Monday, September 29, 2014

Force coupling constant and sphere packing

Alpha here is the called the fine structure constant in physics, e is electric charge. As you can see, its purpose is to estimate the perfect sphere by estimating pi.
This is easy to understand from maximum entropy considerations. The proton needs to match the band limit of light to the quant ratio so that pi is maximally accurate.  It only knows pi as a rational fraction.

The way I work the problem is that there exists an angle x such that tanh(n*x),  are linear for n=1,2,3.

Anyway, these angles are multiples of the log of the exponent in the quant ratio.  The maximum Compton matchi is at wave 91, and 3/2 * 91 = 137.5.   And
4/3 * 137 gets you to 182 where I find a near perfect Compton match with ratio 3/2 at 216.  Coincidence?  Well, I am using the first Lagrange,Phi, and 3/2 is a known ratio for sphere packing. So I am going to find these matches.  The question is, does the vacuum count the same way my spread sheet does? This makes 182 the band limit of light, as an exponent of phi, much higher than I would thing.

As an angle on the hyperbolic then the universe counts from pi/2 (not including pi/2)  by sixths, a quadratic root for each of three cubic roots.

Why I use exponents:

 I cheat, I approximate values using Phi, which nature does not know.  But I cheat and get the property to the left. So when the physicists report a prime number, then they have cancelled common terms, and I know the number they report is a common multi0ple, so I know how far up the Fibonacci chain the universe has gotten.  I do not find Fibonacci numbers, but I look for Fibonacci exponents.  This is primitive, yes, but good enough for me.  The real step is finding the ultimate rational ratio that the unioverse has settled on to approximate pi and e. That is not a Phi thing, but a sphere thing, and the numbers keep coming back to about 2/3 or 3/4, I dunno. The length of the power ies the universe uses seem to be about 6, but as combi8nations of two and three, that comes to a 12-16 bit computer.

How is the universe so stable then? It has matches the relative bubble sizes in the salad bowl to a 12 bit computer. The real stability comes from having the bandwidth of light match the sphere, allowing motion between unit spheres. The real trick of the universe is to recycle the vacuum bubble, polish them up, resize them, and recycle them.

Sunday, September 28, 2014

Wythoff arrays can pack sphere

Here I am working with a rational ratio and arranging packs of quants thare are  exponents of that ratio= r. I am limited to two a few exponents at any level of the sphere, and I have to be hyperbolic.  The circles sum the (1/r) and outside the circle I have motion as r So I have the worm gear evvect, the circles are tracking sum (1/ratio) = 1 and exchange r with the exterior. But volume to area reaches a critical point going from the perimeter to the center, so I jump to the next row of the Whythoff array:
 Anyway, this is where I am at.  I got here my messing with recursions and discovered the Wythoff did that for me.  The balance is matching motion and unit circles going toward the center, and switching to the next row at the critical moment.  The Wythoff work because they are minimally redundant, one can make the most quants with the least redundancy. Hyperbolics and finite differences and finite sums of local aggregates are the way to go.

So, all you sphere packers, carry on! I am trying to stop this, really, but it has become a sort of fun compulsion.

Saturday, September 27, 2014

Local knowledge

Consider a local element in an aggregation with no global knowledge which moves toward minimum density.   It has no uniform knowledge of any dx, and can only make globally stable movements if all elements were of standard size and queued up fairly.  Normally elements avoid queueing and that makes spheres happen as queuing decreases toward the perimeter.

It is not sufficient for all elements to be the same size, they also need to have the same time constant in their movements. Otherwise unnecessary movements starts, and that is not maximum entropy. So element size and time constant have to be equalized.

Now, what happens if there is no optimum move? The element still needs to maintain time constant and will make a random move, I would think. The element itself must always have a self induced imbalance so that its current position in not optimum.

Do these exchanging elements  pack spheres better than the inerts? I would think so, I would think that aggregating empty positions allows more packing by movement, though I cannot prove this. If I had to prove this I would look at a perturbed system, one not in equilibrium. Then, stable configurations must come and go  as it moves toward equilibrium.

If movements are quantized, then is their a background grid system?  Not if the elements are making moves in the second differential and are perturbed.  So empty space is sub optimal, entropy is not always increasing in a static discrete space.

The last point is why any system needs inert nulls plus one active element.  How many other elements does the system need? As many as needed to work the differential order.  If the perturbations are not linear, then a single active element results in repeated movements, entropy is not increasing.

Friday, September 26, 2014

More games with spheres

I am having fun with recursive sequences and their values of cosh and sinh at multiples of the basic angle.  I made a recursion that generate powers of 3/2. And that led me to t6hinking about sphere packing.

My current observation is that is it done in repeats of two actions. A quadratic root adjusts sphere area and a cubic root adjusts volume. So we have the proton doing cubic, then quadratic, then cubic, then quadratic then cubic.  That is why we get this 3*2*3*2*3 = 108. The system is making every grid as a sphere, then adjusting for surface area, then making spheres out of that, and so on.

Each of the adjustments is an additional set of marks on the measuring stick.

Anyway, I am playing with multiples of angles when the starting angle is the solution to a recursion. The sinh and cosh are simply the two components of a quadratic root so solutions at angles by 2 are just a power series in the quadratic root.

Thursday, September 25, 2014

Games I am playing with recursive sequences and hyperbolics

Consider r1=3/2+sqrt(5)/2 and r2=3/2-sqrt(5)/2.   The sum is 3 and their product is one.  I get these values from Phi^2 + Phi^-2.

Take the log(r1) = a and then sinh(a) = sqrt(5)/2. Cosh(a) = 3/2. Tanh(a) = sqrt(5)/3. Hence Phi = 1/2+3*tanh(a)/2 =  1/2 + sinh(a).
Then (Phi)^n = Fn*(1/2+3*tanh(a)/2) + Fn-1 for example. and any (3/2)^n = Cosh(a)^n.

Also the derivative of the tanh will  be 1/cosh(a)^2. Neat stuff.  Is this ratio the reason I got the match at Phi^91 and (3/2)^108? Yes, but I caused the ration because I sorted the powers of Phi and 3/2. Hence I forced this angle, I did a Higgs mechanism on my spread sheet. Now I can work all the orbitals using shifted powers,  just like Higgs, and I can change the ratios r1 and r2.

Thuis idea that light is a bubble exchange with two bubble type swapping places with Higgs nulls is correct.  The shift  I see to get the ratio is charge and spin, and all the properties the Standard model has.

So think of it this way.  Light knows a match radius must be three, it is simply working with Higgs to find a grid size agreement that make the radius three, which it must be at the match point. Higgs and light doing the Lattice QCD for real. These shift in the phase difference between the light bubbles will stabilize then their rate of change can be contained by Higgs bubbles. That determine a unit circle. They can all be calculated using hyperbolic differential wave equations, I think. The phase shifts increase toward the center of  the packed sphere and that add angular motion to the degrees of freedom.

This angle is log(sqrt(5)/2 + 3/2.  That is about ,96 radians. Cannot find it on any map, I am still a little befuddled about where these angles are measured from. And I still think my choice for the Higgs bubble size was mostly luck and guessing. I never actually measured one of them.

