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• CommentRowNumber1.
• CommentAuthorMike Shulman
• CommentTimeMar 5th 2009
How do I make asterisks in itex? Writing "*" gets translated to  tags in the wrong places, but \ast doesn't work either.
• CommentRowNumber2.
• CommentAuthorTobyBartels
• CommentTimeMar 5th 2009
• (edited Mar 5th 2009)
\* in Markdown; $*$ works fine for me already in iTeX. See my tests in the Sandbox.
• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeMar 13th 2009
Try it on the cafe.
• CommentRowNumber4.
• CommentAuthorTobyBartels
• CommentTimeMar 14th 2009
• (edited Mar 14th 2009)
That's pretty weird. What's in the Sandbox:
\* $*$ $\star$ $\ast$ $*f*$ *hi* *
creates an error, but this:
\* $\star$ $\ast$ *hi* *$*$ $*f*$
works just fine. (All with Markdown + iTeX2MML, of course.)

By the way, this software seems to think that  should behave like \verbatim (which it doesn't) and changes my HTML to produce a result as if it did. That is rather annoying, and even makes it impossible to do certain things correctly, but I managed to get this post correct at least.
• CommentRowNumber5.
• CommentAuthorAndrew Stacey
• CommentTimeMar 19th 2009
• (edited Mar 19th 2009)

Hmmm, that might be because the filter 'Html' is slightly misleading. It isn't a 'raw' filter, it does do some processing to strip out "unsafe" html. It says that it doesn't look inside code blocks, is that what you wanted?

I'm not sure if what I see in the database is exactly what you typed as there seem to be lots of escaped characters which I think might just be a feature of how I'm accessing the database (for example, "it's" appears as "it\'s").

I'm also not sure exactly what you mean by the difference between code and verbatium. Is it that if I type & then I ought to see an ampersand whereas what I actually see is &?

Ah, looking at the markdown code I think I see the problem. Inside a code run is the line:

s/&/&/g

which encodes all ampersands to entities regardless of whether or not they are already html entities. Oh, hang on, that's the original perl ... let's look at the php. Yup, it's there:

$_ = str_replace('&', '&',$_);

Right, so that's too crude. What ought to happen is that only ampersands that are not already HTML entity codes should be converted. In some perl code I once wrote I used the modules MathML::Entities and HTML::Entities::Numbered which have conversion strings. Presumably there are PHP equivalents. Either that or I could rewrite the regexp.

Do you think that this ought to be filed as a bug on the Markdown/PHPMarkdown page(s)?

(Okay, I know that you used the Html filter rather than markdown but in testing I encountered the same problem in all the filters and markdown is the one I know best. Actually, the fact that it is a problem in all the filters suggests that there might be another hidden filter somewhere else doing the same thing. I'll investigate.)

• CommentRowNumber6.
• CommentAuthorTobyBartels
• CommentTimeMar 19th 2009
This comment is invalid XHTML+MathML+SVG; displaying source.

Is it that if I type <code>&</code> then I ought to see an ampersand whereas what I actually see is &?
Yep, that's it. I would say that it's a bug in … whatever is doing this.

• CommentRowNumber7.
• CommentAuthorAndrew Stacey
• CommentTimeMar 23rd 2009
• (edited Mar 23rd 2009)
This is really strange. It seems unique to code blocks which are meant to be protected from parsing under all the filters. Compare the following (first is a code block, second a pre block):

&

&
• CommentRowNumber8.
• CommentAuthorAndrew Stacey
• CommentTimeMar 24th 2009
Okay, let's try this now. This is with the Html filter: typing & produces & (for font comparison, an ampersand in normal font is &)
• CommentRowNumber9.
• CommentAuthorAndrew Stacey
• CommentTimeMar 24th 2009
• (edited Mar 24th 2009)

Markdown is a little more subtle. The Markdown code explicitly states:

Encode all ampersands; HTML entities are not entities within a Markdown code span.

So I'm loathe to take out the substitution for Markdown. On the other hand, Markdown ought to allow Html code to go straight through. Thus if I type hello it ought to come out as hello but it actually comes out as hello. The problem seems to be with the wrapper script that calls Markdown on posts. That escapes angle brackets. If I disable that then I can pass code blocks through the Markdown filter without difficulty and so typing & produces & as required.

I'm not sure what that escape routine is for so I'm happy to disable it. I guess that it is intended to protect comments from malicious code.

...

Yes, looking at the comments on the Markdown plugin page on the forum software I see that this is the case. That should be fixable - it oughtn't to be down to a format plugin to oversee malicious code security!

...

Okay, let's try this. I've hacked the Html and Markdown filters so that they work as they ought to and added the kses filter as a non-optional filter to remove malicious (x)html code. So typing <illegal>block</illegal> simply results in whilst & produces & as desired.

Right. You'd better use this functionality!

• CommentRowNumber10.
• CommentAuthorTobyBartels
• CommentTimeMar 24th 2009
This comment is invalid XHTML+MathML+SVG; displaying source.
 I'll use it to complain about the Mardown on the wiki then.
&amp;&rarr; in Instiki produces <code>&amp;amp;?</code> in XHTML, while <code>&amp;&rarr;</code> produces <code>&amp;?</code>.
On the blog, things are still worse (again using Mardown+iTeX): even &rarr; is spelt out between s!
Also, I see from the preview of my first, mistyped, version of this comment that <code> tags still aren't letting through things like <p> or even <b>.
I don't know how important all this is (especially if I can use <tt> instead), but there it is.

`


text:
# .css-4zleql{display:block;}deepboltzer codes

Photo by Elena Popova on Unsplash

# AlphaGo Zero Explained

## Mastering the Game of Go without Human Knowledge

Johannes Loevenich
·Dec 14, 2021·

• The game of Go
• The outstanding idea behind AlphaGo Zero
• Reinforcement Learning in AlphaGo Zero
• Results achieved by AlphaGo Zero
• Exemplary results

In October 2015 DeepMind's AlphaGo Fan beat the European Go champion Fan Hui in the game of Go decisively and just one year later AlphaGo Lee defeated Lee Sedol who is considered to be one of the greatest Go players of the past decade. This demonstrates the potential of RL algorithms to outperform humans in a complex game that requires strategy and forethought. Since then, DeepMind has continued to improve the architecture behind AlphaGo, developing new versions with dramatically greater performance. One such recent variation, dubbed AlphaGo Zero, has surpassed all previous versions of AlphaGo in a relatively short time.

This blog post aims to provide a fundamental knowledge of the AlphaGo Zero architecture and the elegance of its interactions between used algorithms.

# The game of Go

The game of Go (围棋) is an ancient game originating in China more than 3000 years ago. Like chess, Go has very simple rules but requires dramatically different thinking to master. The objective of the game is to surround as much territory on the board as possible by placing white and black stones (one player uses black, the other white) sequentially on intersections (called "points") of a 19x19 grid. Unlike chess, however, there are no restrictions on where to place the first stone.

Additionally, points surrounded wholly by one color are called "territories" and associated with that color's player.

As you can see below only the dark areas are controlled by Black while White controls all bright areas:

Go differs from chess in that the number of moves possible is greater, but the minimum number of moves it takes to capture enemy pieces (a "group") is fewer than in chess.

The game starts with an empty board and both players alternately place stones on unoccupied intersections until they pass their turn. The game ends when both players pass without any territory being captured or one player concedes defeat.

Go is a perfect example for applying RL algorithms since there are no restrictions on where you can place your stone next. Additionally, Go has deep complexity meaning that if you want to beat the best human players you need to think even further ahead than they do! For more details on how AlphaGo Fan achieved this check out their original paper.

# The outstanding idea behind AlphaGo Zero

The main distinction between AlphaGo Zero and its predecessors AlphaGo is that it has no prior human knowledge or supervision, therefore it is essentially starting from the ground up (or as some would say "tabula rasa"), which roughly translates to a blank or clean slate.

