text:
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In Mexico, the popular mariachi band includes a range of guitars, from the small requinto to the guitarrón, a guitar larger than a cello, which is tuned in the bass register. In Colombia, the traditional quartet includes a range of instruments too, from the small bandola (sometimes known as the Deleuze-Guattari, for use when traveling or in confined rooms or spaces), to the slightly larger tiple, to the full-sized classical guitar. The requinto also appears in other Latin-American countries as a complementary member of the guitar family, with its smaller size and scale, permitting more projection for the playing of single-lined melodies. Modern dimensions of the classical instrument were established by the Spaniard Antonio de Torres Jurado (1817–1892).[12]
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# Autocorrelation Computation with MATLAB

I have done the computation of the autocorrelation of a sequence (vector), here called "frame", and I have plotted it. Here the code:

The resulting graph is:

I do not understand the position of the peak in this graph: I know that the peak of the autocorrelation should be in the origin, why is it there?

The x axis is simply an index number, and this is showing your the correlation for leading and trailing shifts. The shift at index 80 is lag = 0, the origin you are interested in. Type help(xcorr) in Matlab to get further detail in how to return the lag. Following that, if you do the following you will get the result you are looking for:

r, lag = xcorr(frame, frame);
figure()
plot(lag, r)

I did it as shown above to highlight how to return lag from xcorr specifically while minimizing change to your code, but xcorr(frame) will return the same result as xcorr(frame,frame) so can be further simplified for the case of autocorrelation.

format: plain


text:
# Teacher Notes

### Why use this resource?

Thinking about the behaviour of functions can help to make sense of identities. Similarly, by looking at identities satisfied by a function we can glean information about the behaviour of the function. In this resource, students are asked to use the half-angle formulae $\cos^2 \frac{\theta}{2} \equiv \frac{1}{2}(1+\cos \theta)$ and $\sin^2 \frac{\theta}{2} \equiv \frac{1}{2}(1-\cos \theta)$ to help them sketch graphs of $y=\cos^2{\frac{\theta}{2}}$ and $y=\sin^{2}\frac{\theta}{2}$.

### Preparation

If printed graphs of $y=\cos \theta$ are available these could be used for students to sketch onto. Otherwise, $\quantity{1}{cm}$ square paper would be useful.

### Possible approach

Students could be asked to start by sketching a $y=\cos \theta$ graph if pre-printed graphs are not available.

Ask students to think individually about what they would expect a $y=\cos^2 \frac{\theta}{2}$ curve to look like using their cosine graph sketch as a starting point. After a minute or so they should add their $y=\cos^2 \frac{\theta}{2}$ curve to their sketch then discuss their ideas with a partner. Pairs can then form small groups to share their ideas and agree on what they think the essential features of $y=\cos^2 \frac{\theta}{2}$ and $y=\cos^2 \frac{\theta}{2}$ are.

If graphing software is available students could check their graphs or this could be done as a whole class plenary, asking students to explain why the graphs have the features shown.

### Key questions

• Are there any points on the graph we know straight away?
• What are the domain and range of the two functions?
• Can you describe $y=\cos^2 \frac{\theta}{2}$ and $y=\sin^2 \frac{\theta}{2}$ in terms of transformations of $y=\cos \theta$?

### Possible support

To support students in thinking about functions and their graphs

• How could you decide whether $\cos^2 \frac{\theta}{2}$ and $\cos \theta$ have the same period?
• What is the range of $\sin^2 \frac{\theta}{2}$ and $\cos \theta$?

To support students in thinking about graph transformtions when looking at $\frac{1}{2}(1+\cos \theta)$ and $\frac{1}{2}(1-\cos \theta)$

• How does the $+1$ transform the graph?
• What effect does the minus sign before the cosine function have?
• What effect does the coefficient of $\frac{1}{2}$ have on the cosine graph?

### Possible extension

• Can you transform a graph of $y=\sin \theta$ into either of these half-angle graphs?
• Can you use that to write down a new identity?

A version of this resource has been featured on the NRICH website.


text:
# The shuffling buffer

1 PRISME - Geometry, Algorithms and Robotics
CRISAM - Inria Sophia Antipolis - Méditerranée
Abstract : The complexity of randomized incremental algorithms is analyzed with the assumption of a random order of the input. To guarantee this hypothesis, the n data have to be known in advance in order to be mixed what contradicts with the on-line nature of the algorithm. We present the shuffling buffer technique to introduce sufficient randomness to guarantee an improvement on the worst case complexity by knowing only k data in advance. Typically, an algorithm with $O(n^2)$ worst-case complexity and O(n) or O(n log n) randomized complexity has an O((n^2 log k)/k) complexity for the shuffling buffer. We illustrate this with binary search trees, the number of Delaunay triangles or the number of trapezoids in a trapezoidal map created during an incremental construction.
Type de document :
Article dans une revue
International Journal of Computational Geometry and Applications, World Scientific Publishing, 2001, 11, pp.555-572. 〈http://www.worldscinet.com/ijcga/11/1105/S021819590100064X.html〉. 〈10.1142/S021819590100064X〉

https://hal.inria.fr/inria-00412567
Contributeur : Olivier Devillers <>
Soumis le : mercredi 2 septembre 2009 - 10:31:46
Dernière modification le : samedi 27 janvier 2018 - 01:31:46
Document(s) archivé(s) le : mardi 15 juin 2010 - 20:22:47

### Fichier

TheShufflingBuffer.pdf
Fichiers produits par l'(les) auteur(s)

### Citation

Olivier Devillers, Philippe Guigue. The shuffling buffer. International Journal of Computational Geometry and Applications, World Scientific Publishing, 2001, 11, pp.555-572. 〈http://www.worldscinet.com/ijcga/11/1105/S021819590100064X.html〉. 〈10.1142/S021819590100064X〉. 〈inria-00412567〉

### Métriques

Consultations de la notice

## 266

Téléchargements de fichiers


text:
# Transformation between Markov Chain, transience and recurrence

Here is a question in my problem set which is quite challenging for me:

Suppose $\{X_n\}$ is a Discrete Markov chain with state space $S$ and transition matrix $P$ and no absorbing state ($P_{ii} < 1, \forall i \in S$). Define a new matrix $P'$ by $P'_{ii} = 0, \forall i \in S$ and $P'_{ij} = \frac{P_{ij}}{1 - P_{ii}}$.

Show that $P'$ is a transition matrix and let $\{Y_n\}$ be a DTMC with transition matrix $P'$, argue that $\{X_n\}$ is transient/recurrent $\iff$ $\{Y_n\}$ is transient/recurrent.

The first part is easy and for the $\iff$ proof, my idea is to show $\sum_{n}^{\infty}P_{ii}^{(n)} = \infty \iff \sum_{n}^{\infty} P_{ii}^{'(n)} = \infty$ and $\sum_{n}^{\infty}P_{ii}^{(n)} < \infty \iff \sum_{n}^{\infty} P_{ii}^{'(n)} < \infty$

But I cannot figure out the relation between $P_{ii}^{(n)}$ and $P_{ii}^{'(n)}$, need some help, thanks!

Notice that $P'_{ij} \geq P_{ij}$ whenever $i \neq j$. Therefore, for any two states $i_0,i_n$ and any path $i_0,\dots,i_n$ such that $i_t \neq i_{t+1}$ for all $t$, the probability to observe that path under $P'$ is at least as high as to observe it under $P$. So now choose some path with $i_0=i_n$ which has positive probability under $P$ and use it to conclude that $P'^{(n)}_{i_0 i_0}>0$.

You can adapt this proof to get the reverse implication.

The intuition is that for recurrence vs. transience, if you started at $i$ and are guaranteed to leave $j$ sooner or later anyway, the time it took to wait there has nothing to do with whether you will eventually get back to $i$. So for studying recurrence it is sufficient to prune all self-loops out of the graph (except for absorbing states of course).

For positive recurrence vs. null recurrence this is no longer the case; it can happen that this "pruned" chain is positive recurrent when the original chain was null recurrent.

• Thank you very much! I think your first paragraph is say : "once there is a path ($i_0,\ldots,i_0$)" in $\{X_n\}$, we have higher probability to reproduce the path in $\{Y_n\}$, but the converse is true? Could you involve more details? – user370220 Oct 10 '17 at 18:30
• @ZHANGYUE The probability to see such a path under $P$ is generally smaller than under $P'$, but you can nevertheless show that it is not zero if it wasn't zero for $P'$. – Ian Oct 10 '17 at 18:30
• @lan Thanks for your interpretation, could you formulize the proof a little bit? Since I am really struggling convert this intuition to rigorous proof :( – user370220 Oct 10 '17 at 18:49
• @ZHANGYUE If you understand the sketch in the first paragraph, the adaptation is not really difficult, you just replace $P'_{ij} \geq P_{ij}$ by $P_{ij} \geq (1-P_{ii})P'_{ij}$ and do the same thing. – Ian Oct 10 '17 at 19:27
• @lan Thank you! I think I understand the intuition here but I still get little confused, for transience, I need to show $\sum_{n=1}^{\infty} P_{i,i}^{(n)} = \infty \iff \sum_{n=1}^{\infty} P_{i,i}^{'(n)} = \infty$. But as you said, for any path $(i_0,i_1\ldots,i_0)$ with $i_n \not = i_{n+1}$, the probability for $P'$ is higher than $P$, but this is not enough to conclude that $P_{i,i}^{'(n)} \ge P_{i,i}^{(n)}$ right? Since for $P$ there may exists path containing consecutive identical entries – user370220 Oct 11 '17 at 3:04


text:
# Chapter 11 - Additional Topics - 11.4 - Complex Numbers - Problem Set 11.4 - Page 493: 29

$-20-8i$

#### Work Step by Step

We first use the distributive property to simplify. Then, we combine like terms to obtain: $(-10-4i)-(10+4i)$ =$-10-4i-10-4i$ =$-10-10-4i-4i$ =$-20-8i$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
# Need help driving N FET gate from LM393 comparator output

I am working on a charge controller circuit idea I had. The circuit 'works' meaning it achieved desired output voltage but the fet doesn't seem to be turning all the way on so lots of heat is generated.

I am using an LM393 dual comparator to measure the voltage at the junction of the zener and series resistor R1, compare it to the voltage at the - pin of the capacitor, and switch the enhancement mode FET on and off accordingly. I built the circuit on a breadboard and was elated when I measured the output and was able to connect a solar panel and battery. The fet needs about 5v at the gate. For some reason i only measure 2 or 3 v at the gate. Can you drive this FET direct from the output pin of the LM393 directly or will I have to amplify the output somehow to make sure the fet turns all the way off and on? Also, I am powering the LM393 directly from the input. I wonder if that could be a problem. Am I going about this all wrong? Any help would be much appreciated. It 'works' but wastes power by the fet presenting resistance. Please take a look.

Input is 16.5 v PV panel
zener is 12V
output is 12v lead acid battery
R1 is 1k, R2 has been 3k and 10k
The FET is a typical enhancement mode N channel, RL2203N or RFP50N06

-
 The link to the pic doesn't work for me. You'll want a hysteresis in your comparator, otherwise it may switch on and off all the time if the input voltage is near the zener's voltage. – stevenvh Apr 11 '12 at 13:36 Thanks steven, link should work now. Is there any harm in waiting until i get the fet turning on all the way first? Also whats the best way to add hysteresis? Im a little new with comparators. – Destin Dorman Apr 11 '12 at 13:44 Adding Extra Hysteresis to Comparators. You add part of the comparator's output to its non-inverting input, so that the voltage at the input is a bit higher or lower, depending on the comparator's output. – stevenvh Apr 11 '12 at 13:54 Looks basically OK. BUT saying "typical ehnancement ..." totally destroys the value of your circuit as it is an unknown "black box" that could be ideal or useless. Part number please. I think hand drawn diagrams are fine (I would :-) ) if neat and tidy and readable BUT use a ruler or square. LM393 – Russell McMahon Apr 11 '12 at 13:57 I tried an RL2203N as well as a RFP50N06, I have a ton of em. – Destin Dorman Apr 11 '12 at 14:01
show 1 more comment

This doesn't answer your question about the low gate voltage, but I want to explain the hysteresis I suggested in comment you might add.

