Saturday, March 28, 2015

Discovery, innovation and counting prime numbers

Few areas of mathematics resonate with so many as those focused on prime numbers which appropriately have a giant role in number theory. And counting prime number offers a way to be introduced into number theory in a big way with surprisingly simple mathematics. Celebrating simple mathematics is something I deeply enjoy, and talking about it, is a pleasure.

So I'm going to talk about how you can maybe have a feel of discovery yourself, using some really simple mathematical ideas for the basic concepts where you just need to know what a prime number is, and simple division with throwing away the remainder. There will be a bit of calculus though thrown in for completeness near the end but don't worry if you don't know calculus as you don't have to understand it! It's just for those folks who find that interesting.

Now let's do some simple math. And start with considering some simple ideas.

For example 3 divides fully into 10 only 3 times. For number theory integers are important, so we don't care about the remainder. And 3 is a prime number. The numbers with 3 as a factor including 3 itself less than 10 are: 3, 6, 9

We'll use [10/3]  = 3, in order to show we don't care about the remainder, where the [] is something called the floor function, which can also be shown as floor(), so floor(10/3) =3 as it's just another way of showing the same thing.

So we have counted 3, 6 and 9, with [10/3]. Similarly 2 is prime, and the numbers with 2 as a factor up to 10, are 5, where we have that exactly: 2, 4, 6, 8, 10

Now these things were noticed a long time ago, and people figured out how to use this approach to count the prime numbers, like we're in the process of doing up to 10, where it matters that you know the numbers that are prime up to the square root of 10. And don't panic with mention of the square root! Again the focus is on integers, so we actually want the biggest integer whose square is less than 10, which is, yup, 3 again.

With 2 and other numbers with it as a factor, called composite, we have a list of even numbers, and with 3 and its composites we have a list of numbers that have 3 as a factor, and have included ALL non-negative integers up to 10, with those lists except, 0, 1, 5 and 7.

In number theory 0, is not considered a positive integer, and doesn't really matter for counting so we will not mention it again. While 1 is not considered prime.

So we've discovered 5 and 7 are prime with this approach.

And if counting, would note the two primes less than square root of 10, as 2 and 3, and the found primes...wait, so how do we count?

Well if we count for the even composites, we know it's [10/2] - 1 = 4, because 2 is prime, and for composites divisible by 3, we have [10/3] - 1 = 2. But there's a problem!

That problem is that 6 is even AND it is divisible by 3, so we counted it twice!

And one approach is to now subtract for that overcount by counting it, which we can do with [10/6] = 1, which is the hard way.

When I considered this problem over a decade ago, I looked at 3, 6, 9 and thought it silly to count 6 again since I had it counted with 2, 4, 6, 8, 10, and I was right! It IS silly, or maybe I should more politely say, naive. However, hundreds of years ago that's how mathematicians did it, which leads to something kind of convoluted.

My approach was to ignore at each prime the counts already made at a lesser prime.

If you wish to play with that concept and figure out a prime counting function from here, go ahead, and I'll continue with it, conceptually before showing the mathematics that follows.

So now we have [10/2] - 1 = 4 composites, and [10/3] - 1 - 1 = 1 composite, for a total of 5 composites, which is correct! That second 1 was subtracted to handle 6, which may seem like cheating at this point as I know to subtract it, but don't worry, you'll see some math in a bit, which shows how it's done.

If you can figure out how to go from here to a complete mathematical function, then you can get a feel for discovery. Remember the idea here is simple: how can I best prevent overcounting by correcting for counts of composites already counted?

The 5 composites up to and including 10 are: 4, 6, 8, 9, 10

So yes! We DID count them, so what's the count of primes?

The prime count is: 10 - 5 - 1 = 4.

Those primes of course are: 2, 3, 5, 7

So we subtract the 5 composites, and then have to subtract 1, for yup, 1, which is not considered prime, and we have the full prime count.

That concludes our first prime count! Easy, right?

One thing I want to talk more about is why we need to know primes up to the square root of 10.