To better understand AlphaGo Zero's concept, let's break it down into simple steps. The appeal of AlphaGo Zero is that the fundamental concept is precisely how we would attempt to play a game without any prior knowledge by just knowing the rules.

## Foresight

Consider a game board in front of you. The first thing we'd do is look at potential scenarios that might affect the game's outcome. Naturally, the most important consideration is which route or state would provide us the greatest advantage and our opponent's reaction to our own actions. We obviously can't always anticipate every move our opponent will make or how the game will develop as a result of taking an action correctly, but it does give us a good sense of what may happen.

## Exploration

We record the results of our operations and compare them to previous actions in order to verify their efficacy. Every time we reach a new state, we want to know how important it is and whether there are any patterns in our path that might be used in later matches or even years down the road. The value of a state can vary during your journey and over time as you play and re-assess.

## Decision

When we've finished analyzing and playing through all of the potential states and paths, we take an action based on our knowledge of the game, which means we'll choose the course of action that we've explored the most since this would be the route with which we have acquired the most information. Our decisions in a given state might differ and alter as a result of learning and increasing expertise.

## Evaluation

In the final analysis, which takes place at the end of the game, we must evaluate our performance in the game by assessing our blunders and misjudgments on a state-by-state basis for the goals of those states. We should also look at where we got it right so that we may repeat this successful step again. After that, we have a greater knowledge of the game and can get ready for the next match with our new awareness.

# Reinforcement Learning in AlphaGo Zero

AlphaGo Zero utilizes two main components, a deep neural network $$f_\theta$$ with parameters $$\theta$$ and a Monte Carlo tree search algorithm (MCTS). The deep neural network takes the raw board representation of the current board combined with previous board states into one state $$s$$ as input to predict a probability for each move of the given board state, encoded as a vector $$p$$, and to predict the chance of victory $$v$$ for the current player, $$(p, v) = f_\theta(s)$$. The resulting combination of move (action) $$a$$ given inboard state $$s$$ is called state-action pair and can be defined by the tuple $$(s, a)$$.

When asked to choose its next move $$a$$ inboard state $$s$$, the algorithm will conduct an MCTS using the deep neural network. This utilizes a tree in such that the nodes of the tree are defined by the board states $$s$$ and the edges represent the state-action pairs $$(s, a)$$.

Moreover, each edge contains a counter $$N(s, a)$$ that counts how frequently the edge was chosen during the search. The edge also contains a value $$P(s, a)$$, calculated from the neural network probability $$(P(s, \cdot), V(s)) = f_\theta(s)$$ . Note that the neural network also provides us with a current board state value function $$V(s)$$. Finally, each edge includes the action value $$Q(s, a)$$ derived from the resulting board state value function $$V(s')$$ after executing action $$a$$.

To summarize, each edge $$(s, a)$$ currently has three values $$N(s, a), P(s, a), Q(s, a)$$ and we can calculate the state-action pair's promise as $$Q(s, a) + U(s, a)$$, in which $$U(s, a) \propto \frac{P(s, a)}{N(s, a) + 1}$$ is the balance between exploration and exploitation.

Assuming that $$s_0$$ is the board state when we start the MCTS, the tree search is composed of the following steps:

1. Initialize the tree with a single node: The current board state $$s_0$$ and its edges are defined by $$(s_0, a)$$.

2. Follow the most promising route, which is to say, always choose to follow the edge with the greatest value of $$Q(s, a) + U(s, a))$$, until an edge $$(s, a)$$ is selected that leads to a board state $$s'$$ that has not yet been recorded in the tree. (Step 1 in the idea diagram)

3. Expand the tree by creating a separate node for the new board state $$s'$$, and increase the counter $$N(s, a)$$ as well as update $$Q(s, a)$$ for each edge leading to this new board position. (Step 2 in the idea diagram)

4. If it is anticipated that the movement will be selected within a limited amount of time, advance to step 2. Else return a stochastic policy $$\pi(a | s_0) \propto N(s_0, a)^{\frac{1}{\tau}}$$ using the hyperparameter $$\tau$$, meaning that the resulting policy $$\pi(a | s_0)$$ recommends moves by assigning probabilities to actions.

5. Derive the next move from policy $$\pi(a | s_0)$$. (Step 3. in the idea diagram)

The resulting policy $$\pi(a | s_0)$$ can be viewed as a refinement of the raw network output probatilites $$p$$ with $$(p, v) = f_\theta$$.

This technique allows the algorithm to play against itself multiple times. After each self-play match has concluded, the match history may be used to construct multiple triples: $$(s, \pi(\cdot| s), z)$$ where $$s$$ is a board state, $$\pi(\cdot| s)$$ is the MCTS move probabilities for this board state, and z is an encoding of the ultimate winner of the match $$z$$ (i.e., did the player whose turn it was in the board state $$s$$ win?).

After multiple of these self-plays have concluded, a batch from the available triples is chosen to be used in training the deep neural network. The network is trained for each triple $$(s, \pi(\cdot| s), z)$$ in the batch. The training minimizes the error between the network output, the MCTS policy and the eventual winner so that $$f_\theta(s) = (p, v) \sim (\pi(\cdot| s), z)$$. (Step 4. in the idea diagram)

The whole process can be visualized as follows:

The deep neural network deployed in AlphaGo Zero is a residual neural network (ResNet). In particular, the neural network in the original paper consisted of 40 residual blocks. These blocks were all initialized randomly, meaning AlphaGo started learning without any prior knowledge and thus learned only by using the self-play mechanism, requiring no human supervision.

# Results achieved by AlphaGo Zero

AlphaGo Zero quickly outperformed previous iterations of AlphaGo, even while beginning with significantly lower performance than a supervised RL approach. This is demonstrated in the graph from the original paper linked below, which compares the training performance to supervised learning and Alpha Go Lee, the version that defeated Sedol, a grandmaster player.

An Elo rating is used to compare the various methods. The new RL approach surpasses the original supervised learning technique after around 20 hours, according to the graph. It surpassed AlphaGo Lee's Elo rating 16 hours later as well. After a total of 72 hours, the learning phase was completed, and AlphaGo Zero beat AlphaGo Lee in a final clash of 100 matches without losing once.

Another interesting side note about the accuracy of professional human players' moves is that AlphaGo Zero learns different gaming strategies and methods, predicting human actions with less accuracy than supervised learning. The graph below demonstrates this fact: While never reaching supervised learning's accuracy of approximately 50%, AlphaGo Zero improves on this aspect. Researchers were also able to discover new enhanced versions of previous methods.

All of this indicates that RL algorithms could be used in a wide range of fields and that RL solutions might compete and even outperform human knowledge acquired over hundreds of years. The research team's efforts, however, are not yet complete. In this instance, the learning technique was utilized to play the popular game Go, but it may also be applied to a variety of games and problems. As a result, the following section will borrow AlphaGo's approach for another type of model: Connect Four - demonstrating the method's flexibility.

# Exemplary results

By exchanging the model - in this case, the rules of Go to Connect Four - and applying the discussed AlphaGo Zero learning approach the following results were observed over 78 iterations with 30 episodes each:

To evaluate the method at different stages of the learning process 28 players competed in a tournament. Each player represents a different iteration and thus a different level of proficiency. During the tournament, each player plays every player four times and gains/loses points equal to its wins/loses respectively (draws result in 0 points). The graphic illustrates the expected result that players representing a later stage of learning generally perform better than those representing earlier stages.

