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### Read Basic College Mathematics (9th Edition) PDF, azw (Kindle)

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With an implementation of this algorithm, we prove ... PMID:25902494 Phytoplankton have attracted increasing attention in climate science due to their impacts on climate systems. In this game, you have to go to work, go shopping, go to the gym, eat and sleep. The effort is the most ambitious the federal government has ever undertaken to track the use of force by the police. Polyadic integers: Greek naming of p-adic integers.

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Ferrimagnetism: With two kinds of dipoles, partial cancellation may occur. Their average yield is 33f bushels to the acre. Since the introduction of the electronic calculator, which can perform the algorithms far more efficiently than humans, an influential school of educators has argued that mechanical mastery of the standard arithmetic algorithms is no longer necessary. Brief video lessons are taught by experienced educators whose instructional styles make the material easy to follow.

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Changes in cell populations are typically described using ecologically based ordinary differential equation models. To determine the effects of music listening style scores and the background music conditions, CLASTST (the test taken with classical background music) and POPTST (the test taken with popular background music) were regressed on their respective music listening style scores and CONTST (the control test with no background music). So ideally we want to have the IQM equal to the mean for symmetric has a mean value xmean = 3, and since it is a symmetric distribution, xIQM = 3 would be desired.

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In regard to (2), it is easy to physically change, say a blue chip, for ten white ones and then have, say, fourteen white ones altogether from which to subtract (if you already had four one's). When a whole is divided into two equal parts, each of these parts is called one Half. You have a very short time to build your muscles; you need not fail, because you will start the game from scratch should you fail to build and tone your muscles. The construction of conceptual knowledge requires identifying the characteristics of concepts, recognizing the similarities and differences among concepts according to these characteristics, and constructing the relations among them.

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Composite numbers are not prime, but the converse need not be true... Juryeonggu: Korean die with 8 hexagonal faces and 6 square ones. Zero itself is not considered to have a sign (though this is context dependent, see below). Landa is also responsible for the destruction of many of the documents that existed of the civilization at that time, having destroyed them as heretical documents. The Newport Member record shows that (a) ecological responses to natural environmental change can be abrupt, (b) productivity can be the primary driver of faunal change during global warming, (c) molluscs had a threshold response to productivity change while foraminifera changed gradually, and (d) changes in bivalve body size and growth rates parallel changes in taxonomic composition at the community level, indicating that, either directly or indirectly through some other biological parameter, the physiological tolerances of species do influence community change.

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Although it is useful to many people for representing numbers and calculating with numbers, it is necessary for neither. This task inspired them to search for new information about this important and timely theme in science. This made the horn-book durable and inexpensive and an important item during the ceducation of children. Adding these values in the machine amounts to calculating a sequence of n- 1 additions, each of which will probably involve a Sn is the computed version of S. Mittheilungen der Naturforschende Gesellschaft in Zürich.

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The construction of conceptual knowledge requires identifying the characteristics of concepts, recognizing the similarities and differences among concepts according to these characteristics, and constructing the relations among them. Casino edge: Gambling beyond the cost of entertainement is foolish. Then multiply the remaining numerators and denominators. However, several years ago, while visiting a French high school, I noticed the teacher used y=sx+b.

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A cake with one quarter (one fourth) removed. Dotted lines indicate where the cake may be cut in order to divide it into equal parts. Second, four process-based diagnostics are calculated for all of the SSW events in each model. For integer $n\gt 1,\;$ let $\displaystyle P(n)=\prod_{k=0}^{n}{n\choose k}\;$ be the product of all the binomial coefficients in the $n\text{-th}\;$ row of the Pascal's triangle. This is the last story in the occasional series on San Francisco's John Muir Elementary School, which, over the past year, has embarked on a process to break its cycle of failure.


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## What's in them?!

Discussion of the recent unfolding of history.

### What's in them?!

Consider:

https://www.nytimes.com/2020/08/20/nyre ... e=Homepage

'A federal judge on Thursday rejected President Trump’s latest effort to block the Manhattan district attorney from obtaining his tax returns, roundly dismissing Mr. Trump’s arguments that the prosecutor’s grand jury subpoena was “wildly overbroad” and issued in bad faith.

'The ruling by Judge Victor Marrero of Federal District Court in Manhattan marked another setback for the president in his yearlong fight to block the subpoena.

'The conflict has already reached the Supreme Court once and could end up there again as Mr. Trump’s lawyers quickly filed papers saying he would appeal. The legal wrangling means a decision is unlikely before the November presidential election.'

Just out of curiosity, why do you suppose Trump is so keen on keeping his returns from being "open to public scrutiny"? What is in them that might be most damaging to him? His connection to foreign governments? The fact that as a businessman he's more a failure than he'd like others to know? That he paid virtually no taxes at all while intent on "draining the swamp" that is crony capitalism in Washington?

And, who knows, maybe someday they actually will be made available for all the rest of us to see.

Then we can compare what we think is spooking him from what he damn well ought to be spooked about.
He was like a man who wanted to change all; and could not; so burned with his impotence; and had only me, an infinitely small microcosm to convert or detest. John Fowles

Start here: viewtopic.php?f=1&t=176529
Then here: viewtopic.php?f=15&t=185296
And here: viewtopic.php?f=1&t=194382

iambiguous
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Posts: 37286
Joined: Tue Nov 16, 2010 8:03 pm
Location: baltimore maryland

### Re: What's in them?!

not to mention that another, another close confidant of
IQ45 has been arrested for yet another crime...

the number of the "best people" been arrested or charged with
a crime or just flat our pleaded guilty of over 30 people.....

how many were arrested or charged during the Obama administration?

less then 5... in eight years compared to IQ45 3 and half years...

just the sheer number of people arrested and charge or just plead
guilty show us the base corruption of the IQ45 administration...

how can you support a president who has had multiple number of
close administration officials charged or arrested for crimes?

if these people are the "best" people IQ45 knew, then he is a criminal
and a thief, for people surround themselves with people they know
and trust and feel comfortable with....and IQ45 only knows criminals and thief's...

whose only goal has been to steal and corrupt and destroy
the America that has been, until now, "the shining city on the hill"

defending IQ45 is to defend criminals and thief's and liars......

I am glad those defenders of IQ45 admit to their own comfort level
of defending criminals, thief's and liars....

to defend IQ45 is to defend corruption on a massive scale, is to defend
over 20,000 lies over 3.5 years in office.....defending IQ45 is to defend breaking
the laws and trampling on the very constitution that IQ45 pledged to
obey and protect......I am glad the supporters of IQ45 are so open
to supporting the violations of the laws and constitution of the United
States of America...

I support the obedience to the laws and the following of constitution,
which makes me in direct contrast to the followers of IQ45....

if you believe it, you must own it.....

so own your support of someone so dedicated to destroying
the constitution and the laws of the United States of America....

Kropotkin
"Those who sacrifice liberty for security
wind up with neither."
"Ben Franklin"
Peter Kropotkin
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Joined: Thu Apr 07, 2005 1:47 am
Location: blue state

### Re: What's in them?!

And around and around it goes...

https://www.washingtonpost.com/election ... OQO33AVYTM

'NEW YORK — A federal appeals court Tuesday blocked the release of Trump’s tax returns and other financial records to the Manhattan District Attorney’s Office — a temporary win for the president in his protracted battle to keep the local prosecutor’s office from obtaining the trove of documents in conjunction with an ongoing criminal probe.

The ruling came as one judge on the three-member panel pressed the district attorney’s team about the scope of its grand-jury probe, saying the request seems “really very broad” as the president’s attorneys have argued.

John Walker Jr., one of three judges hearing arguments at the 2nd Circuit Court of Appeals over whether to prevent the subpoena from being enforced while further litigation in the case continues, asked a lawyer for the district attorney’s office how one would determine whether its request for Trump’s records is, in fact, “very broad and might engage in some fishing.”'

It being the law of course. Or, perhaps, politics?

Now, all we need is a precise philosophical definition of "broad" and "fishing" to finally settle this thing once and for all.

He was like a man who wanted to change all; and could not; so burned with his impotence; and had only me, an infinitely small microcosm to convert or detest. John Fowles

Start here: viewtopic.php?f=1&t=176529
Then here: viewtopic.php?f=15&t=185296
And here: viewtopic.php?f=1&t=194382

iambiguous
ILP Legend

Posts: 37286
Joined: Tue Nov 16, 2010 8:03 pm
Location: baltimore maryland

### Re: What's in them?!

"Trump Could Be Investigated for Tax Fraud, D.A. Says for First Time"

https://www.nytimes.com/2020/09/21/nyre ... e=Homepage

'The Manhattan district attorney’s office, which has been locked in a yearlong legal battle with President Trump over obtaining his tax returns, suggested for the first time in a court filing on Monday that it had grounds to investigate him and his businesses for tax fraud.

'The filing by the office of the district attorney, Cyrus R. Vance Jr., offered rare insight into the office’s investigation of the president and his business dealings, which began more than two years ago.

'Mr. Vance, a Democrat, has never revealed the scope of his office’s criminal inquiry, citing grand jury secrecy. The investigation has been stalled by the fight over a subpoena that the office issued in August 2019 for eight years of the president’s tax returns.

'Lawyers for Mr. Trump have said the subpoena should be blocked, calling it “wildly overbroad” and politically motivated. Mr. Vance responded to that argument in a carefully worded new filing that did not directly accuse Mr. Trump or any of his businesses or associates of wrongdoing and took pains to avoid disclosing details about the inquiry.

'However, prosecutors listed news reports and public testimony that alleged misconduct by Mr. Trump and his businesses. The reports, prosecutors wrote, would justify a grand jury inquiry into a range of possible crimes, including tax and insurance fraud and falsification of business records. It was the first time the office had suggested tax fraud might be among the possible areas of investigation.

Well, we now know that whatever is in them might result in some serious legal problems for the president.