Suppose the diode ladder creates a voltage difference of 10V, then the inverting comparator input is at 6.5V. Also let's start with the situation where the capacitor is fully charged to 16.5V, then the non-inverting input is at 0V. So output is low, FET doesn't conduct, and C1 is allowed to discharge through its load. When the voltage across C1 becomes 9.99V the non-inverting comparator input becomes 6.51V and the FET starts to conduct, so that C1 gets charged. This means $V_{OUT}$ will start to decrease, falling below 6.5V again and the FET will switch off again. Charging will stop, and C1 will again discharge until it reaches 6.51V again. Same things happen: FET switches on, $V_{OUT}$ will descrease and FET will switch off again. And so on.

That's the theory. I made my comment before I had seen your schematic, based on the description. Now since you charge C1 over a very low resistance its voltage may go pretty low before the FET switches off again, but things may also occur as described above.

What can you do about it? This is where the hysteresis comes in. Like I said you add part of the comparator's output voltage to the non-inverting input. Place a resistor between $V_{OUT}$ and the non-inverting input, and one between the output and the input. Now if the FET is off the output is low, and the resistors will form a divider, so that the non-inverting input sees only part of $V_{OUT}$. That's your comparison level. You'll have to adjust the reference voltage on the other input. Say that you're now comparing with +5V. OK, C1 discharges until it reaches 5.01V. Output becomes high and C1 starts to charge. Now this is important, via the resistor divider the high output will set the non-inverting input to a level much higher than the 5.01V. This means that the output will remain active for a longer time (C1 voltage has to charge all the way down to 5V). This will ensure that C1 gets fully charged via the FET's internal resistance, before the FET gets shut off again.

-
 Excellent explanation. – Federico Russo Apr 11 '12 at 15:12 All true, except that because there's no inductor in the circuit, the switch is going to charge the capacitor all the way to the supply (or at least partway to the supply depending on FET on-time) and cause a surge current from the supply, rather than regulating the output smoothly. – Jason S Apr 11 '12 at 15:25 Awesome answer! Thanks. As i said I will post any updates. – Destin Dorman Apr 11 '12 at 15:27 @Jason - Comparators in general are fast (way faster than opamps). How quickly C1 charges will depend on the FET's $R_{DS(ON)}$ and the gate's capacitance. – stevenvh Apr 11 '12 at 15:31 Sure, exactly. But my point is, that this behavior probably isn't what the OP wants. What circuit element limits the current through C1, if the FET gate is turned fully on? If the supply has a very low impedance, the FET is in its constant current mode (NOT Rdson) and dissipating a lot of power. If the supply has moderate impedance, the FET is acting as a low Rdson switch, and will cause the supply to sag while it's charging up the capacitor. Either of those alternatives are undesirable. – Jason S Apr 11 '12 at 15:58

The problem you're running into is that you're using negative feedback around the comparator: the MOSFET effectively inverts its gate signal (increase in gate voltage causes decrease in drain voltage), so feedback into the "+" pin actually creates negative feedback instead of positive feedback.

Negative feedback is great for regulation (probably your idea) but not what you want if you want switching behavior rather than linear behavior.

I suggest you keep negative feedback as an outer loop but use positive feedback as an inner loop.

One way to do this fairly easily is to split up the input into the comparator "+" input as follows:

• For DC/low-frequency negative feedback (outer loop), connect the "-OUT" node to "+" via a fairly high-value resistor Rfb (e.g. 100K or 1M) -- be careful here though because LM393s are bipolar so beware of bias current. You may choose to use a CMOS comparator such as a TLC393 instead, or put a similar resistor in series with the comparator "-" node to balance the input impedances.

• For high-frequency positive feedback (inner loop), connect the comparator output to the "+" node via a series RC circuit. The value of C is chosen so that Rfb*C gives you a crossover time T (= 1/2/pi/frequency); the value of R is just to make sure that the "+" node doesn't get whacked and damaged, so a 1K resistor should be fine.

The crossover time T roughly corresponds to a switching period in PWM -- this is sort of a hysteretic controller rather than a fixed-frequency switching controller, but it's the same idea. (It should be chosen so it's not too high.)

Because you'd be running in a switching mode rather than a linear mode, you'd also need inductance somewhere in this circuit (presumably between FET drain and "OUT" node), otherwise a FET switching fully on across the capacitor on the output will cause large surge currents. And you'd also need a freewheeling diode to conduct the current when the FET switches off.

If you do really want linear behavior, use an op-amp rather than a comparator.

Good luck!

-
Thank you very much. I will post any updates. Also how am i reinventing the charge controller exactly? Is there an easier way to go about this without buying a controller? – Destin Dorman Apr 11 '12 at 14:42
? I didn't mention reinventing the charge controller; that was Russ's comment. – Jason S Apr 11 '12 at 14:45
If by linear you mean the fet would be partially on at times, i want to avoid that. I want it full on or off to avoid a lot of heat dissipation. I figured the circuit would just switch however fast it needed to....im not exactly clear on why the fet gate voltage is low though. – Destin Dorman Apr 11 '12 at 14:51
Because you're apparently presently in linear mode at a point of stable equilibrium. The comparator isn't acting like a comparator, it's acting like an op-amp because there's negative feedback. (increase in comparator output causes increase in gate voltage causes decrease in drain voltage causes decrease at "+" pin which causes decrease in comparator output causes decrease in gate voltage etc.) When FETs are in their linear mode, the gate voltage is small; they're not fully turned on, just on enough to let the appropriate amount of current flow. – Jason S Apr 11 '12 at 15:06
Not a typo. Right now -OUT is connected directly to "+". I'm just saying to create two feedback paths, one analogous to what's happening now, with a resistor inserted, the other feedback path is new. "This is going to be basically a switch-mode converter". Yes, in my opinion, either this circuit (as is) should be fixed to be a linear regulator, or it should be fixed to be a switch-mode converter, or it's a circuit with errors. – Jason S Apr 11 '12 at 16:10

Looks basically OK.
You need to know what you are trying to do.
what this tries to do is limit battery voltage to that of zener + diode string ~= 14.5V. This is OK in some circumstances but too high for normal use.

The fact that FET gate voltage is low suggests that you are not doing as circuit shows. You need some sort of load to test this - battery is OK.

This sort of circuit must either be wasteful or turn on an off hard at limit. Battery will droop when power taken off so will turn on and charge and then power off and drop and.
To add hysteresis add a resistor say R3 = 1k from -out to opamp non inverting and then a resistor R4 from opamp out to opamp non inverting. Value relative to R3 controls hysteresis voltage swing. Maybe 10k- 47k range. Pot may be best.

You are re-re-re-reinventing the solar battery charger wheel. You need to have a good reason to do so. Have you?

Show all component values on diagram. Saying "typical ehnancement ..." totally destroys the value of your circuit as it is an unknown "black box" that could be ideal or useless.

I think hand drawn diagrams are fine (I would :-) ) if neat and tidy and readable BUT use a ruler or square.

providing links to datasheets is desirable. LM393 datasheet here
LM393 is open collector so R2 is gate drive source.
Vin can not go closer than 2V or so to Vdd but your powering looks OK.

Zener plus diodes is strange - unless it's all you have. Vzener+diodes ~= 14.5V+. Could be 15V+. Better to use a zener only or a cheap semi precision reference such as a TL431.

Resistor of maybe 1k from out- to ground may help cct behave when no load present.

~

-
 Thanks everyone. One thing: I didn't think i was 'reinventing' anything. Perhaps someone could enlighten me on another way to do this without using a dummy load. – Destin Dorman Apr 11 '12 at 14:28 I tried a resistor from out to gnd. Thing is the lm393 outputs low only, that is when the output is high the only source for positive voltage is the pull up resistor. So the gate gets grounded everytime the output is low. – Destin Dorman Apr 11 '12 at 14:32 I was using YOUR terminology. Ot- = load-ve terminal = MOSFET drain. If MOSFET is off and there is no load the comparator inverting input floats and may cause mal-operation by floarong into forbidden common mode range near Vcc.Comparator MAY work well when Vin is > Vcommon mode in idle state but maybe not. – Russell McMahon Apr 11 '12 at 14:55


text:
Law of excluded middle

:"This article uses forms of logical notation. For a concise description of the symbols used in this notation, see Table of logic symbols."

In logic, the law of the excluded middle states that the propositional calculus formula "P" ∨ ¬"P" ("P" or not-"P") can be deduced from the calculus under investigation. It is one of the defining properties of classical systems of logic. However, some systems of logic have different but analogous laws, while others reject the law of excluded middle entirely.

The law is also known as the law (or principle) of the excluded third, or, in Latin, principium tertii exclusi. Yet another Latin designation for this law is Tertium non datur: "there is no third (possibility)".

The law of excluded middle is related to the principle of bivalence, which is a semantic principle instead of a law that can be deduced from the calculus.

For some finite "n"-valued logics, there is an analogous law called the law of excluded "n"+1th. If negation is cyclic and '∨' is a "max operator", then the law can be expressed in the object language by (P ∨ ~P ∨ ~~P ∨ ... ∨ ~...~P), where '~...~' represents "n"-1 negation signs and '∨ ... ∨' "n"-1 disjunction signs. It is easy to check that the sentence must receive at least one of the "n" truth values (and not a value that is not one of the "n").

Examples

For example, if "P" is the proposition:

:"Socrates is mortal."

then the law of excluded middle holds that the logical disjunction:

:"Either Socrates is mortal or Socrates is not mortal."

is true by virtue of its form alone. That is, the "middle" position, that Socrates is neither mortal nor immortal, is excluded by logic, and therefore either the first possibility ("Socrates is mortal") or its negation ("Socrates is not mortal") must be true.

An example of an argument that depends on the law of excluded middle follows. [ This well-known example of a non-constructive proof depending on the law of excluded middle can be found in many places, for example: Megill, Norm. "Metamath: A Computer Language for Pure Mathematics", footnote on p. 17, [http://us.metamath.org/index.html#book] and Davis 2000:220, footnote 2.] We seek to prove that there exist two irrational numbers $a$ and $b$ such that

:$a^b$ is rational.

It is known that $sqrt\left\{2\right\}$ is irrational. Consider the number

:$sqrt\left\{2\right\}^\left\{sqrt\left\{2$.

Clearly (excluded middle) this number is either rational or irrational. If it is rational, we are done: $a=sqrt\left\{2\right\}$ and $b=sqrt\left\{2\right\}$. If it is irrational, then let

:$a=sqrt\left\{2\right\}^\left\{sqrt\left\{2$ and $b=sqrt\left\{2\right\}$.

Then

:$a^b = left\left(sqrt\left\{2\right\}^\left\{sqrt\left\{2 ight\right)^\left\{sqrt\left\{2 = sqrt\left\{2\right\}^\left\{left\left(sqrt\left\{2\right\}cdotsqrt\left\{2\right\} ight\right)\right\} = sqrt\left\{2\right\}^2 = 2,$

and 2 is certainly rational. This concludes the proof.

