text:
# Thread: Game of bridge and probability

1. ## Game of bridge and probability

Hello Members,
There are 4 players in game of bridge and thirteen cards out of 52 cards are to be distributed.(A deck of bridge cards are arranged in a four suits of 13 each. The 4 suits are called spades, clubs, hearts, diamonds. The last 2 are red, the first 2 are black. Cards of the same face value are called cards of the same kind. There are 13 face values(2,3,...,10,jack,queen,king,ace)in each suit. Playing bridge means distributing the cards to 4 players namely North, South, East, West so that each player receives 13 cards.) Now what is the probability that some player has complete suit?
Answer: I am referring to various books on statistics to answer this question, if any member knows the answer, he may answer this question

2. ## Re: Game of bridge and probability

For simplicity I will assume that the player is the first dealt to. (It can be shown that it doesn't matter but that proof is more tedious.) The probability the player is dealt, say, a heart on the first deal is 13/52= 1/2. That leaves 51 cards, 39 of them non-spades. So the probability the other three players are dealt non-spades is (39/51)(38/50)(37/49). When we get back to the player, there are 48 cards in the deck, 12 of them spades so the probability this player gets a spade as the second card is 12/48= 1/4. Continue that until all cards have been dealt. That would be the probability that player received all spades. Multiply by 4 to get the probability that player get a hand of all cards from any one of the four suits. Finally, multiply by 4 to get the probability any one of the 4 players has such a hand.

3. ## Re: Game of bridge and probability

There are $\dbinom{52}{13}$ different possible hands.

There are $4$ hands consisting entirely of 1 suit

$p = \dfrac{4}{\dbinom{52}{13}} = \dfrac{1}{158753389900} \approx 6.3 \times 10^{-12}$

4. ## Re: Game of bridge and probability

Originally Posted by romsek
There are $\dbinom{52}{13}$ different possible hands.

There are $4$ hands consisting entirely of 1 suit

$p = \dfrac{4}{\dbinom{52}{13}} = \dfrac{1}{158753389900} \approx 6.3 \times 10^{-12}$
The question does not ask about one hand. It suggests all four hands are dealt, what is the probability that at least one hand is 13 cards of the same suit? Unless I am misreading...

5. ## Re: Game of bridge and probability

The number of ways to deal four hands in bridge is $\dfrac{52!}{(13!)^4}$.

There are $\dfrac{39!}{(13!)^3}$ ways for North to have all spades. Thus, there are $4^2\cdot \dfrac{39!}{(13!)^3}$ ways for one of the directions to have all of one suit.
There are $\dfrac{26!}{(13!)^2}$ ways for North and South to have Spades and Hearts respectively. Thus, there are $2\cdot 6^2 \dfrac{26!}{(13!)^2}$ ways for two of the directions to both have all of one suit.
There are $\dfrac{13!}{13!}$ ways for North, South, and West to have Spades, Hearts, and Clubs respectively. Thus, there are $4!$ ways for three of the direction to have all of one suit.
There is $1$ way for North, South, East, and West to have Spades, Hearts, Clubs, and Diamonds respectively. Thus there are $4!$ ways for all four of the directions to have all of one suit.

The probability for at least one direction to have all of one suit is given by:

$\dfrac{4^2\cdot \dfrac{39!}{(13!)^3}-2\cdot 6^2 \dfrac{26!}{(13!)^2}+4!-4!}{\dfrac{52!}{(13!)^4} \approx 2.52\times 10^{-11}$

6. ## Re: Game of bridge and probability

I have solved this question as follows.
In a game of bridge, let $A_i$ be the event "Player number i has a complete suit". Then $p_i=\frac{4}{\binom{52}{13}}$; the event that both players i and j have complete suits can occur in 4*3 ways and has probability $p_{ij}=\frac{12}{\binom{52}{13}\binom{39}{13}}$. Similarly we get $p_{ijk}=\frac{24}{\binom{52}{13}\binom{39}{13} \binom {26}{13}}$

Finally $p_{ijkl}=p_{ijk}$ since whenever the 3 players have the complete suit so does the fourth.
The probability that some player has a complete suit is therefore $P_1=4*p_i-6*p_{ij}+4*p_{ijk}-p_{ijkl}$.Hence the final answer $2.519631234522642 * 10^{-11}$

7. ## Re: Game of bridge and probability

Originally Posted by Vinod
I have solved this question as follows.
In a game of bridge, let $A_i$ be the event "Player number i has a complete suit". Then $p_i=\frac{4}{\binom{52}{13}}$; the event that both players i and j have complete suits can occur in 4*3 ways and has probability $p_{ij}=\frac{12}{\binom{52}{13}\binom{39}{13}}$. Similarly we get $p_{ijk}=\frac{24}{\binom{52}{13}\binom{39}{13} \binom {26}{13}}$

Finally $p_{ijkl}=p_{ijk}$ since whenever the 3 players have the complete suit so does the fourth.
The probability that some player has a complete suit is therefore $P_1=4*p_i-6*p_{ij}+4*p_{ijk}-p_{ijkl}$.Hence the final answer $2.519631234522642 * 10^{-11}$
Yup, that is exactly what I had in post #5.

8. ## Re: Game of bridge and probability

Hello,
If you put your last calculation in latex in good legible form, it would be better for readers to read and understand.

9. ## Re: Game of bridge and probability

Originally Posted by Vinod
Hello,
If you put your last calculation in latex in good legible form, it would be better for readers to read and understand.
$\dfrac{ 4^2 \cdot \dfrac{ 39! }{ (13!)^3 } - 2\cdot 6^2 \dfrac{ 26! }{ (13!)^2 } + 4! - 4! }{ \dfrac{ 52! }{ (13!)^4 }} \approx 2.52 \times 10^{-11}$


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# What are the intercepts of -x+3y=-3?

Mar 6, 2018

The x-intercept is $\left(3 , 0\right)$.

The y-intercept is $\left(0 , - 1\right)$.

#### Explanation:

Given:

$- x + 3 y = - 3$ is a linear equation in standard form:

$A x + B y = C$.

X-intercept: value of $x$ when $y = 0$.

Substitute $0$ for $y$ and solve for $x$.

$- x + 3 \left(0\right) = - 3$

$- x = - 3$

Multiply both sides by $- 1$.

$x = 3$

The x-intercept is $\left(3 , 0\right)$.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Y-intercept: value of $y$ when $x = 0$

Substitute $0$ for $x$ and solve for $y$.

$0 + 3 y = - 3$

$3 y = - 3$

Divide both sides by $3$.

$\frac{{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}^{1} y}{{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}^{1}} = - {\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}^{1} / {\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}^{1}$

$y = - 1$

The y-intercept is $\left(0 , - 1\right)$.

You can graph this equation by plotting the x- and y-intercepts and drawing a straight line through them.

graph{-x+3y=-3 [-11.25, 11.25, -5.625, 5.625]}

format: plain


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# How-To-Guide: External GPU on OSX - how to use CUDA on your Mac

GROUPS:

The neural network and machine learning framework has become one of the key features of the latest releases of the Wolfram Language. Training neural networks can be very time consuming on a standard CPU. Luckily the Wolfram Language offers an incredible easy way to use a GPU to train networks - and do lots of other cool stuff. The problem with this was/is that most current Macs do not have an NVIDIA graphics card, which is necessary to access this framework within the Wolfram Language. Therefore, Wolfram Inc. had decided to drop support for GPUs on Macs. There is however a way to use GPUs on Macs. For example you can use an external GPU like the one offered by Bizon.

Apart from the BizonBox there a couple of cables and a power supply. You can buy/configure different versions of the BizonBox: there is a range of different graphics cards available and you can buy a the BizonBox 2s which basically connects via Thunderbolt and the BizonBox 3 which connects to USB-C.

Luckily, Wolfram have decided to reintroduce support for GPUs in Mathematica 11.1.1 - see the discussion here.

I have a variety of these BizonBoxes (both 2s and 3) and a range of Macs. I thought it would be a good idea to post a how-to. The essence of what I will be describing in this post should work for most Macs. I ran Sierra on all of them. Here is the recipe to get the thing to work:

## Installation of the BizonBox, the required drivers, and compilers

1. I will assume that you have Sierra installed and that Xcode is running. One of the really important steps if you want to use compilers is to downgrade the command line tools to version 7.3 You will have to log into your Apple Developer account and download the Command Line Tools version 7.3. Install the tools and run the terminal command (not in Mathematica!):

sudo xcode-select  --switch /Library/Developer/CommandLineTools

2. Reboot your Mac into safe mode, i.e. hold CMD+R while rebooting.

3. Open a terminal (under item Utilities at the top of the screen).

4. Enter

csrutil disable

5. Shut the computer down.

6. Connect your BizonBox to the mains and to either the thunderbolt or USB-C port of your Mac.

8. Click on the Apple symbol in the top left. Then "About this Mac" and "System Report". In the Thunderbolt section you should see something like this:

1. In the documentation of the BizonBox you will find a link to a program called bizonboxmac.zip. Download that file and unzip it.

