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# Matrix grammar

A matrix grammar is a formal grammar in which instead of single productions, productions are grouped together into finite sequences. A production cannot be applied separately, it must be applied in sequence. In the application of such a sequence of productions, the rewriting is done in accordance to each production in sequence, the first one, second one etc. till the last production has been used for rewriting. The sequences are referred to as matrices.

Matrix grammar is an extension of context-free grammar, and one instance of a controlled grammar.

## Formal definition

A matrix grammar is an ordered quadruple

${\displaystyle G=(V_{N},V_{T},X_{0},M).}$

where

• ${\displaystyle V_{N}}$ is a finite set of non-terminals
• ${\displaystyle V_{T}}$ is a finite set of terminals
• ${\displaystyle X_{0}}$ is a special element of ${\displaystyle V_{N}}$, viz. the starting symbol
• ${\displaystyle M}$ is a finite set of non-empty sequences whose elements are ordered pairs
${\displaystyle (P,Q),\quad P\in W(V)V_{N}W(V),\quad Q\in W(V),\quad V=V_{N}\cup V_{T}.}$

The pairs are called productions, written as ${\displaystyle P\to Q}$. The sequences are called matrices and can be written as

${\displaystyle m=[P_{1}\to Q_{1},\ldots ,P_{r}\to Q_{r}].}$

Let ${\displaystyle F}$ be the set of all productions appearing in the matrices ${\displaystyle m}$ of a matrix grammar ${\displaystyle G}$. Then the matrix grammar ${\displaystyle G}$ is of type-${\displaystyle i,i=0,1,2,3}$, length-increasing, linear, ${\displaystyle \lambda }$-free, context-free or context-sensitive if and only if the grammar ${\displaystyle G_{1}=(V_{N},V_{T},X_{0},F)}$ has the following property.

For a matrix grammar ${\displaystyle G}$, a binary relation ${\displaystyle \Rightarrow _{G}}$ is defined; also represented as ${\displaystyle \Rightarrow }$. For any ${\displaystyle P,Q\in W(V)}$, ${\displaystyle P\Rightarrow Q}$ holds if and only if there exists an integer ${\displaystyle r\geq 1}$ such that the words

${\displaystyle \alpha _{1},,\ldots ,\alpha _{r+1},\quad P_{1},\ldots ,P_{r},\quad R_{1},\ldots ,R_{r},\quad ,R^{1},\ldots ,R^{r}}$

over V exist and

• ${\displaystyle \alpha _{i}=P}$ and ${\displaystyle \alpha _{r+1}=Q}$
• ${\displaystyle m}$ is one of the matrices of ${\displaystyle G}$
• ${\displaystyle \alpha _{i}=R_{i}P_{i}R^{i}}$ and ${\displaystyle \alpha _{i+1}=R_{i}Q_{i}R^{i}.}$

If the above conditions are satisfied, it is also said that ${\displaystyle P\Rightarrow Q}$ holds with ${\displaystyle (m,R_{1})}$ as the specifications.

Let ${\displaystyle \Rightarrow ^{*}}$ be the reflexive transitive closure of the relation ${\displaystyle \Rightarrow }$. Then, the language generated by the matrix grammar ${\displaystyle G}$ is given by

${\displaystyle L(G)=\{P\in W(V_{T})|X_{0}\Rightarrow ^{*}P\}.}$

## Examples

Consider the matrix grammar

${\displaystyle G=(\{S,X,Y\},\{a,b,c\},S,M)}$

where ${\displaystyle M}$ is a collection containing the following matrices:

${\displaystyle m_{0}:[S\rightarrow XY],\quad m_{1}:[X\rightarrow aXb,Y\rightarrow cY],\quad m_{2}:[X\rightarrow ab,Y\rightarrow c]}$

These matrices, which contain only context-free rules, generate the context-sensitive language

${\displaystyle L=\{a^{n}b^{n}c^{n}|n\geq 1\}.}$ The associate word of ${\displaystyle a^{n}b^{n}c^{n}}$ is ${\displaystyle Aw(a^{n}b^{n}c^{n})=m_{0}m_{1}^{n-2}m_{2},\forall n\geq 2}$ and ${\displaystyle Aw(abc)=m_{0}m_{2}}$.