Why? Well we're pairing primes up with other numbers with composites, and if you get to products of bigger primes, like 5*5 = 25, you are beyond 10. For every composite less than or equal to 10, you will have a prime factor less than the square root of 10, as it takes two to make a composite. That means we can focus on those primes, which are 2 and 3.

The same principle would apply for bigger numbers, like for 100, we need 2, 3, 5 and 7.

When you get to 11, 11*11 = 121, but for anything smaller, one of those primes will have to show up. For instance, 7*11 = 77, gets counted when considering composites with 7 as a factor.

Most of the concepts presented here were quite revolutionary--hundreds of years ago! Where I snuck in something innovative. If you caught it, good for you! It was a new way of looking at things introduced by me in the summer of 2002.

Did you notice it?

If you missed it, see if it jumps out at you in one of the equations I use. If you figured it out on your own before getting here, good for you! If you didn't don't feel bad.

The main workhorse is something I call the dS function, which corrects the count at each prime without overcounting. THAT was my innovation! Just seemed like a smart thing to do.

The corrected count, which I call dS(x,pj) at each composite, where p is prime, where j tells you which one, like j =1 for 2, as it's the first prime, and j = 2, for 3, and so forth is:

dS(x,pj) = [x/pj] - 1 - (j-1) - S(x/pj, pj - 1)

where S(x/pj, pj-1) is the count of composites that multiply times pj to give a product less than or equal to x, where notice that pj must be less than or equal to sqrt(x) or the composite count given by [x/pj] - 1 will not be correct.

That embodies everything we've been doing up to this point. If you figured it out, good for you! If you can read it given our discussion that's great as well.

Now I'll go back through what we did with 10, so you can see it in action.

So for dS(10,2) = [10/2] - 1 - (1-1) - S(10/2, 2-1) = 5 - 1 - S(5,1) = 4

Here S(5,1) = 0, as there are no composites less then 5 that multiply times a prime, as 4 is the only composite less than 5, and it's a product of primes.

Which is what we go before. So it's just saying there are 4 even composites, just like we found out before. That takes no effort for the thing.

I get a little fancy in not showing the floor in the S function, so next is.

dS(10,3) = [10/3] - 1 - (2-1) - S(10/3, 3-1) = 3 - 1 - 1 - S(10/3, 2) = 1

And there's even less room beneath 10/3, as there are no composites beneath it, only 2 and 3 which are both prime.

Here it actually did some work! It correctly threw out another 1 as it corrected for 6.

To see that other piece with the S function pull out other overcounts you have to use bigger numbers. For example S(16/3,2) = S(5,2) = 1, because 2(2)(3) is the only composite case where 3 is multiplied times a composite up to 16.

I don't typically write the floor() inside the S function as I consider it redundant.

If you've followed this far, and maybe wonder what is the point of all of the above, or maybe even still wonder how I got it, never fear! I've discussed these things for over a decade and am concentrating things into a single post for reference. To see it all explained in detail, one of my better posts I think is:

This approach also leads to a nonlinear partial differential equation:

P'y(x,y) = -(P(x/y,y) - P(y, sqrt(y))) P'(y, sqrt(y))

Here's my most recent post which links to a post which steps through the derivation:

To me the research here is most dramatic in showing the simplest, fast prime counter humanly available.

Where pj is the jth prime:

P(x,n) = x - 1 - sum for j=1 to n of {P([x/pj],j-1) - (j-1)}

That summation will count primes if you make sure n equals the count of primes up to sqrt(x), but no higher. And an example of what it gives is P(100,4) = 25.

The summation is built around what I call the dS function, and you can see pieces of it in there, but with it close to full mathematization it's just a lot more compact.

You can actually go even further and have it find primes on its own, where it promptly looks a lot messier:

To see a recent step through of a demonstration I think the following is a good post which is on one of my other blogs:

That is the fastest prime counter available for its size as it completely encapsulates the ideas introduced here, and it turns out that shrinks things down considerable as to presentation.

It's NOT the fastest prime counting algorithm available though. It IS the fastest for its size, with the smallest presentation you will see for a fast prime counter. It is easy to explain, present, and implement, so I think it's a great introduction to the subject.

