Future Value of Savings Account: y Years, j Rate, f Deposit

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Discussion Overview

The discussion revolves around calculating the future value of a savings account with varying deposit amounts over time, specifically focusing on a scenario where deposits increase in a structured manner every three years. Participants explore different mathematical formulations and approaches to derive a general formula for future value based on the number of years, interest rate, and initial deposit.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a detailed example of deposits and interest accumulation over 9 years, leading to a specific future value.
  • Another participant proposes a formula involving a summation to express future value in terms of years, interest rate, and first deposit.
  • Some participants express a desire for a more standard financial formula akin to that of a growing annuity, indicating a preference for established forms.
  • A participant introduces a difference equation to analyze the future value, noting challenges posed by a step function in the deposits.
  • Another participant shares their own formula, which works under the condition that the number of years is a multiple of three, while also seeking a more general solution.
  • Discussions include attempts to derive particular solutions and general solutions based on the established difference equations.
  • Participants acknowledge errors and misunderstandings in earlier posts, indicating a collaborative effort to refine their approaches.

Areas of Agreement / Disagreement

Participants do not reach a consensus on a single formula for future value, with multiple competing views and approaches remaining throughout the discussion. Some participants focus on specific cases, while others seek more general solutions.

Contextual Notes

Limitations include the dependence on the assumption that the number of years is a multiple of three for some formulas, and unresolved mathematical steps in deriving particular solutions. The discussion reflects various interpretations of how to express the future value based on the structure of deposits.

Wilmer
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d is deposited in a savings account for 3 years, then 2d for 3years,
then 3d for 3 years, and so on similarly;
here's an example for 9 years, 1st deposit = $100, rate = 10% annual:
Code: 
YEAR   DEPOSIT   INTEREST    BALANCE 
 0                               .00 
 1     100.00         .00     100.00 
 2     100.00       10.00     210.00 
 3     100.00       21.00     331.00 
 4     200.00       33.10     564.10 
 5     200.00       56.41     820.51 
 6     200.00       82.05   1,102.56 
 7     300.00      110.26   1,512.82 
 8     300.00      151.28   1,964.10 
 9     300.00      196.41   2,460.51
GIVENS:
y = number of years (multiple of 3): 9 in above example
j = annual rate / 100 : .10 in above example
f = first deposit : $100 in above example

What is the future value of the account (2,460.51 in above example) in terms of y, j and f ?
 
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I find:

$$FV(y,j,f)=f\sum_{k=1}^y\left(\left\lceil\frac{k}{3} \right\rceil(1+j)^{y-k} \right)$$

I simply wrote down the first few terms, observed the pattern, then checked the above formula against the provided table.
 
Nice, Mark, but that's not really what I was after;

I meant something like a standard financial formula, a bit like:
FV of growing annuity = P * ((1+r)^n - (1+g)^n) / (r-g)
P=initial payment
r=discount rate or interest rate
g=growth rate
n=number of periods

Guess I should have said so, then(Bandit)
 
Wilmer said:
Nice, Mark, but that's not really what I was after;

I meant something like a standard financial formula, a bit like:
FV of growing annuity = P * ((1+r)^n - (1+g)^n) / (r-g)
P=initial payment
r=discount rate or interest rate
g=growth rate
n=number of periods

Guess I should have said so, then(Bandit)

I have FV on the left of the equal sign and on the right a set of mathematical symbols representing the value of FV...it's just like what you are after. (Clapping)(Sun)
 
MarkFL said:
...it's just like what you are after.
No it's not...the right side is not really a "formula", even if correct,
but more a definition of how the formula is arrived at...
Well, not really important anyway: won't affect price of groceries (Sun)
 
I gather you want a closed form...well, let's look at the difference equation:

$$FV_{y+1}-(1+j)FV_{y}=sf$$

where $s$ is the step-function $$s=\left\lceil\frac{y}{3} \right\rceil$$

The homogeneous solution will take the form:

$$h_y=c_1(1+j)^y$$

But, I have no idea what form the particular solution must take, given that pesky step function, which also thwarts my ability to use symbolic differencing to obtain a homogeneous recurrence. (Emo)
 
Well, this was mine :

n = FLOOR(y/3) + 1 [4 in example]
i = (1 + j)^3 - 1 [.331 in example]
d = f(j^2 + 3j + 3) [331.00 in example]
Code: 
     d[(1 + i)^n - in - 1] 
FV = --------------------- 
             i^2
 
Last edited:
Your formula does work when $y$ is a multiple of 3, and if I had read your original post more carefully, I would have noticed you did state that $y$ is a multiple of 3...I was trying to find a formula that worked for any natural number $y$.

When I have more time later, I will attempt to post a derivation of your formula. (Nerd)
 
MarkFL said:
...and if I had read your original post more carefully, I would have noticed you did state that $y$ is a multiple of 3...
Oh oh: 15 minutes in the corner (Wait)
 
  • #10
The 15 minutes in the corner was nothing compared to the battle I had with silly errors in coming up with the formula...(Sweating)(Rofl)

Let's let $y=3n$ where $n\in\mathbb{N_0}$.

Now, consider the following difference equation obtained from analyzing what goes on in between each value of $n$:

$$FV_{n+1}-(1+j)^3FV_{n}=(n+1)f\left(j^2+3j+3 \right)$$

We see that the homogeneous solution is given by:

$$h_n=c_1(1+j)^{3n}$$

And we seek a particular solution of the form:

$$p_n=An+B$$

Substituting, we find:

$$(A(n+1)+B)-(1+j)^3(An+B)=(n+1)f\left(j^2+3j+3 \right)$$

$$-jAn+\frac{A}{j^2+3j+3}-jB=fn+f$$

Equating coefficients, we find:

$$A=-\frac{f}{j},\,B=-\frac{f(1+j)^3}{j^2\left(j^2+3j+3 \right)}$$

Thus, we have:

$$p_n=-\frac{f}{j}n-\frac{f(1+j)^3}{j^2\left(j^2+3j+3 \right)}$$

And so the general solution is:

$$FV_{n}=h_n+p_n=c_1(1+j)^{3n}-\frac{f}{j}n-\frac{f(1+j)^3}{j^2\left(j^2+3j+3 \right)}$$

Using $FV_0=0$, we find:

$$c_1=\frac{f(1+j)^3}{j^2\left(j^2+3j+3 \right)}$$

Thus, we have (after some simplification):

$$FV_{n}=\frac{f}{j}\left(\frac{(1+j)^{3n}-1}{1-(1+j)^{-3}}-n \right)$$
 
  • #11
I simply converted a "3 periods" to an equivalent "1 period";
in my example, that's 331 deposited periodically , and
increasing by 331 each period, at 33.1%.
Code: 
P  DEPOSIT INTEREST BALANCE
0                        .00
1   331.00            331.00
2   662.00  109.56  1,102.56
3   993.00  364.95  2,460.51