Recurrence Relation of Compound Interest
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So I have the following question:
Suppose that at the end of some month (referred to as month 0), you open a
investment account with an initial investment of I dollars for some given I .The account grows at a fixed interest rate of r% per month, for a given r (note that if the quoted interest rate is an annual interest, then r = r/12). Suppose that at the end of every month you add some fixed amount m dollars for some given m. Denote xn by the amount of money you have in your account at the end of
month n.
a) Write a recurrence relation that relates xn to xn-1, for n$ge$t. The relation should involve and m,r,I,n. Make sure to also indicate the value of x0.
b) Solve the recurrence relation of part (a).
c) Assume that I = $10,000, r = 0.5%, and m = $1,000. Use part (b) to compute the value of the investment after 20 years
I derived the recurrence relation for part (a) as:
xn = (1+r)xn-1 + m x0 = I
Then I used the following definition of a linear recurrence relation to solve for part (b)
xn = a xn-1 + b ==>
xn = (x0 - b⁄(1-a)
) an - b⁄(1-a)
Where:
a = (1+r) & b = m
So:
xn = (I - m⁄r) an - m⁄r
However, when I try to solve for part (c), I get a negative answer (and quite a large one at that).
Where my n = 12 months x 20 years = 240 months
So my assumption is that my initial equation is wrong, but I don't know what else it could be. If anyone has a suggestion of what the proper equation is so I can solve it myself, that would be much appreciated.
discrete-mathematics recurrence-relations
asked Mar 1 '18 at 2:34
willh99willh99
41
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add a comment |
$begingroup$
So I have the following question:
Suppose that at the end of some month (referred to as month 0), you open a
investment account with an initial investment of I dollars for some given I .The account grows at a fixed interest rate of r% per month, for a given r (note that if the quoted interest rate is an annual interest, then r = r/12). Suppose that at the end of every month you add some fixed amount m dollars for some given m. Denote xn by the amount of money you have in your account at the end of
month n.
a) Write a recurrence relation that relates xn to xn-1, for n$ge$t. The relation should involve and m,r,I,n. Make sure to also indicate the value of x0.
b) Solve the recurrence relation of part (a).
c) Assume that I = $10,000, r = 0.5%, and m = $1,000. Use part (b) to compute the value of the investment after 20 years
I derived the recurrence relation for part (a) as:
xn = (1+r)xn-1 + m x0 = I
Then I used the following definition of a linear recurrence relation to solve for part (b)
xn = a xn-1 + b ==>
xn = (x0 - b⁄(1-a)
) an - b⁄(1-a)
Where:
a = (1+r) & b = m
So:
xn = (I - m⁄r) an - m⁄r
However, when I try to solve for part (c), I get a negative answer (and quite a large one at that).
Where my n = 12 months x 20 years = 240 months
So my assumption is that my initial equation is wrong, but I don't know what else it could be. If anyone has a suggestion of what the proper equation is so I can solve it myself, that would be much appreciated.
discrete-mathematics recurrence-relations
asked Mar 1 '18 at 2:34
willh99willh99
41
$endgroup$
add a comment |
$begingroup$
So I have the following question:
Suppose that at the end of some month (referred to as month 0), you open a
investment account with an initial investment of I dollars for some given I .The account grows at a fixed interest rate of r% per month, for a given r (note that if the quoted interest rate is an annual interest, then r = r/12). Suppose that at the end of every month you add some fixed amount m dollars for some given m. Denote xn by the amount of money you have in your account at the end of
month n.
a) Write a recurrence relation that relates xn to xn-1, for n$ge$t. The relation should involve and m,r,I,n. Make sure to also indicate the value of x0.
b) Solve the recurrence relation of part (a).
c) Assume that I = $10,000, r = 0.5%, and m = $1,000. Use part (b) to compute the value of the investment after 20 years
I derived the recurrence relation for part (a) as:
xn = (1+r)xn-1 + m x0 = I
Then I used the following definition of a linear recurrence relation to solve for part (b)
xn = a xn-1 + b ==>
xn = (x0 - b⁄(1-a)
) an - b⁄(1-a)
Where:
a = (1+r) & b = m
So:
xn = (I - m⁄r) an - m⁄r
However, when I try to solve for part (c), I get a negative answer (and quite a large one at that).