This example can be found on pages 8 and 9 of [1] in the following form: Consider the matrix grammar

${\displaystyle G=(\{S,X,Y,Z\},\{a,b,c\},S,M)}$

where ${\displaystyle M}$ is a collection containing the following matrices:

${\displaystyle m_{0}:[S\rightarrow abc],\quad m_{1}:[S\rightarrow aXbYcZ],\quad m_{2}:[X\rightarrow aX,Y\rightarrow bY,Z\rightarrow cZ],\quad m_{3}:[X\rightarrow ab,Y\rightarrow b,Z\rightarrow c]}$

These matrices, which contain only context-regular rules, generate the context-sensitive language

${\displaystyle L=\{a^{n}b^{n}c^{n}|n\geq 1\}.}$

The associate word of ${\displaystyle a^{n}b^{n}c^{n}}$ is ${\displaystyle Aw(a^{n}b^{n}c^{n})=m_{1}m_{2}^{n-2}m_{3},\forall n\geq 2}$ and ${\displaystyle Aw(abc)=m_{0}}$.

## Properties

Let ${\displaystyle {{\ce {MAT^{\lambda}}}}}$ be the class of languages produced by matrix grammars, and MAT the class of languages produced by ${\displaystyle \lambda }$-free matrix grammars.

• Trivially, MAT is included in ${\displaystyle {{\ce {MAT^{\lambda}}}}}$.
• All context-free languages are in MAT, and all languages in ${\displaystyle {{\ce {MAT^{\lambda}}}}}$ are recursively enumerable.
• MAT is closed under union, concatenation, intersection with regular languages and permutation.
• All languages in MAT can be produced by a context-sensitive grammar.
• There exists a context-sensitive language which does not belong to ${\displaystyle {{\ce {MAT^{\lambda}}}}}$ [2].
• Each language produced by a matrix grammar with only one terminal symbol is regular.

## Open problems

It is not known whether there exist languages in ${\displaystyle {{\ce {MAT^{\lambda}}}}}$ which are not in MAT, and it is neither known whether ${\displaystyle {{\ce {MAT^{\lambda}}}}}$ contains languages which are not context-sensitive [3].

## Footnotes

• ^ Ábrahám, S. Some questions of language theory. International Conference on Computational Linguistic, 1965. pp 1–11. [4]
• ^ Gheorghe Păun, Membrane Computing: An Introduction, Springer-Verlag New York, Inc., Secaucus, NJ, USA, 2002. pp 30–32


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Home  >>  CBSE XII  >>  Math  >>  Matrices

# if $F( x ) = \begin{bmatrix} cos\;x & -sin\;x & 0 \\ sin\;x & cos\;x & 0 \\ 0 & 0 & 1 \end{bmatrix},$ show that $F( x )\; F( y ) = F( x + y ).$