There is one better recursive relationship

Phi makes the small and large light bubbles to distant. So I tried this:

r -1/r = 1/2  or r = 1/4+SQRT(17)/4 . This actually worked and makes the proton.electron mass much more accurate, 17*108 = 1836.  It also makes sense since a half shift is r^2 + r.  Log(1836,(r^2+1/r)) = 17/2. Sort of a bingo and a half shift is spin. And the Higgs bubbles do not come out so large, and the light bubble are a reasonable  So, I am looking at these bubble sizes for sphere packing. The hyperbolic angle is .2475, and sinh(a) = 1/4 and cosh(a) = sqrt(17)/4.

So the current plan. We have a measuring stick on the left, measuring the two active bubbles. We design it to count in a way that matches the inactive bubble on the right.  I can make any form of 2/3, 3/2, 3/4; to match some criteria represented by a ratio of grid sizes, using sinh(a). Then have fun with tanh, and read lots of papers by mathematicians doing tanh solutions.

Constraints are boundaries on the null grid, written in cosh as fraction and define null quants at the hyperbolic angle specified by the measuring stick. So my constraint is that null bubble be 3/2 times the measuring stick. We can still formulate (3/2)^n, easy enough, so we still get the dandy match at 108, and now I got a 17, whoopie doo.  The measuring stick is not more accurate at angle .24 then Phi at .96.  But we know Phi needed a shift four to match the density.  Its dense, volume to area favors highly curled action by the measuring stick.  The new ratio will need a 16 shift, is my lucky guess. Phi had the 17s if I remember.


Schizophrenic Persian does not blame the Jews

NRO: Iranian president Hassan Rouhani, whom the Obama administration has considered enlisting in the fight against the Islamic State, blamed the growing threat of the jihadist group on the West’s involvement in the Middle East.
During his speech at the United Nations General Assembly on Thursday, Rouhani said, “Today’s anti-Westernism is the offspring of yesterday’s colonialism; today’s anti-Westernism is a reaction to yesterday’s racism,” according to a live translation. “Certain intelligence agencies have put blades in the hands of madmen who now spare no one.”
I should write for the Onion!

The racist is leaving DOJ

NPR: Eric Holder Jr., the nation's first black U.S. attorney general, is preparing to announce his resignation Thursday after a tumultuous tenure marked by civil rights advances, national security threats, reforms to the criminal justice system and 5 1/2 years of fights with Republicans in Congress.
Two sources familiar with the decision tell NPR that Holder, 63, intends to leave the Justice Department as soon as his successor is confirmed, a process that could run through 2014 and even into next year. A former U.S. government official says Holder has been increasingly "adamant" about his desire to leave soon for fear that he otherwise could be locked in to stay for much of the rest of President Obama's second term.

Now who in their right mind wants to be locked up with a lame duck Obama.

Tuesday, September 23, 2014

Physicists and economists are pissing me off


When was the last time you ever saw a Higgs boson with a wristwatch? Anybody? Never, there is no friggin time in the universe, none, it is completely an artificial variable humans created. The vacuum only knows what is going on in its nearby neighborhood, and I mean nearby, like a quarter of a Planck distance away is all it knows. The vacuum does not know pi of e (Euler's number).  The vacuum does not do integral calculus. One might think physicists would get a clue. 

Economists like time, oh we do intertemporal optimization, they say.  Bull shit, wealthy people watch the the number of investors lined up to buy and sell, they are like the vacuum, looking locally. How did humans get so deluded? Isaac Newton, he and his grammar of calculus, it is nothing more than a scripting language, it uses undefined transendental constants for crying out loud. It has nothing to do with out finite universe or our finite economy, it is simply a short hand way of expressing relationships.

The 1990s had a broken yield curve and a Treasury fraud of mammoth proportions

I have the  federal funds, 2 year, and 3 month treasury yields, YoY, but something is wrong.  See if you can spot the broken system.


If you said that the three month rate is lower than the federal funds rate then you get a star.  The federal funds rate was effectively the six month rate, not the overnight rate.  Why on earth did that happen?

I have  the secondary market three month rate, the private lenders had set that rate, so obviously private lenders and borrowers knew full well the federal funds rate was the half year rate, and they were leaving money in the reserve for half a year.   The curve had effectively been inverted and three month bills were being used as short term cash.

Now Treasury should have been borrowing short term, at three months, and instead they were burying debt at the two to ten year terms.  Why?  Treasury obviously forgot how to count, and they cost the taxpayers a ton of money. So it was Treasury that likely caused the crash, paying to high rates, was it an inside deal? Where is Rubin?
From November to December 2007, he served temporarily as chairman of Citigroup[3][4] and resigned from the company on January 9, 2009. He received more than $126 million in cash and stock during his tenure at Citigroup,[5] up through and including Citigroup's bailout by the U.S. Treasury. He is currently engaged actively as a founder of The Hamilton Project, an economic policy think tank which produces research and proposals on how to create a growing economy that benefits more Americans.[4] 

So we have our culprit for the final bonehead move of the Clinton presidency, Robert Rubin. Which numbskull followed Rubin into Treasury? Yep, you guessed, chief bonehead of Harvard, lead sec stagger in chief. What was the total cost to the taxpayers? Looks like we got ripped of for about 90 billion a year, for nearly five years; or 450 billion. Citigroup made out like a bandit, earning the six month rate on overnight funds and earning an inflated rate on two to ten year bonds.  There was no excuse, Clinton was wioping out debt, we had no obiligation to the debt cartel.  This was clearly an sider fraud by Rubin and Summers.

Who was debuty secretary when this started? Yes, that is from the Larry Summers wiki entry. He was in on the deal from the beginning.

United States Deputy Secretary of the Treasury
In office
1995–1999
President Bill Clinton
Preceded by Frank Newman
Succeeded by Stuart Eizenstat

I call for the immediate arrest and prosecution of both Rubin and Summers.

Andrew Sullivan, incompetent hack

Andrew Sullivan: It’s been a remarkable aspect of the foreign policy “debate” over the last month that I haven’t heard a single leading Republican express misgivings about a new Iraq war’s impact on fiscal policy. And yet, for a few years now, we have been subjected to endless drama about the mounting debt when it comes to anything the government wants to do. Cost was one (ludicrous) reason to oppose Obamacare;

Andrew is bitching about the $15 billion a year we will spend shooting Islamics. He backs up his rhetoric by say the 1.5% of GDP costs for Obamacare, now arriving at Treasury is nothing.  That cost is about $170 billion this year, assuming middle class taxpayers cover about $80 billion up front.

I am all for paying the costs of the Islamic shootout, and I am all in favor of California covering the Obamacare costs of all California citizens even if we are not all citizens of the Swamp. But to compare a $170 billion dollar hit on the American middle class with a $15 billion shooting spree is ridiculous.

Monday, September 22, 2014

Entropy encoding again

Years ago I Huffman encoded the SP500, assuming the investor wanted his regular investments to be normally distributed. I got a tree like this:
The weighted nodes told the investor how much money to invest in a particular stock at some price. Probably not optimum but it minimizes transactions and maximizes use of pricing information. Under the assumptions, (−wi log2 wi), contribution to entropy, was maximally equal for each stock.

Anyway, I am going to do more with this, mainly to try out my favorite tool, wxMaxima.

Did the Fed lower rates during the crash?

