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# Tag Info

62

What you're missing here is the chance of playing at all, given that the game ends when someone finds the prize. (or, chance of finding a prize goes to 0, which is the same thing) Person 1 has a 100/100 chance of playing, and a 1/100 chance of winning. Person 2 has a 99/100 chance of playing, and a 1/99 chance of winning. Person 3 has a 98/100 chance of ...

37

Actually,

23

Other people have already given correct answers, but I wanted to suggest a different way of thinking about the question that involves less calculation:

21

A difference between the Monty Hall scenario and the train platform scenario is that After you remember that the train will definitely not be leaving from platform 2,

18

You switch your door, because you have double the chance of getting the car Explanation: When you pick a door, there is a 1 in 3 chance that the door contains the car. Let's examine that case of when you originally picked the car, and when you originally picked a blank separately. Originally picked the car(1/3 chance) In this case, the host can open ...

14

Switch. There's a $2/3$ chance that you'll choose a goat, and a $1/3$ chance you'll choose the car. If you chose a goat, after switching you will win a car. (Guaranteed) If you chose a car, after switching you will win a goat. (Guaranteed) Therefore, the chance you'll win a car is $2/3$ if you switch. Simply put, the chance you picked the car before the ...

13

For simplicity, I'm going to refer to the associates as knight, knave, and joker. Also, I'm assuming that they won't reveal the location of the money. If they did, then obviously you would switch to that once they reveal it. Assuming that there is an equal chance of any of the three associates coming up and talking to you, we have a two in three chance of ...

12

I've answered this over on Math.SE, so I'll just quote most of that. Suppose we have $n$ doors, with a car behind $1$ of them. The probability of choosing the door with the car behind it on your first pick, is $\frac{1}{n}$. Monty then opens $k$ doors, where $0\leq k\leq n-2$ (he has to leave your original door and at least one other door closed). ...

12

Here's the formal expression of what we seek. Let $C$ be the event "chose the door with the car" and $R$ be the event "revealed the door with the car". Now, assuming that the door opened is random (but not the one the player chose), we have  Pr(C) = \frac13\\ Pr(R|C) = 0\\ Pr(R|\bar C) = \frac12\\ Pr(C|\bar R) = \frac{Pr(\bar R|C)Pr(C)}{Pr(\bar R|C)Pr(C)+...

10

Here's how the original Monty Hall worked: You pick a door. (2/3 chance of it being a goat, 1/3 chance of it being a car) Monty Hall opens a door that he knows has a goat in it, and then asks you if you'd like to switch to the other door. 2/3 of the time, the door Monty Hall opened is the only other door with a goat - because 2/3 of the time, you picked a ...

9

Without assuming anything more? No, we have absolutely no idea whether it is beneficial to switch or not. If the host knows what is behind each door and is malicious, then he will try to trick us. But if we know he will try to trick us then we can use that against him. Unless he knows that we know, and so on and so on. In the end that means there is no ...

7

If I want, I can always win my money back! Not switching is academic: I have a 1/3 chance of picking the 20k, 1/3 of picking the -3k and 1/3 of picking the +7.5k. My expected win is 6.6k-1k+2.5k = 8.1k. Blindly switching to the closed door raises my chances. The original Month Hall problem teaches me that I have 2/3 chance of picking the 20k by switching ...

7

I will stay with the door I have. I am banking on a malicious host who takes advantage of my Monty Hall familiarity and only offers me a switch when I am pointing to the car.

6

Since It should be clear that

6

6

You have a: EDIT: Now I have better understanding of what you've after. You have a robot:

5

Let me give it a try. Attempt at the bonus

5

The accepted answer conflates "the host's knowledge and motivations are unknown" with "the host's actions are uniform and random", which is at best a debatable assumption. I actually think the correct answer to this question is that there is not enough information to answer it as phrased. For example, if I flip a coin outside of your view and ask what are ...

5

The question is vague, and assumptions must be made that drastically change the possible answers. EDIT: Analysis added below. Am I missing something?

4

Let's assume that the Host randomly decides whether to reveal a goat and offer a switch: if your first door has a car, he reveals with probability $p$, and if it was first a goat, then with probability $q$. This is the best we can do: we don't know whether the Host planned to open a door or not. Note that $p=q=1$ gives the original Monty Hall scenario. Let \$...

4

Well in this case the game becomes more or a Poker game because you don't know if the host followed a rule or not. There are two (already well-known) cases: The host chose the other door randomly (he didn't know what was behind that one): in this case it doesn't change anything if you change or not: each door has a 1/2 chance. The host chose on purpose a ...

4

Original answer (now Part 1) If we assume each team has an equal chance of winning any particular match and that you know A plays B and C plays D in the first round. Then Second answer (now Part 2),

3

When doing Bayesian updating on new information, we should change our probability assignment for a hypothesis only if the probability of seeing the new information depends on whether the hypothesis is true. Suppose we choose Door A, and Monty Hall reveals a goat behind Door B. If Monty Hall always randomly chooses between the unchosen doors when we ...

3

If he always opens one of the doors you did not pick at random, then the door he opens should contain a car 1/3 of the time, a sheep 1/3 of the time, and a goat 1/3 of the time. If the animals and car were placed at random, we had the following possible cases to consider at the start, each of which had equal probability: S G C S C G G S C G C S C S G C G ...

3

The other answers are correct. I just wanted to add here the "long way" of doing it. I mean all the possible cases. Case 1. Car Goat Goat Case 1.1. You choose door 1. Monty opens| switch | don't switch 2 | Lose | Win 3 | Lose | Win Case 1.2. You choose door 2. Monty opens| switch | don't switch 3 | Win | ...

2

I always start with box 1. I'm assuming he picks randomly if neither box 2 or 3 contains the 20,000. If he shows the -3000 box, you're better off switching. Case 1: Probability 33% - Box 1 contains 20k, remaining box contains 7500. I win 7500 by switching. Case 2: Probability 66% - Box 1 contains 7500, remaining box contains 20k. I win 20,000 by ...

2

EDIT: to make the question a little more specific, what is the ideal strategy for you, that maximizes your chances of getting the car in the worst case, no matter what the hosts motives are. If I understood your question correctly, "in the worst case" is the key here. So the correct answer is a strategy where you maximize your winning chance over all ...

2

EDIT: to make the question a little more specific, what is the ideal strategy for you, that maximizes your chances of getting the car in the worst case, no matter what the hosts motives are. You can't sweep the host's motives under the rug like some small, unimportant detail. The player's strategy is completely dependent on the host's strategy. HOST ...

2

I will answer your edited question. In case that you have prior knowledge that the ones before you didn't find the price then the better turn is...the last. It's a little underwhelming but basically you have a 1*100/n% chance in getting the price with any number left of cake pieces (being "n" the number of pieces) so 100/1 is 100%.

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Only top voted, non community-wiki answers of a minimum length are eligible

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MathSciNet bibliographic data MR2054530 46E35 (26D10 26D15 31B15) Diening, Lars Riesz potential and Sobolev embeddings on generalized Lebesgue and Sobolev spaces $L\sp {p(\cdot)}$$L\sp {p(\cdot)}$ and $W\sp {k,p(\cdot)}$$W\sp {k,p(\cdot)}$. Math. Nachr. 268 (2004), 31–43. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


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# Electromotive force

EMF is defined as: "The potential difference across the terminals of a battery or dry cell when it is giving no current to the circuit."

So, if current starts to flow from the battery, will the potential difference across the ends of the battery not called EMF....?

-

Though the terminology of electromotive force (not a force) is actually the external work done per unit charge, there's another definition (which you've already mentioned) in current electricity. So, It will not be called EMF if the circuit is closed. It's then called the voltage. Both are similar where EMF is the cause and the latter is the effect.

The problem with EMF that it's measured in an open circuit is because there's always an internal resistance associated with a cell which depends on the electrolyte inside. Though the current flows from positive terminal to negative terminal in external circuit, there's a necessity for it to flow from negative terminal to positive terminal inside the cell. The internal resistance $r$ increases with the age of cell.

If $E$ is the EMF of the cell, $I$ is the current, $V$ is the voltage drop across the circuit and $R$ is the effective resistance associated with the external circuit, then the internal resistance could be measured by including a resistance in the circuit and comparing the voltage drop across them. The voltage drop across the resistor would be $V=E-Ir$ and the internal resistance could be calculated as $$r=\frac{(E-V)}{V}R$$

-
thanks a lot..:) –  Rafique Nov 18 '12 at 17:33


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# Question About Electric Aircraft Propulsion

#### cjl

The compressor /burner behind the duct ed fan spin at high rate. The fan must keep the tip of blades below sound speed.
Pretty much all modern turbofans run with a blade tip speed at full throttle of mach 1.5 or so.

Latest jet engines have a reducing gear from the compressor shaft to the fan ( just like turboprops).An electric motor will spin the fan ,with practically 100% efficiency, without any gear.A Dreamliner burns 1.3 Kg/sec.That is 50 Mj/sec ( or Mw). My guess is that 4 Mw electric will be needed operating the fan from an electric motor from a battery. weighing 120 Tons (6 hours flight). The plane will burn 28 tons of fuel.280 passengers @60 Kg/passenger is 17 Tons. Conclusions: 1)I am not suggesting to convert a Dreamliner to e drive,2)batteries are no more than an order of magnitude away from intermediate range, subsonic aircraft for passenger flight.
4MW isn't anywhere close to enough. Propulsive power at cruise for a modern jetliner the size of a 787 is more on the order of 40MW. At takeoff, each engine has to be making more like 50-60MW of shaft power just to run the front fan. Rerun the electric numbers knowing that and you'll see why running jetliners on electric power is a pipedream without a massive breakthrough in technology.

#### gmax137

Pratt & Whitney said:
Overall efficiency here refers to the efficiency with which the engine converts the power in the fuel flow to propulsive power. It is the product of thermodynamic efficiency of the process that converts fuel flow power to shaft power (herein called motor thermodynamic efficiency) and propulsive efficiency (the conversion of shaft power to propulsive power).

The most efficient commercial aircraft gas turbines in service or entering service in this decade have takeoff thrusts of 20,000 lb and above. These turbines operate at cruise, with motor thermodynamic efficiencies of up to 55 percent and propulsive efficiencies of well over 70 percent, yielding an overall efficiency (the product of the two) of about 40 percent

#### essenmein

Just for reference we work with reasonably state of the art electric machines (automotive), so a stake in the ground:
55kW peak, 35kW continuous liquid cooled PM 6ph machine, 22k rpm max, 135mm stator OD, 120mm stator stack length, weight about 12-15kg (estimated weight, too much other stuff connected to it to measure machine on its own...). This is at a reasonable limit for air gap, air gap flux and demag on relatively cost effective PM material.

If a number mentioned earlier is correct (100k Hp for dream liner turbo fan), then this is about 75MW e machine, and if built using similar tech as the PM machine mentioned, then you're looking at about 22500kg machine. I don't know what a core of a turbofan weighs, but I somehow doubt its 22 tones... Thats using the 55kW peak number not the 35kw continuous.

In cars you can play games with big short term numbers, unlikely for example you'd sit at full throttle for more than 10-20sec continuously, compared to a plane or boat, the power is needed for much longer periods (eg climbing to altitude) so relying on thermal mass is not really going to work, at least the air is cold up there...