And then this:

'The president has said he expects the dispute over the subpoena will end up in the Supreme Court. “This is a continuation of the witch hunt, the greatest witch hunt in history,” Mr. Trump said last month.

Maklng a 6 to 3 conservative majority all the more vital for him.

Finally...

'The appeals court scheduled oral arguments on the matter for Friday. However the court rules, either party could take the case back to the Supreme Court, making it unlikely the dispute will be decided before the presidential election on Nov. 3.

'Even if Mr. Vance’s prosecutors ultimately obtain Mr. Trump’s tax records, grand jury secrecy rules make it unlikely the materials will become public anytime soon. They might only surface if Mr. Vance’s office brings charges and the tax returns are introduced as evidence in court.'

Really, what are the odds that the American people will be apprised of what is in them anytime soon. If at all.
He was like a man who wanted to change all; and could not; so burned with his impotence; and had only me, an infinitely small microcosm to convert or detest. John Fowles

Start here: viewtopic.php?f=1&t=176529
Then here: viewtopic.php?f=15&t=185296
And here: viewtopic.php?f=1&t=194382

iambiguous
ILP Legend

Posts: 37286
Joined: Tue Nov 16, 2010 8:03 pm
Location: baltimore maryland

### Re: What's in them?!

This is in them:

https://www.nytimes.com/interactive/202 ... e=Homepage

'Donald J. Trump paid $750 in federal income taxes the year he won the presidency. In his first year in the White House, he paid another$750.

'He had paid no income taxes at all in 10 of the previous 15 years — largely because he reported losing much more money than he made.

'As the president wages a re-election campaign that polls say he is in danger of losing, his finances are under stress, beset by losses and hundreds of millions of dollars in debt coming due that he has personally guaranteed. Also hanging over him is a decade-long audit battle with the Internal Revenue Service over the legitimacy of a $72.9 million tax refund that he claimed, and received, after declaring huge losses. An adverse ruling could cost him more than$100 million.'

I'm not really clear as to whether these revelations encompass the entirety of the returns, or just a part of them.

Thus:

'The returns are some of the most sought-after, and speculated-about, records in recent memory. In Mr. Trump’s nearly four years in office — and across his endlessly hyped decades in the public eye — journalists, prosecutors, opposition politicians and conspiracists have, with limited success, sought to excavate the enigmas of his finances. By their very nature, the filings will leave many questions unanswered, many questioners unfulfilled. They comprise information that Mr. Trump has disclosed to the I.R.S., not the findings of an independent financial examination. They report that Mr. Trump owns hundreds of millions of dollars in valuable assets, but they do not reveal his true wealth. Nor do they reveal any previously unreported connections to Russia.'

That Donald "art of the deal" Trump is as shitty a businessman as he is a president has got to be embarrassing for him. But it's what might send him to jail that his most fierce opponents are intent on bringing to light.

Uh, right?
He was like a man who wanted to change all; and could not; so burned with his impotence; and had only me, an infinitely small microcosm to convert or detest. John Fowles

Start here: viewtopic.php?f=1&t=176529
Then here: viewtopic.php?f=15&t=185296
And here: viewtopic.php?f=1&t=194382

iambiguous
ILP Legend

Posts: 37286
Joined: Tue Nov 16, 2010 8:03 pm
Location: baltimore maryland

### Re: What's in them?!

iambiguous wrote:This is in them:

https://www.nytimes.com/interactive/202 ... e=Homepage

'Donald J. Trump paid $750 in federal income taxes the year he won the presidency. In his first year in the White House, he paid another$750.

'He had paid no income taxes at all in 10 of the previous 15 years — largely because he reported losing much more money than he made.

'As the president wages a re-election campaign that polls say he is in danger of losing, his finances are under stress, beset by losses and hundreds of millions of dollars in debt coming due that he has personally guaranteed. Also hanging over him is a decade-long audit battle with the Internal Revenue Service over the legitimacy of a $72.9 million tax refund that he claimed, and received, after declaring huge losses. An adverse ruling could cost him more than$100 million.'

I'm not really clear as to whether these revelations encompass the entirety of the returns, or just a part of them.

Thus:

'The returns are some of the most sought-after, and speculated-about, records in recent memory. In Mr. Trump’s nearly four years in office — and across his endlessly hyped decades in the public eye — journalists, prosecutors, opposition politicians and conspiracists have, with limited success, sought to excavate the enigmas of his finances. By their very nature, the filings will leave many questions unanswered, many questioners unfulfilled. They comprise information that Mr. Trump has disclosed to the I.R.S., not the findings of an independent financial examination. They report that Mr. Trump owns hundreds of millions of dollars in valuable assets, but they do not reveal his true wealth. Nor do they reveal any previously unreported connections to Russia.'

That Donald "art of the deal" Trump is as shitty a businessman as he is a president has got to be embarrassing for him. But it's what might send him to jail that his most fierce opponents are intent on bringing to light.

Uh, right?

Yeah, but he is like a Houdini , he is always able to get out of tight spots. Of coursed Houdini failed in his final trick.
Meno_
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Posts: 7118
Joined: Tue Dec 08, 2015 2:39 am
Location: Mysterium Tremendum

### Re: What's in them?!

Meno_ wrote:
Yeah, but he is like a Houdini , he is always able to get out of tight spots. Of coursed Houdini failed in his final trick.

Yes, but Houdini was successful only insofar as the audience couldn't figure out how he did the tricks. With Trump, the rich and the powerful couldn't care less how he does them. As for his blue collar "base", most of them are not even aware that they are being tricked at all. Keep them distracted with issues revolving around race and religion and law and order and guns and sexual perversions and, sure, you might stay in office four more years.

Anyway, with his taxes, most liberals are only concerned with the part where his "tricks" might have been illegal. Will the "deep state" crony capitalists in Washington and New York actually go so far as to arrest him, to try him, to put him in prison?!

Not in a million years if my own understanding of "the system" is correct.
He was like a man who wanted to change all; and could not; so burned with his impotence; and had only me, an infinitely small microcosm to convert or detest. John Fowles

Start here: viewtopic.php?f=1&t=176529
Then here: viewtopic.php?f=15&t=185296
And here: viewtopic.php?f=1&t=194382

iambiguous
ILP Legend

Posts: 37286
Joined: Tue Nov 16, 2010 8:03 pm
Location: baltimore maryland

### Re: What's in them?!

iambiguous wrote:
Meno_ wrote:
Yeah, but he is like a Houdini , he is always able to get out of tight spots. Of coursed Houdini failed in his final trick.

Yes, but Houdini was successful only insofar as the audience couldn't figure out how he did the tricks. With Trump, the rich and the powerful couldn't care less how he does them. As for his blue collar "base", most of them are not even aware that they are being tricked at all. Keep them distracted with issues revolving around race and religion and law and order and guns and sexual perversions and, sure, you might stay in office four more years.

Anyway, with his taxes, most liberals are only concerned with the part where his "tricks" might have been illegal. Will the "deep state" crony capitalists in Washington and New York actually go so far as to arrest him, to try him, to put him in prison?!

Not in a million years if my own understanding of "the system" is correct.

It depends how Trumps masters perceive the pulse. If it’s too soon and the backlash of the public is severe... they’ll throw Trump under the bus in a heartbeat to look good.
Ecmandu
ILP Legend

Posts: 10662
Joined: Thu Dec 11, 2014 1:22 am

### Re: What's in them?!

Ecmandu wrote:
iambiguous wrote:
Meno_ wrote:
Yeah, but he is like a Houdini , he is always able to get out of tight spots. Of coursed Houdini failed in his final trick.

Yes, but Houdini was successful only insofar as the audience couldn't figure out how he did the tricks. With Trump, the rich and the powerful couldn't care less how he does them. As for his blue collar "base", most of them are not even aware that they are being tricked at all. Keep them distracted with issues revolving around race and religion and law and order and guns and sexual perversions and, sure, you might stay in office four more years.

Anyway, with his taxes, most liberals are only concerned with the part where his "tricks" might have been illegal. Will the "deep state" crony capitalists in Washington and New York actually go so far as to arrest him, to try him, to put him in prison?!

Not in a million years if my own understanding of "the system" is correct.

It depends how Trumps masters perceive the pulse. If it’s too soon and the backlash of the public is severe... they’ll throw Trump under the bus in a heartbeat to look good.

Ec'

Biggy and me have been going on about the reductive process below the contextual issues surrounding self valuing , ....

It may be a remote possibility, but like Houdini, admittedly. Trump , at this late stage, may even have precursed his puppetmasters' reaction if they fail to perceive him as a manageable genius.

He may even have a plan to trick the masters, in a last ditch effort to exclude any and everybody, that stands in his way.

The masters singularly can be exorcised the same way , by the process that works best for him : divide and conquer

Even now, the shark-tank is seething with turbulence, for the swamp blinds everything

Besides the puppet masters may consist of a wider. International group that is less consistent than supposed.

What if, the genius characterization stands , but the manageable part is converted to psychologically impaired.

The borderline may shift, away from the norm to such a significant difference, that would imunge into public sentiment

Books have been written on the subject, but he appears to be able to get out from under the constraints, that ex-prez Nixon could not wash in his dirty tricks attempt.
Meno_
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format: plain


text:
# The value of a force of 100 N on a system based upon the metre, the kilogram and the minute as the fundamental units is

$36 \times 10^4$


text:
# Math Help - Diagonalizable transformation question

1. ## Diagonalizable transformation question

In another forum (in Spanish : Aplicación diagonalizable - Foro fmat.cl), one asked if $T:\mathbb{R}^2 \to \mathbb{R}^2$ , $T(x,y)=(x+y,y)$ is diagonalizable.
Another guy answered that no, because the geometric multiplicity and the algebraic multiplicity are not equal.
Forming a matrix with T(1,0) and T(0,1) as column-vectors gives eigenvalues of 1. (2 eigenvalues that is).
Then the guy says that it implies that the algebraic multiplicity of 1 is 2. I don't understand the "of 1", and also in another forum I read that the algebraic multiplicity is the multiplicity of the root of the characteristic polynomial, so I understand why here it is 2.
But then the guy says that the geometric multiplicity is 0 because the dimension of the eigenspace is 0. But he never explained why it is 0. I don't see why. Can you help me?