Here we have the Fed target rate in red and the effective rate in blue. Look at the path going down from 2007 to 2009.  Ask yourself, was the Fed following the market down or was the Fed leading? Other than the summer of 2008, it simply looks like the Fed was chasing private lenders down hill. Notice the Fed set the upper limit at 1% even though the market had the effective funds at .1% in 2009.

I simply do not see any point at which the Fed stopped and intervened to force rates lower. Yet stupid economists think the Fed determined these low rates.  It was not until 2009 that the Fed got permission to set interest on reserves at .25, and still the market rate for one year treasuries stayed at .1%.  Economists are simply a stupid lot, unable to learn a damn thing.

Here is some duimbshit CEO of a hedge fund:
Carlyle CEO William Conway:
Before the Federal Reserve lowered rates to keep money flowing during the recession...

I mean, this nuthead never even looked to see if that really happened, he just repeated some old truism he learned somewhere.

Black Swan swooping


AP: EAST PORTERVILLE, Calif. (AP) -- Hundreds of domestic wells in California's drought-parched Central Valley farming region have run dry, leaving many residents to rely on donated bottles of drinking water to get by.
Girl Scouts have set up collection points while local charities are searching for money to install tanks next to homes. Officials truck in water for families in greatest need and put a large tank in front of the local firehouse for residents to fill up with water for bathing and flushing toilets.

Sunday, September 21, 2014

Obamacare, still unpopular

PollDateSampleFor/Favor Against/Oppose Spread
RCP Average9/2 - 9/15--41.051.4Against/Oppose +10.4
CBS News/NY Times9/12 - 9/151009 A4151Against/Oppose +10
Rasmussen Reports*9/13 - 9/141000 LV4353Against/Oppose +10
ABC News/Wash Post9/4 - 9/7RV4353Against/Oppose +10
NBC News/Wall St. Jrnl9/3 - 9/71000 RV3448Against/Oppose +14
Pew Research9/2 - 9/92002 A4452Against/Oppose +8

Bizarre trading language

Reverse repurchase agreement, repurchase agreement.  One party gives an asset to the other party who attempts to make a profit on the asset and returns the asset, with fee.  Why all the complication? 

The little fat guy and the New Yawk dingbat, let's make a deal

My proposal.  California just give our votes to the Yawker dingbat, Texas to the Joisy blob; and we call it even.  Does the SouthWest really need these two bimbos running around having making bizarre speeches? Why bother, these two nutheads have one common goal, lie and cheat their way through California and Texas taxes.  We can at least skip the bullshit, and just make a tax deal.

Saturday, September 20, 2014

Who wants to secede?

Texas, mainly. But California, you are cowards, California has already seceded and we are better off for it. I have no idea why some 78% of Californians want to be part of some corrupt swamp nation 3,000 miles away. Shame on Californians. If Californians want some foreign New Yawker dingbat setting the rules, then Californians will pay some hefty taxes back to the Swamp.

What does Brad DeLong call lack of demand?

Here I have two lines, the blue is inflation expectations, the red is the ten year Treasury rate.  Inflation expectations (Ten year minus inflation indexed ten year) is what the market expects inflation to be, and it has gotten lower.  Expected inflation lower along with the 1.7% inflation rate, down from 2%. But the red line is the ten year Treasury going up, (see that jump at the end).  Why is the Ten year DC borrowing costs going up when we have disinflation?

Brad calls it lack of demand, but clearly there is a demand for Congress to pay higher rates.  This is our UC Professor making up stories to protect the Keynesian idea.  We have lower growth now because DC is crowding out the private sector, DC is having a debt crisis and causing a recession.

Look here, 30 year mortgage rates jumping up, because of the ten year yeild jumping up.  Read the horseshit  from Frank Nothaft.  He claims the Fed did it. BS, government debt is due, Obamacare is due and all these bills have piled up causing DC to increase borrowings. It hasn't shown up yet because it is done quarterly. Recession time is drawing near.
WASHINGTON (MarketWatch) -- The average rate for a 30-year fixed-rate mortgage rose to 4.23% in the week that ended Sept. 18, hitting the highest rate since early May, in a large jump from the prior week's reading of 4.12%, according to a Thursday report from federally controlled mortgage-buyer Freddie Mac FMCC, -3.05% "Fixed-rate mortgage rates rose this week following the increase in 10-year Treasury yields being partially fueled by market speculation the Federal Reserve might change its interest rate guidance," said Frank Nothaft
Here is the almost honest CBO:
The nonpartisan Congressional Budget Office (CBO) on Wednesday raised its projection for this year’s federal deficit to $506 billion. The budget office’s last report in April had projected the deficit for fiscal 2014 would top out at $492 billion on Sept. 30.

That would be an understatment as the CBO is reading unreliable corporate taxes.
But the CBO said it is increasing the deficit figure now, in part, because receipts from corporate income taxes are turning out to be $37 billion less than expected.
CBO Director Douglas Elmendorf told repoters it's difficult to assess what is causing the drop in corporate receipts.
“It’s hard to know what to make of that because we don’t have detailed data,” he said. “Companies have put off paying some of the taxes they owe in legal ways. … We think there’s more deferral for payments in the next year than we anticipated.”
Elmendorf added the corporate tax rate has been volatile in recent years, making receipts difficult to forecast. He said the expiration of tax provisions last year has allowed companies to defer their payments.
But more than that, the bond market is piling money to match next months debt demands, which have been well advertised.  It is not expectations, it is calculated debt demands rising and actual debt demands rising; all from DC.

 Here is debt to GDP, up a half point in the last six months.  That is what causes rates and interest payments to rise, that is called crowding out.

But rates and debt are not quite aligned you say? Not quite, but this is a cartel, there are only about 20 CEOs who run the cartel. They manage rates in the near term to maximize income for their clients.

Friday, September 19, 2014

Italy in trouble with debt

They are in a recession, but the debt burden keeps on growing.

Thursday, September 18, 2014

Republicans vote massive increase in deficit

DC cannot cover interest payments, large deficit expansion in the works, mortgage rates to rise, housing to crash, imports to soar. Republicans to be blamed.
(CNSNews.com) - House Speaker John Boehner (R.-Ohio) and House Minority Leader Nancy Pelosi (D.-Calif.) joined forces early Wednesday evening as the House passed a continuing resolution that will fund the government after the end of the fiscal year on Sept. 30, and that will permit funding for Planned Parenthood (the nation's largest abortion provider), the entirety of Obamacare, and an amendment requested by President Barack Obama "to train and equip appropriately vetted elements of the Syrian opposition."
The bill passed 319 to 108 with four members not voting. But there were not enough Republican members to pass the bill without significant support from Democrats. While Pelosi sided with the Republican leadership and voted for the bill, 53 Republicans joined with 55 Democrats in voting against it.
In addition to Pelosi, some of the other Democrats voting for the Republican leadership's bill, included Rep. John Conyers (D.-Mich.), Rep. Debbie Wasserman Schlultz (D.-Fla.), Rep. Xavier Becerra (D.-Calif.), Rep. Earl Blumenauer (D.-Ore.), and Rep. Jan Schakowsky (D.-Ill.).

