Sorta weird coincidence I found.

Question

kainui · Answer

I've shortened this to get to the point while leaving out some details, so don't nitpick me on them haha. I was thinking about the interesting case that
$$2*2=2+2$$ and was wondering about "higher versions" of this, for instance,
$$k*k*k=k+k+k$$
$$k^3=3k$$
$$k=\sqrt{3}$$
More generally even, (for convenience of simplification in a minute, I've added +1)
$$k^{n+1}=(n+1)k$$ So we get:
$$k=(n+1)^{1/n}$$

However the thing that struck me as fascinating is I have seen this exact thing somewhere else before, specifically in solving this finite difference equation, if I define:
$$\Delta f(t) = \frac{f(t+a)-f(t)}{a}$$
Then if I attempt to solve:
$$\Delta f(t)=f(t)$$
We find:
$$f(t+a)=(a+1)f(t)$$
If we make a guess that
$$f(t)=b^t$$ and plug it in we get:
$$f(t+a)=b^a f(t)$$
equating the two pieces to solve for b in our guess:
$$b^a = (a+1)$$
$$b=(a+1)^{1/a}$$

So perhaps this is a complete coincidence, however it seems like there might be some sorta fishy relationship going on. Any ideas?

kainui · Answer

I realize I might have left out some crucial details here, so if anyone's curious about any of it I'll happily get into the details / motivations of stuff.

ganeshie8 · Answer

Interesting... I remember @myininaya  or @freckles  working on this some time ago...

ganeshie8 · Answer

in your question ofcourse
but back then we hadn't talked about the relationship between the two

kainui · Answer

Yeah, I didn't state it here but this equation is basically like a sorta generalization of the $e^x$ formula to other stuff. Just sorta funky how this sorta popped up.

I was originally thinking about how 0 and 1 are related in the sense that 0 is the additive identity and 1 is the multiplicative identity and 0 has no multiplicative inverse and this relationship seemed to be kind of odd... So I wondered if the number 2 is also somehow a result of this kind of relationship since I know we've talked about its relationship to the Knuth's up arrow notation being a sorta... fixed point I guess you could say. $2+2=2*2=2^2=\cdots =2\uparrow^k2$ So overall just sorta wondering what's going on. The fact that it seems to also pull in e^x makes this all the more fascinating and I don't even see why this should be related haha.

thomas5267 · Answer

There is something really fishy going on in here.$$\lim_{a\to\infty}b=e$$ and the difference equation is the derivative quotient.

thomas5267 · Answer

Actually $\displaystyle\lim_{a\to0}b=e$.  $\Delta f(t)=f(t)$ reduces to $f'(t)=f(t)$ in the limit of a to 0 and hence we get $e$ appearing in the equation.

kainui · Answer

Yeah exactly, that's the fun of it! :D

thomas5267 · Answer

I guess e appears in this coincidence because $e^x$ is a fix point of the derivative operator?

kainui · Answer

Maybe?

Here's something I'm looking at currently,

$$\lim_{a \to \infty} (a+1)^{1/a}=1$$
$$\lim_{a \to0} (a+1)^{1/a}=e$$

kainui · Answer

I think it's kind of funny since in this scheme of looking at:
$$k*k=k+k$$ there are only two solutions, k=0 and k=2
$$k*k*k=k+k+k$$ there are three solutions, k=0, $k=\pm \sqrt{3}$.
But there are infinitely many solutions for 
$$k=k$$
But when we write it as the generalization is $k^{n+1}=(n+1)k$ then setting n=0 gives that result. However if we algebraically rearrange it and then take the limit as n approaches 0 we get:
$$k=(n+1)^{1/n} \to e$$ 

Uhhh...?

kainui · Answer

Am I making a simple arithmetic mistake or am I allowed to feel like my brain is exploding lol.

ganeshie8 · Answer

The sequences $k^{n+1}$ and $(n+1)k$ happen to have the same limit $k$ as $n\to 0$

and the sequence $(n+1)^{1/n}$  has a different limit as $n\to 0$

ganeshie8 · Answer

I don't see any arithmetic mistake... but why must it be strange ?

kainui · Answer

In the first case, k is left undetermined, but in the second case it determines k. It seems that taking both limits simultaneously forces k to be a single solution.

e is only a single solution to k=k out of infinitely many, I just don't see a particular reason why it would end up choosing e.

For instance, look at any other case:

$$k*k=k+k$$
$$k^2=2k$$
Now "unsubstitute" n=1 and we'll take the limit after rearranging:
$$k^{n+1}=(n+1)k$$
$$k=(n+1)^{1/n}$$
$$2=\lim_{n\to 1} (n+1)^{1/n}$$

Maybe you don't understand what I'm getting at and that's why you don't think it's strange or you do see what I'm getting at and you just don't feel like it's strange in which case I can't convince your emotions to feel the same way as I do haha.

thomas5267 · Answer

The "problem" is that $k^{n+1}=(n+1)k$ should be logically equivalent to $k=(1+n)^{1/n}$ yet taking the limit of n to 0 of the second expression fixes k to e but if we just substitute n=0 to the first expression any k would work.  Are the two expression logically equivalent though?

ganeshie8 · Answer

I see what you're saying now... we do get indeterminate value for $k$ if we simply plugin $n=0$ in $(1+n)^{1/n}$ hmm

kainui · Answer

@thomas5267 I think that's a good point I'm trying to find out if there's something like that floating around.

