? operator

I was wondering (not for the first time) what comes next in the progression $a^b$ , $a\times b$ , $a+b$ … $a?b$
Probably this is a thing already. Anyway…
What is $a?b$ ?
Well, my idea was that $\ln ab=\ln a+\ln b$ , so $\ln (a+b)=\ln a ? \ln b$
So then $a?b=\ln(e^a+a^b)$ . Sounds reasonable. (and silly.)
Hmm so what’s the “I”, like $a\times 1=a$ and $a+0=a$ ?
$a?x=a \to \ln(e^a+e^x)=a \to e^x=0 \to x=\ln 0 \to x=-\infty$
Hmm interesting hehe.
And $a?a$ ?
$2e^a=e^x\to x=\ln 2e^a =a+\ln 2$
Hmm so $a?0=\ln 2$ .. that’s weird.
Turns out that $a?b$ is, for $a<b$ , very close to $b$ .
(And obviously for $a>b$ , very close to $a$ .)
It seems once $a<\frac{4}{5}b$ , $a?b$ is extremely close to $b$ .
So it’s almost a max function! (just equals the higher value) How strange.
Except if $a=b$ then $a?a=a+\ln 2$ .
And $a?0$ is $\ln(e^a+1)$
e.g. $10?0$ is $10.0000453988992\dots$
$1?0=1.31326168751822\dots$
Well, I guess $10+0>10\times 0$ , so it’s only right that $10?0$ be $>10$ .

And going the other way, $a+b\to a\times b\to a^b$ – why is $a^b$ not symmetrical in $a$ and $b$ like $+$ and $\times$ ? That doesn’t seem right.
So (you guessed it).. maybe call this $\circ$ .

$\begin{align*} \ln(ab)&=\ln a+\ln b \\ \ln(a\circ b)&= \ln a \times \ln b \\ a\circ b &= e^{\ln a \ln b} \\ 2\circ 3 &\approx 2.1414?! \\ 100\circ 100 &= 1623081665.62\dots \\ &= e^{(\ln 100)^2} \\ \intertext{So, find I.} a\circ x &= a \\ e^{\ln a \ln x}&=a \\ \ln x \ln a&=\ln a \\ \ln x&=1\\ x &= e \to a\circ e=a\\ \text{and }e\circ e&=e \\ \text{hmm but } 1\circ \text{ anything } =e^{0}=1 \end{align*}$

operator1

And the next one up? (and down?!)
Call the next one up $\infty$ maybe.. $a\infty b$

$\begin{align*} \ln (ab)&=\ln a+\ln b \\ \ln(a\circ b)&=\ln a \ln b \\ \ln(a\infty b)&= \ln a \circ \ln b \\ a\infty b &= e^{\ln a \circ \ln b} \\ &= e^{e^{\ln \ln a \ln \ln b}} \\ 2\infty 3 &\approx 2.62772268\dots \\ 1000\infty 1000 &\approx 1.562581\times 10^{18} \\ &= e^{e^{\ln \ln 1000)^2}} \end{align*}$

So, find I.

$\begin{align*} a\infty x &= a \\ e^{e^{\ln \ln a \ln \ln x}}&=a \\ \ln \ln x \ln \ln a&=\ln \ln a \\ \ln \ln x&=1\\ x &= e^e \to a\infty e^e=a\\ \text{and }e\infty e^e&=e^{e^{0}}=e \\ \text{hmm but} $e\infty$ \text{ anything }&=e \end{align*}$

adamponting.com

? operator

Leave a Reply Cancel reply