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# looking for flying wires on a biplane

### Help Support Homebuilt Aircraft & Kit Plane Forum:

#### crazymike

##### Member
HBA Supporter
does anyone know where i can get some cables " flying wires " that support the wings built for a aerosoprt scamp biplane ? like what you see on a stearman

#### don january

##### Well-Known Member
Log Member
I would contact www,steenaero.com also known as Bruntons flying wires. and also check aircraft spruce and specialties.

#### Dana

##### Super Moderator
Staff member
On most biplanes, the flying wires are not cables, but streamline wires. Bruntons is the only remaining manufacturer in the world. All wires are custom ordered and they're expensive. Due to the cost, some people do use braided steel cable (my Hatz has cables instead of wires). Round cables have significantly more drag, so there will be an effect on the aircraft's speed.

#### TFF

##### Well-Known Member
Someone is making wires in Australia or New Zealand, also.

#### crazymike

##### Member
HBA Supporter
Someone is making wires in Australia or New Zealand, also.
thanks for the reply's i already checked out bruntons and the new zealand both are way up there in price like $300 per wire x 6 like ouch i googled your hatz and checked out the 50 years on you tube . what do you mean by cables are you using like hardware store style for the turn buckles and cable u-bolt clamps ? and will faa give me an airwothiness cert. with the cables ? can you explain a little more . i was looking at grainger they had 1/4" cable with a 2000 lb. rating but did not know if faa would approve of them . #### Dana ##### Super Moderator Staff member thanks for the reply's i already checked out bruntons and the new zealand both are way up there in price like$300 per wire x 6 like ouch i googled your hatz and checked out the 50 years on you tube . what do you mean by cables are you using like hardware store style for the turn buckles and cable u-bolt clamps ? and will faa give me an airwothiness cert. with the cables ? can you explain a little more . i was looking at grainger they had 1/4" cable with a 2000 lb. rating but did not know if faa would approve of them .
Hardware store stuff and u-bolt clamps, absolutely not. It has to be aircraft grade cable, more properly called "wire rope". There are various constructions with different degrees of flexibility. 1X19, which has 19 strands, is used for bracing. 7x7 (7 bundles of 7 strands each) and 7x19 are used for control cables. Mine are 1/4" and 3/16", and the tail wires are 5/32". At the ends, you use machine swaged terminals like steveair posted above (you order them cut and swaged to the length you specify, the hydraulic roller swaging machine is very expensive), or Nicopress sleeves with thimbles (which you can do yourself with a relatively inexpensive tool), which is what mine are. Here's one end of mine:

What you see here is a shackle pinned to the painted tab, then the cable wrapped around a "thimble" where it goes through the shackle, and the Nicopress sleeve swaged around the cable. Beyond that, it's wrapped with wire and soldered, which is not necessary, it's just for appearance. At the other end is another thimble and Nicopress sleeve, with a turnbuckle for adjustment:

Cable: https://www.aircraftspruce.com/catalog/appages/controlcable.php

Thimbles: https://www.aircraftspruce.com/catalog/appages/an100.php

Nicopress sleeves: https://www.aircraftspruce.com/catalog/appages/nicopress.php

Turnbuckles: https://www.aircraftspruce.com/catalog/appages/turnbuckles.php

Swaging tool: https://www.aircraftspruce.com/catalog/topages/no3swageit.php

#### TFF

##### Well-Known Member
The difference between the ends on Dana’s and Steveair ‘s is Dana’s can be done at home with a relatively cheap aviation nicopress swaging tool. Steveair’s is much cleaner but requires a very expensive swaging tool that must be used or it comes apart. Don’t mix parts without the right tools.

#### crazymike

##### Member
HBA Supporter
I should of posted the spec's I guess . this thing looks to me like it is using 1/4" and 5/16" cables . aircraft spruce link i also looked at previously was a maximum of 3/16" dia.
AN674-1/4x28x71.88 PLC AN665-34 R&L 2 EACH
AN674-1/4x2871.062 PLC AN665-34 R&L 2 EACH
AN675-5/16x24x78.50 PLC AN665-46 R&L 2 EACH

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#### crazymike

##### Member
HBA Supporter
CONTROL CABLE 1/4 7X19 GALV. has listed 7000 strength is this strong enough to support all 6 wires ?
TURNBUCKLE ASSEMBLY AN130-46S,AN130-32S,AN130-32L rated 4600 and 3200 and 3600
cable shackle AN115-32 thimble AN100-5 clevis pin MS20392-3C15

could not locate any sleeves for the 1/4" cable

#### imacfii

##### Active Member
How much us your life worth?
Do you want me to quote in these streamline wires:
AN674-1/4x28x71.88 PLC AN665-34 R&L 2 EACH
AN674-1/4x2871.062 PLC AN665-34 R&L 2 EACH
AN675-5/16x24x78.50 PLC AN665-46 R&L 2 EACH
Additionally the -46 fittings are normally used for landing wires only and not reccommended for flying wires, the correct clevis would be an AN665-61 fitting.
Yes, we are down here in New Zealand, approved by the Shuttleworth trust, the only approved supplier to the French Air & Space Museum at Le Bourget. We have been building and supplying streamline flying wires and tie rods for almost 15 years and have never had a failure. Your choice - your life.
[email protected] or www.vintageaero.com

#### Dana

##### Super Moderator
Staff member
Russ, have you made a wire set for a Hatz? I'd be interested in knowing (ballpark) what that would cost. I know some Hatz builders have gotten a group order to Steen for wires.

Nothing inherently unsafe about cables, if they're properly sized. My plane (I didn't build it) has been flying for over 30 years with cables, but no doubt it would be faster with streamline wires.

#### Turd Ferguson

##### Well-Known Member
does anyone know where i can get some cables " flying wires " that support the wings built for a aerosoprt scamp biplane ? like what you see on a stearman
What was the plane designed to use? Harris Woods built economical planes so I can't believe he would use streamlined flying wires.

#### crazymike

##### Member
HBA Supporter
just a simple single seat biplane that was designed for a 60hp v.w. motor but i do not like the thought of a prop bolted on to the crankshaft snout so i am going with a 1.3L 16v suzuki motor with a gear reduction prop drive unit.and as for the cables those were the part #'s and spec's in the plans the mounting points are the same sizes as the cables so 1/4" and 5/16" it looked like it was going to be easy but i could not find a shackle that would accept 1/4" cable with a 1/4" hole and it did not look any better for the 5/16" .

#### Dana

##### Super Moderator
Staff member
just a simple single seat biplane that was designed for a 60hp v.w. motor but i do not like the thought of a prop bolted on to the crankshaft snout so i am going with a 1.3L 16v suzuki motor with a gear reduction prop drive unit.and as for the cables those were the part #'s and spec's in the plans the mounting points are the same sizes as the cables so 1/4" and 5/16" it looked like it was going to be easy but i could not find a shackle that would accept 1/4" cable with a 1/4" hole and it did not look any better for the 5/16" .
People have been bolting propellers to crankshafts for over 100 years, including on VW engines.

As for the flying wires, the standard advice is to use what the plans call for unless you have the engineering skills to analyze any proposed alternate. People here can offer advice, but the final responsibility is yours.

#### TFF

##### Well-Known Member
Sounds like way more power. Prop clearance might be an issue. I would put that plane is in the Tennie and VP class of small simple planes. The T tail is interesting for a biplane.

#### GeeZee

##### Well-Known Member
HBA Supporter
Hey crazymike, the Scamp looks like a neat bipe, have you considered a build thread here on HBA? I’d love to see construction details.

#### Wanttaja

##### Well-Known Member
thanks for the reply's i already checked out bruntons and the new zealand both are way up there in price like \$300 per wire x 6 like ouch i googled your hatz and checked out the 50 years on you tube . what do you mean by cables are you using like hardware store style for the turn buckles and cable u-bolt clamps ? and will faa give me an airwothiness cert. with the cables ? can you explain a little more . i was looking at grainger they had 1/4" cable with a 2000 lb. rating but did not know if faa would approve of them .
See the Cable and Turnbuckle information starting on Page 7:

http://www.bowersflybaby.com/PB100/nuts_bolts.pdf

Ron Wanttaja

#### imacfii

##### Active Member
Russ, have you made a wire set for a Hatz? I'd be interested in knowing (ballpark) what that would cost. I know some Hatz builders have gotten a group order to Steen for wires.

Nothing inherently unsafe about cables, if they're properly sized. My plane (I didn't build it) has been flying for over 30 years with cables, but no doubt it would be faster with streamline wires.
Hi Dana,
Not too sure as mostly I get a request for 4 of AN674AC-7200, 6 of AN703AC-4800 etc. We've made just about everything from AN674AC-11000 (yes 11 inches long, rear fuselage streamline wires for a Sopwith Snipe) to 4.6 meters long 8mm wires for a 1923 Potez 25 for the French Air & Space Museum at Le Bourget.(Paris, France). Currently just finishing some AGS 307 tie rods for a WW1 SE5A. These are a 4ba thread (138.5 thousandths). .125" 304 is swaged in the center down to 99.5 thou then the thread is rolled up to .1385". Next you stamp a flat on the .125" shank, so you can tighten the tie rod with a wrench. The .0995" tie rods will hold over 1200lbs. There are over 100 of these cute little critters on the fuselage alone.
We've built lots of Pitts wires all the way from S1C to Model 12 wires and have replaced quite a few wires from 'the other guys' which have failed in disassembly due to galling. We've done lots of Bearhawk tail wires. We also manufacture clevis fittings in 420SS. If you can give me a list of what you need, I can give you a price. If you need unsolicited testimonials I can provide those as well.
cheers,
Russ Ward
www.vintageaero.com [email protected]

#### PTAirco

##### Well-Known Member
Have you checked the price of turnbuckles lately? It's getting pretty astronomical too. I'm taking about AN stuff here.

The flying wire question had been bugging me for years. Of course we all would like "proper" streamline wires, but when it adds up to a substantial portion of the cost of the airframe, I would like an alternative. There is nothing special about the material, but getting threads rolled is pricey. You can come you with terminal fittings with a bit of ingenuity. Or eat the cost of AN terminals. I'd be happy with round wires if I can do it at one tenth the cost. I looked into spoke manufacturers and I believe there is potential there for the lighter gauge wires.


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# New formulas for the Laplacian of distance functions and applications

created by cavallett on 26 Mar 2018

[BibTeX]

Preprint

Inserted: 26 mar 2018
Last Updated: 26 mar 2018

Year: 2018

Abstract:

The goal of the paper is to prove an exact representation formula for the Laplacian of the distance (and more generally for an arbitrary 1-Lipschitz function) in the framework of metric measure spaces satisfying Ricci curvature lower bounds in a synthetic sense (more precisely in essentially non-branching MCP(K,N)-spaces). Such a representation formula makes apparent the classical upper bounds and also some new lower bounds, together with a precise description of the singular part. The exact representation formula for the Laplacian of 1-Lipschitz functions (in particular for distance functions) holds also (and seems new) in a general complete Riemannian manifold. We apply these results to prove the equivalence of CD(K, N ) and a dimensional Bochner inequality on signed distance functions. Moreover we obtain a measure-theoretic Splitting Theorem for infinitesimally Hilbertian, essentially non-branching spaces verifying MCP(0,N).

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# Prove a certain homomorphism in the category of rings is an epimorphism

I've been reading the Queen Mary Note of Maurice Auslander 's paper Representation dimension of artin algebras. However I met some difficulties that are not so relevant to the topic. The following is such one.

On page six he describes the image of a morphism between two commutative rings in the category of rings with identities.

Let R and T be commutative rings and $f:R\rightarrow T$ a ring homomorphism. Let $f(R)=S$. We have $f_1:T\rightarrow T\otimes_ST$ defined by $f_1(t)=1\otimes t$ and $f_2(t)=t\otimes 1$.