#### cjl

You could probably get away with more like 50MW, and you only really need about a minute of full throttle capability, so you might be able to pull that mass down a bit. That having been said, an entire GEnx engine (including the fan and nacelle, which you'd still need for the electric) only weighs about 6 tons, so you're still at a massive weight disadvantage there.

#### essenmein

Yeah and thats just the metal in the machine, stator steel, PM, wire, bearings etc, you'd still need to add the control electronics etc, not huge numbers there, but they are not zero weight.

#### essenmein

And when people say "the technology needs to catch up", a lot of the time that is just wishful thinking, the laws of physics just get in the way.

If we are talking electric machines, then the tech is pretty stable, they are basically determined by F=BIL, B is fundamentally limited to about 1T in the air gap due to the magnetic material limits (B sat of stator steel, demag of the PM material, bearing tolerance etc), I is limited by the capacity to cool a wire and if PM then demag also will limit current, then L is a physical dimension (length), so if B and I are limited by existing materials/physics then all you can do is change L, ie larger size. I think its highly unlikely we'll discover a magic bullet around the B sat limit and the demag issue with permanent magnets...

#### merriam

The key to getting a flyable aircraft is the power/weight ratio of the engine. For Jets, the power outlet depends a lot on the speed, but roughly speaking a Boeing 777 engine (GE 90) puts out about 6 hp/lb (10 Kw/Kg). Very good electric motors can manage power/weight in this neighborhood. Viability depends on having light batteries that hold a lot of energy.

Electric aircraft exist, generally as small, experimental, short range aircraft. However, the Solar Impulse 2 has circumnavigated the globe (powered by solar cells). A number of projects are exploring hybrid aircraft, using the electric motors to increase the takeoff thrust, but turning them off in flight.

The power required for supersonic flight more or less precludes electric supersonic aircraft irrespective of the jet/propeller question.

Regarding jets and flight speed, the key is to look at the diameter of the column of gas coming out. You can have a large diameter at low speed or a small diameter at high speed. For the same level of thrust, low speed/large diameter takes less power. Example: A Harrier jump jet (hovering) vs any helicopter (hovering). The former has (4) small diameter jets of gas and requires a lot of power to hover. The latter has a big blade circle (and a big jet of air going down) and requires a lot less power. Top speed of helicopters is under 200 mph though, while the Harrier is over 600 mph.

That's why commercial aircraft have high bypass engines. Effectively they increase the diameter of the blade circle. This improves the gas mileage and lowers the noise. However, the top speed in level flight is limited to the speed at which the air comes out the back - the higher the bypass, the lower the top speed. Since all commercial aircraft (since Concorde) are subsonic (M = 0.85), the bypass is optimized for this speed. The fans are powered by the middle of the jet (so called hot section) where all the fuel burning happens. This de-energizes the exhaust from that part and makes it a lot quieter.

In summary, there is a long way to go in electric batteries and motors, before they enter commercial service (let alone military service).

#### OCR

Top speed of helicopters is under 200 mph. . .

You are real close, and for all practical purposes I agree. . . .

But, to get every thing right on the money. . . lets say 249.09 mph . .

It's rather amazing. . . the record still stands as of this year!

.

#### merriam

I'd really like to see a calculation on the volume of lifting gas required to carry 100,000 tons, to replace a cargo ship.

Then I'd like to see the kinetic energy of impact if one breaks apart at 10,000 ft.
Lift is about 1 gram/liter. Lets call it 100,000 metric tons or 100 million Kg = 100 billion grams = 100 billion liters. Roughly speaking, a cube that is 1500 feet on a side.

#### merriam

The key to getting a flyable aircraft is the power/weight ratio of the engine. For Jets, the power outlet depends a lot on the speed, but roughly speaking a Boeing 777 engine (GE 90) puts out about 6 hp/lb (10 Kw/Kg). Very good electric motors can manage power/weight in this neighborhood. Viability depends on having light batteries that hold a lot of energy.

Electric aircraft exist, generally as small, experimental, short range aircraft. However, the Solar Impulse 2 has circumnavigated the globe (powered by solar cells). A number of projects are exploring hybrid aircraft, using the electric motors to increase the takeoff thrust, but turning them off in flight.

The power required for supersonic flight more or less precludes electric supersonic aircraft irrespective of the jet/propeller question.

Regarding jets and flight speed, the key is to look at the diameter of the column of gas coming out. You can have a large diameter at low speed or a small diameter at high speed. For the same level of thrust, low speed/large diameter takes less power. Example: A Harrier jump jet (hovering) vs any helicopter (hovering). The former has (4) small diameter jets of gas and requires a lot of power to hover. The latter has a big blade circle (and a big jet of air going down) and requires a lot less power. Top speed of helicopters is under 200 mph though, while the Harrier is over 600 mph.

That's why commercial aircraft have high bypass engines. Effectively they increase the diameter of the blade circle. This improves the gas mileage and lowers the noise. However, the top speed in level flight is limited to the speed at which the air comes out the back - the higher the bypass, the lower the top speed. Since all commercial aircraft (since Concorde) are subsonic (M = 0.85), the bypass is optimized for this speed. The fans are powered by the middle of the jet (so called hot section) where all the fuel burning happens. This de-energizes the exhaust from that part and makes it a lot quieter.

In summary, there is a long way to go in electric batteries and motors, before they enter commercial service (let alone military service).
I stand corrected. At the Paris Airshow (june 2019) an electric aircraft was offered for sale. 3 engines, propeller driven, 650 miles at 500 miles/hour. 9 passengers. $4 million each. Roughly a dozen orders. #### russ_watters Mentor I stand corrected. At the Paris Airshow (june 2019) an electric aircraft was offered for sale. 3 engines, propeller driven, 650 miles at 500 miles/hour. 9 passengers.$4 million each. Roughly a dozen orders.
Don't stand corrected until it happens. I flat-out don't believe those specs are possible.

Edit: searching finds some badly written articles giving the impression it could fly 500 mph, but the actual spec calls for 240kt. Still won't believe it until I see it, but it at least it passes the laugh test at that speed claim.

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cjl

#### cjl

Don't stand corrected until it happens. I flat-out don't believe those specs are possible.

Edit: searching finds some badly written articles giving the impression it could fly 500 mph, but the actual spec calls for 240kt. Still won't believe it until I see it, but it at least it passes the laugh test at that speed claim.
That makes sense - I was about to express disbelief at that cruise speed myself. I don't believe a 500mph cruise is possible with current electric motors, at least not in a remotely economically viable design.

Gold Member
The key to getting a flyable aircraft is the power/weight ratio of the engine. For Jets, the power outlet depends a lot on the speed, but roughly speaking a Boeing 777 engine (GE 90) puts out about 6 hp/lb (10 Kw/Kg). Very good electric motors can manage power/weight in this neighborhood. Viability depends on having light batteries that hold a lot of energy.
You're forgetting energy storage. Hydrocarbon fuels are currently more efficient than batteries for storing maximum energy in a small volume and weight.

#### merriam

You're forgetting energy storage. Hydrocarbon fuels are currently more efficient than batteries for storing maximum energy in a small volume and weight.
Energy storage affects range. Jet fuel has an energy density of about 43 MJ/kg. A fully charged lithium battery can manage about 1MJ/Kg on a good day (maybe half that on an average day). There is a big difference. That means that if your aircraft allocates 1000 Kg for "fuel", you can go farther if you use jet fuel. The electric aircraft referenced above has a range of 650 miles (compare to 777 range of 5000 to 8500 miles) limited strictly by battery capacity (and rules about amount of reserve that must be carried).

But to get off the ground you have to overcome drag (and inertia) and that takes engine thrust. Power is thrust times speed. A typical aircraft has lift = 10 x drag (gliders more, fighter planes less) and the lift has to at least equal the weight of the engine. That's why thrust/weight is so important. Even if the plane is made of super material that weighs nothing, the engines still have to get off the ground.

Gold Member
That's all true but was not the point I was making. My point is that just having enough power to lift a plane off the ground is not all it takes to be flyable. So yes, you need electric engines with an appropriate thrust to weight ratio so that you can actually lift off, but you also need to be able to carry enough stored energy that the vehicle can fly a useful distance. If you don't solve both of these problems simultaneously, then what you have is essentially a new age Wright Flyer: intellectually interesting but not particularly useful without substantial continued technological development.

(Note: I am not trying to denigrate the Wright Flyer. I am merely pointing out that an electric airplane is not practically useful unless it can solve both problems, much like the Wright Flyer wasn't practically useful except for demonstrating that powered flight was possible.)

#### gleem

Two items have come to my attention that may expedite the development of larger (non hover) electric aircraft.

First, higher energy density LI-S batteries with energy densities of up to 500W/kg. They are cheaper, lighter than current Li batteries and non flammable. They are currently limited in discharge rate as well as cycle life.

Second, superconducting electric motors with power densities of 20kW/kg. Current test motor is 1 MW and scalable to at least 10MW.

#### Anand Sivaram

That is the reason why many people are doing research into renewable biofuel based aviation fuel.

That's all true but was not the point I was making. My point is that just having enough power to lift a plane off the ground is not all it takes to be flyable. So yes, you need electric engines with an appropriate thrust to weight ratio so that you can actually lift off, but you also need to be able to carry enough stored energy that the vehicle can fly a useful distance. If you don't solve both of these problems simultaneously, then what you have is essentially a new age Wright Flyer: intellectually interesting but not particularly useful without substantial continued technological development.

(Note: I am not trying to denigrate the Wright Flyer. I am merely pointing out that an electric airplane is not practically useful unless it can solve both problems, much like the Wright Flyer wasn't practically useful except for demonstrating that powered flight was possible.)

#### cjl

Energy storage affects range. Jet fuel has an energy density of about 43 MJ/kg. A fully charged lithium battery can manage about 1MJ/Kg on a good day (maybe half that on an average day). There is a big difference. That means that if your aircraft allocates 1000 Kg for "fuel", you can go farther if you use jet fuel. The electric aircraft referenced above has a range of 650 miles (compare to 777 range of 5000 to 8500 miles) limited strictly by battery capacity (and rules about amount of reserve that must be carried).

But to get off the ground you have to overcome drag (and inertia) and that takes engine thrust. Power is thrust times speed. A typical aircraft has lift = 10 x drag (gliders more, fighter planes less) and the lift has to at least equal the weight of the engine. That's why thrust/weight is so important. Even if the plane is made of super material that weighs nothing, the engines still have to get off the ground.
It's worth noting that engine thrust and power are not directly interchangeable concepts. Modern jetliners have vastly more power than they need to get off the ground safely, and the reason for this is the need to fly at a high cruise speed. If you eliminate the high cruising speed requirement, you can change over to propellers with a significantly larger disk area than the exit area of the jet engines, and by doing so, you substantially improve the power to thrust ratio.

This is also why all the electric concepts have multiple motors driving fairly large props along with a relatively low cruising speed - all of those things reduce the total power requirement.

