$



      This result is incorrect, since the correct result should be
      $$
      i^{(p)} = p[(1+i)^frac{1}{p}-1]
      $$



      Where in my attempted derivation did I go wrong?







      finance






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      asked Nov 23 at 15:33









      J. Chapman

      132




      132






















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          $$
          frac{i^{(p)}}{p} cdot frac{1 - left( (1+i)^frac{1}{p} right)^p}{1 - (1+i)^frac{1}{p}} = frac{i^{(p)}}{p} cdot frac{-i}{1 - (1+i)^frac{1}{p}} = i
          $$






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            up vote
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            You've forgotten a minus sign:



            $$
            frac{i^{(p)}}{p} cdot frac{1 - left( (1+i)^frac{1}{p} right)^p}{1 - (1+i)^frac{1}{p}} = frac{i^{(p)}}{p} cdot frac{-i}{1 - (1+i)^frac{1}{p}} = i
            $$






            share|cite|improve this answer

























              up vote
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              down vote



              accepted










              You've forgotten a minus sign:



              $$
              frac{i^{(p)}}{p} cdot frac{1 - left( (1+i)^frac{1}{p} right)^p}{1 - (1+i)^frac{1}{p}} = frac{i^{(p)}}{p} cdot frac{-i}{1 - (1+i)^frac{1}{p}} = i
              $$






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                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                You've forgotten a minus sign:



                $$
                frac{i^{(p)}}{p} cdot frac{1 - left( (1+i)^frac{1}{p} right)^p}{1 - (1+i)^frac{1}{p}} = frac{i^{(p)}}{p} cdot frac{-i}{1 - (1+i)^frac{1}{p}} = i
                $$






                share|cite|improve this answer












                You've forgotten a minus sign:



                $$
                frac{i^{(p)}}{p} cdot frac{1 - left( (1+i)^frac{1}{p} right)^p}{1 - (1+i)^frac{1}{p}} = frac{i^{(p)}}{p} cdot frac{-i}{1 - (1+i)^frac{1}{p}} = i
                $$







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                answered Nov 23 at 20:25









                alexjo

                12.2k1329




                12.2k1329






























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