2. Open the folder and click on "bizonbox.prefPane" to install. (If prompted to, do update!)

3. You should see this window:

1. Click on Activate. Type in password if required to do so. It should give something like this:

Then restart.

what you download should be something like cuda8.0.61mac.dmg and it should be more or less 1.44 GB worth.

1. Install the toolkit with all its elements.

## First tests

Now you should be good to go. Open Mathematica 11.1.1. Execute

Needs["CUDALink"]
Needs["CCompilerDriver"]
CUDAResourcesInstall[]

Then try:

CUDAResourcesInformation[]

which should look somewhat like this:

Then you should check

SystemInformation[]

So far so good. Next is the really crucial thing:

CUDAQ[]

should give TRUE. If that's what you see you are good to go. Be more daring and try

CUDAImageConvolve[ExampleData[{"TestImage","Lena"}], N[BoxMatrix[1]/9]] // AbsoluteTiming

You might notice that the non-GPU version of this command runs faster:

ImageConvolve[ExampleData[{"TestImage","Lena"}], N[BoxMatrix[1]/9]] // AbsoluteTiming

runs in something like 0.0824 seconds, but that's ok.

## Benchmarking (training neural networks)

obj = ResourceObject["CIFAR-10"];
trainingData = ResourceData[obj, "TrainingData"];
RandomSample[trainingData, 5]

You can check whether it worked:

RandomSample[trainingData, 5]

should give something like this:

These are the classes of the 50000 images:

classes = Union@Values[trainingData]

Let's build a network

module = NetChain[{ConvolutionLayer[100, {3, 3}],
BatchNormalizationLayer[], ElementwiseLayer[Ramp],

net = NetChain[{module, module, module, module, FlattenLayer[], 500,
Ramp, 10, SoftmaxLayer[]},
"Input" -> NetEncoder[{"Image", {32, 32}}],
"Output" -> NetDecoder[{"Class", classes}]]

When you train the network:

{time, trained} = AbsoluteTiming@NetTrain[net, trainingData, Automatic, "TargetDevice" -> "GPU"];

you should see something like this:

So the thing started 45 secs ago and it supposed to finish in 2m54s. In fact, it finished after 3m30s. If we run the same on the CPU we get:

The estimate kept changing a bit, but it settled down at about 18h20m.That is slower by a factor of about 315, which is quite substantial.

## Use of compiler

Up to now we have not needed the actual compiler. Let's try this, too. Let's grow a Mandelbulb:

width = 4*640;
height = 4*480;
iconfig = {width, height, 1, 0, 1, 6};
config = {0.001, 0.0, 0.0, 0.0, 8.0, 15.0, 10.0, 5.0};
camera = {{2.0, 2.0, 2.0}, {0.0, 0.0, 0.0}};
AppendTo[camera, Normalize[camera[[2]] - camera[[1]]]];
AppendTo[camera,
0.75*Normalize[Cross[camera[[3]], {0.0, 1.0, 0.0}]]];
AppendTo[camera, 0.75*Normalize[Cross[camera[[4]], camera[[3]]]]];
config = Join[{config, Flatten[camera]}];

pixelsMem = CUDAMemoryAllocate["Float", {height, width, 3}]

Now this should work:

mandelbulb =
CUDAFunctionLoad[File[srcf], "MandelbulbGPU", {{"Float", _, "Output"}, {"Float", _, "Input"}, {"Integer32", _, "Input"}, "Integer32", "Float", "Float"}, {16}, "UnmangleCode" -> False, "CompileOptions" -> "--Wno-deprecated-gpu-targets ", "ShellOutputFunction" -> Print]

Under certain circumstances you might want to specify the location of the compiler like so:

mandelbulb =
CUDAFunctionLoad[File[srcf], "MandelbulbGPU", {{"Float", _, "Output"}, {"Float", _, "Input"}, {"Integer32", _, "Input"}, "Integer32", "Float",
"Float"}, {16}, "UnmangleCode" -> False, "CompileOptions" -> "--Wno-deprecated-gpu-targets ", "ShellOutputFunction" -> Print,
"CompilerInstallation" -> "/Developer/NVIDIA/CUDA-8.0/bin/"]

This should give:

Now

mandelbulb[pixelsMem, Flatten[config], iconfig, 0, 0.0, 0.0, {width*height*3}];
pixels = CUDAMemoryGet[pixelsMem];
Image[pixels]

gives

So it appears that all is working fine.

## Problems

I did come up with some problems though. There is quite a number of CUDA functions:

Names["CUDALink*"]

Many work just fine.

res = RandomReal[1, 5000];
ListLinePlot[res]

ListLinePlot[First@CUDAImageConvolve[{res}, {GaussianMatrix[{{10}, 10}]}]]

The thing is that some don't and I am not sure why (I have a hypothesis though). Here are some functions that do not appear to work:

CUDAColorNegate CUDAClamp CUDAFold CUDAVolumetricRender CUDAFluidDynamics

and some more. I would be very grateful if someone could check these on OSX (and perhaps Windows?). I am not sure if the this is due to some particularity of my systems or something that could be flagged up to Wolfram Inc for checking.

When I wanted to try that systematically I wanted to use the function

WolframLanguageData

to look for the first example in the documentation of the CUDA functions, but it appears that no CUDA function is in the WolframLanguageData. I think tit would be great to have them there, too, and am not sure why they wouldn't be there.

In spite of these problems I hope that this post will help some Mac users to get CUDA going. It is a great framework and simple to use in the Wolfram Language. With the BizonBox and Mathematica 11.1.1 Mac users are no longer excluded from accessing this feature.

Cheers,

Marco

PS: Note, that there is anecdotal evidence that one can even use the BizonBox under Windows running in a virtual box under OSX. I don't have Windows, but I'd like to hear if anyone get this running.

2 months ago
15 Replies
 Sander Huisman 2 Votes That looks really neat! I had no idea that there was such a large speed-up! Which GPU do you have inside your bizon box? nevermind I see it in the screenshot I'm thinking about buying one...
2 months ago
 Neil Singer 5 Votes Marco,Awesome post! I was just looking into doing this. What is the reason for downgrading the command line tools? If you do not downgrade can you still run the built in Neural net functions (without using the compiler)? Thanks
2 months ago
 Arno Bosse 4 Votes Besides the Bizon Box (which comes with support), there are also a couple of other, cheaper DIY options available which have been reviewed on https://egpu.io/news/ For more eGPU benchmarks (not Mathematica) see http://barefeats.com.
2 months ago
 Marco Thiel 3 Votes Hi Sander,yes, I've got the TitanX. I do not have comparative benchmarks with the other ones though. For me it was definitely worth buying the boxes - and I am lucky that Wolfram reintroduced the support for them. I wouldn't say that I am particularly good at CUDA (quite the opposite), but I could make some code run substantially faster, which was really important for a project I have. Note, that you can also buy the BizonBox without the GPU, so if you have a spare one flying around you can (most likely) use that one.Cheers,Marco
2 months ago
 Marco Thiel 6 Votes Dear Neil,the downgrading is strictly speaking not necessary if you only want the Wolfram Language's Machine Learning and functions that do not require compilation. If you have the latest you see something like this, but with "The Version ('80300')" or so. It is a warning that the compilation failed. It is not a Mathematica/WolframLanguage problem. If you followed the instructions in the OP you would have generated a folder /Developer/NVIDIA/CUDA-8.0/samples/2_Graphics/Mandelbrot/ `you could try to use "make" to compile and that will fail unless you have downgraded the command line tools. See also this discussion here. The process needs the command line c-compilers and there is an incompatibility, I think.Best wishes,Marco
2 months ago
 Moderation Team 3 Votes - Congratulations! This post is now a Staff Pick! Thank you for your wonderful contributions. Please, keep them coming!
2 months ago
 Marco Thiel 2 Votes Dear Wolfram Team,I am very glad and thankful that you reacted so quickly to the comments about GPU access on Macs. Having access to this framework opens up many possibilities in research and teaching. I appreciate it that you sorted this out so swiftly and efficiently.Thank you,Marco
2 months ago
 Diego Zviovich 3 Votes Has anyone set up an eGPU with Windows?
2 months ago
 David Proffer 2 Votes A good source of current information on eGPU's: http://barefeats.com/
2 months ago
 Dear Marco, Thank you for a very informative post. You had responded to a question about downgrading the command line tools with: "the downgrading is strictly speaking not necessary if you only want the Wolfram Language's Machine Learning and functions that do not require compilation."I was wondering if you were had tried NetTrain without the downgrade of command line tools to 7.2. Thanks..Jan
15 days ago
 Updating Name 3 Votes Thank you Marco! I used your instructions to get a BizonBox 2S successfully working on my MacBook Pro.Some glitches I ran into that I'll mention in case they come up for others: I had trouble getting the Bizon to activate, and then to have the Nvidia control panel find the Bizon. Sometimes actions would work and other times not. After experiments and consultation with support, I replaced the long Thunderbolt cable supplied with the Bizon with a shorter one (three foot). That solved many of the problems. I still had difficulty getting the mac to see the box. Support said that the Mac should be off when the box is connected or disconnected. That helped. Finally I couldn't get cuda recognized as installed by the Nvidia app or Mm. The last thing I did before it worked was plug in a display to the Bizon - then everything started working. The display didn't stay plugged in, and I don't need it now, but it seemed that it needed to be there to initialize something. All of this was with a lot of on/off, rebooting, trying different things so I'm not sure the above is necessary, but in the end it got mine working following your instructions.Thanks again Marco - I'm not sure I would have stuck it out without knowing there was light at the end of the tunnel.Mike
14 days ago
 Eric Smith 2 Votes Marco, thank you for this post. I assume the instructions will get a lot simpler once everyone switches to macOS High Sierra which natively supports eGPU'shttps://9to5mac.com/2017/06/07/hands-on-macos-high-sierra-native-egpu-support-shows-promise-video/


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Chapter 17. Kinetics

# Rate Laws

Jessie A. Key

### Learning Objectives

• To gain an understanding of rate laws and determine rate laws from initial rates.
• To gain an understanding of and the ability to determine reaction orders (including units).