But here is the full prime counting function fully mathematized where it figures out which numbers are prime below square root of x on its own. Figured I should include it, though it's not nearly as pretty as when you hand it those primes, and it's not as fast:

P(x,y) = [x] - 1 - sum for j=2 to y of {(P([x/j],j-1) - P(j-1, sqrt(j-1)))(P(j, sqrt(j)) - P(j-1, sqrt(j-1)))}

where if y is greater than sqrt(x), then P(x,y) = P(x,sqrt(x)).

It's actually a little prettier with the summation sign and square root symbol though it's still all integers. But Blogger doesn't make that easy for me to do, so I've just used the format above.

There's nothing else like it in number theory. It actually involves something called a difference equation. If that means nothing to you, don't worry! Just covering all the bases and this thing links to some of the most complicated mathematics human beings have ever discovered. It is one of my greatest discoveries.

Whew! What a journey. That was SO much fun too. From simple to advanced, we journeyed from very easy concepts to mathematics touching on the limits of human knowledge.

And I've stepped through the concepts it uses in this post! If you can understand what I said previously about counting composites at each prime without counting ones previously counted then you have the principles in place! And maybe can kind of see things within it that echo back to what was above.

If you've skimmed through this post to the end, maybe you can challenge yourself by trying to derive the prime counter JUST using the principles explained. And if you get stuck you can check out the various links to find more detail on how it is done.

James Harris

Sunday, March 15, 2015

Two conics connected

One of my more popular research areas explores a connection between ellipses and hyperbolas--two of the three conic sections.

The connection starts with the equation x2 - Dy2 = 1.

It is a hyperbola if D is a positive integer, which is the assumption throughout this post.

Rare situation where I did things in rationals, as I'm going to copy a LOT from a prior post and expanding on it a bit:

(D-1)j2 + (j+1)2 = (x+y)2 where j = ((x+Dy) -1)/D


(D-1)j2 + (j-1)2 = (x+y)2 where j = ((x+Dy) +1)/D

And those are ellipses.

And going to give an example where you get all integers. Why that happens is explained further down, when I include more research from a second post on the subject.

With D=2, and x=17, y=12, 172 - 2(12)2 = 1, and using the first one:

j = ((17+2(12)-1)/2 = 20 is a solution giving:

202 + 212 = 292

And it's like, who knew? But now you know that:

172 - 2(12)2 = 1 is connected to: 202 + 212 = 292

Oh yeah, despite being in rationals, noticed this weird thing that the result means that you can always find integer solutions for an equation of the form:

af2 + g2 = h2

using a rational solution for x2 - Dy2 = 1 when a = D - 1. That's just kind of freaky cool.

For more details including links to derivations check out the full post:

Later I focused exclusively on integers and found more rules.

The ellipse is: (D-1)j2 + (j +1)2 = (x+y)2 where j = ((x+Dy)-1)/D, when x = 1 mod D.

Or: (D-1)j2 + (j -1)2 = (x+y)2 where j = ((x+Dy)+1)/D, when x = -1 mod D.

And I used a famous large solution to show an example:

Here is an example of the result with the famous case of D=61:

17663190492 - 61(226153980)2 = 1

and x = 1766319049 = -1 mod 61, so the second equations apply.

(D-1)j2 + (j - 1)2 = (x+y)2, j = (x+Dy+1)/D

Which gives j = 255110030.

60*2551100302 + 2551100292 = 19924730292

That post, which also includes a proof, is:

So I've copied key points from two related posts.

The connection between ellipses and hyperbolas in rationals is a fascinating one to me. It means that one can be used to generate the other and these are infinity connections.

Despite it being a rational connection, or because of it, there is a Diophantine variant, where the residue of x modulo D is important. The importance of that residue modulo D emerges elsewhere as key to the size of the fundamental solution x2 - Dy2 = 1.

So things may be kind of convoluted for us (or is it just me?), but perfectly logical mathematically where these things are just profoundly connected.

It's kind of interesting then that this connection may be HUGE for the conics involved. It links their solutions, so I guess that makes sense.

James Harris

Thursday, March 12, 2015

For fans I guess

Last year I kind of had a theme of sharing certainty, and this year I've been moving in a direction I finally accepted was showing pity on anyone who is actually a fan of me or my research. If you're a fan, um, sorry.

