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Friday, November 30, 2012

More detail on weird construction

Here I want to explain a good bit more about a weird mathematical construction.

That construction is as follows.

The start is very importantly in the complex plane:

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

where the a's are roots of

a2 - (7x-1)a + (49x2 - 14x) = 0

At this point there is just quadratic construction and the information that the a's are kind of bizarre in that they are roots of a quadratic! Not something you see every day I think, but also VERY IMPORTANT that at this point you're in the complex plane. There is a good reason for that being true which comes up soon.

And now normalize the functions, that is, have functions that equal 0, when x=0.

Looking at x=0, gives

a2 + a = 0, so a1(0) = 0, or -1, and a2(0) = -1, or 0.

So I can let a1(0) = 0, and then introduce a new function b2(x), where:

b2(x) = a2(x) + 1, so a2(x) = b2(x) - 1, and making that substitution:

7(175x2 - 15x + 2) = (5a1(x) + 7)(5b2(x)+ 2)

where now a1(0) = b2(0) = 0.

So with normalized functions I have:

7(175x2 - 15x + 2) = (5a1(x) + 7)(5b2(x)+ 2)

And a lot of the idea here is that it's starting to look clear from what we can see on the complex plane how the 7 distributed, where it looks easy.

So now I have easily that the 7 multiplied in a trivial way--which makes sense as I made up this example, why would I make it too hard?

So now it looks like we need one more functional substitution which is:

a1(x) = 7b1(x)

Which gives me:

7(175x2 - 15x + 2) = (5(7)b1(x) + 7)(5b2(x)+ 2)

And I can divide off the trivial factor of 7 to find:

175x2 - 15x + 2 = (5b1(x) + 1)(5b2(x)+ 2)

And in the complex plane, we're done!

Easy.

So at this point I've removed an extraneous factor of 7 in a fairly straightforward way, using basic algebraic techniques, but I need a LOT more, which is what's coming up, where I'm moving from the field of complex numbers as I need a factor result.

But I don't exactly state it upfront which may seem sneaky. I'll explain why in a bit.

But um, we have this result now that one of the a's results by multiplying one of the b's by 7, and the a's are roots of:

a2 - (7x-1)a + (49x2 - 14x) = 0

But now if we want to play with our result, say with x=1, we find that the a's are: 3+sqrt(-26) and 3-sqrt(-26), and in fact if you try:

7(175(1)2 - 15(1) + 2) = (5(3+sqrt(-26))  +  7)(5(3-sqrt(-26))  +  7)

You'll see it is correct! That math worked!

But it implies that one of the a's has 7 as a factor. But established number theory says neither of the a's have 7 as a factor?

And I've made a seemingly HUGE leap! In the complex plane factors are meaningless, so why make this sudden claim about 3+sqrt(-26)?

So what did I do wrong?

Turns out, I didn't do anything wrong. I simply forced you to rely on the field of complex numbers telling you what happened at the top level, so you can figure out what happens if you go to a ring where factors are meaningful in a way that they aren't in the field.

So let's look at a seeming counter-example to this approach!

Consider:

a2 - (x-1)a + (49x2 - 14x) = 0

Here I've removed a 7 so that I can force a case where we can seemingly see the 7 split up into square roots, as at x=1, I have: a2  + 35 = 0

Then the a's are sqrt(-35) and -sqrt(-35), and clearly sqrt(7) is a factor, not 7.

But what about the complex plane? The complex plane encompasses lesser rings, right?

But in the complex plane with my earlier example:

7(175x2 - 15x + 2) = (5a1(x) + 7)(5b2(x)+ 2)

with normalized functions, requiring that the 7 distributed in only one way, and yup, the mathematical linchpin here is the distributive property!

Quite simply there is no way for sqrt(7) to have distributed, as then you'd see something like:

7(175x2 - 15x + 2) = (5a1(x) + sqrt(7))(5b2(x)+ 2sqrt(7))

And notice, it's basic math at this point. If the 7 split up into squares then they'd multiply through, and you'd actually see them in the construction.

So how do we maintain mathematical consistency at this point? Or is it time to throw away algebra?