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#### sophiecentaur

Gold Member
Sophie . . . I don't think that there's no hope for us, just that it mostly lies with the thinking people coming up with technical solutions.
Present experience of the Politics of the world do not support the theory that "thinking people" will be allowed the power to affect things. Extremist rulers tend not to think very far into the future and there will always (however bad things get) be some group of people (robber barons warlords etc. ) who will step in and profit at the expense of a defenceless population. Those kinds of régimes think in terms of just one lifetime.
Even the second amendment would not help in that respect - just speeding up the process of decline. But I do not need to point that out to the majority 'thinking' members of PF.

#### sophiecentaur

Gold Member
You're forgetting energy storage. Hydrocarbon fuels are currently more efficient than batteries for storing maximum energy in a small volume and weight.
In the overall picture, this is one of the most relevant facts. It's only when all electric transport energy produces almost no climate effect that the quoted massive ratio can be ignored. An intermediate solution would be to store energy in the form of Hydrogen, which sits somewhere in between. Hydrogen is something that seems to vary in popularity over the years.

Gold Member
In the overall picture, this is one of the most relevant facts. It's only when all electric transport energy produces almost no climate effect that the quoted massive ratio can be ignored. An intermediate solution would be to store energy in the form of Hydrogen, which sits somewhere in between. Hydrogen is something that seems to vary in popularity over the years.
Hydrogen and manned aviation have a fraught history, as it turns out.

#### sophiecentaur

Gold Member
Hydrogen and manned aviation have a fraught history, as it turns out.

#### cjl

It also has a very poor density, so even though it has the advantage of being very light, you need massive tanks (and the associated drag and weight penalty) to carry very much of it.

#### sophiecentaur

Gold Member
It also has a very poor density, so even though it has the advantage of being very light, you need massive tanks (and the associated drag and weight penalty) to carry very much of it.
The timescale is a bit different for rockets but why not store it cryogenically? That must have ben considered for planes.

#### cjl

Even cryogenically the density sucks. LH2 has a density of 71kg/m^3, while kerosene (or Jet A) is around 810 kg/m^3. Jet fuel has 42.8 MJ/kg, so the volumetric energy content of jet fuel is 34.7 GJ/m^3. Hydrogen has 130MJ/kg, but combine this with the low density and it only has 9.2 GJ/m^3, so you need nearly 4 times the tank volume to store identical energy compared to jet fuel.

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カテゴリー

# Machine Learning [Week 04/11] Neural Networks: Representation

[mathjax]
Machine Learning (Stanford University) の受講メモ

• 内容
• Neural network forward propagation
• Multiple classification using logistic regression and the one-vs-all technique
• メモ
• この講座だと、$$x_0$$ としてb(ias)定義し、$$\Theta$$のDimensionを求めている。
• $$z=wx+b$$
• $$w.shape:(n^{[l]},n^{[l-1]})$$
• $$\Theta.shape:(s_{j+1},s_j+1)$$: +1はbias

## Motivations

### Non-linear Hypotheses

Picture: feature N is too large.

### Neurons and the Brain

Origins: Algorithms that try to mimic the brain. 80s-early 90s; popularity diminished in late 90s.
Recent resurgence: State-of-the-art technique for many applications.

• The “one learning algorithm” hypothesis
• Auditory cortex learns to see.
• Somatosensory cortex learns to see.
• Sensor representations in the brain
• Human echolocation (sonar).
• Haptic belt: Direction sense.
• Implanting a 3rd eye.
• センサーは脳のたいていの場所に繋げて、扱い方学習する
• コースではNNは人工知能のためではなく、機械学習の応用として扱う

## Neural Networks

### Model Representation I

Let’s examine how we will represent a hypothesis function using neural networks. At a very simple level, neurons are basically computational units that take inputs (dendrites) as electrical inputs (called “spikes”) that are channeled to outputs (axons). In our model, our dendrites are like the input features $$x_1\cdots x_n$$, and the output is the result of our hypothesis function. In this model our $$x_0$$ input node is sometimes called the “bias unit.” It is always equal to 1. In neural networks, we use the same logistic function as in classification, $$\frac{1}{1 + e^{-\theta^Tx}}$$, yet we sometimes call it a sigmoid (logistic) activation function. In this situation, our “theta” parameters are sometimes called “weights”.

• Neurons in the brain
• Neuron model: Logistic unit
• $$x_0=1$$: bias unit
• 便利だから
• $$g(z)$$: Sigmoid(logistic)activation function.
• $$\theta$$: weight
• label
• \begin{align*}& a_i^{(j)} = \text{“activation” of unit i in layer j} \newline& \Theta^{(j)} = \text{matrix of weights controlling function mapping from layer j to layer j+1}\end{align*}
• one hidden layer
• $\begin{bmatrix}x_0 \newline x_1 \newline x_2 \newline x_3\end{bmatrix}\rightarrow\begin{bmatrix}a_1^{(2)} \newline a_2^{(2)} \newline a_3^{(2)} \newline \end{bmatrix}\rightarrow h_\theta(x)$
• The values for each the “activation” nodes
• \begin{align*} a_1^{(2)} = g(\Theta_{10}^{(1)}x_0 + \Theta_{11}^{(1)}x_1 + \Theta_{12}^{(1)}x_2 + \Theta_{13}^{(1)}x_3) \newline a_2^{(2)} = g(\Theta_{20}^{(1)}x_0 + \Theta_{21}^{(1)}x_1 + \Theta_{22}^{(1)}x_2 + \Theta_{23}^{(1)}x_3) \newline a_3^{(2)} = g(\Theta_{30}^{(1)}x_0 + \Theta_{31}^{(1)}x_1 + \Theta_{32}^{(1)}x_2 + \Theta_{33}^{(1)}x_3) \newline h_\Theta(x) = a_1^{(3)} = g(\Theta_{10}^{(2)}a_0^{(2)} + \Theta_{11}^{(2)}a_1^{(2)} + \Theta_{12}^{(2)}a_2^{(2)} + \Theta_{13}^{(2)}a_3^{(2)}) \newline \end{align*}
• This is saying that we compute our activation nodes by using a 3×4 matrix of parameters. We apply each row of the parameters to our inputs to obtain the value for one activation node. Our hypothesis output is the logistic function applied to the sum of the values of our activation nodes, which have been multiplied by yet another parameter matrix $$\Theta^{(2)}$$ containing the weights for our second layer of nodes.
• The dimensions of these matrices of weights
• If network layer $$j$$ unit $$s_j$$ to layer $$j+1$$ unit $$s_{j+1}$$ will be of dimension:$\Theta^{(j)}\in\Bbb R^{( s_{j+1} \times (s_j + 1) )}$
• Example: If layer 1 has 2 input nodes and layer 2 has 4 activation nodes. Dimension of $$\Theta^{(1)}$$ is going to be 4×3 where $$s_j = 2$$ and $$s_{j+1} = 4$$, so $$s_{j+1} \times (s_j + 1) = 4 \times 3$$.

### Model Representation II

• Forward propagation: Vectorized implementation
• In this section we’ll do a vectorized implementation of the above functions. We’re going to define a new variable $$z_k^{(j)}$$ that encompasses the parameters inside our g function. In our previous example if we replaced by the variable z for all the parameters we would get:\begin{align*}a_1^{(2)} = g(z_1^{(2)}) \newline a_2^{(2)} = g(z_2^{(2)}) \newline a_3^{(2)} = g(z_3^{(2)}) \newline \end{align*}
• layer j=2 and node k, the variable z will be: $z_k^{(2)} = \Theta_{k,0}^{(1)}x_0 + \Theta_{k,1}^{(1)}x_1 + \cdots + \Theta_{k,n}^{(1)}x_n$
• Vector representation of x and z^j\begin{align*}x = \begin{bmatrix}x_0 \newline x_1 \newline\cdots \newline x_n\end{bmatrix} &z^{(j)} = \begin{bmatrix}z_1^{(j)} \newline z_2^{(j)} \newline\cdots \newline z_n^{(j)}\end{bmatrix}\end{align*}
• $z^{(j)} = \Theta^{(j-1)}a^{(j-1)}$
• We are multiplying our matrix $$\Theta^{(j-1)}$$ with dimensions $$s_j\times (n+1)$$ (where $$s_j$$ is the number of our activation nodes) by our vector $$a^{(j-1)}$$ with height (n+1). This gives us our vector $$z^{(j)}$$ with height $$s_j$$. Now we can get a vector of our activation nodes for layer j as follows:$a^{(j)} = g(z^{(j)})$
• We can then add a bias unit (equal to 1) to layer j after we have computed $$a^{(j)}$$. This will be element $$a_0^{(j)}$$ and will be equal to 1. To compute our final hypothesis, let’s first compute another z vector:$z^{(j+1)} = \Theta^{(j)}a^{(j)}$
• We get this final z vector by multiplying the next theta matrix after $$\Theta^{(j-1)}$$ with the values of all the activation nodes we just got. This last theta matrix $$\Theta^{(j)}$$ will have only one row which is multiplied by one column $$a^{(j)}$$ so that our result is a single number. We then get our final result with:$h_\Theta(x) = a^{(j+1)} = g(z^{(j+1)})$

## Applications

### Examples and Intuitions I

• Function: x1 and x2 then true;
• The graph of our functions will look like:\begin{align*}\begin{bmatrix}x_0 \newline x_1 \newline x_2\end{bmatrix} \rightarrow\begin{bmatrix}g(z^{(2)})\end{bmatrix} \rightarrow h_\Theta(x)\end{align*}
• first matrix:$\Theta^{(1)} =\begin{bmatrix}-30 & 20 & 20\end{bmatrix}$
• \begin{align*}& h_\Theta(x) = g(-30 + 20x_1 + 20x_2) \newline \newline & x_1 = 0 \ \ and \ \ x_2 = 0 \ \ then \ \ g(-30) \approx 0 \newline & x_1 = 0 \ \ and \ \ x_2 = 1 \ \ then \ \ g(-10) \approx 0 \newline & x_1 = 1 \ \ and \ \ x_2 = 0 \ \ then \ \ g(-10) \approx 0 \newline & x_1 = 1 \ \ and \ \ x_2 = 1 \ \ then \ \ g(10) \approx 1\end{align*}

So we have constructed one of the fundamental operations in computers by using a small neural network rather than using an actual AND gate. Neural networks can also be used to simulate all the other logical gates. The following is an example of the logical operator ‘OR’, meaning either $$x_1$$ is true or $$x_2$$ is true, or both:

### Examples and Intuitions II

• FunctionをAND,NOR,ORとして直感的な理解を

### Multiclass Classification

To classify data into multiple classes, we let our hypothesis function return a vector of values. Say we wanted to classify our data into one of four categories. We will use the following example to see how this classification is done. This algorithm takes as input an image and classifies it accordingly:

• We can define our set of resulting classes as y:$y^{(i)}=\begin{bmatrix}1\\0\\0\\0\end{bmatrix}, \begin{bmatrix}0\\1\\0\\0\end{bmatrix}, \begin{bmatrix}0\\0\\1\\0\end{bmatrix}, \begin{bmatrix}0\\0\\0\\1\end{bmatrix}$
• Our final hypothesis function setup:$\begin{bmatrix}x_0\\x_1\\x_2\\x_3\end{bmatrix}\rightarrow \begin{bmatrix}a_0^{[2]}\\ a_1^{[2]}\\a_2^{[2]}\\ a_3^{[2]}\end{bmatrix} \rightarrow \begin{bmatrix} a_0^{[3]}\\ a_1^{[3]} \\ a_2^{[3]}\\ a_3^{[3]}\end{bmatrix} \rightarrow\dots\rightarrow\begin{bmatrix}h_\Theta (x)_1\\ h_\Theta (x)_2 \\ h_\Theta (x)_3 \\ h_\Theta (x)_4 \end{bmatrix}$
• Our resulting hypothesis for one set of inputs may look like:$h_\Theta(x) =\begin{bmatrix}0 \newline 0 \newline 1 \newline 0 \newline\end{bmatrix}$


text:
# How do you find the sum of the infinite geometric series 8+6+9/2+27/8+...?