Toolbox:
• Given $F( x ) = \begin{bmatrix} cos\;x & -sin\;x & 0 \\ sin\;x & cos\;x & 0 \\ 0 & 0 & 1 \end{bmatrix}$, to arrive at $F(y)$, we replace $x$ with $y$, i.e., $F( y ) = \begin{bmatrix} cos\;y & -sin\;y & 0 \\ sin\;y & cos\;y & 0 \\ 0 & 0 & 1 \end{bmatrix}$
• Given $F( x ) = \begin{bmatrix} cos\;x & -sin\;x & 0 \\ sin\;x & cos\;x & 0 \\ 0 & 0 & 1 \end{bmatrix}$, to arrive at $F(x+y)$, we replace $x$ with $x+y$, i.e., $F(x+y)= \begin{bmatrix} cos (x+y) & -sin(x+y) &0\\ sin(x+y) & cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix}$
• Multiplication of two matrices is defined only if the number of columns of the left matrix is the same as the number of rows of the right matrix.
• If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B:
• $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
• $sin (x+y) = sinx.cosy + cosx.siny$
• $cos(x+y) = cosx.cosy - sinx.siny$
We have to show that:$F(x)F(y)=F(x+y). LHS: F(x)F(y)$ and $RHS: F(x+y).$
Evaluating $LHS:F(x)F(y)=\begin{bmatrix} cos\;x & -sin\;x & 0 \\ sin\;x & cos\;x & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} cos\;y & -sin\;y & 0 \\ sin\;y & cos\;y & 0 \\ 0 & 0 & 1 \end{bmatrix}$:
$\Rightarrow \begin{bmatrix} cos\;xcos\;y-sin\;xsin\;y+0 & cos\;x(-sin\;y)+(-sin\;xcos\;y)+0 &0+0+0 \\ sin\;xcos\;y+cos\;xsin\;y+0 & -sin\;xsin\;y+cos\;xcos\;y+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+1 \end{bmatrix}$
$\Rightarrow \begin{bmatrix} cos\;xcos\;y-sin\;xsin\;y & cos\;x(-sin\;y)+(-sin\;xcos\;y) &0\\ sin\;xcos\;y+cos\;xsin\;y & -sin\;xsin\;y+cos\;xcos\;y & 0 \\ 0 & 0 & 1 \end{bmatrix}$
We know that $sin (x+y) = sinx.cosy + cosx.siny$ and $cos(x+y) = cosx.cosy - sinx.siny$.
$\Rightarrow LHS: \begin{bmatrix} cos (x+y) & -sin(x+y) &0\\ sin(x+y) & cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$RHS: F(x+y)= \begin{bmatrix} cos (x+y) & -sin(x+y) &0\\ sin(x+y) & cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix}$, which is the equal to $LHS$.
edited Feb 27, 2013

format: plain


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# Assume that $N\subset \mathbb{R}$ is not Lebesgue measurable. Is then the set $N\times \mathbb{R}$ also not Lebesgue measurable?

Assume that $N\subset \mathbb{R}$ is not Lebesgue measurable. Is then the subset $N\times \mathbb{R} \subset \mathbb{R}^2$ also not Lebesgue measurable?

Im not sure whether this is true or not. The problem is that $\mathcal{L}(\mathbb{R}^2) \neq \mathcal{L}(\mathbb{R})\otimes \mathcal{L}(\mathbb{R})$ but $\mathcal{L}(\mathbb{R}^2)$ is equal to the completion of $\mathcal{L}(\mathbb{R})\otimes \mathcal{L}(\mathbb{R})$, so we may find a Lebesgue null set $M\in \mathcal{L}(\mathbb{R})\otimes \mathcal{L}(\mathbb{R})$ such that $N\times \mathbb{R} \subset M$ and therefore we may obtain that $N\times \mathbb{R}$ is Lebesgue measurable.

HINT: Suppose $N \times \mathbb{R}$ is Lebesgue measurable. What can you say about $(N \times \mathbb{R}) \cap ( \mathbb{R} \times \{ 0 \} )$?
• The intersection is equal to $N\times \{0\}$ and is also Lebesgue measurable with respect to the 2 dimensional Lebesgue measure. But I dont't see the contradiction. Is it true that $\lambda_2( N\times \{0\})=\lambda_1(N)$ ? – Giuliano Basso Nov 19 '14 at 18:59


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# Percents

## Write percents as ratios and ratios as percents

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Practice Percents
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Recognize and Write Percents

Have you ever tried to figure out scores for a football team? Take a look at this dilemma.

Carla knows that the Springstead Raiders, the high school football team won 80% of their games this year. She is wondering how to understand that percent.

What is a percent? Do you know?

This Concept will help you to understand percents so that you will be able to help Carla by the end of the Concept.

### Guidance

We see percents all around us every day. If you walk into a store of any kind, you will often see a percent sign. Whether it is a sale sign for 10% off or a sign on a bank for a new mortgage rate, by looking all around you, you will see percents.

What is a percent?

Simply, a percent is a part of a whole.

It is a part of a whole that is being compared to 100. In this way, we can also think of a percent as a ratio. Remember that a ratio is a comparison between two quantities.

Fractions and decimals are also parts of a whole.

We can say that fractions, decimals and percents are all related.