Answer is, easy, as look at:

175x2 - 15x + 2 = (5b1(x) + 1)(5b2(x)+ 2)

Here I don't see a 7 at all. And I've used the assumption:

a1(x) = 7b1(x)

So if we extrapolate from here to something that results from removing the 7, if we say the first of the a's is sqrt(-35) we'd have a solution for one of the b's of:

b1(1) = sqrt(-35)/7 = sqrt(-5)/sqrt(7)

So a 1/sqrt(7) forced its way into the picture.

And note that the above is not direct from my original construction but would follow if I made one with a key 7 removed, as I'm talking about what happens if you try to force a factor of 7 to move around, when the field of complex numbers tells you it can't. The math simply shifts, as it IS consistent. It has no choice.

The field of complex numbers says how the 7 distributed which includes all lesser rings. If you have an example where that appears to not be true, then you are wrong, as the field of complex numbers is right!

Here is a giant area and it's critical to hammer home the point that the field of complex numbers tells you how the 7 distributed and it MUST be correct. From there certain things logically follow--or you lose something called mathematical consistency.

That is, otherwise I've demonstrated that human mathematics falls apart here.

(I actually considered that carefully for some time. Whether our mathematics is consistent or not is actually a really big deal.)

The construction is built using special techniques designed to make it valid in what I call the ring of objects which I figured out to handle this problem.

So now it's time to focus on how I built my construction!!!

It's suddenly of HUMONGOUS importance, as the field of complex numbers will only allow one thing, but we broke my construction by just removing a 7, which forced 1/sqrt(7) to be included in whatever ring or field, or where are we now?

Traditionally to talk about factors with non-rationals I'd be using the ring of algebraic integers.

However, it turns out in that ring you can prove that neither of the a's can have 7 as a factor!!!

Here things get a little more complicated as if you don't remember your number theory or never had much of it--like I hadn't as I was a physics student--you might wish to see that proof. And I have to say, go look for it.

For the math people who are up on their number theory though, it's a standard result which is well established, and they can see my conundrum: I'd proven a factor result using the complex plane by the distributive property, which MUST include lesser rings, only to find that the ring of algebraic integers didn't wish to agree, with the field of complex numbers!!!

And I thought about it, and thought about, and thought about it some more and came up with the ring of objects which doesn't have the problem. In the ring of objects 7 is indeed a factor of 3+sqrt(-26).

One thing that emerges though is that you cannot see the 7, as it's permanently entangled, because the square root cannot be resolved.

Like imagine you couldn't resolve sqrt(9), how would you know with the following?

It's kind of like 10 + sqrt(9) has 7 as a factor for one of its two solutions.

If you think 10 + sqrt(9) is 13, you're wrong, as it's 13 or 7 because sqrt(9) = 3 or -3.

For the truly curious and mathematically sophisticated, you may now wonder how the ring of algebraic integers pulls off the trick of not allowing 7 to be a factor. Turns out I puzzled that one out years ago, where the answer is, it wraps it up in unit factors, which are units in the ring of objects but NOT in the ring of algebraic integers!

Tricky, but not really. Unit factors CAN move in the field of complex numbers with the original example, with ease, without being forced to show themselves in the same way as the 7 is forced.

What's wild is that these are not unit factors within the ring of algebraic integers, but must be in the ring of objects.

And notice I rely heavily on the result from the field of complex numbers because there is where you can see what actually happens with the 7.

And the weird thing for those who'd like to try and start from the ring of algebraic integers is, it won't allow you to see the result. In fact, it won't even let you divide the 7 off, blocking entirely the final result:

175x2 - 15x + 2 = (5b1(x) + 1)(5b2(x)+ 2)

with normalized functions.

So could this have been figured out before? Well, read through all of the above and it's easy to see why it's hard to figure out. The ring of algebraic integers works you can say, to hide the correct result. Only by use of the field of complex numbers and belief in the distributive property was I able to tease out the correct answer, and then needed to figure out a ring which worked ok!

Whew! That was actually quite a bit of work. Had never figured out my own ring before, but I had no choice.