The mathematical relationship of reaction rate with reactant concentrations is known as the rate law. This relationship may rely more heavily on the concentration of one particular reactant, and the resulting rate law may include some, all, or none of the reactant species involved in the reaction.

For the following hypothetical reaction

a A + b B → c C

the rate law can be expressed as:

Rate = k[A]y[B]z

The proportionality constant, k, is known as the rate constant and is specific for the reaction shown at a particular temperature. The rate constant changes with temperature, and its units depend on the sum of the concentration term exponents in the rate law. The exponents (y and z) must be experimentally determined and do not necessarily correspond to the coefficients in the balanced chemical equation.

## Reaction Order

The sum of the concentration term exponents in a rate law equation is known as its reaction order.  We can also refer to the relationship for each reactant in terms of its exponent as an order.

For the following reaction between nitrogen dioxide and carbon monoxide:

NO2(g) + CO(g) → NO(g) + CO2(g)

The rate law is experimentally determined to be:  rate = k[NO2]2

Therefore, we would say that the overall reaction order for this reaction is second-order (the sum of all exponents in the rate law is 2), but zero-order for [CO] and second-order for [NO2].

The reaction order is most often a whole number such as 0, 1, or 2; however, there are instances where the reaction order may be a fraction or even a negative value.

Earlier it was mentioned that the units of the rate constant depend on the order of the reaction. Let’s quickly examine why this occurs. A simplified rate law can be expressed generically in the following way:

Rate = k[reactant]y

Units of rate = (units of rate constant)(units of concentration)y

Units of rate constant = $\frac{Units\ of\ rate}{{(Units\ of\ concentration)}^y}\$ = $\frac{M/s}{M^y}\$

Therefore, the units of the rate constant should be:

 Reaction Order Units of rate constant Zero-order M s-1 First-order s-1 Second-order L mol-1 s-1

## Determining Rate Laws from Initial Rates

The rate law can be determined experimentally using the method of initial rates, where the instantaneous reaction rate is measured immediately on mixing the reactants. The process is repeated over several runs or trials, varying the concentration one reactant at a time. These runs can then be compared to elucidate how changing the concentration of each reactant affects the initial rate.

Example 3

The initial rate of reaction for the reaction E + F → G was measured at three different initial concentrations of reactants as shown in the table.

1. Determine the rate law of the reaction.

2. Determine the rate constant.

 Trial Initial Rate (mole L-1 s-1) [E] (mole L-1) [F] (mole L-1) 1 2.73 x 10-5 0.100 0.100 2 5.47 x 10-5 0.200 0.100 3 2.71 x 10-5 0.100 0.200

Solution

1. Comparing trials 1 and 2, [E] is doubled, while [F] and the rate constant are held constant. This comparison will allow us to determine the order of reactant E:

$\frac{initial\ rate\ 2\ }{initial\ rate\ 1}\$ = ${\left(\frac{{{\mathbf [}{\mathbf E}{\mathbf ]\ }}_{{\mathbf 2}}}{{{\mathbf [}{\mathbf E}{\mathbf ]\ }}_{{\mathbf 1}}}\right)}^y\$

$\frac{{\rm 5.47\ x\ }{{\rm 10}}^{-5}\ Ms^{-1}\ }{{\rm 2.73\ x\ }{{\rm 10}}^{-5\ \ Ms^{-1}}\ }\$ = ${\left(\frac{{\mathbf 0}.{\mathbf 200}{\mathbf \ }{\mathbf M}}{{\mathbf 0}.{\mathbf 100}{\mathbf \ }{\mathbf M}}\right)}^y\$

2.00 = 2.00y

y = 1

Therefore, the reaction is first order with respect to [E].

Comparing trials 1 and 3, [F] is doubled, while [E] and the rate constant are held constant. This comparison will allow us to determine the order of reactant F:

$\frac{initial\ rate\ 3\ }{initial\ rate\ 1}\$ = ${\left(\frac{{{\mathbf [}{\mathbf F}{\mathbf ]\ }}_{{\mathbf 3}}}{{{\mathbf [}{\mathbf F}{\mathbf ]\ }}_{{\mathbf 1}}}\right)}^z\$

$\frac{{\rm 2.71\ x\ }{{\rm 10}}^{-5}\ Ms^{-1}\ }{{\rm 2.73\ x\ }{{\rm 10}}^{-5\ \ Ms^{-1}}\ }\$ = ${\left(\frac{{\mathbf 0}.{\mathbf 200}{\mathbf \ }{\mathbf M}}{{\mathbf 0}.{\mathbf 100}{\mathbf \ }{\mathbf M}}\right)}^z\$

0.993 = 2.00z

z = 0

Therefore, the reaction is zero order with respect to [F].

The rate law can now be written as:

Rate = k[E]1

2. Using the rate law we have just determined, substitute in the initial concentration values and initial rate for any trial and solve for the rate constant:

Rate = k[E]1

Using Trial 1: ${\rm 2.73\ x\ }{{\rm 10}}^{-5\ }M\ s^{-1} = k\left({\rm 0.100}{\rm M}\right)\$

$= \frac{{\rm 2.73\ x\ }{{\rm 10}}^{-5\ }Ms^{-1}}{\left({\rm 0.100}{\rm M}\right)}\$

k = 2.73 x 10-4 s-1

### Key Takeaways

• The rate law is a mathematical relationship obtained by comparing reaction rates with reactant concentrations.
• The reaction order is the sum of the concentration term exponents in a rate law equation.
• A reaction’s rate law may be determined by the initial rates method.


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Browse Questions

# Show that semi-vertical angle of the cone of maximum volume and of given slant height is $\tan^{-1} \sqrt 2$.

let ø be the semivertical angle of the right circular cone we know in a right circular cone l²=r²+h²............1 V = 1/3 πr²h V(h)= 1/3 π (l²-h²) h V'(h) = 1/3π (l²-3h²) V'(h)= 0 then l²/3 = h² V''(h) = -2πh V"(h)


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# Resistor wattage?

I'm new to electronics, what types of resistors should I get to make this circuit?

I have 6 IR LEDs (High output infrared LED 5mm 1.2VDC 100mA 940nm) that will be connected to an Ardunio Uno.

Going to be plugging this into the cigarette lighter which I think goes up to 13.5vdc?

• That circuit you've drawn looks suspiciously like an SCR. If that's the case, you'll be able to turn it on once, but it will then latch on, and you'll have to reset the supply to turn it off. – Thomas O Oct 4 '11 at 6:58
• @Thomas - no it's not an SCR. An SCR is a combination of a PNP and an NPN. The BC337 is a current limiter. – stevenvh Oct 4 '11 at 8:04
• Be warned that the power in a car can spike to in excess of 40V - often up in the 60V region. You should use a properly rated regulator to limit the spikes. – Majenko Oct 4 '11 at 8:53

The 330$\Omega$ will see a very low current, so that can be a standard 1/4W.
Right, that's spoken from experience. Let's do the calculation anyway. You always have to calculate for worst case, and worst case here means the highest voltage drop across the resistor. Therefore let's take 5V in (don't know if Arduino works at 3.3V or 5V) and 0V across the 6.8$\Omega$ resistor. Then

$P = \dfrac{V^2}{R} = \dfrac{(5V - 0.6V)^2}{330\Omega} = 0.059 W$

which confirms what we thought. So a leaded 1/4W resistor is possible, but also an SMT resistor down to 0402 size will do. A 0402 will allow 100mW.

The 6.8$\Omega$ resistor will see slightly more than the collector current $I_C$, which will be limited to 100mA by the current limiting transistor BC337. Then

$P = I^2 \cdot R = (0.1A)^2 \cdot 6.8\Omega = 0.068 W$

So a 1/4W will do here as well.

edit
About the BC337. This works as a current limiter. If the current through the 6.8$\Omega$ resistor is greater than about 100mA the voltage drop will be 0.7V and the BC337 will start to conduct, reducing the base voltage of the BD135. This way the emitter current (and hence the collector current) of the BD135 can't go higher than 100mA.