It's a very frustrating situation, and it occurs to me I should show more concern other than explaining yet again how I don't need mathematicians, not helpful for fans I'm sure. Or explain even more why I'm not so much interested in fans personally but would like fans of my research. Yet do I really help fans of my research out, if they exist, by for the most part simply explaining things over and over again in blog posts here?

But I'm not sure what's helpful for someone who wants to cheer this research, or, um, me. It just seems so awkward.

Yeah, I've been sitting here tapping keys nervously for a bit, and got nothing.

It IS hard. I don't know what to say.

My idea over a decade ago was just you find some interesting math, show it to people who are interested and things happen.

Now over a decade later I think you need a marketing plan or something. Also it would help if you didn't spend a lot of time insulting experts in the field where you supposedly found something, relentlessly explaining how you are NOT an expert in that field, and keep talking about how you get little support from the community of that field.

You know, I was thinking this post would go a different way, but it is a harsh situation, if I DO have any fans, and I'm not sure I do! And that's not a big deal.

I prefer BEING a fan, and it's so awesome!!! As a fan of mathematics I've had the best adventure of all time. And it just keeps getting better and better. I keep finding things!!!

So if you're in some way a fan--who may not actually exist so I may be talking to no one--I don't think I have much comfort or hope to offer.

To me it's all about the math. And as a fan of mathematics I am beyond giddy!!!

It just doesn't get any better. From my fan perspective, I am just right in the middle of it, on the greatest fan experience of all time. This post wasn't what I thought it would be, but yeah, I'm the fan. Others gotta fend for themselves.

Woo hoo!!!

James Harris

Wednesday, March 11, 2015

Utility and issue of mathematics fans

Spent a couple of posts recently explaining my supposed independence from the opinion of mathematicians, which I think raises the question of what delusion really drives me if I think I can talk about "some math" without mathematicians?

And I say the primary issue is utility: mathematics DOES something right?

It's like, can you appreciate electricity without physicists?


Mathematicians are the academic specialists in an area of great utility, like physicists are in a HUGE area which includes electricity which I just picked because it's so exciting. I could pick other things, like chemistry, which still requires physics, as well as chemists, but definitely less electric of a topic, and I needed that pun. And most people don't know who any modern mathematicians are anyway, and don't need them to use most of the mathematics they need.

Most people no more need to know any modern mathematicians to do lots of important research and work needing mathematics than they need to know modern physicists to use--electricity.

It's not complicated.

To me the MOST important thing is finding useful mathematics, while for most academic mathematicians usefulness is a strange topic. Many of them focus on something they call pure mathematics.

However, beyond the usefulness where you get that fun thing of helping bring value to the human species, for solutions to problems you can't even imagine, as consider what mathematics helps solve beyond the imaginations of say, Euclid, there is the sheer joy of learning.

So you suppose there will be fans of mathematics.

The issue of math fans is completely different from utility, just like sports fans are a separate category from the sport.

Those are the people who cheer a result.

Which puts it all together! For instance Euclid doesn't need to worry about people's opinions about his work any more. He's been dead for a few thousand years, so he's past worrying anyway. And it's not like he necessarily has a ton of fans with kids forced to learn in school about results he collected. Yet the utility of that mathematics is unaffected. And there ARE fans of Euclid, who are people who get excited about such things, but more importantly there are fans of the results he put in one place--for the good of humanity.

Now does it make sense?

So for instance with any particular result of mine, I'm most interested in its utility, where I get the most positive feedback, which comes from search results. It's the MOST important thing of all, as that is what means your efforts will endure. Nothing else does.

Next I'm curious about fans of the mathematics itself. Where there is a lot less indication to me that there are any. Which is interesting.

Next is the issue of fans of myself, and I tell myself I don't want any. So that one is easy. One wonders of course if that is a lie. Is it just something I want to believe?

Now shift to the modern academic world of mathematics, and I'm sure you could find mathematicians who clearly DO want fans. I'm not a mathematician and I tell myself I don't want fans, which is remarkably easy to achieve. (Not surprisingly, eh?)

