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Short Version: Find $r$

$$61.051 = \frac{(1-(1+r)^5)}{(1-(1+r))} * 10$$

(The answer is $0.1$. I am looking for the steps/process to find the answer)

Detailed Version:

Example Scenario: Purchasing Widgets

The first widget costs $x. Each subsequent widget costs r times more than the last widget purchased (x2 = x1 * (1 + r)).

Illustration

The following table illustrates the scenario for a base widget cost (x1) of $10 that increases 10% (r) with each additional unit purchased.

N (units purchased) xn (unit cost) T (cumulative cost for N units)
1 10 10
2 11 21
3 12.1 33.1
4 13.31 46.31
5 14.641 61.051

As shown, the first widget costs \$10, the second costs \$11 (\$10 * 1.1), and so on. Such that the total cumulative cost for five widgets is \$61.051.

The total cumulative cost for any N units may be found using:

$$T = \frac{(1-(1+r)^N)}{(1-(1+r))} * x_1$$

So for the example above:

$$61.051 = \frac{(1-(1+0.1)^5)}{(1-(1+0.1))} * 10$$

In other words, purchasing five widgets would cost $61.051.

The Question

If the first widget's cost x1, total cumulative cost T and number of widgets purchased N are known, how do we find the value of r.

** Work so far **

$$61.051 = \frac{(1-(1+r)^5)}{(1-(1+r))} * 10$$

$$\frac{61.051}{10} = \frac{\frac{(1-(1+r)^5)}{(1-(1+r))} * 10}{10}$$

$$6.1051 = \frac{(1-(1+r)^5)}{(1-(1+r))}$$

Here's where I start to derail...

$$log(6.1051) = \frac{1 - 5*log(1+r)}{1-(1+r)}$$

$$0.785693 = \frac{1 - \frac{5*log(1+r)}{5}}{1-\frac{(1+r)}{5}}$$

$$0.785693 = \frac{1-log((1+r)}{1-\frac{(1+r)}{5}}$$

...and now I am off the rails completely.

What are the steps to isolate $r$?

Thanks

TShiong
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AWaddington
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    I would start by $x=1-r$, and $1-x^5=(1-x)(1+x+x^2+x^3+x^4)$. Then I think it is unlikely that you will find a simple closed formula for the solution. Although a closed formula exists for 4th degree equations. – ECL Feb 26 '22 at 21:32
  • Thank you ECL - If the problem were for 100 units, not 5, would the second parenthesis in your comment then become $(1 + x + x^2 + ... x^{100})$? I will have to study this a bit more, I'm having difficulty visualizing finding $r$ from $x$ if $x$ includes $r$ in its definition. Looks like I have some reading in my future :) – AWaddington Feb 26 '22 at 22:46

1 Answers1

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Starting with the formula you wrote $$T = \frac{(1-(1+r)^n)}{(1-(1+r))} \, x_1$$ let $A=\frac T{x_1}$,simplify the denominator to make $$A=\frac{(1+r)^n-1} r$$ If $n>4$, no explicit solution ; then either numerical methods or approximations.

Since we know that $r \ll 1$, using $$A=\sum_{k=0}^\infty \binom{n}{k+1} \,r^k $$ truncate to some order and use series reversion. This would give $$r=t-\frac{n-2}{3} t^2+\frac{(n-2) (5 n-7)}{36} t^3-\frac{(n-2) \left(17 n^2-44 n+29\right) }{270} t^4+O(t^5)$$ where $t=\frac{2 (A-n)}{(n-1) n}$.

Trying with your numbers $n=5$ and $A=6.1051$, this will give $r=0.0999342$ which does not seem too bad.

Adding two more terms in the expansion would give $r=0.0999969$

Claude Leibovici
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  • What I thought was a simple question has become a fascinating discussion. Ultimately I was hoping for a closed formula (for use in programming) but I see now this is not possible. I tested your comment that n>4 cannot be solved using the microsoft math solver, sure enough, it stops trying at n=5. That's interesting. Your response gives me some terminology to dig deeper, thank you! – AWaddington Feb 27 '22 at 17:51
  • You say "Since we know that r≪1,..." I don't beleive we do know that. We know that only for this simplified example, but it is possible widgets increase by more than 100% with each purchase. Does that change the approach? – AWaddington Feb 27 '22 at 17:52