• So what's the advantage of using this type of circuit (an extra transistor) to limit the current across the LEDs as opposed to a simple resistor? – m.Alin Oct 4 '11 at 11:33
• @m.Alin - A series resistor will drop the voltage difference between your supply and the sum of the LEDs, so the current will depend on the supply voltage. In the given schematic it's independent of that. – stevenvh Oct 4 '11 at 11:36
• Dissipation in the BD135 will be 100 mA * (Vsuppy-Vleds -0.7) or 0.1*(13.5-7.2-0.7) or 560 mW. BD135 junction/ambient is 100 C/W, so the junction will rise 56C. No problem for reasonable ambient temperatures, will never reach the abs. max temperature of 150C. – markrages Oct 4 '11 at 22:34

I can explain the calculation here.

EDIT: (Please note that it is a calculation in which the transistor BC337 is omitted)

Your input voltage Vdc = 13.5V

Voltage drop across a single IR LED Vled = 1.2V

Total voltage across 6 LEDs 6X1.2V = 7.2V

Now the Voltage that must be dropped across the Resistor Vr = 13.5-7.2 V = 6.3V

The current Flows through the LED I = 100 mA

The current Through the resistor is same, I = 100mA

Now the Value of resistor R = V/I = 6.3V/100mA = 63Ohm.

(Puting a standered value, R = 6.8 Ohm)

Now The Power Calculation

The power dissipated in the Resistor P = V X I = 6.3V X 100mA = 0.63W

Now you can't use ordinary 1/4W or 1/2W resistors.

It is better to use a 1W resistor of 68 Ohms.

• You're ignoring the current limiting BC337! – stevenvh Oct 4 '11 at 7:53
• @stevenvh How does the BC337 transistor work in this circuit? – m.Alin Oct 4 '11 at 7:55
• Also, I think you're ignoring the voltage drop across the BD135 transistor. – m.Alin Oct 4 '11 at 7:58
• @m.Alin - I'll add it to my answer. – stevenvh Oct 4 '11 at 8:07
• Saneesh's answer applies if onlt the resistors is used as a current limiter. Here the main voltagedropping is done by the BD135 transistor so Saneesh's answer does not apply in this case. – Russell McMahon Oct 4 '11 at 8:15


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Article

# Ehrhart polynomials of integral simplices with prime volumes

07/2011;
Source: arXiv

ABSTRACT For an integral convex polytope $\Pc \subset \RR^N$ of dimension $d$, we call
$\delta(\Pc)=(\delta_0, \delta_1,..., \delta_d)$ the $\delta$-vector of $\Pc$
and $\vol(\Pc)=\sum_{i=0}^d\delta_i$ its normalized volume. In this paper, we
will establish the new equalities and inequalities on $\delta$-vectors for
integral simplices whose normalized volumes are prime. Moreover, by using
those, we will classify all the possible $\delta$-vectors of integral simplices
with normalized volume 5 and 7.

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• Source
##### Article: Inequalities and Ehrhart -vectors
[hide abstract]
ABSTRACT: -vector of a lattice polytope. We also provide combinatorial proofs of two results of Stanley that were previously established using techniques from commutative algebra. Finally, we give a necessary numerical criterion for the existence of a regular unimodular lattice triangulation of the boundary of a lattice polytope.
• Decompositions of rational convex polytopes. R P Stanley . 333-342.


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# Equations of Asymptotes

The results and formulas related to asymptotes are listed below.

Asymptotes
a.  Obtain ${\phi _n}\left( m \right)$ by putting $x = 1$, $y = m$ in the highest degree terms of the equation of the curve.

b.  Obtain ${\phi _{n – 1}}\left( m \right)$ by putting $x = 1$, $y = m$ in the ${\left( {n – 1} \right)^{{\text{th}}}}$ degree terms of the equation of the curve.

c. Obtain ${\phi _{n – 2}}\left( m \right)$ by putting $x = 1$, $y = m$ in the ${\left( {n – 2} \right)^{{\text{th}}}}$ degree terms of the equation of the curve.

d. Put ${\phi _n}\left( m \right) = 0$ and solve it for $m$. Let ${m_1},{m_2},{m_3}$ etc., be its roots, then:

e. $C = – \frac{{{\phi _{n – 1}}\left( m \right)}}{{{{\phi ‘}_n}\left( m \right)}}$ where ${\phi ‘_n}\left( m \right) \ne 0$.

f.  If ${\phi ‘_n}\left( m \right) = 0$, then $c$ is obtained by $\frac{{{c^2}}}{{2!}}{\phi ”_n}\left( m \right) + c{\phi ‘_{n – 1}}\left( m \right) + {\phi _{n – 2}}\left( m \right) = 0$ asymptotes are then $y = mx + c$ asymptotes parallel to $x$-axis.

1. The asymptotes parallel to the $x$-axis are obtained by equating to zero, the coefficients of the highest power of $x$ in the equation of curve.
2. The asymptotes parallel to the $y$-axis are obtained by equating to zero, the coefficients of the highest power of $y$ in the equation of curve.
3. The number of asymptotes to an algebraic curve of the $n$th degree cannot exceed $n$.


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# Problem: In a naturally occurring sample, 19.8% of boron atoms have five neutrons and 80.2% have six neutrons.What is the average mass of the atoms in the drawing?

###### Problem Details
In a naturally occurring sample, 19.8% of boron atoms have five neutrons and 80.2% have six neutrons.

What is the average mass of the atoms in the drawing?


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ch_regime_plot {CSHShydRology} R Documentation

## Plots the regime of daily streamflows using quantiles

### Description

Produces a regime hydrograph similar to that in the reference. It shows the flow quantiles for each day of the year and the maximum and minimum. Parameters can be set to change colours and set the y-scale to allow plots of same scale to be produced.

### Usage

ch_regime_plot(
DF,
wyear = 1,
colour = TRUE,
mx = 1,
quant = c(0.95, 0.9, 0.75, 0.5, 0.25, 0.1, 0.05)
)

### Arguments

 DF data frame of daily flow data wyear set wyear = 10 for October, water year = 1 for calendar year, can be any month colour if TRUE plot is in colour, if FALSE plot is grayscale. mx set the maximum y value; if = 1 then maximum value of the flows is used to set metadata a data frame of metadata, defaults to HYDAT_list. the y-axis value. The value of mx can be specified to produce a series of plots with the same scale. quant quantiles; default is quant = c(0.95,0.9,0.75,0.5,0.25,0.1,0.05). Can be changed but the length must be 7 and the 4th value must be 0.5 (median)

### Value

No value is returned; a standard R graphic is created.

Paul Whitfield

### References

MacCulloch, G. and P. H. Whitfield (2012). Towards a Stream Classification System for the Canadian Prairie Provinces. Canadian Water Resources Journal 37: 311-332.

### Examples

data(CAN05AA008)
ch_regime_plot(CAN05AA008, colour = TRUE, wyear = 1)

[Package CSHShydRology version 1.3.0 Index]


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##### Definition2.1.5

The linear combination of the vectors $$\vvec_1,\vvec_2,\ldots,\vvec_n$$ with scalars $$c_1,c_2,\ldots,c_n$$ is the vector

\begin{equation*} c_1\vvec_1 + c_2\vvec_2 + \ldots + c_n\vvec_n\text{.} \end{equation*}

The scalars $$c_1,c_2,\ldots,c_n$$ are called the weights of the linear combination.

in-context


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# How do you write a complete balanced reaction of aqueous lead (II) nitrate, Pb(NO_3)_2, and aqueous potassium chloride, KCL?

$P b {\left(N {O}_{3}\right)}_{2} \left(a q\right) + 2 K C l \left(a q\right) \rightarrow P b C {l}_{2} \left(s\right) \downarrow + 2 K N {O}_{3} \left(a q\right)$
How did I know that $P b C {l}_{2}$ was insoluble? Chemistry is an experimental science and this is something that you must learn. All halides are soluble, EXCEPT for lead(II) chloride, mercurous chloride ($H {g}_{2} C {l}_{2}$), and silver chloride, $A g C l$.


text:
## Top new questions this week:

### Why do differentiation rules work? What's the intuition behind them? (Not asking for proofs)

Differentiation rules have been bugging me ever since I took Basic Calculus. I thought I'd develop some intuitive understanding of them eventually, but so far all my other math courses (including ...

(calculus) (real-analysis) (derivatives)

### Is it possible to represent every irrational number as a (limit of) an infinite sum of rational numbers?

For instance, we can certainly represent π in this fashion. $$\frac{\pi}{4} \;=\; \sum_{n=0}^\infty \, \frac{(-1)^n}{2n+1} .\!$$ $\ln(2)$ is also irrational. And even that can be represented as an ...

(irrational-numbers)

### Expected number of people to not get shot?

Suppose $n$ gangsters are randomly positioned in a square room such that the positions of any three gangsters do not form an isosceles triangle. At midnight, each gangster shoots the person that is ...

(probability) (recreational-mathematics) (puzzle) (average) (geometric-probability)

### Have there been (successful) attempts to use something other than spheres for homotopy groups?

Homotopy groups are famous invariants in algebraic topology. They have a myriad of wonderful properties: For $n \ge 1$, $\pi_n(X,*)$ is a group; for $n \ge 2$, this group is abelian. $\pi_n$ defines ...

