Weird relation that I found: Any relation to Geometric Series?

Question

I was doing some chemistry homework, and I happened to find the following relation between some numbers:

$$(-273.15 + 121.4)\times\frac95 = -273.15$$

Weird coincidence, right? You add a number to an unrelated number, multiply that new number by another unrelated number, and you get the number you started with! This intrigued me enough to check a general case, for rational numbers $a, b, r \in\mathbb{Q}$:

$$(a + b)\cdot r = a$$

Some simple rearranging gives the following:

$$ \frac{a}{b} = \frac{r}{1-r}$$

So, trivially, it seems that, by pure coincidence, I happened to find two rational numbers, $a$ and $b$, whose ratio happens to be the same as the ratio between another rational number and itself subtracted from $1$. This alone wouldn't have motivated me to write this post, had it not been for the fact that the right hand side of the above exactly is exactly equal to an expression for a geometric series, with the first term being $r$!

This makes me wonder what I just stumbled upon, if anything. Is this just an isolated case in which the geometric series happened to show up, or did I find something that's related to something deeper?

Edit: this question seems related: Is it just a coincidence that the solution to $y=x$, $y=mx+b$ ($m<1$) is the sum of an infinite geometric series with first term $b$ and ratio $m$?

Answer 1

If $f(x)=(1+x)r,$ with $|r|<1,$ then $f(x)$ is a contraction, so we can find the value $x_*$ where $f(x_*)=x_*$ by starting with any $x_0$ and computing $x_{n+1}=f(x_n)$ to get a sequence $x_n\to x_*$ with $f(x_*)=x_*.$

If we start with $x_0=0,$ then $$\begin{align} x_1&=r\\x_2& =(1+r)r=r+r^2\\ &\vdots\\ x_n&=r+r^2+\cdots+r^n\\ &\vdots \end{align} $$

So $$x_n\to \sum_{k=1}^\infty r^k=\frac r{1-r}$$

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Of course, $\frac r{1-r}$ is still the answer when $|r|>1$ or $r=-1,$ while the function isn't a contraction in that case, and the geometric series doesn't converge.

If $|r|>1,$ you'd want to write the equation as:

$$x=\frac 1r x-1.$$

Then $g(x)=\frac1rx-1$ is a contraction.

Starting with $x_0=0, x_{n+1}=g(x_n)$ you get:

$$x_n=-1-\frac1r-\cdots-\frac1{r^{n-1}}$$

And $$x_n\to \frac{-1}{1-\frac1r}=\frac{-r}{r-1}=\frac r{1-r}.$$

Answer 2

You wrote $$ (a + b)\cdot r = a. $$

This is related to a convergent infinite geometric series as follows:

If the sum of the entire series is $a,$ and the first term of the series is $br,$ you can also take $ar$ ($r$ times the original series sum), which is the sum of the series shifted by one term (summing the series starting at the second term instead of the first), and adding the first term ($br$) to the sum of the other terms you have the sum of the entire series again:

$$ br + ar = a. $$

Factor out the common factor of $r$ and you have $(a + b)\cdot r = a$ again.

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More explicitly, the geometric series is

$$ br + br^2 + br^3 + \cdots . $$

We say that $a$ is the sum of the infinite series:

$$ a = br + br^2 + br^3 + \cdots . \tag1 $$

Multiplying by $r$ on both sides:

$$ ar = br^2 + br^3 + br^4 + \cdots . $$

Adding $br$ to both sides:

$$ ar + br = br + br^2 + br^3 + br^4 + \cdots . $$

Now the right-hand side is just the original geometric series, so

$$ ar + br = a. $$

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Alternatively, if we divide a factor $b$ out of the geometric series shown in Equation $(1)$, we get

$$ \frac ab = r + r^2 + r^3 + \cdots . $$

That is,

$$ \frac ab = \frac{r}{1 - r}, $$

using the usual formula for such a geometric series.

What your formula does not have in common with a geometric series is that your formula gives you well-defined results even when $\lvert r\rvert \geq 1,$ which is not true for the geometric series.