James Harris

Sunday, November 25, 2012

Some weird math

My most unsettling mathematical result is also one of the hardest to explain even though I can use only simple algebra. Thing is, I constructed this balancing act, where I forced a function to reveal more, by doing things a little stranger than others had tried.

In the complex plane, here's the wacky construction:

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

where the a's are roots of

a2 - (7x-1)a + (49x2 - 14x) = 0

And now normalize the functions, that is, have functions that equal 0, when x=0.

Looking at x=0, gives

a2 + a = 0, so a1(0) = 0, or -1, and a2(0) = -1, or 0.

So I can let a1(0) = 0, and then introduce a new function b2(x), where:

b2(x) = a2(x) + 1, so a2(x) = b2(x) - 1, and making that substitution:

7(175x2 - 15x + 2) = (5a1(x) + 7)(5b2(x)+ 2)

where now a1(0) = b2(0) = 0.

So now I have easily that the 7 multiplied in a trivial way--which makes sense as I made up this example, why would I make it too hard?

So now it looks like we need one more functional substitution which is:

a1(x) = 7b1(x)

Which gives me:

7(175x2 - 15x + 2) = (5(7)b1(x) + 7)(5b2(x)+ 2)

And I can divide off the trivial factor of 7 to find:

175x2 - 15x + 2 = (5b1(x) + 1)(5b2(x)+ 2)

And in the complex plane, we're done!

Easy.

But um, we have this result now that one of the a's results by multiplying one of the b's by 7, and the a's are roots of:

a2 - (7x-1)a + (49x2 - 14x) = 0

But now if we want to play with our result, say with x=1, we find that the a's are: 3+sqrt(-26) and 3-sqrt(-26), and in fact if you try:

7(175(1)2 - 15(1) + 2) = (5(3+sqrt(-26))  +  7)(5(3-sqrt(-26))  +  7)

You'll see it is correct! That math worked!

But it implies that one of the a's has 7 as a factor. But established number theory says neither of the a's have 7 as a factor?

So what did I do wrong?

The construction is built using special techniques designed to make it valid in what I call the ring of objects which I figured out to handle this problem.

In the ring of objects then--as I figured it out BECAUSE of this problem--there is no problem at all. It says that 7 is indeed a factor.

It's kind of like 10 + sqrt(9) has 7 as a factor for one of its two solutions.

If you think 10 + sqrt(9) is 13, you're wrong, as it's 13 or 7 because sqrt(9) = 3 or -3.


James Harris

Saturday, November 03, 2012

Differential equation and prime counting

A bit over a decade ago in the summer of 2002 I found a difference equation which can be constrained to count prime numbers. The derivation is remarkably simple requiring only very elementary techniques, but weirdly enough leads to this difference equation which in turn leads to a partial differential equation.

My pivotal idea was to count composites at each prime excluding counts from smaller primes. And that seemingly small shift leads to dramatic mathematical consequences.

The count at each composite is given by:

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

That is the count of composites for a particular prime excluding composites that are products of lesser primes.

First it gets a count of integers with a prime as a factor, subtracts 1 for the prime itself, and then subtracts the count of primes less than it. And finally it subtracts the count of composites multiplied times that prime.

Here's an example:

ΔS(16,3) = [16/3] - 1 - 1 - S(16/3,2) = 5 - 1 - 1 - 1 = 2

There are 5 numbers with 3 as a factor below 16, but you subtract 1 for 3, then 1 for 2, to handle 2(3), and then 1 more for the count of composites times 3, where that is for 12.

The S function is just a complete count of composites, so it is a sum of ΔS functions, where of course, if you have the complete count of composites you can count primes.

The form of the ΔS function leads to a very compact prime counting function.

I call the prime counting function P:

P(x,pj) = [x] - S(x,pj) - 1

which allows me to simplify ΔS(x, pj) to

ΔS(x,pj) = P(x/pj,pj-1) - (j-1)

where again pj has to be less than or equal to sqrt(x).

And I can make things a little simpler with the P function by not writing the prime itself but just using its indice, so then P(x,j) is equivalent to P(x,pj).

And the entire thing is:

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

where if n is greater than the count of primes up to and including sqrt(x) then n is reset to that count.