Let $T'=\{t\in T| f_1(t)=f_2(t)\}$. Clearly $T'$ contains S, so that $f$ maps $R$ into $T'$. Prove that the inclusion map $S\rightarrow T'$ is an epimorphism in the category of rings and the inclusion map $T'\rightarrow T$ is a strong monomorphism.

I tried to use the notion dominion and pushout to prove the epimorphism part but failed.

• Concerning the second part: Regular monomorphisms are strong. And $T' \to T$ is the equalizer of $f_1,f_2$. – HeinrichD Apr 6 '17 at 14:15
• I am not confident that $S \to T'$ is always an epimorphism of comm. rings. Notice that $T'$ is the dominion aka epicenter of $S \hookrightarrow T$, whose elements have a matrix-like representation (see P. Mazet, Characterization des epimorphismes par generators and relateurs, Seminaire Samuel; or stacks.math.columbia.edu/tag/04VY), but these matrices have coefficients in $T$, not necessarily in $T'$. Also, in the literature on epimorphisms of commutative rings (which I am quite familiar with) I cannot find the statement that $S \to T'$ is an epimorphism. – HeinrichD Apr 6 '17 at 14:30
• @HeinrichD I will post this paper if it helps – user12580 Apr 6 '17 at 14:38
• If you mean a link to Auslander's paper, this would be appreciated. I could not find it online. – HeinrichD Apr 6 '17 at 14:44
• @HeinrichD I find the paper through my university's data base, and I could not find it online either – user12580 Apr 6 '17 at 14:53


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The application to the 2017 NBA Hackathon featured several interesting questions, two of which I would like to unpack in this post:

1. How likely is a team to lose consecutive games in an 82-game season?
2. How much money is the NBA likely to generate from playoff ticket sales?

All of the R code for both questions can viewed in this repo.

## Probability of Consecutive Losses

Imagine that the Warriors have a $80$% of winning each game during their 82-game regular season. What’s the chance they don’t lose back-to-back games? This was a popular question after Kevin Durant joined the Warriors before the 2016-2017 season.

### Approximation Approach

Monte Carlo simulations are a simple method to approximate probability distributions; once you have an approximation of a distribution, generating a point estimate, such as an expected value, is straight forward. The plot below shows the first $10000$ samples from such a simulation, which will eventually converge to the expected value of the distribution.

The simulation predicts that the Warriors have a $\underline{5.041\%}$ chance of not losing consecutive games (code). It would would be helpful to know the precision of this approximation. The variance of an expected value from a Monte Carlo simulation, with samples ${\phi_1, …, \phi_n}$, is described by:

$\text{Var}_{MC}[\overline{\phi}] = \frac{\text{Var}[\phi]}{S}$

We can compute $\text{Var}[\phi]$ since our samples are drawn from a Binomial distribution

$Var[\phi] = np(1 - p) = 82 * 0.8 (1 - 0.8) = 13.12$

With $200000$ samples, $Var_{MC}[\overline{\phi}] = 6.56 \times 10^{-5}$, which means we can expect the true value to be contained within the following interval for 95% of all MC simulations:

$\overline{\phi} \pm 2 \sqrt{Var_{MC}[\overline{\phi}]} = 5.041\% \pm 1.62\% = (3.421\%, 6.661\%)$

Now let’s calculate the exact answer and see how our approximation compares.

### Exact Approach

Let’s consider three distinct groups of games and their total counts for the Warriors 82-game season, where $l$ represents the number of losses:

• losses that are followed by a win (LW), of which there are $(l - 1)$ but represent $2(l-1)$ games.
• a season-ending loss (L), of which there is $1$.
• single, standalone wins (W), of which there are $82 - 2(l - 1) - 1 = 83 - 2l$.

We arrive at the total number of units by adding all three groups together: $83 - 2l + (l - 1) + 1 = 83 - l$. These three groups can be placed in any order and the season can be guaranteed to not contained any back-to-back losses. We just need to figure out how to enumerate these combinations.

If we could lump two of the units together, we would be choosing the ordering of two groups and could use the binomial formula; and lucky for us, we can do just that! If we group the two types of units with losses together (of which there are $l$), we don’t have to designate which loss is the season-ending one. With this insight, we have $83-l$ total slots to put a loss, and have $l$ losses to place, which gives us the following equation for the number of $l$-loss seasons that do not contain consecutive losses:

$\binom{83 - l}{l} (1-p)^l(p)^{82-l}$

If we consider the Warriors probability of winning a game is 80%, and we want to get the total probability over all possible seasons, we get the following equation:

$\sum_{k=0}^{82} \binom{83 - l}{l} (1-p)^l(p)^{82-l} = \sum_{k=0}^{82} \binom{83 - l}{l} (0.2)^l(0.8)^{82-l} = \underline{0.058817}$

So the Warriors had exactly a $\underline{5.88}$% chance of not losing consecutive games if indeed they had an 80% chance of winning every game. And the exact answer falls within the range given by our Monte Carlo simulation; yes, all is right with the world. If we wanted to make this problem more interesting, we could take into account the Warriors schedule and calculate the win probability of each game. But now it’s time to talk about cash flow!

## Expected Playoff Gate Revenue

How much money from ticket sales can the NBA expect to bring in during the Playoffs? The League provided two datasets as a part of the Hackathon application:

1. The win probabilities for home and away games for every combination of teams (view data).
2. The expected gate revenue at each NBA stadium for all four playoff rounds (view data).

First, we need to figure out the expected revenue from each matchup, and this is dependent on how many games are played of the 7-game series.

### 7-Game Series Math

Let’s start simply by considering the probabilities that a 7-game series ends in either $N = {4,5,6,7}$ games if the two teams have equal probabilities of winning:

• N = 4. $2 \binom{3}{3} / 2^3 = 0.125$
• N = 5. $2 \binom{4}{3} / 2^4 =0.25$
• N = 6. $2 \binom{5}{3} / 2^5 = 0.3125$
• N = 7. $2 \binom{6}{3} / 2^6 = 0.3125$

Notice how these four numbers sum to $1$, which confirms that we are working with a valid probability distribution. In this equations above, the denominator gives the total number of possible win/loss combinations and the numerator describes the number of outcomes in which 4 games is won by either team (hence the doubling factor in the numerator).

Now let’s make the problem more complex: we expect one team (Warriors) to beat the other team (Spurs) with probability $p = 0.55$. Now we have to split up the probability of the Warriors winning and the Spurs winning:

• N = 4. $\binom{3}{3} (p^4 + (1-p)^4) = 0.133$
• N = 5. $\binom{4}{3} (p^4(1-p) + (1-p)^4p) = 0.255$
• N = 6. $\binom{5}{3} (p^4(1-p)^2 + (1-p)^4p^2) = 0.309$
• N = 7. $\binom{6}{3} (p^4(1-p)^3 + (1-p)^4p^3) = 0.303$

These calculations show that the more lopsided a matchup, the more likely the series is to last fewer games! But we can still make the problem more complicated by taking into account home and away games; for this discussion, we will assume the 2-2-1-1-1 playoff series format. If we assume that the Warriors have a $p_h$ chance of winning at home against the Spurs and a $p_a$ chance of winning away against the Spurs, then the following expressions describe the probability that the series ends in $N$ games:

#### N = 4, 2 Home games, 2 Away games.

$p_h^2p_a^2 + (1-p_h)^2(1-p_a)^2$

#### N = 5, 3 Home games, 2 Away games.

\begin{align} \binom{2}{1} &\left[ p_h^2p_a^2(1-p_h) + (1-p_h)^2(1-p_a)^2p_h\right] +\\\\ \binom{2}{1} &\left[ p_h^3p_a(1-p_a) + (1-p_h)^3(1-p_a)p_a) \right]\\\\ \end{align}

#### N = 6, 3 Home games, 3 Away games.

\begin{align} \binom{2}{1} \binom{3}{1} &\left[ p_h^2p_a^2(1-p_h)(1-p_a) + (1-p_h)^2(1-p_a)^2p_hp_a\right] +\\\\ &\left[ p_h^3p_a(1-p_a)^2 + (1-p_h)^3(1-p_a)p_a^2 \right] + \\\\ \binom{3}{2} &\left[ p_hp_a^3(1-p_h)^2 + (1-p_h)(1-p_a)^3p_h^2 \right] \\\\ \end{align}

#### N = 7, 4 Home games, 3 Away games.

\begin{align} \binom{3}{2} \binom{3}{1} &\left[ p_h^2p_a^2(1-p_h)^2(1-p_a) + (1-p_h)^2(1-p_a)^2p_h^2p_a\right] + \\\\ \binom{3}{1} \binom{3}{2} &\left[ p_h^3p_a(1-p_a)^2 + (1-p_h)^3(1-p_a)p_a^2 \right] + \\\\ &\left[ p_hp_a^3(1-p_h)^3 + (1-p_h)(1-p_a)^3p_h^3 \right] + \\\\ &\left[ p_h^4(1-p_a)^3 + (1-p_h)^4p_a^3 \right] \\\\ \end{align}

The difficulty in deriving the above equations is keeping track of the number of home and away games that have been played before the last game. Some of the expressions have two binomial coefficients because both the away games and home games can occur in different orders.

### Simulations and Results

These equations will help us compute the expected gate revenue from any matchup between NBA teams with the two provided datasets. The expected gate revenue at NBA Stadiums is displayed below.

After running $1000$ playoff simulations (view code), we arrive at the following probability distribution:

This distribution is bimodal, with the chance of generating $$160 million or$$120$million equal to about$75$%. If I were in the NBA Finance department, I would want to know which teams need to win in order to maximize playoff ticket sales. We can arrive at that answer by comparing which teams win at each round in simulations where the total revenue is $$160 million and$$120$ million.

No huge discrepancies are present in the Round 2 and 3 plots, but the Round 4 plot, which shows us the NBA Finals participants, tells us everything we need to know: the top seed in the West (the 2017 Warriors), must make it to the Finals in order for the NBA to maximize their profit from ticket sales. Yes, most fans would suspect this to be true, but there’s nothing like proving intuitive suspicions with some math.


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Distinguishing the two irreducible representations of odd-dimensional complex Clifford-Algebras

The complex Clifford algebra $A$ of a complex, non-degenerate quadratic space $(V,q)$ of odd dimension $2k+1$ admits up to isomorphism exactly two non-trivial, irreducible and finite-dimensional representations on a complex vector space $S$. If $\Phi:A\to \mathrm{End} (S)$ represents one, then $\Phi\circ\chi$ represents the other, where $\chi: A\to A$ is the automorphism defined by $\chi(v)=-v$ for $v\in V$.

My question is now: how can one decide whether two generic (irreducible, non-zero, finite-dimensional complex) representations $\Phi_1,\Phi_2$ of $A$ belong to the same isomorphism class? From what I have seen so far, I suspect that it should have something to do with the volume element of $A$...?