In the above argument, the assertion "this number is either rational or irrational" invokes the law of excluded middle. An intuitionist, for example, would not accept this argument without further support for that statement. This might come in the form of a proof that the number in question is in fact irrational (or rational, as the case may be); or a finitistic algorithm that could determine whether the number is rational or not.

The Law in non-constructive proofs over the infinite:The above proof is an example of a "non-constructive" proof disallowed by intuitionists::"The proof is nonconstructive because it doesn't give specific numbers a and b that satisfy the theorem but only two separate possibilities, one of which must work. (Actually $a=sqrt\left\{2\right\}^\left\{sqrt\left\{2$ is irrational but there is no known easy proof of that fact.)" (Davis 2000:220)

By "non-constructive" Davis means that "a proof that there actually are mathematic entities satisfying certain conditions would have to provide a method to exhibit explicitly the entities in question." (p. 85). Such proofs presume the existence of a totality that is complete, a notion disallowed by intuitionists when extended to the "infinite" -- for them the infinite can never be completed::"In classical mathematics there occur "non-constructive" or "indirect" existence proofs, which intuitionists do not accept. For example, to prove "there exists an n such that P(n)", the classical mathematician may deduce a contradiction from the assumption for all n, not P(n). Under both the classical and the intuitionistic logic, by reductio ad absurdum this gives "not for all n, not P(n)". The classical logic allows this result to be transformed into "there exists an n such that P(n)", but not in general the intuitionistic... the classical meaning, that somewhere in the completed infinite totality of the natural numbers there occurs an "n" such that "P(n)", is not available to him, since he does not conceive the natural numbers as a completed totality." [In a comparative analysis (pp. 43-59) of the three "-isms" (and their foremost spokesmen) -- Logicism (Russel and Whitehead), Intuitionism (Brouwer) and Formalism (Hilbert) -- Kleene turns his thorough eye toward intuitionism, its "founder" Brouwer, and the intuitionists' complaints with respect to the law of excluded middle as applied to arguments over the "completed infinite".] (Kleene 1952:49-50)

Indeed, Hilbert and Brouwer both give examples of the law of excluded middle extended to the infinite. Hilbert's example: "the assertion that either there are only finitely many prime numbers or there are infinitely many" (quoted in Davis 2000:97); and Brouwer's: "Every mathematical species is either finite or infinite." (Brouwer 1923 in van Heijenoort 1967:336).

In general, intuitionists allow the use of the law of excluded middle when it is confined to discourse over finite collections (sets), but not when it is used in discourse over infinite sets (e.g. the natural numbers). Thus intuitionists absolutely disallow the blanket assertion: "For all propositions P concerning infinite sets D: P or ~P" (Kleene 1952:48).

:"For more about the conflict between the intuitionists (e.g. Brouwer) and the formalists (Hilbert) see Foundations of mathematics and Intuitionism."

Putative counterexamples to the law of excluded middle include the liar paradox or Quine's Paradox. Certain resolutions of these paradoxes, particularly Graham Priest's dialetheism as formalised in LP, have the law of excluded middle as a theorem, but resolve out the Liar as both true and false. In this way, the law of excluded middle is true, but because truth itself, and therefore disjunction, is not exclusive, it says next to nothing if one of the disjuncts is paradoxical, or both true and false.

One more counterexample to LEM may be noted: Gödel's First Incompleteness Theorem [1931] provides a constructive example of a contingent proposition, which is neither deductively (syntactically) true nor false. An intuitionist would point out that the negation of the "G-sentence" has only infinite (non-standard) models, and is thus another demonstration of the unsoundness of LEM with respect to infinite domains.

History

Aristotle

Aristotle wrote that ambiguity can arise from the use of ambiguous names, but cannot exist in the "facts" themselves:

It is impossible, then, that 'being a man' should mean precisely 'not being a man', if 'man' not only signifies something about one subject but also has one significance. …And it will not be possible to be and not to be the same thing, except in virtue of an ambiguity, just as if one whom we call 'man', and others were to call 'not-man'; but the point in question is not this, whether the same thing can at the same time be and not be a man in name, but whether it can be in fact. ("Metaphysics" 4.4, W.D. Ross (trans.), GBWW 8, 525–526).

Aristotle's assertion that "...it will not be possible to be and not to be the same thing", which would be written in propositional logic as ¬ ("P" ∧ ¬"P"), is not the statement a modern logician would call the law of excluded middle ("P" ∨ ¬"P"). The former claims that no statement is "both" true and false; the latter requires that no statement is "neither" true nor false.

However, Aristotle also writes, "since it is impossible that contradictories should be at the same time true of the same thing, obviously contraries also cannot belong at the same time to the same thing" (Book IV, CH 6, p. 531). He then proposes that "there cannot be an intermediate between contradictories, but of one subject we must either affirm or deny any one predicate" (Book IV, CH 7, p. 531). In the context of Aristotle's traditional logic, this is a remarkably precise statement of the law of excluded middle, "P" ∨ ¬"P".

Leibniz

Its usual form, "Every judgment is either true or false" [footnote 9] ..."(from Kolmogorov in van Heijenoort, p. 421)footnote 9: "This is Leibniz's very simple formulation (see "Nouveaux Essais", IV,2)...." (ibid p 421)

Bertrand Russell and "Principia Mathematica"

Bertrand Russell asserts a distinction between the "law of excluded middle" and the "law of contradiction". In "The Problems of Philosophy", he cites three "Laws of Thought" as more or less "self evident" or "a priori" in the sense of Aristotle:

# Law of identity: 'Whatever is, is.'
# Law of noncontradiction: 'Nothing can both be and not be.'
# Law of excluded middle: 'Everything must either be or not be.' These three laws are samples of self-evident logical principles... (p. 72)

It is correct, at least for bivalent logic — i.e. it can be seen with a Karnaugh map — that Russell's Law (2) removes "the middle" of the inclusive-or used in his law (3). And this is the point of Reichenbach's demonstration that some believe the "exclusive"-or should take the place of the "inclusive"-or.

The tertium non datur
1. ("x") ["f"("x") ∨ ~"f"("x")]
is not exhaustive in its major terms and is therefore an inflated formula. This fact may perhaps explain why some people consider it unreasonable to write (29) with the inclusive-'or', and want to have it written with the sign of the "exclusive"-'or'
1. ("x") ["f"("x") ⊕ ~"f"("x")] , where the symbol " ⊕ " signifies exclusive-or [ the original symbol as used by Reichenbach is an upside down V, nowadays used for AND. The AND for Reichenbach is the same as that used in Principia Mathematica -- a "dot" cf p. 27 where he shows a truth table where he defines "a.b". Reichenbach defines the exclusive-or on p. 35 as 'the negation of the equivalence'. One sign used nowadays is a circle with a + in it, i.e. ⊕ (because in binary, a ⊕ b yields modulo-2 addition -- addition without carry). Other signs are ≢ (not identical to), or ≠ (not equal to).]
in which form it would be fully exhaustive and therefore nomological in the narrower sense. (Reichenbach, p. 376)

In line (30) the "(x)" means "for all" or "for every", thus an example of the expression would look like this:

* (For all "Q"): ("P" ⊕ ~"P")
* (For all instances of "pig" seen and unseen): ("Pig does fly" or "Pig does not fly" but not both simultaneously)

What Aristotle and Russell believed is characteristic of traditional logic, but this view implicitly depends on a particular notion of truth in which every statement is either true or false.

A formal definition from "Principia Mathematica"

"Principia Mathematica" ("PM") defines the law of excluded middle formally:

:*2.1 : ~p ∨ p ("PM" p. 101)Example: Either it is true that “this is red”, or it is true that “this is not red”. Hence it is true that “this is red or this is not red”. (See below for more about how this is derived from the primitive axioms).

So just what is "truth" and "falsehood"? At the opening "PM" quickly announces some definitions:

"Truth-values". The “truth-values” of a proposition is "truth" if it is true and "falsehood" if it is false* [*This phrase is due to Frege] ...the truth-value of “p ∨ q” is truth if the truth-value of either p or q is truth, and is falsehood otherwise ... that of “~ p” is the opposite of that of p...” (p. 7-8)

This is not much help. But later, in a much deeper discussion, ("Definition and systematic ambiguity of Truth and Falsehood" Chapter II part III, p. 41 ff ) "PM" defines truth and falsehood in terms of a relationship between the "a" and the "b" and the "percipient". For example "This 'a' is 'b'" (e.g. "This 'object a' is 'red'") really means "'object a' is a sense-datum" and "'red' is a sense-datum", and they "stand in relation" to one another and in relation to "I". Thus what we really mean is: "I perceive that 'This object a is red'" and this is an undeniable-by-3rd-party "truth".

"PM" further defines a distinction between a "sense-datum" and a "sensation":

That is, when we judge (say) “this is red”, what occurs is a relation of three terms, the mind, and “this”, and “red“. On the other hand, when we perceive “the redness of this”, there is a relation of two terms, namely the mind and the complex object “the redness of this” (p. 43-44).

Russell reiterated his distinction between "sense-datum" and "sensation" in his book "The Problems of Philosophy" (1912) published at the same time as "PM" (1910 – 1913):

Let us give the name of ‘sense-data’ to the things that are immediately known in sensation: such things as colours, sounds, smells, hardnesses, roughnesses, and so on. We shall give the name ‘sensation’ to the experience of being immediately aware of these things… The colour itself is a sense-datum, not a sensation. (p. 12)

Russell further described his reasoning behind his definitions of "truth" and "falsehood" in the same book (Chapter XII "Truth and Falsehood").

Consequences of the law of excluded middle in "Principia Mathematica"

From the law of the excluded middle, formula *2.1 in "Principia Mathematica," Whitehead and Russell derive some of the most powerful tools in the logician's argumentation toolkit.

*2.1 ~"p" ∨ "p" “This is the Law of excluded middle” (PM, p. 101).

The proof of *2.1 is roughly as follows: “primitive idea” *1.08 defines "p" → "q" = ~"p" ∨ "q". Substituting "p" for "q" in this rule yields "p" → "p" = ~"p" ∨ "p". Since "p" → "p" is true (this is Theorem 2.08, which is proved separately), then ~"p" ∨ "p" must be true.

*2.11 "p" ∨ ~"p" [Permutation of the assertions is allowed by axiom 1.4]
*2.12 "p" → ~(~"p") [Principle of double negation, part 1]
*:If “ This rose is red ” is true then it's not true that “ ‘This rose is not-red’ is true”.]
*2.13 "p" ∨ ~{~(~"p")} [Lemma together with *2.12 used to derive *2.14]
*2.14 ~(~"p") → "p" [Principle of double negation, part 2]
*2.15 (~"p" → "q") → (~"q" → "p") [One of the four “Principles of transposition”. Similar to 1.03, 1.16 and 1.17. A very long demonstration was required here.]
*2.16 ("p" → "q") → (~"q" → ~"p") [If it's true that “If this rose is red then this pig flies” then it's true that “If this pig doesn’t fly then this rose isn’t red.”]
*2.17 ( ~"p" → ~"q" ) → ("q" → "p") [Another of the 'Principles of transposition']
*2.18 (~"p" → "p") → "p" [Called “The complement of "reductio ad absurdum". It states that a proposition which follows from the hypothesis of its own falsehood is true” (p. 103-104).]

Most of these theorems--in particular *2.1, *2.11, and *2.14--are rejected by intuitionism. These tools are recast into another form that Kolmogorov cites as "Hilbert's four axioms of implication" and "Hilbert's two axioms of negation" (Kolmogorov in van Heijenoort, p. 335).