Let’s think about percents in a little more detail.

Look at the word “percent.” Its root “cent” means one hundred and “per” implies division or means “for each.” If you have three candies per person, it means three candies for each person. So “percent” means one for every one hundred.

If 25% of the people want chocolate cake, then 25 out of every 100 want chocolate cake. For this reason, percent can be written as a fraction with the denominator being 100.

\begin{align*}25\% = \frac{25}{100}\end{align*}

Now think about this again. Anytime you see a percent, you know that the amount is being compared to 100, or is “out of” 100.

18% means 18 out of 100

We are comparing the quantity of 18 to the whole of 100.

125% means 125 out of 100

Here we have a percent that is greater than 100. This means that we have greater than the total whole included in our percent.

That’s a great question. It is possible because we are thinking about what’s possible not the actual number. It is possible to have 100 percent on a test. However, if there are bonus questions, then you could also have greater than 100 percent. Sales is like this too. If a car salesman needs to sell 5 cars in a week that is 100 percent for him. However, if he sells 8 cars, then he sold greater than 100%.

Remember that a percent is a ratio compared to 100! Write this down in your notebook.

Write each example as a percent.

23 out of 100

Solution: 23%

18.5 out of 100

Solution: 18.5%

#### Example C

97 out of 100

Solution: 97%

Now let's go back to the dilemma from the beginning of the Concept.

If the team won 80% of their games, then Carla can think of this as out of 100. It means that the team won 80 out of 100 games. If the team did not play 100 games, then you can get an idea of how they did by simplifying the ratio.

Take a look at the ratio now.

\begin{align*}80\% = \frac{80}{100} = \frac{8}{10}\end{align*}

This means that the team won 8 out of 10 games. A pretty great season!

### Vocabulary

Ratio
a comparison between two quantities.
Percent
a ratio that is being compared to the quantity of 100. Percent means out of 100.

### Guided Practice

Here is one for you to try on your own.

Write this as a percent.

Carmen saw 100 movies in one year. She chose 60 of the movies as her favorites. The other 40 movies were not her favorites. Write her favorites as a percent and the other movies as a percent too.

Solution

To start, let's write the favorites as a percent.

Carmen chose 60 out of 100 as favorites.

\begin{align*}60\%\end{align*}

Carmen did not choose 40 movies as favorites.

\begin{align*}40\%\end{align*}

### Practice

Directions: Write the following percents as a ratio with a denominator of 100.

1. 64%
2. 3%
3. 119%
4. 4.7%
5. 88%
6. 99.5%
7. 12%
8. 14%

Directions: Write the following as percents.

1. 12 out of 100
2. 13.5 out of 100
3. 87 out of 100
4. 99 out of 100
5. 5 out of 100
6. 3.5 out of 100
7. 130 out of 100
8. 175 out of 100

### Vocabulary Language: English

Decimal

Decimal

In common use, a decimal refers to part of a whole number. The numbers to the left of a decimal point represent whole numbers, and each number to the right of a decimal point represents a fractional part of a power of one-tenth. For instance: The decimal value 1.24 indicates 1 whole unit, 2 tenths, and 4 hundredths (commonly described as 24 hundredths).
fraction

fraction

A fraction is a part of a whole. A fraction is written mathematically as one value on top of another, separated by a fraction bar. It is also called a rational number.
Percent

Percent

Percent means out of 100. It is a quantity written with a % sign.


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# Botany technician needed: Summer 2019 at Isle Royale National Park

Number of Positions: 2

Duration: May 14 – August 28, 2019 (dates are firm)

Pay: $20.00-$21.00/hour

Description and Duties: The incumbent will be employed by Northland College, but will work directly with, and report to, the Great Lakes Network of the National Park Service Inventory & Monitoring (I&M) program. The goals of the I&M program include monitoring key environmental variables so that ecological change can be detected and addressed in a timely manner.

This is largely a field-based position. The incumbent will assist with the sampling of permanent forest monitoring plots at Isle Royale National Park in Michigan. The incumbent will collect extensive data on the herbaceous layer and browse. He/she will be expected to identify at least 350 herbaceous species, and have considerable experience identifying members of the family Poaceae and genus Carex. Field work will be performed in teams of 3-4 people. The incumbent may also spend approximately 5-10% of time performing computer work.