Wednesday, September 17, 2014

Barbara Boxer, Californias affirmative action ambassador to the Swamp

Barbara, you are nothing but an ambassador to a foreign nation. If DC wants a pompous ass hole for foreign secretary, then that is their business. You missed the boat, Barbara, we Californians are now a the nation of Pacific America, in federation with Central America. You are a nothing in the DC swampland.

BNews: Wednesday at the Senate Foreign Relation Committee hearing on U.S. strategy for combating ISIS, after Sen. Bob Corker (R-TN) criticized Secretary of State John Kerry, Sen. Barbara Boxer (D-CA) was left "shaking and trembling" in shock.

Boxer said, "I think it is shocking and a sad state of affairs that we heard just now, such angry comments aimed at you, Mr. Secretary, and through you, at our president instead of at ISIS, a savage group who decapitated two Americans and have warned, and I quote, that their thirst for more American blood is right out there."
"I think it's shocking," she continued. "I'm actually shaking and trembling. This is not the time to show anger at the people who are working night and day, whether you agree with them or not, to protect our people."

Why we have poor people in California

Robert Reich, idiot of UC California makes more than the average CEO for a two hour a week class in Bullshit. When Jerry Brown talks about high paid, out of work liberals going on welfare, Robert Reich is the one he has in mind. Robert lives off of big government in California and thinks all solutions to life come from some Swampland three thousand miles away. No kid should ever think of going to UC, it needs to have its funding vut.
Mark Perry: Former Labor Secretary Robert Reich is currently a professor of public policy at the University of California-Berkeley and he was paid $242,613 in 2013 according to this University of California database. According to this link provided in an article by the Daily Caller, Professor Reich is scheduled to teach only one undergraduate class this fall semester – Public Policy 260 – that meets only one day a week (Monday) for two hours (12 – 2 p.m.). That works out to about $2,500 for each hour of lecture time that Professor Reich will spend with UC-Berkeley students this semester, or about the same amount as the average adjunct college professor get paid for teaching an entire one-semester 15-week class ($2,700 according to this AAUP report)!
As the Daily Caller reported yesterday, Professor Reich took time off from his hectic one-course, two-hour a week teaching schedule to berate and excoriate American CEOs and Harvard Business School in a post on the Harvard Business Review blog for allowing “a pay gap between CEOs and ordinary workers that’s gone from 20-to-1 fifty years ago to almost 300-to-1 today.”

Drop drawers for Clemson

The faculty wants to peak at your junk:
Campus Reform: Clemson University is requiring students to reveal how many times they’ve had sex in the past month and with how many partners.
In screenshots obtained exclusively by Campus Reform, the South Carolina university is asking students invasive and personal questions about their drinking habits and sex life as part of what they’ve billed as an online Title IX training course.

Tuesday, September 16, 2014

The Nevada legislature is a stupid bunch

Kenneth Thomas: Via an email from Greg LeRoy of Good Jobs First, we learn that the Tesla deal, as enacted by the Nevada Legislature, is even worse than announced. Aside from the widely touted 6500 jobs only being 6000 jobs for which the state is paying for, it turns out that Tesla doesn’t even have to create the jobs itself!
You read that right. Tesla gets to receive tax credits for investment and job creation not only for itself, but for any of its suppliers (“participants,” in the law’s language) that locate on the project’s huge location. Theoretically, Telsa does not even have to create half the jobs for which it will receive subsidies.

After mentioning that some stupid legislature will step forward, it become an embarrassment  when they are next door to my own stupid legislature here in California. 

Inflation is an odd measurment

mainly because there are millions of things measured in price, none of them measured simultaneously, and few of them falling into neat categories. Hence, price theory is difficult.

Monetary inflation, sort of a meaningless statement. Are dollar/dollar going up or down? I think monetary inflation is log(one).

Consumer inflation, high or low? Depends on wages, employment and savings.

Nominal GDP is down, has been going down, and looks to go down in the future. But all that also applies to M1 Velocity, a measure of cash transaction rates.  So if consumers buy less often, do they buy more quantity per transaction? A drop in transaction rates causes a trend in inflation, probably up. We prefer the comfort of as needed purchases, I think. The price of convenience is too high. How well does the BLS calibrate this?


Monday, September 15, 2014

The jobs picture is all cyclical

Here the Yahoo reporter repeats all the old rules about structural vs cyclical unemployment.  Demographics is cyclical, cyclical on a 30 year period.  Still caused by DC where the tax and benefit system causes demographic changes over time.  Why the phase alignment of the baby boomers? Simple, the social security system incentivizes cycles. After an eight year recession cycle, retirement become counted in units of eight years, plus or minus. That, in turn, reinforces the eight year. What is different is  the thirty year cycle, counted in units of eight, has reached retirement age and the two cycles coincide.
Yahoo: The Federal Reserve has had its foot on the gas for so long that the conventional wisdom says a tap of the brakes can’t be far away. But just when and how hard the Fed should hit those brakes is still the subject of furious debate. The right answer — what’s really best for the economy — depends in large part on how much “slack” there is in the labor market, and economists readily admit that’s a hard measure to gauge. 
Since the Great Recession, the labor force participation rate — the percentage of the population gainfully employed or looking for work — has dropped significantly. Part of the reason is demographic, because the baby boom generation is hitting retirement age. As a result, labor force participation started declining before the recession and is projected to continue that decline for at least another decade. Demographic changes and other similar issues not specifically related to economic conditions are referred to as “structural” effects. 
Part of the change clearly has to do with the economic cycle — “cyclical” effects. It hasn’t just been older Americans dropping out of the workforce; the percentage of men between ages 25 and 54 has fallen to multi-decade lows. The aftermath of the Great Recession saw extraordinarily high levels of long-term unemployment. Many of those who were unemployed for more than 27 months (the rough definition of “long-term” joblessness) became “discouraged” and left the labor force, which meant that they were no longer counted among the unemployed for purposes of official recordkeeping.

Wacked out central planning

Not only do the DC Swamp Creatures make some 3 trillion in new debt roll over in Q3, but they saddle the states with a 1.5% GDP jump in health care costs, all due today, then make that worse by putting all of it on subsidized Obamacare, boosting the federal debt payable another 1.5% of GDP. All of this crap hitting the New York debt machine in the next few months. And the boneheads wonder why we have recessions.
Yahoo: WASHINGTON (AP) — Income inequality is taking a toll on state governments.
The widening gap between the wealthiest Americans and everyone else has been matched by a slowdown in state tax revenue, according to a report being released Monday by Standard & Poor's.
Even as income for the affluent has accelerated, it's barely kept pace with inflation for most other people. That trend can mean a double-whammy for states: The wealthy often manage to shield much of their income from taxes. And they tend to spend less of it than others do, thereby limiting sales tax revenue.

Saturday, September 13, 2014

Ben Leubsdorf at the WSJ says


Health-Care Revenue Rebound Could Boost U.S. Economic Growth
Total revenue at health-care and social-assistance firms rose 3% in the second quarter from the first three months of the year, the Commerce Department said Thursday in its Quarterly Services Survey. Hospital revenue rose 2.8% from the first quarter and revenue at physician offices jumped 4.1%.
But how does this surge affect other sectors? He does not say.  Here is how it affects government debt service:
 The blue line, straight up.  It is Obamacare bills and roll over of debt all piling up now. And that is driving up the ten year.  Now up to 2.6, but should be 2.8 with the two year at .56. Likely to rise to 3.0 plus much faster than economists think.