Propositions *2.12 and *2.14, "double negation":The intuitionist writings of L. E. J. Brouwer refer to what he calls "the "principle of the reciprocity of the multiple species", that is, the principle that for every system the correctness of a property follows from the impossibility of the impossibility of this property" (Brouwer, ibid, p. 335).

This principle is commonly called "the principle of double negation" (cf PM p. 101-102). From the law of excluded middle *2.1 and *2.11 PM derives principle *2.12 immediately. We substitute ~"p" for "p" in *2.11 to yield ~"p" ∨ ~(~"p"), and by the definition of implication (i.e. *1.01 p → q = ~p ∨ q) then ~p ∨ ~(~p)= p → ~(~p). QED (The derivation of 2.14 is a bit more involved.)

Footnotes

References

* Aquinas, Thomas, "Summa Theologica", Fathers of the English Dominican Province (trans.), Daniel J. Sullivan (ed.), vols. 19–20 in Robert Maynard Hutchins (ed.), "Great Books of the Western World", Encyclopedia Britannica, Inc., Chicago, IL, 1952. Cited as GB 19–20.
* Aristotle, "Metaphysics", W.D. Ross (trans.), vol. 8 in Robert Maynard Hutchins (ed.), "Great Books of the Western World", Encyclopedia Britannica, Inc., Chicago, IL, 1952. Cited as GB 8. 1st published, W.D. Ross (trans.), "The Works of Aristotle", Oxford University Press, Oxford, UK.
* Martin Davis 2000, "Engines of Logic: Mathematicians and the Origin of the Computer", W. W. Norton & Company, NY, ISBN 0-393-32229-7 pbk.
* Dawson, J., "Logical Dilemmas, The Life and Work of Kurt Gödel", A.K. Peters, Wellesley, MA, 1997.
* van Heijenoort, J., "From Frege to Gödel, A Source Book in Mathematical Logic, 1879-1931", Harvard University Press, Cambridge, MA, 1967. Reprinted with corrections, 1977.
* Luitzen Egbertus Jan Brouwer, 1923, "On the significance of the principle of excluded middle in mathematics, especially in function theory" [reprinted with commentary, p. 334, van Heijenoort]
* Andrei Nikolaevich Kolmogorov, 1925, "On the principle of excluded middle", [reprinted with commentary, p. 414, van Heijenoort]
* Luitzen Egbertus Jan Brouwer, 1927, "On the domains of definitions of functions", [reprinted with commentary, p. 446, van Heijenoort] Although not directly germane, in his (1923) Brouwer uses certain words defined in this paper.
* Luitzen Egbertus Jan Brouwer, 1927(2), "Intuitionistic reflections on formalism", [reprinted with commentary, p. 490, van Heijenoort]
* Stephen C. Kleene 1952 original printing, 1971 6th printing with corrections, 10th printing 1991, "Introduction to Metamathematics", North-Holland Publishing Company, Amsterdam NY, ISBN 0 7204 2103 9.
* Kneale, W. and Kneale, M., "The Development of Logic", Oxford University Press, Oxford, UK, 1962. Reprinted with corrections, 1975.
* Alfred North Whitehead and Bertrand Russell, "Principia Mathematica to *56", Cambridge at the University Press 1962 (Second Edition of 1927, reprinted). Extremely difficult because of arcane symbolism, but a must-have for serious logicians.
* Bertrand Russell, "The Problems of Philosophy, With a New Introduction by John Perry", Oxford University Press, New York, 1997 edition (first published 1912). Very easy to read: Russell was a wonderful writer.
* Bertrand Russell, "The Art of Philosophizing and Other Essays", Littlefield, Adams & Co., Totowa, NJ, 1974 edition (first published 1968). Includes a wonderful essay on "The Art of drawing Inferences".
* Hans Reichenbach, "Elements of Symbolic Logic", Dover, New York, 1947, 1975.
* Tom Mitchell, "Machine Learning", WCB McGraw-Hill, 1997.
* Constance Reid, "Hilbert", Copernicus: Springer-Verlag New York, Inc. 1996, first published 1969. Contains a wealth of biographical information, much derived from interviews.
* Bart Kosko, "Fuzzy Thinking: The New Science of Fuzzy Logic", Hyperion, New York, 1993. Fuzzy thinking at its finest. But a good introduction to the concepts.
* David Hume, "An Inquiry Concerning Human Understanding", reprinted in Great Books of the Western World Encyclopedia Britannica, Volume 35, 1952, p. 449 ff. This work was published by Hume in 1758 as his rewrite of his "juvenile" "Treatise of Human Nature: Being An attempt to introduce the experimental method of Reasoning into Moral Subjects Vol. I, Of The Understanding" first published 1739, reprinted as: David Hume, "A Treatise of Human Nature", Penguin Classics, 1985. Also see: David Applebaum, "The Vision of Hume", Vega, London, 2001: a reprint of a portion of "An Inquiry" starts on p. 94 ff

ee also

* Laws of thought
* Logical graphs — Graphical syntax for propositional logic
* Peirce's law — Another way of turning intuition classical
* Law of bivalence

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### Look at other dictionaries:

• law of excluded middle — noun An axiom of classical logic which states that, given a proposition P, either P is true or P is false, i.e. in symbols: . This law is rejected by intuitionistic logic …   Wiktionary

• law of excluded middle — a principle in logic: if one of two contradictory statements is denied the other must be affirmed * * * Logic. the principle that any proposition must be either true or false …   Useful english dictionary

• law of excluded middle — Logic. the principle that any proposition must be either true or false. * * * …   Universalium

• principle of excluded middle — law of excluded middle …   Useful english dictionary

• Law of noncontradiction — This article uses forms of logical notation. For a concise description of the symbols used in this notation, see List of logic symbols. In classical logic, the law of non contradiction (LNC) (or the principle of non contradiction (PNC), or the… …   Wikipedia

• Law of thought — The laws of thought are fundamental logical rules, with a long tradition in the history of philosophy, which collectively prescribe how a rational mind must think. To break any of the laws of thought (for example, to contradict oneself) is to be… …   Wikipedia

• law of thought — any of several principles in logic: a. : law of contradiction b. : law of excluded middle c. : law of identity * * * any of the three basic laws of traditional logic: the law of contradiction …   Useful english dictionary

• law of thought — any of the three basic laws of traditional logic: the law of contradiction, the law of excluded middle, and the law of identity. * * * …   Universalium

• law (or principle) of the excluded middle Logic — the principle that one (and one only) of two contradictory propositions must be true. → exclude …   English new terms dictionary

• law of the excluded middle — law (or principle) of the excluded middle Logic the principle that one (and one only) of two contradictory propositions must be true …   Useful english dictionary


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Author: Subject: Open-source Raman spectrometer looking for academic partners
Luc
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Open-source Raman spectrometer looking for academic partners

Hi all,

Before I start, I know that there has been an extensive post about DIY Raman spectroscopy in the past but I decided to create a new post because what I would like to discuss here is not related to the instrument itself but on a different matter that I will detail below just after a short intro.

Long things short, I spent the last few years working almost exclusively on developing a DIY Raman spectrometer and got very nice results that I decided to share as an open-source & modular platform. I opened a website and very recently a first fully functional version.

Here is a picture of the setup:

All the assembly instructions & CAD files to create copies are detailed on the website : http://www.open-raman.org/build/

I have foreseen two versions at the moment that I call "Starter Edition" and "Performance Edition". Only the Starter Edition is available at the moment. It is essentially designed for teaching applications as it features a low power laser and a reduced resolution (<35 cm-1 tested). The Performance Edition will come shortly and has a 12 cm-1 tested resolution but it is still too expensive and I'm trying to take the cost down. This last version does work but it's about ~4500€ at the moment and I would like to make it go below 3000€. The Starter/EDU version is about 2500€. Please note that I am not selling these instruments and am NOT making any money on this; people who would like to create their own version have to download the BOM and place orders themselves at the various suppliers.

Here are typical spectra obtained with iso-propanol:

There are more spectra here for those who are interested and you can even download them in .csv: http://www.open-raman.org/performances/

I have already performed the following reactions that you can read on the website:
- Quantitative analysis of MeOH in EtOH with a detection threshold of 0.6%
- Tracking of Ethyl Acetate kinetics over 35 hours where I evaluated every minute the concentration of ethyl acetate, ethanol, and acetic acid to find the kinetic equation of the reaction
- Proof of the addition of the bisulfite ion to formaldehyde; the aim was to show that Raman spectroscopy can unambiguously show that a reaction took place when your eyes see nothing

There are many more reactions I'm still working on but you know what it is...

So I'm writing this post because the thing works great and I'm confident that it can find applications in Universities so I would like to find academics who would be excited in a partnership. Non-commercial applications only, I'm looking for people who like science and would like to share something with others.

I'm looking for both partners for teaching applications and for research applications although I'm currently focusing on the teaching applications first since the Performance Edition is not available yet.

The idea is that the academics would develop a concept such as a lab session for their students and work that with them. They would own 100% of the application and I will not interfere with what they are doing. I will provide support in assembling and operating the instrument to my best as a trained professional optical engineer with a strong background in chemistry. In exchange, I would like to advertise on my website that "XYZ University" is using the spectrometer for "ABC" type of activity and provide a link to the webpage of the teacher/lab facility. If a paper is issued, I would like to be credited (either as an author or just cited in the references depending on the amount of work done -- it is obvious that I'm not asking to be a co-author if I just exchanged a few e-mails but that I would be if I start procuring custom CAD files to extend the instrument for that particular application; but this is common sense I guess).

Since the idea is to create an "open" framework, I would strongly encourage people to share notes if they create a lab that uses the instrument (like protocols, what teachers should focus on etc.) but it is in no way compulsory. Again, academics remain 100% owners of what they produce and I'm not interefering with that.