This position operates on an 8 day on/6 day off schedule. During the 8-day sampling periods, incumbent will be expected to carry field equipment weighing up to 50 lbs. He/she will be exposed to high temperatures, stinging and biting insects, including extensive mosquitoes, and poisonous plants. Hikes to and from plots will be as long as six miles, each direction. The position will also involve camping, sometimes up to seven consecutive nights.

The incumbents are responsible for securing and paying for their own housing while at the park. Isle Royale is remote and opportunities for travel to the mainland will be limited. Housing accommodations will be shared with others and employees will be expected to be flexible and accommodating, as well as team players. Logistics will be explained in full, during interviews.

To Apply: Please send a resume, cover letter, references, and unofficial transcripts to Suzanne Sanders (suzanne_sanders@nps.gov) by January 21, 2019. If you are selected for an interview, you will be contacted shortly thereafter. No phone calls or emails, other than applications, please.


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54 questions linked to/from Plotting implicitly-defined space curves
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128 views

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How to get the output graph which consider only x^2+y^2=4 (the boundary of the region only)? I found this on the Help. ...
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I'm trying to find the graphic of the curve of the intersection of the hyperbolic paraboloid $z=y^2-x^2$ and the cylinder $x^2+y^2=1$. I could show them together: I want to see the curve of ...
I have two functions, let's call them $F1$ and $F2$. Both take the same three arguments: $x1, x2$ and $\epsilon$. With every $\epsilon^* \in [0,1]$, there should be either zero or one unique point $(... 1answer 236 views ### Contour3Dplot more graphs in one picture [duplicate] How can I plot two equations in one 3D graph. I have to plot x^2+y^2=1 and x^2+y^2+z^2=2 but I want tu plot only their intersection :( Thank you 6answers 2k views ### Finding length of intersection of two surfaces I would like to know how we find the length of the intersection of two surfaces. For instance, in the following example,a surface intersects with a plane: How do we find the length of intersection ... 3answers 7k views ### I'd like to display field lines for a point charge in 3 dimensions I'd like to display field lines for a point charge in 3 dimensions. Not a force field (short arrows) but continuous field lines that start on the charge. 4answers 30k views ### How to plot an ellipse? I'm new to Mathematica, and I'm finding it difficult to plot an ellipse. I tried using Plot[(x/5)^2 + (y/3)^2 == 1, {x, -5, 5}, {y, -3, 3}] but I'm getting ... 1answer 3k views ### Implementing a first person view of 3D objects in a scene I've created the following scene with a Chinese-style building surrounded by trees, and a horse and a rabbit grazing on the grass in Mathematica (don't ask me why there's a bust of Beethoven in there..... 3answers 6k views ### Plotting contours plot for f(x,y,z)=c I have the following question: I have a file that has structure: x1 y1 z1 f1 x2 y2 z2 f2 ... xn yn zn fn I can easily visualize it with Mathematica using ... 2answers 2k views ### Getting an InterpolatingFunction from a ContourPlot I have a function, say minimizeme[Ω_][β_][ϵ_] = ϵ^2 Ω - Log[2 (Cosh[2 β] + Cosh[2 β ϵ])]/(2 β); I want to find its critical points in$\epsilon\$ for a given ...
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# Can Gödel's theorem be proved within PA?

Gödel proves his theorem informally by using natural languages. However, is there a way to carry out his proof in PA itself? (so that maybe PA could prove that itself could not prove its own consistency) If there isn't (and we have to carry out his proof in some stronger background theory such as ZFC), what makes his proof inexpressible in PA?