The high rates are driving up mortgage costs, and mortgage rates are the single important determinant of the housing market.



We are also seeing the dollar rise as a result of higher rates. That in turn drives up imports, further dropping domestic growth. And capital gains taxes from the market are running low after the one time payout.  The result of the Obamacare bunching will be several downward revisions og GDP growth.


How about California home sales?
SST September 11, 2014 - An estimated 37,228 new and resale houses and condos sold statewide in August. That was down 6.0 percent from 39,608 in July, and down 12.5 percent from 42,546 sales in August 2013, according to CoreLogic DataQuick data.

August sales have varied from a low of 29,764 in 1992 to a high of 73,285 in 2005. Last month's sales were 21.6 percent below the average of 47,456 sales for all the months of August since 1988, when Irvine-based CoreLogic DataQuick's statistics begin. California sales haven’t been above average for any month in more than eight years.
And part time work?
 Part time work shoots up just when healthcare spending goes up. Bad indicator, it means fewer taxes and fewer Obamacare taxes especially. Employment growth over all is stable, no batch of new jobs are coming down the line.


How much new debt?
Optimistically, Obamacare adds 1% of GDP to the deficit. Realistically, make that 2%.  DC is stable when the debt service, including the social security payments, are at the ten year rate. The deficit is now 3.56% of GDP, but if that jumps to 5% then the ten year goes to 3.5%.

But consumer spending is up. Yes, byt 2.5%, was the number I think.  But over half of that will go to imports and be a negative on growth.

How bad a recession? Maybe be a simple slog, a few quarters of struggle as the interest payments work their way through. But it could also be a continuing build up of Obamacare bills, we do not know. Watch California, the previous recession started there, the current slow down will likely start there.

Voters threaten end of the world because of Obama

Voter approval down to 42% with disapproval up to 51%. Evidently voters would prefer Pelosi to threaten civilization rather than support Obama and his Senators.

My favorite secessionist threatens the end of the world

But why would she care about the Senate in some swamp back east, she has already declared Federal law null and void while Jerry Brown has completed out independence from the USA and merger with Mexico. President Nieto needs to keep his political underlings in line.
Daikly Caller: On the one hand, California U.S. Rep. Nancy Pelosi claims that Democrats are not “fear-mongers;” on the other hand, she believes civilization is doomed if Republicans take control of the Senate from Democrats in November. The former speaker of the House made those dramatic, incongruous statements on “Real Time with Bill Maher,” which aired live from Washington, D.C. Friday. Maher asked Pelosi about recent polling which shows that the GOP is likely to take over the upper chamber and asked, given gridlock in Washingon, why it matters that Democrats keep control. “It would be very important for the Democrats to retain control of the Senate,” Pelosi told Maher. “Civilization as we know it today would be in jeopardy if the Republicans win the Senate.”

Friday, September 12, 2014

The recession is not quite here yet

But we are getting close.
(CNSNews.com) - Inflation-adjusted federal tax revenues hit a record $2,663,426,000,000 for the first 11 months of the fiscal year this August, but the federal government still ran a $589,185,000,000 deficit during that time, according to the latest Monthly Treasury Statement.
Each month, the Treasury publishes the government’s “total receipts,” including all revenue from individual income taxes, corporate income taxes, social insurance and retirement taxes (including Social Security and Medicare taxes), unemployment insurance taxes, excise taxes, estate and gift taxes, customs duties, and “miscellaneous receipts.”
The largest share of the tax revenue so far this year has come from individual income taxes, which totaled $1,233,274,000,000 in the first 11 months of fiscal 2014.

Did Mexico grant amnesty to all Californians?

I missed the exact wording of President Nieto when he visited. As I understand it, we Californians are now the other Mexico.  So how do I apply for an absentee ballot from Mexico?  Did anyone get the exact procedure or is this still being worked out?

How's the vacuum theory coming, you ask?

I am looking at tanh'(n)/tanh(n) as the determining factor in quant values for quants of integer n. We all know these form a two binary system with a trinomial; (perfect prime).  These forms will be the standard finite elements for spherical differentials, sphere packing.

How close am I? I am so far away that I expect a smart mathematician to have the answer well before I can say wxMaxima, my favorite math tool.

Wednesday, September 10, 2014

John Cochrane nails it

There are economists who know statistics and stability. Chris Sims, Roger Farmer and John Cochrane are a few examples, there are many more. So the Keynesian idea that some magical expectation operatorss us in the head and we undergo personality change is horse manure, frankly.

Grumpy Economist:
The Friedman Optimal Quantity and Financial Stability Milton Friedman long ago wrote a very nice article, showing that the optimum state of monetary affairs is a zero short-term rate, with slow deflation giving rise to a small positive short-term real interest rate. Friedman explained the optimal quantity in terms of "shoe-leather" costs of inflation. Interest rates are above zero, people go to the bank more often and hold less cash, to avoid lost interest. This is a socially unproductive activity. Bob Lucas once added up the area under the money demand curve to get a sense of this social cost, and came up with about 1% of GDP. Not bad, but not earth-shattering.
Simple really. Keynesians contend that having a Fed that always acts as the short term monopoly lender causes NGDP growth, the total money circulating increases relative to reak growth. Yet a steady state analysis tells us that the Fed, in doing so, actually causes deflation because the Fed is earning unencumbered money with each interest payment it receives.

How did Keynes miss such an obvious problem of connecting two dots on a flat paper? I dunno, I suspect he intended to be fraudulent from the start.

Mad magazine and its cultural influence

Courtesy of Fellowship of The Minds blog.