Thank you for your attention and for reading our blog post! Our team, Jonas Bode, Tobias Hürten, Luca Liberto, Florian Spelter and Johannes Loevenich would appreciate your inspiring feedback!

format: plain


text:
Tensors are generalizations of vectors. These manifestate as

1. product of vector spaces
2. product of linear maps
3. multi-indexed arrangements of numbers or scalar functions

In the first case we take two vector spaces , and construct a new one by mean of their basis: if V has as basic vectors and W has then

is the vector space generated by the symbols which are all the linear combinations of pairs .

In another hand, if we want to construct a bilinear map then we pick up a pair of covectors and then we manufacture

defined via

It is also know that for finite dimensional vector spaces the tensor of rank one are the elements of and its dual space

In the multi-indexed versions tensors are expressions like , , , which actualy are the components of tensors and are in nature scalars or scalar functions. In the multi-indexed version tensor are arrays of quantities that obeys certain rules of transformations when we change systems of coordinates.

## Comments on how to learn this

• First of all you have to be sure that you understand linear algebra
• A key concept is that the set of all linear transformations among vector spaces it is also a vector space
• This vector space is representable thru matrices
• Some good experience on calculus on several variables are ideal
• Tensors are objects in the branch of math called multilinear algebra which can be developed in several levels
• It is better to understand multilinear algebra over the real numbers at the beginnings
• At the same time you are studying multilinear algebra, study differential geometry at least of curves and of surfaces
• There are discrepancies among the many notations of many different authors, so compare, contrast
• You shouldn't be scare of the intense use of symbols and relations inter symbols


text:
# Application of Integrals

### Integrals

In mathematics, Integration and Differentiation are the most significant ways to solve complex mathematical problems. Besides mathematics, Integration and Differentiation play key roles in Science, Engineering, and various other facets of our life. We have outlined what's, ifs, and how's related to integrals and their application. But how? Let us take an example.

To determine the area of a rectangle, the formulae 'length × breadth' is used. But what if you are given to calculate the area of the shaded portion, of a shaded rectangle with an unshaded circle within?

### Integrals

In mathematics, the application of Integrals is applied to find the area under a curve, areas bounded by any curve, and so on.

### Definition of Integrals

An integral is a function, of which a given function is a derivative. It is also known as the anti-derivative or reverse of a derivative.

Integrals are used to determine the area of 2D objects and the volume of 3D objects in real life.

### Types of Integrals

There are two types of Integrals.

• Definite Integrals

• Indefinite Integrals

### Definite Integrals:

An integral which has a start and an end value is known as a definite integral. In simple words, the function is restricted within an interval [a,b], where a and b are upper limit and lower limit, respectively. It is represented as $y=\int_{b}^{a}f(x)dx$

Example: y = ∫42 6x dx

Sol:  y= $\int_{2}^{4}6x dx$  here a = 4, b = 2

$y = 6 [\frac{x^{2}}{2}]_{2}^{4} = 6[\frac{4^{2}}{2}] - [\frac{2^{2}}{2}] = 6[8-2]=36$

### Indefinite Integrals:

An integral which does not have an upper and lower limit is known as a definite integral. It is represented as

$y=\int_{b}^{a}f(x)dx$ = F(x)+C, where ‘C’ is a constant

Example: $y=\int_{2}^{4}6x dx$

Sol: $y=\int 6xdx$

$y=6[\frac{x^{2}}{2}]$ = 3x + C, where C is a constant

Solved Examples

Question 1: Determine the Area Enclosed By a Circle  x2 + y2 =a2

Sol: It is observed that the area enclosed by the given circle is ‘4 x area of the region AOBA bounded by the curve, x-axis and the ordinates x=0 and x=a’.

[ as the circle is symmetrical about both x-axis and y-axis]

= $4\int_{0}^{a}ydx$ (taking vertical strips)

=  $4\int_{0}^{a}\sqrt{a^{2}-x^{2}}dx$

Since x2 + y2 =a2 gives $y=\pm \sqrt{a^{2}-x^{2}}$

The quarter AOBA lies in the first quadrant, hence 'y' is taken as positive. On integrating, we get the entire area enclosed by the given circle.

= $4[[\frac{x}{2}]\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\frac{x}{a}]$

= $4[([\frac{x}{2}]\times 0+\frac{a^{2}}{2}sin^{-1}1)-0]$

= $4(\frac{a^{2}}{2})(\frac{\pi}{2})$

=$\pi a^{2}$

Question 2: Determine the Area of the Region Bounded b the Curve y = x2 and the line Y = 4

Sol: Since the given curve expressed by the equation y = x2is a parabola symmetric about y-axis only, therefore, the required area of the region AOBA is given bt

= $2\int_{0}^{4}x dy$

= 2 area of the region BONB bounded by the curve, y-axis and the lines y=0 and y =4

= 2 x ( area of the region BONB bounced by the curve, y-axis and the lines y = 0 and y = 4 )

=$2\int_{0}^{4}\sqrt{y}dy$

= $2\times\frac{2}{3}[y^{\frac{3}{2}}]_{0}^{4}$

= $\frac{4}{3}\times8$

= $\frac{32}{3}$

### Application of Integrals

Integrals have their application in both science and maths. In maths, the application of integral is made to determine the area under a curve, the area between two curves, the center of mass of a body, and so on. Whereas in science (Physics in particular), the application of integrals is made to calculate the Centre of Gravity, Mass, Momentum, Work done, Kinetic Energy, Velocity, Trajectory, and Thrust.

### Application of Integrals in Engineering Fields:

There’s a vast application of integration in the fields of engineering.

### In Architecture:

• To determine the amount of material required in a curved surface. For instance, take the construction of a dome.

### In Electrical Engineering:

• Integrals are used in Electrical Engineering to calculate the length of a power cable required for transmission between two power stations.

### Application of Integrals in Different Fields:

In Medical Science:

• Integrals are used to determine the growth of bacterias in the laboratory by keeping variables such as a change in temperature and foodstuff.

In Medicine:

• To study the rate of spread of infectious disease, the field of epidemiology uses medical seine to determine how fast a disease is spreading, its origin, and how to best treat it.

In Statistics:

• To estimate survey data to help improve marketing plans for different companies because a survey requires many different questions with a range of possible answers.


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# 5th Grade Data Handling Worksheet

In 5th grade data handling worksheet, students can practice the questions on bar graph, line graph and pie chart. We represent the data in many ways. A graph is a pictorial representation of information. In pictograph the information is presented by using a picture as symbol. Bar graph is a method of presenting information by drawing rectangular bars of equal width and with equal space between the bars.

1. Choose the right answer to show the type of graph that will be used for the following information.

I. Number of students in the 5 sections of Grade 5.

(i) Double Bar graph                 (ii) Bar Graph

II. Favorite color of the students of class 5.

(i) Bar graph                           (ii) Line Graph

III. The sale of product of a company in last 6 months.

(i) Pictograph                          (ii) Line Graph

IV. Marks obtained by Michael in different objects in 2 terms of the year.

(i) Double Bar graph                (ii) Bar Graph

2. The face of the clock is like a pie chart with 12 parts. Then:

(i) What is the angle of each part?

(ii) How many hours will represent $$\frac{1}{3}$$ of the Pie Chart?

3. Given below is the amount of money withdrawn by Mr Matthew from his bank account in last 6 months. Tick the most suitable scale to be used for making a Bar Graph.

(i) $1,00,000 (ii)$ 70,000          (iii) $25,000 (iv)$ 50,000

4. Tick the right Pie Chart for the given information.

 Choice of Restaurant Number of Students Mc Donald's (MC)Pizza Hut (PH)Dominos (Do)Haldiram's 153610

5. Fill in the blanks.

(i) _________ graph is suitable when we need to compare data elements.

(ii) _________ is a collection of numbers gathered to give some meaningful information.

(iii) A __________ represents data through pictures of objects.

(iv) On the scale of 1 unit length = 5 kg, the bar of length 7 units will represent _______ kg.

(v) __________ graphs are used to show change or trend over time.