• While I understand the question you're asking, I think it's a bit misleading to characterize Gödel's proof as "informal": it was as rigorous as any mathematical proof created by humans. Mar 19 at 16:52
• PA is more than enough to prove "If PA is consistent then PA does not prove that PA is consistent." Similarly, PA is more than enough to prove "If ZFC is consistent then ZFC does not prove that ZFC is consistent." Mar 19 at 17:32
• Wow, thank you a lot! Mar 20 at 2:20
• @AndreasBlass it sounds like PA is not enough to prove the full Gödel incompleteness theorem. If so, do you have a reference that specifically discusses the valid choices of X and Y in "X is enough to prove 'If Y is consistent, then Y does not prove that Y is consistent'"? Mar 20 at 8:58
• @Bananach When X in your comment is PA, Y can be any computably axiomatized theory, with a $\Sigma^0_1$ definition of Y being used in the formal statement about Y. The same works for weaker X; I'm confident that I$\Sigma_1$ suffices, but I'm not sure how much weaker one can make X (maybe I$\Delta_0+\exp$) Mar 20 at 14:19


text: This chapter features a lot of third-party tools; most of them are installed independently of your TeX distribution. A friend asked to convert file in pdf and I started reading about methods. And have tried to convert file using the suggested methods, like first convert to DVI, then ps and then to pdf. I am able to view file with different editors like Texworks and WinShell. change the  type of the figure from EPS or PS Option 2: Tried this method with TeXworks but it does not do anything. This page gives you hints on extracting stuff from a PDF file (again, individual figures from a PDF, not just using a PDF as a figure), but it seems pretty Windows/Mac-centric. If you want to display TikZ images only, use \PreviewEnvironment{tikzpicture}.If you are interested, you can find the sample code here.. An online LaTeX editor that's easy to use. I just tried to help you out and compile it. I have figures and I am using the command below to save it as pdf. 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Since we are talking about exporting to PDFs I can save graphs to PDF without any problems. xpdf Report.pdf. command convert a figure in EPS or PS into a figure It is one of the most common file formats in use today and extensively used in all areas. To generate the pdf file from latex The purpose of this command is to generate a report including the Diagram, equations, solution, tables, and plots. You're missing a file tcilatex.tex which is needed to compile. In this respect, this command is similar to the Print command. 2. Compile your ''tex'' file via LaTeX > open pdf file in ''Adobe Acrobat Pro'' > go to ''file'' option > select ''save as'' > go to ''save as type'' > choose ''.png/.jpeg'' I 1. If you'd like to download a copy for offline use, you can download a pdf via the icon above the preview pane, or download the full set of source files from the Overleaf menu in the upper left. in PDF format. in the place where the figures should be included in EPS or Generate PDF files from Latex Files: To generate the pdf file from latex file just following these tips: If you are including any figure then change the type of the figure from EPS or PS to PDF. Instead of compiling your .tex with the “latex” program, you just compile it with the “pdflatex” program, which outputs straight to .pdf. 3. to fig6.pdf only. But I have failed altogether. If your latex file includes postscript files (such as in figures), then pdflatex will not be able to generate a proper pdf file. From the pop-up window, press the "Change" button under the "Destination" tab to convert the document into a PDF file. You can upload … This issimilar to style sheets (CSS) in HTML and styl… Option 1: Running the program latex on your .tex file creates a .dvi file. But this time I need to use the .fig files in a Word document. Click inside the file drop area to upload PDF files or drag & drop PDF files. to PDF. https://drive.google.com/file/d/1i3QSe2zTDUU6GCo-EIxLdWyjof59NHud/view?usp=sharing, Most people haven’t actually followed the .tex -> .dvi -> .pdf process in a LONG time (think: decades). I suggest you spend the time now and figure it out because it's vital and will pay off. To open a new file in TeXnicCenter, click File in the upper left, then click New. You can then either convert a .dvi directly to a .pdf or to a .ps first and then on to a .pdf. LaTeX is a package that allows the use of markup in the document to structure itlogically in a way that users would never have to think about formatting. from PS to PDF format. Inkscape will allow you to open the figures, remove the page, and save as an eps file if you wish or as a pdf file. For example, I can copy graphics from my own LaTeX generated PDFs in Mac OS X's Preview, and paste them into Photoshop as vector graphics, and then save them for inclusion into another LaTeX document. So you could generate a "plain old" PDF file from your LaTeX sources, and then convert it