Before moving on, let's note another variation of the recentering approach we just took.
What's the need for another variation?
Well, near the end of our last discussion, we found the sum $ \lim_{d \to 0} \sum_{1 \le k \le \log_{1+d} x} \frac{(1+d)^k}{k} $didn't converge.
Right, we addressed that in two different ways.
True. We rectified that outside of the recursive sum... which was fine and useful. Nevertheless, there is another, more involved way of addressing that same issue internal to the recursive definition.
So to do that, let's return to our basic identity again.
And now introduce a different helper function, one closely related to our original alternating series function. Do you recall the form our original alternating series function $(-1)^{k+1}$ took after we had generalized it?
We wrote it as $ a_2(n) = 1 - 2 \cdot ( \lfloor \frac{n}{2} \rfloor - \lfloor \frac{n-1}{2} \rfloor ) $, didn't we?
That's right. In fact, that was also $ 1 - 2 \cdot ( 1 \textrm{ if } k \textrm{ if } 2 | n, 0 \textrm{ otherwise } )$, right? That's conceptually the role of those floor functions being subtracted from each other.
Oh, true. That makes sense.
Well now, we begin with another function with a distinct family resemblance.
Fine, but what is the point of bringing in this function exactly?
Rather than discuss this, let's just take a look at how it behaves.
Ah, okay. Seeing it graphed out that way certainly clarifies its behavior. The graph bears a striking resemblance to a ruler when zoomed out, or so it seems to me.
Perhaps so. Anyway, with this helper function defined, our new modified prime counting function is
Graph it and compare it to our original weighted prime counting function, and we see this:
Wait, what? The graph of the difference here looks identical to the graph from our previous discussion, the original alternating series one...
It does bear an extremely striking similarity, doesn't it? But don't let a cursory glance be deceiving - it isn't quite identical. If you actually work through the difference between the two functions here, you'll find they differ by $$\displaystyle \sum_{1 \le k \le \log_2 x} \frac{2^k-1}{k}$$
Ah ha! So using this function indeed incorporates the $-1$ term in the sum organically. And so we don't have to account for it externally. I can see where that is interesting, in addition to the helper function itself being intriguing.
Right. As usual, here is an animation of that function in action.
And as in the prior example, we can immediately generalize to whole numbers, too. So, for example, the following definition works too:
Okay. And this function also looks like a ruler?
Well, see for yourself... but yes, obviously.
You know, the relationship between this generalization and the last is striking in its own right.
How so?
Well, our original alternating series and all its generalizations were cyclical. They had all fixed lengths at which they started their cycles over.
True.
But here, there seem to be, let's say, cycles inscribed in cycles. So the helper function here doesn't loop, but instead it has behavior that reminds me of fractals, varying at different scales.
I think that's a nice observation, yes. And once again, if we graph $f_1(x, 2)$ and $g_1(x, 2)$ we see
That's certainly a familiar look at this point.
True. And in fact, with this definition of $h(x)$,
$$ g_1(x,2)= f_1(x,2) \rr{- \sum_{1 \le k \le \log_3 x} \frac{3^k-1}{k}} $$And animated, it looks like this.
A bit of experimentation shows this generalizes, and in so doing produces a nice connection to the p-adic valuation for integers, $\nu_m(n)$. If we replace the 3 with some whole number $m > 1$ as
then we'll see that
$$ g_1(x,2)=f_1(x,2) \rr{ - \sum_{1 \le k \le \log_m n} \frac{m^k-1}{k}} $$Exactly as we'd hoped.
So do we generalize to the rationals again, to make a new recursive form that we can take the limit of?
Well, you are welcome to go down that road yourself... but no. I won't be doing that here.
Oh really? Why not?
Let me generalize further, but without putting it into that particular recursive form. If we wanted to generalize this style of alternating series beyond whole numbers, one way to do that would be to define the following function, where $d$ is some real number $>0$.
$$G'_0(n)=1 \textrm{ if } n \ge 1 \textrm{, otherwise } 0$$ $$G'_k(n) = \sum_{j>1} G'_{k-1}(\frac{n}{j}) - d \cdot \sum_{ \substack{ j>0 \\ m > 0 }} G'_{k-1}( \frac{n}{ j \cdot (1+d)^m } ) $$And using this generalization, we could then express the difference between the weighted count of prime powers and that sum we've been investigating as
$$\sum_{k>0} \frac{(-1)^{k+1}}{k} G'_k(n) = \Pi(n) - \sum_{1 \le k \le \log_{1+d} x} \frac{(1+d)^k-1}{k}$$And then, because $d$ can be a real variable $>0$, we could take the limit as $d$ goes to $0$, finally arriving at:
$$\lim_{d \to 0}{} \sum_{k>0} \frac{(-1)^{k+1}}{k} G'_k(n) = \Pi(n) - \mathrm{li}(n) + \log \log n + \gamma$$So why aren't you including a nice recursive formula for this?