6. Observe the double bar graph given below and answer the questions.

(i) How many students are there in grade 1 in the year 2018?

(ii) What was the least strength of Grade 2 in the given four years?

(iii) What was the difference in the strength of Grade 1 and 2 in the years 2016?

(iv) In which year the difference in the strength of both the grades was maximum?

7. Number of cars sold by a dealer in a week is given below. Draw a line graph for the given data.

8. 48 students were surveyed to find out about their favorite drink. The information collected has been tabulated. Observe the given data and draw a pie chart.

 Favourite Drink Number of Students Cold DrinkFresh Fruit JuiceLemonadeApple Juice 624126

(i) Which is the most favorite drink of the students?

(ii) How many more students like Lemonade than Apple Juice?

(iii) Which is your favorite drink?

1. I. (ii) Bar Graph

II. (i) Bar graph

III.  (ii) Line Graph

IV. (i) Double Bar graph

2. (i) 30

(ii) 4 hours

3.  (iii) \$ 25,000

4. (ii)

5. (i) Bar

(ii) Data

(iii) Pictograph

(iv) 35

(v) Line

6. (i) 140

(ii) 115

(iii) 15

(iv) 2018

8. (i) Fresh Fruit Juice

(ii) 6

## You might like these

• ### Construction of Pie Chart | Steps of Construction | Solved Examples

The pie chart is a pictorial representation of data relative to a whole. Each portion in the circle represent an element of the collected data. The pie chart represents the composition of various elements in a whole. The total value of the pie chart is always 100%.

• ### Line Graph | How to Construct a Line Graph? | Solve Examples

Bar graphs are used to compare changes between different groups. Line graphs are used to compare two variables. The data which changes over a period of time can be displayed through a line graph. In line graph: ● Points are plotted on the graph related to two variables

• ### Double Bar Graph | Bar Graph |Solved Examples | Construction

Double bar Graph helps us to compare two data groups. For example the graph given below compares the number of boys and girls in classes I – V. Solve examples on Double bar Graph Observe the graph and answer the questions given below. Answer the given questions

• ### Bar Graph | Bar Chart | Interpret Bar Graphs | Represent the Data

We learn how to represent the data on the bar graph. Data can be represented by bars (like rectangle) whose lengths represent numerical values. One can use horizontal or vertical bars. Instead of rectangles one way even use line segments in such graphs.

• ### Reading Pictographs | Collection and Recording of Data | Tabulating of

Pictograph: Any information represented through pictures is called pictograph. Collection and Recording of Data: There are different ways of collecting and recording of information (data). We shall take up two examples of collection of data.


text:
# calculate surface area of a spherical ball's portion [duplicate]

Imagine a spherical ball with radius r that equally coated into two hemisphere P(grey) and S(white) is thrown at the pool. Consider the ball can float (up-down) and rotate. There is an interface separate two different fluid A(air) and B(water). Assume that the interface is flat. α defines the position of the ball and β defines the orientation of the ball. So α larger means ball is more in zone A and vice versa. Range of β and α is $$[0,360^0]$$ and $$[0,180^0]$$. The Schematic representation of this is below.

To calculate the surface area of P(grey) coated portion of the ball into A(air) is given by following equation, $$\mathrm{Area}_{P@A}= r^2\int_{\theta=\frac{\pi}{2}-\beta}^\alpha \int_{\phi=\arcsin(1/(\tan\theta \tan \beta))} ^{\pi -\arcsin(1/(\tan\theta \tan \beta)} \sin\theta\; d\theta d\phi.$$

After solving this equation, we can get the following solution. $$\textrm{Area}_{P@A}= 2 r^2 \left\{ \cos (\alpha) \sin ^{-1}(\cot (\alpha) \cot (\beta)) -\tan ^{-1}\left(\frac{\cos (\beta)}{\sqrt{\sin ^2(\beta)-\cos ^2(\alpha)}}\right)\right\}+\pi r^2 (1-\cos (\alpha)).$$

Note that the solution is defined as a function of α and β. If we change the value of α and β, the area of P coated portion into A(air) will change eventually. Thus we can calculate the area for all valid condition of $$\alpha$$ and $$\beta$$.

Now consider a different situation where P and S are not equally coated rather S>P. To show that I have introduced another angle γ also $$r1$$ the radius of the base of the cap.

Now how can i write $$z$$ as an increased angle? what will be the limits of integration for a spherical cap? Is there any particular analytical solution exists to calculate the area P(grey) into A(air) as a function of $$α,γ,β$$?


text:
# Show that [A, F(B)] = [A, B]F'(B)

Tags:
1. Feb 29, 2016

### Jon.G

1. The problem statement, all variables and given/known data
A and B commute with their commutator, C=[A,B]
Show that [A, F(B)] = [A, B]F'(B)
F(B) = ∑n=0fnBn

2. Relevant equations
[A,B] = AB - BA
[A,BC] = [A,B]C + B[A,C]

3. The attempt at a solution
So all I can think to do is:
[A,Bn] = [A,BBn-1] = [A,B]Bn-1 + B[A,Bn-1]
=ABBn-1-BABn-1+BABn-1-BBn-1A
=ABBn-1 - BBn-1A
and this seems to be a dead end? Just loops back around to ABn - BnA
If this is along the right track, whereabouts would I go next? If this is not the way to start it, how would I go about beginning this problem?
And even if this is along the right lines, I've ignored the sum here as I'm not quite sure how that affects the whole thing.

2. Feb 29, 2016

### Dick

Start by showing the case where f(B)=B^2. You should get [A,f(B)]=[A,B]2B. Which is [A,B]f'(B). Then try f(B)=B^3. For the general case use induction to prove if f(B)=B^n, then [A,B^n]=[A,B]nB^(n-1). Which is [A,B]f'(B). Now deal with the sum.

3. Mar 1, 2016

### Jon.G

Ooh ok thanks.
So I went through it with n=2,3 and 4 and [A, F(B)] = [A, B]F'(B) was shown for each of them.
For the sum is it simply that when n is any value, fn is a constant so that
(for example)
[A,f2B2] = [A, f2B]B + f2B[A, B]
f2 a constant and can be taken outside of [] as it effect both sides the same so (and as B commutes with [A, B])
[A, f2B]B + f2B[A, B] = [A,B]f2B + [A,B]f2B
=[A,B]2f2B = [A, B]F'(B)
?
Is this how it's shown?

4. Mar 1, 2016

### Dick

Yes, the Lie bracket is linear over scalars, so $[A,cf(B)+dg(B)]=c[A,f(B)]+d[A,g(B)]$. That's how you take care of the sum. It's good that you've shown it works for n=2,3 and 4. To show it's true for general $n$ that $[A,B^n]=[A,B]nB^{n-1}$ there's a general procedure called mathematical induction. You assume it's true for $n$ and then show that it must be true for $n+1$. Since you've already shown that it works for small values of n, that shows it works for all values of n. A less formal argument is to show, for example, that $[A,B^3]=[A,BBB]=[A,B]BB+B[A,B]B+BB[A,B]$ and then say this pattern holds for all n.

Last edited: Mar 1, 2016


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# Conversion of benzenediazonium chloride to benzene using lithium aluminium hydride

• Why does benzenediazonium chloride get converted to benzene upon reacting with $$\ce{H3PO2}$$ and not $$\ce{LiAlH4}$$ ?
• Also, what is the reaction mechanism of the reaction with $$\ce{H3PO2}$$ ?

This Wikipedia page list all kind of nucleophilic substitution reaction on aromatic compound why can't $$\ce{LiAlH4}$$ / $$\ce{H-}$$ react like one of these to give benzene and expel nitrogen.

I just wanted to add that it's up to my knowledge that $$\ce{LiAlH4 / H-}$$ can't reduce the double bond of Benzene, because when styrene reacts with $$\ce{LiAlH4 / H-}$$ it gives ethyl benzene.