But I would like fans of my mathematical research. I'm a fan of it, which is why I cheer about it so much! That's what fans do. You know, like with sports teams.

See? Simple.

And I find myself adding more over the question of do I want fans of me, and my feeling is that can bring out the worst in you and is something you don't control. If people consider themselves fans of someone that's their business. No one elses. Almost wish I could have left that subject out entirely but think it's pertinent to the overall picture.

Maybe more important, I think for some people their own need for fans can skew their perception.

For me fans are great. I like fans. But I don't try to make fans. I prefer to be one myself.

James Harris

Monday, March 09, 2015

Loss of the gatekeepers

Was wondering a bit about why I feel like I don't actually need much if any participation from established mathematicians with regard to my research, as it is math, and pondered: what would happen if I didn't have the web?

Well I'd type things up into some papers and try to figure out some people who understand math, who might appreciate it, who I could mail. Those would most likely be mathematicians.

Probably would maybe lobby a bit as well, as in, visit some college campus, or make friends with people on some college campuses as I worked to get to some level of trust and mutual accord where I could show them some of my research.

But with the web I don't need any of those things. And in fact, they'd just get in the way.

That's HUGE. In essence the scenario without the web involves gatekeepers.

With the web though I can just put my mathematical research online and people who are interested can find it, without needing all that extra.

The web can simply give desired information to the person who wants it, without as many middle-people.

The web can just cut-out the gatekeepers, which it has also done in other areas. So fascinating. So new.

For me it's a major issue which I study quite a bit, and I guess some might be worried about noise gaining traction, where without experts to clarify and critique someone might gain progress with bogus results. However in number theory it's hard to see how that is possible. You'd need people all over the world valuing highly results that don't actually work, and it turns out that's easier to have happen for people with status!

Why would anyone bother with mathematical research from some person without any esteemed social status if it didn't work? It pushes the limits of our understanding of human behavior.

There are areas of course where some people usually members of the public who desperately want something to be true will tend to choose bogus research, but generally there is some wish, which defies scientific reality. It's a need for fantasy. With concrete mathematics like number theory there is very little room for such fantasy.

But if mathematics works very well then it does make sense for it to be valued, regardless of the source, so people would simply need the opportunity to get it.

So the web seeking out and finding math research from a non-established source is I think more evidence in favor of that source than if from an established expert, or other potential gatekeepers, as the motivation is more basic.

People MIGHT seek out research, say, from a famous mathematician, even if they didn't understand it, but from some person without status? Makes sense only if it works and they need it.

Simple. Which means I don't need mathematicians. Worse, if I convinced mathematicians early that my approaches might have merit, I could have a group of people along with me by now--sharing credit.

So the web not only can remove a role for the gatekeepers, it can put you in a situation where using prior ones can feel like a waste of your time.

Will ponder this thing more, but I am sure it's new! My guess is lots of people labor under the belief that if they have some great math idea they should get mathematicians to acknowledge it, rather than seeing them as competitors.

Wonder how that will shift things over time?

Maybe I should give one example of the power of the web: claim from me is that I have the BEST way to reduce binary quadratic Diophantine equations EVER found by humanity?

Doubt me?

Search: reduce binary quadratic Diophantine equations

What need I gatekeepers then?

Just did that search for myself on Google and came up only #5, which is lots less dramatic. I'd prefer #1 but I don't control it. Oh well.

Still don't need, or want, gatekeepers.

Of course I could edit to tone done the post, but the reality of web search is that it IS dynamic. Even if I came up #1 in my search that's not necessarily what someone else would see.

Worse, these type posts can actually, yup, drive down such search results! The web tends to react, and not in a good way.

So why do them?

Well I'm not as interested in having top search results as one might think. The competitive side of me loves them but the pragmatic side is less thrilled.

But the real point is: I do have the best way to reduce binary quadratic Diophantine equations ever found by humanity.

So it should be #1.

Which means to me, I just noticed search making what I think is a mistake, as it should return the best not at #5 but at #1. Looks like Google is trying to give some undeserved stature to established sources. And that's of interest.

If someone doubts me it's an objective area where they can compare my method against all others known.