Rick Newman on Texas Independence

Rick Newman: Global markets are suddenly jittery about the prospect that Scotland, after 307 years as part of the United Kingdom, could become its own country if Scots vote for independence in a Sept. 18 referendum. If proud but tiny Scotland can do it — which polls suggest is a distinct possibility — then America’s 28th state, Texas, will certainly take notice.
If any state is fed up with the rest of America, it’s Texas. Republican Gov. Rick Perry floated the idea of seceding from the United States in 2009, though he later backpedaled. A petition for Texas to “withdraw" from the United States, lodged on the White House’s “We the People” Web page, gathered 125,000 signatures before voting closed in 2013. A group called the Texas Nationalist Movement has nearly 190,000 likes on Facebook.
Even as a state, Texas has strong anti-federal leanings. It’s a hotbed of Tea Party activity and has declined, so far, to participate in the Affordable Care Act. Perry has called Social Security, the cherished American retirement program, a Ponzi scheme. Freshman Sen. Ted Cruz, also a Republican, wants to abolish the IRS. In lieu of a strong federal overlord, secessionists want to form — or rather, recreate — the Republic of Texas, which was an independent nation for a decade before Texas joined the union in 1845.
The case for Texas existing as an independent nation is considerably stronger than it is for Scotland. Here are some of the reasons Texas might thrive as an independent nation:
It’s big. With a population of nearly 27 million and GDP of $1.6 trillion, an independent Texas would be the 13th biggest economy in the world, between Australia and Spain. That’s plenty of heft to play in the big leagues. Scotland, by comparison, is puny, with 5.1 million people and GDP equivalent to about $210 billion--which would rank around 50th.
Texas could lure companies from America. The corporate tax rate in Texas is 0, which would instantly make Texas the most tax-friendly country in the developed world if it became a country. Instead of fleeing to Canada or Ireland, U.S. firms seeking a better deal than the federal government’s 35% corporate rate could just head to Dallas or Houston. Scotland, by contrast, would have no particular tax advantages as a nation, since its tax rate — 21% for big firms — is the same as in the U.K. overall.
Texas has a healthy, diverse economy. It has energy galore, along with Big Ag, a tech hub centered on Austin and a few corporate giants such as Exxon Mobil (XOM), AT&T (T) and American Airlines (AAL). Scotland also enjoys oil wealth due to long-established wells in the North Sea, but oil extraction is declining and Scotland has little of the oil infrastructure or home-grown energy firms Texas does.
Adios, Federal Reserve. Splitting from the United States would allow Texas to wriggle free of the Fed’s loose-money policies, which have rankled Perry and other prominent Texans. If Texas adopted a new currency, meanwhile, it could make it as weak (good for exports) or as strong (good for egos) as Texans wanted. Scotland will have to wean itself off the Bank of England if it becomes independent, which is more problematic since the financial sector is a bigger part of the economy in Scotland than in Texas, and Scottish financial firms could suffer without the BOE’s implicit backing.
Independence would produce a few disadvantages for Texas, too. Here are the cons:
No more federal funds. Texas gets a good deal from Washington, receiving about 43% more from the federal government than its citizens pay in federal taxes. If it were to become independent and lose highway funding, U.S. military establishments and other types of federal spending, it might have to impose corporate taxes after all. Scotland is in a similar position, since it accounts for more public spending per person than in other parts of the U.K. and would suffer a net loss if it became sovereign.
The Texas Dept. of Defense. Texas would have to establish its own national security force to deal with problems such as illegal immigration, coastal defense, terrorist threats and of course any territorial incursions from New Mexico, Oklahoma or Louisiana. The good news is Texas has a well-armed citizenry it can tap to form local militias. (Scotland doesn’t.)
Political opposition. Texas has voted Republican in every presidential election since 1980. Losing the state’s 38 electoral votes would severely impair Republican chances of retaking the White House in future elections, which could make the kind of small-government Republicans who run Texas intent on keeping the state in the union. In the U.K., leading politicians want Scotland to stay, too. Threatening to secede is one way to find out who really cares about you.

Tuesday, September 9, 2014

More sphere packing

This is the equal incircles theory.

I call it a minimum redundancy finite log network. It is also a condition for Stokes theorem I think. A unit sphere in nature can sum the 1/F for some collection and compute the finite log. Stokes theorem is mostly about maintaining more degrees of freedom within the surface than outside. It relies on symmetry across the surface to volume phase boundary, and so related to the number of ways motion can be combined. Physicists use the principle to count up probability in near vacuum states. The idea is closely related to the finite Zeta function. Lately I am a bit slogged down getting the 3 D plotter working and going through linear combinations of tanh and how they relate to quant shifts and roots relative to the proton. Inching along.


Monday, September 8, 2014

Interest payment crisis in DC

The US Treasury suffer a bit of volatility in interest payments.  The capital gains tax surge is drying up and the two year rate is now .53%. The ten year is at 2.47%.  The central banks fees to member banks compounds to .5%. So the debt cartel is now back in actions collecting Keynesian debt service premiums. During the swing in tax income, the interest costs vary from 7 - 13% of the federal budget, which rarely carries more than a few days cash.

The actual one year bond is .09%, so Treasury pays, over the whole yields curve, about 1% of the economy in debt cartel fees. The debt service crisis comes right where the Keynesian cycle says it should come, right after the mid term elections.

Using the right log

The debt carter counts log 2 from the reserve account to the two year:
(1+.0025)^2 . Using the natural log results in a 17% error.  The next log up from 2 is base 5. So the cartel target is 2.52%.

Anyway, the 20% error bounds after two years when using the natural log tells us how good the Keynesian model is.  Rarely are they good for longer than 2 years. The relationship etween logs comes from computing the natural exponent to some base.  It is (1+x/N)^N as N goes to infinity.  Compounding in five years chunks makes a 1/5 as a multiplier in x, and that gets e^log(1/5) = 1/5, or the finite base 5.  So, as N goes to infinity, and you are compounding five year terms, then just use five  1/5, as the base.

So, the debt cartel needs to solve for some mix of ten and two year purchases that match the one year rate, compounded yearly.  The normal method os to convert everything to natural log, then solve for a relationship between the mix of ten and two years.  The particular value that satisfy the relations come fro external conditions placed on choices. If we assume minimum redundancy, then the external condition is simply to minimize the number of trades. This leads to a -pLog(p) condition. That assumption is the same assumption in the Poisson distribution, no trades happening two or three at a time. It is also the assumption in the general Shannon-Nyquist sampling rate theory. That latter theory can be generalized to any partition using a prime number.

So, queuing theory is fundamental. The system limits flow by adding unit circles, and this keeps the queue length down but causes the network to grow by Nlog(N). 


Saturday, September 6, 2014

We have proved that gravity is not constant

I think our connection between combinatorics and calculus leave us with some facts. In a spherical world, equations greater than the third differential will not work with a universal constant.  The combinatorics limit Stokes theorem; hyperbolic functions of greater than third order will not have the combinations to meet Stokes.

So, the N-Body orbits, with a constant gravity, have no solution when N is greater than four because gravity cannot be confined to a value less than some rational bound.  Hence, we have to use local wave methods, and these methods maintain a local gravity with finite travel time. This is all related to the limits of sphere packing.

This value: 5*(1−tanh(n)^2) *tanh(n)^4 gives us the coefficient on the fourth differential in combinatorics. When we subdivide n into a more divisible 'ruler', we reach the point where tanh(n)^2 does not change more than n. So the difference in the order differentials is not noticeable.  So it is fundamental, to counting which is simply a method of integer indexing values. At some point the values change faster slower than the integers are subdivided. Differentials are integer indexed. Minimal redundancy tells us that at some point delta log(n)/n is less than 1/n. Another way of stating the problem is that the natural log does not gain accuracy with n as fast as other variables. So adding another quant level makes the value of some differential the same with an increase in quant. That ends up being a divide by zero somewhere in calculus. Let's reformulate the natural process. The fundamental goal of nature, anywhere, is to estimate the value e, or log, such that the system is stable. But that estimation is limited by the differential order of the system. An N-body system where N > system order, cannot find a global estimate of e more accurate than the local estimate.

There are whole classes of problems that we can determine solvability if we can bound the combinatorics via some quantization method. Tanh is a very powerful indicator of measurement bounds. It is fundamental to counting.

Bounded, finite order polynomial wave equations

This chart has tanh(n)^k, for several k with n on the x axis.  I use this chart because I only consider polynomial wave equations with Tanh(n) solutions. Polynomials meet the condition when their differential goes to zero within some bounded range. The range is specified by a Tanh(x)^k, where k is one more than the order of the polynomial. Stability requires that the nth derivative go to zero faster then the nth +1 derivative. Thus, the range of n, the argument in tanh(n) will look like:
n1 < n2 < n3 <... where each ni is the bandstop for the ith derivative.  It is the coefficients of the polynomial terms which limit the bandwidth, but the ranges are easily determined.