Oct 26, 2017

${S}_{\infty} = 32$

#### Explanation:

The sum of infinite GP is given by

${S}_{\infty} = \frac{a}{1 - r}$

where

$a =$the first term

$r =$the common ratio

AND$\text{ } | r | < 1$" "if $\text{ "S_oo" }$is to exist

we need to fin teh common ratio . We do this by

$r = \frac{{u}_{n + 1}}{u} _ n$

$r = \frac{6}{8} = \frac{3}{4}$

$| r | < 1 \text{ ":.S_oo" exists}$

${S}_{\infty} = \frac{8}{1 - \frac{3}{4}}$

${S}_{\infty} = \frac{8}{\frac{1}{4}} = 8 \times \frac{4}{1}$

${S}_{\infty} = 32$


text:
## February 28, 2018

### yes

Books to Read While the Algae Grow in Your Fur, February 2018

Attention conservation notice: I have no taste. I also have no qualifications to opine on the theory of measurement in psychology, the philosophy of science, the law, race relations, or criminology.

Joel Michell, Measurement in Psychology: A Critical History of a Methodological Concept
Comments having passed the 1500 word mark, including long quotations, this will have to be a separate review.
H. P. Lovecraft, At the Mountains of Madness
This is an umpteenth re-read, of course. (I tend to do them in the winter.) This one made me want to read a history of subsequent Elder Thing archaeology, where the mountains and the city were revisited during the International Geophysical Year, and it's become obvious that 99% of this is as much a product of the discoverers' imagination and preconceptions as, say, Arthur Evans's views of the Minoans. (But that 1%...)
Lauren Willig, The English Wife
Mind candy historical mystery, set in New York and London just a bit before 1900. An interesting aspect of the writing is that here, as in her historical romance novels, Willig uses two time-lines, where the characters in one time-line are trying to discover what happened in the other. But in the romances the time-lines are parallel, whereas here they converge. What this signifies, I couldn't say.
Peter Godfrey-Smith, Theory and Reality: An Introduction to the Philosophy of Science
I can easily say that this is the one of the best modern introductory books on the philosophy of science I've ever read. (Another, of a very different sort, is William Poundstone's Labyrinths of Reason.) It's presented roughly historically, beginning with Logical Positivism and moving forward, through Popper, Kuhn, such post-Kuhnians as Lakatos, Feyerabend and Laudan, and classic 1970s/1980s "sociology of scientific knowledge", before ending with a range of contemporary topics. Throughout, Godfrey-Smith strikes a good balance between persuading the reader that there are problems worth wrestling with, and that they're not hopeless.
To the former: too many scientists, encountering issues from the philosophy of science, find them pointless, or at most things which could be cleared up in an afternoon with a little clear thinking and maybe some algebra. (Occasionally this results in weird little cults like self-styled "strong inference", which is firmly put in its place here.) Godfrey-Smith is very good at conveying how there are real issues here, which very smart people have wrestled with, without coming to any truly satisfactory answers.
This then raises the possibility that the exercise is futile, not because it's unimportant but because it's doomed, that the problems are just too hard for us. Against this, Godfrey-Smith is good at conveying how, if we're still confused about questions like "When does observing something that a theory predicts confirm the theory?", or "How can the social organization of a scientific community support its cognitive goals?", we're at least understanding the issues much better. (For example, it's become very clear that social organization does matter.)
This book is worthwhile reading for any scientist interested in philosophical issues. It might be even more worthwhile for those who aren't interested, but...
--- Two thoughts which occurred to me while reading Godfrey-Smith's discussion of how "naturalistic" philosophy of science is anti-foundationalist, in the sense of eschewing the search for philosophical foundations for the sciences which are somehow prior to the sciences themselves.
1. Strong forms of this would say that such foundations are impossible or undesirable. A weaker form, however, would compare the track-records of philosophy and science, and say that it's rash to expect philosophy to be more secure than (say) neurophysiology any time soon. (Where this would leave, say, social psychology is a nice question.) I am not sure whether anyone has taken this position within the philosophical literature, or even what it would be called.
2. Saying that we will use the results of scientific inquiry to understand the process of scientific inquiry can sound like a vicious circle, but can also, more reasonably, be just a self-consistency check. If our best scientific understanding of the world and ourselves implied that scientific inquiry was unreliable, we would have a real problem. Worries about science being self-undermining are a a long-running theme in the history of the sort of philosophy of science that Godfrey-Smith writes about, going back before the Logical Positivists into the nineteenth century (see, e.g., Leszek Kolakowski's The Alienation of Reason / History of Positivist Thought from Hume to the Vienna Circle and his Husserl and the Search for Certitude), and continues on today (naturally in meme format). Even if all naturalistic philosophy of science achieves is showing that science doesn't undermine itself the way that the more ambitious and outrageous forms of sociology of knowledge do, this would be a real accomplishment.)
§
Richard Thompson Ford, The Race Card: How Bluffing about Bias Makes Race Relations Worse
Let me spoil the ending:
No doubt some readers will wish to ask whether I really think playing the race card is now the biggest racial justice issue this society faces. No, I don't. I hope it's clear that I believe old-school bigotry remains a severe social problem and that subtler and systemic racial disadvantages --- even when they can't be blamed on "racists" --- are profound social evils that demand redress. These are bigger problems than playing the race card. But the race card is an impediment to dealing with these problems. It distracts attention from larger social injustices. It encourages vindictiveness and provokes defensiveness when open-mindedness and sympathy are needed. It leads to an adversarial, tit-for-tat mind-set ("You're a bigot!" "No, you're just playing the race card!") when a cooperative spirit of dialogue is required.
The race card is symptomatic of a real crisis in the way we currently think and talk about race: a crisis borne of our failure to keep up with a changing social world, a crisis of social change and of intellectual stasis. We need new intellectual tools and new language to deal with the new realities of American racism. Thus far we've failed to develop them, so we find ourselves increasingly unable to discuss issues of race intelligently and convincingly. We find ourselves listening to and repeating the slogans and catch-phrases of the past, whether or not they apply, like a catechism that's long since lost its power to invoke or inspire, or like a curse that damns guilty perpetrator and innocent bystander with indiscriminate contempt. [p. 349]
And this was in 2008! (Ford's skepticism about the Implicit Association Test is looking pretty good these days. His confidence that open expressions of outright racism have been driven to the fringes of American public life, maybe not so much.)
More constructively, I found chapter 2's discussion of "racism by analogy" thought-provoking, and chapter 3 on legal criteria for discrimination and disparate impact quite eye-opening. §
John Pfaff, Locked In: The True Causes of Mass Incarceration --- and How to Achieve Real Reform
This is a thoughtful book about the causes of mass incarceration, and what can and should be done to reverse it. I should say at the beginning that Pfaff is as outraged as anyone about how many people we have in prison (or otherwise subject to "corrections"), so that when Pfaff challenges elements of what he calls the "standard story", it's not to minimize the disaster and disgrace, it's to help efforts at reform actually improve things. I found a lot of it convincing, but I should say up-front that I haven't tried to independently check any of Pfaff's figures or calculations.
The most convincing parts of the preliminary de-bunking are as follows:
1. Private prisons are awful, but they are quantitatively too small to account for mass incarceration. Also, the lobbying efforts of private prison corporations are too small, and come too late in the surge in incarceration, to explain it.
2. Most of our prison population isn't there for drug offenses, or non-violent offenses in general, but for violent crimes, and so undoing mass incarceration will mean changing how we deal with those convicted of violence. Pfaff presents this as a refutation of the idea that mass incarceration is due to the war on drugs, which I think is a bit too hasty (as I will explain below).
3. Maximum legal prison sentences have gone way up, and longer prison terms would naturally lead to more people being in prison. But this can't explain most of the growth in incarceration, because the actual average length of time served hasn't increase very much.
It then behooves Pfaff to explain why, in his view, we have so many more people in prison than we used to, even adjusting for population. Implicitly --- this is a popular book and he does no explicit models here --- he works with a "compartment" model, where the compartments or stages are something like: $[\text{Commit crime}] \Rightarrow [\text{Arrested}] \Rightarrow [\text{Charged}] \Rightarrow [\text{Convicted}] \Rightarrow [\text{Prison}] \Rightarrow [\text{Release/Parole}]$ where at each stage before prison one might be diverted away (e.g., arrested but not charged), and prison is of course of variable duration. The advantage of approaching the question "why are so many people in prison?" this way is that if you can track the number of people in each stage, and the flows of people from one stage to the next, they have to add up: the number of people in prison on 1 July 2018 will be the number who were in prison on 1 July 2017, plus those convicted and sentenced over the year, minus those released over the year. (At the risk of being dis-respectful, I am counting deaths in prison under "release".) Changing the proportions who go on from one stage to the next changes the flows, and hence will accumulate over time to the number of those in prison.
Pfaff claims that the big change which drove up the number of people in prison wasn't at the stage of being arrested, or convicted, or even the length of time spent in prison, but rather in the proportion of those arrested who are actually charged with a crime. This is a decision made by local public prosecutors. If we believe Pfaff's numbers, this locates a key source of the problem.
Unfortunately, as he is at pains to say, we have very little systematic information on prosecutors' offices and how they make their decisions. We do know that they face a somewhat perverse set of incentives, in that declining to charge someone who goes on to do something bad is electoral poison, but charging someone who's really harmless has almost no downside (for the prosecutor; it has plenty for the person charged, and their family and community). Prosecutors also face little opposition from public defenders, which is a big part of why almost all criminal charges are settled by plea deals, not brought to trial. The whole business is a mess, with almost no accountability (either to hierarchical superiors or to the democratic public), and scarcely any systematic reporting. Pfaff does not attempt to say why any of these issues should have gotten worse during this period, however.
Popular books about policy or social problems usually have a last chapter which talks about what to do about the issue. Pfaff follows this practice, and, as usual, it's the weakest part of the book, because his proposals are so much smaller than his own account of the scale of the problems. (Whether this is better or worse than the alternative tradition, of proposing measures which would solve the problem but also be totally unworkable, is a nice question.) In no small part this is because he has fairly convincingly localized the problem, but he's localized it not so much to a black box as to a mob of 3,000-odd ill-coordinated black boxes.
--- I said above that I am not sure Pfaff is entirely fair to the blame-the-drug-war camp; in particular, I think he ignores a fairly obvious counter-argument. He attacks the idea that the growth in incarceration is a result of the war on drugs, by pointing out that only a minority of those in prison are there for non-violent drug offenses, while the majority are there for violent crimes. Grant that this is true (as I said, I haven't checked his figures*.) How much of that violence is due to the war on drugs? Legal businesses get robbed, of course, and from time to time one even reads of, say, dentists conspiring to assassinate rival dentists, but this sort of thing is rare in trades where the law is available to settle disputes and protect property. A criminalized but lucrative drug trade, on the other hand, seems conducive to violence. Localizing the trade to specific neighborhoods make those dangerous, law-less places, further inciting violence (cf. Allen and Leovy). Effects like these are hard to quantify --- we can't just read them off from administrative data, as Pfaff likes to do --- but they could be very important. I'm not sure where this leaves us. §
*: One point which would be good to check is how possessing of a firearm while committing another crime gets coded in these records. If every drug-dealer who gets busted while also carrying a gun counts as "violent", for example, that might make a substantial difference. (Or it might not; that's why someone should check.) ^
Danielle Allen, Cuz: The Life and Times of Michael A.
A memoir of the life, imprisonment and death of Allen's cousin Michael. It's at once the specific story of a unique person and their family, and a slice through what's gone wrong with our country*, that someone could be thrown in prison for eleven years for some stupid crimes committed at fifteen (where Michael was the only one hurt), ultimately setting his life on a path where, at age 29, his corpse was found in a shot-up car on the street. Michael made bad choices, which Danielle never shies from, but he made them in a foolishly, evilly un-forgiving context, in a society which essentially threw his life away for no good reason, and that is messed up. It's horribly, horribly sad, but beautifully told. §
Disclaimer: I know Prof. Allen, and have participated in a series of workshops she organized and contributed to a book she edited, but I feel under no obligation to write a positive notice of her books.
*: One of the things which makes this a complicated book is that it is also, implicitly and in glimpses, the story of what has gone right with our country that it now creates people like its author. ^