(algebraic-topology) (homotopy-theory)

### why $\rm{lcm}[1,2,3,\cdots,n]\in (2^n,4^n)$

Let $n\ge 7$ be positive integers,show that $$f(n)=\rm{lcm}[1,2,3,\cdots,n]\in (2^n,4^n)$$ Anyone know this problem background?or maybe have best proof or best result?

(number-theory) (inequality) (prime-numbers) (approximation) (least-common-multiple)

### Can we recover a space from its continuous functions?

Let $X$ be a topological space and let $\mathcal{F}(X,\Bbb R)$ be the set of continuous function from $X$ to $\Bbb R$. Can we recover the topology of $X$ by only the knowledge of ...

(general-topology)

### Group with two generators of order 3 is finite

A group $G$ is generated by two elements $a$ and $b$ such that for any $g\in G:$ $g^3=e$. Show that $G$ is finite. I don't understand how this is so. I don't know if $G$ is abelian, so I can ...

(abstract-algebra) (finite-groups)

(riemann-zeta)


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# existence and uniqueness solution Cauchy problem $y'=y(y+1)e^{-y}$

I have this Cauchy problem ($\alpha \in \mathbb{R}$)

\begin{cases} y'(t) = y(y+1)e^{-y} \\ y(0) = \alpha \end{cases}

$f(t,y)=y(y+1)e^{-y} \in C^1(\mathbb{R}^2)$ so for Cauchy-Lipschitz theorem exists and is unique the solution of locally $\forall \alpha$.

I know $y=0$ and $y=-1$ are stationary solutions.

How can I prove the other solutions are defined on all $\mathbb{R}$? Can I use existence and global theorem?

If I consider $\alpha>0$, $y_\alpha>0$ and $\partial_yf=e^{-y}(-y^2+y+1)$.

|$\partial_yf|=e^{-y}|-y^2+y+1|\rightarrow 0$ for $y\rightarrow+\infty$.

So $\partial_yf$ is limited and I have global existence and uniqueness on all $\mathbb{R}$ for $\alpha>0$?

Also for $-1<\alpha<0$ $\partial_yf$ is limited but for $\alpha<-1$ the solution exists for all $R$?

• If $\alpha<-1$, the maximal solution $y$ is defined on the interval $(t_\alpha,+\infty)$ for some finite negative $t_\alpha$. The situation is similar to the case $z'=e^z$ with $z(0)=\beta$ for some $\beta>0$, which can be solved explicitely as $e^{-\beta}-e^{-z(t)}=t$, hence the maximal solution is defined on $(-\infty,t_\beta)$ with $t_\beta=e^{-\beta}$. – Did Jun 22 '16 at 18:54

For $t < 0$, there is no problem. The reason is that $y$ in increasing for $y(0) > 0$ and decreasing for $y(0) < -1$. Hence we only consider the case $t > 0$.
For $y(0) > 0$, $y(t) > 0$. Since $f(t,y)$ is Lipschitz in $y$ for $y\geq 0$, you have $$y'(t) \leq Cy(t)$$ for some $C$. This implies that $y(t)\leq y(0)e^{Ct}$.
For $y(0) < 1$, $y(t) < -1$. Put $u = -y-1$. Then $$u'(t) = u(u+1) e^{u+1} \geq u^2e^u \geq u^4.$$ By integration, $$\frac{1}{u(0)^3} - \frac{1}{u(t)^3} \geq 3t.$$ Which means the solution can't be global (otherwise $u(t) < 0$ for large $t$).
• for $\alpha<-1$ I want calculate $lim_{t \rightarrow t_{\alpha}}y(t)$. It's $-\infty$? – Giulia B. Jun 23 '16 at 9:52


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Prove or Disprove: For all $x \in \mathbb{R}$, there exists a positive real number$\epsilon$ such that either $x \leq (-\epsilon) or x \geq \epsilon$

Can you guys help me prove or disprove this? Also this is what I got for the negation of this statement: There exists $x \in \mathbb{R}$ for all positive real numbers $\epsilon$ such that either $x > (-\epsilon)$ or $x < \epsilon$

• No, the negation is not correct. Negating an "or" produces an "and" (and vice versa). Start with $0$. – Daniel Fischer Sep 24 '13 at 23:27
• What do you mean by start with 0? – Jake Park Sep 25 '13 at 0:01

Your negation is not quite correct, because the negation of

either $x\le-\epsilon$ or $x\ge\epsilon$

is not

either $x>-\epsilon$ or $x<\epsilon$;

it’s

$x>-\epsilon$ and $x<\epsilon$.

However, the original statement is not true: it has one (and only one) exception. What real number is that exception?

• How do you know the original statement is not true? – Jake Park Sep 25 '13 at 0:07
• @Jake: Because I know a counterexample. Note that the original statement is equivalent to saying that for each real number $x$ there is an $\epsilon>0$ such that $|x|\ge\epsilon$; there’s one real number for which that’s clearly false. – Brian M. Scott Sep 25 '13 at 0:09
• Would that be 0? – Jake Park Sep 25 '13 at 0:51
• @Jake: It would indeed. For any non-zero $x$ you can take $\epsilon=|x|$, but if $\epsilon>0$, it’s never true that $0\le-\epsilon$ or $0\ge\epsilon$. – Brian M. Scott Sep 25 '13 at 0:57

Let $x \in R$.

Let $\epsilon \in R : 0 < \epsilon < x$ or $\epsilon < -x < 0$.

Since $x + \epsilon \in R$, and $0 < \epsilon$, your statement is true.

• why $x>0$? surely $\exists x\in\mathbb{R}$ such that $x\not>0$... – Tim Sep 24 '13 at 23:47
• And even x = 0, but I thought that was all too clear. – Don Larynx Sep 24 '13 at 23:57


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### ICT 2nd ESO AB

This course is aimed at students of 2nd ESO A and B to carry out active methodology in ICT subject.


text:
# User:ANI

Its my favourite letter.
Natalie Isaac
Occupation:Educator
email
 This user was certified a Wiki Apprentice Level 2 by ASnieckus .
 This user was a proud participant of eL4C41.
My Learning4Content course home page

# My profile

I am a  Trinbagonian,which really means I am from Trinidad and Tobago.. I have always enjoyed learning and I try to have my students enjoy their own experiences of learning.

# Professional background

I am an educator at the primary or elementary level. This means that as teachers we are responsible for all subject areas. I have been teaching for over ten years . I have experienced different levels of training and certification in an effort to be current in the craft of teaching. I consider this workshop to be another opportunity to experience learning and teaching in a different way.

# Education

I am currently pursuing a Masters in Education.

# Professional

I believe that to stop learning or trying to improve on the craft of teaching is to lose sight of what it means to teach.

# Personal

I enjoy reading,good music and dancing.

## My Book List

### My optional community service (learning contract) project

 Agreement By signing this optional learning contract I will try to complete my training in basic wiki editing skills to achieve the status of a Wikibuddy. In return for this free training opportunity, I will give the gift of knowledge by donating or developing at least one free content resource licensed under a CC-BY-SA or CC-BY license which can be used by myself (and others) on WikiEducator. Brief description of project Target date for completion . Signature

## Feedback and notes from my WikiNeighbours

Hullo Natalie. This is Cynthia from India. Glad to meet you thro Wikieducator. You will surely find this to be a lovely learning experience. Feel free to visit my user page by clicking on the highlighted name(Agnes) you see. Happy learning!--Agnes 00:45, 21 July 2010 (UTC)

Dear Natalie, Welcome to WikiEducator. Glad you are taking advantage of this wonderful learning opportunity. Thanks for your introduction--it's good to "Be bold and just try it". This page is your user page. We've added some typical headings and such, but feel free to revise/adjust as you see fit. If you need any help, feel free to contact me (click on my name at the end of this message to access my userpage where you can leave a note). Your WikiNeighbor, --Alison Snieckus 18:22, 21 July 2010 (UTC)