### Update

I was just researching (on google) for the mechanism where I came across this site suggesting the mechanism to be radical and as follow :

But still there is one problem the side products don't match this site states the side product to be $$\ce{H3PO3}$$ and $$\ce{HCl}$$ but as stated above the side products must be $$\ce{H2PXO2}$$.

• Consider the possibility that the PCl(OH)2 undegoes hydrolysis - how stable is PCl3 in water? – Waylander Mar 9 '19 at 20:30
• @Waylander I know PCl3 reacts with water vigourously to give H3PO4 and HCl – Advil Sell Mar 9 '19 at 20:34
• So do you think it likely that other P-Cl species mat hydrolyse easily? – Waylander Mar 9 '19 at 21:31
• @Waylander yeah they may ........ – Advil Sell Mar 9 '19 at 21:44

I just wanted to add that it's up to my knowledge that $$\ce{LiAlH4 / H^{−}}$$ can't reduce the double bond of Benzene, because when styrene reacts with $$\ce{LiAlH4 / H^{−}}$$ it gives ethyl benzene.
That answer is as good as any for the first part of your question. It demonstrates that LAH will react preferentially with aliphatic pi bonds rather than the aromatic ring. Similarly there are aliphatic pi bonds in $$\ce{Ar - N^+\equiv N}$$, and so the LAH will reduce the diazonium group rather than displacing it.


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# Example of a function in Sobolev-Slobodeskji space

How do i show that $u(r,\theta) = r^\lambda sin(\lambda \theta)$ belongs to $H^{1+\lambda}(\Omega),$ with $\lambda \in (0,1)$ for $\Omega \subset \mathbb{R}^2$?

• $\Omega$ is probably an open set containing the origin. – Giuseppe Negro Apr 27 '17 at 21:16
• @GiuseppeNegro yes. – uli.xu Apr 27 '17 at 22:04


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$B$-$L$ global symmetry in the grand unification theories

+ 2 like - 0 dislike
420 views

Does the precise $$B$$-$$L$$ global symmetry present in:

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart

+ 7 like - 0 dislike

In the standard model lagrangian, B and L are separately conserved global charges, and B-L, a vector like symmetry, is anomaly-free. GUTs, like the G-G SU(5) violate B and L, but preserve B-L.

Wikipedia effectively defines the SU(5)-model U(1) symmetry X as $$X = 5(B − L) -2Y_W,$$ introduced by Wilczek & Zee in 1979. It is not a generator of SU(5), of course.

So you may readily compute for the left-chiral $$\bar {\bf{5}}$$, $$\overline{ d_R} : ~~(2Q=)~~~ Y=2/3 , ~~~~ B-L= -1/3 ~~~ \leadsto X=-3 \\ e^-_L, \nu_L : ~~~~~~~ Y= -1 , ~~~~~~ B-L= -1 \qquad \leadsto X=-3.$$ So the entire multiplet possesses a common X-charge: -3.

Proceed to verify for the left-chiral 10, X= 1. $$\overline{ u_R} : ~~(2Q=)~~~~~ Y=-4/3 , ~~~~ B-L= -1/3 ~~~\leadsto X=1 \\ d_L,u_L : ~~~~~~~~~ Y=1/3 , ~~~~ B-L= 1/3 \qquad \leadsto X=1 \\ e^+_L : ~~(2Q=)~~~~~ Y= 2 , ~~~~~~~ B-L= 1 \qquad \leadsto X=1.$$

Thence, for the $$\langle \phi ^* \rangle$$, X=2, so the Yukawa (mass) term is chargeless.

Note that B-L is vectorlike, but Y is not, so, ipso facto, X is not!

Similarly for SO(10), except here X is now a generator of SO(10) and is then gauged (hence SSBroken), hence violated by small amounts.

Responses to comment questions.

1) B-L is a good global symmetry for the SM and SU(5) and local for SO(10). So, e.g., in SU(5) proton decay to a pion and a positron, it is visibly preserved!

2) X is fine in the SM and SU(5) as a global symmetry, as a linear combination of good quantum numbers. As defined, it has a unique eigenvalue for each SU(5) rep, not necessarily the same for all reps, as you observe. Same for the SM which has smaller reps, several of which entered into each SU(5) rep. That means that, even for SU(5) which mixes baryons and leptons, it is straightforward to match X eigenvalues and monitor the symmetry of the coupling terms, like the Yukawas.

3) SO(10) largely follows suit, paralleling SU(5), and likewise lacks exotic particles, but gauges X, so SSBreaks it. But now the above two reps of SU(5) plus an extra singlet (the R-chiral neutrino) fit into a spinor 16 of SO(10). If your wrote down the multiplet, it might be easier to parse out. This review, eqn. (3.3), has X as a traceless diagonal generator, so you can check its eigenvalue on the fermion 16-plet (4.2).

Now, in most models, even though X is SSBroken and its corresponding gauge boson made massive, this happens at a high scale and results in a GUT with SU(5)-like symmetry, with effective B-L violating operators of dimension 6, i.e. suppressed by the square of such heavy scales, so very small. Thus, in such models, for all intents and purposes, proton decay is still approximately respectful of B-L!

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user Cosmas Zachos
answered Apr 15, 2020 by (370 points)
Thanks, I vote up. So B, L and B-L are all global symmetry in standard model. Agree?

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
Otherwise, should we say $G=U(1)_{B-L}$ is gauged in SU(5) g.u.t? Thank you~

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
No! Global means not gauged! In G-G, the U(1) charges in a rep sum to not 0, so not traceless, so not part of the SU(5) gauge group.

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user Cosmas Zachos
Yes, thanks, but 5¯ and 10 are also gauged under SU(5). Should the best way to describe the
p.s. thus 5¯ and 10 are not the best to describe the
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ys$\varnothing$csOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.


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# Principles of Reactivity: Chemical Kinetics

## Presentation on theme: "Principles of Reactivity: Chemical Kinetics"— Presentation transcript:

Principles of Reactivity: Chemical Kinetics
Chapter 15

Kinetics Chemical kinetics is the study of the rates of chemical reactions. Macroscopic level: rates of reactions, reaction rate, factors affecting rates Particulate level: reaction mechanism (the detailed pathway taken by atoms and molecules as a reaction proceeds)

15.1 Rates of Chemical Reactions
reaction rate is a measure of the speed of a reaction rate of a chemical reaction refers to the change in concentration of substance per unit of time amounts of reactants decrease and amounts of products increase often in units of M/s

Rates of Chemical Reactions
2N2O5  4NO2 + O2 rate of rxn = - Δ[N2O5]/Δt rate is expressed as a positive value; the sign on the rate indicates a decrease (reactant concentration) or increase (product concentration)

Rates of Chemical Reactions
2N2O5  4NO2 + O2 The graph of concentration vs. time is not straight because the rate of reaction changes during the course of the reaction. (largest at time = 0) Over a time period, the average rate can be calculated. For a single point in time, an instantaneous rate can also be calculated from the slope of the tangent line.

Practice Problem Compare the rates of appearance or disappearance of each product and reactant, in the decomposition of nitrosyl chloride, NOCl. 2NOCl(g)  2NO(g) + Cl2(g)

Practice Problem Sucrose decomposes to fructose and glucose in acid solution. A plot of the concentration of sucrose as a function of time is given on page What is the rate of change of the sucrose concentration over the first 2 h? What is the rate of change over the last 2 h? Estimate the instantaneous rate at 4 h.