The spectrometer can be used with optics or chemistry students as well. It can be a low-cost alternative to expensive lab hardware (nobody like students to "accidentally" drop conc HCl on a 20,000 spectrometer), a possibility for the students to align the spectrometer themselves or even let them develop custom parts for the spectrometer. This last possibility (upgrade) is the "modular" part of the design. By default the spectrometer comes naked and you have to add a "cuvette" to it to analyze a certain type of sample. By default I propose what I call the "standard cuvette" which allows to test liquids in test tubes. But one can imagine custom cuvettes to study solids, liquids flow of reaction, and so on! You just plug off the type of cuvette and replace it with the new one So I strongly encourage academics to design their own custom cuvettes and I will provide all the info to help them do that. So if there are people here who are interested in the idea, just reply to the post or contact me so that we see together how we can partner. Again, I would like to stress that this is not like a "prototype version" or anything else, it's working for real and it's complete. I'm ready to offer support to my best to anybody who would like to use the spectrometer for his own applications Also, even though it is not mentioned so far, if you are just an enthusiast and would like to do Raman at home for youtube videos or anything else I will gladly help you too Metacelsus International Hazard Posts: 2502 Registered: 26-12-2012 Location: Boston, MA Member Is Offline Mood: Double, double, toil and trouble Very cool! I noticed the most expensive item on the Bill of Materials was the camera. In high school I built a Raman spectroscope using this guide: https://hackaday.io/project/1279-ramanpi-raman-spectrometer which used a linear CCD array. The CCD array (TCD1304DG) was much cheaper and performed reasonably well. Do you think it would be possible to substitute this for the camera? As below, so above. Luc Harmless Posts: 5 Registered: 23-12-2017 Member Is Offline Mood: No Mood  Quote: Originally posted by Metacelsus Very cool! I noticed the most expensive item on the Bill of Materials was the camera. In high school I built a Raman spectroscope using this guide: https://hackaday.io/project/1279-ramanpi-raman-spectrometer which used a linear CCD array. The CCD array (TCD1304DG) was much cheaper and performed reasonably well. Do you think it would be possible to substitute this for the camera? No, unfortunately it is not possible. These CCD arrays are too noisy and require you to build custom ultra-low-noise read-out electronics which are extremely complex to design. It is true that professional Raman spectrometers use linear CCD array but these (CCDs) are in the 1-2k€ price range without the readout electronic (typically TEC cooled expensive Hamamatsu sensors). The spectrometer that uses cheap sensor like the ILX511 in the USB2000 spectrometer from OceanOptics do not work in Raman applications due to their high noise level. When I designed the spectrometer, I knew that the noise of the sensor would be a critical feature (as well as light collection efficiency) so I made an educated choice for the sensor. I chose to use CMOS machine vision camera because these sensors are produced at high volumes and therefore achieves very good "bang 4 bucks" ratio *plus* you get the electronics enclosed and the SW to operate them. To give you an idea, I tested the FLIR camera I'm using on a test bench at the office and I experimentally confirmed that the sensor was able to detect as low as 4 photons! This is a truly impressive performance for less than 400€ (all-included). Experimentally, I'm reaching SNR on the order of 1:1000 for only 3-5 seconds exposures for common lab solvents so the spectra are really clean even with short exposures snapshots. Even at 10 Hz / 100 ms exposure you get good real-time Raman spectra. I also wrote an article about how to choose the best camera sensor for spectroscopy based on decision-aid methods. You can check it here: http://www.thepulsar.be/article/choosing-the-best-camera-sen... earpain Hazard to Others Posts: 102 Registered: 11-9-2019 Member Is Offline Quote: Originally posted by Luc  Quote: Originally posted by Metacelsus Very cool! I noticed the most expensive item on the Bill of Materials was the camera. In high school I built a Raman spectroscope using this guide: https://hackaday.io/project/1279-ramanpi-raman-spectrometer which used a linear CCD array. The CCD array (TCD1304DG) was much cheaper and performed reasonably well. Do you think it would be possible to substitute this for the camera? No, unfortunately it is not possible. These CCD arrays are too noisy and require you to build custom ultra-low-noise read-out electronics which are extremely complex to design. It is true that professional Raman spectrometers use linear CCD array but these (CCDs) are in the 1-2k€ price range without the readout electronic (typically TEC cooled expensive Hamamatsu sensors). The spectrometer that uses cheap sensor like the ILX511 in the USB2000 spectrometer from OceanOptics do not work in Raman applications due to their high noise level. When I designed the spectrometer, I knew that the noise of the sensor would be a critical feature (as well as light collection efficiency) so I made an educated choice for the sensor. I chose to use CMOS machine vision camera because these sensors are produced at high volumes and therefore achieves very good "bang 4 bucks" ratio *plus* you get the electronics enclosed and the SW to operate them. To give you an idea, I tested the FLIR camera I'm using on a test bench at the office and I experimentally confirmed that the sensor was able to detect as low as 4 photons! This is a truly impressive performance for less than 400€ (all-included). Experimentally, I'm reaching SNR on the order of 1:1000 for only 3-5 seconds exposures for common lab solvents so the spectra are really clean even with short exposures snapshots. Even at 10 Hz / 100 ms exposure you get good real-time Raman spectra. I also wrote an article about how to choose the best camera sensor for spectroscopy based on decision-aid methods. You can check it here: http://www.thepulsar.be/article/choosing-the-best-camera-sen... I salute your work! My background is in Information Technology and I am open-source to the bone. I see that you are offering CAD files and source code, you are the real thing. I am no where near your level so I doubt any collaboration would be of interest. But I do plan to build a spectrophotometer via a Raspberry PI NOIR camera. I'm pretty sure this would not even be in the same universe of resolution and precision as your device, but people have had impressive results. Especially considering the cost of the parts is ~50

Initially I was inspired by this professor teaching first year Chemistry, and authoring a lesson which many other teachers are doing: He uses an iphone to determine the exact concentration of sugar in kool-aid:
http://faculty.ccbcmd.edu/~cyau/124%2003%20Concentration%20o...

Initial talks of PI Noir: https://www.raspberrypi.org/forums/viewtopic.php?t=40614

@Luc you probably know all this already, and I didn't mean to derail your very specific request, but just wanted to share my excitement about your project, and similar projects i've seen. And I believe I can get the pi noir device to work and hopefully will be posting about it soon.
Thank you!
andy1988
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Here is possible application if it can be packaged in a hand held form factor, with this research group:
https://phys.org/news/2019-11-quickly-accurately-scan-nutrie...
https://kurouskilab.com/

Their work brings to mind Gregor Mendel, the "Father of Genetics".

Luc
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Mood: No Mood

Just wanted to make a quick update on the project;

I just released today the "full" performance edition. Link is still here: http://www.open-raman.org/build/performance-edition/

Briefly said:
- 500-3,500 cm-1
- 12 cm-1 resolution
- high snr on common solvent (acetone at 100 Hz is not a problem!)
- less than 3,300€ for a complete spectrometer (with "standard cuvette" sample holder for liquids)

I do not sell instruments so it's 100% DIY, 100% open stuff access to all CAD & STEP files!

I'm also looking for academics to develop application around the spectrometer so don't hesitate to contact me or share the info with colleagues/teachers, that would be really helpful

I'll also be a bit more on the forum during the next months because now that the hardware & software is fully mature I want to do a ton of chemistry experiments to test what is actually going on. I know too well how frustrating it can be to make a synth and then wonder if the actual product has been produced or not, as well as the (almost) infinite discussions on the forum about it for some mystery reactions
ChemTalk
Hazard to Self

Posts: 65
Registered: 13-12-2018
Location: United States
Member Is Offline

Mood: colloidal

Thanks Luc, we are in the market for a spectrometer at some point this year, and are very interested in the performance edition, I'm going to discuss it with the team during our meeting tomorrow. Stay tuned!

-------------------------------------
ChemTalk
chemistrytalk.org


text:
# What is the range of the function y=sqrt(1-cosxsqrt(1-cosx(sqrt(1-cosx ......oo ?

Apr 9, 2018

I need double-check.

#### Explanation:

Apr 9, 2018

$\left[\frac{- 1 + \sqrt{5}}{2} , \frac{1 + \sqrt{5}}{2}\right]$

#### Explanation:

Given:

$y = \sqrt{1 - \cos x \sqrt{1 - \cos x \sqrt{1 - \cos x \sqrt{\ldots}}}}$

write $t$ for $\cos x$ to get:

$y = \sqrt{1 - t \sqrt{1 - t \sqrt{1 - t \sqrt{\ldots}}}}$

Square both sides to get:

${y}^{2} = 1 - t \sqrt{1 - t \sqrt{1 - t \sqrt{\ldots}}} = 1 - t y$

Add $t y - 1$ to both sides to get:

${y}^{2} + t y - 1 = 0$

This quadratic in $y$ has roots given by the quadratic formula:

$y = \frac{- t \pm \sqrt{{t}^{2} + 4}}{2}$

Note that we need to choose the $+$ sign of $\pm$, since the principal square root defining $y$ is non-negative.

So:

$y = \frac{- t + \sqrt{{t}^{2} + 4}}{2}$

Then:

$\frac{\mathrm{dy}}{\mathrm{dt}} = - \frac{1}{2} + \frac{t}{2 \sqrt{{t}^{2} + 4}}$

This is $0$ when:

$\frac{t}{\sqrt{{t}^{2} + 4}} = 1$

That is:

$t = \sqrt{{t}^{2} + 4}$

Squaring both sides:

${t}^{2} = {t}^{2} + 4$

So the derivative is never $0$, always negative.

So the maximum and minimum values of $y$ are attained when $t = \pm 1$, being the range of $t = \cos x$.

When $t = - 1$:

$y = \frac{1 + \sqrt{5}}{2}$

When $t = 1$

$y = \frac{- 1 + \sqrt{5}}{2}$

So the range of $y$ is:

$\left[\frac{- 1 + \sqrt{5}}{2} , \frac{1 + \sqrt{5}}{2}\right]$

graph{(y-(-(cos x)+sqrt((cos x)^2+4))/2) = 0 [-15, 15, -0.63, 1.87]}

Apr 9, 2018

See below.

#### Explanation:

We have

${y}_{\min} = \sqrt{1 - {y}_{\min}}$
${y}_{\max} = \sqrt{1 + {y}_{\max}}$

Here

${y}_{\min}$ is associated to the value $\cos x = 1$ and
${y}_{\max}$ is associated to $\cos x = - 1$

Now

${y}_{\min} = \frac{1}{2} \left(- 1 \pm \sqrt{5}\right)$ and
${y}_{\max} = \frac{1}{2} \left(1 \pm \sqrt{5}\right)$

then the feasible limits are

$\frac{1}{2} \left(- 1 + \sqrt{5}\right) \le y \le \frac{1}{2} \left(1 + \sqrt{5}\right)$

NOTE

With $y = \sqrt{1 + \alpha y}$

we have that $y$ is an increasing function of $\alpha$


text:
Here's the question you clicked on:

## ehamer2990 if tanΦ=-1/3 find sinΦ and cosΦ 8 months ago 8 months ago

• This Question is Closed
1. itsmylife

|dw:1375486500757:dw| now you can find anything u like ;) always use this method

2. jdoe0001

$$\bf tan(\theta) = -\cfrac{1}{3} \implies \cfrac{opposite}{adjacent} \implies \cfrac{b}{a}\\ c^2 = a^2+b^2 \implies c = \sqrt{a^2+b^2}$$

3. jdoe0001

once you have all 3 guys, a, b and c you can find the others


text:
#### Question

A1-m3 insulated, rigid tank contains air at 800 kPa, 25$^{\circ} \mathrm{C}$. A valve on the tank is opened, and the pressure inside quickly drops to 150 kPa, at which point the valve is closed. Assuming that the air remaining inside has undergone reversible adiabatic expansion, calculate the mass withdrawn during the process.

Verified

#### Step 1

1 of 2

We are given following data for Air in rigid tank in reversible adiabatic expansion:

$T_1=298\text{ K}$

$P_1=800\text{ kPa}$

$P_2=150\text{ kPa}$

$V=1\text{ m}^3$

From Properties table A.5. we can find Air gas constant and heat capacity ratio:

$R=0.287\frac{\text{ kJ}}{\text{ kg K}}$

$k=1.4$

Calculating final temperature:

\begin{align*} T_2&= T_1\cdot \left(\dfrac{P_2}{P_1} \right)^{\dfrac{k-1}{k}}=298\cdot \left(\dfrac{150}{800} \right)^{\dfrac{1.4-1}{1.4}}\\\\ &=185\text{ K} \end{align*}

Mass withdrawn during the process is equal to:

\begin{align*} m&=m_1-m_2=\dfrac{P_1\cdot V}{R\cdot T_1}-\dfrac{P_2\cdot V}{R\cdot T_2}\\\\ &=\dfrac{800\cdot 1}{0.287\cdot 298}-\dfrac{150\cdot 1}{0.287\cdot 185}\\\\ &=\boxed{6.53\text{ kg}} \end{align*}

## Create an account to view solutions

2,116 solutions

#### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

4th EditionRandall D. Knight
3,509 solutions


text:
# Integrating a simple equation

I am trying to integrate a simple function:

$\frac{dA}{dt} = j - kA(t)$

where j and k are constants, and $A(0)=A_0$ is also a constant. I got to

$A(t) = jt + C - k\int A(t)dt$

but I dont know how to proceed now. Could you help me out please? I know the answer is

$A(t) = e^{-k*t} (A_0+ \frac{j}{k}e^{k*t} -1)$

How can I get there? thanks so much

-

If $k=0$, this is is easy. Suppose $k\neq 0$.