Oh, I know! Maybe I should give one where currently I DO have #1 when I search. So here's another search. I have lots of them actually.

Search: count quadratic residue pairs

That should be more impressive. Again, there is a good reason for me to be at the top of the heap. It is the one that I think best makes sense too!

James Harris

Saturday, March 07, 2015

Now THAT is support

Should mention I do have heroes in the modern mathematical community, which I guess is in a very understandably selfish way, as I'm a mathematical author, and know of one organization which has preserved my work, along with that of lots of other mathematical authors online and that is the European Mathematical Information Service.

And no, they were NOT doing it for me personally as a favor, but it's just something that they do: preserve mathematical papers.

And it's so fascinating that for me my own story revealed things I never cared to know before.

So here is a link to the preserved archives of the mathematical journal that published my paper:

EMIS Archives--Southwest Journal of Pure and Applied Mathematics

And here is a link to the table of contents of the edition which had my paper, which the chief editor changed after the fact to "Withdrawn", I did NOT withdraw it, so here is what came later:

EMIS Archive--SWJPAM Issue 2, December 2013 Table of Contents

And you can get the actual paper regardless, as EMIS preserved it, and I'm going to include a link, but should add it is to Postscript format so don't just assume you can download and read directly. If you click the link it will download:

So the chief editor's actions were rather worthless. And it's also kind of a sad situation. I HAD informed the editorial staff I was NOT a mathematician from the outset. And had been in contact with them for nine months prior to publication. And I think they had some sense of the importance of the paper. You can see in the Table of Contents it was published as the second paper of that edition. But maybe were shocked when the mathematical community did not react as expected.

It should have been a historical triumph.

I'm still considered to be a published mathematical author, as I am. You can't unpublish, as that's logically impossible. What you can do is issue a retraction or a refutation. And neither of those things were done.

So what does it matter? Well it was a formally peer reviewed and published paper, where I have since made things simpler in explaining what it was discussing, so that now you can just consider:

P(x) = (g1(x) + 1)(g2(x) + 2)

is blocked if P(x) is a primitive quadratic irreducible over Q, and g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x.

I used cubics in the original paper but realized quadratics would work just as well and were simpler to explore.

But the point is that it IS an established result--by the rules.

To me the people at the European Mathematical Information Service are great in that they have rules and they follow them.

They are heroes to me. And I was so glad to find out they existed. Of course I found out for some rather stunning reasons, but regardless it's so much better to learn our world has support in ways you don't even realize--until you need it.

Those other mathematical authors are probably grateful as well. You can see the journal had a bit of history, and consider all those other papers that were preserved.

For me, of course I really lost nothing. Stories like mine can quite simply change the world, but you do worry about others who can be collateral damage. So I'm very glad their research did not simply get lost in the futile effort to hold down mine.

James Harris

Thursday, March 05, 2015

Why harder to convince

It occurs to me that noting it has been over 10 years since I found my innovative tweak to old ways of counting prime numbers, which leads to a simpler, fast approach, and a partial differential equation, may not help with those who may wonder, but what do the experts say?

And in retrospect I've thought that I made one basic mistake early on, which is I found and successfully contacted the leading mathematical experts in this area.

So yes, the result is verified.

But one of those men simply said I should put it on arxiv, and I replied as a non-academic I didn't have access to it, but could put it up if he helped me. He never replied further.

The other told me he didn't find it to be of interest.

But did refer me to a colleague who wrote the first implementation in C. But then begged off saying he had other research to do.

Which left me arguing with nasty people on the web, and as a lark just to see what they'd do, I posted the C program from the mathematician, and they ripped on it relentlessly. Which gave me an odd grim satisfaction. They thought I'd written it, not one of the top mathematicians on the planet.

But the leading mathematicians knew what I didn't then understand: without support from established mathematicians I would be a lonely figure making big claims, in a world where people see that as indication that you must be wrong.

I've contacted other mathematicians through the years, and noted reactions where my favorite was the mathematician who insisted my prime counting function didn't exist! I noted I had sent it to him, so he could see that it did exist. He did not reply further.

So why would they react that way?

I finally realized over time that in one swoop I could be seen to have topped them all, some guy, without a math degree even, who just happened to notice something.