15.2 Reaction Conditions and Rate
Reaction rate increases with more collisions and more energy. Factors affecting the speed of reaction: concentration of reactants – increasing [reactant] causes more collisions therefore increasing reaction rate surface area – more surface area results in more collisions (powdered medicine works more quickly than pill forms) physical state of reactants – homogeneous mixtures of liquids and gases tend to react faster (tied to surface area) temperature – higher temp increases KE therefore increases collisions catalysts – accelerate reaction but are not transformed themselves; changes reaction mechanism bio catalysts are called enzymes (BEANO!)

Homework After reading sections , you should be able to do the following… p. 712 (1-5)

15.3 Effect of Concentration on Rate
the rate of reaction is proportional to the reactant concentration may be proportional to the concentrations of more than one reactant catalyst can affect the rate

Rate Equations A rate equation or rate law expresses the relationship between reactant concentration and rate rate of reaction = k[N2O5] where k is the rate constant.

Rate Equations For a reaction such as aA + bB  xX
the rate equation is rate = k[A]m[B]n where m and n are experimentally determined exponents NOT necessarily the stoichiometric coefficients (usually positive whole numbers)

Order of a Reaction The order with respect to a reactant is the exponent of its concentration term in the rate expression. Total reaction order is the sum of the exponents in the concentration terms.

Order of a Reaction A reactant is first order if doubling it (while holding the other reactants constant) doubles the rate. A reactant is second order if doubling it quadruples the rate. A reactant is third order if doubling it increases the rate by 8. A reactant is zero order if doubling it has no effect on the rate.

Order of a Reaction 2NO(g) + Cl2(g)  2NOCl(g) Rate = k[NO]2[Cl2]
This reaction is second order in NO, first order in Cl2, and third order overall. Compare experimental evidence to figure out the order respective to reactants. Exp. [NO] mol/L [Cl2] mol/L Rate mol/L∙s 1 0.250 1.43e-6 2 0.500 5.72e-6 3 2.86e-6 4 11.4e-6

Rate Constant, k The rate constant, k, is a proportionality constant relating rate and concentrations at a given temp. Units of the rate constant have to match; depends on order of reaction. 0 order = mol/L∙time 1st order = time-1 2nd order = L/mol∙time 3rd order = L2/mol2∙time

Determining a Rate Equation
A rate equation must be determined experimentally. “method of initial rates” initial rate is the instantaneous reaction rate at the start of the reaction (at t = 0)

Practice Problem The initial rate of the reaction of nitrogen monoxide and oxygen 2NO(g) + O2(g)  2NO2(g) was measured at 25oC for various initial concentrations of NO and O2. Data are collected in the table on p Determine the rate equation from these data. What is the value of the rate constant and what are the appropriate units of k?

Practice Problem The rate constant k is 0.090 h-1 for the reaction
Pt(NH3)2Cl2(aq) + H2O(l)  [Pt(NH3)2(H2O)Cl]+(aq) + Cl-(aq) and the rate equation is rate = k[Pt(NH3)2Cl2] Calculate the rate of reaction when the concentration of Pt(NH3)2Cl2 is 0.020M. What is the rate of change in the concentration of Cl- under those conditions?

Homework After reading section 15.3, you should be able to do the following… P. 713 (9-13)

15.4 Concentration-Time Relationships
Integrated rate equation: ln([R]t/[R]o) = -kt [R]o and [R]t are concentrations of reactant at time t = 0 and at a later time, t. The ratio of concentrations is the fraction of reactant that remains after a given time has elapsed.

First Order Reactions ln[R]t/[R]o = -kt
For a first order reaction, k has units of time -1, which means that you can choose any unit for [R] such as M, moles, grams, atoms, molecules, or pressure.

Practice Problem Sucrose, a sugar, decomposes in acid solution to glucose and fructose. The reaction is first order in sucrose, and the rate constant at 25oC is k=0.21h-1. If the initial concentration of sucrose is M, what is its concentration after 5.0 h.

Practice Problem Gaseous NO2 decomposes when heated 2NO2(g)  2NO(g) + O2(g) The disappearance of NO2 is a first order reaction with k = 3.6x10-3 s-1 at 300oC. A sample of gaseous NO2 is placed in a flask and heated at 300 oC for 150 s. What fraction of the initial sample remains after this time? How long must a sample be heated so that 99% of the sample has decomposed?

Second-Order Reaction
For second-order reactions, the rate equation becomes 1/[R]t – 1/[R]o = kt same symbolism applies and k has units of L/mol∙time.

Practice Problem Using the rate constant for HI as k = 30. L/mol∙min, calculate the concentration of HI after 12 min if [HI]0 = mol/L.

Zero-Order Reactions For a zero-order reaction, the integrated rate equation is [R]0 – [R]t = kt where the units of k are mol/(L∙time).

Graphical Methods Each of the integrated rate laws can be rearranged to take the form of a straight line y = mx + b where m is the slope and b is the y-intercept.

Graphical Methods Zero Order First Order Second Order
[R]t = -kt + [R]o First Order ln[R]t = -kt + ln[R]o Second Order 1/[R]t = kt + 1/[R]o

Half-Life and First-Order Reactions
The half-life, t1/2, of a reaction is the time required for the concentration of a reactant to decrease to one-half its initial value. Longer half-life equals slower reaction. For first order reactions: t1/2 = 0.693/k half life is independent of concentration!

Half-Life and Other Reactions
Zero Order t1/2 = [R]o/2k Second Order t1/2 = 1/(k[R]o)

Practice Problem Americium is used in smoke detectors and in medicine for the treatment of certain malignancies. One isotope of americium, 241Am, has a rate constant, k, for radioactive decay of year -1. In contrast, radioactive iodine-125, which is used for studies of thyroid functioning, has a rate constant for decay of day-1. a. What are the half-lives of these isotopes? b. Which element decays faster? c. If you begin a treatment with iodine-125, and have 1.6e15 atoms, how many remain after 2.0 days?

Homework After reading section 15.4, you should be able to do the following… P. 714 (19-23)

15.5 Particulate View of Reaction Rates
For a reaction to occur: Collision Theory Reacting molecules must collide with each other. Reacting molecules must collide with sufficient energy. Molecules must collide in an orientation that can lead to rearrangement of the atoms.

Collision Theory Increasing the concentration of the reactant particles will increase the number of collisions. The number of collisions between the reactant molecules is directly proportional to the concentrations of each reactant.

Activation Energy An “energy barrier” must be overcome for a reaction to occur. The energy required to overcome this barrier is called activation energy, Ea. Fast reactions usually have low activation energies, and slow reactions have high activation energies.

Activation Energy As a reaction passes over the activation energy barrier, the transition state (also called an activated complex) is the arrangement of reactant molecules and atoms at the maximum point in the reaction coordinate diagram. (see page 695) Must be inferred!

Effect of Temperature Increase
Raising temperature increases the reaction rate by increasing the fraction of molecules with enough energy to surmount the activation energy barrier.

Effect of Molecular Orientation
“steric factor” The lower the probability of achieving the proper alignment, the lower the value of k and the slower the reaction.

Arrhenius Equation The observation that reaction rates depend on the energy and frequency of collisions between reacting molecules, on the temperature, and on whether the collisions have the correct geometry is summarized by the Arrhenius equation: k = Ae-Ea/RT

Arrhenius Equation k = Ae-Ea/RT
where R is the gas constant ( e-3 kJ/Kmol) and T is the temperature in kelvins. The parameter A is called the frequency factor and is related to the number of collisions that have correct geometry. The other factor refers to the fraction of molecules having at least the minimum energy required for reaction.

Arrhenius Equation Can be used to calculate the value of activation energy from the temperature dependence of the rate constant AND calculate the rate constant for a given temperature if the activation energy and A are known. Conceptual aspects of the Arrhenius equation and interpreting graphs are tested on the National Exam, but calculations will not be.