Another question : if I take the identity matrix $3 \times 3$. Clearly the eigenvalues of it are 1, 1 and 1. So the algebraic multiplicity is 3. But the geometric multiplicity seems to be 1. Hence the identity matrix is not diagonalizable... I don't think it's right. I'm having trouble understanding how do you find the dimension of the eigenspace. Thanks in advance.

2. Originally Posted by arbolis

But then the guy says that the geometric multiplicity is 0 because the dimension of the eigenspace is 0. But he never explained why it is 0. I don't see why. Can you help me?
that is not correct! the dimesnion of the eigenspace in this example is 1 not 0. the dimesnion of the eigenspace corresponding to an eigenvalue is never 0. it's always at least 1.

the eigenspace corresponding to the eigenvalue $\lambda$ of a transformation $T$ is the solution set of $Tv=\lambda v.$ for example, in the above example, you need to solve $T(x,y)=(x,y),$

because $\lambda=1.$ so we'll have $(x+y,y)=(x,y),$ which gives us $y=0$ and $x$ can be anything. so the eigenspace corresponding to $\lambda=1$ is the set $\{(x,0): \ x \in \mathbb{R} \},$ which is

obviously generated by $(1,0).$ so {(1,0)} is a basis for the eigenspace and thus the dimension of the eigenspace is 1. i think you can now answer your second question yourself.

3. Thanks, that was very clear.
However I'm stuck to answer my second question. My intuition says that the geometric multiplicity is 3 because a basis for the eigenspace is $\{ (1,0,0), (0,1,0), (0,0,1) \}$. I'm stuck to see what is the transformation associated to the identity matrix. I guess it's something like $T(x,y,z)=(x,y,z)$. If it is right then the basis I just gave is right and I understand everything.

4. Originally Posted by arbolis
Thanks, that was very clear.
However I'm stuck to answer my second question. My intuition says that the geometric multiplicity is 3 because a basis for the eigenspace is $\{ (1,0,0), (0,1,0), (0,0,1) \}$. I'm stuck to see what is the transformation associated to the identity matrix. I guess it's something like $T(x,y,z)=(x,y,z)$. If it is right then the basis I just gave is right and I understand everything.
correct! in general, if a matrix $A$ is given, the corresponding transformation is defined by: $Tv=Av.$ so if $A=I,$ the identity matrix, then $Tv=Iv=v,$ which is what you guessed.


text:
# Corrfun.m

Rotational correlation function, normalized to be the correlation func- tion between second rank Wigner functions. Syntax:

             [weights,rates]=corrfun(spin_system,k,m,p,q)

where the indices k,m,p,q correspond to the four indices found in the ensemble-averaged Wigner function product:

                       G=<D2(k,m)'*D2(p,q)>

the function requires spin_system.rlx.tau_c to be either a single cor- relation time (in which case the isotropic rotational diffusion model is used) or vector with two correlation times (in which case they are assumed to be correlation times for rotation around and perpendicular- ly to the main axis respectively) or vector with three correlation ti- mes (in which case those are assumed to be the correlation times for the rotation around the XX, YY and ZZ direction respectively of the rotational diffusion tensor).

The function returns the weights and decay rates of the individual ex- ponentials in the correlation function. Note that the decay rates re- turned are negative numbers.

Note: Wigner function indices are sorted in descending order, that is, [1 2 3 4 5] in the input maps onto [2 1 0 -1 -2].


text:
# Why is coax loss logarithmic?

With DC hookup wire, losses are directly proportional to length. If the resistance of 100m of wire is 10 ohms then the resistance of 200m of wire will be 20 ohms.

But we measure coax loss in decibels per given length. For example, this chart lists RG-8X at 10MHz as having a 1dB loss per 100ft. So 200ft of this coax would have only 2dB of loss. This is not twice as much loss! In fact in this case it happens to be "half" as much (in decibels) as twice the loss [in "loss"ables?].

What I mean is: if we start with 1 dB of loss for the first hundred feet, twice that loss would be 4 dB (since adding approximately 3 dB is mathematically the same as multiplying by two). But instead of 4 dB of loss, we only have 2 dB of loss.

What's even weirder about this is that coax gets less lossy as the frequency decreases. Going back to the chart we see a 2.5dB/100ft loss at 50MHz, 1.0dB/100ft at 10MHz, 0.5dB at 1MHz. Surely at 50/60Hz coax must be some sort of superconductor, then?!

How do we get this scenario where at UHF frequencies a transmission line is so much worse than double the power loss (basically wherever the loss per 100ft exceeds 3dB…but wait, why should the unit length matter… hmmm…), but at HF frequencies it's so much better, when comparing the logarithmic losses to the linear resistance of a conductor?

• Because twice the resistance doesn't mean twice the loss. – user253751 Mar 16 '17 at 1:03
• This is an excellent question. It seems pretty simple on the surface, but as I think about it more it cuts really deep and I still can't find a complete answer. – Phil Frost - W8II Mar 16 '17 at 1:53

Firstly I would note that your initial statement isn't true. Indeed, the resistance of a wire increases with length, but resistance isn't loss in the same sense of transmission lines where loss is measured in power or a ratio of powers. Consider:

simulate this circuit – Schematic created using CircuitLab

$$I_\text{source} = I_{R1} = I_{R2} = \sqrt{1 \over R_1 + R_2}$$

$$P_{R2} = I_{R2}^2 R_{R2} = {R_2 \over R_1 + R_2}$$

Replacing the source with a constant current or a constant voltage source we can make some linear relationships to length, but under those conditions source power changes with length also. This isn't the same sense of "losses" used in transmission lines which compare how much power from the source makes it to the load.

But this doesn't answer your question: although the relationship isn't linear, neither is it what's expected of a transmission line:

$$P_\text{out} \propto 10^{-l}$$

It's a basic property of transmission lines that if the line is infinitely long, or terminated in a matched load, then the source will see the same impedance no matter how long, lossy, or not lossy the transmission line is. This holds for all frequencies, even DC.

There's no way to do that with simply a series resistor. But it can be done with a T attenuator. Here's an example of a 50Ω, 1dB attenuator:

simulate this circuit

The impedance seen by V1 is:

$$(R_2 + R_4) || R_3 + R_1 = 50\:\Omega$$

Since V1 sees 50Ω, input power is:

$$(1\:\mathrm V)^2 / 50\:\Omega = 0.02\:\mathrm W$$

I'll not go through the math, but the power in R4 is 1dB less, or:

$$0.02\:\mathrm W \cdot 10^{-1/10} = 0.0159\:\mathrm W$$

This property of not changing the impedance seen by the source holds true only

You could then imagine a transmission line as any number of these T attenuators connected together. As more are connected the total conductor resistance (R1 + R2 + ...) increases linearly, but each additional attenuator adds another 1dB of attenuation.

Alternately, you can take that 1dB attenuator and divide it into two 0.5dB attenuators. And then keep subdividing infinitely until each attenuator along the length of the line is an infinitesimal section of the line.

At this point, we are getting close to a decent model of a transmission line. Missing though is any way to transfer power efficiently since the whole thing is a resistor, and there can't be a wave in it, since a wave requires a notion of time and there's no time term in the definition of resistance.

Solve that by adding to the model the inductance of the conductor, and the capacitance between the conductors:

simulate this circuit

Since these inductors and capacitors are lossless, as long as their contribution to impedance is much larger than the resistive elements the loss will be low. Furthermore, the definition of capacitance and inductance are differential equations with time, the mathematics can support a wave.

Losses increase with frequency for (at least) three reasons:

1. Skin effect
2. Dielectric losses

The resistive elements in the transmission line model are more accurately functions of frequency, due to these effects.

As frequency increases, skin effect constrains the current to a thinner cross-section of the conductor. This means less conductivity overall, increasing resistive losses. This is why larger-diameter coax has lower loss: there's more conductor surface area, and thus less resistive loss.

Dielectric losses occur as the voltage alternates between opposite polarizations. Reversing the polarity of some dielectric (say, PTFE) takes some fixed amount of energy. With increased frequency this reversal occurs more times per second, meaning it consumes more power. This is why air-dielectric transmission lines have lower loss: it takes much less energy to reverse the polarity of air than something like PTFE.

Radiation losses occur in coax for example when the shield is not a perfect, solid conductor. Consider coax with a braided shield: it does not provide perfect coverage, but has holes in it. At low frequencies, these holes are minuscule relative to the wavelength, making them negligible. As frequency increases the holes appear increasingly large, leading to a less complete containment of the fields within the coax, and more radiation. This is why coax intended for high-frequency use tends to incorporate foil shielding, sometimes multiple layers of it.

• I'm still not entirely satisfied with this answer :-/ While it explains how the attenuation works, I can't explain how the attenuation stops working that way for hookup wire and a DC load. – Phil Frost - W8II Mar 16 '17 at 0:18

You are mixing units.

In the case of the wire, you specify resistance. But dB refers to power, not resistance.

In a DC (or low frequency) power line, current is constant and the voltage drop is due mainly to resistive losses. So voltage drop is proportional to wire length. Constant current and voltage drop proportional to distance gives power loss proportional to distance.

In a transmission line (assuming no standing waves), the relationship between current and voltage is given by the characteristic impedance of the line. Losses occur due to series resistance and parallel conductance across the dielectric.

So both voltage and current are reduced along the line, and thus the non linearity.

If, say, you get 3dB loss in the 1st 100m section of line (i.e. half the power), then you'll only have 1/2 the power entering the 2nd 100m section. The 2nd section will dissipate 1/2 of that (1/4 of the total power), and so on.