I feel like the +1 seems to be causing potential trouble but at the same time since the +1 is arbitrary, we could just as well solve:

$$n^k=kn$$$$n^{k-1}=k$$$$n=k^{1/(k-1)}$$
and we'd get the exact same thing if we take the limit as k approaches 1 and we'd still get e. 
Hmm.

kainui · Answer

Alright so even though looking at this it seems that k-1=0 so taking the 0th root gives us this weird limit... I don't know...

At the end of the day I'm really more concerned with why there's a relationship between:

The solutions of $$\Delta f(n)=f(n)$$ and 
$$n*\cdots * n = n+\cdots +n$$$$n^k=kn$$

Looking at this specific limit might give us an idea as to why this relationship exists I just don't want to lose sight by focusing on this limit too much.

ganeshie8 · Answer

Maybe lets interpret each using some physical situation like dumping waste etc

kainui · Answer

Just to show that there's some consistency between everything https://www.desmos.com/calculator/ygnawc0pbz Yeah that's a good idea, I'm trying to think of some geometric or other picture. It seems like we have two different perspectives colliding here so I think there's some general way to understand these. I was looking at perhaps other similar relationships like $n^k = n!$ or something like $n*n*n=n^2+n^2+n^2$ to try to wiggle into a more general perspective idk. Just lost and confused and having fun being interested haha.

kainui · Answer

But I think I'll try to go your route, just wanted to sorta say that as my stopping place so I'd know roughly where I was so I could resume.

ganeshie8 · Answer

Suppose a virus is $f(n)$ on a particular day and the difference in growth between any two consecutive days days is given by 
$$\Delta f(n) = f(n+1)-f(n) $$
I'm fixing $a=1$ for now hoping we can add it later...

ganeshie8 · Answer

And your equation is 
$$\Delta f(n) = f(n)$$

kainui · Answer

Yup so then $f(n)=C2^n$  and the population doubles each time. Unless there are perhaps other solutions we've missed with this.

ganeshie8 · Answer

with that eqn basically you want the virus to grow by an variable amount $f(n)$  on the $n+1$ th day

ganeshie8 · Answer

Yeah here $b=2$.
I'm trying to relate this situation to the other eqn
$$k*k*k*\ldots = k+k+k+\ldots $$

kainui · Answer

Come to think of it, there is some sorta, idk what to call it "internal freedom" on the interval $[0,1)$ in which we can define C(n) as any arbitrary function.

kainui · Answer

When moving towards a=0 the case of $e^x$ we will lose that extra bit, I don't know if it matters I just thought of it and seems possibly useful later.

kainui · Answer

@ganeshie8 So the amount you add with each generation is the amount the total population multiplies by. I feel like I am not sure if that makes it obvious or if that confuses me more now haha.

kainui · Answer

Interesting, I solved for a general solution given an initial function g(n) on the interval [0,a)
So starting with this general solution with the C(n) function, which is only defined on [0,a), I represent that by plugging in $n \mod a$ so that it restricts the domain without me worrying about it.
$$f(n)= C(n \mod a) (a+1)^{n/a}$$

Now what I'd really like is for some initial setup, that the f function aligns with my initial function g, in equations:
$$f(n \mod a) = g(n\mod a)$$ 
So I solve for C:
$$g(n \mod a) = C(n \mod a) (a+1)^{\frac{n \mod a}{a}}$$
$$ C(n \mod a) = g(n \mod a) (a+1)^{-\frac{n \mod a}{a}}$$
Plugging back in we get:
$$f(n) =  g(n \mod a) (a+1)^{-\frac{n \mod a}{a}} (a+1)^{n/a}$$
$$\large f(n) =  g(n \mod a) (a+1)^{\frac{n-(n \mod a)}{a}}$$
$$f(n) =  g(n \mod a) (a+1)^{\lfloor n \rfloor/a}$$
So I guess this is off the beaten path but I just thought this was cool anyways that it simplified down like this haha.

thomas5267 · Answer

$$
\begin{align*}
k^{n+1}&=(n+1)k\
k^n&=n+1\
\frac{k^n-1}{n}&=1
\end{align*}
$$$$
\begin{align*}
f(x)&=a^x\
f'(x)&=\lim_{h\to 0}\frac{a^{x+h}-a^x}{h}\
&=a^x\lim_{h\to 0}\frac{a^h-1}{h}\
&=a^xf'(0)
\end{align*}\
\lim_{n\to 0}\frac{k^n-1}{n}=e
$$
e is only number such that the derivative of $a^x$ at x=0 equals 1.

anonymous · Answer

the eigenfunctions of the finite difference $\frac{\delta f}{\delta x}=\frac1h(f(x+h)-f(x))$ 'continuously' turn into the eigenfunctions of the derivative $\frac{df}{dx}$ with $h\to0$, yes

anonymous · Answer

the analogous 'property' with exponentials here is that they are the functions that satisfy
$$(e^{kx})^n=e^{nkx}$$