No way they'd simply hand me what they might see as, the crown. I'd have to fight for it.

I don't even want it. I'm NOT a mathematician, and I don't want to be one, ever.

Still, turns out that leads to MUCH more adventure.

It is kind of weird though.

Just because you have a major mathematical result, doesn't mean you get social approval for it, as one thing is about facts, but the other is about choice.

Oh yeah, so now I would not have gone to mathematicians considered the leading experts in the field! But then again, one wonders where could I have gone?

Come to think of it, thankfully, I started with the mathematicians considered tops in this area in the world! And should be glad I was able to get feedback from them. Whew!

Talk about luck! I'm glad they did talk to me. Could have tried them and not received a response and then always wondered if that is what was needed. But with one also, later on down the line with other mathematicians could compare.

Of course now, I know, and no, would not take any new research to the top person or persons in that research field! Have the benefit of experience now.

So where could I have gone? Where can I go now?

Guess maybe I should put that out there.

The answer of course is easy: there is nowhere to go. It's a social problem. Which gets fixed over time by attrition. As more and more people become aware of better research, and as the established people are replaced by new people, it becomes acceptable and accepted.

It's worked that way for thousands of years.

James Harris

Wednesday, March 04, 2015

My prime counting confidence booster

Mathematical discovery can give you confidence boosters impossible for anyone to remove, like one of my favorites is MY prime number counting function in its sieve form.

Where pj is the jth prime:

P(x,n) = x - 1 - sum for j=1 to n of {P([x/pj],j-1) - (j-1)}

That summation will count primes if you make sure n equals the count of primes up to sqrt(x), but no higher. And it is integers only in there. I'm using the bracket form of the floor() function so for instance [10/3] = 3.

And an example of what it gives is P(100,4) = 25.

Which is the count of primes up to 100, where there are 25. And in its sieve form it needs to be given the primes up to sqrt(100) = 10, and there are 4 of them.

And those primes are: 2, 3, 5 and 7.

It needs those same primes up until 121, so they will count up to 120.

That's using the form where it needs to be told the primes up to sqrt(x), but you can fully mathematicize it into a form where it finds them on its own. But that slows it down and it looks a little more complicated.

Turns out you cannot express a prime number counting function as simply that works as fast.

What I said there is very specific and highly checkable.

For instance, you can write a computer program implementing it, and do timing tests, as well as check lines of code needed versus any other algorithms for counting primes.

It's the best by those criteria ever found which is verifiable by objective measures.

That's one of the best things about mathematics, of course, the ability to check! Opinions? Not needed.

Which is the kind of thing that can make me feel better if things get me down.

It's also great for figuring out social stuff, as everything I've written here is true, and the harder thing is accepting that I can confidently claim to have one of the greatest finds in mathematical history. It just feels wrong to admit it. I work at that feeling.

But how other people react, what they believe? Now that's another subject entirely.

I think some people rely more on community than facts: trusting that if enough people believe something it must be true.

That's why reality testing is so important.

And that's just one of my mathematical confidence boosters. It is not my greatest discovery either.

Which raises that question, what makes a mathematical discovery "great"?

In this case it's an assertion based entirely on merit: simpler, fastest for its size, smaller, does more.

Each of those assertions--checkable.

Nothing else for counting prime numbers is even close when objectively compared.


James Harris

Sunday, March 01, 2015

Beyond polynomial factorization

Some things can be so familiar until we take it for granted that there's one way to do things, and can miss what happens if you break those assumptions. So for instance, consider a simple polynomial factorization:

P(x) = x2 + 3x + 2 = (x+1)(x+2)

That is such an obvious thing to do it's hard to think of factoring any other way, but of course can have quadratics irreducible over Q, which just means they do not so simply factor, like:

P(x) = x2 + 5x + 2

Though if P(x) = 0, you can solve for x, but what about just not factoring into a polynomial at all?

What I discovered was that nonpolynomial factorizations were a wide open territory, which I know because they indicated weird things, like consider the following result.

In the ring of algebraic integers the expression:

P(x) = (g1(x) + 1)(g2(x) + 2)

is blocked if P(x) is a primitive quadratic irreducible over Q, and g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x.