As frequency increases, the losses on the dielectric increase significantly.


text:
Volume 291 - 9th International Workshop on the CKM Unitarity Triangle (CKM2016) - Parallel WG4
Measurements of CKM angles at BELLE
B. Pal* On behalf of the BELLE Collaboration
*corresponding author
Full text: pdf
Pre-published on: 2017 June 05
Published on: 2017 June 13
Abstract
In this review recent studies on $CP$ violation and related hadronic $B$ decays by the Belle experiment, in particular measurements of CKM angles $\phi_1$ and $\phi_2$ are reported.
Open Access
Copyright owned by the author(s) under the term of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.


text:
# Reflection on the coloring and chromatic numbers

Joint work with Chris Lambie-Hanson.

Abstract.  We prove that reflection of the coloring number of graphs is consistent with non-reflection of the chromatic number.  Moreover, it is proved that incompactness for the chromatic number of graphs (with arbitrarily large gaps) is compatible with each of the following compactness principles: Rado’s conjecture, Fodor-type reflection, $\Delta$-reflection, Stationary-sets reflection, Martin’s Maximum,  and a generalized Chang’s conjecture.

This is accomplished by showing that, under GCH-type assumptions, instances of incompactness for the chromatic number can be derived from square-like principles that are compatible with large amounts of compactness.

This entry was posted in and tagged , , , , , , , , , , , , . Bookmark the permalink.

### 2 Responses to Reflection on the coloring and chromatic numbers

1. saf says:

Update Feb/2017: Added a new section, entitled “Realizing all closed intervals”.

0 likes

2. saf says:

Submitted to Combinatorica, December 2016.
Accepted, June 2017.

0 likes


text:
# ENDPOINT OF rp PROCESS USING RELATIVISTIC MEAN FIELD APPROACH AND A NEW MASS FORMULA

@article{Lahiri2012ENDPOINTOR,
title={ENDPOINT OF rp PROCESS USING RELATIVISTIC MEAN FIELD APPROACH AND A NEW MASS FORMULA},
author={C. Lahiri and G. Gangopadhyay},
journal={International Journal of Modern Physics E-nuclear Physics},
year={2012},
volume={21},
pages={1250074}
}
• Published 2012
• Physics
• International Journal of Modern Physics E-nuclear Physics
Densities from relativistic mean field calculations are applied to construct the optical potential and, hence calculate the endpoint of the rapid proton capture (rp) process. Mass values are taken from a new phenomenological mass formula. Endpoints are calculated for different temperature-density profiles of various X-ray bursters. We find that the rp process can produce significant quantities of nuclei upto around mass 95. Our results differ from existing works to some extent.
4 Citations

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#### References

SHOWING 1-10 OF 38 REFERENCES
Relativistic Mean Field in $A\approx$80 nuclei and low energy proton reactions
• Physics
• 2011
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Microscopic calculation of proton capture reactions in the mass 60–80 region and its astrophysical implications
• Physics
• 2011
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NEUTRON RICH NUCLEI IN A NEW BINDING ENERGY FORMULA AND THE ASTROPHYSICAL r-PROCESS
• Physics
• 2012
Neutron rich nuclei has been studied with a new phenomenological mass formula. Predictions of different mass formulas for the location of the neutron drip line are compared with those from theExpand
The Endpoint of the rp process on accreting neutron stars
We have studied the evolution of the shell structure of nuclei near the neutron drip line in the Relativistic Hartree-Bogoliubov (RHB) theory with the vector self-coupling of omega meson. TheExpand
Helium nuclei around the neutron drip line
• Physics
• 2012
Neutron rich He nuclei have been investigated using relativistic mean field approach in co-ordinate space. Elastic partial scattering cross sections for proton scattering in inverse kinematics haveExpand
Proton dripline in a new formula for nuclear binding energy
• Physics
• 2011
The location of the proton dripline in a new phenomenological mass formula is calculated. Predictions of different mass formulas for the dripline are compared. The implications of the new massExpand
A density-dependent interaction in the folded alpha-nucleus potential
Abstract Assuming that the density dependence of the effective nulceon-nucleon interaction can be factorized into a target and a projectile term, an alpha-nucleus potential is obtained. It is shownExpand
Microscopic mass formulas.
• Physics, Medicine
• Physical review. C, Nuclear physics
• 1995
The Bethe-Weizsaecker formula is shown to be the asymptotic limit of the present one(s), and the surface energy of nuclear matter turns out to be probably smaller than currently accepted. Expand
A NEW PHENOMENOLOGICAL FORMULA FOR GROUND-STATE BINDING ENERGIES
A phenomenological formula based on liquid drop model has been proposed for ground-state binding energies of nuclei. The effect due to bunching of single particle levels has been incorporated throughExpand
Optical-model potential in finite nuclei from Reid's hard core interaction
• Physics
• 1977
Starting from the Brueckner-Hartree-Fock approximation and Reid's hard core nucleon-nucleon interaction, we calculate and parametrize the energy and the density dependence of the isoscalar,Expand


text:
# How to correctly use the palatino font (in memoir)?

I'm preparing a PhD thesis (using memoir and pdfLaTeX), and I find the palatino font a nice body font.

Reading the font related questions and answers on this awesome site, I think I have pieced together the correct combination of commands and packages but it feels like a lot of hacking. So I want to ask if there is a better way than my current solution.

\RequirePackage[sc]{mathpazo} % use mathpazo to get the math fonts working correctly, why can't we scale them?
\RequirePackage[scale=0.95]{tgpagella}\normalfont % but use tgpagella as main font (it's newer than mathpazo?}
\newcommand*{\memfontfamily}{qpl} % tgpagella as main memoir font
\newcommand*{\memfontenc}{T1}
\newcommand*{\memfontpack}{tgpagella}
\documentclass[a4paper, oneside, 10pt, extrafontsizes, showtrims, draft]{memoir}
\RequirePackage[scaled=0.75]{luximono}  % we can load the correct tt font after the document class
% tgpagella contains no slanted shapes, so lets borrow them from mathpazo
\DeclareFontShape{T1}{qpl}{m}{sl} { <-> ssub * pplj/m/sl }{}
\DeclareFontShape{T1}{qpl}{b}{sl} { <-> ssub * pplj/b/sl }{}

%just some stuff for the demo page
\setlrmarginsandblock  {30mm}{*}{*}
\setlxvchars \setxlvchars[\small\sffamily]
\checkandfixthelayout
\fixpdflayout

\usepackage[final, babel=true]{microtype}
\usepackage[english]{babel}
\usepackage{amsmath} % amsmath which also loads fonts?

\usepackage{lipsum}

\begin{document}
\mainmatter

\chapter{Regular font families}

\textrm{roman text: (ffi --- --) \\ \lipsum[4] }

\textit{italic text: (ffi --- --) \\ \lipsum[4]}

\textbf{boldface text: (ffi --- --) \\ \lipsum[4] }

\textsl{slanted text: (ffi --- --) \\ \lipsum[4] }

\texttt{typewriter text: (ffi --- --) \\ \lipsum[4] }

\chapter{Lesser used types}
\textit{\textbf{italic bold  text: (ffi --- --) \\ \lipsum[4]}}

\textsl{\textbf{slanted bold  text: (ffi --- --) \\ \lipsum[4]}}

\textsc{smallcaps: (ffi --- --) \\ \lipsum[4] }

\textsc{\textbf{smallcaps bold  text: (ffi --- --) \\ \lipsum[4]}}

\chapter{Other test cases}

\textrm{Regular text (\texttt{http://tttext}) and \textbf{bold text} and \textit{italic text} and \textsl{slanted text}}

\textrm{\LARGE Regular text (\texttt{http://tttext}) and \textbf{bold text} and \textit{italic text} and \textsl{slanted text}}

\textrm{ }
\textrm{ffi --- }

$$d=Xh+S\phi+\varepsilon \label{eq:fMRIAnalysisChapter3}$$

$$h=\left( X^TP_sX\right)^{-1} X^T P_s d \label{eq:Eq2Chapter3}$$

\end{document}

So I want to use the newest tgpagella font, but it does not offer a math font, so we first load mathpazo to later only overwrite the body font with the tgpagella fonts. But we have to use luximono as a nice companion typewriter font. Then to solve all the warnings we borrow the slanted shapes from the mathpazo font.

At the moment, I think it looks fairly good, although I'm not entirely happy about the spacing/sizes in the math example and the fact that the mathpazo shapes are slightly larger then tgpagella.

So the question, is this really the best way? And what would you do to improve this?

-
Why exactly are you trying to use Pagella and Pazo at the same time? When it comes to fonts "newer" does not always translate to "better". –  Sharpie Feb 8 '12 at 21:03
well, Pagella does nothing for the math of a document. and I like Pagella a bit more since it allows me to slightly scale it down. –  Davy Landman Feb 8 '12 at 21:08
Your "slanted roman" font does not seem to be set up correctly: When I run your code, on p. 3 of the output file I definitely get italics and not slanted-roman. Separately: If you're going to use pdflatex, you might as well stick with the mathpazo package for both text and math, unless you need lots of accented characters (for which tgpagella will be useful). If you wish to use xelatex instead of pdflatex, you could use the Asana Math package and together with Hermann Zapf's fully-updated (2005) Palatino nova text font (from Linotype, in otf format); not free, though. –  Mico Feb 8 '12 at 22:26
@Mico I agree, that was wrong, it just looked differenlty, but that was off course because mathpazo also doesn't define a slanted roman font. Could you clarify why you feel tgpagella provides no advantage? –  Davy Landman Feb 9 '12 at 13:19
Davy, I'm sorry if I implied that the tgpagella font package provides no advantages (relative to mathpazo). If you have lots of words with accented characters, you're certainly better off loading tgpagella as the (text) font. Conversely, if you have no need to typeset accented characters, I don't see what advantages that the non-opentype version of tgpagella may have for you -- better to keep your setup as simple as possible. Moreover, I like the "small caps" font shape of mathpazo more than tgpagella's version. –  Mico Feb 9 '12 at 13:55

The pplj family has no slanted font: indeed the relevant line in t1pplj.fd reads

\DeclareFontShape{T1}{pplj}{m}{sl}{<->ssub * pplj/m/it}{}

If you want to avoid the messages, you can write

\DeclareFontShape{T1}{qpl}{m}{sl} { <-> ssub * qpl/m/it }{}
\DeclareFontShape{T1}{qpl}{b}{sl} { <-> ssub * qpl/b/it }{}

so getting Pagella Italic. The most convenient way is just not using slanted, because it doesn't exist.