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You are looking for a method to determine whether or not two representations are isomorphic? Did you intend to assume any conditions on them at all? Are we assuming the form is nondegenerate? – rschwieb Sep 10 '12 at 15:31
Oh yes, $q$ is non-degenerate. The representations $\Phi_1,\Phi_2$ under consideration are required to be non-zero, irreducible and finite-dimensional. – Robert Rauch Sep 10 '12 at 15:48
Put all your requirements in the question, please. – rschwieb Sep 10 '12 at 18:20

I'm sorry for not reacting for a long time. I have to admit my question was a bit vague, but I have found the solution, I was looking for, now. There is a complete isomorphism-invariant $\theta(\Phi)\in\mathbb{Z}_2$ of the irreducible, complex, finite dimensional representations $\Phi$ of the Clifford-Algebra $Cl(V)$ of an odd-dimensional, complex, non-degenerate quadratic space $(V,q)$ defined as follows. Assume we have chosen a preferred orientation in $V$ and consider an oriented orthonormal basis $b_1\dots,b_n$ of $V$. Then we call the element $$\eta=i^{n(n-1)/2}b_1\cdots b_n\in Cl(V)$$ the volume element of $Cl(V)$, and the Clifford Relations imply that $\eta$ does not depend on the chosen basis. The factor is chosen such that $\eta^2=1$, hence $\Gamma=\Gamma(\Phi):=\Phi(\eta)$ also satisfies $\Gamma^2=1$ and decomposes the representation space $S$ of $\Phi$ into $S=S_+\oplus S_-$ with $S_\pm:=\ker \Gamma\mp 1$. Since $n$ is odd, $\eta$ lies in the center of $Cl(V)$ and thus the $S_\pm$ are $Cl(V)$-invariant subspaces of $S$. By irreducibility, we thus have either $S_+=S$ or $S_-=S$, i.e. $\Gamma=\theta(\Phi)\cdot\operatorname{id}_S$ for some $\theta(\Phi)\in\mathbb{Z}_2$. One then easily computes that $\theta(\Phi)$ only depends on the isomorphism class of $\Phi$ and that $\theta(\Phi\circ\chi)=-\theta(\Phi)$, where $\chi\in\operatorname{Aut}(Cl(V))$ is the canonical involution on $Cl(V)$, defining the $\mathbb{Z}_2$-grading. Therefore $\Phi\not\cong\Phi\circ\chi$ and by the structure theorem, these are the only irreducible representations (upto isomorphism).

To summarize, two irreducible (complex, finite-dimensional) representations $\Phi,\Psi$ of $Cl(V)$ are isomorphic if and only if both $\Phi(\eta)$ and $\Psi(\eta)$ both act as either plus or minus the identity on the corresponding representation space.

Cheers, Robert

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By the structure classification theorem, such Clifford algebras are actually two copies of a matrix ring over $\mathbb{C}$.

Being semismple, all of its modules are direct sums of copies of the simple modules.

The irreducible ones would just be the simple ones. (right?)

(When talking about representations as opposed to modules I always feel like there is a subtle difference I haven't mastered...)

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What you write is correct but doesn't answer my question, i guess. Regarding your concerns about representations opposed to modules: whenever you have an associative unital $k$-algebra $A$, then the category of $k$-linear representations of $A$ is equivalent to the category of unital left $A$-modules. Under this equvalence, irreducible corresponds to simple. However, if you, for example, want to classify complex representations of a real algebra up to real isomorphims, things become complicated from the module point of view. – Robert Rauch Sep 10 '12 at 16:49
@RobertRauch Well again, if you want to use real representations only, you need to mention that in your question. From what is currently written, one can only assume you are interested in $\mathbb{C}$ modules. This is mainly because you mentioned complex representations! If you indeed are only interested in complex representations, then what remains to be answered? – rschwieb Sep 10 '12 at 18:19
What I just wrote about real representations in my comment was to illustrate, that there are in fact some traps in identifying representations of an algebra $A$ with modules over $A$, considered as ring. Back to the original question: As you have mentioned, the fact that there are exactly two irreducible (non-zero, complex, finite-dimensional) representations of $Cl(V)$ is a consequence of $Cl(V)\cong End(S)\oplus End(S)$. What I asked above is a different thing: we want to know, how one can decide whether two given irreducible rep. $\Phi_1,\Phi_2$ of $Cl(V)$ lie in the same isomrphism class. – Robert Rauch Sep 10 '12 at 21:07
@RobertRauch Yes, you did write some comments, but they are about real representations which continue to be irrelevant to your question. From what have told me, you agree this algebra has exactly two isoclasses of simple (=irreducible) modules as a complex algebra, which you have also told me correspond exactly to the irreducible complex representations of this algebra. I must be missing something, because to me it looks like the question is answered. :( – rschwieb Sep 11 '12 at 1:30
As I already said, my comments on real representations were only for your pleasure, as you mentioned that you feel unsure about the identifications of representations with modules. And Yes, these comments are "irrelevant" for the original question (sorry for trying to help you with your difficulties), which is now formulated perfectly clear in the original post: given two concrete irreducible (complex, finite-dimensional, nonzero) reps of $Cl(V)$, how can i decide if they are isomorphic?? – Robert Rauch Sep 11 '12 at 11:49


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# ltePSBCHPRBS

PSBCH pseudorandom binary scrambling sequence

## Syntax

``[seq,cinit] = ltePSBCHPRBS(ue,n)``
``[seq,cinit] = ltePSBCHPRBS(ue,n,mapping)``
``[subseq,cinit] = ltePSBCHPRBS(ue,pn)``
``[subseq,cinit] = ltePSBCHPRBS(ue,pn,mapping)``

## Description

example

````[seq,cinit] = ltePSBCHPRBS(ue,n)` returns a column vector containing the first `n` outputs of the PSBCH pseudorandom binary scrambling sequence (PRBS) for the specified UE settings structure. It also returns an initialization value `cinit` for the PRBS generator.The scrambling sequence generated should be applied to the coded PSBCH data carried by the associated subframe. The PRBS generator is initialized with cinit = `ue`.`NSLID`.```

example

````[seq,cinit] = ltePSBCHPRBS(ue,n,mapping)` specifies the format of the returned sequence, `seq`, through the `mapping` input.```
````[subseq,cinit] = ltePSBCHPRBS(ue,pn)` returns a subsequence of a full PRBS sequence, specified by `pn`.```
````[subseq,cinit] = ltePSBCHPRBS(ue,pn,mapping)` specifies the format of the returned subsequence, `subseq`, through the `mapping` input.```

## Examples

collapse all

Scramble a PSBCH codeword by generating the PSBCH pseudorandom binary sequence (PRBS) and applying an exclusive OR operation on the two sequences.

Create a UE configuration structure and SL-BCH codeword. Generate the required length of the PRBS and scramble the PSBCH codeword with the PRBS sequence using `xor`.

```ue = struct('NSLID',2); codeword = lteSLBCH(ue,ones(40,1)); psbchPrbs = ltePSBCHPRBS(ue,length(codeword)); scrambled = xor(psbchPrbs,codeword);```

Scramble PSBCH Codeword

• Create a UE configuration structure and SL-BCH codeword.

• Generate the required length of the PRBS and scramble the PSBCH codeword with the PRBS sequence using `xor`.

• Modulate the logical scrambled data.

```ue = struct('NSLID',2); codeword = lteSLBCH(ue,ones(40,1)); psbchPrbs = ltePSBCHPRBS(ue,length(codeword)); scrambled = xor(psbchPrbs,codeword); txsym = lteSymbolModulate(scrambled,'QPSK');```

Descramble Recovered Codeword

• Generate the PSBCH PRBS in signed form.

• Descramble a vector of noisy demodulated symbols representing a sequence of soft bits. To do so, perform a pointwise multiplication between the PRBS sequence and the recovered data.

• Compare the transmitted codeword to the recovered codeword.

```rxsym = awgn(double(txsym),30,'measured'); softdata = lteSymbolDemodulate(rxsym,'QPSK'); scramblingSeq = ltePSBCHPRBS(ue,length(softdata),'signed'); descrambled = softdata.*scramblingSeq; isequal(codeword,descrambled > 0)```
```ans = logical 1 ```

The transmitted codeword matches the hard decision on the descrambled data.

## Input Arguments

collapse all

User equipment settings, specified as a parameter structure containing this field:

Physical layer sidelink synchronization identity, specified as an integer from 0 to 355. (${N}_{\text{ID}}^{\text{SL}}$)

Data Types: `double`

Data Types: `struct`

Number of elements in returned sequence, `seq`, specified as a nonnegative integer.

Data Types: `double`

Range of elements in returned subsequence, `subseq`, specified as a row vector of `[p n]`. The subsequence returns `n` values of the PRBS generator, starting at position `p` (0-based).

Data Types: `double`

Output sequence formatting, specified as `'binary'` or `'signed'`.

• `'binary'` maps `true` to 1 and `false` to 0.

• `'signed'` maps `true` to –1 and `false` to 1.

Data Types: `char` | `string`

## Output Arguments

collapse all

PSBCH pseudorandom scrambling sequence, returned as a logical column vector or a numeric column vector. `seq` contains the first `n` outputs of the physical sidelink broadcast channel (PSBCH) scrambling sequence. If you set `mapping` to `'signed'`, the output data type is `double`. Otherwise, the output data type is `logical`.

Data Types: `logical` | `double`

PSBCH pseudorandom scrambling subsequence, returned as a logical column vector or a numeric column vector. `subseq` contains the values of the PRBS generator specified by `pn`. If you set `mapping` to `'signed'`, the output data type is `double`. Otherwise, the output data type is `logical`.

Data Types: `logical` | `double`

Initialization value for PRBS generator, returned as a numeric scalar.

Data Types: `uint32`


text:
1. ## Integrate this:

$\int \frac{5x+3}{x^2+8x+25}$
I really dont know how to start this integral

$\int \frac{5x+3}{x^2+8x+25}$
Note that $\frac {5x + 3 }{x^2 + 8x + 25} = \frac {5x + 20 - 17}{x^2 + 8x + 25} = \frac {5x + 20}{x^2 + 8x + 25} - \frac {17}{x^2 + 8x + 25}$
thus, $\int \frac {5x + 3 }{x^2 + 8x + 25}~dx = \int \frac {5x + 20}{x^2 + 8x + 25}~dx - \int \frac {17}{x^2 + 8x + 25}~dx$
for the first integral, use the substitution, $u = x^2 + 8x + 25$
for the second, complete the square of the denominator and simplify so you can get it in the form to get the arctan integral, that is, try to get it in the form of $c \frac 1{u^2 + 1}$, where $c$ is a constant, and $u$ is a function of x. (it is easier than you think, try it)


text:
Stone duality

(0,1)-category

(0,1)-topos

duality

# Stone duality

## Idea

Stone duality is a subject comprising various dualities between space and quantity in the area of general topology and topological algebra.

## Particular cases

### Locales and frames

Perhaps the most general duality falling under this heading is that between locales (on the space side) and frames (on the quantity side). Of course, this duality is not very deep at all; the category Loc of locales is simply defined to be the opposite of the category Frm of frames. But there are several interesting dualities between subcategories of these.

### Topological spaces

Stone duality is often described for topological spaces rather than for locales. In this case, the most general duality is that between sober spaces and frames with enough points (which correspond to topological locales). In many cases, one requires the ultrafilter theorem (or other forms of the axiom of choice) in order for the duality to hold when applied to topological spaces, while the duality holds for locales even in constructive mathematics.

### Coherent spaces and distributive lattices

Any distributive lattice generates a free frame. The locales which arise in this way can be characterized as the coherent locales, and this gives a duality between distributive lattices and coherent locales. Note that one must additionally restrict to “coherent maps” between coherent locales. Also, at least assuming the axiom of choice, every coherent locale is topological, so we may say “coherent space” instead.

### Stone spaces and Boolean algebras

The duality which is due to Marshall Stone, and which gives its name to the subject, is the duality between Stone spaces and Boolean algebras. Specifically, a distributive lattice is a Boolean algebra precisely when the free frame it generates is the topology of a Stone space, and any continuous map of Stone spaces is coherent. Therefore, the category of Stone spaces is dual to the category of Boolean algebras. The Boolean algebra corresponding to a Stone space consists of its clopen sets.

This duality may be realized via a dualizing object as follows. The two-element Boolean algebra may be regarded as a Boolean algebra object $\mathbf{2}$ internal to the category of compact Hausdorff spaces $CH$. Thus, for each finitary Boolean algebra operation $\theta\colon \mathbf{2}^n \to \mathbf{2}$, there is a corresponding operation on the representable functor $CH(-, \mathbf{2}): CH^{op} \to Set$ given by

$CH(-, \mathbf{2})^n \cong CH(-, \mathbf{2}^n) \stackrel{CH(-, \theta)}{\to} CH(-, \mathbf{2})$

and therefore we obtain a lift

$CH(-, \mathbf{2}): CH^{op} \to BoolAlg$

A Stone space is by definition a totally disconnected compact Hausdorff topological space. Let $Stone \hookrightarrow CH$ denote the full subcategory of Stone spaces.