• Welcome Natalie!! A good beginning. Please feel free and go ahead and fill in formation. See you tomorrow to check on your progress!! Savithri Singh 18:51, 21 July 2010 (UTC)
• Hi Natalie, Your userpage is really coming along, a testament to your increasing wiki skills. Good work. I noticed that you tried to edit the "infobox" that displays at the top right of the page. This box is based on a template that requires that the attributes be specified in a certain way. No worries, I reset the box, and you are to be commended for "being bold"...this is the wiki way :D. So here's a quick explanation of how to fill out your infobox...
1. In the rich editor, highlight the {T} that is at the very top left corner of the screen.
2. Click on the {T} button on the top row of the toolbar.
3. In the template properties dialog, type in your personal info following the equal sign, '=', for each attribute.
Your info should now display in the box when you preview or save. Note that some attributes may be a bit more tricky, but you will make a lot of progress using this simple method.
You are almost ready for the first level wiki apprentice certification. I encourage you to try out a few more heading levels, and demonstrate bullets and auto-numbering. I'll be back to check on your progress. <smile>,  --Alison Snieckus 01:20, 26 July 2010 (UTC)
• Hi Natalie, Great work. I see Wayne (Mackiwg) has already added your Wiki Apprentice Level 1 certification -- see new box in the infobox. Congrats. --Alison Snieckus 01:22, 27 July 2010 (UTC)
• Natalie! Lovely. And congrats. I hope you are now ready to add your picture. Savithri Singh 18:29, 27 July 2010 (UTC)
• Natalie, The file upload issue is likely because the file you are uploading does not have one of the allowed extentions (the three characters after the (.) in the file name. Permitted file types are: png, gif, jpg, jpeg, svg, elp, swf, flv, pdf, odt, odp, ods, mp3, ogg, ogv, oga. These three letters each indicate the kind of file that is being uploaded, which WikiEducator needs to know so it can interpret the code in the file. Most image files are .jpg. I noticed a file listed in red at the top of your page that ends with .jpg1 -- WikiEducator would not allow this file to be uploaded because it doesn't know what to do with this file type.
The key in uploading the file is to leave the extension that same as in the original, but rename the beginning part to be something unique, like your full name if it's a picture of you. If you are still having problems, post on my page with the name of the file that you are trying to upload. Good luck, --Alison Snieckus 02:50, 29 July 2010 (UTC)
• Hi Natalie, Glad you enjoyed the workshop. I noticed in your introduction that you are a primary school teacher and interested in how WikiEducator can impact your work in the classroom. One way is to create and share resources. A resource created by a group of teachers in New Zealand is a good example of the possibilities. It's about Deforestation. The navigator will likely take you to all of the pages, but you can also browse the pages in Category:Deforestation for primary students. Feel free to leave me a note or ask a question on my user page or talk page. I've enjoyed working with you in the workshop. See you around the wiki. Your WikiNeighbour, --Alison Snieckus 22:35, 6 August 2010 (UTC)


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# why should we define "the magnetic flux is the dot product of the magnetic field and the plane area , instead of perimeter of the plane

When a uniform magnetic field $$\\mathbf B\$$ passes a plane at any angle, the magnetic flux $$\\phi_B\$$ is the dot product of the magnetic field $$\\mathbf B\$$ and the plane area $$\\mathbf{a}\$$,that is ,$$\\phi_B=\mathbf B \cdot \mathbf{a} \$$

Honestly,this explanation of magnetic flux is really abstract to me,i can write the formula above,but i don't really understand that,why should we define "the magnetic flux $$\\phi_B\$$ is the dot product of the magnetic field $$\\mathbf B\$$ and the plane area $$\\mathbf{a}\$$" instead of perimeter ,$$\p\$$ , of the plane ,that is ,$$\\phi_B=\mathbf B \cdot p\$$ ?

Can anyone explain this in detail?

• Why should we use the perimeter of the plane? – Hearth Apr 11 '20 at 15:48
• Total flux is flux density (B) x area! – Andy aka Apr 11 '20 at 16:33
• Because when we do the experiment with lots of different coils, we find the voltage is proportional to the area of the coil rather than the perimeter. That's what came first, the observation. Then we dreamt up mathematical formulae to describe what happened. Then we tried to add intuition as to why the formulae worked, and we invented 'lines of field' and 'flux' models to help visualise it. They're models, and all models are wrong, but they're useful, at non-relativistic speeds anyway. But always remember, the observations came first. – Neil_UK Apr 11 '20 at 19:02

"ϕB is the dot product of the magnetic field B and the plane area a"

No, it's: $$\Phi_A = \iint_A\vec{B}\cdot d\vec{A}$$ where B and dA are both vectors. You can't have a dot product between scalars.

B is a vector that points to the direction of the field and its magnitude is the field strength.

dA is the normal vector of the surface infinitesimal (an infinitely small part of the surface). That is, it points perpendicularly away from the surface and its magnitude is the area of the surface.

The dot product gives you the magnitude of the component of one vector that is parallel to the the other vector multiplied by the magnitude of the other vector.

Flux is the amount of magnetic field that is "flowing" through a given surface. So you only want to calculate with the component of B that is perpendicular to the surface. Hence the dot product.

If B is uniform in magnitude and is perpendicular to the surface at each point than the equation simplifies to just $$\Phi_A=B*A$$ where B is the magnitude of B and A is the total are of the surface. Because then B gives you the "flow rate" per unit area.

If B is uniform in magnitude but has a uniform angle with the surface at each point, than you have to take the part of B that is "flowing" through the surface, the perpendicular component that is. That's why you need the dot product. In this case the equation is $$\Phi_A=\vec{B}\cdot\vec{A}=B_\perp*A$$

"instead of perimeter ,p , of the plane"

Why would you possibly want to multiply by the perimeter? That is just simply not what flux is.

Given magnetic fields are just Efields with slight distortion by relativistic time shifts, the math we are taught is IMHO "because it works".

But I don't (yet) have a Masters in E&M so as to properly evaluate the history and evolution of the concepts and maths.

A friend of mine was delving into E&M, and noted the FORCE was orthogonal to the FLUX, "how convenient", he thought.

Ultimately its all about the electric fields.

Study on the A_VECTOR maths.


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# Use induction in a non standard way

1. Jul 17, 2013

### Dmobb Jr.

So I have to do an induction but I am not quite sure how to set it up. I have already proven that at each step I have either my intended result or I can advance one more step. I have also proven that there are a finite number of steps.

Intuitively I have essentially completed the proof. I just can't figure out how to present this in a way that is completely rigorous.

I get the feeling that "Well there can only be finitely many steps so eventually it will happen" is not good enough.

2. Jul 17, 2013

### HallsofIvy

Staff Emeritus
I don't see how anyone can suggest anything when what you are trying to do is so vague.

3. Jul 18, 2013

### Dmobb Jr.

Well I feel a little bit guilty for even posting a homework question at all (It's cheeting). So I am not going to post specifics. I will clarify certain things if people have questions. I think that someone could figure this out with the information I have given. If not then I will just have to figure it out myself which is what I should be doing anyway.

4. Jul 18, 2013

### rideabike

Sounds like a good "dot-dot-dot" proof, but maybe your professor doesn't like those. I.e. show the process of a couple steps, then "dot-dot-dot", then show the final step!

5. Jul 18, 2013

### verty

No, I was going to ask questions but this is so vague, it isn't in the ballpark of meaning anything.

6. Jul 23, 2013

### Dmobb Jr.

Yeah i wish I could dot dot dot this one but that's definitely not allowed. I reallized that I had messed up at an earlier part of this problem anyway. Thanks though.

@verty While what I said was not 100% rigourus I could easily make it that way.

For all natural numbers n, if x is not $\geq$ n, then x < n. Also there exists Y$\in$N such that x $\leq$ Y.

Prove that x exists and is a natural number.

Edit: We must assume x is a natural number not prove it.

Last edited: Jul 23, 2013


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# American Institute of Mathematical Sciences

• Previous Article
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• DCDS-B Home
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Double zero singularity and spatiotemporal patterns in a diffusive predator-prey model with nonlocal prey competition
September  2020, 25(9): 3491-3521. doi: 10.3934/dcdsb.2020070

## Dynamic analysis and optimal control of a three-age-class HIV/AIDS epidemic model in China

 1 Department of Mathematics, Nanjing University of Aeronautics and Astronautics, Nanjing 210016, China 2 Department of Mathematics, University of Miami, Coral Gables, FL 33146-4250, USA

* Corresponding author: Hongyong Zhao, Email: Hyzho1967@126.com

Received  July 2019 Revised  October 2019 Published  September 2020 Early access  April 2020

Fund Project: The first author is supported by NSFC grants [11571170, 11971013]

Based on the fact that HIV/AIDS manifests different transmission characteristics and pathogenesis in different age groups, and the proportions of youth and elderly HIV infected cases in total are increasing in China, we classify the whole population into three age groups, youth (15-24), adult (25-49), and elderly ($\geqslant$50), and establish a three-age-class HIV/AIDS epidemic model to investigate the transmission dynamics of HIV/AIDS in China. We derive the explicit expression for the basic reproduction number via the next generation matrix approach. Qualitative analysis of the model including the local, global behavior and permanence is carried out. In particular, numerical simulations are presented to reinforce these analytical results and demonstrate HIV epidemiological discrepancy among different age groups. We also formulate an optimal control problem and solve it using Pontryagin's Maximum Principle and an efficient iterative numerical methods. Our numerical results of optimal controls for the elderly group indicate that increasing the condom use and decreasing the rate of the formerly HIV infected persons converted to AIDS patients are important measures to control HIV/AIDS epidemic among elderly population.

