Accumulated value of force of interest












0












$begingroup$



For the period from time 0 to time 2, the force of interest is defined as follows:
$$text{force of interest at }(t) = begin{cases}
0.05 & 0 lt t le 1\
0.05+0.02(t-1) & 1lt t le 2
end{cases}$$
$10,!000$ is invested at time $0$. Find the accumulated value at time $1$ and at time $2$.




I got the correct answer for accumulated value at time $1$ using the accumulation function
$$A(t)=A(0)expleft(int_0^Ttext{force of interest} mathrm dtright),$$
by taking the integral of $0.05$ at $1$ and $0$, and then I ended up with $A(t)=10000(e^{0.05-0})$, which gave me $10512.7109$, the correct answer.



However, when I try the same steps but with $T=2$, and the only thing changing is that I take the integral of the second equation and then plug in the same formula, I am getting the wrong answer. The answer in the book is $11162.78$. Any help would be appreciated.






integration finance







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$endgroup$












  • $begingroup$
    The integral should include both the first and second equation. You have to break the integral up as $int_0^2 = int_0^1 + int_1^2$ using the appropriate part of the definition in each integral. Generally speaking, that's the only way to integrate a function that is defined piecewise if the interval of integration laps over different pieces of the definition.
    $endgroup$
    – MPW
    Oct 6 '15 at 20:21


















0












$begingroup$



For the period from time 0 to time 2, the force of interest is defined as follows:
$$text{force of interest at }(t) = begin{cases}
0.05 & 0 lt t le 1\
0.05+0.02(t-1) & 1lt t le 2
end{cases}$$
$10,!000$ is invested at time $0$. Find the accumulated value at time $1$ and at time $2$.




I got the correct answer for accumulated value at time $1$ using the accumulation function
$$A(t)=A(0)expleft(int_0^Ttext{force of interest} mathrm dtright),$$
by taking the integral of $0.05$ at $1$ and $0$, and then I ended up with $A(t)=10000(e^{0.05-0})$, which gave me $10512.7109$, the correct answer.



However, when I try the same steps but with $T=2$, and the only thing changing is that I take the integral of the second equation and then plug in the same formula, I am getting the wrong answer. The answer in the book is $11162.78$. Any help would be appreciated.






integration finance







share|cite|improve this question











$endgroup$












  • $begingroup$
    The integral should include both the first and second equation. You have to break the integral up as $int_0^2 = int_0^1 + int_1^2$ using the appropriate part of the definition in each integral. Generally speaking, that's the only way to integrate a function that is defined piecewise if the interval of integration laps over different pieces of the definition.
    $endgroup$
    – MPW
    Oct 6 '15 at 20:21
















0












0








0





$begingroup$



For the period from time 0 to time 2, the force of interest is defined as follows:
$$text{force of interest at }(t) = begin{cases}
0.05 & 0 lt t le 1\
0.05+0.02(t-1) & 1lt t le 2
end{cases}$$
$10,!000$ is invested at time $0$. Find the accumulated value at time $1$ and at time $2$.




I got the correct answer for accumulated value at time $1$ using the accumulation function
$$A(t)=A(0)expleft(int_0^Ttext{force of interest} mathrm dtright),$$
by taking the integral of $0.05$ at $1$ and $0$, and then I ended up with $A(t)=10000(e^{0.05-0})$, which gave me $10512.7109$, the correct answer.



However, when I try the same steps but with $T=2$, and the only thing changing is that I take the integral of the second equation and then plug in the same formula, I am getting the wrong answer. The answer in the book is $11162.78$. Any help would be appreciated.






integration finance







share|cite|improve this question











$endgroup$





For the period from time 0 to time 2, the force of interest is defined as follows:
$$text{force of interest at }(t) = begin{cases}
0.05 & 0 lt t le 1\
0.05+0.02(t-1) & 1lt t le 2
end{cases}$$
$10,!000$ is invested at time $0$. Find the accumulated value at time $1$ and at time $2$.




I got the correct answer for accumulated value at time $1$ using the accumulation function
$$A(t)=A(0)expleft(int_0^Ttext{force of interest} mathrm dtright),$$
by taking the integral of $0.05$ at $1$ and $0$, and then I ended up with $A(t)=10000(e^{0.05-0})$, which gave me $10512.7109$, the correct answer.