###### Theorem (Stone representation)

The representable functor restricts to an equivalence of categories $Stone^{op} \to BoolAlg$.

This important theorem can be exploited to give a third description of the free Boolean algebra on a set $X$:

$Bool(X) \cong CH(2^X, \mathbf{2})$

where $2$ denotes the 2-element compact Hausdorff space, and $2^X$ the product space $\prod_X 2$. Indeed, the inverse equivalence

$BoolAlg^{op} \to Stone$

takes a Boolean algebra $B$ to its spectrum, i.e., the space of Boolean algebra maps $Bool(B, 2)$ (this $2$ is the two-element Boolean algebra $\mathbb{Z}_2$!) equipped with the Zariski topology. Applied to $B = Bool(X)$, we have

$BoolAlg(B, 2) \cong Set(X, 2) = 2^X$

where the Zariski topology coincides with the product topology on $2^X$. By the equivalence, we therefore retrieve $Bool(X)$ as $CH(2^X, \mathbf{2})$. This in turn is identified with the Boolean algebra of clopen subsets of the generalised Cantor space $2^X$.

A second description of the inverse equivalence $BoolAlg^{op} \to Stone$ comes about through the yoga of ambimorphic objects. Namely, the Boolean compact Hausdorff space $\mathbf{2}$ can equally well be seen as a compact Hausdorff object in the category of Boolean algebras. Thus, the representable functor $Bool(-, \mathbf{2}): Bool^{op} \to Set$ lifts canonically to a functor

$Bool^{op} \to CH$

and in fact part of the Stone representation theorem is that this factors through the inclusion $Stone \hookrightarrow CH$ as the inverse equivalence $Bool^{op} \to Stone$. In particular this lift determines the topology, providing an description alternative to the description in terms of the Zariski topology (although they are of course the same).

An extension of the classical Stone duality to the category of Boolean spaces (= zero-dimensional locally compact Hausdorff spaces) and continuous maps (respectively, perfect maps) was obtained by G. D. Dimov (respectively, by H. P. Doctor) (see the references below).

### Stone spaces and profinite sets

Note that a finite Stone space is necessarily discrete, and these correspond to the finite Boolean algebras, i.e. $FinSet \simeq FinStoneTop \simeq FinBool^{op}$. However, since Boolean algebras form a locally finitely presentable category, we have $Bool \simeq Ind(FinBool) \simeq Pro(FinSet)^{op}$ (see ind-object and pro-object). In consequence, $StoneTop \simeq Pro(FinSet)$: i.e. Stone spaces are equivalent to profinite sets, in this context then often called profinite spaces.

One way of explaining this classical Stone duality is hence via the following sequence of equivalences of categories

$Bool \simeq Ind(FinBool) \simeq Ind(FinSet^{op}) \simeq Pro(FinSet)^{op} \,,$

where “FinSet” is the category of finite sets, “$Ind$” stands for ind-objects, “$Pro$” for pro-objects and ${}^{op}$ for the opposite category and the equivalence $FinSet^{op} \simeq FinBool$ is that discussed at FinSet – Opposite category.

### Stonean spaces and complete Boolean algebras

The category of Stonean spaces, i.e., compact extremally disconnected Hausdorff topological spaces equipped with open continuous maps as morphisms, is contravariantly equivalent to the category of complete Boolean algebras and continuous Boolean homomorphisms as morphisms.

### Profinite algebras

If $T$ is a Lawvere theory on $Set$, we can talk about Stone $T$-algebras, i.e. $T$-algebras with a compatible Stone topology, and compare the resulting category $T Alg(Stone)$ with the category $Pro(Fin T Alg)$ of pro-(finite $T$-algebras). The previous duality says that these categories are equivalent when $T$ is the identity theory. It is also true in many other cases, such as:

• the theory of groups, resulting in the rich theory of profinite groups
• the theories of semigroups and monoids
• the theory of rings (with or without 1)
• the theories of distributive lattices, Heyting algebras, and Boolean algebras
• the theory of $M$-sets, where $M$ is a finite monoid
• the theories of $R$-modules and $R$-algebras, where $R$ is a finite ring

However it is false for some $T$, such as:

• the theory of $\mathbb{N}$-sets, i.e. sets equipped with an endomorphism
• the theory of Jónsson-Tarski algebras
• the theory of lattices

All of these can be found in chapter VI of Johnstone’s book cited below.

The corresponding fact is also notably false for groupoids, i.e. $Gpd(Stone)$ is not equivalent to $Pro(FinGpd)$, in contrast to the case for groups. (Of course, groupoids are not described by a Lawvere theory.)

## References

• Olivia Caramello, A topos-theoretic approach to Stone-type dualities, arxiv/1103.3493 158 pp.

• G. D. Dimov, Some generalizations of the Stone Duality Theorem, Publ. Math. Debrecen 80/3-4 (2012), 255–293.

• H. P. Doctor, The categories of Boolean lattices, Boolean rings and Boolean spaces, Canad. Math. Bulletin 7 (1964), 245–252.

• Jacob Lurie, section A.1.1 of Spectral Algebraic Geometry.

There is a version in model theory, Makkai duality,

• M. Makkai, Stone duality for first-order logic, Adv. Math. 65 (1987) no. 2, 97–170, doi, MR89h:03067; Duality and definability in first order logic, Mem. Amer. Math. Soc. 105 (1993), no. 503

Other variants are in

• Henrik Forssell, First-order logical duality, Ph.D. thesis, Carnegie Mellon U. 2008, pdf

• Spencer Breiner, Scheme representation for first-order logic, Ph.D. thesis, Carnegie Mellon U. 2014, pdf

Discussion in E-∞ geometry is in

Discussion of an $(\infty, 1)$-version of Stone duality is in

Last revised on April 20, 2020 at 04:53:00. See the history of this page for a list of all contributions to it.


text:
# What is the probability/likelihood of a sample being drawn from a probability distribution over binary values

Suppose we have a known discrete probability distribution $$X$$ over $$\{0,1\}^k$$.
Given a sequence of binary values $$e = (e_1, ..., e_n)\text{, where } e_i\in \{0,1\}^k$$, what is the probability (or the likelihood) that $$e$$ was drawn from the Distribution $$X$$?
Is there for example a bound that tells us that such a sequence of samples $$e$$ is very unlikely to be drawn from $$X$$, and can thus be ignored?

• So you're asking for the density function of all binary sequences of length $n$? Seems like they would all be equiprobable from how you have described the problem. Are you thinking of a particular type of sequences (maybe sequences which have some sort of autocorrelation structure)? – Demetri Pananos Nov 13 '18 at 6:14
• @DemetriPananos Actually X is a noise distribution which is not uniform. So the values of $e_i$ are drawn non uniformally. X would have maybe some distribution function, but I'm aiming more at a general case. – abdul rahman taleb Nov 13 '18 at 7:29

At the moment you have not imposed any conditions on the distribution other than saying that it is a distribution on the set of binary vectors of length $$k$$. Your distribution can therefore be expressed as a probability vector over the $$2^k$$ possible states of the random variable. If the distribution is known, as you state, then presumably you can see exactly which outcomes are very unlikely.


text:
# Thread: Graphing with Parametric Equations

1. ## Graphing with Parametric Equations

I was looking at a chapter in my calculus book on parametric equations and I found the following exercise.

Graph:
$x=y-3y^3+y^5$

Usually I would just use the standard methods of calculus to graph the function $y=x-3x^3+x^5$, and then I would flip it over.

However, since it is in the chapeter on parametric equations, I figured there may be a clever way to parameterize this and graph it. However, I'm not seeing it.

2. Originally Posted by adkinsjr
I was looking at a chapter in my calculus book on parametric equations and I found the following exercise.

Graph:
$x=y-3y^3+y^5$

Usually I would just use the standard methods of calculus to graph the function $y=x-3x^3+x^5$, and then I would flip it over.

However, since it is in the chapeter on parametric equations, I figured there may be a clever way to parameterize this and graph it. However, I'm not seeing it.
$x = t - 3t^2 + t^5$

$y = t$

3. Originally Posted by skeeter
$x = t - 3t^2 + t^5$

$y = t$
I definitely noticed that one. But the parametric equations $x=t-3t^3+t^5$, $y=t$ don't make the shape of the curve any more obvious than $x=y-3y^3+y^5$. I was speculating that there may be a parametrization that would allow me to easily graph
y(t) and x(t) and then draw the curve based on those graphs. Otherwise, it would be pointless to parameterize at all.

4. I don't believe there is a magic bullet for an easy parametric in this case.

If someone can come up with one, they'll be sure to point it out.


text:
Alternating Current Important Questions

Question 1
Why it is not possible to have electrolysis by A.C?

Question 2
What is the value of the inductive reactance XL in a DC circuit?

Multiple Choice Questions

Question 3
Which of the these works on the principle of self-induction?
(a) Transformer
(b) Transistor
(c) choke
(d) None of these

Question 4
If the velocity of the object is zero,the acceleration would also be zero
(a) True
(b) False

Question 5
What is true of any AC circuit?
(a) None of these
(b) The power dissipation happens in resistance only in AC circuit
(c) The algebraic sum of the rms voltage across the different element of the circuit is equal to the applied rms voltage
(d) The algebraic sum of the instantaneous voltage  across the different element of the circuit  is equal to the applied instantaneous voltage

Question 6
Transformers are used
(a) To step up or step down ac voltage
(b) To convert a.c to d.c
(c) To step up d.c voltage
(d) To convert d.c to a.c

Question 7
What is true of AC GENERATOR?
(a) It works on the principle of electromagnetic induction
(b) The mechanical energy of the rotation is converted into electrical energy
(c) A pair of slip rings are used  in the AC generator
(d) All the above

Question 8
An alternating voltage V=200sin300t is applied across a series combination of R=10Ω and L=800mH.
(a) The impedence of the circuit is 240.2Ω
(b) Power factor is given by .05
(c) The peak current in the circuit is .8326 A
(d) None of these

Question 9
In an series LC circuit, the voltage across  the L and C are
(a) 180 out of phase
(b) 90 out of phase
(c) are in same phase
(d) 45 out of phase

Question 10
Let A and B are the heat production of the electric heater in AC and DC of same voltage.
(a) A=B
(b) A> B
(c) A <B
(d) none of the above

Question 11
An inductor having a reactance of 25 and a resistance R gives off heat at the rate of 10W when it carries a current of 0.50A (rms). What is the impedance of the inductor?

Question 12
At the moment depicted in the LC circuit the current is non-zero and the capacitor plates are charged (as shown in the figure below). The energy in the circuit is stored

(a) only in the electric field and is decreasing.
(b) only in the electric field and is constant.
(c) only in the magnetic field and is decreasing.
(d) only in the magnetic field and is constant.
(e) in both the electric and magnetic field and is constant.
(f) in both the electric and magnetic field and is decreasing.

Question 13
When an AC voltage of 220 V is applied to the capacitor C
(a) the maximum voltage between plates is 220 V.
(b) the current is in phase with the applied voltage.
(c) the charge on the plates is in phase with the applied voltage.
(d) power delivered to the capacitor is zero.

Question 14
A resistor of 50 ohm ,an inductor of 20/π H and a capacitor of 5/π μ F are connected in series to a voltage supply of 230 V-50 Hz, The impedance of the circuit is
(a) 10 ohm.
(b) 50 ohm.
(c) 100 ohm.
(d) 1000 ohm.