Where my n = 12 months x 20 years = 240 months
So my assumption is that my initial equation is wrong, but I don't know what else it could be. If anyone has a suggestion of what the proper equation is so I can solve it myself, that would be much appreciated.
discrete-mathematics recurrence-relations
asked Mar 1 '18 at 2:34
willh99willh99
41
$endgroup$
So I have the following question:
Suppose that at the end of some month (referred to as month 0), you open a
investment account with an initial investment of I dollars for some given I .The account grows at a fixed interest rate of r% per month, for a given r (note that if the quoted interest rate is an annual interest, then r = r/12). Suppose that at the end of every month you add some fixed amount m dollars for some given m. Denote xn by the amount of money you have in your account at the end of
month n.
a) Write a recurrence relation that relates xn to xn-1, for n$ge$t. The relation should involve and m,r,I,n. Make sure to also indicate the value of x0.
b) Solve the recurrence relation of part (a).
c) Assume that I = $10,000, r = 0.5%, and m = $1,000. Use part (b) to compute the value of the investment after 20 years
I derived the recurrence relation for part (a) as:
xn = (1+r)xn-1 + m x0 = I
Then I used the following definition of a linear recurrence relation to solve for part (b)
xn = a xn-1 + b ==>
xn = (x0 - b⁄(1-a)
) an - b⁄(1-a)
Where:
a = (1+r) & b = m
So:
xn = (I - m⁄r) an - m⁄r
However, when I try to solve for part (c), I get a negative answer (and quite a large one at that).
Where my n = 12 months x 20 years = 240 months
So my assumption is that my initial equation is wrong, but I don't know what else it could be. If anyone has a suggestion of what the proper equation is so I can solve it myself, that would be much appreciated.
discrete-mathematics recurrence-relations
discrete-mathematics recurrence-relations
asked Mar 1 '18 at 2:34
willh99willh99
41
asked Mar 1 '18 at 2:34
willh99willh99
41
asked Mar 1 '18 at 2:34
willh99willh99
41
asked Mar 1 '18 at 2:34
willh99willh99
41
asked Mar 1 '18 at 2:34
willh99willh99
41
41
add a comment |
add a comment |
1 Answer
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The formula you must use is:
$$x_n=left(x_0-frac{b}{1-a}right)cdot a^n color{red}{+} frac{b}{1-a}.$$
answered Mar 1 '18 at 2:57
farruhotafarruhota
20.5k2739
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$begingroup$
Ahh, thank you. I must've copied it down wrong in my notes
$endgroup$
– willh99
Mar 1 '18 at 3:42
add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The formula you must use is:
$$x_n=left(x_0-frac{b}{1-a}right)cdot a^n color{red}{+} frac{b}{1-a}.$$
answered Mar 1 '18 at 2:57
farruhotafarruhota
20.5k2739
$endgroup$
$begingroup$
Ahh, thank you. I must've copied it down wrong in my notes
$endgroup$
– willh99
Mar 1 '18 at 3:42
add a comment |
$begingroup$
The formula you must use is:
$$x_n=left(x_0-frac{b}{1-a}right)cdot a^n color{red}{+} frac{b}{1-a}.$$
answered Mar 1 '18 at 2:57
farruhotafarruhota
20.5k2739
$endgroup$
$begingroup$
Ahh, thank you. I must've copied it down wrong in my notes
$endgroup$
– willh99
Mar 1 '18 at 3:42
add a comment |
$begingroup$
The formula you must use is:
$$x_n=left(x_0-frac{b}{1-a}right)cdot a^n color{red}{+} frac{b}{1-a}.$$
answered Mar 1 '18 at 2:57
farruhotafarruhota
20.5k2739
$endgroup$
The formula you must use is:
$$x_n=left(x_0-frac{b}{1-a}right)cdot a^n color{red}{+} frac{b}{1-a}.$$
answered Mar 1 '18 at 2:57
farruhotafarruhota
20.5k2739
answered Mar 1 '18 at 2:57
farruhotafarruhota
20.5k2739
answered Mar 1 '18 at 2:57
farruhotafarruhota
20.5k2739
answered Mar 1 '18 at 2:57
farruhotafarruhota
20.5k2739
20.5k2739
$begingroup$
Ahh, thank you. I must've copied it down wrong in my notes
$endgroup$
– willh99
Mar 1 '18 at 3:42
add a comment |
$begingroup$
Ahh, thank you. I must've copied it down wrong in my notes
$endgroup$
– willh99
Mar 1 '18 at 3:42
$begingroup$
Ahh, thank you. I must've copied it down wrong in my notes
$endgroup$
– willh99
Mar 1 '18 at 3:42
$begingroup$
Ahh, thank you. I must've copied it down wrong in my notes
$endgroup$
– willh99
Mar 1 '18 at 3:42
add a comment |
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