So, there is some range of quant,n, which satisfies the differential, But the solutions are linear, so if S(Nmax) solves the equation for some n < Nmax, then we know that the argument is periodic. Thus  S(j*Nmax)  will solve it where j >= Nmax.  (I leave out the proof) So, we get a hierarchical system of digits. Further, we know there is a range of rational bases such that abs(e - (p/q)) is bounded, so the solution need not use transcendental symbols.

In other words, this is the general solution method for systems like the proton. Once we know the arrangement of allowable quants, then we can derive the equations of motion in quant variables, without reference to time or space.

Differentials of x = Tanh(n)^i, for example when i=4 we get:  5*x^4*(1−x^2)

So we have a relationship between the coefficients of the polynomial in powers of Tanh(n), and 1-tanh(x)^2. We can use the relationship to directly derive the bounds on n in tanh(n), for each term of the polynomial.

Games to play

 If two ranges of quants overlap, then we know we have an unfactorable polynomial in the system, and the quants have to have a relationship over some degree of freedom. We can estimate the roots because we know the quant n when the differential goes to zero.

We can generate local time. If polynomial has few factors, then its operation count is smaller and so we can compute time. If we know the density of the vacuum, then we can derive time/space equation by using time and space as output.  If we know, for example, the wave equations for light in one of the atomic orbitals, then we can derive a momentum operator that removes one degree of freedom and generates the free light wave equation.

This method is the standard tool for quantum physics. It works because quants tell us the number of actions possible from N total, taken k at a time over m recursions.  That is how the Tanh(n) is constructed. So we have an algebra to do combinatorics and differentials.

 Example. This is an F(n)*Tanh(n)^3, ot something like that from my spreadsheet.  It is the quantum solution to something, I can tell because the quants are sparse and the lines not smooth.  Could this be a particle in a box? Sure could, or a dog in a doghouse, or deliveries of eggs. Who knows since it was a constructed example.  But it is all going to work this way, start with an estimation of the quants and their relationships, because, I think, that is how nature works the problem.  Time and space are not really part of the deal.

Sphere packing:

I think sphere packers are bound to a fourth order spectrum, so the solutions are all third order polynomials in the variable quant.  But, at some point, subdividing the quant resolution will not exceed the bounds on p/k, the rational approximation to the natural base.  So, I have sort of hand waved the proof of existance for a maximum Avogadro's number.  There is a maximum finite integer in the sphere packing world.

Anyway, I call this part a wrap, mathematicians all know what to do from here, and most of them smarter than I.

Friday, September 5, 2014

If we lived in a watermelon world

We could skip the multiples of three and go right to five. Everything would be torpedo shaped if the universe had a different scaler for curvature. Why not? A number system that simply ignored groups of three. Would it work?

Bounding makes it simple

This arrangement, for example. The range of motion for three spherical things bounded by another spherical thing. So the actual unit sphere is centered in each of the circles and we have four of them.  I don't care how it does motion, I care that the enclosed error bounds are stable. If the three blue circles are identical, then they are held in place by some unit force, set to one and placed in the center of the yellow. So each blue has a force relationship with the center, they do not overlap. I know right away I have a simple polynomial with three roots, each root identified relative to the center of the big yellow.
The Efimov condition says the blue circles only interact with their neighbors, so the polynomial looks like:

x^2 -3*x -1 = 0.   No x^3 component. If the blue circles have cross coupling between all three then the equation is: x^3-3*x^2-3^3-1 = 0. The blue will be spread out and I have two degrees of motion.
Here the circles overlap, so their bounds are coupled. It really all boils down to makes circles, mostly.  Each degree of freedom generates another finite log, so the system has a sequence of them.

The key point is one needs to have unit spheres which always accumulate the 1/F to approximate the unit value.  That locks in a specific p/q that acts as the base for the system, and from that the finite log is determined.. Relativity is no problem because time is an output. The variable x is an integer quant index. The accumulated finite log is a quant value, and the digit base the quant base.  No time, and the only distance is relative to the smallest unit sphere.

Thursday, September 4, 2014

Approximating roots

I have a polynomial to the N th degree, P(n), and need to approximate the roots, bound them. Then I know that x^n - x must be bound, so that the nth derivative is within some bound error of x. That gives the finite log value used up to the nth derivative. The solutions to the approximations are N equally space angles from -pi/2 to pi/2 along the hyperbolic curve, when the bound is one. All points where the sum of tanh(n) = pi, and n are equally spaced, that gives a geodesic approximation of the unit circle.
I can repeat the process recursively, as long as I subselect the range of valid x such that the next fractional errors are bound by the first fractional error. decrease as I apply the process repeatedly. Each time I break my error sphere into a sum of smaller error sphere having one less degree of freedom.

There is a simpler method.  Consider some r and its finite log approximation:

 Where we truncate the series. We want the approximation to go uniformly lower, so we limit ourselves to x such that: abs((x-1)^n)/n goes uniformly smaller as n goes from 1 to Nmax. so:
 x-1 < root(2), and (x-1) < root(3)... and they also go from large to small. Limiting the angle from 0 to pi, then I can construct the roots such that they are within some finite bound. The angle of any root for an nth degree polynomial is:

k* Pi/(n) + j*pi/(n-1) + i*pi/(n-2).... the coefficients limited to n-k+1.
The radius are all bounded by the previous error bounds. So:
 r(n-1)^2 = 1/2-(1/4+r(n))^2, I think this works, or some similar. This gets the real component of all the roots. The imaginary component defines the error function, the equations of motion. This works when the polynomial coefficients are all 1.
If we normalize the polynomial for each recursion we get a complicated result. But consider each coefficient in the from r^(cn/n) for n = 1,Nmax. That makes the the polynomial have a dot product of Pk(n)  dot Ck(N), where the Ck are the sequenc of coefficients written as powers of the base,r. That limits out set of coefficients.
But the coefficients set a bound on error if the error functions obeys a triangle rule.  So the error bound becomes 1/sum(Cn^(1/n)) fior n = Nmax to 1. Just use that error bound and we get a hyperbolic with the narrow angle.


This works because the finite log is always a rational fraction.  So when the coefficients narrow the error bounds, the integer p and q making p/q become very large.

Error bounding polynomials

If I start with 3/2 as an estimat of LogF(r), then my error is bound by 3/2.  If I have another degree of freedom then I can remover more error.

My equatin is r^2 - my previous error bounds = 1. The previous error bounds, in the new degree of freedom is (3/2)^2. So adding one more degree of freedom makes my new r a sqr funfion of 3/2 or 9/4. Compute the root and get the sqrt(2). Adding new root in jumps of one make the Lagrange set, I thi8nk. But we can make then in jumps of any amount. We always have to keep the hyperbolic condition: (T(q)^n -1) binds the available error spectrum.
So my 5-body equation has three digit types, 5,3,2.  Hence my quants count modulo 30. And I find my roots such that
LogF(5) / r5 = LogF(3)/r3 = LogF(2)/ r2; more or less, for each root.    

So we can count iinteger spheres in the proton

We got the sphere packing thing going.
The proton counts integers 3,2,1.  So its got six unit spheres going on. Lets count, 3 quarks, an electron and the quark bag. It gets a free pass on the 1 digit.

