What is the difference between simple interest and simple discounting?

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I have been given the following statements:



"Simple interest: $C$ now $equiv (1+in)C$ in $n$ years; $C$ in $n$ years $equiv fracC1+in$ now.



Simple discounting:
$C$ in $n$ years $equiv (1−dn)C$ now; $C$ now $equiv fracC1-dn$ in $n$ years.



Where $d=fraci1+i$"



These statements imply that simple interest and simple discounting are not equivalent, since the statements do not equate to one another. Why is this, and what is the difference between simple interest and simple discounting?



Thanks







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    I have been given the following statements:



    "Simple interest: $C$ now $equiv (1+in)C$ in $n$ years; $C$ in $n$ years $equiv fracC1+in$ now.



    Simple discounting:
    $C$ in $n$ years $equiv (1−dn)C$ now; $C$ now $equiv fracC1-dn$ in $n$ years.



    Where $d=fraci1+i$"



    These statements imply that simple interest and simple discounting are not equivalent, since the statements do not equate to one another. Why is this, and what is the difference between simple interest and simple discounting?



    Thanks







    share|cite|improve this question






















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I have been given the following statements:



      "Simple interest: $C$ now $equiv (1+in)C$ in $n$ years; $C$ in $n$ years $equiv fracC1+in$ now.



      Simple discounting:
      $C$ in $n$ years $equiv (1−dn)C$ now; $C$ now $equiv fracC1-dn$ in $n$ years.



      Where $d=fraci1+i$"



      These statements imply that simple interest and simple discounting are not equivalent, since the statements do not equate to one another. Why is this, and what is the difference between simple interest and simple discounting?



      Thanks







      share|cite|improve this question












      I have been given the following statements:



      "Simple interest: $C$ now $equiv (1+in)C$ in $n$ years; $C$ in $n$ years $equiv fracC1+in$ now.



      Simple discounting:
      $C$ in $n$ years $equiv (1−dn)C$ now; $C$ now $equiv fracC1-dn$ in $n$ years.



      Where $d=fraci1+i$"



      These statements imply that simple interest and simple discounting are not equivalent, since the statements do not equate to one another. Why is this, and what is the difference between simple interest and simple discounting?



      Thanks









      share|cite|improve this question











      share|cite|improve this question




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      asked May 18 '15 at 9:48









      M Smith

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          I think you have the wrong formula for d.



          $C_0c$: present value (simple compound)



          $C_0d$: present value (simple discount)



          $C_nd$: future value (simple discount)



          $C_nc$: future value (simple compound)



          $C_nc=C_0ccdot (1+in)$



          $C_0d=C_ndcdot frac11-dcdot n=C_ndcdot frac11-fracicdot n1+icdot n$



          with $boxedd=fraci1+icdot n$



          Here I have a different expression for d.



          The future value of the simple compound has to be equal to the present value of the simple discount. This is your asked connection.



          $C_0ccdot (1+in)=C_ndcdot frac11-fracicdot n1+icdot n$



          $C_0c$ and $C_nd$ can be cancelled, because they are equal, too.



          $(1+in)= frac11-fracicdot n1+icdot n$



          Multiplying the equation by the denominator of the RHS.



          $(1+in)cdot left( 1-fracicdot n1+icdot n right)=1$



          Now you can proof, if the equation is true.






          share|cite|improve this answer






















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            I think you have the wrong formula for d.



            $C_0c$: present value (simple compound)



            $C_0d$: present value (simple discount)



            $C_nd$: future value (simple discount)



            $C_nc$: future value (simple compound)



            $C_nc=C_0ccdot (1+in)$



            $C_0d=C_ndcdot frac11-dcdot n=C_ndcdot frac11-fracicdot n1+icdot n$



            with $boxedd=fraci1+icdot n$



            Here I have a different expression for d.



            The future value of the simple compound has to be equal to the present value of the simple discount. This is your asked connection.



            $C_0ccdot (1+in)=C_ndcdot frac11-fracicdot n1+icdot n$



            $C_0c$ and $C_nd$ can be cancelled, because they are equal, too.



            $(1+in)= frac11-fracicdot n1+icdot n$



            Multiplying the equation by the denominator of the RHS.



            $(1+in)cdot left( 1-fracicdot n1+icdot n right)=1$



            Now you can proof, if the equation is true.






            share|cite|improve this answer


























              up vote
              0
              down vote













              I think you have the wrong formula for d.



              $C_0c$: present value (simple compound)



              $C_0d$: present value (simple discount)



              $C_nd$: future value (simple discount)



              $C_nc$: future value (simple compound)



              $C_nc=C_0ccdot (1+in)$



              $C_0d=C_ndcdot frac11-dcdot n=C_ndcdot frac11-fracicdot n1+icdot n$



              with $boxedd=fraci1+icdot n$



              Here I have a different expression for d.



              The future value of the simple compound has to be equal to the present value of the simple discount. This is your asked connection.



              $C_0ccdot (1+in)=C_ndcdot frac11-fracicdot n1+icdot n$



              $C_0c$ and $C_nd$ can be cancelled, because they are equal, too.



              $(1+in)= frac11-fracicdot n1+icdot n$



              Multiplying the equation by the denominator of the RHS.



              $(1+in)cdot left( 1-fracicdot n1+icdot n right)=1$



              Now you can proof, if the equation is true.






              share|cite|improve this answer
























                up vote
                0
                down vote










                up vote
                0
                down vote









                I think you have the wrong formula for d.



                $C_0c$: present value (simple compound)



                $C_0d$: present value (simple discount)



                $C_nd$: future value (simple discount)



                $C_nc$: future value (simple compound)



                $C_nc=C_0ccdot (1+in)$



                $C_0d=C_ndcdot frac11-dcdot n=C_ndcdot frac11-fracicdot n1+icdot n$



                with $boxedd=fraci1+icdot n$



                Here I have a different expression for d.



                The future value of the simple compound has to be equal to the present value of the simple discount. This is your asked connection.



                $C_0ccdot (1+in)=C_ndcdot frac11-fracicdot n1+icdot n$



                $C_0c$ and $C_nd$ can be cancelled, because they are equal, too.



                $(1+in)= frac11-fracicdot n1+icdot n$



                Multiplying the equation by the denominator of the RHS.



                $(1+in)cdot left( 1-fracicdot n1+icdot n right)=1$



                Now you can proof, if the equation is true.






                share|cite|improve this answer














                I think you have the wrong formula for d.



                $C_0c$: present value (simple compound)



                $C_0d$: present value (simple discount)



                $C_nd$: future value (simple discount)



                $C_nc$: future value (simple compound)



                $C_nc=C_0ccdot (1+in)$



                $C_0d=C_ndcdot frac11-dcdot n=C_ndcdot frac11-fracicdot n1+icdot n$



                with $boxedd=fraci1+icdot n$



                Here I have a different expression for d.



                The future value of the simple compound has to be equal to the present value of the simple discount. This is your asked connection.



                $C_0ccdot (1+in)=C_ndcdot frac11-fracicdot n1+icdot n$



                $C_0c$ and $C_nd$ can be cancelled, because they are equal, too.



                $(1+in)= frac11-fracicdot n1+icdot n$



                Multiplying the equation by the denominator of the RHS.



                $(1+in)cdot left( 1-fracicdot n1+icdot n right)=1$



                Now you can proof, if the equation is true.







                share|cite|improve this answer














                share|cite|improve this answer



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                edited May 18 '15 at 23:57

























                answered May 18 '15 at 23:25









                callculus

                16.6k31427




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