Citation: Hongyong Zhao, Peng Wu, Shigui Ruan. Dynamic analysis and optimal control of a three-age-class HIV/AIDS epidemic model in China. Discrete & Continuous Dynamical Systems - B, 2020, 25 (9) : 3491-3521. doi: 10.3934/dcdsb.2020070
##### References:
 [1] A. Babiker, S. Darby, et al., Time from HIV-1 seroconversion to AIDS and death before widespread use of highly-active antiretroviral therapy: A collaborative re-analysis, Lancet, 355 (2000), 1131–1137. doi: 10.1016/S0140-6736(00)02061-4.  Google Scholar [2] N. Bacaër, X. Abdurahman and J. Ye, Modelling the HIV/AIDS epidemic among injecting drug users and sex workers in Kunming, China, Bull. Math. Biol., 68 (2006), 525-550.  doi: 10.1007/s11538-005-9051-y.  Google Scholar [3] CCDC, HIV Among people aged 50 and over last update: February 12, 2018., Available from: https://www.cdc.gov/hiv/group/age/olderamericans/index.html. Google Scholar [4] C. C. Chavez, Z. Feng and W. Huang, On the computation of ${R}_0$ and its role on global stability, in Mathematical Approaches for Emerging and Reemerging Infectious Disease: An Introduction, IMA Vol. Math. Appl., 125, Springer, New York, 2002, 229–250. doi: 10.1007/978-1-4757-3667-0_13.  Google Scholar [5] S. Chen, X. Yang, H. Li and Y. Qiu, AIDSand syphilis infection in the elderly and status of knowledge and behavior, Occup. Health, 31 (2015), 272-275.   Google Scholar [6] China Association for the Prevention and Control of STDs and AIDS, Increasing by 14% of New HIV/AIDS Infection Cases Were Reported in 2018, China, 2018. Available from: http://2018aids.medmeeting.org/cn. Google Scholar [7] CHPC, China Statistical Yearbook 1998-2012, 2003. Available from: http://www.stats.gov.cn. Google Scholar [8] E. A. Coddington and N. Levinson, Theory of Ordinary Differential Equations, McGraw-Hill Book Company, Inc., New York-Toronto-London, 1955.  Google Scholar [9] O. Diekman, J. A. P. Heesterbeek and J. A. J. Metz, On the definition and computation of the basic reproduction ratio $R_0$ in models for infectious diseases in heterogeneous populations, J. Math. Biol., 28 (1990), 365-382.  doi: 10.1007/BF00178324.  Google Scholar [10] P. Driessche and R. 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show all references