However, when I try the same steps but with $T=2$, and the only thing changing is that I take the integral of the second equation and then plug in the same formula, I am getting the wrong answer. The answer in the book is $11162.78$. Any help would be appreciated.






integration finance




integration finance






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edited Oct 6 '15 at 20:56









rubik

6,76132661




6,76132661










asked Oct 6 '15 at 20:05









Alex RochAlex Roch

11




11












  • $begingroup$
    The integral should include both the first and second equation. You have to break the integral up as $int_0^2 = int_0^1 + int_1^2$ using the appropriate part of the definition in each integral. Generally speaking, that's the only way to integrate a function that is defined piecewise if the interval of integration laps over different pieces of the definition.
    $endgroup$
    – MPW
    Oct 6 '15 at 20:21




















  • $begingroup$
    The integral should include both the first and second equation. You have to break the integral up as $int_0^2 = int_0^1 + int_1^2$ using the appropriate part of the definition in each integral. Generally speaking, that's the only way to integrate a function that is defined piecewise if the interval of integration laps over different pieces of the definition.
    $endgroup$
    – MPW
    Oct 6 '15 at 20:21


















$begingroup$
The integral should include both the first and second equation. You have to break the integral up as $int_0^2 = int_0^1 + int_1^2$ using the appropriate part of the definition in each integral. Generally speaking, that's the only way to integrate a function that is defined piecewise if the interval of integration laps over different pieces of the definition.
$endgroup$
– MPW
Oct 6 '15 at 20:21






$begingroup$
The integral should include both the first and second equation. You have to break the integral up as $int_0^2 = int_0^1 + int_1^2$ using the appropriate part of the definition in each integral. Generally speaking, that's the only way to integrate a function that is defined piecewise if the interval of integration laps over different pieces of the definition.
$endgroup$
– MPW
Oct 6 '15 at 20:21












2 Answers
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oldest

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$begingroup$

If you are getting the wrong answer it means that you're not integrating correctly. When computing
$$int_0^2text{force of interest} mathrm dt$$
you have to remember that the force of interest is defined by a piecewise function, and therefore you must split the integral. We get (writing $f(t)$ instead of $text{force of interest}$):
$$begin{align}
int_0^2 f(t),mathrm dt &= int_0^1 f(t),mathrm dt + int_1^2 f(t),mathrm dt =\
&= int_0^1 (0.05),mathrm dt + int_1^2left(0.05 + 0.02(t - 1)right),mathrm dt =\
&= 0.11
end{align}$$



And indeed, $10000e^{0.11} = 11162.78$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Let be the force of interest defined by
    $$
    delta(t) = begin{cases}
    0.05 & 0 lt t le 1\
    0.05+0.02(t-1) & 1lt t le 2
    end{cases}
    $$
    Integrating $frac{A'(t)}{A(t)}=delta(t)$ from $t_1$ to $t_2$ we have
    $$
    A(t_2)=A(t_1)mathrm e^{int_{t_1}^{t_2}delta(u)mathrm d u}
    $$
    So we have
    $$
    A(1)=A(0)expleft(int_{0}^{1}0.05,mathrm d uright)=A(0)mathrm e^{0.05}=10,000,mathrm e^{0.05}approx 10,512.71
    $$
    where $A(0)=10,000$.



    And
    $$
    A(2)=A(1)expleft(int_{1}^{2}[0.05+0.02(u-1)],mathrm d uright)=A(1)mathrm e^{0.06}=10,512.71,mathrm e^{0.06}approx 11,162.78
    $$






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      If you are getting the wrong answer it means that you're not integrating correctly. When computing
      $$int_0^2text{force of interest} mathrm dt$$
      you have to remember that the force of interest is defined by a piecewise function, and therefore you must split the integral. We get (writing $f(t)$ instead of $text{force of interest}$):
      $$begin{align}
      int_0^2 f(t),mathrm dt &= int_0^1 f(t),mathrm dt + int_1^2 f(t),mathrm dt =\
      &= int_0^1 (0.05),mathrm dt + int_1^2left(0.05 + 0.02(t - 1)right),mathrm dt =\
      &= 0.11
      end{align}$$



      And indeed, $10000e^{0.11} = 11162.78$.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        If you are getting the wrong answer it means that you're not integrating correctly. When computing
        $$int_0^2text{force of interest} mathrm dt$$
        you have to remember that the force of interest is defined by a piecewise function, and therefore you must split the integral. We get (writing $f(t)$ instead of $text{force of interest}$):
        $$begin{align}
        int_0^2 f(t),mathrm dt &= int_0^1 f(t),mathrm dt + int_1^2 f(t),mathrm dt =\
        &= int_0^1 (0.05),mathrm dt + int_1^2left(0.05 + 0.02(t - 1)right),mathrm dt =\
        &= 0.11
        end{align}$$



        And indeed, $10000e^{0.11} = 11162.78$.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          If you are getting the wrong answer it means that you're not integrating correctly. When computing
          $$int_0^2text{force of interest} mathrm dt$$
          you have to remember that the force of interest is defined by a piecewise function, and therefore you must split the integral. We get (writing $f(t)$ instead of $text{force of interest}$):
          $$begin{align}
          int_0^2 f(t),mathrm dt &= int_0^1 f(t),mathrm dt + int_1^2 f(t),mathrm dt =\
          &= int_0^1 (0.05),mathrm dt + int_1^2left(0.05 + 0.02(t - 1)right),mathrm dt =\
          &= 0.11
          end{align}$$



          And indeed, $10000e^{0.11} = 11162.78$.






