Falling factorial and a functional equation.

Question

kainui · Answer

If we define the falling factorial as:
$$n^{(k)} = n(n-1)\cdots(n-k+1)$$
For instance, $5^{(3)} = 5*4*3$ then we have the finite difference $\Delta f(n) = f(n+1)-f(n)$ applied to it as:

$$\Delta n^{(k)} = k*n^{(k-1)}$$

Pretty like the power rule! :P

So now I'd like to switch this around; that's the forward finite difference and I'd like to look at the backwards finite difference now,

$$\Delta f(n) := f(n)-f(n-1)$$
What function fulfills this functional equation:

$$f(n,k)-f(n-1,k)=k*f(n,k-1)$$

kainui · Answer

Ultimately all of that is just motivation. I'm really just looking for an f that satisfies:

$$f(n,k)-f(n-1,k)=k*f(n,k-1)$$

But it's still fun either way :P

kainui · Answer

Feel free to modify this any way you like to make it "more natural" like right now I think just asserting that $f(n,1)=n$ and $f(n,0)=1$ is probably a good start. I have no idea where to go with this, but those are true for the falling factorial as well as the regular exponents, $n^{(1)}=n$ and $n^{(0)} =1$.

ganeshie8 · Answer

Your falling factorial seems to represent a permutation : n^(k) = number of ways of arranging k objects choosing from n objects

kainui · Answer

Yeah, you could alternatively say from a kind of opposite perspective of "what is more fundamental" and say that you can represent the factorial in terms of the falling factorial as this which is kinda a cute thing too
$n^{(n)} = n!$

kainui · Answer

Also, it appears that when $k>n$ and n is a positive integer, then $n^{(k)}=0$. Just sorta acquired an interest for this function solely because it looks like the power rule in derivatives when you take its finite difference.

kainui · Answer

Just as a guess, I tried this, which I guess is probably called the rising factorial,

$$x^{[k]} = x(x+1)\cdots(x+k-1)$$
Which ends up, for the backwards finite difference being:

$$\Delta x^{[k]} = kx^{[k-1]}$$

So I feel pretty content with this. I am really more interested now in two separate things, what about the split difference:

$$\Delta f(x) = f(x+1/2)-f(x-1/2)$$
and perhaps more general in that direction but more importantly since the Dirichlet convolution in general for any function of a power of a prime is:

$$(\mu \star g(p^n) = g(p^n)-g(p^{n-1})$$
I'm ready to try to see what I can do to apply what I've learned here about the backwards finite difference to this to find some function of an argument which I'll just denote hopefully intuitively as $f^{(k)}(p^n)$ such that we also get the power rule,

$$(\mu \star f^{(k)})(p^n) = k * f^{(k-1)}(p^n)$$
Or perhaps, instead of multiplication, it should involve a convolution with a constant function k maybe, 
$$(\mu \star f^{(k)})(p^n) = (k \star f^{(k-1)})(p^n)$$

Whatever ends up feeling most natural, if anything lol.

kainui · Answer

Wishful thinking, I had tried:

$$f^{(k)}(p^n) = p^n p^{n+1}\cdots p^{n+k-1}$$
and I ended up with:

$$(\mu\star f^{(k)})(p^n) = p^{n-1}(p^k-1) f^{(k-1)}(p^n)$$
So close, and these other extra terms seem kinda interesting so I might play with this a bit to see if it can be fixed somehow.

kainui · Answer

I'm considering taking "the easy way out". No, not suicide lol, I mean I can define a multiplicative function that throws away the prime to just be exactly the rising factorial. Then it just collapses down to the regular backwards difference. Lame imo but maybe it's interesting for something.

$$f^{(k)}(p^n) = n^{[k]}$$
$$(\mu\star f^{(k)})(p^n) = k f^{(k-1)}(p^n)$$

ganeshie8 · Answer

If we define $n^{\{k\}}=\dfrac{n^(k)}{k! }$, we have
$$\Delta n^{\{k\}}=n^{\{k-1\}} =\dfrac{\Delta n^{(k)}}{k}$$

kainui · Answer

Excitingly, there are three cool consequences of this!
In order to insure that it's multiplicative we require:

$$f^{(k)}(1) = 1$$
We also have this pretty simple fact for the empty product:
$$f^{(0)}(p^n)=n^{[0]}=1$$

This means we can write the Riemann zeta function as a Dirichlet series,

$$D(f^{(0)};s) = \zeta(s)$$
And that implies that since $(\mu \star f^{(1)})(p^n) = 1f^{(0)}(p^n)$ that we have their corresponding Dirichlet products (since the convolution in the functions relates to a product in the series of course):

$$D(\mu;s)D(f^{(1)};s) = \zeta(s)$$
Strangely enough, since $\frac{1}{\zeta(s)} = D(\mu;s)$ it would appear that we have:
$$D(f^{(1)};s) = \zeta^2(s) = D(\tau;s)$$

So perhaps I've gone wrong somewhere.

kainui · Answer

Or maybe I discovered some interesting thing. I dunno yet lol.

kainui · Answer

Weird so...

$f^{(1)}(n)$ is equal to the product of all nonzero exponents on the prime factorization of n.

This convolved with the mobius function is the unit function, $u(n)=1$ however the divisor counting function $\tau(n)$ also is convolved with the unit function to get the same unit function.

So either the product of the nonzero prime factor's exponents is equal to the number of divisors of the number, or the Dirichlet convolution with multiplicative functions doesn't form a group...? Seems fishy. Proof:

By multiplicativity, 
$$f^{(1)} (n) = f^{(1)} \left(\prod p^k \right) =\prod f^{(1)} \left(p^k \right) $$
By definition of rising factorial $f^{(1)} (p^k) = k^{[1]} = k$

$$f^{(1)} (n) =\prod k  $$

Now, let's start fresh and show that taking the convolution with the mobius function gives:

$$(\mu \star f^{(1)})(n) = \sum_{d|n}  \mu(d)f^{(1)} (n/d) = \prod_p f^{(1)}(p^k)-f^{(1)}(p^{k-1})=\prod k -k+1 = 1$$

That middle step is due to the fact that it's multiplicative and the mobius function is 0 for most terms.

kainui · Answer

Interesting. It is a mistake in the derivation, the derivation is almost identical since tau is essentially shifted by 1.

$$\tau(12) = \tau(4)\tau(3) = (2+1)*(1+1)$$
$$f^{(1)}(12) = f^{(1)}(4)f^{(1)}(3) = (2+0)*(1+0)$$

In the mobius convolution, the added +1 cancels itself out interestingly.

Here's the problem with it, in my derivation I said this was 1, but the exception is here.
$$f^{(1)}(p)-f^{(1)}(p^0) = 1-1 = 0\ne 1$$