Let's first rewrite it as $$\frac{dA}{dt}+kA=j.$$ Multiplying both sides by the positive function $e^{kt}$ gives us $$e^{kt}\frac{dA}{dt}+ke^{kt}A=je^{kt}.$$ Using the product rule on the left side, we have $$\frac{d}{dt}\left[e^{kt}A\right]=je^{kt}.$$ Integrating with respect to $t$, we then have $$e^{kt}A=j\int e^{kt}\,dt=\frac{j}{k}e^{kt}+C,$$ and so $$A(t)=\frac{j}{k}+Ce^{-kt}.$$ Using our initial value to find $C$, we get $$A(t)=\frac{j}{k}+\left(A_0-\frac{j}{k}\right)e^{-kt}.$$

-
Thanks Cameron! –  Sosi Nov 13 '12 at 11:32
by the way, how can I write down all the expressions in the correct formatting in here? do you use latex or something? –  Sosi Nov 13 '12 at 11:42
LaTeX is the preferred method to use. If you're ever curious about a particular formatted bit, right-click on it, then choose Show Math As-->TeX Commands. –  Cameron Buie Nov 13 '12 at 11:48

Do you know that the solution to $\frac {dA}{dt}=-kA$ is $A(0)\exp^{-kt}$? You can see that (not a proof) by differentiating and plugging in. If so, define $B(t)=A(t)-\frac jk$. Then the answer you give just pulls out the factor $e^{-kt}$

-
thanks Ross! I'll check it out! –  Sosi Nov 12 '12 at 16:46


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# R Recursion - Part 1

## Intro

I’ve been thinking about recursion lately. It’s very powerful but it can be hard to reason about sometimes. Through this article and the following ones, hopefully I can make it easier for you to understand it and to write more recursive functions.

## Find the first 3 prime numbers

Let’s start with a very simple task: find the first 3 prime numbers. How’d you do it? The intuitive way is to start with 1, and ask if 1 is a prime. The answer is no. So we move to 2 and ask if 2 is a prime, and the answer is yes. We now have our first prime number. We then move onto 3 and ask if 3 is a prime and we repeat the procedure until we find the 3rd prime number. The answer, of course, is the sequence (2, 3, 5). This way of thinking is recursion. Keep reading to find out why.

## Find the first N primes that are greater than M

The task is to find the first 3 prime numbers. What’s unsaid but self-evident is that the numbers must be greater than 0 because all prime numbers are greater than 0. So we can rewrite the task: find the first 3 primes that are greater than 0. Further more, let’s be ambitious and consider its general version: find the first N primes that are greater than M.

The basic structure of any recursion involves the base case and the non-base case. In the general task, the base case is when N = 0 and the result will be the empty sequence (). The non-base case is When N > 0, and when N > 0, if M+1 is a prime, the result will be (M+1, the first N-1 primes that are greater than M+1), otherwise, (the first N primes that are greater than M+1)

You probably realized it by now we can view the task as a math function f(N, M) = (the first N primes > M) with input parameters N and M and outputs a sequence of first N primes > M. We can then translate the above recursive description into R code.

The function is_prime() is taken from this stackoverflow answer, you can read the logics behind there. Alternatively, you can implement it using recursion. Consider it as today’s homework. :)

Let’s use f to find the first 3 prime numbers and first 5 primes greater than 100.


text:
## Algebra 1: Common Core (15th Edition)

When x is set equal to a constant, this means that the line has no run to it. That means that the line is infinitely steep, for y (the rise) changes, but x (the run) does not change. This means that the line $x=-1.8$ is vertical.


text:
Research ContributionArnold Mathematical Journal

Received: 7 December 2015 / Revised: 28 July 2016 / Accepted: 2 August 2016

q-Polynomial Invariant of Rooted Trees

Józef H. PrzytyckiDepartment of Mathematics The George Washington University Washington, DC 20052 USA, University of Maryland College Park, College Park USA,
University of Gdańsk, Gdańsk Poland
przytyck@gwu.edu

Abstract

We describe in this note a new invariant of rooted trees. We argue that the invariant is interesting on it own, and that it has connections to knot theory and homological algebra. However, the real reason that we propose this invariant to readers of Arnold Journal of Mathematics is that we deal here with an elementary, interesting, new mathematics, and after reading this essay readers can take part in developing the topic, inventing new results and connections to other disciplines of mathematics, and likely, statistical mechanics, and combinatorial biology.

Keywords
Tree, Invariant, Quantum plane, UnimodalityKauffman bracket, Gaussian polynomial
$\newcommand\Z{{\mathbb Z}}$

1 Introduction

1.1 Quantum Plane

We know well the Newton binomial formula:

 $\displaystyle \begin{eqnarray*} (x+y)^n= \sum_{i=0}^n {n \choose i} x^iy^{n-i}. \end{eqnarray*}$

Of course we assume here that the variables commute, that is $yx=xy$.

We can also visualize formulas for coefficients of $(x+y)^n$ by considering the tree $T_{b,a}$ of two long branches of length $b$ and $a$ respectively, coming from the root (Fig. 1). We can ask now how many different ways there are to “pluck” the tree $T_{b,a}$ one leaf at a time. Of course we can tear (like in the child’s play “likes not like”) the left or right branch. In total: the left should be torn $b$ times and the right $a$ times so the answer about the number of different pluckings is clearly ${a+b \choose a}$. Our notes present a dramatic generalization of this example.

We can ask now what happens with the binomial formula if we weaken commutativity by replacing it with $yx=qxy$ ($q$ commutes with $x$ and $y$). In applications in physics, $q$ is often taken to be a complex number but it is better to work generally with the ring $Z[q]$ and q-commutative11By $q$-commutative we understand exactly $yx=qxy$. polynomials of variables $x$ and $y$ over this ring (we call this ring of polynomials the quantum plane or noncommutative plane). The formula for $(x+y)^n$ in the quantum plane was known already in the XIX century, but if we do not know the result it is good to first work out small examples: $$(x+y)^2= y^2+ xy +yx + x^2= y^2 + (1+q)xy +x^2,$$ \begin{multline*} (x+y)^3= y^3+ xy^2+ yxy +y^2x + x^2y+ xyx + yx^2 +x^3\\ = y^3 + (1+q+q^2)xy^2 + (1+q+q^2)x^2y+ y^3, \end{multline*} \begin{multline*} (x+y)^4= y^4 + xy^3+ yxy^2 +y^2xy + y^3x + x^2y^2 + xyxy\\ + xy^2x + yx^2y + yxyx + y^2x^2 + x^3y+ x^2yx + xyx^2 + yx^3 + x^4 \\= y^4 + (1+q+q^2+q^3)xy^3 + (1+ q + 2q^2+ q^3 + q^4)x^2y^2 + (1+q+q^2+q^3)x^3y +x^4 \\= y^4+(1+q+q^2+q^3)xy^3 + (1+q^2)(1+q+q^2)x^2y^2+ (1+q+q^2+q^3)x^3y +x^4 .\end{multline*} Observe that we can think of $1+q+\cdots+q^{n-1}$ as a $q$-analogue of the number $n$. We define formally $[n]_q=1+q+\cdots+q^{n-1}$. In particular, for $q=1$, $[n]_q=n$. In the same vein we define $q$ factorial of the $q$-number $[n]_q$ as: $[n]_q!=[n]_q[n-1]_q\cdots [2]_q[1]_q$, and $q$-analogue of the binomial coefficient22Called also the Gaussian polynomial, $q$-polynomial of Gauss, Gaussian binomial coefficient, Gaussian coefficient, or $q$-binomial coefficient. by $\binom{n}{i}_q = \frac{[n]_q!}{[i]_q! [n-i]_q!}$. Sometimes to stress symmetry, $i \leftrightarrow n-i$, of the Gaussian polynomial we write $\binom{n}{i,n-i}_q$. We also observe that the coefficient of $x^2y^2$ in $(x+y)^4$ is $(1+q^2)(1+q+q^2)= \frac{1+q+q^2+q^3}{1+q}(1+q+q^2)=\frac{[4]_q[3]_q}{[2]_q}= \binom{4}{2}_q$.

In the notation we introduced, our calculations can be concisely written as:

 $\displaystyle \begin{eqnarray*} &\displaystyle (x+y)^2= y^2 +[2]_qxy +x^2,\\ &\displaystyle (x+y)^3= y^3 + [3]_qxy^2 + [3]_qx^2y+x^3,\\ &\displaystyle (x+y)^4= y^4+ [4]_qxy^3 + \binom{4}{2}_qx^2y^2 + [4]_qx^3y+ x^4. \end{eqnarray*}$

Now it is not difficult to guess the general formula for $(x+y)^n$ in the quantum plane:

Proposition 1.1.

([MacMahon1916]; see e.g. [Kac and Cheung2002]; [Mansour and Schork2011]; [Mansour and Schork2016])  If $yx=qxy$ then

 $\displaystyle \begin{eqnarray*} (x+y)^n= \sum_{i=0}^n {n \choose i}_qx^iy^{n-i}. \end{eqnarray*}$
Proof.

The simplest proof is by induction on $n$ however one can also find proofs without words, interpreting combinatorially left and right sides of the equation.

Inductive proof  We already checked the formula for $n\leq 4$, and we should add that we need the convention (as in the classical case). that $[0]_q!=1$ and consequently ${n \choose 0}_q=1 ={n \choose n}_q$. Now we perform the inductive step (from $n-1$ to $n$):

 $\displaystyle \begin{eqnarray*} (x+y)^n=& (x+y)(x+y)^{n-1}=(x+y)\sum_{i=0}^{n-1} {n-1 \choose i,n-i-1}_qx^iy^{n-i-1}\\ =&\sum_{i=0}^{n-1} {n-1 \choose i,n-i-1}_qx^{i+1}y^{n-i-1} +\sum_{i=0}^{n-1}q^i{n-1 \choose i,n-i-1}_qx^iy^{n-i}\\ =&\sum_{i=0}^n\left({n-1 \choose i-1,n-i}_q +q^i{n-1 \choose i,n-i-1}_q\right)x^iy^{n-i}. \end{eqnarray*}$

We use here the convention that ${n-1 \choose -1,n}_q=0$.

We are left now to check that

 $\displaystyle \begin{eqnarray*} {n-1 \choose i-1,n-i}_q +q^i{n-1\choose i,n-i-1}_q)={n \choose i}_q. \end{eqnarray*}$

We encourage the reader to check it by themselves before looking at the following calculation.