Question 15
Dimension of the impedance is
(a) $[M L^2 T^{-3} A^{-2}]$
(b) $[M L^2 T^{-3} A^{-1}]$
(c) $[M L^2 T^{-2} A^{-2}]$
(d) $[M L^2 T^{-1} A^{-2}]$

Question 16
In pure inductive circuit, the curves between frequency f and reciprocal of inductive reactance $\frac {1}{X_L}$ is

Question 17
if C and R denote Capacitance and Resistance, then dimension of CR is
(a) $[M^0 L^0 T^{-1} A^{0}]$
(b) $[M^0 L^0 T^{1} A^{0}]$
(c) $[M^0 L^0 T^{1} A^{-2}]$
(d) $[M^0 L^1 T^{1} A^{0}]$

Question 18
In a AC circuit ,the current is expressed as
$I= 200 \ sin 100 \pi t$.
In this circuit , the current rises from zero to peak value in time
(a) 1/200 sec
(b) 1/100 sec
(c) 1/300 sec
(d) 1/400 sec

Question 19 In LCR series circuit connected with an AC voltage source of frequency f, the current leads the voltage by 45°, The value of C is
(a) $\frac {1}{\pi f (2 \pi fL -R)}$
(b) $\frac {1}{2\pi f (2 \pi fL -R)}$
(c) $\frac {1}{\pi f (2 \pi fL +R)}$
(d) $\frac {1}{2\pi f (2 \pi fL +R)}$

Question 20
Quantity that remains unchanged in a transformer is
(a) Voltage
(b) Current
(c) Frequency
(d) none of these

Text Based Questions

Question 1
Show that in the free oscillations of an LC circuit, the sum of energies stored in the capacitor and the inductor is constant in time.
Question 2
What is the principle of transformer? . Explain how laminating the core of a transformer helps to reduce eddy current losses in it. Why the primary and secondary coils of a transformer are preferably wound on the same core

Question 3
A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced?

Question 4
Prove that an ideal capacitor in an A.C circuit does not dissipate power?
Question 5
(i) Explain briefy with the help if labelled diagram , the basic principle of the working of an AC generator
(ii) In an A.C generator , coil of N turns and Area A is rotataed at 'v' revolutions per sec in a uniform magnetic field B. Write the expression for the EMF produced
Question 6
(i) Derive the expression for the average power consumed in a series LCR circuit connected to AC source where phase difference between voltage and current is φ
(ii) Define the quality factor in an AC circuit?
Question 7
Why is choke coil neeed in the use of fluorescent tubles with AC mains?

Numerical Questions

Question 1
A 100 μF capacitor in series with a 40 &Omega resistance is connected to 100 V 60Hz supply
Calculate the following
a. Reactance
b. Impedance
c. Maximum Current in the circuit
d. rms voltage across the resistor and capacitor.Is the algebraic sum of these voltages more than the source voltage?. if Yes resolve the paradox

Question 2
An A.C voltage given by $V= 70 \sin 100 \pi t$ is connected across a pure resistor of 25 Ω. Find the following
a. Frequency of the source
b. rms current through the resistor

Question 3
An inductor L of inductance $X_L$ is connected in series with a bulb B and an ac source.
How would brightness of the bulb change when (i) number of turn in the inductor is reduced, (ii) an iron rod is inserted in the inductor and (iii) a capacitor of reactance $X_C = X_L$ is inserted in series in the circuit. Justify your answer in each case.

Question 4
The given circuit diagram shows a series LCR circuit connected to a variable frequency 230 V source. Here L=5.0 H, C=80 μ F, R=40 Ω

(a) Determine the source frequency, which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
(c) Determine the rms potential drops across the three elements of the circuit.
(d) How do you explain the observation that the algebraic sum of the voltage of the three elements obtained in (c) is greater than the supplied voltage?

Question 5
A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.


text:
# Tag Info

0

Diff-in-Diff is a parametric model. A rule of thumb for parametric models is that you should have at least 25-30 observations (different authors might disagree but it is around 30) per independent regressor used in your model (see discussion in Verbeek A Guide to Modern Econometrics pp 36). However, note this is a rule of thumb and as Verbeek points out (my ...

1

Elasticity of demand is normally considered in relation to market demand for a good, that is, the sum of individual customer demands. Different customers will probably respond in different ways to a price increase. If the increase is 5%, especially on a low value item, very likely many customers will not notice, some will notice but not change their buying ...

2

The standard elasticity still makes sense. Elasticity is not necessarily constant. Also, I am not sure what sort of formula you learned at your college but elasticity is rigorously defined (and also taught at college level for econ majors) as $EL = \frac{df(x)}{dx}\frac{x}{f(x)}$ for some function $f(x)$ (see Essential Mathematics for Economic Analysis by ...

1

According to advocates of MMT, the primary risk once the economy reaches full employment is inflation, which can be addressed by gathering taxes to reduce the spending capacity of the private sector. This statement is in accord with MMT, and it can be traced back to the concept of Functional Finance. One could do a search for Abba Lerner’s articles on ...

6

Does Modern Monetary Theory (MMT) provide a useful insight into how to manage the economy? That depends on your definition of MMT, because it is not generally agreed on what it even is. You will find some arguing it is just a macro/monetary theory (such as the Wikipedia page) but then I seen MMT proponents on this site arguing it is a whole new paradigm ...

1

Price elasticity of demand of $-0.5$ means that if price increases by $1\%$ demand decreases by $0.5\%$ (and vice versa in case of decrease). Consequently, if the equilibrium quantity with floor is $100$ and the price of lettuce is $25\%$ that means that eliminating the price flow - which will offset the $25\%$ will increase the quantity demanded by: $$0.5\... 0 There could be many answers, but the one that comes to my mind immediately is product differentiation. If you're selling exactly the same thing as everyone else, buyers are going to flock to the lowest price buyer, so you everyone will have to follow a price cut. But say you are selling ice cream cones on the boardwalk at the beach. Any little thing, such as ... 5 The \gamma on the RHS comes from applying chain rule when differentiating the second term with respect to a_i. Regarding elasticity, note that with a differentiable function f, the ratio f'(x)/f(x) can be interpreted as the percentage change in the value of f around x. So isolating \gamma from the FOC, you'll get an expression for elasticity, ... 6 It just comes from the derivative of profit function. I assume that a_i is the choice variable here so the derivative of \pi wrt a_i is (step by step):$$\frac{\partial \pi}{ \partial a_i} = \frac{\partial \pi}{ \partial a_i} [ \ln R(a_i) ] + \frac{\partial \pi}{ \partial a_i} [ \ln N_i(\gamma a_i, \gamma a_{-i}) ] \\ = \frac{1}{ R(a_i)} R'(a_i) ...

Top 50 recent answers are included


text:
# How to make a “function”?

I dropped out of school early when I was still a teenager and now I'm trying to take my GED. I'm really close to passing but I'm still having trouble understanding some concepts.

In the pre-test, there is this question:

Add one number to each column of the table so that it shows a function. Do not repeat an ordered pair that is in the table.

$$\begin{array}{c|c} x & y \\ \hline 6 & 6 \\ 3 & 8 \\ 9 & 12 \\ 7 & 8 \\ \fbox{?} & \fbox{?} \end{array}$$

$$\fbox{ 3 }\quad\fbox{ 6 }\quad\fbox{ 7 }\quad\fbox{ 8 }\quad\fbox{ 9 }\quad\fbox{ 12 }$$

I'm not entirely sure what a function is, and I have found some questions on SE explaining it, I'm not really getting any of it. It's all confusing.

So I was hoping if someone could explain this question to me, and what's the answer, and why. I'm hoping this could help me understand the concept.

• Ever heard of the vertical line test? – Daniel W. Farlow Jan 29 '15 at 21:15
• No, @induktio.. – mathnoob757 Jan 29 '15 at 21:18
• Check it out: en.wikipedia.org/wiki/Vertical_line_test. It will give you a pictorial way of thinking about Peter's answer. – Daniel W. Farlow Jan 29 '15 at 21:19
• Lots of good answers already, so I'll just leave a comment. My high school math teacher used to say "A function is like a vending machine. You push a button, it gives you food. Different buttons might give you the same food, but if you push the same button, it's going to give you the same thing every time." – hasnohat Jan 30 '15 at 12:08
• I'm voting up just in support of your efforts to earn a GED. – Matthew Leingang Jan 30 '15 at 20:36

There are enough answers explaining what a function is in general, so instead, I'd like to look at a particular function and discuss that.

Consider the graph of me walking to the grocery store to buy some pickled herring. I don't buy it all of the time, so when I go to the store, I really have to look for it.

On the horizontal axis is time ($t$), which represents the number of minutes since I left home, and on the vertical axis is the distance $(d)$, in meters, which represents the distance that I am from my house.

Initially, the distance (from home) is increasing. This represent my walk to the store.

Then it alternates between increasing and decreasing as I go up and down the aisles because the west side of the store is closer to my house than the east side of the store is. What is interesting about this part of my trip is that I can be the same distance from my house at different points in time. In fact, I am $350$m from home no fewer than $6$ times. At times, I may even be in the same location, but I can never be in $2$ places at once (this is really what the vertical line test will tell me).

Finally, are the last two legs of my trip. There is a horizontal leg to this function, which represents me waiting in line at the cashier, and not moving at all, hence not going further away nor closer to home. And lastly, I gradually decrease my distance until it reaches $0$m (at about the $28$ minute mark).

There are a multitude of other scientific relationships where one quantity depends on another that are modeled as functions. Think about throwing a ball up in the air. The ball can be at the same altitude at two different times (on the way up, and on the way down), but cannot be at two different altitudes at the same time (there's that vertical line test again).

By the way, the pickled herring was delicious.

• Stop lying. Pickled herring is never delicious. – KSmarts Jan 30 '15 at 21:07
• inlanders ಠ_ಠ sheesh – John Joy Jan 31 '15 at 12:51
• i am not sure if you should use distance or position in this instant. i always think of the distance as increasing even when you backtrack as you are in the aisles. – abel Feb 7 '15 at 21:45
• Certainly, were this answer presented in a university physics course, I agree, "position", or "displacement" would definitely have been more appropriate than "distance". As it is, the OP is just learning the concept of what a function is, and I don't think he would find that getting hung up on the semantics of "position" vs. "distance" would be very helpful. – John Joy Feb 7 '15 at 22:12

The other answers are good. I thought I would include these pictures from Wikipedia because, while you can visualize a function as a graph on $x,y$ axes, I also like visualizing functions as objects connected by arrows.

Here we have an example of a function that turns $x$'s into $y$'s just by traveling along the arrows.

This is a perfectly good function because there is no ambiguity. We can see that $f(1)$ gives us $D$, $f(2)$ gives us $C$, and $f(3)$ gives us $C$ as well.

This next example is not a function, because one of the inputs ($2$) has more than one output.

We can see that $f(1)$ is $D$, but what about $f(2)$? We can't decide what it is because there is more than one output. So this is not a function. Rather, it is a relation.

(In case you're disturbed that the outputs are letters and not numbers... functions can connect any two collections of "things". These things don't have to be numbers, but they often are.)

• A question: Is y = $x^{sin(x+y^2)}$ a function too? I don't get what the difference between functions and equations is. – HeWhoMustBeNamed Oct 18 '17 at 10:28

The key point here is that by definition, if $y$ is a function of $x$, then for each value of $x$ there is a single value of $y$ (Thanks to N.F. Taussig for phrasing).

Four $x$ values have already been determined, so we cannot reuse any of those. There are just two values left from our answer choices: $8$ and $12$. We can pick either of these for our $x$ value and anything we want for our $y$ value and we will have a function.