##### References:
 [1] A. Babiker, S. Darby, et al., Time from HIV-1 seroconversion to AIDS and death before widespread use of highly-active antiretroviral therapy: A collaborative re-analysis, Lancet, 355 (2000), 1131–1137. doi: 10.1016/S0140-6736(00)02061-4.  Google Scholar [2] N. Bacaër, X. Abdurahman and J. Ye, Modelling the HIV/AIDS epidemic among injecting drug users and sex workers in Kunming, China, Bull. Math. Biol., 68 (2006), 525-550.  doi: 10.1007/s11538-005-9051-y.  Google Scholar [3] CCDC, HIV Among people aged 50 and over last update: February 12, 2018., Available from: https://www.cdc.gov/hiv/group/age/olderamericans/index.html. Google Scholar [4] C. C. Chavez, Z. Feng and W. Huang, On the computation of ${R}_0$ and its role on global stability, in Mathematical Approaches for Emerging and Reemerging Infectious Disease: An Introduction, IMA Vol. Math. Appl., 125, Springer, New York, 2002, 229–250. doi: 10.1007/978-1-4757-3667-0_13.  Google Scholar [5] S. Chen, X. Yang, H. Li and Y. 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Biosci., 247 (2014), 1–12. doi: 10.1016/j.mbs.2013.10.006.  Google Scholar [35] S. P. Sethi and G. L. Thompson, Optimal Control Theory. Applications to Management Science and Economics, Kluwer Academic Publishers, Boston, MA, 2000.  Google Scholar [36] Shangli County CCDC, HIV/AIDS epidemic and control strategy, Shangli County CCDC, 2016. Google Scholar [37] S. Siringi, HIV/AIDS on the rise in young people in Kenya, Lancet HIV, 10 (2010). doi: 10.1016/S1473-3099(10)70037-2.  Google Scholar [38] H. L. Smith, Monotone Dynamical Systems. An Introduction to the Theory of Competitive and Cooperative Systems, Mathematical Surveys and Monographs, 41, American Mathematical Society, Providence, RI, 1995.  Google Scholar [39] UNAIDS, Report on the Global Epidemic, WHO Geneva, 2006. Google Scholar [40] H. Wang, R. Xu, Z. Wang and H. Chen, Global dynamics of a class of HIV-1 infection models with latently infected cells, Nonlinear Anal. Model. Control, 20 (2015), 21-37.  doi: 10.15388/NA.2015.1.2.  Google Scholar [41] L. Wang, The Elder Infected HIV/AIDS, A High-Risk Population That May Be Neglected, 2017. Available from: http://www.cas.cn/kx/kpwz/201704/t20170421_4597865.shtml. Google Scholar [42] L. Wang, Z. Ding, R. Xuan and D. Li, HIVprevalence among population at risk, using sentinel surveillance date from 1995 to 2009 in China, Chin. J. Epidemiol., 32 (2011), 20–24. Google Scholar [43] L. Wang, Q. Qin, W. Zheng, L. Li and X. Hei, Current case reporting of HIV/AIDS epidemic in 2010, China, Chin. J. AIDS STD., 17 (2014), 275-278.   Google Scholar [44] Y. Wang, Q. Qin and L. Ge, Characteristics of HIV infection among over 50-year olds population in China, Chin. J. Epidemiol., 24 (2018). Google Scholar [45] Y. Wang, Q. Qian, W. Zheng, C. Cai and Y. Cui, Analysis of epidemiological characteristics of AIDS cases in children under 15 years of age in China, Dis Surveil, 32 (2017), 227–231. Google Scholar [46] World Health Organization, 10 Facts About HIV/AIDS, HIV Fact Sheet No. 204, 2018. Available from: https://www.who.int/news-room/facts-in-pictures/detail/hiv-aids. Google Scholar [47] Y. Xiao, S. Tang, Y. Zhou, R. Smith, J. Wu and N. Wang, Predicting the HIV/AIDS epidemic and measuring the effect of mobility in mainland China, J. Theoret. Biol., 317 (2013), 271-285.  doi: 10.1016/j.jtbi.2012.09.037.  Google Scholar [48] J. Xing, Y. Li, W. Tang, W. Guo and et al., HIV/AIDS epidemic among older adults in China during 2005-2012: Results from trend and spatial analysis, Clin. Infect. Dis., 59 (2014), 53-60.  doi: 10.1093/cid/ciu214.  Google Scholar [49] J. Xu, The Number of HIV Infected Cases Among Adolescents Increased Rapidly in China, Which 80 Percents Infected Cased by Homosexual Transmission, 2016. 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), (b): The new HIV/AIDS infection cases among elderly ($\geq50$) from 2005 to 2017 (see Table 2)">Figure 1.  (a): The new HIV/AIDS infection cases among youth (15-24) from 2005 to 2017 (see Table 2), (b): The new HIV/AIDS infection cases among elderly ($\geq50$) from 2005 to 2017 (see Table 2)
The diagram of transmission among three epidemiological classes
">Figure 3.  Dynamics of the solutions of system $(1)$ when $R_0 = 1.73415$, where the variables and parameters are given in Table 1
The fitting curves of total new HIV/AIDS infection cases
The fitting curves of new HIV/AIDS infection cases among youth, elderly, adult, respectively
$PRCC-R^{(2)}_0$
">Figure 7.  $R^{(2)}_0$ trends with respect to different parameters, the other parameters are listed in Table 1
">Figure 8.  $R^{(2)}_0$ trends with respect to different parameters, the other parameters are listed in Table 1
">Figure 9.  (a): Optimal solutions for the system $(1)$, $u^{*}_2$ in green full line and $u^{*}_1$ in purple full line. (b): $A_e(t)$ trends in regard to the optimal controls, where $A_e$ without control strategy in red solid line, $A_e$ under the optimal control in blue dashed line. Other parameters are the same in Table 1
">Figure 10.  (a): Optimal control with respect to $I_a(0)$, (b): Optimal control with respect to $I_y(0)$, (c): Optimal control with respect to $I_e(0)$. Other parameters are the same in Table 1
">Figure 11.  (a): The influence of $R^{(1)}_0$ on $u^*_1$. (b): The influence of $R^{(1)}_0$ on $u^*_2$, other parameters are the same in Table 1
">Figure 12.  (a): Optimal control strategy $u_2(t)\neq 0, \ u_1(t) = 0$. (b): The impact on $A_e(t)$ with the optimal control strategy $u_2(t)\neq 0, \ u_1(t) = 0$, other parameters are the same in Table 1
">Figure 13.  (a): Optimal control strategy with $u_1(t)\neq 0, \ u_2(t) = 0$. (b): The impact on $A_e(t)$ with the optimal control strategy $u_1(t)\neq 0, \ u_2(t) = 0$, other parameters are the same in Table 1
The parameters and numerical values
 Parameters Description Range $(\%)$ Value $(year^{-1})$ Source $\Lambda$ Recruitment of the youth class - 3075405 Assume $\mu_y$ Natural death rate of youth [0.066-0.087] 0.765‰ [7] $\mu_a$ Natural death rate of adult [0.1-0.327] 1.852‰ [7] $\mu_e$ Natural death rate of elderly [0.19-0.45] 2.07‰ [7] $r_{y}$ Average remove rate from $I_y$ to $A_y$ - 1/12.5 [46] $r_{a}$ Average remove rate from $I_a$ to $A_a$ - 1/10 [46] $r_{e}$ Average remove rate from $I_e$ to $A_e$ - 1/7.9 [46] $d_y$ AIDS-related death rate among youth [1.0-4.3] 2.3% [12] $d_a$ AIDS-related death rate among adult [5.03-12.1] 9.7% [12] $d_e$ AIDS-related death rate among elderly [12.1-17.5] 16% [52] $c_{11}, c_{12}$ Condom use rate of youth [34-68] 57.5% [51] $c_{21}, c_{22}, c_{23}$ Condom use rate of adult [30-40] 34.5% [51] $c_{33}$ Condom use rate among elderly [17.5-37.5] 20.0% [17] $\alpha_y$ Transfer rate of youth group - 0.043 Fit $\alpha_a$ Transfer rate of adult group - 0.031 Fit $\delta_{11}, \delta_{12}$ Infected rate among youth [0.012-0.065] 0.014% [23] $\delta_{21}$ Infected rate from adult to young $[0.13-9.9]$ 2.8% [42] $\delta_{22}$ Infected rate among adult $[0.13-9.9]$ 4.2% [42] $\delta_{23}$ Infected rate from adult to elderly $[0.13-9.9]$ 6.6% [42] $\delta_{33}$ Infected rate among elderly [2.7-30.6] 25% [21] $\beta_{y}$ Transfer rate from $I_y$ to $A_a$ - 0.036 Fit $\beta_{a}$ Transfer rate from $I_a$ to $A_e$ - 0.041 Fit
 Parameters Description Range $(\%)$ Value $(year^{-1})$ Source $\Lambda$ Recruitment of the youth class - 3075405 Assume $\mu_y$ Natural death rate of youth [0.066-0.087] 0.765‰ [7] $\mu_a$ Natural death rate of adult [0.1-0.327] 1.852‰ [7] $\mu_e$ Natural death rate of elderly [0.19-0.45] 2.07‰ [7] $r_{y}$ Average remove rate from $I_y$ to $A_y$ - 1/12.5 [46] $r_{a}$ Average remove rate from $I_a$ to $A_a$ - 1/10 [46] $r_{e}$ Average remove rate from $I_e$ to $A_e$ - 1/7.9 [46] $d_y$ AIDS-related death rate among youth [1.0-4.3] 2.3% [12] $d_a$ AIDS-related death rate among adult [5.03-12.1] 9.7% [12] $d_e$ AIDS-related death rate among elderly [12.1-17.5] 16% [52] $c_{11}, c_{12}$ Condom use rate of youth [34-68] 57.5% [51] $c_{21}, c_{22}, c_{23}$ Condom use rate of adult [30-40] 34.5% [51] $c_{33}$ Condom use rate among elderly [17.5-37.5] 20.0% [17] $\alpha_y$ Transfer rate of youth group - 0.043 Fit $\alpha_a$ Transfer rate of adult group - 0.031 Fit $\delta_{11}, \delta_{12}$ Infected rate among youth [0.012-0.065] 0.014% [23] $\delta_{21}$ Infected rate from adult to young $[0.13-9.9]$ 2.8% [42] $\delta_{22}$ Infected rate among adult $[0.13-9.9]$ 4.2% [42] $\delta_{23}$ Infected rate from adult to elderly $[0.13-9.9]$ 6.6% [42] $\delta_{33}$ Infected rate among elderly [2.7-30.6] 25% [21] $\beta_{y}$ Transfer rate from $I_y$ to $A_a$ - 0.036 Fit $\beta_{a}$ Transfer rate from $I_a$ to $A_e$ - 0.041 Fit
Numbers of youth, adult, the elderly and total new reporting HIV/AIDS cases in China $(2005-2017)$. Adult cases are calculated by the total cases minus other groups cases, where the data on children under 15 years old from 2005 to 2017 are counted in [45]
 Year Total cases Source The youth cases (proportion) Source The adult cases (proportion) The elderly cases (proportion) Source 2005 40711 [43] 4186 (10.28%) [55] 33430 (82.12%) 2563 (6.30%) [48] 2006 44070 [43] 4872 (11.06%) [55] 34227 (77.67%) 3437 (7.80%) [16] 2007 45151 [43] 5524 (12.23%) [55] 34449 (76.30%) 4515 (10.06%) [16] 2008 50081 [43] 6628 (13.23%) [55] 36064 (72.01%) 6599 (13.18%) [44] 2009 53249 [43] 7416 (13.93%) [55] 35916 (67.45%) 9016 (16.93%) [44] 2010 64108 [43] 7875 (13.28%) [55] 44696 (69.72%) 11537 (18.00%) [44] 2011 74517 [36] 8925 (11.98%) [55] 48983 (65.73%) 16609 (22.30%) [44] 2012 82434 [28] 10195 (12.37%) [55] 52718 (63.95%) 19521 (23.68%) [44] 2013 90119 [29] 10800 (13.49%) [36] 56253 (62.42%) 23066 (25.61%) [44] 2014 103501 [30] 15000 (14.61%) [36] 60139 (58.10%) 27520 (26.60%) [44] 2015 114656 [31] 16986 (14.81%) [49] 63308 (55.22%) 33522 (29.24%) [41] 2016 124555 [26] 18437 (15.00%) [49] 70356 (56.49%) 35762 (28.71%) [41] 2017 134551 [27] 21250 (15.79%) [49] 72468 (53.86%) 40833 (30.35%) [41]
 Year Total cases Source The youth cases (proportion) Source The adult cases (proportion) The elderly cases (proportion) Source 2005 40711 [43] 4186 (10.28%) [55] 33430 (82.12%) 2563 (6.30%) [48] 2006 44070 [43] 4872 (11.06%) [55] 34227 (77.67%) 3437 (7.80%) [16] 2007 45151 [43] 5524 (12.23%) [55] 34449 (76.30%) 4515 (10.06%) [16] 2008 50081 [43] 6628 (13.23%) [55] 36064 (72.01%) 6599 (13.18%) [44] 2009 53249 [43] 7416 (13.93%) [55] 35916 (67.45%) 9016 (16.93%) [44] 2010 64108 [43] 7875 (13.28%) [55] 44696 (69.72%) 11537 (18.00%) [44] 2011 74517 [36] 8925 (11.98%) [55] 48983 (65.73%) 16609 (22.30%) [44] 2012 82434 [28] 10195 (12.37%) [55] 52718 (63.95%) 19521 (23.68%) [44] 2013 90119 [29] 10800 (13.49%) [36] 56253 (62.42%) 23066 (25.61%) [44] 2014 103501 [30] 15000 (14.61%) [36] 60139 (58.10%) 27520 (26.60%) [44] 2015 114656 [31] 16986 (14.81%) [49] 63308 (55.22%) 33522 (29.24%) [41] 2016 124555 [26] 18437 (15.00%) [49] 70356 (56.49%) 35762 (28.71%) [41] 2017 134551 [27] 21250 (15.79%) [49] 72468 (53.86%) 40833 (30.35%) [41]
The PRCC of the parameters in model $(1)$
 Parameters p value PRCC Parameters p value PRCC $c_{11}$ 0.3721 -0.1622 $\delta_{22}$ 0.6886 0.7230 $c_{12}$ 0.6879 -0.2595 $\alpha_{y}$ 0.6152 0.3462 $c_{21}$ 0.2277 -0.4457 $\alpha_{a}$ 0.7502 -0.2868 $c_{22}$ 0.8075 -0.6452 $\beta_{y}$ 0.5915 -0.0771 $\delta_{11}$ 0.4428 0.0172 $\beta_{a}$ 0 -0.8105 $\delta_{21}$ 0.3971 0.5004 $r_{y}$ 0.2674 -0.3981 $\delta_{12}$ 0.8916 -0.0008 $r_{a}$ 0.2412 -0.3063 $\delta_{23}$ 0.8646 -0.0036 $c_{23}$ 0.1928 -0.0292 $\delta_{33}$ 0.3956 0.0190 $c_{33}$ 0.5424 -0.0135
 Parameters p value PRCC Parameters p value PRCC $c_{11}$ 0.3721 -0.1622 $\delta_{22}$ 0.6886 0.7230 $c_{12}$ 0.6879 -0.2595 $\alpha_{y}$ 0.6152 0.3462 $c_{21}$ 0.2277 -0.4457 $\alpha_{a}$ 0.7502 -0.2868 $c_{22}$ 0.8075 -0.6452 $\beta_{y}$ 0.5915 -0.0771 $\delta_{11}$ 0.4428 0.0172 $\beta_{a}$ 0 -0.8105 $\delta_{21}$ 0.3971 0.5004 $r_{y}$ 0.2674 -0.3981 $\delta_{12}$ 0.8916 -0.0008 $r_{a}$ 0.2412 -0.3063 $\delta_{23}$ 0.8646 -0.0036 $c_{23}$ 0.1928 -0.0292 $\delta_{33}$ 0.3956 0.0190 $c_{33}$ 0.5424 -0.0135
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# What Is the Beta Scission Rule?

## Answer

The Beta Scission Rule is an important rule that governs the reaction in chemistry known as the Beta scission. Beta scission is the thermal cracking of hydrocarbons leading to the splitting of the carbon-carbon bond and thus forming free radicals. Free radicals are extremely reactive and short lived.
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