Arrenius Equation Can be rearranged to form an equation for a straight line relating ln k to 1/T. k = Ae-Ea/RT ln k = ln A – (Ea/R)(1/T) y = mx + b

Arrhenius Equation We can also solve for Ea by comparing k at two different temperatures… ln k2/k1 = - Ea/R[1/T2 – 1/T1]

Practice Problem The colorless gas N2O4 decomposes to the brown gas NO2 in a first-order reaction: N2O4(g)  2NO2(g) The rate constant k = 4.5 x 103 s-1 at 274K and 1.00 x 104 s-1 at 283K. What is the energy of activation?

Effect of Catalysts Catalysts speed up the rate of a chemical reaction but are not consumed in the reaction. Reaction intermediates are species formed in one step of a reaction that are then consumed in a later step.

Catalysis Acid-base catalysis: a reactant either gains or loses a proton; changing the rate of the reaction Surface catalysis: either a new reaction intermediate is formed, or the probability of successful collisions is modified Some enzymes accelerate reactions by binding to the reactants in a way that lowers activation energy; others react with reactant species to form a new intermediate

Homework After reading section 15.5, you should be able to do the following… P. 715 (32-36)

15.6 Reaction Mechanisms Reaction mechanisms refer to the sequence of bond-making and bond-breaking steps that occurs during the conversion of reactants to products.

Reaction Mechanisms Step 1: Br2 + NO  Br2NO
Step 2: Br2NO + NO  2BrNO Overall: Br2 + 2NO  2BrNO Each step is an elementary step. Each step has its own Ea and k. Intermediates are substances formed in one step and then consumed in another (don’t appear in overall reaction).

Elementary Steps Although the rate constant for a reaction MUST be determined experimentally, for an elementary step it is determined stoichiometrically. The molecularity (unimolecular, bimolecular, termolecular) is the same as its order. Classified by the number of reactant molecules that come together. NOT true for the overall reaction!

Practice Problem Nitric oxide is reduced by hydrogen to give nitrogen and water, 2NO(g) + 2H2(g)  N2(g) + 2H2O(g) and one possible mechanism is 2NO(g) N2O2(g) N2O2(g) + H2(g)  N2O(g) + H2O(g) N2O(g) + H2(g)  N2(g) + H2O(g) What is the molecularity of each of the three steps? What is the rate equation for the third step? Show that the sum of these elementary steps gives the net equation.

Rate-Determining Steps
Products can never be produced faster than the rate of the slowest elementary step, which is therefore called the rate-determining step.

Practice Problem The Raschig reaction produces the industrially important reducing agent hydrazine, N2H4, from NH3 and OCl- in basic, aqueous solution. A proposed mechanism is given on page 707. What is the overall stoichiometric equation? Which step is rate-determining? Write the rate equation for the rate-determining step. What reaction intermediates are involved?

Homework After reading section 15.6, you should be able to do the following… P. 716 (39-43)


text:
# php7下的filesize

```
echo round(filesize(basename(\$path))/1024,2).'KB';

```

```
echo filesize(basename(\$path)).'B'; //结果是正确文件字节大小。

```

```
\$path = 'c:\www\htdocs\index.php';
\$path2 = 'c:\www\htdocs\index.html';
echo filesize(basename(\$path)).'B'; //结果是正确文件字节大小。
echo '------';
echo round(filesize(basename(\$path2))/1024,2).'KB'; //结果是KB大小。