• I would suggest rephrasing your first sentence slightly to emphasize that if $f$ is a function of $x$, then for each value of $x$, there is exactly one value of $y$. – N. F. Taussig Jan 29 '15 at 21:20
• @N.F.Taussig Thanks! – Peter Woolfitt Jan 30 '15 at 1:40
• Actually we can also pick any of the $x$ values already used, say $7$ or $9$—provided that the corresponding $y$ value is the same as before, i.e. $(7,8)$ or $(9,12)$, respectively... – CiaPan Jan 30 '15 at 9:41
• @CiaPan The question prohibits that. – Peter Woolfitt Jan 30 '15 at 11:45
• @PeterWoolfitt Right, I must have missed that. I apologize for that, the text size was too small for me, I suppose. – CiaPan Jan 30 '15 at 11:48

None of the other answers are wrong, but given you are self-taught, it might be illuminating to see some wrong answers.

## $x=6$, $y=6$

This is wrong because you're not allowed to give an ordered pair that's already provided. This isn't helpful for learning what a function is, but it's probably something you already understood.

## $x=6$, $y=7$

This too is wrong. The reason it's wrong is that the other rows in the table have already defined the function's $y$ value for $x=6$. And it's not $y=7$. The essence of a function (as opposed to a relation) is that there is only one $y$ for each $x$.

In fact, the only wrong answers are those which

• give a row that's already provided
• give an $x$ that's already provided, and give a $y$ that is different to the one next to the corresponding $x$

So the correct answer is: any $x$ that's not already in the $x$ column, with literally any $y$ in the other column. The first bullet point is just to prevent "cheating"; the second point is really what functions are about.

The picture below may help you think about this in a different light. As Peter pointed out, a function can only have a particular $x$-value paired with a single $y$-value. In the picture, the points $(6,6), (3,8),(9,12),(7,8)$ have been plotted.

Given the definition of a function, you need to choose an $x$-value that corresponds with a $y$-value "not already taken". For example, the point $(3,6)$, where $x=3$ and $y=6$, will not work because $(3,8)$ is already a point and $6\neq 8$. In the context of the vertical line test and the picture below, this would mean a blue dot would appear directly below the first blue dot. Since a vertical line would intersect both dots, then we would not have a function (this is the vertical line test). Of course, you could do something trivial like list the point $(3,6)$ in the table again and you would have a function, but the test makers anticipated this "cheat" by saying, "Do not repeat an order pair that is in the table."

As Peter pointed out, using an $x$-value of $8$ or $12$ and any correspondent $y$-value will do. In the context of the picture, this works because there is no dot on the line $x=8$ or $x=12$.

Dirty explanation, but maybe it will give you more intuition.

It can't be wrong if all the values under $x$ in the table are different. That's because the $y$-value to a certain $x$-value must be unique.

Consider a simple analogy- "A person has to reach home from office by using his car as a transportation medium. Now the faster he drives his car the sooner he reaches his home i.e.. it takes lesser time to reach home. Here 'the speed of the car' becomes the FUNCTION of 'time taken to reach home' ".

If any value(say X) that depends on another value(say Y), then X is said to be a function of Y since the value of X depends on the value of Y. In the analogy above the time taken to reach home depends on the speed at which the car is traveling, hence time becomes a function of speed.

Let me talk mathematically,

Consider an equation: y=4+7x.

If I put the value of x as 2 then evaluating y, I get 4+7(2)=18=y. Here the value of y becomes 18 when the value of x is 2. Similarly the value of y is 39 when the value of x is 5. What we infer is that as the value of x changes the value of y also changes, in other words the value of y depends on the value of x. In other words y is said to be a function of x, as the value of y depends on that of x.

Mathematically it is written as f(x)=y or f(x)=4+7x(as given in the above example).

Where f=function, f(x)=function of x.

That's what 'function' means. Simple isn't it.

• you switched X and Y from those that are typically used and didn't highlight the difference between a function and a relation which is what the pre-test question is asking about. – Rick Jan 30 '15 at 18:38


text:
### The generalised binomial theorem

The generalised binomial theorem was discovered by Isaac Newton (1642–1727). It states that, for $$|x|<1$$,

$(1+x)^r = \sum_{k=1}^{ \infty}\dfrac{r(r-1)\dotsm(r-k+1)}{k!} x^k$

where $$r$$ is any real number.

For $$r = -1$$, this is

$\dfrac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 - x^5 + \dotsb.$

The expansion for $$(1-x)^{-1}$$ is

$\dfrac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + \dotsb.$

These are examples of infinite geometric series. We know that they are only convergent for $$|x| < 1$$.

Next page - History and applications - Applications


text:
I need some help with an electricity problem

1. Feb 9, 2005

benji

So I thought I understood this electricity stuff pretty well--I've read through the chapter once in full and skimmed over it a couple times. I've studied all the examples and still my problems aren't coming out correct. Here is one problem in particular that is giving me a lot of trouble:

What I did with this problem is draw a free-body diagram for the forces that act upon the sphere attatched to the string. I found the force of gravity by simply using F=mg, so I have the left side of the right triangle. Then I solved for the electrical force between the two spheres by using F=(k*q1*q2)/(r^2). That gives me the bottom of the right triangle. So shouldn't I just be able to use simple trig to find the hypotenuse of the triangle, thus finding tension on the string? I did this and my answer came out to be 0.615 N--the correct answer is 0.813 N. Where did I go wrong? I've checked and re-checked my math and I still can't figure out why. So that was part (b), for part (a) of the problem I just used trig--inverse tangent multiplied by the bottom side of the trianle over the tension. For this answer I'm WAY off--I get -2.3X10^-21 degrees when the correct answer is 15.4 degrees.

So I know I wrote a little novel here, but I could really use the help!

Thanks.

2. Feb 9, 2005

StatusX

I don't know if this is the problem, but did you keep in mind that the electrical force will increase as the ball swings closer to the charge?

3. Feb 9, 2005

MathStudent

Have you tried using newton's second law for the horizontal and vertical components of force on the sphere?

$$\Sigma F_x = ma_x$$

$$\Sigma F_y = ma_y$$

its hard to tell what went wrong because we can't see your equations, post any work you've done so far

Last edited: Feb 9, 2005
4. Feb 10, 2005

benji

I think I'll just ask about this one tomorrow in class, thanks anyways though. I should have just scanned my paper ;) .


text:
# Math Help - Evaluate limit

1. ## Evaluate limit

Evaluate: $\lim_t\in 1\int_{t}^{t^{\frac{1}{b+1}}\frac{1}{ln x}dx$

2. ## Re: Evaluate limit

Originally Posted by pankaj
Evaluate: $\lim_t\in 1\int_{t}^{t^{\frac{1}{b+1}}\frac{1}{ln x}dx$
The link doesn't work. Can you write it out?

-Dan

3. ## Re: Evaluate limit

Originally Posted by topsquark
The link doesn't work. Can you write it out?
That is due to the fact that the LaTeX compiler on this site is down.


text:
# Opel Corsa Diesel Usage

June 24, 2013
By

(This article was first published on Wiekvoet, and kindly contributed to R-bloggers)

I wanted to extend my car weight distribution calculation of June 16 from only 2000 to years 2000 to 2013. Unfortunately, come Sunday afternoon the code seemed too slow and not even the beginning of a post. So, I went on to another calculation I was working on; scraping Diesel usage data from the web and examining how it varied. The background of this is the repeated news items one reads, which state that actual fuel usage is much higher than reported. Since I may be in the market for an Opel Corsa Diesel, this was the car of choice.

### Sourcing the data

In contrast to my normal posts I won't be putting code for data scraping here. The reason is purely ethical; should I make it all too easy to get these data? I am not so sure webmasters like that idea. I took the data from http://www.spritmonitor.de/. The scraping techniques I took from r-bloggers, original was on statistical-research.com. At the end I had data where first records and relevant variables looked like shown below. The data was selected from 2010 to now with engine sizes 55, 70 and 96 kW. Engine sizes which were one kW off, were merged to the stated levels, because I had the feeling these were errors. In the end this resulted in 3713 records and 106 drivers.
usage
Driver Month lp100 buildyear         engine
1     63    06  4.38   BJ 2011  70 kW (95 PS)
2     63    06  3.55   BJ 2011  70 kW (95 PS)
3     63    05  4.16   BJ 2011  70 kW (95 PS)
4     63    05  3.50   BJ 2011  70 kW (95 PS)
5     63    05  3.90   BJ 2011  70 kW (95 PS)
6     63    04  3.54   BJ 2011  70 kW (95 PS)

### Model

The model used makes usage, liters per 100 km (lp100), a function of engine type, Month, driver and buildyear. It was also chosen to make the driver's standard deviation part of the model. The model is estimated in JAGS. The code is at the bottom of the post. Below the plot from jagsfit. From my point of view this looks reasonable. The trend at drivers is because the data was sorted when scraped.

### Engines

The data shown are conditional on their own drivers. It seems the strongest engine has the highest usage, while the weakest engine has second worst performance. This almost suggests that the 55 kW engine is just too small.

### Months

The data shown is the average usage over all drivers and cars. Summer is cheaper to drive than winter. However, the difference is smaller than between engines.

### Drivers

Drivers' means are conditional on their own engines, averaged over months and build years. In a sense, these are the expected long term usages, with shrinkage to the population means for drivers with limited data. It is easy possible to have a style for more than 6 liter per 100 km, one extreme driver expects more than 7 liter per 100 km, with the most economic car (upon examining the data notes, it may be this driver has loads of within city driving, average low speed, maybe loads of start/stop with air conditioning on). On the other hand, some drivers manage clearly less than 4.5 liter per 100 km.  3.8 l/100 km is what the best driver observes as his/her personal average, in the plot it is slightly shrunk upwards.

### Factory results and driver results

Based on these data one can certainly expect differences between what a factory presents as data and what is observed. If September weather gives 0.2 l/100 km less usage than December, the lower value will be given. The driving cycle for fuel usage has, as I read it, a fairly sedate speed. So, close or better than the best driver. The average driver by definition has a much higher usage than the best driver. A lease driver might be among the worst performers. After all, not paying your own fuel or repairs does not make for austere driving.

### R code

library(R2jags)
data_list <- list(NDriver=nlevels(usage$Driver), NBY=nlevels(usage$buildyear),
NEngine=nlevels(usage$engine), NMonth=nlevels(usage$Month),
N=nrow(usage))

