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The question arised when I was thinking about population growth. If a population grows by 1% we can simply plug into to the formula: $$ \text{final population}= P \cdot (1+p)^n $$ where $P$ is the initial population, $p=0.01$ in the example and $n$ is the number of years.

Is there a similar (explicit) formula if $p$ shrinks at a constant rate, let's say for example after $n$ years $p_n = 0.99 p_{n-1}$ and $p_0 = 0.01$?

Of course, the formula $P \cdot \prod_{i=0}^{n}(1+p_i)$ would describe this, if we would also define the sequence $p_n$ but I want something that only takes $p$ as an input.

timtam
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    If you use exponential interest instead of geometric compound interest, you get a closed form solution – Snoop Jul 18 '22 at 11:54
  • Nice question. Do any banks actually offer this kind of interest rate? Why or why not? – Dan Jul 19 '22 at 04:01

1 Answers1

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Suppose $1$ dollar is deposited, and the interest rate decreases over time, such that after $n$ years the interest rate is $s^{n-1}r$ where $0<s<1.$

Assuming interest is compounded annually, the amount after $N$ years is:

$$A_N=\prod_{n=1}^N{(1+s^{n-1}r)}$$

This can be expressed in terms of the q-Pochhammer Symbol, but that is not helpful if we want to calculate or approximate its value with a simple calculator.

Here is how we can approximate it with a simple calculator:

$$A_N=\prod_{n=0}^{N-1}{(1+s^{n}r)}$$

$$=\exp{\sum_{n=0}^{N-1}\ln{{(1+s^{n}r)}}}$$

Use the Maclaurin series for $\ln{(1+s^{n}r)}$:

$$=\exp{\sum_{n=0}^{N-1}\sum_{k=1}^\infty{\frac{(-1)^{k+1}(s^nr)^k}{k}}}$$

Switch the order of summation:

$$=\exp{\sum_{k=1}^\infty\frac{(-1)^{k+1}r^k}{k}}\sum_{n=0}^{N-1}{(s^k)^n}$$

The second series is a geometric series:

$$=\exp{\sum_{k=1}^\infty\frac{(-1)^{k+1}r^k}{k}}\left(\frac{1-s^{kN}}{1-s^k}\right)$$

This series converges by the alternating series test. $r$ is an interest rate, so $r$ is usually close to $0$, so the series can be approximated by the sum of the first few terms. In fact, using only the first term gives a decent approximation (unless $r$ is large and/or $s$ is very close to $1$).

$$A_N\approx e^{r\left(\frac{1-s^N}{1-s}\right)}$$

Since $e^r\approx 1+r$:

$$A_N\approx (1+r)^{\frac{1-s^N}{1-s}}$$

Notice that $A_N$ converges as $N\to\infty$:

$$\lim_{N\to\infty}A_N \approx (1+r)^{\frac{1}{1-s}}$$

Dan
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  • Why not simply begin by using $\log(1+x)=x+O(x^2)$ and write $$e^{\sum_{n=0}^{\infty} (rs^n)+O(r^2s^{2n})}=e^{r/(1-s)}+O(r^2) =(1+r)^{1/(1-s)}+O(r^2)$$ – Mark Viola Jul 19 '22 at 03:26
  • @MarkViola I included just enough detail so that, if $r$ is large and/or $S$ is very close to $1$, the reader can conveniently include more terms of the series to get a better approximation for $A_N$ and/or $\lim_{N\to\infty}{A_N}$. – Dan Jul 19 '22 at 03:51