Well, previously, with the generalized alternating series, near the end of the discussion, I generalized to reals as $E'_k(n) = \sum_{j>1} E'_{k-1}(\frac{n}{j}) - (1+d) \cdot \sum_{ \substack{ j>0 }} E'_{k-1}( \frac{n}{ j \cdot (1+ d) } ) $.
Okay.
In that context, to go from that to a tidy recursive formula, I needed to count over elements of size $d$, and $d$ needed to be a rational number. That's a consequence of this term in that definition: $E'_{k-1}( \frac{n}{ j \cdot (1+ d) } )$
Go on...
But in our multi-scale generalization to the reals here, the size of the elements I need to count over is much, much thornier. That's because of the sum $G'_{k-1}( \frac{n}{ j \cdot (1+d)^m } )$ - the term $(1+d)^m$ specifically makes things more complicated, and indeed more trouble that I will bother with here.
Fair enough.
So again... you can do it, but I won't endure the trouble.
But just as in the last discussion, this gets us back to the logarithmic integral in the form of $\mathrm{li}(x) -\log \log x - \gamma$.
Done and done.
Let's take a step back for a second and examine what we've just worked through from a slightly higher vantage point.
Why is that?
There's an idea I want to draw attention to, about both our generalizations. This observation is intended to be conceptual. So, in our previous discussion, how did we rewrite our alternating series function, which had begun initially as $(-1)^{j+1}$?
We rewrote it as $1 - 2 \cdot ( \lfloor \frac{n}{2} \rfloor - \lfloor \frac{n-1}{2} \rfloor )$.
Good. Now, what happens if you remove those floor functions?
Well, I suppose you would have $1 - 2 \cdot ( \frac{n}{2} - \frac{n-1}{2} )$... which simplifies down to $0$, actually. That's interesting.
Jump ahead a bit. What was our final generalization for the alternating series, to rationals?
That was $a_{\frac{c}{d}} = d \cdot ( \lfloor \frac{n}{d} \rfloor - \lfloor \frac{n-1}{d} \rfloor ) - c \cdot ( \lfloor \frac{n}{c} \rfloor - \lfloor \frac{n-1}{c} \rfloor )$
And if you remove the floor functions there?
You would have ... $d \cdot ( \frac{n}{d} - \frac{n-1}{d} ) - c \cdot ( \frac{n}{c} - \frac{n-1}{c} )$, which after simplifying, is also $0$, it seems.
Great. Now, what about above? What was our generalized, multi-scale alternating series function?
Well, we stopped generalizing at whole numbers rather than rationals there.
Fair.
But yes, by the time we stopped, we had $ 1-(m-1) \cdot \sum_{k=1}^{\infty} \lfloor \frac{n}{m^k} \rfloor - \lfloor \frac{n-1}{m^k} \rfloor $.
And if you remove the floor function there?
Then you would have $ 1-(m-1) \cdot \sum_{k=1}^{\infty} \frac{n}{m^k} - \frac{n-1}{m^k} = 1-(m-1) \cdot \sum_{k=1}^{\infty} \frac{1}{m^k}$.
And that last term is just a geometric series, equaling $\frac{1}{m-1}$...
Right. And so that entire equation, after a bit of work, simplifies down to $0$, too.
So, putting this all together...
It means all the various generalized alternating series functions here reduce to $0$ if you remove the floor function. Which seems striking somehow.
Another way to put this would be to say that the real equivalents of all these functions end up being equal to $0$, essentially, right? The whole number versions of these functions have all sorts of lively, interesting behavior, but the real versions reduce to a flat $0$.
That seems right.
Meanwhile, what, would you say, has been the major theme of these last two discussions?
I suppose we've been trying to find alternative ways, with our recursive functions, to separate the smooth part of the weighted count of prime powers from its unpredictable, high frequency part. And we've been trying to extract the high frequency part, because its long term behavior is the question that we find so perplexing. It reminds me of the way that integer division, which gives rise to remainders and can't be easily reversed, is so much messier than real division, honestly. All these operations we just looked at, and our weighted prime counting here too, seem to have this fundamental split between their simple, smooth, real parts, and their messy remainders.
That does seem like an interesting conceptual connection, doesn't it? At any rate, we should move on, but I think this property of generalized alternating series helps explain why they play the role they do in separating out the smooth from the high frequency elements in these recursive forms...
Okay. And I'm sure we can connect all this to the Riemann zeta function as well, right?
Sure. If we take our identity from above and generalize it by adding a $s$ parameter, we have
And then if we unroll this definition and take a limit as $n \to \infty$ what the sums converge, we will arrive at $$\log (\frac{1-2^{1-s}}{1-2^{-s}}\zeta(s)) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}(\frac{1-2^{1-s}}{1-2^{-s}}\zeta(s)-1)^k$$ for our original variation, and at $$\log (\frac{1-m^{1-s}}{1-m^{-s}}\zeta(s)) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}(\frac{1-m^{1-s}}{1-m^{-s}}\zeta(s)-1)^k$$ more generally.
Good enough.