So I can guess the electron two root system, which makes the bag a conjugate root of the electron, in radius? The proton polynomial and electron polynomial separated (except for bounding error squared). That puts the entire third order polynomial inside the bag (except fractional error), with bounding error cubed.

The exponent quant have modifiers, the value log(2) for threes and log(2) for the twos, which is a finite p/q.

Wednesday, September 3, 2014

Its about approximating the natural base,e

Finite systems are always the case.  So, any condition where some finite derivatives of F are within some bound of F itself. Each derivative is a degree o freedom in finite systems, in the limit, the natural log has infinite degrees of freedom.

Take my N-body, for example. For N bodies, the local gravity vecor is a polynomial of Nth degree. The first N derivatives are bound and real with respect to the zeroeth derivative. So each body will have a root equadistant from the dominate root, Gravity.  The polynomial is: (1-Gx^N)*(Vx+Px)*(Vx+Gx) = 0

Using my 'everything is a unit' method. Implicit in Gx is a p/q, a finite approximation of e. So ln(r) is the root of an Nth degree polynomial, and is approximated by the sum of (f^ -n), keep one (and sometimes 2) of the inverses of a set F being the finite sum of exponents in p/q. Tanh bounds the error of p/q.

So any prototype system of finite order is to define an Nth recursive approximation of the natural log.  That  equation of motion with the N+1 derivative must be zero. The wave otion is finite, and local. The system process is local exchanges of units 1/F and F for some F(n) n = quant number. The spectrum is then bound by f^k and f, where k is small. The number of root determined by coupling between the different differential from k=0,1,2,..N.
The component used in estimating ln(r), in the N-Body come from mostly orthogonal polynomials: (Gx -1)^n * (px-1)*(vx-1) = approximation of e. There is a limited set of local spectrum needed to solve.

Time is simple. local time = local time^ -1 + 1, it is phi when you want the most accurate time.  It means the approximation p/q is a power of Phi. Less acurate time is Tl = Tl^ -3 + 1 is within bonds. So time is a count of quants. Quants are markers on the yard stick, and the yard stick is appropriately curved. So in my N-body, the solution, in units of time, is going to be: T(logF(q) + P(3*logF(q).

But you can see time  can use a different bas, as can G, as can the third degree os motion. They are independent.

The Efimov state.
That is a ste where the approximation of LogF(p/q) has higher order terms within the approximating error. Those terms are eliminated in Tanh functions. So the three body Efimov solution is a square root, there are two root among the three bodies and all equantions can be referenced to that one and use the p/q approximation. This means you can use a cube root as one partition in the finite log network, and stay within the unit^n finite band bandstops. So, log(p/q) has two recursive moments but is a cubic. That means a free degree of freedom for motion in the solution. The Efimov bodies get a region of simultaneous action. Hence the quarks and gluons, they have a range of motion within some band that is white noise, well encoded.

So, you take these quarks, within the proton. Well the band stops are whit, the largest shift in exponent. It is broken into three groups by the quark cubit root, whihc is brokent in a polynomial root.  The color white is likely a +3 to -3. That root reduces in error faster than some square root. So, within the cubic band stops, a quark can operate about some square root, fewer transactions needed to maintain stability.

The error, then, the band width allocated for the quarks will be as follows:

e3 = 1-(p/q)^3, e2 = 1-(p/q)^2, e1 = (p/q)^1 The error decrease from right to left, higher order powers have less approximation error. That means the quants are separable by common multiples. Quants, the exponents, for threes digits go as 3*k, and for two digits as 2*i.  Any linear combination of twos and threes digits is accommodated as long as all linear combinations of error are less than p/q - e^ln(p/q).

In the quark, the quants digits count as a threes digit and twodigit: [3^n * 2^m] in shifts of n + m.  n is 3*k and m is 2*i. The i and j represent independent degrees of freedom. So the canonical form of a system is a system that produces the optimum integer set. It simply counts things with an integer output counting machine. These integer are the optimum markers in our  yardstick manufacturing house.

So, yes indeed, I can write the polynomial solution to the N body in  a polynomial, P(i), i 1 to N. i can be units of time, p(i) being some position in an independent vectors of dimension N. But you have to use the integer counting system natural to the system.

This works in proton/electron mass, 17*108.  That means the count size is 1^17, and it counts b^(2*17) of them, and produces 108 equally distinguishable integers. The proton is making a yard stick. A universe wide plot to get a better approximation to the log. The idea is to generate as many distinguishable integers as possible. How accurate? The proton can count the largest integer to 9e-5. I think it manages 16(2^16) groups of nulls, altogether.
The system returns an integer count of all actions possible, in increasing order.  This is arcTanh, actually, the quant number. Each digits can count unique actions relative to the degrees of freedom. It is a not quite invertable counter for groups and their relationships. It has an almost inverse. It approximates roots using groups of Higgle mass spread around as band stops.

So we see the proton count 3^3 * 2^2 * 1^17 = 27*4 = 108 times 17 electrons.

Spectral function
Consider only S(n) = S(n-1)^(-k) + 1 = 0;  Any S(n) is a recursive function of S(n-k).  However, by making my N into subsets of n*3 and n*2; I can find all the solution for the roots of r.  I need only solve for the S(n) as a quadric root and S(n) as a cubic root. Then I can mix and match and get all the roots for the system.

Aggregate charge?

Charge is simply a phase shift in quant between two Compton bands. It needs to be large to make sparks. so you get: g^5 - g^1 = 0, somewhere, charge time. The gluon has is, about a charge step of four to eight, in exponent. It is the band stops. it likely has a triple root in there.

But the gluon has to be, it is the bandwidth reservoir. This is sphere, there is the maximum point about a third way out where the greatest degrees of freedom make many more separable groups. So, the gluon is actually a bit devoid of nulls and cannot make a unit circle. It can hold bandwidth because it holds a large exponent with a large phse shift. So multiple combinations of polynomial within a up to five roots, great bandwidth for doing more thing most of the time.

So, whatever polynomial runs the gluon, say P(n), then (r^6  -  1) * r^P(n), become a bandstop.  Het into high frequency and you hit this gluon bandstop.

Regular charge seems to be a shift 1 1/2 at the electron. Spin 1/2, but the quark spin up way up in exponent, like everything up in the proton.

Polynomials have multiple roots and can have delta n, the maximum shift, of four.  We get the equations of magneto,charge, mass,momentum, position, and global constant. This system solves it by quant number, is stable because the unit circle can move. The discrete model is bound, it has the unit one.

Why?
A class of polynomials can count out steps in a cyclic graph, they have a unit one. So you can make a finite log, and run in exponents with the optimum base. Did I say that? So, for my N bodies I get:

x =  x + vx, vx = vx + Gx, then Gx is updated. remove last: dGx/n, and add in new dGx/d(n+1). Basically all  bodies have the same polynomial. They all keep Gx updates, but mostly are dominated by their own root. Its all initialization.

So, the simple recursion also generates x(t), which is a known polynomial in say, four roots, at (-1,1),(-1,-1),(1,-1),(1,1), like they didn't even try. I have one polynomial, running four times faster, covering all four bodies. But, absent the collective updating og G, they have no connection.