Accumulated value of force of interest
$begingroup$
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
edited Oct 6 '15 at 20:56
rubik
6,74632661
asked Oct 6 '15 at 20:05
Alex RochAlex Roch
11
$endgroup$
add a comment |
$begingroup$
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
edited Oct 6 '15 at 20:56
rubik
6,74632661
asked Oct 6 '15 at 20:05
Alex RochAlex Roch
11
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Oct 6 '15 at 20:56
rubik
6,74632661
asked Oct 6 '15 at 20:05
Alex RochAlex Roch
11
$endgroup$
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
edited Oct 6 '15 at 20:56
rubik
6,74632661
asked Oct 6 '15 at 20:05
Alex RochAlex Roch
11
edited Oct 6 '15 at 20:56
rubik
6,74632661
asked Oct 6 '15 at 20:05
Alex RochAlex Roch
11
edited Oct 6 '15 at 20:56
rubik
6,74632661
edited Oct 6 '15 at 20:56
rubik
6,74632661
edited Oct 6 '15 at 20:56
rubik
6,74632661
6,74632661
asked Oct 6 '15 at 20:05
Alex RochAlex Roch
11
asked Oct 6 '15 at 20:05
Alex RochAlex Roch
11
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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, $10000e^{0.11} = 11162.78$.
answered Oct 6 '15 at 20:23
rubikrubik
6,74632661
$endgroup$
add a comment |
$begingroup$
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
$$
A(2)=A(1)expleft(int_{1}^{2}[0.05+0.02(u-1)],mathrm d uright)=A(1)mathrm e^{0.06}=10,512.71,mathrm e^{0.06}approx 11,162.78
$$
answered Sep 4 '17 at 20:36
alexjoalexjo
12.4k1430
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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, $10000e^{0.11} = 11162.78$.
answered Oct 6 '15 at 20:23
rubikrubik
6,74632661
$endgroup$
add a comment |
$begingroup$
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, $10000e^{0.11} = 11162.78$.
answered Oct 6 '15 at 20:23
rubikrubik
6,74632661
$endgroup$
add a comment |
$begingroup$
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, $10000e^{0.11} = 11162.78$.
answered Oct 6 '15 at 20:23
rubikrubik
6,74632661
$endgroup$
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, $10000e^{0.11} = 11162.78$.
answered Oct 6 '15 at 20:23
rubikrubik
6,74632661
answered Oct 6 '15 at 20:23
rubikrubik
6,74632661
answered Oct 6 '15 at 20:23
rubikrubik
6,74632661
answered Oct 6 '15 at 20:23
rubikrubik
6,74632661
6,74632661
add a comment |
add a comment |
$begingroup$
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
$$
A(2)=A(1)expleft(int_{1}^{2}[0.05+0.02(u-1)],mathrm d uright)=A(1)mathrm e^{0.06}=10,512.71,mathrm e^{0.06}approx 11,162.78
$$
answered Sep 4 '17 at 20:36
alexjoalexjo
12.4k1430
$endgroup$
add a comment |
$begingroup$
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
$$
A(2)=A(1)expleft(int_{1}^{2}[0.05+0.02(u-1)],mathrm d uright)=A(1)mathrm e^{0.06}=10,512.71,mathrm e^{0.06}approx 11,162.78
$$
answered Sep 4 '17 at 20:36
alexjoalexjo
12.4k1430
$endgroup$
add a comment |
$begingroup$
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
$$
A(2)=A(1)expleft(int_{1}^{2}[0.05+0.02(u-1)],mathrm d uright)=A(1)mathrm e^{0.06}=10,512.71,mathrm e^{0.06}approx 11,162.78
$$
answered Sep 4 '17 at 20:36
alexjoalexjo
12.4k1430
$endgroup$
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
$$
A(2)=A(1)expleft(int_{1}^{2}[0.05+0.02(u-1)],mathrm d uright)=A(1)mathrm e^{0.06}=10,512.71,mathrm e^{0.06}approx 11,162.78
$$
answered Sep 4 '17 at 20:36
alexjoalexjo
12.4k1430
edited Sep 4 '17 at 20:45
answered Sep 4 '17 at 20:36
alexjoalexjo
12.4k1430
answered Sep 4 '17 at 20:36
alexjoalexjo
12.4k1430
answered Sep 4 '17 at 20:36
alexjoalexjo
12.4k1430
12.4k1430
add a comment |
add a comment |
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$begingroup$
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