If I remove the scale=0.95 option from the call to tgpagella, I get characters that are the same width as the normal Palatino provided by mathpazo. If you want smaller fonts, just use 9pt as the main size.

I see bad spacing in the second math formula only because you say \left(...\right), which in this case is wrong.

There is no need to do \RequirePackage before \documentclass:

\documentclass[a4paper, oneside, 9pt, extrafontsizes, showtrims, draft]{memoir}
\usepackage[T1]{fontenc} % choose the default encoding
\usepackage[english]{babel} % choose the language

\usepackage[sc]{mathpazo} % use mathpazo for math fonts
\usepackage{tgpagella} % but use tgpagella as main font
\usepackage[scaled=0.75]{luximono}

% Are the following really necessary? I don't think so
%\renewcommand*{\memfontfamily}{qpl} % tgpagella as main memoir font
%\renewcommand*{\memfontenc}{T1}
%\renewcommand*{\memfontpack}{tgpagella}

\normalfont % we want to avoid annoying warnings
\DeclareFontShape{T1}{qpl}{m}{sl} { <-> ssub * qpl/m/it }{}
\DeclareFontShape{T1}{qpl}{b}{sl} { <-> ssub * qpl/b/it }{}

\setlrmarginsandblock  {30mm}{*}{*}
\setlxvchars \setxlvchars[\small\sffamily]
\checkandfixthelayout
\fixpdflayout

\usepackage[final, babel=true]{microtype}
\usepackage{amsmath} % amsmath which also loads fonts?

hi egreg, thank you for your answer, I thought I read in the memoir manual (p15) that for extrafontsizes and other calculations to work you would have to load the fonts before the \documentclass. Why is using \left(...\right) wrong here? I thought you always had to do that? –  Davy Landman Feb 9 '12 at 8:49
I get the same results with or without mathpazo; extrafontsizes is needed only if you need to pass an option such as 25pt; it is evident that \left and \right choose too large a parenthesis, in that case. –  egreg Feb 9 '12 at 9:03


text:
# How to make \ref references include the word 'Figure'? [duplicate]

I'm using the package hyperref. When referencing my figure, I currently use code like

Figure \ref{fig:chorus-overview} gives an overview of the basic concept.

because \ref will just yield a single integer 1, as opposed to Figure 1.

However, the word Figure is auto-generated for the figure environment caption, so when using a package such as [ngerman]{babel}, it's Abbildung instead, and I need to fix the text before my reference.

Is there any way to directly get the Figure 1 text via \ref or a similar command?

EDIT

Additionally to the steps described in Sveinung's answer (using \autoref from hyperref or \cref from cleveref), I needed to make sure to place the label before the caption in the figure environment, as pointed out by Werner, and to include babel before hyperref/cleveref.

\nameref does not do what I need, since it copies the text of the caption instead.

-

## marked as duplicate by Werner, Thorsten, lockstep, barbara beeton, Torbjørn T.Jul 8 '12 at 14:49

Since the question most likely has something to do with the placement of labels, and/or babel options, it is probably not an exact duplicate. –  Sveinung Jul 8 '12 at 14:32
+1 for mentioning that the label needs to go before the caption. –  Will May 2 at 17:07

Load the package cleveref in your preamble and change your \ref{label} to \cref{label}. As egreg emphasizes, the package cleveref must be loaded after hyperref.

Or you may change your \ref{label} to \autoref{label}. See the hyperref manual page 15.

If I misunderstood you, and the problem is that you want the English word Figure instead of the German word Abbildung, you have two possibilities:

1. Either load babel with the option UKenglish instead of ngerman if your text is in English.
2. If your text is in German, but you still want to use the English word figure, you can use the babel-environment \begin{otherlanguage} before your figure, see the babel-manual page 6 (use texdoc babel at a command prompt).
-
The package cleveref must be loaded after hyperref. Just a reminder, but it's important because usually it's the other way around and people don't read manuals. :) –  egreg Jul 8 '12 at 14:11
@mic_e: This may be because you have your \label and \caption command inverted. See Where to put the \label on a figure environment? –  Werner Jul 8 '12 at 14:20
Then it is most likely a misplaced \label, see @Werner's comment. I suggest you try to make an minimal working example (MWE) –  Sveinung Jul 8 '12 at 14:33
I've solved the problem now. I made two mistakes: First, I included hyperref before including babel, and second, I placed \label before \caption. Thanks for your help. –  mic_e Jul 8 '12 at 14:37
@mic_e Note that, with some exceptions such as cleveref, hyperref should be loaded as the last package. See Which packages should be loaded after hyperref instead of before? –  Torbjørn T. Jul 8 '12 at 14:52


text:
What determines how transparent a material is?

Every time light hits a material, both reflection and refraction occurs. How does a material determine how much light gets reflected/ refracted e.g. glass vs silver? So far what I could find is light gets scattered by atoms in the material in all directions, but they all cancel out except in these 2 directions. But that does not explain the different degrees of reflection and refraction.

• I recommend you to read Feynman’s QED. It addresses exactly these questions in a very readable form – Superfast Jellyfish Feb 28 at 6:12

Classical light is the super position of zillions of photons with the energy of $$hν$$, mathematically this means their wave functions are added to create light and the images carried by its variations.

Lets take glass:

In order for the material to be transparent, it means that the image information carried by the superposition goes through carrying the same phases and energy photons, so that the colors and shapes pass through unchanged. This means that the photons should scatter elastically with the whole solid state crystal lattice that composes the glass. i.e. individual wave function solutions are photon +lattice elastic scattering.

There are variations on this, from color material, which absorbs some frequency photons and leaves others to scatter thus changing the balance, to opaque which do not carry images through but light goes through in a combination of absorption and reemission scaters. Completely opaque materials absorb all the photons, or reflect them, the energy turning into lattice vibrations and in the end macroscopically heat.

The same goes for reflection, for glass the elastic scattering in the back direction generates reflection. For a silver backed mirror the reflection is elastic scattering of photons at the first atomic level of the silver lattice, but the forward part is absorbed into lattice vibrations.

I believe your question has basically two main parts:

1. why does a material let light through, that is translucent, which is more then just transparent. Transparent materials are not only translucent, but they have a single index of refraction throughout the material

Translucency (also called translucence or translucidity) allows light to pass through, but does not necessarily (again, on the macroscopic scale) follow Snell's law; the photons can be scattered at either of the two interfaces, or internally, where there is a change in index of refraction. In other words, a translucent material is made up of components with different indices of refraction. A transparent material is made up of components with a uniform index of refraction.1 Transparent materials appear clear, with the overall appearance of one color, or any combination leading up to a brilliant spectrum of every color. The opposite property of translucency is opacity.

Basically this part of your question is about the material being translucent or opaque. The answer to this question is that the photons as they interact with the material, can either be:

1. reflected, that is, if a material reflects most of the incident photons, like metals, or shiny materials,

2. refracted, that is certain materials refract most of the photons, like glass, certain plastics, crystals, water etc.

3. absorbed (heating up the material)

Now it is very important to understand that all three happen with all materials, and it is all probabilities. Metals reflect most of the photons, and will only refract very little number of photons, and absorb some (that will heat up the material).

Glass does both, refracts most photons, and reflects some, and absorbs some too. If you see a glass window, you can see through, and a reflection too, and some will heat up the glass too.

You are asking how a material decides how much of these photons will be either reflected or refracted or absorbed. The answer is that this is a QM process, and the structure of the material's atoms and molecules decides the interaction, and the ratios.

https://en.wikipedia.org/wiki/Transparency_and_translucency

1. and in certain cases, why is this refraction through the material so that it keeps images recognizable (like glass)

This question is answered in the other answer, and that is elastic scattering, throughout the material, as photons refract, if this refraction keeps the relative energy level of the photons, and the relative angle, and phase, then the material will be able to transfer a recognizable image, like certain glasses.

It is very important to understand that certain materials are translucent, will let light through, or even transparent, having a single index of refraction throughout the material, but they will still not be able to transfer a recognizable image. This is because the structure of the material is so, that the photons' relative energy levels, relative angle, or phase will not be kept. There are certain crystals like that, certain plastics etc.

If you look through certain materials (crystals, plastics), you see they let light through, but the images are not recignizable, because the photons relative energy levels, angle and phase is not kept.


text:
## Functiones et Approximatio Commentarii Mathematici

### Products of consecutive values of some quartic polynomials

Artūras Dubickas

#### Abstract

In this paper, we investigate some special quartic polynomials $P$ whose coefficients for $x^4,x^3,\dots,1$ are $a^2, 2a(a+b), a^2+b^2+3ab+2ac, (a+b)(b+2c), (a+b+c)c$, where $a,b,c \in \mathbb{Z}$, and consider the question whether the product $\prod_{k=1}^m P(k)$ is a perfect square for infinitely many $m \in \mathbb{N}$ or for only finitely many $m \in \mathbb{N}$. The answer depends on the solutions of the Pell type diophantine equation $(a+b+c)(ax^2+bx+c)=y^2$. Our results imply, for example, that the product $\prod_{k=1}^m (4k^4+8k^2+9)$ is a perfect square for infinitely many $m \in \mathbb{N}$, whereas the product $\prod_{k=1}^m (k^4+7k^2+16)$ is a perfect square for $m=3$ only, when it equals $230400=480^2$.