data_list$Driver <- (1:data_list$NDriver)[usage$Driver] data_list$BY    <- (1:data_list$NBY)[usage$buildyear]
data_list$Engine<- (1:data_list$NEngine)[usage$engine] data_list$Month  <- (1:data_list$NMonth)[usage$Month]
data_list$lp100 <- usage$lp100
uu <- unique(data.frame(engine=usage$engine,cDriver=usage$Driver,Driver=data_list$Driver,Engine=data_list$Engine))
data_list$uDriver <- uu$Driver
data_list$uEngine <- uu$Engine
data_list$Driver1 <- uu$Driver[uu$Engine==1] data_list$Driver2 <- uu$Driver[uu$Engine==2]
data_list$Driver3 <- uu$Driver[uu$Engine==3] data_list$NDriver1 <- length(data_list$Driver1) data_list$NDriver2 <- length(data_list$Driver2) data_list$NDriver3 <- length(data_list$Driver3) mymod <- function() { for (i in 1:N) { fit[i] <- grandmean + MMonth[Month[i]] + MEngine[Engine[i]] + MBY[BY[i]] + MDriver[Driver[i]] lp100[i] ~dnorm(fit[i],tauOfDriver[Driver[i]]) } grandmean ~ dnorm(0,.001) MMonth[1] <- 0 for (i in 2:NMonth) { MMonth[i] ~ dnorm(0,tauMonth) } tauMonth ~ dgamma(0.001,0.001) sdMonth <- sqrt(1/tauMonth) MBY[1] <- 0 for (i in 2:NBY) { MBY[i] ~ dnorm(0,tauBY) } tauBY ~ dgamma(0.001,0.001) sdBY <- sqrt(1/tauBY) MEngine[1] <- 0 for (i in 2:NEngine) { MEngine[i] ~ dnorm(0,tauEngine) } tauEngine ~ dgamma(0.001,0.001) sdEngine <- sqrt(1/tauEngine) MDriver[1] <- 0 for (i in 2:NDriver) { MDriver[i] ~ dnorm(0,tauDriver) } tauDriver ~ dgamma(0.001,0.001) sdDriver <- sqrt(1/tauDriver) for (i in 1:NDriver) { tauOfDriver[i] <- pow(sdOfDriver[i],-2) sdOfDriver[i] ~ dlnorm(mu.lsdD,tau.lsdD) } mu.lsdD ~ dnorm(0,.0001) tau.lsdD <- pow(sigma.lsdD,-2) sigma.lsdD ~ dunif(0,100) for (i in 1:NDriver) { meanDriver[i] <- grandmean + MDriver[uDriver[i]] + mean(MMonth[1:NMonth]) + mean(MBY[1:NBY]) + MEngine[uEngine[i]] } for (i in 1:NMonth) { meanMonth[i] <- grandmean + mean(MDriver[1:NDriver]) + MMonth[i] + mean(MBY[1:NBY]) + mean(MEngine[1:NEngine]) } for (i in 1:NBY) { meanBY[i] <- grandmean + mean(MDriver[1:NDriver]) + mean(MMonth[1:NMonth]) +MBY[i]+ mean(MEngine[1:NEngine]) } for ( i in 1:NDriver1) { MDriver1[i] <- MDriver[Driver1[i]] } for ( i in 1:NDriver2) { MDriver2[i] <- MDriver[Driver2[i]] } for ( i in 1:NDriver3) { MDriver3[i] <- MDriver[Driver3[i]] } meanEngine[1] <- grandmean + mean(MDriver1) + mean(MMonth[1:NMonth])+ mean(MBY[1:NBY]) +MEngine[1] meanEngine[2] <- grandmean + mean(MDriver2) + mean(MMonth[1:NMonth])+ mean(MBY[1:NBY]) +MEngine[2] meanEngine[3] <- grandmean + mean(MDriver3) + mean(MMonth[1:NMonth])+ mean(MBY[1:NBY]) +MEngine[3] } model.file <- file.path(tempdir(),'mijnmodel.txt') write.model(mymod,con=model.file) inits <- function() list( grandmean = rnorm(1,5,1), MDriver = c(NA,rnorm(data_list$NDriver-1)) ,
MBY = c(NA,rnorm(data_list$NBY-1)) , MEngine = c(NA,rnorm(data_list$NEngine-1)) ,
MMonth = c(NA,rnorm(data_list$NMonth-1)) , tauDriver = runif(1,1,3), tauBY = runif(1,1,3), tauEngine=runif(1,1,3), tauMonth=runif(1,1,3), sdOfDriver = runif(data_list$NDriver,1,3)
)

parameters <- c('sdOfDriver','meanMonth','meanDriver','meanBY',
'meanEngine','sdDriver','sdMonth','sdEngine')
jagsfit <- jags(data=data_list,inits=inits,model.file=model.file,
parameters.to.save=parameters,n.chains=4,DIC=FALSE,n.iter=20000,
progress.bar='none')
png('jagsfit.png')
plot(jagsfit)
dev.off()

jagsfit.mc <- as.mcmc(jagsfit)
fitsummary <- summary(jagsfit.mc)
stats <- fitsummary$statistics quants <- fitsummary$quantiles

library(ggplot2)

Engines <- as.data.frame(cbind(stats[grepl('meanEngine',rownames(stats)),],
quants[grepl('meanEngine',rownames(quants)),]))
Engines$type <- levels(usage$engine)
Engines
limits <- aes(ymax = 25%, ymin=75%)
p <- ggplot(Engines, aes(y=Mean, x=type))
dodge <- position_dodge(width=0.9)
png('engines.png')
p + geom_point(position=dodge) +
geom_errorbar(limits, position=dodge, width=0.25) +
coord_flip()
dev.off()

Months <- as.data.frame(cbind(stats[grepl('meanMonth',rownames(stats)),],
quants[grepl('meanMonth',rownames(quants)),]))
Months$Month <- levels(usage$Month)
limits <- aes(ymax = 25%, ymin=75%)
p <- ggplot(Months, aes(y=Mean, x=Month))
dodge <- position_dodge(width=0.9)
png('Months.png')
p + geom_point(position=dodge) +
geom_errorbar(limits, position=dodge, width=0.25) +
coord_flip()
dev.off()

Driver <- as.data.frame(cbind(stats[grepl('meanDriver',rownames(stats)),],
quants[grepl('meanDriver',rownames(quants)),]))
Driver <- cbind(uu,Driver)
limits <- aes(ymax = 25%, ymin=75%)
p <- ggplot(Driver, aes(y=Mean, x=Driver,colour=engine))
dodge <- position_dodge(width=0.9)
png('driver.png')
p + geom_point(position=dodge) +
geom_errorbar(limits, position=dodge, width=0.25) +
coord_flip()
dev.off()

ag <- aggregate(usage$lp100,by=list(Driver=usage$Driver),mean)
min(ag\$x)


text:
+0

# Simplify this.

0
215
2
+183

Simplify this equation: x^2+2x-1+x2+6y^9

PsychoBaker  Jun 21, 2017
Sort:

#1
+1807
+1

I will assume for this problem that $$x2\hspace{1mm}\text{means}\hspace{1mm}{x^2}$$. In this problem, there is only one like term in the expression $$x^2+2x-1+x^2+6y^9$$; the 2 x^2-terms. Let's combine them:

$$x^2+2x-1+x^2+6y^9=2x^2+2x-1+6y^9$$

Normally, I would stop here, but sometimes you should order your terms in term of its individual degree. If this is the case, your final answer is:

$$6y^9+2x^2+2x-1$$

TheXSquaredFactor  Jun 22, 2017
#2
+183
0

Thank you.

PsychoBaker  Jun 22, 2017

### 8 Online Users

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text:
# Describe the level surfaces of the function. f(x,y,z)=x+3y+5z

Describe the level surfaces of the function.
$$\displaystyle{f{{\left({x},{y},{z}\right)}}}={x}+{3}{y}+{5}{z}$$

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

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mylouscrapza
$$\displaystyle{f{{\left({x},{y},{z}\right)}}}={x}+{3}{y}+{5}{z}$$
So if we have some constant k, then
$$\displaystyle{x}+{3}{y}+{5}{z}={k}$$
Recall that the equation of a plane is $$\displaystyle{a}{x}+{b}{y}+{c}{z}={d}$$, with a surface normal of $$\displaystyle{<}{a},{b},{c}{>}$$. Varying l wi;; result in different parallel planes. So f describes a familly of parallel planes.
Result: family of parallel planes


text:
# Continuous Function

## Homework Statement

find k for the function so it is continuous and differentiable.
x^2-1 x<=1
k(x-1) x>1

## The Attempt at a Solution

k(x-1)=0 for x=1
k(0)=0
k = 0/0?

How do I know if the function is differentiable?

Homework Helper
Work on making it differentiable at x = 1 first - if you can, then you know it will be continuous there. (what are the left / right-hand derivatives)

Actually, that function is continuous for any value of k. The only point you have to worry about, both for continuity and differentiability, is the point x=1. You can explicitly show that the function is continuous at 1 by computing the limit of f as x approaches 1 and show that it equals f(1). You'll see it doesn't matter what k is for continuity.

To worry about differentiability, all you need to check is that the slope of the function k(x-1) at x=1 matches up with the slope of x2-1 at x=1. Remember that for a function to be differentiable at a point, the limit of the difference quotient at that point must exist. By checking that the slopes match up, you are checking that the left and right handed limits of the difference quotient equal each other (i.e., checking that the limit exists). The function is differentiable everywhere else since it is defined there by polynomials.

f(x) = x^2-1
f'(1) = 2(1) = 2

f(x) = k(x-1)
f'(1) = k = 2

Therefore k=2 and is differentiable at this point now, correct?


text:
## Stream: general

### Topic: cancel add + canonically ordered?

#### Peter Nelson (Apr 01 2021 at 12:22):

I'm trying to prove something about an ordered add_comm monoid which is both canonically ordered and has additive cancellation. But when I include both instances as hypotheses, it seems like lean is getting confused about which zero and which has_le.leI mean. I'm not sure whether this is just some sort of diamond problem, or I am actually handling two unrelated structures on the same set.

For example, I get an error when I try to invoke the theorem zero_le from canonically_ordered_monoid; see below. It works fine if I remove the ordered_cancel_add_comm_monoid instance. Can someone advise me how to set things up so I have the set of assumptions from both instances on a common ordered_add_comm_monoid?

import tactic

example {M : Type*}
(x : M):
false :=
begin
have : 0 ≤ x := zero_le x,
-- fails, but works ok if the first instance is removed from hypotheses.
end

#### Peter Nelson (Apr 01 2021 at 12:24):

Here is the error in all its glory.

invalid type ascription, term has type
@has_le.le M
(@preorder.to_has_le M
(@partial_order.to_preorder M
0
x
but is expected to have type
@has_le.le M
(@preorder.to_has_le M (@partial_order.to_preorder M (@ordered_cancel_add_comm_monoid.to_partial_order M _inst_1)))
0
x

#### Kevin Buzzard (Apr 01 2021 at 12:24):

Yeah you're putting two unrelated monoid structures on M

#### Peter Nelson (Apr 01 2021 at 12:25):

I thought so - is there a way to take the union of both sets of axioms for these structures without doing this?

#### Kevin Buzzard (Apr 01 2021 at 12:26):

You either make a new typeclass with instances implying that it's both of the other typeclass, or just use one typeclass and add the extra axioms as assumptions and then make an instance of the other typeclass manually

#### Kevin Buzzard (Apr 01 2021 at 12:27):

Your error in all its glory just says "I have _inst_1 on the left hand side and _inst_2 on the right hand side and these are unrelated" (look at the end of each term on either side of the =)

#### Kevin Buzzard (Apr 01 2021 at 12:29):

Didn't we decide that one typeclass around here was a load of old nonsense recently?

#### Eric Wieser (Apr 01 2021 at 12:30):

was docs#ordered_semiring that nonsense?

#### Peter Nelson (Apr 01 2021 at 12:57):

What is the value of the extends that is used in defining various strengthenings of ordered_add_comm_monoid? Would this kind of problem not go away if canonically_ordered_add_monoid was defined like this:

class canonically_ordered_add_monoid (α : Type*) [ordered_add_comm_monoid α] [order_bot α] :=
(le_iff_exists_add : ∀ a b : α, a ≤ b ↔ ∃ c, b = a + c)

class canonically_ordered_add_monoid (α : Type*) extends ordered_add_comm_monoid α, order_bot α :=
(le_iff_exists_add : ∀ a b : α, a ≤ b ↔ ∃ c, b = a + c)

?

#### Peter Nelson (Apr 01 2021 at 13:00):

By way of comparison, in my one and only PR, I added a typeclass defined as follows:

class has_exists_add_of_le (α : Type u) [ordered_add_comm_monoid α] : Prop :=
(exists_add_of_le : ∀ {a b : α}, a ≤ b → ∃ (c : α), b = a + c)

and this allowed the instances to fire seamlessly in settings where we wanted some structure plus this one extra axiom. This made it through the PR process - is there some reason it should have been done the other way?

#### Peter Nelson (Apr 01 2021 at 13:01):

(using extends, I mean)

#### Peter Nelson (Apr 01 2021 at 13:08):

Is this because canonically_ordered_add_monoid is extending two different typeclasses?