```

```
\$path = 'c:\www\htdocs\index.php';
\$path2 = 'c:\www\htdocs\index.html';
echo filesize(basename(\$path));  //精确后的大小
echo '----';
echo filesize(basename(\$path2)); //未精确大小

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


text:
# Khan/khan-exercises

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63 lines (58 sloc) 2.61 KB
 Factoring difference of squares 1
randRange(2, 9) -A * A

Factor the expression below completely. All coefficients should be integers.

x^2 +- abs(CONSTANT)

(x + A)(x - A)

The expression is in the form \pink{a^2} - \blue{b^2}, which is a difference of two squares so we can factor it as (\pink{a} + \blue{b})(\pink{a} - \blue{b}).

$("#question-x").addClass("hint_pink");$("#question-c").addClass("hint_blue");

What are the values of \pink{a} and \blue{b}?

\qquad \pink{a = \sqrt{x^2} = x}

\qquad \blue{b = \sqrt{A * A} = A}

Use the values we found for \pink{a} and \blue{b} to complete the factored expression, (\pink{a} + \blue{b})(\pink{a} - \blue{b}).

So we can factor the expression as: (\pink{x} + \blue{A})(\pink{x} - \blue{A})


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# Marine Phytoplankton – Nutrient-Rich Algae Superfood Health Benefits?

on

## What is Marine Phytoplankton?

It is a single-celled organism that is originally derived from the ocean. The micro-algae organism is rich in various minerals, including trace minerals, essential amino acids, chlorophyll, carotenoid, nucleic acid, and necessary vitamins. It is believed that marine phytoplankton is responsible for creating more than 90 percent of the oxygen supply on the earth. NASA has referred to marine phytoplankton as one of the most beneficial plants on earth. The Marine Phytoplankton is estimated to occupy approximately 25 percent of the earth's vegetation. The plant is unique in how rapidly it can reproduce itself; this makes it a sustainable food source for marine life as well as humans.

Pure forms of Marine Phytoplankton are grown in specialized research labs. The growing of these plants in the labs enables researchers to maximize the levels of important phytochemicals and avoid heavy metal contamination. When ingested, the ocean plant oxygenates and detoxifies the body system as it would in the sea. Researchers and scientists refer to this plant as the future of natural medicine due to the hundreds of synergistic elements it contains. The plant is rich in nano-sized nutrients that are readily absorbable in the human body.

However, it is important to note that not all phytoplankton are grown equally. Some forms of the microalgae may be inactive, contaminated, or diluted. When looking for the microalgae, it is important to look for high-quality phytoplankton.

### The Health Benefits of Marine Phytoplankton

#### 1. Mood lifter

The addition of the phytoplankton supplement to the diet helps in uplifting an individual’s mood. The health benefit has made the microalgae important in the management of depression. A study that was done by scientists from Utah University revealed significant improvements in participant’s depression symptoms after being administered with the plant’s supplement.

The participants claimed to feel relaxed, calm, and respectful after taking the supplement. The participants who were given the phytoplankton supplement during the study had significant improvements in scores of the SF36.

#### 2. Detoxification

The effects of the ocean microalgae on the human body resemble those observed in the sea. The plasma, which is the liquid part of the blood, contains salts together with other ions that make it resemble the ocean water. Scientists have added marine phytoplankton to the category of superfoods, which includes chlorella and spirulina. It is believed that phytoplankton also provides the body with Superoxide dismutase . The superoxide dismutase is an important antioxidant enzyme in the body. The enzyme lowers inflammatory responses in the body. Studies have also revealed it as a powerful detoxifying agent of heavy metals.

#### 3. Effects on the body’s defense system and liver

The body does not have to depend on the digestive process or the liver for processing the phytoplankton supplements. The microalgae are available in an absorbable form. This property enables the body to acquire essential nutrients even when the liver is defective. It can also improve the functioning of the liver without straining the other detoxification pathways of the body.

Studies have revealed that individuals who are administered with high levels of phytoplankton supplement have increased blood levels of CD3. The CD3 levels in blood indicate the presence of T-cells or the T-lymphocytes. The lymphocytes play a major role in the protection of the body against infections. The T-lymphocytes get rid of viruses and bacteria that may cause infection in the human body.

#### 4. Cellular regeneration

Supplementation of the body with the ocean microalgae strengthens the cell membranes. It triggers the human body to reproduce the current cells and give rise to new cells. A group of researchers comprising of botanists, microbiologists, and doctors from Europe discovered a strain of the marine phytoplankton that has the highest percentage of nutrients that are important to human health.

The Nannochloropsis gaditana has an incredible health-giving power. The strain is rich in more than 65 nutritional compounds, including amino acids, trace elements, vital minerals, and electrolytes. It is also rich in phospholipids, together with enzymes and coenzymes.

The strain is capable of helping humans to restore their health rapidly due to its ability to multiply explosively. The ocean microalgae improves the health of cells by providing electrons that are necessary for cell regeneration. It also provides the cells with essential amino acids that are necessary for building healthy cell walls.

#### 5. Improved cardiovascular system

The ocean plant is rich in antioxidants, minerals, and amino acids that are essential to keeping the heart healthy. Studies reveal that a daily intake of these nutrients is essential for keeping the heart healthy and functional. Omega-3 fatty acids are crucial for heart health; however, the body cannot synthesize the fatty acids on their own.

A letter from Harvard University entitled “marine fatty acids,” outlines the potential benefits of these fatty acids to the heart. The fatty acids are crucial for reducing inflammation as well as improving the performance of the heart. Marine phytoplankton supplements also prevent the formation of clots that may cause harmful effects to the heart. Individuals who consume phytoplankton supplements have a reduced risk of developing coronary heart diseases.

### Marine Phytoplankton Nutritional Facts

· The microalgae are loaded with complete protein: the marine phytoplankton is rich in nine amino acids that cannot be produced naturally. The amino acids are building blocks of proteins that have beneficial effects in the body. Phytoplankton is among the few plant foods that contains complete protein. Others include soybeans.

· The omega-3 fatty acids: phytoplankton is loaded with EPA and DHA. The marine algae are the only sources of EPA and DHA. A serving of phytoplankton supplement may contain up to 600 milligrams of the omega-3 fatty acids. The fatty acids have an array of health benefits, including proper functioning of the brain and heart.

· Trace minerals: the ocean microalgae are rich in trace minerals that include iodine, selenium, iron, and magnesium. According to experts, a serving of the marine phytoplankton contains 375 milligrams of magnesium. The minerals have several functions in the body, including proper functioning of the nerves and regulation of fluid balance. Iron is required by the body in the synthesis of hemoglobin.

· Source of vitamin B complex: Marine phytoplankton is a major source of the B complex vitamins. The B vitamins play a crucial role in ensuring proper functioning of the nervous system. The vitamins also boost the energy levels in the body through the breakdown of carbohydrates. The marine algae are a major source of vitamin B-12 because it lacks in other plant foods. Strict vegetarians are advised to consume phytoplankton supplements to lower the possibility of developing anemia.

### Top Marine Phytoplankton Products

#### 1. Marine Phytoplankton 90 Capsules Phyto4life

Mr Ros manufactures the phytoplankton supplement. The marine phytoplankton health supplement is rich in various nutrients, including the omega-3 fatty acids and COQ9. The supplement is also rich in minerals that are not available in freshwater algae such as spirulina and chlorella. Some of the health benefits of this product include the elimination of free radicals from the body and prevention of cellular damage. Its powerful antioxidant function provides a healthy and glowing skin. The supplement enhances the breakdown of carbohydrates in the body, boosting the body’s energy levels. Other health benefits include repair of damaged cells, anti-inflammatory effects, and improved cardiovascular performance.

#### 3. UMAC-Core Marine Phytoplankton Liquid

The phytoplankton supplement offers an ultimate source for cellular health. A study that was done on UMAC revealed its ability to boost the body's energy levels and vitality. It also supports the body's immune system function. According to Amazon, UMAC is wild-harvested, and it is approved for human use. The product is fast-acting, and it provides great cell health.

#### 5. Marine Phytoplankton Powder Raw Omega 3 EPA Vegan ATP Energy Superfood

The Longevity Herbs and Superfoods manufacture the phytoplankton supplement. The product is rich in various amino acids and trace elements that are crucial for maintaining the body's health. The product is indicated for individuals who need to work extra hard in stressful environments with no sleep. The primary ingredient in this product is pure marine phytoplankton. The serving size of the supplement is between three and six grams as directed by the health practitioner. The phytoplankton supplement can be taken with protein drinks, juice, water, or smoothies.

### How To Use Marine Phytoplankton Supplements

The supplements are available in different dosage forms, including liquid, capsule, or powder. The supplement can be purchased from a local drug store or online. It is advisable to read the ingredient label before making a purchase. Experts recommend the purchase of non-GMO, raw, and unprocessed marine phytoplankton. If one can tolerate the taste of the concentrated green supplements, it is advisable to take it directly. However, for those individuals who cannot tolerate the strong taste, they can add the supplement to water, juice, or coconut water. It can also be taken with smoothies and salad dressings.

## Marine Phytoplankton Conclusion

The marine phytoplankton are amazing algae that are found in the ocean. The plant has supported life on earth for billions of years by ensuring adequate oxygen supply. It is currently available in supplement form to boost internal, mental, and external health. Hopefully, further research will reveal more therapeutic potential of the ocean microalgae. Currently, available evidence points towards the potential of the plant to kill cancer cells, improve heart health, and increase overall immunity among others.

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text:
# How to apply Neuman boundary condition to Finite-Element-Method problems?

I have a 2D rectangular domain. The governing equation on this domain is Laplace equation:

$\nabla^2 f = 0$

In the left edge there is Neumann boundary conditon : $\frac{\partial f}{\partial n} = -a$

n is the normal vector to the domain's boundary(here on the left edge it's equal to the negative direction of x axis) and 'a' is a given date and it's a constant.

There is a Dirichlet boundary condition at the bottom edge and there is no boundary condition on right and top edge.

My problem is how to apply that Neumann boundary condition. I'm using finite element method (with first order triangulation)

As you may know, in finite element method first we make stiffness matrix (or global coefficient matrix from local coefficient matrix). Then we apply our governing equation(here the Laplace equation).

-
to scott morrison: why do you think my question is too localized and closed it? I just want to know how to apply neumann boundary condition to FEM problems –  kami Nov 11 '09 at 19:10
I also think the question is too localized: it looks like you are asking the MO community to solve a specific differential equation for you. I'd recommend that you clean up the question (capitalization, etc.), take out the specific details, give a short background on FEM, and ask a theoretical question; then someone with more reputation points than I have might re-open it. –  Theo Johnson-Freyd Nov 12 '09 at 1:50
As per Theo's comment. Please flag for moderator attention is you've edited and would like someone to re-open. –  Scott Morrison Nov 12 '09 at 5:34

You need to modify right-hand vector b of an equation Kx = b, where K is your stiffness matrix.

Here's how to do it, depending on which edge is on von Neumann boundary:

• edge 1-2 (i.e. connecting local nodes 1 and 2),

l = sqrt(x21*x21 + y21*y21);
b[node1] += a * l / 2.0f;
b[node2] += a * l / 2.0f;
b[node3] += 0;

• edge 1-3,

l = sqrt(x13*x13 + y13*y13);
b[node1] += a * l / 2.0f;
b[node2] += 0;
b[node3] += a * l / 2.0f;

• edge 2-3,

l = sqrt(x32*x32 + y32*y32);
b[node1] += 0;
b[node2] += a * l / 2.0f;
b[node3] += a * l / 2.0f;

-
Check "Introduction To The Finite Element Method In Electromagnetics" by Anastasis Polycarpou. There is no better introduction to FEM. –  Limal Nov 16 '09 at 17:56


text:
# Hypothesis,ANOVA,t-test problem

• November 26th 2009, 05:47 PM
dhammikai
Hypothesis problem
Please give me a help to find the answer for the following question

$
\begin{array}{|c|c|c|c|} S-male & S-female & T-male & T-Female \\
\hline
16&16&16&14\\
18&12&12&10\\
14&11&14&12\\
14&14&14&13\\
16&11&11\\
16&11\\
\end{array}
$

What can be say about the hypothesis

I need some word about that only null-hypothesis and alternative hypothesis