Posted at February 28, 2018 23:59 | permanent link


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# m06 Q 37

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Kudos [?]: 6 [0], given: 10

m06 Q 37 [#permalink]  31 Jul 2010, 14:58
There are 6 points on the plain. Any 3 points of these 6 don't lie on the same line. How many unique triangles can be drawn using these 6 points as vertices?

5
10
20
30
60

Why isn't the answer 60. We can choose 1 vertice and there are 5C2 = 10 ways to 2 choose the remain 2 vertices. 10*6 choices = 60 (10 choices each for each vertex)?
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Kudos [?]: 1248 [0], given: 235

Re: m06 Q 37 [#permalink]  31 Jul 2010, 15:23
suhasrao wrote:
There are 6 points on the plain. Any 3 points of these 6 don't lie on the same line. How many unique triangles can be drawn using these 6 points as vertices?

5
10
20
30
60

Why isn't the answer 60. We can choose 1 vertice and there are 5C2 = 10 ways to 2 choose the remain 2 vertices. 10*6 choices = 60 (10 choices each for each vertex)?

$$6C_3 =$$ Number of triangles that can be formed using 6 points = Number of ways in which you can select 3 points out of 6.
Since it is clearly stated that " Any 3 points of these 6 don't lie on the same line" , the above statement holds true.

thus answer is $$6C_3 = 20$$

Your logic is wrong as you are counting the items repeatedly.

Suppose a b c d e f are 6 points
you can select 1 out of 6 in 6 ways, and remaining 2 in 10 ways.
Let the first selection is a , and the later two are b and c.
Now when you will select b as first selection and remaining 2 in 10 ways, a and c are counted again.
The selection becomes b , a , c.
But the triangle formed by a b c should be counted once but you have counted it 3 times.

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Re: m06 Q 37 [#permalink]  30 May 2011, 10:36
why not pick 2 points out of 6 ?
=6C2

now since every 2 point will form a triangle with the 4 remaining points.

hence answer = 6C2*4 = 60

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# Help add feedback to Sine Wave Generation using ATmega16

#### AnwarB

##### Newbie level 4
Hi all
On the subject of Tahmid
tahmidmc.blogspot.com/2012/10/generation-of-sine-wave-using-spwm-in_10.html
Talk about generating a sine wave with SPWM in the PIC16F684
tahmidmc.blogspot.com/2012/11/feedback-in-sine-wave-inverter-pic16f.html
To be like this

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const unsigned char sin_table[416]=
{
0,16,33,49,64,79,93,106,118,129,138,146,153,158,161,163,163,161,158,153,146,138,129,118,106,93,79,64,49,33,16,0, //65%
0,17,34,51,67,82,97,110,122,134,143,152,158,164,167,169,169,167,164,158,152,143,134,122,110,97,82,67,51,34,17,0, //67.5%
0,18,35,52,69,85,100,114,127,138,149,157,164,169,173,175,175,173,169,164,157,149,138,127,114,100,85,69,52,35,18,0, //70%
0,18,36,54,71,88,103,118,131,143,154,163,170,175,179,181,181,179,175,170,163,154,143,131,118,103,88,71,54,36,18,0, //72.5%
0,19,38,56,74,91,107,122,136,149,160,169,176,182,186,188,188,186,182,176,169,160,149,136,122,107,91,74,56,38,19,0, //75%
0,20,39,58,77,94,111,126,141,153,165,174,182,188,192,194,194,192,188,182,174,165,153,141,126,111,94,77,58,39,20,0, //77.5%
0,20,40,60,79,97,114,130,145,158,170,180,188,194,198,200,200,198,194,188,180,170,158,145,130,114,97,79,60,40,20,0, //80%
0,21,41,62,81,100,118,134,149,163,175,185,193,199,204,206,206,204,199,193,185,175,163,149,134,118,100,81,62,41,21,0, //82.5%
0,22,43,64,84,103,122,139,154,168,181,191,200,206,211,213,213,211,206,200,191,181,168,154,139,122,103,84,64,43,22,0, //85
0,22,44,66,86,106,125,143,159,173,186,197,205,212,216,219,219,216,212,205,197,186,173,159,143,125,106,86,66,44,22,0, //87.5%
0,23,45,67,89,109,129,147,163,178,191,202,211,218,222,225,225,222,218,211,202,191,178,163,147,129,109,89,67,45,23,0, //90%
0,23,46,69,91,112,132,150,167,183,196,207,217,224,228,231,231,228,224,217,207,196,183,167,150,132,112,91,69,46,23,0, //92.5%
0,24,48,71,94,116,136,155,173,188,202,214,223,230,235,238,238,235,230,223,214,202,188,173,155,136,116,94,71,48,24,0, //95%
};

unsigned int TBL_POINTER_NEW, TBL_POINTER_OLD,

TBL_POINTER_SHIFT, SET_FREQ;
unsigned int DUTY_CYCLE, FBV, FB_Step, adder;

void interrupt()
{
if (TMR2IF_bit == 1){TBL_POINTER_NEW =

TBL_POINTER_OLD + SET_FREQ;
if (TBL_POINTER_NEW < TBL_POINTER_OLD){P1M1_bit =

~P1M1_bit;}
TBL_POINTER_SHIFT = TBL_POINTER_NEW >> 11;
CCPR1L = sin_table[DUTY_CYCLE];
TBL_POINTER_OLD = TBL_POINTER_NEW;
TMR2IF_bit = 0;
}}

void main()
{
SET_FREQ = 410;
TBL_POINTER_SHIFT = 0;
TBL_POINTER_NEW = 0;
TBL_POINTER_OLD = 0;
DUTY_CYCLE = 0;
FB_Step = 0;
T1CON = 0b00001001;//1:1 Prescale Value, Enables Timer1
ANSEL = 0b00000001;//RA0/AN0 as analogue input.
TRISA = 0b00000001;//PortA as output except RA0 as input
CMCON0 = 7;
PR2 = 249;
TRISC = 0x3F;
CCP1CON = 0x4C;
TMR2IF_bit = 0;
T2CON = 4;
while (TMR2IF_bit == 0);
TMR2IF_bit = 0;
TRISC = 0;
TMR2IE_bit = 1;
GIE_bit = 1;
PEIE_bit = 1;

while(1)
{
if (FBV < 512){FB_Step++;
if (FB_Step > 12) FB_Step = 12;
}
else
{
if (FB_Step > 0){FB_Step--;}}
TMR1L = 0;
TMR1H = 0;
T1IF_bit = 0;
}
}

And on the topic of Thamed
Talk about Sine Wave Generation with "Fast PWM Mode" of AVR - using ATmega16
I tried adding a feedback to it and it stopped at ADC pins
This is what I modified, is it correct?
I am doing an experiment on the ATmega8
Thank

#### Attachments

• 6.3 KB Views: 9
Last edited by a moderator:

##### Full Member level 6
Just an observation but un-commented code makes it doubly difficult to
understand what code is doing.

Also helps when posting code to use code tags so indentation preserved hence
logic block flow more visually apparent.

Regards, Dana.

#### andre_teprom

##### Super Moderator
Staff member
At a glance, once you did not eable peripherals interrupt yet, seems weird this part of the code:

Code:
TMR2IF_bit = 0;
...
while (TMR2IF_bit == 0);

#### AnwarB

##### Newbie level 4
Is there a problem with this?

#### betwixt

##### Super Moderator
Staff member
while (TMR2IF_bit == 0);
Stops the program progressing until the TMR2 interrupt occurs. Usually, interrupts are designed to 'interrupt' the program flow rather than suspend it!
Brian.

#### AnwarB

##### Newbie level 4
Sorry, I'm inexperienced with this
This is from the topic of Tahamid
Someone on this date October 2, 2013 at 1:28 AM
Put this

#### betwixt

##### Super Moderator
Staff member
I haven't gone back to see Tahamid's original postings but the software doesn't look correct to me and some of it seems to be missing. I assume the intention of having 13 sine tables is to give it voltage regulation capability but exactly how it does it isn't obvious. I'm not sure why TMR1 is used at all, it doesn't seem to do anything.

Brian.

#### andre_teprom

##### Super Moderator
Staff member
In addition, the code could be optimized in memory usage, since the entire table is composed of values symmetrically distributed starting from the middle of each array. Frankly, considering that this code was not wrote by the author of the Blog, but rather from someone in the comment section, I would throw it away and take another from a reliable source, or read datasheet to do it yourself.

#### AnwarB

##### Newbie level 4
I mentioned that I have little experience in this area
Just want to add feedback to avr-atmega8

#### betwixt

##### Super Moderator
Staff member
We can't help with partial code, especially porting it from on device to another.
My guess is the way it is supposed to work is you step through ONE line of the table at a rate of (number of entries on one line multiplied by the frequency you want to produce) and use that value to feed the PWM generator. To regulate the output voltage you periodically read the ADC and use it's output to select which line in the table you use. As Andre pointed out, you can halve the size of the table by using each value twice, utilizing its symmetry. If I'm right about how the regulation works, personally I would use a different array for each set of values, it would make the code far simpler.

Brian.


text:
MathSciNet bibliographic data MR313190 10B35 (12CXX) Webb, William A. Waring's problem in ${\rm GF}[q,\,x]$${\rm GF}[q,\,x]$. Acta Arith. 22 (1973), 207–220. Links to the journal or article are not yet available

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.