1. (i)

\begin{multline*}[a+b]_q= 1+q+\cdots+q^{a+b-1}= (1+q+\cdots+q^{a-1}) + q^a(1+q+\cdots+q^{b-1})\\ = [a]_q+q^a[b]_q= [b]_q+q^b[a]_q\end{multline*}

2. (ii)
 $\displaystyle \begin{eqnarray*} \binom{a+b}{a,b}_q=&\frac{[a+b]_q!}{[a]_q! [b]_q!}=[a+b]_q\frac{[a+b-1]_q!}{[a]_q! [b]_q!}\\ =&([a]_q+q^a[b]_q)\frac{[a+b-1]_q!}{[a]_q! [b]_q!}= \frac{[a+b-1]_q!}{[a-1]_q! [b]_q!} + q^a \frac{[a+b-1]_q!}{[a]_q! [b-1]_q!}\\ =&\binom{a+b-1}{a-1,b}_q + q^a\binom{a+b-1}{a,b-1}_q= \binom{a+b-1}{a,b-1}_q\\& +\, q^b\binom{a+b-1}{a-1,b}_q. \end{eqnarray*}$

$\square$

We can now go back to the tree $T_{b,a}$ and play the game in the $q$-fashion that is the leaf taken from the right ($a$) branch is counted as $1$ but the leaf taken from the left is counted with the weight $q^a$. Thus we eventually get not the plucking number but the plucking polynomial $Q(T_{b,a})$ which by definition satisfies the recursive relation $Q(T_{b,a})= Q(T_{b,a-1}) + q^aQ(T_{b-1,a})$. We also notice that $Q(T_{0,a})=Q(T_{b,0})=1$, thus we immediately recognize that the plucking polynomial is $Q(T_{b,a})= \binom{a+b}{a,b}_q$. This is the starting point to our definition of the polynomial of plane rooted trees mentioned in the title of the note.

We can repeat our considerations with many variables, $x_1,x_2,\ldots,x_k$. If variables commute we get the familiar multinomial formula (e.g. familiar to every student taking multivariable calculus):

 $\displaystyle \begin{eqnarray*} (x_1+x_2+\cdots+x_k)^n= \sum_{a_1,\ldots,a_k; \sum a_i=n}^n {n \choose a_1,\ldots,a_k}x_1^{a_1} x_2^{a_2}\cdots x_k^{a_k}, \end{eqnarray*}$

where ${n \choose a_1,\ldots,a_k} = \frac{[n]!}{[a_1]!\ldots[a_k]!}$. As before the pleasant interpretation of ${n \choose a_1,\ldots,a_k}$ is the number of pluckings of a tree $T_{a_k,\ldots,a_2,a_1}$ of $k$ long branches of length $a_k,\ldots,a_2,a_1$, respectively, as illustrated in Fig. 2.

We can now consider a noncommutative space with $x_jx_i=qx_ix_j$, for $i $\displaystyle \begin{eqnarray*} (x_1+x_2+\cdots+x_k)^n= \sum_{a_1,\ldots,a_k; \sum a_i=n}^n {n \choose a_1,\ldots,a_k}_qx_1^{a_1} x_2^{a_2}\cdots x_k^{a_k}, \end{eqnarray*}$where$ {n \choose a_1,\ldots,a_k}_q = \frac{[n]_q!}{[a_1]_q!\ldots[a_k]_q!}$. To prove the formula we can again use an induction on$ n$and again the following two properties are of value: 1. (i)$ [a_1+a_2+\cdots+a_k]_q= [a_1]_q + q^{a_1}[a_2]_q + q^{a_1+a_2}[a_2]_q+\cdots+q^{a_1+a_2+\cdots+a_{k-1}}[a_k]_q.$2. (ii) $\displaystyle \begin{eqnarray*} {a_1+a_2+\cdots+a_k \choose a_1,a_2,\ldots,a_k}_q =& {a_1+a_2+\cdots+a_k-1 \choose a_1-1,a_2,\ldots,a_k}_q\\ &+\, q^{a_1}{a_1+a_2+\cdots+a_k-1 \choose a_1,a_2-1,\ldots,a_k}_q\\ &+\cdots + q^{a_1+a_2+\cdots+a_{k-1}} {a_1+a_2+\cdots+a_k-1 \choose a_1,a_2,\ldots,a_k-1}_q. \end{eqnarray*}$As in the$ q$-binomial case, we can interpret the$ q$-multinomial coefficients by$ q$-plucking the tree$ T_{a_k,\ldots,a_2,a_1}$, that is assuming the following plucking formula $\displaystyle \begin{eqnarray*} Q(T_{a_k,\ldots,a_2,a_1}) =& Q(T_{a_k,\ldots,a_2,a_1-1}) + q^{a_1}Q(T_{a_k,\ldots,a_2-1,a_1})\\ &+\,q^{a_1+a_2}Q(T_{a_k,\ldots,a_3-1,a_2,a_1}) + \cdots + q^{a_1+a_2+\cdots+a_{k-1}}Q(T_{a_k-1,\ldots,a_2,a_1}). \end{eqnarray*}$The recursion is the same as for the$ q$-multinomial coefficient so we conclude that $\displaystyle \begin{eqnarray*} Q(T_{a_k,\ldots,a_2,a_1}) = {a_1+a_2+\cdots+a_k \choose a_1,a_2,\ldots,a_k}_q. \end{eqnarray*}$One can find more properties of$ q$-binomial coefficients in [Kac and Cheung2002] and [Mansour and Schork2016]. The quantum plane relation$ yx=qxy$can be generalized to the$ q$-deformed or quantized Weyl algebra where$ yx-qxy=h$; see Chapter 7 of [Mansour and Schork2016]. The generalization to rooted trees is not clear, or at least it is not unique. Our$ q$-polynomial is motivated by the Kauffman bracket relation in knot theory ([Dabkowski et al.2015]; [Dabkowski and Przytycki2016]). Possibly a connection of$ q$-deformed Weyl algebra to other quantum invariants of knots, developed by [Fenn and Turaev2007], can lead to other$ q$-polynomial invariants of trees. I would encourage readers to look in this direction. In [Asakly et al.2013] the relation of the Weyl algebra to labeled rooted trees is given but I do not see a clear relation with our work. 2 Recursive Definition of$ q$-Polynomial of Plane Rooted Tree Our description of the quantum plane and noncommutative space is already over one hundred years old (e.g. MacMahon 1854–1929, [MacMahon1916]). It may, therefore, look surprising that a simple generalization to rooted trees, which we present here, is totally new. Maybe the explanation lies in the observation that, because there are so many$ q$-analogues, a new one is studied only if there is an outside reason to do so (like the Jones revolution in knot theory, in my case). Perhaps our$ q$-polynomial is buried somewhere in a work of MacMahon contemporaries? We start our definition from the polynomial of a plane (that is embedded in the plane) rooted tree. In Corollary 2.3(iii) we show that the polynomial does not depend on the plane embedding and is therefore an invariant of rooted trees. In our work we use the convention that trees are growing up (like in) (Fig. 3). Definition 2.1. Consider the plane rooted tree$ T$(compare ) (Fig. 3). We associate to$ T$the polynomial$ Q(T,q)$(or succinctly$ Q(T)$) in the variable$ q$as follows. 1. (i) If$ T$is the one vertex tree, then$ Q(T,q)=1$. 2. (ii) If$ T$has some edges (i.e.$ |E(T)|>0$) then $\displaystyle \begin{eqnarray*} Q(T,q)=\sum_{v \in \mbox{ leaves }}q^{r(T,v)}Q(T-v,q), \end{eqnarray*}$where the sum is taken over all leaves, that is vertices of degree$ 1$(not a root), of$ T$and$ r(T,v)$is the number of edges of$ T$to the right of the unique path connecting$ v$with the root (an example is given in) (Fig. 3). In Fig. 4 we give an example of the expansion by our formula and if we complete the calculation we get the polynomial$ Q(T)$: $\displaystyle \begin{eqnarray*} &1+4q+9q^2+17q^3+28q^4+41q^5+56q^6+ 71q^7+83q^8+91q^9+94q^{10}\\ &\quad+\,91q^{11}+ 83q^{12}+ 71q^{13}+ 56q^{14}+ 41q^{15}+ 28q^{16}+ 17q^{17}+9q^{18}\\&\quad+\,4q^{19}+q^{20}. \end{eqnarray*}$33We show later that it is in fact equal to$ [2]_q^2[4]_q{8\choose 3,5}_q;$Corollary 2.3 (ii). As an example we can check that for$ T_{1,1,\ldots,1}$a star with$ n$rays we get$ Q(T_{1,1,\ldots,1}) = [n]_q!$. In particular,$ Q(\bigvee)= (1+q)=[2]_q$. One can look at a proof of the formula by direct induction on$ n$, but of course it is a very special case of the last formula of Sect. 1. The first important result in our theory of$ Q(T)$polynomials is the product formula for trees glued along their roots (wedge or pointed product). Theorem 2.2. Let$ T_1 \vee T_2$be a wedge product of trees$ T_1$and$ T_2$. Then: $\displaystyle \begin{eqnarray*} Q(T_1 \vee T_2)= \binom{|E(T_1)|+|E(T_2)|}{|E(T_1)|}_qQ(T_1)Q(T_2). \end{eqnarray*}$Proof. We proceed by induction on the number of edges of$ T$,$ |E(T)|$, with the obvious initial case of no edges in one of the trees, that is$ |E(T_1)|=0$or$ |E(T_2)|=0$. For simplicity we write$ E_i$for$ |E(T_i)|$. Let$ T$be a rooted plane tree with$ E_1E_2\neq 0$, then we have: $\displaystyle \begin{eqnarray*} &Q(T)= \sum_{v\in L(T)} q^{r(T,v)}Q(T-v)\\ &\quad=\sum_{v\in L(T_1)} q^{r(T_1,v)+E_2 }Q(T_1-v)\vee T_2)+ \sum_{v\in L(T_2)} q^{r(T_2,v)}Q(T_1\vee (T_2-v))\\ &\quad\stackrel{inductive\ assumption}{=} \sum_{v\in L(T_1)} q^{r(T_1,v)+E_2 }{E_1+E_2-1 \choose E_1-1,E_2}_q Q(T_1-v)Q(T_2)\\ &\qquad+ \sum_{v\in L(T_2)} q^{r(T_2,v)}{E_1+E_2-1 \choose E_1,E_2-1}_qQ(T_1)Q(T_2-v)\\ &\quad= Q(T_2)q^{E_2}{E_1+E_2-1\choose E_1-1,E_2}_q\sum_{v\in L(T_1)} q^{r(T_1,v)}Q(T_1-v)\\ &\qquad+ Q(T_1){E_1+E_2-1 \choose E_1,E_2-1}_q\sum_{v\in L(T_2)} q^{r(T_2,v)}Q(T_2-v)\\ &\quad\stackrel{definition}{=} Q(T_1) Q(T_2)\left(q^{E_2}{E_1+E_2-1 \choose E_1-1,E_2}_q +{E_1+E_2-1\choose E_1,E_2-1}_q\right)\\ &\quad= Q(T_1) Q(T_2){E_1+E_2\choose E_1,E_2}_q \mbox{ as needed}. \end{eqnarray*}$46One can modify the polynomial$ Q(T)$so that the formula of Theorem 2.2 can be interpreted as a homomorphism. For this we take$ Q'(T)=\frac{Q(T)}{[|E(T)|]_q!}$. With this definition we have$ Q'(T_1\vee T_2)= Q'(T_1)Q'(T_2)$. A disadvantage of such an approach is that$ Q'(T)$is not ncessarily a polynomial but only a rational function.$ \square$Directly from Theorem 2.2 we conclude several properties of the$ q$-polynomial,$ Q(T)$, which by the nature of its definition, as pointed up before, we propose to be called the plucking polynomial. We should stress, in particular, part (iii) which establishes the independence of the polynomial of its plane embedding. Corollary 2.3. 1. (i) Let a plane tree be a wedge product of$ k$trees that is $\displaystyle \begin{eqnarray*} T=T_{k} \vee \cdots \vee T_2 \vee T_1, \; \mbox{ then } \end{eqnarray*}$$\displaystyle \begin{eqnarray*} Q(T)= \binom{E_k+E_{k-1}+\cdots+E_1}{E_k,E_{k-1},\ldots,E_1}_qQ(T_k)Q(T_{k-1})\cdots Q(T_1). \end{eqnarray*}$where$ E_i=|E(T_i)|$is the number of edges in$ T_i$. 2. (ii) (State product formula) $\displaystyle \begin{eqnarray*} Q(T) = \prod_{v\in V(T)}W(v), \end{eqnarray*}$where$ W(v)$is a weight of a vertex (we can call it a Boltzmann weight) defined by: $\displaystyle \begin{eqnarray*} W(v)= \binom{|E(T^v)|}{|E(T^v_{k_v})|,\ldots,|E(T^v_{1})|}_q, \end{eqnarray*}$where$ T^v$is a subtree of$ T$with root$ v$(part of$ T$above$ v$, in other words$ T^v$grows from$ v$) and$ T^v$may be decomposed into wedge of trees as follows:$ T^v= T^v_{k_v} \vee \cdots \vee T^v_2 \vee T^v_{1}.$3. (iii) (Independence from a plane embedding) Plucking polynomial,$ Q(T)$does not depend on a plane embedding, it is therefore an invariant of rooted trees. 4. (iv) (Change of root). Let$ e$be be an edge of a tree$ T$with the endpoin ts$ v_1$and$ v_2$. Denote by$ E_1$the number of edges of the tree$ T_1$with the root$ v_1$and$ E_2$the number of edges of the tree$ T_2$with the root$ v_2$, where Proof. (i) The formula (i) follows by using$ (k-1)$-times the product formula of Theorem 2.2 and the fact that the$ q$-multinomial coefficient is a product of binomial coefficients: $\displaystyle \begin{eqnarray*} &\binom{a_k+a_{k-1}+\cdots+a_2+a_1}{a_k,a_{k-1},\ldots,a_2,a_1}_q\\ &\quad\stackrel{def.}{=} \frac{[a_k+a_{k-1}+\cdots+a_2+a_1]_q!}{[a_k]_q![a_{k-1}]_q!\ldots[a_2]_q![a_1]_q!}\\ &\quad=\frac{[a_2+a_1]_q!}{[a_2]_q! [a_1]_q!}\cdot\frac{[a_3+a_2+a_1]_q!}{[a_3]_q! [a_2+a_1]_q!}\cdot\, \cdots \,\cdot \frac{[a_k+a_{k-1}+\cdots+a_2+a_1]_q!}{[a_k]_q! [a_{k-1}+\cdots+a_2+a_1]_q!}\\ &\quad= \binom{a_2+a_1}{a_2,a_1}_q\binom{a_3+a_2 + a_1}{a_3,a_2+a_1}_q\binom{a_4+a_3+a_2 + a_1}{a_4,a_3+a_2+a_1}_q\\ &\qquad\cdots \binom{a_k+a_{k-1}+\cdots+a_2+a_1}{a_k,a_{k-1}+\cdots+a_2+a_1}_q \end{eqnarray*}$(ii) The formula (ii) follows by using (i) several times. (iii) Independence of embedding follows from the fact that the state product formula (ii) does not depend on the embedding. (iv) We compare product formulas of Theorem 2.2 for$ v_1$and$ v_2$and we get: $\displaystyle \begin{eqnarray*} Q(T,v_1)=&{E_1+E_2+1 \choose E_1, E_2+1}_q Q(T_1)Q(T_2)\quad \mbox{ and }\\ Q(T,v_2)=& {E_1+E_2+1 \choose E_1+1, E_2}_q Q(T_1)Q(T_2), \end{eqnarray*}$from which the formula of (iv) follows directly.$ \square$There are other nice properties of$ Q(T)$(e.g. the observation that it has polynomial time complexity) and I am sure readers will discover more. Here we give a few properties which have some importance in knot theory. Corollary 2.4. 1. 1.$ Q(T)$is of the form$ c_0+c_1q+\cdots+c_Nq^N$where: (i)$ c_0=1=c_N$,$ c_i>0$for every$ i\leq N$, (ii)$ c_i=c_{N-i}$, that is$ Q(T)$is a palindromic polynomial (often, less precisely we say symmetric polynomial), (iii) the sequence$ c_0,c_1,\ldots,c_N$is unimodal, that is for some$ j$: $\displaystyle \begin{eqnarray*} c_0\leq c_1 \leq \cdots \leq c_j \geq c_{j+1} \geq \cdots \geq c_N \end{eqnarray*}$(in our case$ j= \lfloor\frac{N}{2}\rfloor$or$ \lceil\frac{N}{2}\rceil$). (iv) For a nontrivial tree$ T$, that is a tree with at least one edge, we have: $\displaystyle \begin{eqnarray*} c_1 = \sum_{v\in V(T)} (k_v-1), \end{eqnarray*}$where$ k_v= deg_{T^v}(v)$is the number of edges growing up from$ v$, that is the degree of$ v$in the tree$ T^v$growing from$ v$(as in Corollary 2.3 (ii)). In particular, if$ T$is a binary tree,$ c_1(T)$is the number of vertices of$ T$which are not leaves. 2. 2. 1. (i)$ Q(T)$is a product of$ q$-binomial coefficients (of type$ {a+b\choose a}_q$). 2. (ii)$ Q(T)$is a product of cyclotomic polynomials. 77 Recall that$ n$th cyclotomic polynomial is a minimal polynomial which has as a root$ e^{2\pi i/n}$. We can write this polynomial as:$ \Psi_n(q)=\prod_{\omega^n=1,\omega^k\neq 1, k. For example $\Psi_4(q)=1+q^2$ , $\Psi_6(q)=1-q+q^2$ .