That is quite compact!

So much from a simple shift in an idea, where excluding counts of composites with smaller primes as factors yields a surprisingly efficient result.

Actually using it to count primes is very straightforward, for instance to count the primes up to 100, you need primes up to sqrt(100) = 10, which is 4 primes. And those primes of course are 2, 3, 5, and 7.

And the prime count is given by:

P(100,4) = 100 - 1 - (P(50, 0) - 0)  - (P(33,1) - 1) - (P(20,2) - 2) - (P(14,3) - 3)

Except P(14,3) needs a correction because the 3rd prime is 5 which squares to 25, so that is going too high as it's bigger than 14 and has to be reset to P(14,2). Then I have:

P(100,4) = 100  - 1 - 49 - 16 - 6 - 3 = 25

So there are 49 even composites, 16 odd composites with 3 as a factor, 6 composites with 5 as a factor but no smaller prime as a factor. And 3 that have 7 as a factor with no smaller prime, and I'll give those as it's just three numbers so easy to do and they are of course: 49, 77 and 91.

(It's fascinating to me that the symbols "S" and "ΔS" disappear at this point though the sum is the S function and the ΔS is what is being summed so they are still there.)

And for a comparison to other techniques you can look up "prime counting function" on the web.

However, we're going to push forward as there is a difference equation yet to be found, which remarkably enough is lurking within that form!

Notice that P(x,pj) is the full count of primes up to and including x, if pj is greater than or equal to the last prime less than sqrt(x), so I can remove all mention of primes with the following

ΔS(x,y) = (P(x/y,y-1) - P(y-1, sqrt(y-1)))(P(y, sqrt(y)) - P(y-1, sqrt(y-1)))

as if y is not prime then

P(y, sqrt(y)) - P(y-1, sqrt(y-1)) = 0

with the constraint that if y>sqrt(x), then P(x,y) = P(x,sqrt(x)) to keep the correct count.

And of course now I finally have my difference equation.

Notice here that sqrt(x) is the integer square root, so it means to find the largest integer less than or equal to sqrt(x), for instance, the integer square root of 10 is 3.

I also now have

P(x,y) = [x] - S(x,y) - 1

as I can remove mention of primes themselves!

Here's the new function then with all mention of primes removed:

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>sqrt(x), then P(x,y) = P(x,sqrt(x)).

Not as efficient a way to actually count, but in this fully mathematized form we can see a reason for a connection between the count of primes and non-discrete functions that intrigued notables from Euler, to Chebyshev, Gauss, and of course Riemann along with others.

From the difference equation in that summation I can get to a partial differential equation:

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

And our connection between the count of prime numbers and a differential form is complete.

One of the coolest things to me that jumps out is the x/y within it. With the integration x would be constant, and of course 1/y integrates as the logarithm, but it's still a bit of a reach I think as x/ln x is one of the known approximations to the prime count. And I'm just really out of my area of math comfort at this point.

There is more to guess here though, as notice above I ended up with a reset if y is greater than sqrt(x), which comes in from the original ΔS function and how it counts composites:

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

You only need primes up to or equal to the sqrt(x) or the count is off! For instance, [100/11] = 9 numbers with 11 as a factor, but all composites have already been counted, so it screws everything up if you keep going.

But if you integrate the partial differential that rule is not in sight. That would add more ΔS values. Continuing on with additional ΔS values with the discrete equations decreases the size of P(x,y) as you're subtracting more. So one more guess is that the integration would tend to be below the prime count.

And that is what's seen with what we can assume are approximations, with x/ln x, for instance 100/ln 100 equals approximately 21.71, which lags behind the prime count of 25. And 1000/ln 1000 is approximately 144.76 which lags behind the count of primes up to 1000, which is 168. Guessing then it would never catch it.

But now there is the possibility of a direct answer to one of the more intriguing mathematical questions ever presented. Evaluating that partial differential equation could answer quite a few questions and again I admit it is out of my league to move well there.

While I did take numerical methods for my physics degree it was over two decades ago, and besides there are some serious experts in this area all over the world who could do the best job. I just wish I could get some help here, as I will admit I'm a bit curious!


James Harris