 American Mathematical Society 201 Charles Street Providence, RI 02904-6248 USA


text:
## Quicknote: LogInsight not recieving host data via cfapi

On a recent, fresh installation of LogInsight I observed that no data from a set of hosts was coming into the LogInsight server.
Debugging showed not issues related to the usual suspects of DNS, NTP, firewall or network which left me a bit puzzled.
I found the answer by going through the logs of the loginsight agent (usually in /var/log/loginsight-agent/liagent_<date>):

## VMware EPIC2 Award - Integrity

Wow, what a day:
Today I have received the Germany FY22/Q1 EPIC2 award at VMware for “Integrity”!

## Tagging the vCenter (object)

Today I learned...

I am working on enhancing my vSphere release tagging script with the ability to tag the vCenter with a human-readable release name. As shown in the screenshot, it’s easy enough in the UI.

## Tagging ESXi hosts with canonical release names

Save some time looking up build numbers


text:
## Introduction

STO is a complex oxide semiconductor with the perovskite structure and an indirect band gap of 3.2 eV1, 2. STO has been used as a photo-catalyst to hydrolyze water3, a material for high-temperature oxygen sensors4, and a substrate for oxide thin films5. The electrical properties are tunable from an insulator to metallic-like conduction6, 7, and the interface between SrTiO3 and insulating oxides can form a highly conductive layer8,9,10. Cr-doped STO has been shown to have bi-stable resistance that could potentially be used for resistance change memory11.

Annealing STO at 1200 °C produces defects that give rise to persistent photoconductivity (PPC), with sub-gap light exposure causing the free-electron concentration to increase by a factor of several hundred12. The free carriers persist for days or weeks after the light is turned off, at room temperature. The specific defect that causes PPC has not been identified, but the results are consistent with a native acceptor defect such as a V TiV O pair13. Annealing in vacuum prior to the 1200 °C anneal yields optimal PPC behavior14. The vacuum anneal introduces oxygen vacancies15, which are necessary for PPC, consistent with the V TiV O model.

Prior work on large PPC has focused on DX centers, deep-level defects that involve the displacement of a donor impurity16, 17. Their metastable behavior stimulated interest in using DX centers for holographic data storage, where bits are optically written and read within the volume of a crystal18, 19. Because DX centers exhibit PPC, optical exposure increases the free-carrier density and thus changes the refractive index20. The change in refractive index is then read out optically. In addition to holographic memory, 3D electronic architectures could be defined optically, with current pathways traveling throughout the bulk of a crystal. The main practical problem, however, is that the DX barrier is too small (~0.2 eV). At room temperature, the barrier is surmounted quickly, so the sample does not show PPC.

In the present work, room temperature PPC was utilized to selectively reduce two-point resistance values. Because secondary luminescence from the sample might affect this process, we performed PL spectroscopy to characterize the samples. A typical PL spectrum is displayed in Fig. 1, which shows a peak at 3.2 eV and a broad band at 2.9 eV. These features have been attributed to the radiative recombination of an exciton bound to a defect such as an oxygen vacancy21, 22. The 2.9 eV peak, in particular, is relevant to this study because its photon energy can induce PPC.

Figure 2 shows the effect of light exposure on resistance. Light impinged on the side opposite to the contacts (Fig. 2 inset). This geometry prevents the complication of shadowing by the contacts. Light exposure started with a wavelength of 325 nm for 10 s, and proceeded by successively increasing the wavelength in steps of 10 nm. Significant resistance change did not occur until 385 nm (3.22 eV), which is near the band gap. The penetration depth at 325 nm is tens of nm, while at 355 nm it is ~1 μm2. Therefore, the crystal acts as a long-pass filter and blocks above-gap photons. Since the light is absorbed well before it reaches the region near the contacts, the measured resistance remains high. The PL (Fig. 1) does appear to induce some PPC. As shown in Fig. 2, however, this secondary effect causes only a minor drop in resistance.

Figure 2 also shows the light exposure threshold starting from the low-energy side. Exposure started at a wavelength of 500 nm for 10 s, decreasing by steps of 10 nm. Only small changes were observed in the resistance values until 420 nm. With 410 nm light, the change is very significant (3 orders of magnitude). This threshold for PPC, 2.9–3.0 eV, is below the band gap (3.2 eV) and consistent with the value obtained by IR absorption12.

The strength of mid-IR absorption is proportional to the carrier density23. Successively measuring the IR spectrum after incremental light exposure allowed us to correlate the change in electrical resistance with the free-electron density. First, a reference IR spectrum was obtained prior to light exposure. This sample had a resistance of 2.46 MΩ. A 405 nm light-emitting diode was then turned on for several seconds and another spectrum was taken. The first 5 s of light exposure dropped the resistance from 2.46 MΩ to 1.33 KΩ, while only a small change in free-carrier absorption was observed (Fig. 3). This procedure was repeated until the absorbance reached a saturation value.

Plots of absorbance versus exposure time are shown for two different wavenumber values, 2500 and 5000 cm−1 (Fig. 4). They both have the same single-exponential time constant of 43 s, consistent with excitation from a single defect level to the conduction band. Figure 4 also shows a plot of conductance (1/R) versus exposure time. These data were fit with two exponential functions, with time constants of 7 and 122 s. While the free-carrier absorption and conductance both show monotonically increasing behavior, they do not correlate exactly. Specifically, the contact resistance has a rapid initial decrease (short time constant). This observation suggests that the contact resistance has a nonlinear dependence on free-carrier concentration or that interface traps are filled during the first few seconds of light exposure.

To further explore this effect, we plotted resistance R versus carrier density n, where n was determined from the IR absorbance24. The resistivity is given by

$${\rm{\rho }}=\frac{1}{ne{\rm{\mu }}},$$
(1)

where μ = 5 cm2/Vs is the mobility at room temperature25. In the absence of contact resistance, the measured two-point resistance can be estimated as (Supplementary Information)

$${R}_{{\rm{bulk}}}=\frac{{\rm{\rho }}}{{\rm{\pi }}{\rm{\delta }}}\,\mathrm{ln}(\frac{d-a}{a}),$$
(2)

where d is the distance between the contacts, a is the contact radius, and δ is the sample thickness. Using d = 5 mm, a = 0.5 mm, and δ = 0.5 mm, Eq. (2) yields R bulk ≈ (14 cm−1)ρ. This estimate is significantly below the measured resistance (Fig. 5), suggesting that contact resistance is important. The contact resistance may involve a defective, resistive layer at the STO surface26, 27, perhaps introduced during the thermal treatments. Regardless of the mechanism, it is clear that the contact resistance shows the same overall trend as the bulk; namely, a monotonic, persistent drop versus light exposure.

To demonstrate the possibility of lithography, PPC was induced by focusing a 405 nm laser on specific regions of a sample. We first exposed contact 2 (from the back) for 3 s, turned off the laser, and measured the resistance (Fig. 6). Then contact 1 was exposed for 3 s and resistance was measured again. Finally, a path was traced from contact 1 to contact 2. To test the persistence, measurements were made 1 and 5 days (in the dark) after the exposures. Values are given in Table 1.

By successively exposing each contact, then writing a path, one can see that the two-point resistance is affected by contact and bulk resistance. Exposing the contacts resulted in a 400-fold decrease in resistance, although light scattering probably reduced the resistivity of the material between the contacts as well. Drawing a path between the contacts further reduced the resistance by a factor of 4. The resistance between contacts 3 and 4 changed very little when the path was created between contacts 1 and 2, verifying electrical isolation. The 3 order-of-magnitude change in resistance persists for 1 day and is still significant after 5 days.

In conclusion, we have demonstrated that PPC can be exploited to write a high-conductivity pathway using an optical pen. The defined path persists for several days at room temperature and does not affect the transparency of the crystal. In principle, this technique could be used to create transparent electronic circuits that could be erased (e.g., by resistive heating) and re-written.

## Methods

All measurements were performed at room temperature. (100) oriented STO crystals with one side finely polished were obtained from MTI. Samples were first placed in an evacuated ampoule (50–100 mTorr) and vacuum annealed at 800 °C for 1 hr. Then samples were placed in an evacuated ampoule with 0.6 g SrO powder and annealed at 1200 °C for 1 hr. After the two-step annealing process, crystals were cut into two pieces. Contacts of pressed indium were applied to the rough side of the annealed sample and copper wires were attached with silver paint. These wires were attached to a Keithley model 2400 source meter, which is capable of measuring resistance up to 200 MΩ. Current-voltage plots taken before and after light exposure confirmed that the pressed indium contacts showed nearly ohmic behavior (Supplementary Information). Prior studies of pressed indium contacts on heavily Nb-doped STO also showed nearly ohmic behavior but with lower contact resistance values28.

For the monochromatic light source and PL measurements, we used a JY-Horiba FluoroLog-3 spectrofluorometer system equipped with double grating excitation and emission monochromators. Different pieces of the same crystal were used for the IR experiment and the photo-lithography experiment. To measure IR spectra and resistance simultaneously, a 405 nm high brightness light emitting diode (LED) was placed approximately 2.5 cm from the polished side of the sample inside the compartment of a Bomen DA8 vacuum Fourier transform spectrometer (FTIR). When on, the LED was powered with approximately 20 mA and 3.4 V. The exposure geometry was the same as that shown in Fig. 2 (inset).

For the photo-lithographic experiment, 4 indium contacts were pressed on the rough side of the annealed STO crystal, and wires were attached with silver paint. The crystal was attached to a glass microscope slide with the contacts facing downwards with double-sided carbon sticky tape. The tape was insulating and did not affect the electrical measurements. The slide was placed in a PL microscope (Klar Scientific, 10x objective, numerical aperture 0.2) with an XY stage and a 405 nm laser diode (5 mW) as the excitation source. The laser spot at the back of the contact had a diameter of ~5 μm and was surrounded by a weaker spot (~1 mm diameter) of scattered light.

### Data Availability Statement

The datasets generated during and/or analysed during the current study are available from the corresponding author on reasonable request.


text:
### A Quantitative Analysis of Security, Anonymity and Scalability for the Lightning Network

Sergei Tikhomirov, Pedro Moreno-Sanchez, and Matteo Maffei

##### Abstract

Payment channel networks have been introduced to mitigate the scalability issues inherent to permissionless decentralized cryptocurrencies such as Bitcoin. Launched in 2018, the Lightning Network (LN) has been gaining popularity and consists today of more than 5000 nodes and 30000 payment channels that jointly hold 895 bitcoins (7.6M USD as of February 2020). This adoption has motivated research from both academia and industry. Payment channels suffer from security vulnerabilities, such as the wormhole attack, anonymity issues, and scalability limitations related to the upper bound on the number of concurrent payments per channel, which have been pointed out by the scientific community but never quantitatively analyzed. In this work, we first analyze the proneness of the LN to the wormhole attack and attacks against anonymity. We observe that an adversary needs to control only 2% of LN nodes to learn sensitive payment information (e.g., sender, receiver and payment amount) or to carry out the wormhole attack. Second, we study the management of concurrent payments in the LN and quantify its negative effect on scalability. We observe that for micropayments, the forwarding capability of up to 50% of channels is restricted to a value smaller than the overall channel capacity. This phenomenon not only hinders scalability but also opens the door for DoS attacks: We estimate that a network-wide DoS attack costs within 1.5M USD, while isolating the biggest community from the rest of the network costs only 225k USD. Our findings should prompt the LN community to consider the security, privacy and scalability issues of the network studied in this work when educating users about path selection algorithms, as well as to adopt multi-hop payment protocols that provide stronger security, privacy and scalability guarantees.