          share|cite|improve this answer









          $endgroup$



          If you are getting the wrong answer it means that you're not integrating correctly. When computing
          $$int_0^2text{force of interest} mathrm dt$$
          you have to remember that the force of interest is defined by a piecewise function, and therefore you must split the integral. We get (writing $f(t)$ instead of $text{force of interest}$):
          $$begin{align}
          int_0^2 f(t),mathrm dt &= int_0^1 f(t),mathrm dt + int_1^2 f(t),mathrm dt =\
          &= int_0^1 (0.05),mathrm dt + int_1^2left(0.05 + 0.02(t - 1)right),mathrm dt =\
          &= 0.11
          end{align}$$



          And indeed, $10000e^{0.11} = 11162.78$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 6 '15 at 20:23









          rubikrubik

          6,76132661




          6,76132661























              0












              $begingroup$

              Let be the force of interest defined by
              $$
              delta(t) = begin{cases}
              0.05 & 0 lt t le 1\
              0.05+0.02(t-1) & 1lt t le 2
              end{cases}
              $$
              Integrating $frac{A'(t)}{A(t)}=delta(t)$ from $t_1$ to $t_2$ we have
              $$
              A(t_2)=A(t_1)mathrm e^{int_{t_1}^{t_2}delta(u)mathrm d u}
              $$
              So we have
              $$
              A(1)=A(0)expleft(int_{0}^{1}0.05,mathrm d uright)=A(0)mathrm e^{0.05}=10,000,mathrm e^{0.05}approx 10,512.71
              $$
              where $A(0)=10,000$.



              And
              $$
              A(2)=A(1)expleft(int_{1}^{2}[0.05+0.02(u-1)],mathrm d uright)=A(1)mathrm e^{0.06}=10,512.71,mathrm e^{0.06}approx 11,162.78
              $$






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Let be the force of interest defined by
                $$
                delta(t) = begin{cases}
                0.05 & 0 lt t le 1\
                0.05+0.02(t-1) & 1lt t le 2
                end{cases}
                $$
                Integrating $frac{A'(t)}{A(t)}=delta(t)$ from $t_1$ to $t_2$ we have
                $$
                A(t_2)=A(t_1)mathrm e^{int_{t_1}^{t_2}delta(u)mathrm d u}
                $$
                So we have
                $$
                A(1)=A(0)expleft(int_{0}^{1}0.05,mathrm d uright)=A(0)mathrm e^{0.05}=10,000,mathrm e^{0.05}approx 10,512.71
                $$
                where $A(0)=10,000$.



                And
                $$
                A(2)=A(1)expleft(int_{1}^{2}[0.05+0.02(u-1)],mathrm d uright)=A(1)mathrm e^{0.06}=10,512.71,mathrm e^{0.06}approx 11,162.78
                $$






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Let be the force of interest defined by
                  $$
                  delta(t) = begin{cases}
                  0.05 & 0 lt t le 1\
                  0.05+0.02(t-1) & 1lt t le 2
                  end{cases}
                  $$
                  Integrating $frac{A'(t)}{A(t)}=delta(t)$ from $t_1$ to $t_2$ we have
                  $$
                  A(t_2)=A(t_1)mathrm e^{int_{t_1}^{t_2}delta(u)mathrm d u}
                  $$
                  So we have
                  $$
                  A(1)=A(0)expleft(int_{0}^{1}0.05,mathrm d uright)=A(0)mathrm e^{0.05}=10,000,mathrm e^{0.05}approx 10,512.71
                  $$
                  where $A(0)=10,000$.



                  And
                  $$
                  A(2)=A(1)expleft(int_{1}^{2}[0.05+0.02(u-1)],mathrm d uright)=A(1)mathrm e^{0.06}=10,512.71,mathrm e^{0.06}approx 11,162.78
                  $$






                  share|cite|improve this answer











                  $endgroup$



                  Let be the force of interest defined by
                  $$
                  delta(t) = begin{cases}
                  0.05 & 0 lt t le 1\
                  0.05+0.02(t-1) & 1lt t le 2
                  end{cases}
                  $$
                  Integrating $frac{A'(t)}{A(t)}=delta(t)$ from $t_1$ to $t_2$ we have
                  $$
                  A(t_2)=A(t_1)mathrm e^{int_{t_1}^{t_2}delta(u)mathrm d u}
                  $$
                  So we have
                  $$
                  A(1)=A(0)expleft(int_{0}^{1}0.05,mathrm d uright)=A(0)mathrm e^{0.05}=10,000,mathrm e^{0.05}approx 10,512.71
                  $$
                  where $A(0)=10,000$.



                  And
                  $$
                  A(2)=A(1)expleft(int_{1}^{2}[0.05+0.02(u-1)],mathrm d uright)=A(1)mathrm e^{0.06}=10,512.71,mathrm e^{0.06}approx 11,162.78
                  $$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Sep 4 '17 at 20:45

























                  answered Sep 4 '17 at 20:36









                  alexjoalexjo

                  12.4k1430




                  12.4k1430






























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