#### Eric Wieser (Apr 01 2021 at 13:37):

I think there are two arguments against that approach

1. Something something exponential blowup of term length
2. People don't like writing [ordered_add_comm_monoid α] [order_bot α] [canonically_ordered_add_monoid α] in their lemma statements because it's long. A more extreme example would be [has_mul α] [has_one α] [mul_one_class α] [mul_assoc_class α] instead of [monoid α].

#### Greg Price (Apr 01 2021 at 16:52):

Eric Wieser said:

1. People don't like writing [ordered_add_comm_monoid α] [order_bot α] [canonically_ordered_add_monoid α] in their lemma statements because it's long. A more extreme example would be [has_mul α] [has_one α] [mul_one_class α] [mul_assoc_class α] instead of [monoid α].

Or to expand out what that approach might mean for a canonically ordered, cancellative, additive monoid: something like [has_add α] [has_zero α] [add_zero_class α] [add_assoc_class α] [add_comm_class α] [preorder α] [partial_order α] [ordered_add_comm_monoid α] [has_bot α] [order_bot α] [canonically_ordered_add_monoid α] [add_left_cancel_class α] [add_right_cancel_class α] [ordered_cancel_add_comm_monoid α].

#### Greg Price (Apr 01 2021 at 16:54):

I.e. that's what you get if not only is canonically_ordered_add_monoid defined like this:

class canonically_ordered_add_monoid (α : Type*) [ordered_add_comm_monoid α] [order_bot α] :=
(le_iff_exists_add : ∀ a b : α, a ≤ b ↔ ∃ c, b = a + c)

but also ordered_add_comm_monoid is defined in a similar way, and so on.

#### Eric Wieser (Apr 01 2021 at 17:06):

Right, exactly. Having said that, in lean 4 it's looking like you could define something like

class monoid extends has_mul α, has_one α, mul_one_class α, mul_assoc_class α

instance monoid.of_parts [p : has_mul α] [q : has_one α] [r : mul_one_class α] [s : mul_assoc_class α] : monoid α := {..p, ..q, ..r, ..s}

and it won't cause a loop, unlike lean 3. That means that you are free to switch back and forth between the shorthand names when you don't need to be precise, and the long list when you need something with no shorthand name

#### Greg Price (Apr 01 2021 at 17:36):

Here, I think it would mean you could write something like [add_comm_monoid α] [order_bot α] [is_canonically_ordered_add_monoid α] [is_ordered_cancel_add_comm_monoid α].

I.e. you'd rewrite a definition like the existing

class canonically_ordered_add_monoid (α : Type*) extends ordered_add_comm_monoid α, order_bot α :=
(le_iff_exists_add : ∀ a b : α, a ≤ b ↔ ∃ c, b = a + c)

to something like

class is_canonically_ordered_add_monoid (α : Type*) [ordered_add_comm_monoid α] [order_bot α] :=
(le_iff_exists_add : ∀ a b : α, a ≤ b ↔ ∃ c, b = a + c)

and similarly ordered_add_comm_monoid and so on.
And both the is_canonically_ordered_add_monoid and is_ordered_cancel_add_comm_monoid would end up using the common [add_comm_monoid α] and [order_bot α], and at the same time you'd get to invoke things that needed a [canonically_ordered_add_monoid α], because Lean would be able to produce the needed instance.


text:
# DatagramOutputFile¶

from panda3d.core import DatagramOutputFile

class DatagramOutputFile

Bases: DatagramSink

This class can be used to write a binary file that consists of an arbitrary header followed by a number of datagrams.

Inheritance diagram

__init__()
close()None

Closes the file. This is also implicitly done when the DatagramOutputFile destructs.

open(file: FileReference)bool

Opens the indicated filename for writing. Returns true if successful, false on failure.

open(filename: Filename)bool

Opens the indicated filename for writing. Returns true on success, false on failure.

open(out: ostream, filename: Filename)bool

Starts writing to the indicated stream. Returns true on success, false on failure. The DatagramOutputFile does not take ownership of the stream; you are responsible for closing or deleting it when you are done.

property streamostream

Returns the ostream represented by the output file.

writeHeader(header: str)bool

Writes a sequence of bytes to the beginning of the datagram file. This may be called any number of times after the file has been opened and before the first datagram is written. It may not be called once the first datagram is written.


text:
### Author Topic: Jump  (Read 14745 times)

0 Members and 1 Guest are viewing this topic.

#### mrmprog

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##### Jump
« on: July 27, 2011, 02:51:18 am »
This is my first attempt at something big in Axe. You jump around and try not to fall. Yay! I am going to try and make this a full game.

It looks better on calc. Thanks to Eeems, Darl, Hayleia, FinaleTI, Builderboy,and p2 for help.  I need to somehow make it so impossible places can't happen. If anyone has a solution to this problem, please let me know. I will be fine tuning things, improving code and the such meanwhile. Please give me feedback, as this is my first Axe project.

Controls:2nd to jump, left and right to move, enter to select stuff, clear to exit or skip stuff.

« Last Edit: November 15, 2011, 09:08:25 pm by mrmprog »

#### Sitarknight

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##### Re: mrmprog's game
« Reply #1 on: July 27, 2011, 04:57:47 am »
Looks pretty good man, I like it
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#### mrmprog

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##### Re: mrmprog's game
« Reply #2 on: July 27, 2011, 09:28:06 am »
Glad you like it Yay, updates! I made platforms more "regular", a new sprite for falling/jumping, you can jump further, and dying is... different.

Next: Highscores, a menu, and a death animation.
Maybe Next: Title screen.
« Last Edit: July 29, 2011, 05:36:56 am by mrmprog »

#### p2

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##### Re: mrmprog's game
« Reply #3 on: July 27, 2011, 10:02:13 am »
looks not bad - it looks good!
You may add, that the lines are just possible to be at every trind line of the screen, or something like this. (then they are not like giant, grey blocks of 4 or 5 lines)

I think, A ball or a smiley would be a better than a stickfigure. (you can't really see, what the pic is; ih's hard in cause of the grey lines)
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#### mrmprog

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##### Re: mrmprog's game
« Reply #4 on: July 27, 2011, 10:03:59 am »
Yeah, I might change that. What would look best in your opinion?

Edit: I made some smiley sprites and they look much better, thanks

Tommorow, I will post another screenshot with the new sprites.
« Last Edit: July 27, 2011, 10:11:24 am by mrmprog »

#### p2

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##### Re: mrmprog's game
« Reply #5 on: July 27, 2011, 10:30:45 am »
I thought of something like this:

Quote
--  -  - -  - --- -

-- - ---- - --- --
-- - - - -  --  --

- - -- - ---- --
----- -- - ---- -

- - --  --- -- -

--- --- --  -- -

--- --- - -- --
-- ---- - - ----

If you make it like this:
Code: [Select]
:rand^16 -> A           //A is the Y-coord of the line4*(A) -> A
the game would look like that:
Quote
--  -  - -  - --- -

-- - ---- - --- --

-- - - - -  --  --

- - -- - ---- --

----- -- - ---- -

- - --  --- -- -

- - --  --- -- -

--- --- --  -- -

--- --- - -- --

-- ---- - - ----

It will be better like that, because else there are giant grey blocks of 3 or 4 lines, each one intersecting each other.

Another idea:
Try to use
:pt-off(X,Y,[FFFFFFFFFFFFFFFF])
:pt-on(X,Y,[code of smiley])
It'll make a white space behind the smiley, so that you can see it better.
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#### Hayleia

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##### Re: mrmprog's game
« Reply #6 on: July 27, 2011, 11:36:56 am »
Another idea:
Try to use
:pt-off(X,Y,[FFFFFFFFFFFFFFFF])
:pt-on(X,Y,[code of smiley])
It'll make a white space behind the smiley, so that you can see it better.
For that idea, a
Pt-Off(X,Y,[code of smiley or pointer]
would be enough as "Pt-Off" puts a sprite but erases what there is at the location before.
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Sorry if I answer with something that seems unrelated, English is not my primary language and I might not have understood well. Sorry if I make English mistakes too.

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##### Re: mrmprog's game
« Reply #7 on: July 27, 2011, 12:32:52 pm »
Another idea:
Try to use
:pt-off(X,Y,[FFFFFFFFFFFFFFFF])
:pt-on(X,Y,[code of smiley])
It'll make a white space behind the smiley, so that you can see it better.
For that idea, a
Pt-Off(X,Y,[code of smiley or pointer]
would be enough as "Pt-Off" puts a sprite but erases what there is at the location before.

No it doesn't.  Pt-Off() simply XORs the sprite in parameter slot 3 against the said place on the buffer, it doesn't clear the space behind it first.

#### Hayleia

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##### Re: mrmprog's game
« Reply #8 on: July 27, 2011, 02:18:46 pm »
No it doesn't.  Pt-Off() simply XORs the sprite in parameter slot 3 against the said place on the buffer, it doesn't clear the space behind it first.
Wut ? Weren't you confused with Pt-Change ?
Here is a screenie of Commands.htm
Or maybe I'm wrong because of a grayscale story. I don't know anything to grayscale.
« Last Edit: July 27, 2011, 02:20:06 pm by Hayleia »
I own: 83+ ; 84+SE ; 76.fr ; CX CAS ; Prizm ; 84+CSE
Sorry if I answer with something that seems unrelated, English is not my primary language and I might not have understood well. Sorry if I make English mistakes too.

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##### Re: mrmprog's game
« Reply #9 on: July 27, 2011, 04:01:37 pm »
Oh, my bad, you are correct.  I remember that Pt-Off used to mean it XORed it against the buffer,  I guess that has recently changed.
« Last Edit: July 27, 2011, 04:02:46 pm by Ashbad »

#### calcdude84se

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##### Re: mrmprog's game
« Reply #10 on: July 27, 2011, 04:37:14 pm »
Oh, my bad, you are correct.  I remember that Pt-Off used to mean it XORed it against the buffer,  I guess that has recently changed.
I don't think it's ever been like that. Pt-Change( has always been used for XOR.
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I'll put it online when it does something.

#### LincolnB

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##### Re: mrmprog's game
« Reply #11 on: July 27, 2011, 06:20:43 pm »
I need to ... make a counter of some sort so that it shifts the screen every few times through the loop, instead of every time.

Try this:

Code: [Select]
.use the variable A for the counter you need.Main game loop:1->ARepeat getkey(15)A+1->AIf A=41->AEndIf A=1Vertical +Vertical +{r}End.Rest of your code goes here..End
Let me know how that works out. Also, if you wanted to have multiple difficulty levels, you could have another variable , like B or something, and instead of writing:

Code: [Select]
If A=41->AEnd
, you could write:

Code: [Select]
If A=B1->AEnd
, where B is the difficulty level, 4 being easiest / slower because it makes it wait longer to scroll, and 2 being the fastest (where it scrolls every iteration of the main game loop).

Let me know if I didn't explain that very well.
« Last Edit: July 27, 2011, 06:27:14 pm by buttsfredkin »
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#### mrmprog

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##### Re: mrmprog's game
« Reply #12 on: July 27, 2011, 10:46:55 pm »
@buttsfredkin Thank you for the code, but I already did that using the score counter. You will see it soon.
Also, about the pixeloff(, if I do that, wont it erase the lines behind the sprite? The lines are just drawn every few cycles of the loop, so I don't think I can redraw them if they are overwritten.
« Last Edit: July 28, 2011, 03:42:53 am by mrmprog »

#### Hayleia

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##### Re: mrmprog's game
« Reply #13 on: July 28, 2011, 04:00:13 am »
Yes, the lines would be erased
But what do you use ? Do you use Pt-On or Pt-Change ? Because the guy's pixels seem to be inverted if there is a line behind him, so it looks like a Pt-Change.
I own: 83+ ; 84+SE ; 76.fr ; CX CAS ; Prizm ; 84+CSE
Sorry if I answer with something that seems unrelated, English is not my primary language and I might not have understood well. Sorry if I make English mistakes too.