Available format(s)
Category
Applications
Publication info
Published elsewhere. Security & Privacy on the Blockchain (IEEE S&B) 2020
Keywords
bitcoinblockchainpayment channel networks
Contact author(s)
sergey s tikhomirov @ gmail com
History
2021-03-31: last of 3 revisions
See all versions
Short URL
https://ia.cr/2020/303

CC BY

BibTeX

@misc{cryptoeprint:2020/303,
author = {Sergei Tikhomirov and Pedro Moreno-Sanchez and Matteo Maffei},
title = {A Quantitative Analysis of Security, Anonymity and Scalability for the Lightning Network},
howpublished = {Cryptology ePrint Archive, Paper 2020/303},
year = {2020},
note = {\url{https://eprint.iacr.org/2020/303}},
url = {https://eprint.iacr.org/2020/303}
}

Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.


text:
# Are the following expressions for base dissociation constant equivalent?

Given a weak base such as $\ce{C2H5NH2}$, I usually see the expression for $K_\text{b}$ as

$K_\text{b} = \frac{[\ce{C2H5NH3+}][\ce{OH-}]}{[\ce{C2H5NH2}]}$.

Are the expressions

$K_\text{b} = \frac{[\ce{C2H5NH3+}]}{[\ce{C2H5NH2}][\ce{H+}]}$, $K_\text{b} = \frac{[\ce{C2H5NH3+}][\ce{H2O}]}{[\ce{C2H5NH2}][\ce{H3O+}]}$

equivalent? Do they represent the same process (if so, why do the latter two not work with calculations?) or fundamentally different ways of dissociation? Which one represents what's happening molecularly most accurately? Given the Brønsted-Lowry definition of a base as a proton acceptor I'd think that the second one is most accurate, but I rarely see it used.

I try to say it clearer and simpler.

The second expression respects to the equilibrium

$\ce {C2H5NH2 + H+ <=> C2H5NH3+}$.

And the third

$\ce {C2H5NH2 + H3O+ <=> C2H5NH3+ + H2O}$.

However, what you want to know is the equilibrium constant of

$\ce {C2H5NH2 + H2O <=> C2H5NH3+ + OH-}$.

The former two both describe how $\ce {C2H5NH3+}$ behaving as a weak acid in aqueous solution, and the third one describes how $\ce {C2H5NH2}$ behaving as a weak base in aqueous solution (hydrolyzation). They are different because in the former two equilibrium $\ce {C2H5NH2}$ are not receiving proton directly from $\ce {H2O}$.

Or you may look it by this view: the last one can be seen as the result of two equilibrium:

1. $\ce {C2H5NH2 + H+ <=> C2H5NH3+}$
2. $\ce {H2O <=> H+ + OH-}$

Or you may write as

1. $\ce {C2H5NH2 + H3O+ <=> C2H5NH3+ + H2O}$
2. $\ce {2H2O <=> H3O+ + OH-}$

So you can see there is obvious different. The equilibrium constant of $\ce {2H2O <=> H3O+ + OH-}$ is not 1, so the base dissociation constant $K_b$ will not be the same as what you said.

And there is an important convention that in an aqueous solution system, the equilibrium constant should omit [$\ce {H2O}$] because it is an insignificant constant in the most cases.

$$K_\mathrm b = \frac{[\ce{C2H5NH3+}]}{[\ce{C2H5NH2}][\ce{H+}]}$$ This expression is infact $1/K_\mathrm a$ which is related with $K_\mathrm a K_\mathrm b=K_\mathrm w$

$$K_\mathrm b = \frac{[\ce{C2H5NH3+}][\ce{H2O}]}{[\ce{C2H5NH2}][\ce{H3O+}]}$$ here $K_\mathrm b$ is correct but nowadays any constant concentration (here water) is not written in the expression. As it is assumed that concentration of water is not changing when reaction proceed.


text:
## Results (1-50 of 82 matches)

Next
Label $\alpha$ $A$ $d$ $N$ $\chi$ $\mu$ $\nu$ $w$ prim $\epsilon$ $r$ First zero Origin
4-1440e2-1.1-c0e2-0-0 $0.847$ $0.516$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$0.0, 0.0 0 1 0 0.848296 Modular form 1440.1.bh.b 4-1440e2-1.1-c0e2-0-1 0.847 0.516 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $0.0, 0.0$ $0$ $1$ $0$ $1.05868$ Modular form 1440.1.bh.a
4-1440e2-1.1-c0e2-0-2 $0.847$ $0.516$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$0.0, 0.0 0 1 0 1.32639 Modular form 1440.1.bh.c 4-1440e2-1.1-c1e2-0-0 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $0$ $0.155592$ Modular form 1440.2.f.b
4-1440e2-1.1-c1e2-0-1 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 1 0 0.216340 Modular form 1440.2.x.b 4-1440e2-1.1-c1e2-0-10 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $0$ $0.536994$ Modular form 1440.2.x.d
4-1440e2-1.1-c1e2-0-11 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 1 0 0.568764 Bianchi modular form 2.0.8.1-32400.3-g Elliptic curve 2.0.8.1-32400.3-g 4-1440e2-1.1-c1e2-0-12 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $0$ $0.585507$ Bianchi modular form 2.0.8.1-32400.2-f Bianchi modular form 2.0.8.1-32400.4-g Elliptic curve 2.0.8.1-32400.2-f Elliptic curve 2.0.8.1-32400.4-g
4-1440e2-1.1-c1e2-0-13 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 1 0 0.586999 Modular form 1440.2.q.b 4-1440e2-1.1-c1e2-0-14 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $0$ $0.593812$ Bianchi modular form 2.0.8.1-32400.2-d Bianchi modular form 2.0.8.1-32400.4-a Elliptic curve 2.0.8.1-32400.2-d Elliptic curve 2.0.8.1-32400.4-a
4-1440e2-1.1-c1e2-0-15 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 1 0 0.599925 Bianchi modular form 2.0.8.1-32400.2-a Bianchi modular form 2.0.8.1-32400.4-d Elliptic curve 2.0.8.1-32400.2-a Elliptic curve 2.0.8.1-32400.4-d 4-1440e2-1.1-c1e2-0-16 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $0$ $0.630250$ Modular form 1440.2.x.f
4-1440e2-1.1-c1e2-0-17 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 1 0 0.632443 Modular form 1440.2.q.a 4-1440e2-1.1-c1e2-0-18 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $0$ $0.687166$ Modular form 1440.2.f.e
4-1440e2-1.1-c1e2-0-19 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 1 0 0.700816 Modular form 1440.2.f.c 4-1440e2-1.1-c1e2-0-2 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $0$ $0.310573$ Modular form 1440.2.q.d
4-1440e2-1.1-c1e2-0-20 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 1 0 0.701396 Modular form 1440.2.x.i 4-1440e2-1.1-c1e2-0-21 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $0$ $0.706816$ Modular form 1440.2.x.a
4-1440e2-1.1-c1e2-0-22 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 1 0 0.715435 Bianchi modular form 2.0.8.1-32400.1-a Bianchi modular form 2.0.8.1-32400.5-a Elliptic curve 2.0.8.1-32400.1-a Elliptic curve 2.0.8.1-32400.5-a 4-1440e2-1.1-c1e2-0-23 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $0$ $0.722251$ Modular form 1440.2.a.p
4-1440e2-1.1-c1e2-0-24 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 1 0 0.731932 Bianchi modular form 2.0.8.1-32400.2-e Bianchi modular form 2.0.8.1-32400.4-c Elliptic curve 2.0.8.1-32400.2-e Elliptic curve 2.0.8.1-32400.4-c 4-1440e2-1.1-c1e2-0-25 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $0$ $0.791800$ Modular form 1440.2.f.f
4-1440e2-1.1-c1e2-0-26 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 1 0 0.802418 Modular form 1440.2.x.h 4-1440e2-1.1-c1e2-0-27 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $0$ $0.811234$ Modular form 1440.2.k.b
4-1440e2-1.1-c1e2-0-28 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 1 0 0.862970 Bianchi modular form 2.0.8.1-32400.3-i Elliptic curve 2.0.8.1-32400.3-i 4-1440e2-1.1-c1e2-0-29 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $-1$ $1$ $0.885365$ Bianchi modular form 2.0.8.1-32400.3-f Elliptic curve 2.0.8.1-32400.3-f
4-1440e2-1.1-c1e2-0-3 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 1 0 0.320189 Modular form 1440.2.b.a 4-1440e2-1.1-c1e2-0-30 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $0$ $0.905665$ Modular form 1440.2.f.g
4-1440e2-1.1-c1e2-0-31 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 1 0 0.918514 Bianchi modular form 2.0.8.1-32400.2-g Bianchi modular form 2.0.8.1-32400.4-h Elliptic curve 2.0.8.1-32400.2-g Elliptic curve 2.0.8.1-32400.4-h 4-1440e2-1.1-c1e2-0-32 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $0$ $0.925447$ Modular form 1440.2.k.c
4-1440e2-1.1-c1e2-0-33 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 1 0 0.926426 Modular form 1440.2.b.b 4-1440e2-1.1-c1e2-0-34 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $0$ $0.956692$ Modular form 1440.2.a.q
4-1440e2-1.1-c1e2-0-35 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 -1 1 0.962955 Bianchi modular form 2.0.8.1-32400.3-b Bianchi modular form 2.0.8.1-32400.3-e Elliptic curve 2.0.8.1-32400.3-b Elliptic curve 2.0.8.1-32400.3-e 4-1440e2-1.1-c1e2-0-36 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $0$ $0.984168$ Modular form 1440.2.q.f
4-1440e2-1.1-c1e2-0-37 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 -1 1 1.01518 Bianchi modular form 2.0.8.1-32400.3-h Elliptic curve 2.0.8.1-32400.3-h 4-1440e2-1.1-c1e2-0-38 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $0$ $1.02074$ Modular form 1440.2.q.e
4-1440e2-1.1-c1e2-0-39 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 1 0 1.06503 Bianchi modular form 2.0.8.1-32400.3-n Elliptic curve 2.0.8.1-32400.3-n 4-1440e2-1.1-c1e2-0-4 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $0$ $0.331720$ Modular form 1440.2.k.a
4-1440e2-1.1-c1e2-0-40 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 -1 1 1.08829 Bianchi modular form 2.0.8.1-32400.3-d Elliptic curve 2.0.8.1-32400.3-d 4-1440e2-1.1-c1e2-0-41 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $0$ $1.10234$ Modular form 1440.2.x.j
4-1440e2-1.1-c1e2-0-42 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 -1 1 1.12197 Bianchi modular form 2.0.8.1-32400.3-k Bianchi modular form 2.0.8.1-32400.3-l Elliptic curve 2.0.8.1-32400.3-k Elliptic curve 2.0.8.1-32400.3-l 4-1440e2-1.1-c1e2-0-43 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $-1$ $1$ $1.12756$ Bianchi modular form 2.0.8.1-32400.2-c Bianchi modular form 2.0.8.1-32400.4-b Elliptic curve 2.0.8.1-32400.2-c Elliptic curve 2.0.8.1-32400.4-b
4-1440e2-1.1-c1e2-0-44 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 -1 1 1.18512 Bianchi modular form 2.0.8.1-32400.3-j Elliptic curve 2.0.8.1-32400.3-j 4-1440e2-1.1-c1e2-0-45 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $-1$ $1$ $1.23011$ Bianchi modular form 2.0.8.1-32400.2-h Bianchi modular form 2.0.8.1-32400.4-f Elliptic curve 2.0.8.1-32400.2-h Elliptic curve 2.0.8.1-32400.4-f
4-1440e2-1.1-c1e2-0-46 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 -1 1 1.23519 Bianchi modular form 2.0.8.1-32400.3-m Elliptic curve 2.0.8.1-32400.3-m 4-1440e2-1.1-c1e2-0-47 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $0$ $1.26966$ Modular form 1440.2.q.h
4-1440e2-1.1-c1e2-0-48 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 1 0 1.27146 Modular form 1440.2.q.g 4-1440e2-1.1-c1e2-0-49 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $2$ $1.30436$ Modular form 1440.2.q.c
4-1440e2-1.1-c1e2-0-5 $3.39$ $132.$ $4$ $2^{10} \cdot 3^{4} \cdot 5^{2}$ 1.1 $$1.0, 1.0 1 1 0 0.361034 Bianchi modular form 2.0.8.1-32400.2-b Bianchi modular form 2.0.8.1-32400.4-e Elliptic curve 2.0.8.1-32400.2-b Elliptic curve 2.0.8.1-32400.4-e 4-1440e2-1.1-c1e2-0-50 3.39 132. 4 2^{10} \cdot 3^{4} \cdot 5^{2} 1.1$$ $1.0, 1.0$ $1$ $1$ $2$ $1.49852$ Modular form 1440.2.x.c
Next