#### Article information

Source
Funct. Approx. Comment. Math., Volume 60, Number 2 (2019), 237-244.

Dates
First available in Project Euclid: 26 October 2018

https://projecteuclid.org/euclid.facm/1540519331

Digital Object Identifier
doi:10.7169/facm/1733

Mathematical Reviews number (MathSciNet)
MR3964262

Zentralblatt MATH identifier
07068533

#### Citation

Dubickas, Artūras. Products of consecutive values of some quartic polynomials. Funct. Approx. Comment. Math. 60 (2019), no. 2, 237--244. doi:10.7169/facm/1733. https://projecteuclid.org/euclid.facm/1540519331

#### References

• [1] T. Amdeberhan, L.A. Medina and V.H. Moll, Arithmetical properties of a sequence arising from an arctangent sum, J. Number Theory 128 (2008), 1807–1846.
• [2] Y.-G. Chen and M.-L. Gong, On the products $(1^\ell+1)(2^\ell+1)\dots(n^\ell+1)$ II, J. Number Theory 144 (2014) 176–187.
• [3] Y.-G. Chen, M.-L. Gong and X.-Z. Ren, On the products $(1^\ell+1)\break(2^\ell+1)\dots(n^\ell+1)$, J. Number Theory 133 (2013), 2470–2474.
• [4] Y.-G. Chen, M.-L. Gong and X.-Z. Ren, On the products $(1^\ell+1)\break(2^\ell+1)\dots(n^\ell+1)$, II, J. Number Theory 144 (2014), 176–187.
• [5] J. Cilleruello, Squares in $(1^2+1)\dots(n^2+1)$, J. Number Theory 128 (2008), 2488–2491.
• [6] J. Cilleruello, F. Luca, A. Quirós and I.E. Shparlinski, On squares in polynomial products, Monatsh. Math. 159 (2010), 215–223.
• [7] A. Dubickas, Multiplicative dependence of quadratic polynomials, Lith. Math. J. 38 (1998), 225–231.
• [8] P. Erdös and J.L. Selfridge, The product of consecutive integers is never a power, Illinois J. Math. 19 (1975), 292–301.
• [9] J.-H. Fang, Neither $\prod_k=1^n(4k^2+1)$ nor $\prod_k=1^n(2k(k-1)+1)$ is a perfect square, Integers 9 (2009), paper #A16, 177–180.
• [10] M. Filaseta, S. Laishram and N. Saradha, Solving $n(n+d)\dots (n+(k-1)d)=by^2$ with $P(b) \leq Ck$, Intern. J. Number Theory 8 (2012), 161–173.
• [11] K. Gaitanas, An infinite family of quartic polynomials whose products of consecutive values are infinitely often perfect squares, Integers 17 (2017), paper #A32, 3 pp.
• [12] E. Gürel, A note on the products $((m + 1)^2 + 1)\dots(n^2 + 1)$ and $((m + 1)^3 + 1)\break\dots(n^3 + 1)$, Math. Commun. 21 (2016) 109–114.
• [13] E. Gürel, On the occurrence of perfect squares among values of certain polynomial products, Amer. Math. Monthly 123 (2016), 597–599.
• [14] K. Györy, L. Hajdu and A. Pintér, Perfect powers from products of consecutive terms in arithmetic progression, Compos. Math. 145 (2009), 845–864.
• [15] C. Niu and W. Liu, On the products $(1^3+q^3)(2^3+q^3)\dots(n^3+q^3)$, J. Number Theory 180 (2017), 403–409.
• [16] C.L. Siegel, The integer solutions of the equation $y^2 = ax^n + bx^n-1 + \dots + k$, J. Lond. Math. Soc. 1 (1926), 66–68.
• [17] W. Zhang and T. Wang, Powerful numbers in $(1^k +1)(2^k + 1)\dots(n^k + 1)$, J. Number Theory 132 (2012), 2630–2635.


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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# Use Square Roots to Solve Quadratic Equations

## Isolate the squared term, take the root of both sides

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Use Square Roots to Solve Quadratic Equations

What if a bird were flying through the air carrying a worm in its beak? All of the sudden, the worm slips from the bird's beak, and the poor bird no longer has his dinner! If the bird were flying at a height of 2000 feet, how long would it take the worm to hit the ground, assuming there were no air resistance? In this Concept, you'll learn to solve quadratic equations like the one representing this scenario by using square roots.

### Guidance

Suppose you needed to find the value of x\begin{align*}x\end{align*} such that x2=81\begin{align*}x^2=81\end{align*}. How could you solve this equation?

• Make a table of values.
• Graph this equation as a system.
• Cancel the square using its inverse operation.

The inverse of a square is a square root.

By applying the square root to each side of the equation, you get:

xx=±81=9 or x=9

In general, the solution to a quadratic equation of the form 0=ax2c\begin{align*}0=ax^2-c\end{align*} is:

x=ca or x=ca

Another way to write this is x=±ca.\begin{align*}x=\pm \sqrt{\frac{c}{a}}.\end{align*}

#### Example A

Solve for x when 0=2x218\begin{align*}0=2x^2-18\end{align*}.

Solution:

0=2x21818=2x29=x2x=±3.\begin{align*}0=2x^2-18 \Rightarrow 18=2x^2\Rightarrow 9=x^2 \Rightarrow x=\pm 3.\end{align*}

This is the same result you would get if you use the formula above:

x=±ca=±182=±9=±3.\begin{align*}x=\pm \sqrt{\frac{c}{a}}=\pm \sqrt{\frac{18}{2}}=\pm \sqrt{9}=\pm 3.\end{align*}

#### Example B

Solve (x4)29=0\begin{align*}(x-4)^2-9=0\end{align*}.

Solution:

Begin by adding 9 to each side of the equation.

(x4)2=9

Cancel the square by taking the square root.

x4=3 or x4=3

Solve both equations for x:x=7 or x=1\begin{align*}x: x=7 \ or \ x=1\end{align*}

Real-World Problems Using Square Roots

In the previous Concept, you learned Newton’s formula for projectile motion. Let’s examine a situation in which there is no initial velocity. When a ball is dropped, there is no outward force placed on its path; therefore, there is no initial velocity.

#### Example C

A ball is dropped from a 40-foot building. When does the ball reach the ground?

Solution:

Using the equation from the previous Concept, h(t)=12(g)t2+v0t+h0\begin{align*}h(t)=-\frac{1}{2} (g) t^2+v_0 t+h_0\end{align*}, and substituting the appropriate information, you get:

Simplify:Solve for x:00402.5t=12(32)t2+(0)t+40=16t2+40=16t2=t21.58 and t1.58

Because t\begin{align*}t\end{align*} is in seconds, it does not make much sense for the answer to be negative. So the ball will reach the ground at approximately 1.58 seconds.

-->

### Guided Practice

A rock is dropped from the top of a cliff and strikes the ground 7.2 seconds later. How high is the cliff in meters?

Solution:

Using Newton’s formula, substitute the appropriate information.

Simplify:Solve for h0:00h0=12(9.8)(7.2)2+(0)(7.2)+h0=254.016+h0=254.016

The cliff is approximately 254 meters tall.

### Explore More

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Solving Quadratic Equations by Square Roots (11:03)

1. x2=196\begin{align*}x^2=196\end{align*}
2. x21=0\begin{align*}x^2-1=0\end{align*}
3. x2100=0\begin{align*}x^2-100=0\end{align*}
4. x2+16=0\begin{align*}x^2+16=0\end{align*}
5. 9x21=0\begin{align*}9x^2-1=0\end{align*}
6. 4x249=0\begin{align*}4x^2-49=0\end{align*}
7. 64x29=0\begin{align*}64x^2-9=0\end{align*}
8. x281=0\begin{align*}x^2-81=0\end{align*}
9. 25x236=0\begin{align*}25x^2-36=0\end{align*}
10. x2+9=0\begin{align*}x^2+9=0\end{align*}
11. x216=0\begin{align*}x^2-16=0\end{align*}
12. x236=0\begin{align*}x^2-36=0\end{align*}
13. 16x249=0\begin{align*}16x^2-49=0\end{align*}
14. (x2)2=1\begin{align*}(x-2)^2=1\end{align*}
15. (x+5)2=16\begin{align*}(x+5)^2=16\end{align*}
16. (2x1)24=0\begin{align*}(2x-1)^2-4=0\end{align*}
17. (3x+4)2=9\begin{align*}(3x+4)^2=9\end{align*}
18. (x3)2+25=0\begin{align*}(x-3)^2+25=0\end{align*}
19. x26=0\begin{align*}x^2-6=0\end{align*}
20. x220=0\begin{align*}x^2-20=0\end{align*}
21. 3x2+14=0\begin{align*}3x^2+14=0\end{align*}
22. (x6)2=5\begin{align*}(x-6)^2=5\end{align*}
23. (4x+1)28=0\begin{align*}(4x+1)^2-8=0\end{align*}
24. (x+10)2=2\begin{align*}(x+10)^2=2\end{align*}
25. 2(x+3)2=8\begin{align*}2(x+3)^2=8\end{align*}
26. How long does it take a ball to fall from a roof to the ground 25 feet below?
27. Susan drops her camera in the river from a bridge that is 400 feet high. How long is it before she hears the splash?
28. It takes a rock 5.3 seconds to splash in the water when it is dropped from the top of a cliff. How high is the cliff in meters?
29. Nisha drops a rock from the roof of a building 50 feet high. Ashaan drops a quarter from the top-story window, which is 40 feet high, exactly half a second after Nisha drops the rock. Which hits the ground first?
30. Victor drops an apple out of a window on the 10th\begin{align*}10^{th}\end{align*} floor, which is 120 feet above ground. One second later, Juan drops an orange out of a 6th\begin{align*}6^{th}\end{align*}-floor window, which is 72 feet above the ground. Which fruit reaches the ground first? What is the time difference between the fruits’ arrival to the ground?