That is something where I recently posted a proof, and had to make a bit of an adjustment to what I had originally as that last bit, is necessary to force a factorization versus having:

P(x) = (g2(x) + 2) if g1(x) does equal 0 for all x.

Where of course then there's no factorization! And I was thinking factorization but had to write that down so that it was clear. And the result is weird because it shows algebra being blocked for a particular ring, which is the ring of algebraic integers.

That result has important consequences as it means you cannot display an example of:

P(x) = (g1(x) + 1)(g2(x) + 2)

if P(x) is irreducible, the g's equal 0 at 0, and g1(x) does not equal 0 for all x.

Looking at what you can show:

P(x) = x2 + 3x + 2 = (x+1)(x+2)

It's clear that polynomial factorizations work just fine. And notice here all the other requirements are met, as you have: g1(x) = x, g2(x) = x

So you can't show that factorization when P(x) is irreducible, but guess what? If you multiply by some nonzero integer other than 1 or -1, and do some simple algebraic manipulations to force symmetry, then you can show the result of that!!!

c1*P(x) =  (f1(x) + c1)(f2(x) + c1)

Here c1 is some nonzero integer not 1 or -1, and you can force that factorization, which is not polynomial when P(x) is irreducible over Q, and actually see the results!

And it's worth emphasizing that forcing symmetry is important as that is what makes it possible to get close as you can visually see the f's expressed.

So what does such a nonpolynomial factorization look like? I've used one basic example for years, where I have c1 = 7:

7*P(x) = 7(175x2 - 15x + 2) = (5a1(x) + 7)(5a2(x) + 7)

So f1(x) = 5a1(x), and f2(x) = 5a2(x), where the a's are roots of

a2 - (7x-1)a + (49x2 - 14x) = 0

so the f's are algebraic integer functions, since the polynomial expression is monic, as long as x is an algebraic integer.

And if you want you can solve for the a's using the quadratic formula. So yes, you can see them, though I think it looks a bit messy or I'd show it here.

In contrast g1(x) cannot be seen directly, but 7*g1(x) = 5a1(x), which can be seen, as multiplying by 7 makes it visible. And I matched indices but there's no way really to pick which root is which! So it's more of something that works for human eyes.

The f's are roots of the same quadratic expression, which is forced on them as they're indistinguishable.

So you have this weird thing: the base expression cannot be shown, with actual values though you can show it algebraically in abstract, but you can show with actual equations if you multiply by an integer constant.

So we can always see:

P(x) = (g1(x) + 1)(g2(x) + 2)

as an abstraction, but we can't fill in the g's, if P(x) has integer coefficients but doesn't factor into two polynomial factors with integer coefficients, if we don't allow g1(x) equal 0 for all x.

But we can multiply P(x) by some integer and display the f's, when:

c1*P(x) =  (f1(x) + c1)(f2(x) + c1)

As long as c1 is not zero, 1 or -1.

However, what we can see displayed in a way familiar to our human eyes does not change what exists! And of course in the field of complex numbers it exists, but we still can't see it displayed.

So the ring of algebraic integers doesn't tell us what doesn't exist, only what we can see in a familiar way or not.

So you may wonder: what happens if we try to force the issue and see it anyway?

Well the multiplying integer is entangled. It's not visually separable.

For example with:

a2 - (7x-1)a + (49x2 - 14x) = 0

If you try x = 1, that is:

a2 - 6a + 35 = 0, which solves as a = 3 +/- sqrt(-26)

The 7 is entangled in there and NOT separable to our human eyes. But mathematically it is possible to remove it. So it's something we can't see that the math can do anyway.

That the math can remove that 7 as easily as we can from polynomials is fascinating to me, as what our minds can visualize is not a limitation mathematically, as the math is not human.

This result shifted my thinking about mathematics as it reveals a fundamental limitation of creatures like us to visualize mathematics. It also shows how our biases for the familiar can lead us away from surprising answers!

For most people doing polynomial factorizations is so obvious it might never occur to them to do a nonpolynomial one, whereas for the math, one thing is hardly different from the other.

James Harris