text:
1. May 24, 2006

### Natalya18

Hi,
Im studying year 12 physics and im a bit stuck with this question:

The Earth is in its orbit around the sun. The mass of the earth is 6*10^24kg and the radius of its orbit is 1.5*10^11 metres

Question wants me to find the magnitude of the momentum
Im missing something, and i cant seem to remember how to do it,
any help is appreciated,

thanks, :uhh:

2. May 24, 2006

### Curious3141

Start with the formulae for gravitational force and centripetal force.

3. May 24, 2006

### Exotic-Matter

It's a bit simple, as Curious said, start with equating the gravitational and centripetal forces, then substitute in the momentum formula.

4. Jul 12, 2006

### thiotimoline

Need to know the mass of the Sun so that the equation of centripetal force with Sun's gravitational force gives you Earth's orbital velocity, which then can be used to find Earth's momentum.


text:
Change the chapter
Question
One half the $\gamma$ rays from ${}^{99\textrm{m}}\textrm{Tc}$ are absorbed by a 0.170-mm-thick lead shielding. Half of the $\gamma$ rays that pass through the first layer of lead are absorbed in a second layer of equal thickness. What thickness of lead will absorb all but one in 1000 of these $\gamma$ rays?
$1.69\textrm{ mm}$
Solution Video

# OpenStax College Physics for AP® Courses Solution, Chapter 32, Problem 12 (Problems & Exercises) (3:13)

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Video Transcript
This is College Physics Answers with Shaun Dychko. We are told that a 0.170 millimeter thickness of lead blocks half of the gamma rays that are passing through it created by technetium-99 decay so the question is what total thickness is needed then to have 1 in a 1000 gamma rays pass through the total thickness? So we have one thickness here of width d subscript one-half and then another thickness here again d subscript one-half and these gamma rays are being emitted by some technetium-99 and half of them will get absorbed by this layer and then half of the ones that remain will get absorbed by the next layer and then so on and so on and then by the time you get to the end here, you have only one out of the original 1000 making it this far. Okay! So the total number that end up at the end... so this is N is one-one thousandth of what we had originally or you can say that the number that pass through is the original number that hit the beginning multiplied by one-half as many times as there are thicknesses of lead and that's what this fraction is. This fraction tells how many layers of 0.170 millimeters of lead do you have so this t is what we are solving for— this total thickness. Okay! So we can divide both sides by N naught and we get the total number that pass through divided by the original number that hit equals one-half to the power of total thickness divided by the thickness for absorbing half. This N over N naught is 1 over a thousand and we can take the natural logarithm of both sides here because we want to solve for this factor in the exponent and so taking natural logarithm of both sides we have ln of N over N naught on the left and on the right side, one of the logarithm rules is that when you take the log of something with an exponent, you can turn that exponent into a factor in front so we have t over d one-half times natural logarithm of a half on the right side. And then we multiply both sides by d one-half divided by natural log of 2 or natural log of a half I should say and then we solve for t that way. So t is the thickness that blocks half the gamma rays times natural logarithm of the fraction that penetrates divided by natural logarithm a half. So that's 0.170 millimeters times natural log of 1 over a 1000 divided by natural log of a half and that's 1.69 millimeters.


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# Finding the Point of Tangency Theoretically

1. Mar 30, 2006

### SwAnK

Hey I was given the question:
A (x1,y1) and B (x2,y2) are two points on the parabola y=ax^2+bx+c. At what point is the tangent to the parabola parallel to the secant AB.
Here are the steps I took, im just wanting to know if im heading in the right direction.

First I just made the slope of AB = m.
then found the derivative of ax^2+bx+c (2ax+1) and made m=2ax+1.
Then isolated x, which would give an x coordinate of m/2a+1.
Took the x value and subed it into the equation ax^2+bx+c to get a y value.

2. Mar 30, 2006

### PPonte

Why 2ax + 1, instead of 2ax + b ?

3. Mar 30, 2006

### SwAnK

ahh, yes good point, i wasn't thinking. Thanx for catching that PPonte!

4. Apr 1, 2006

### arildno

Nowhere have you related your "x" to the x1 and x2; what you have done, is to parametrize the parabola in terms of the local slope m which, by the way, you haven't done correctly.

$$m=2ax+b\to{x}=\frac{m-b}{2a}$$


text:
Python笔记 ＋

# 神经元激活函数

\begin{align} z & = w^{T}a + b \\ & = \sum_{j}w_{j}a_{j} + b \end{align}

a是前一层神经元激活值，作为这一层的输入，或者就是最外层的输入，w是对应的weight。以上公式所所有值都是vector。激活函数使用 $$\sigma$$ 来表示。

## sigmoid

sigmoid函数公式如下：

$$\sigma (z) = \text{sigmoid}(z) = \frac{1}{1+e^{-z}}$$

$$\sigma'(z) = \sigma(z) \cdot (1-\sigma(z))$$

## tanh（读tanch）

$$\sigma(z) = \text{tanh}(z) = \frac{e^z-e^{-z}}{e^z+e^{-z}}$$

tanh函数对z求导的结果为：

$$\sigma'(z) = 1 - \sigma(z)^2$$

tanh激活函数与sigmoid函数有着相似的曲线，不同的是导数，tanh的导数最大值为1，输出也不同，sigmoid输出在0到1之间，tanh的输出在-1到1之间，因此使用tanh对数据的normailization会有所不同。

Which type of neuron should you use in your networks, the tanh or sigmoid? A priori the answer is not obvious, to put it mildly! However, there are theoretical arguments and some empirical evidence to suggest that the tanh sometimes performs better* （*See, for example, Efficient BackProp, by Yann LeCun, Léon Bottou, Genevieve Orr and Klaus-Robert Müller (1998), and Understanding the difficulty of training deep feedforward networks, by Xavier Glorot and Yoshua Bengio (2010)..） Let me briefly give you the flavor of one of the theoretical arguments for tanh neurons. Suppose we're using sigmoid neurons, so all activations in our network are positive. Let's consider the weights $$w^{l+1}_{jk}$$ input to the jth neuron in the (l+1)th layer. The rules for backpropagation tell us that the associated gradient will be $$a^l_{k}\delta^{l+1}_j$$. Because the activations are positive the sign of this gradient will be the same as the sign of $$\delta^{l+1}_j$$. What this means is that if $$\delta^{l+1}_j$$ is positive then all the weights $$w^{l+1}_{jk}$$ will decrease during gradient descent, while if $$\delta^{l+1}_j$$ is negative then all the weights $$w^{l+1}_{jk}$$ will increase during gradient descent. In other words, all weights to the same neuron must either increase together or decrease together. That's a problem, since some of the weights may need to increase while others need to decrease. That can only happen if some of the input activations have different signs. That suggests replacing the sigmoid by an activation function, such as tanh, which allows both positive and negative activations. Indeed, because tanh is symmetric about zero, tanh(−z)=−tanh(z), we might even expect that, roughly speaking, the activations in hidden layers would be equally balanced between positive and negative. That would help ensure that there is no systematic bias for the weight updates to be one way or the other.

How seriously should we take this argument? While the argument is suggestive, it's a heuristic, not a rigorous proof that tanh neurons outperform sigmoid neurons. Perhaps there are other properties of the sigmoid neuron which compensate for this problem? Indeed, for many tasks the tanh is found empirically to provide only a small or no improvement in performance over sigmoid neurons. Unfortunately, we don't yet have hard-and-fast rules to know which neuron types will learn fastest, or give the best generalization performance, for any particular application.

$$\text{sigmoid}(z) = \frac{1+\text{tanh}(\frac{z}{2})}{2}$$

## ReLU

Rectified Linear Unit，公式为：

$$\sigma(z) = \text{relu}(z) = \text{max}(0, z)$$

$$\sigma'(z) = \begin{cases} 1, & \text{if } z > 0 \\ 0, & \text{if } z \le 0 \end{cases}$$

relu很不一样，跟sigmoid和tanh相比，因为它会“死掉”。

ReLU的输入层也是ReLU，因此输入一定大于等于0，如果输出为0，可以想象很多w都是负数，此时w和b得不到更新，输入符号也不会改变，因此，可能永远活不过来......

## softmax

softmax神经元常用在网络的最后一层，在分类问题中，它可以输出一个概率分布，计算公式为：

$$a_j^L=\frac{e^{z_j^L}}{\sum_k e^{z_k^L}}$$

softmax的输出既然是概率分布，那必然就是在 0-1 之间了！

\begin{align} \frac{\partial a_j}{\partial z_j} &= \frac{e^{z_j} (\sum_k e^{z_k}) - {e^{z_j}}^2}{(\sum_k e^{z_k})^2} \\ &= \frac{e^{z_j}}{\sum_k e^{z_k}} - \left(\frac{e^{z_j}}{\sum_k e^{z_k}}\right)^2 \\ &= a_j - (a_j)^2 \\ &= a_j(1-a_j) \end{align}

\begin{align} \frac{\partial a_j}{\partial z_i} &= \frac{- e^{z_j} e^{z_i}}{(\sum_k e^{z_k})^2} \\ &= - a_i a_j \end{align}

-- EOF --

### 留言区

《神经元激活函数》有1条留言

• 麦新杰

tanh拥有更深的饱和程度，从学习卡壳的状态走出来更加缓慢... [回复]

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