Mixed Review

1. Graph y=2x2+6x+4\begin{align*}y=2x^2+6x+4\end{align*}. Identify the following:
1. Vertex
2. x\begin{align*}x-\end{align*}intercepts
3. y\begin{align*}y-\end{align*}intercepts
4. axis of symmetry
1. What is the difference between y=x+3\begin{align*}y=x+3\end{align*} and \begin{align*}y=x^2+3\end{align*}?
2. Determine the domain and range of \begin{align*}y=-(x-2)^2+7\end{align*}.
3. The Glee Club is selling hot dogs and sodas for a fundraiser. On Friday the club sold 112 hot dogs and 70 sodas and made $154.00. On Saturday the club sold 240 hot dogs and 120 sodas and made$300.00. How much is each soda? Each hot dog?

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 10.4.

### Vocabulary Language: English Spanish

The $x-$intercept of a quadratic equation is also called a root, solution, or zero.
Perfect Square

Perfect Square

A perfect square is a number whose square root is an integer.
Roots

Roots

The roots of a function are the values of x that make y equal to zero.
Zeroes

Zeroes

The zeroes of a function $f(x)$ are the values of $x$ that cause $f(x)$ to be equal to zero.


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# Cantilever

Wooden cantilever bridge that led from the hamlet of Eye over the Vispa to St. Niklaus Dorf.
The two arms of this bridge to the right and left of the central pillar are - statically speaking - cantilever arms

A cantilever beam (also: cantilever beam or cantilever ) is in the engineering mechanics (in particular in the structural analysis ) is a unilaterally mounted (in the strict sense firmly clamped ), is often the horizontal bar , the free at its end or is charged up over its entire length transversely.

As a basic system of statics , it is the idealization of a simple component (other simple static systems are the beam on two supports and the three-hinged frame ). As a beam, the length of a cantilever is significantly greater than its height and width.

The cantilever is subjected to shear , bending and sometimes torsion .

The support must be a restraint in which all six degrees of freedom are fixed. The support reactions are a bearing force and a clamping moment or - respectively decomposed into the three spatial directions - three force - and three torque components . ${\ displaystyle {\ vec {F}}}$ ${\ displaystyle {\ vec {M}}}$${\ displaystyle (F_ {x}, ~ F_ {y}, ~ F_ {z})}$${\ displaystyle (M_ {x}, ~ M_ {y}, ~ M_ {z})}$

Cantilever beam for transporting loads on a church tower ( Marienkirche in Pirna )
Cantilever in road traffic ( Marl )

The clamping torque prevents the beam from rotating around the clamping point. A bending moment acts in the beam .

A cantilever is statically determined . In mathematical terms, the clamping torque is equal to the sum of the torques of all forces acting on the carrier ( torque equilibrium ).

In the simplest case, the moment only has to absorb the dead weight of the girder. Just as easily is the case that the carrier is regarded as weightless and a force acts on the free end.

Basic functionality of a cantilever:

Voltages in a cantilever beam blank with the Kragträgerverfahren calculated.

Historically, the name comes from the extended ridge of the roof of a house to which a cable is attached to pull goods up to the upper floors.


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26453 articles – 20338 Notices  [english version]
 Fiche détaillée Preprint, Working Paper, Document sans référence, etc.
 Versions disponibles : v1 (17-01-2011) v2 (06-10-2011)
Liste des fichiers attachés à ce document :
 PDF
 singleindex5.pdf(387.3 KB)
 PS
 singleindex5.ps(913.1 KB)
 Sparse single-index model
 Pierre Alquier1, 2, Gérard Biau1, 3, 4
 The single-index model is known to offer a flexible way to model a variety of high-dimensional real-world phenomena. However, despite its relative implicity, this dimension reduction scheme is faced with severe complications as soon as the underlying dimension becomes larger than the number of observations (p larger than n'' paradigm). To circumvent this difficulty, we consider the single-index model estimation problem from a sparsity perspective using a PAC-Bayesian approach. On the theoretical side, we offer a sharp oracle inequality, which is more powerful than the best known oracle inequality for other common procedures of single-index recovery. The proposed method is implemented by means of the reversible jump Markov chain Monte Carlo technique and its performance is compared with that of standard procedures.
 Mots Clés : Nonparametric statistics – single-index model – sparsity – PAC-Bayesian inequalities – oracle inequalities – MCMC.
 hal-00556652, version 1 http://hal.archives-ouvertes.fr/hal-00556652 oai:hal.archives-ouvertes.fr:hal-00556652 Contributeur : Pierre Alquier <> Soumis le : Lundi 17 Janvier 2011, 14:53:13 Dernière modification le : Lundi 17 Janvier 2011, 15:37:02


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# Has anyone tabulated 2-knots? Would anyone like to try?

I'd love to have a list of 'small' 2-knots, for some sense of small. It's not clear what one should filter by, but there are two obvious candidates

• Write a movie presentation, and count the frames.
• Project the 2-knot to R^3, and count the triple points.

Does anyone know if this has been attempted? Such a list could be quite useful.

-

There is a table in Yoshikowa's paper based upon the representation as graphs with certain markings of vertices. The markings indicate which directions are the A and B smoothings. The table is quite small.

Kamada's book on Surface braids has a nice table of 2-knots that don't have triple points. There are tons of these. So your candidate for small is sort of strange. Braid index might be better.

Quandle cocycles are the only known ways to get lower bounds for triple points, and there are not nearly enough 3-cocycles that have been calculated.

Last spring Dennis Roseman told me a neat way to generate "random" 2-knots. I don't think he has written anything down about it though.

So if you look at movies, you have to be careful about what you define as an event. In the CRS point of view each of Reid. I,II,III, birth, death, saddle, switchback, and psi (pitchfork) moves is an event, but exchanges of distant critical points are also events. Any such event leaves its trace on the chart.

There are other notions of simplicity. For example, there is a notion of thickness. Project into 3-space and subsequently onto the plane (that gives the chart). Take a spear perpendicular to the plane and see how many times it generically passes through the surface. Find the maximum of these, and then minimize over all projections.

Probably the real reason for making a census would be to discover new invariants. So it is a good problem in that sense.

-
A problem with the Yoshikawa technique is that having one of those decorated knotted 4-valent graphs is not enough to specify an embedding $S^2 \to S^4$ -- when you resolve the 4-valent graph either way you get a trivial link. One also has to know the spanning discs for those two trivial links. Change your spanning discs and (in general) you change the knot. –  Ryan Budney Aug 8 '10 at 23:55
As far as I know, the assumption is not that one adds a generic spanning disk to the trivial link $L_t$. The assumption is that one adds a trivial disk system to $L_t$, for the definition of a trivial disk system see section 8.5 of "Braids and Knot Theory in Dimension Four" by Kamada. Any two trivial disk systems bounding a given trivial link $L_t$ are ambient isotopic, Proposition 8.6 of Kamada. Thus, the 2-knot type is independent of the trivial disk system which closes the trivial link $L_t$, Kamada Proposition 9.11. –  Kelly Davis Aug 3 '11 at 10:30

Another complexity filter could be the number of 4-dimensional simplices in a triangulation of the complement.

2-knots complements have simple ideal triangulations. The proof is very similar to the proof for 1-knots, viewed through the lens of Morse theory on manifolds with boundary. ie: the Wirtinger presentation is viewed as the fundamental group computation coming from the cell decomposition of a Morse height function (in the stratified sense) on the knot complement.

It'd be nice if there was a SnapPea-like algorithm for triangulating 2-knot complements. As far as I know those kinds of details haven't been worked out by anyone. But it should be reasonably doable.

-


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Some cablebot parts are milled in a composite panel consisting of two aluminium cover sheets and a polymer core, this material can also be bended if cutted at specific depth with a 90 degrees V end mill. In the cablebot repository you can find all the parts as DXF files. Gcode has to be generated with specific settings for the used CNC machine.

Feeds and speeds for V - Curve milling bit to bend alucobond

In our case we are using a high revolution spindle that works in between 18k-24k RPM. After some tests, we have found that this feeds and speeds work us well for cutting 3mm alucobond composite:

Tool RPM Feed Feed plunge Stepdown
V carve - 90º 24,000 rpm 5,000 mm/min 3,000 mm/min 0.22 mm/step
3 mm flat tool 24,000 rpm 5,000 mm/min 3,000 mm/min 1.2 mm/step
6 mm flat tool 18,000 rpm 5,000 mm/min 3,000 mm/min 1.7 mm/step

In the Romi Cablebot github repository you can find all the drawings (in DXF format) to generate the needed gcode. In the files you will find different layers depending on the operation and the depth, in this example you can see the layer called vcut for the folding marks, pocket-0.95mm indicates a pocketing operation with a depth of 0.95 mm and profiling to cut the piece. Keep in mind that you will need to add bridges on your piece, so it doesn't move during machining, this process is different depending on the CAM software you use.

Blendercam is a free/libre addon that allows gcode generation inside blender, in this way we avoid the use of extra software and model exporting. Don't forget to check their documentation.

To use it you need to clone the addon repository to some place in your computer:

git clone https://github.com/vilemduha/blendercam

Some python dependencies sould also be installed, you can do it with pip from the command line:

$./pip3 install shapely$ ./pip3 install vtk
\$ ./pip3 install Equation

OpenCamLib is an optional dependency, but based on our tests we recommend its installation.

To activate the addon, in Blender, open the Preferences window (edit → preferences).
Clik on File Paths button and enter the path where you cloned the blender CAM repository in the Scripts field.

Save perferences and restart blender. Now enable it in Add-ons section (preferences window).