3. 3.

The degree of the polynomial $N=deg Q(T)$ can be described by the formula:

 $\displaystyle \begin{eqnarray*} N=deg Q(T)=\sum_{v\in V(T)}\left(\sum_{1\leq i < j \leq k_v}E_i^vE_j^v\right), \end{eqnarray*}$

where, as in Corollary 2.3(ii) $T^v$ is a subtree of $T$ with the root $v$ (part of $T$ above $v$, in other words $T^v$ is growing from $v$) and $T^v$ can be presented as a bouquet of trees: $T^v= T^v_{k_v} \vee \cdots \vee T^v_2 \vee T^v_{1}.$

Proof.

1(i) follows easily from the definition of the plucking polynomial. Namely we see that the constant term is obtained in a unique way by always plucking the most rightmost leaf from the tree (repeating this each time in the calculation). Thus $c_0=1$. Similarly, the highest power of $q$ is obtained uniquely by always plucking the leftmost leaf of the tree. The condition that $c_i>0$ for any $i\leq N$ requires a more careful look at the recursive computation of $Q(T)$ and the proof is absolutely elementary; we leave it to the reader because in (iii) we provide much a stronger condition (but using a nontrivial fact proven by [Sylvester1878]). (1)(ii) We start from observing the symmetry of $q$-binomial coefficients; namely we have:

 $\displaystyle \begin{eqnarray*} {a+b \choose a,b}_{q^{-1}} = q^{-ab} {a+b \choose a,b}_q. \end{eqnarray*}$

Then we use an easy observation that a product of symmetric polynomials is symmetric, and the fact we proved already and formulated in (2) that the polynomial $Q(T)$ is always a product of binomial coefficients. (1) (iii) Follows from the nontrivial fact, originally proved by Sylvester that $q$-binomial coefficients are unimodal and from a simple observation that a product of symmetric (i.e. palindromic) positive unimodal polynomials is symmetric unimodal (see [Stanley1989]; [Wintner1938]). (1) (iv) The formula for $c_1$ follows from the observation that ${a+b \choose a, b}_q= 1+q+\cdots$ for $a,b >0$ and more generally, ${a_1+a_2+\cdots+a_k \choose a_1,a_2,\ldots,a_k}_q = 1+ (k-1)q + \cdots$ for $a_1,a_2,\ldots,a_k >0$. Then the conclusion follows from the product formula of Corollary 2.3. (2) and (3): These conditions follow directly from the product formula of Corollary 2.3(ii). $\square$

With regard to Corollary 2.4(iii) we can ask: for which trees, is $Q(T)$ strictly unimodal, that is

 $\displaystyle \begin{eqnarray*} c_0 < c_1 < \cdots < c_j > c_{j+1} > \cdots > c_N \end{eqnarray*}$

for some $j$; compare the computation for the tree of Fig. 3 (see also [Pak and Panova2013]).

As I was stressing from the beginning, and by now many readers may agree, the $q$-polynomial (plucking polynomial) is interesting on its own. However, I would have never constructed or discovered it if I had not observed its shadow in my knot theory research. Concretely it was my work with M.Dabkowski and his student C.Li concerning skein modules of generalized (lattice) crossings (Fig. 5), which gave some initial motivation ([Dabkowski et al.2015]). In that paper we do not use $Q(T)$, as it was observed after the paper was completed. We will use it, however, in future research ([Dabkowski and Przytycki2016]). Knot theory also motivates specific generalizations of the plucking polynomial by enhancing it with a delay function, $f: L(V) \to {\mathcal N}=\{n\in \Z\ | \ n \geq 1\}$. This function regulates which leaves are used in the recursive formula (Definition 2.1) and which are “delayed”. In fact we have much flexibility in the definition so we challenge a reader to play with possibilities.

The relation of the plucking polynomial to the Kauffman bracket skein modules is precise but difficult to describe succinctly. To have a concrete idea we can say that it concerns the study of the generalized (lattice) crossing (Fig. 5), under the assumption that we resolve every crossing using the Kauffman bracket skein relation , and replace every trivial component by the Laurent polynomial $-A^2 - A^{-2}$.

One can generalize the plucking polynomial to any graph. If a graph $G$ has a base-point $b$ then the invariant $Q(G,b)$ is a collection (multiset) of plucking polynomials of spanning trees $G$ with the root $b$ ([Przytycki2016]). I challenge the reader to study connections of this invariant with known invariants of graphs.

Our polynomial also has relations to homological algebra: let $\mathcal C$ be a chain complex, that is a sequence of abelian groups, $C_n$, and homomorphisms $\partial_n:C_n \to C_{n-1}$, so that $\partial_{n-1} \partial_n = 0$. On the basis of a chain complex we build homology groups by defining $H_n(\mathcal C)= \ker \partial_n/im(\partial_{n+1})$. Very often in topology and homological algebra the boundary operation $\partial_n$ is an alternating sum of homomorphisms, called face maps: $\partial_n= \sum_{i=0}^n (-1)^id_i$. M. Kapranov asked about what happens if $(-1)^i$ is replaced by $q^n$, that is we define

 $\displaystyle \begin{eqnarray*} \partial^q_n= \sum_{i=0}^n q^id_i. \end{eqnarray*}$

He noticed that if $q$ is a $k$th root of unity different from $1$ (i.e. $q^k=1, q\neq 1$) then the $k$th iteration of $\partial^q$ is the zero map ($\partial^q_{n-k+1}\ldots\partial^q_{n-1} \partial^q_n = 0$) ([Kapranov2016]). The fact that Kapranov’s idea is related to our $q$-polynomial is clear, however deep connections require careful study: possibly you—the reader—could make a breakthrough.

Acknowledgements
J. H. Przytycki was partially supported by the GWU REF Grant, and Simons Collaboration Grant-316446. The author would like to thank a referee for directing his attention to [Asakly et al.2013] and [Mansour and Schork2011]; [Mansour and Schork2016].

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