After installing the addon you will need a postprocessor script that works with your specific CNC machine. If none of the included ones works for you, you can easily create your own:

1. Modify scripts/addons/cam/__init__.py and add a new item on the machineSettings class (around line 125):

2. Create a new file in scripts/addons/cam/nc/ directory with your post processor name. (ej. raptor.py). You can copy an existing postprocessor and modify it to fit your needs.

3. Modify scripts/addons/cam/gcodepath.py, search for the exportGcodePath() function and add a condition for your post processor where you specify the extension of the file and the name of the module you just created on step 2.

There is example commit on what's needed to add a postprocessor here. It is a little outdated (use gcodepath.py instead of utils.py) but can be used as a general guide.

### Steps to get folding tracesLink

As an example on how to get the proper traces for alucobond milling with folding parts.

1. With the object cutting side pointing up, duplicate it and rotate 90º with the bottom corner as rotation point.

2. Displace the duplicated part 2mm towards the center. That's one millimeter per side of the folding axis.

3. Repeat both steps on the other side

4. Create a line at 1mm from the part border (centered between both pieces)

5. Repeat the process for the other side, now you have the V cutt milling traces

6. Join the two parts, remove the vertices outside the bottom layer and create a bridge to join both parts. This paths should be -2.2mm from the surface of the material (leaving a thikness of 0.8mm after cutting)

Now you can process the part with blendercam to get the gcode.


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# Adaptive Control using Multiple Models

This post has been adapted from a university project to understand and simulate the findings of [1]

## Preamble

The evolution of a dynamical system is mathematically modeled as differential equations. Properties of the system like position and velocity at any given moment are the State of the system. Differential equations capture how these states change and influence each other over time.

The mathematical models imply that a given system (or Plant) would follow a deterministic trajectory for a given initial state. For example, an airplane would remain at rest if it is parked on level ground. It would glide down if it is released at 1000 feet above in the air with a particular initial velocity and orientation.

But we often want to do more with systems than to let them follow their natural trajectories. And we cannot always precisely set or measure initial conditions. We want more Control.

Thus we influence some properties of the dynamical system by applying a control input. For example applying torque to the wheels of your car, or by propelling the motors of an airplane to create thrust that pushes it forward. This control input directly or indirectly alters the state of our system.

A control designer’s job is to ensure that the system follows a trajectory of our choice, which is often represented as a Reference Model; and to ensure stability of the overall system, i.e. it does not behave erratically on small deviations.

## PID Controller

For simple single-order plants, an often effective strategy is to change the control input in proportion to the difference (or error) between the plant state and reference state.

The error term is often integrated over time to capture the past, and differentiated to capture the instantaneous future of the error. The three terms (Proportional, Integral, Derivative) are used together in different proportions to design a control input, resulting in a PID Controller.

The performance of simple PID controllers isn’t always good enough, and the control law is not designed to the specification of plant dynamics and parameters, but only to its output.

Especially for complex multi-order plants, a lot of information is lost if we use only the plant output to control it and not the known information about its internal dynamics.

Many times the parameters of the dynamical system are not completely known, or they are changing as the system evolves. In such scenarios, we need to adaptively change the control law itself with time.

Adaptive control primarily focuses on system identification (i.e. estimating the parameters of the system) to design the control input; and using Lyapunov analysis to ensure stability of the system.

### Plant Model

Let’s start with the simple case of a multi-order linear plant. Its dynamics can be described in state-space form as:

Where $x_p$ is the n-dimensional state space vector, $A_p$ are the parameters as a $n \times n$ companion matrix. $u$ is the control input, and $b = [0, \cdots, 0, 1]^T$

Side Note: More details on how to represent an nth order d.e. in state-space form with companion matrices is given here

The last row of the companion matrix $A_p$ is the parameter vector $\theta_p$, which can also be represented as:

### Reference Model

A reference model that this plant should track can be described via a similar set of equations:

Where $x_m$ and $A_m$ are described similarly, and $r$ is a known reference signal.

### Identification Model

Before we proceed to designing our control law, we need to estimate the plant parameters $\theta_p$, as they are unknown to us.

To achieve that we set up an identification model as follows:

Where $A_I$ is a matrix in companion form with its last row as the plant parameter estimate $\theta_I$. The identification error $e_I$ is defined as:

Let $P$ be the positive definite matrix solution to the lyapunov equation for the reference model:

Where our Lyapunov function candidate is defined as:

#### Theorem 1

If we define our parameter adaptive law as:

we end up with a stable system under Lyapunov criteria:

### Control Law

Using the Lyapunov stability analysis, we can define our control input as:

Where $k(t) = \theta_I(t) - \theta_m$. From Barbalat’s lemma it follows that the tracking error $e_c$ approaches 0 $\lim_{t \rightarrow \infty} e_c(t) = 0$, or in other words that $x_p$ tracks $x_m$.

#### Simulation 1

For a second order plant, with parameters $\theta_p = [-7, -2]^T$ tracking a reference model with $\theta_m = [-5, -6]^T$. We initialize our identification model with $\theta_i = [-8, -4]^T$. The tracking signal is $r(t) = 5\sin(0.2\pi t)$.

(d/l multidim_singlemodel.slx)

The tracking error $e_c$ dies down slowly. This is because the parameter estimates $\theta_i$ converge to the plant values $\theta_p$ very slowly (as seen in the following figure).

## Multiple Identification Models

### Level 1

Let us use $N = n + 1$ identification models, defined similarly as before, with same initial state as $x_p$, but different initial values for the parameter estimate $\theta_i$:

Despite having $N$ models, we can still only have a single control input. We achieve that by taking a convex combination of the parameter estimates as:

Where $\sum_{i=1}^N \alpha_i = 1$. We can restate this in matrix-vector notation, with $\theta_i$ as columns of $\Theta_I$, and $\alpha = [\alpha_1, \alpha_2, \cdots, \alpha_N]^T$:

#### Theorem 2

If $N$ adaptive identification models are initialized as $\theta_i(t_0)$ such that $x_i(t_0) = x_p(t_0)$, and plant parameter vector $\theta_p$ lies in the convex hull of $\theta_i(t_0)$, then $\theta_p$ continues to lie in the convex hull of $\theta_i(t), t > t_0$.

#### Simulation 2

For a second order plant, with parameters $\theta_p = [-2, 9]^T$ tracking same reference model as before $\theta_m = [-5, -6]^T$. We initialize our identification model with $\alpha = [0.7, 0.1, 0.2]^T$ and

(d/l multidim_multimodel.slx)

Tracking error $e_c$ still dies down slowly in the later phase. Despite having multiple models, we’re still estimating only one value in the convex hull of the multiple models. And the individual parameter errors of the models are still converging slowly to the actual value.

### Level 2

Level 1 leads to an equivalent performing system as before, albeit using multiple models. The key insight is to adaptively estimate $\alpha$ itself, along with each model’s parameters.

The conclusion of theorem 2 can be restated as:

Subtracting $\theta_p$ from both sides:

The previous equation can be simplified down using:

Substituting $\bar{\alpha} = [\alpha_1, \alpha_2, \cdots, \alpha_{N-1}]^T$ leads to:

#### Estimate $\bar{\alpha}$

We can now use a standard estimation process for $\bar{\alpha}$ like we did for parameters before. Let $\hat{\bar{\alpha}}(t)$ be its estimation, then the adaptive law is given by:

#### Theorem 3

Using analysis similar to Theorem 1, we can show that the second level of adaptation also leads to an overall stable system. And when convergence is slow in first level (i.e. large parametric errors), convergence of $\hat{\bar{\alpha}}$ to $\bar{\alpha}$ is rapid. (and vice versa)

#### Simulation 3

For the same second order plant as in Simulation 2. That is $\theta_p = [-7, -2]^T$, $\theta_m = [-5, -6]^T$. This time we initialize our identification model with $\hat{\bar{\alpha}}(t_0) = [0.7, 0.1, 0.2]^T$ estimating it on the fly along with plant parameters with

(d/l multidim_multimodel_lvl2.slx)

The tracking error is converging to 0 rapidly.

The individual parameter estimates are still converging to the plant values very slowly. But due to fast second level estimation of $\alpha$ (as seen in next fig.), the system as a whole is converging down rapidly.

## Result

Through qualitative analysis and simulations we have demonstrated the effectiveness of using multiple models for adaptive control, as introduced by the original paper [1].

## References

[1] Z. Han and K. S. Narendra, “New Concepts in Adaptive Control Using Multiple Models,” in IEEE Transactions on Automatic Control, vol. 57, no. 1, pp. 78-89, Jan. 2012. doi: 10.1109/TAC.2011.2152470


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# Homework Help: Consequence of Cauchy's Integral Formula

1. Mar 13, 2009

### Nebula

1. The problem statement, all variables and given/known data
Suppose f is analytic inside and on the unit circle |z|=1. Prove that if |f(z)|$$\leq$$M for |z|=1, then |f(0)|$$\leq$$M and |f'(0)|$$\leq$$M.

Finally, what estimate can you give for |f^n(0)|?

2. Relevant equations
This problem is in a section dealing with Cauchy's Integral Formula and it's generalized form.
http://en.wikipedia.org/wiki/Cauchy_integral_formula

3. The attempt at a solution
Im not quite sure how to get started here. Obviously Cauchy's formulas apply to our function f considering its analiticity. I guess I don't really understand the if part of our statement. IE if |f(z)|$$\leq$$M for |z|=1. The magnitude of the function f is bounded??? I don't know what it means "for |z|=1." I'm missing something here. A nudge in the right direction would be appreciated. Thanks!

2. Mar 13, 2009

### Dick

The absolute value of the integral of a function is less or equal to the integral of the absolute value of the function times the length of the curve you are integrating over. The 'a' the Cauchy formula is zero. |z|=1. The function f is bounded on |z|=1 by M. That's a nudge.