to PDF/A using Acrobat Pro. In this case you have two options: You can convert your Postscript graphics to PDF graphics, or you can use dvips and convert the output to PDF. I am able to view file with different editors like Texworks and WinShell. You can view the files in PDF format on linux systems by My main Latex file was While one can sometimes get away with this method, the problem is that the resulting PDF/A-1b is of poor quality: if the input document did not contain metadata or Unicode mappings, then the conversion software will not be able to create this data out of thin air. It can be done using the following command: e.g. Comment: save.image function saves current workspace not graph! Open your web browser and open LaTeXtoPDF. fig6.pdf. The function prints the file in current buffer as a PDF in the same folder by default. Both of these The generated file was So I tried saving them and exporting them as .pdf files but when I opened the pdf file, the figure is cropped. A friend asked to convert file in pdf and I started reading about methods. By using our Services or clicking I agree, you agree to our use of cookies. You can put the below in your init.el and bind the function to a binding of your choice.. PDF stands for Portable Document Format. The LaTeX option is much more appealing to me. Step 2. Overview. It says some unknown error occurs. Convert ASPX File to PDF It is now time to convert the file into PDF. Traditionally, you converted your LaTeX source file to a DVI file, whichcould then be converted to PostScript with dvips. However, the Create LaTeX/PDF Report command does not directly print but instead creates a TeX document in LaTeX2e. We save your project automatically every few seconds, as you type. e.g. Option 2: Running the program pdflatex on your .tex file creates the .pdf file directly. used the command ps2pdf  fig6.ps and it generated the file Instead, you, like me and like all other weenies, will be working with LaTeX, whoseauthor Leslie Lamport has tamed the beast for the rest of us. This is shown below by an example: A figure can be so the creator have sent me single file (incomplete data). This way you create a pdf that can have hyperlinks, selectable text, fillable form fields, etc. Option 1: I have installed WinShell but failed to convert. Is there any way to save it directly as pdf instead of saving as .ps and convert to .pdf. LaTeX is semantic markup, where the same input can give you multiple outputs depending on what style you load, parameters you set. Alternatively, you can use NotePad++ to open TEX file but you are only allowed to view and edit the source codes rather than compile or save as PDF files. Open the .pdf of your picture in inkscape (free). Start TeXnicCenter if you haven't already done so. If you have the choice, it is often easier with Unix systems for command line tools. Alternatively, you can use Latex to compile a pdf which will keep the text as text - although there's a fair bit of work in using Latex. Onlythe most seasoned computer veterans with a death wish would descend to that level. This, in turn,can be converted to a PDF file by ps2pdf: latex dvips ps2pdftext.tex -------> text.dvi -------> text.ps --------> text.pdf. But I have failed altogether. Type the following lines (or copy and paste them): Note that it is crimped brackets surrounding the words "article" and "document" in the above example, not parentheses. 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text:
# 3.3 Electrical potential due to a point charge  (Page 2/2)

 Page 2 / 2

The voltages in both of these examples could be measured with a meter that compares the measured potential with ground potential. Ground potential is often taken to be zero (instead of taking the potential at infinity to be zero). It is the potential difference between two points that is of importance, and very often there is a tacit assumption that some reference point, such as Earth or a very distant point, is at zero potential. As noted in Electric Potential Energy: Potential Difference , this is analogous to taking sea level as $h=0$ when considering gravitational potential energy, ${\text{PE}}_{g}=\text{mgh}$ .

## Section summary

• Electric potential of a point charge is $V=\text{kQ}/r$ .
• Electric potential is a scalar, and electric field is a vector. Addition of voltages as numbers gives the voltage due to a combination of point charges, whereas addition of individual fields as vectors gives the total electric field.

## Conceptual questions

In what region of space is the potential due to a uniformly charged sphere the same as that of a point charge? In what region does it differ from that of a point charge?

Can the potential of a non-uniformly charged sphere be the same as that of a point charge? Explain.

## Problems&Exercises

A 0.500 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 40.0 pC charge on its surface. What is the potential near its surface?

144 V

What is the potential $0\text{.}\text{530}×{\text{10}}^{–10}\phantom{\rule{0.25em}{0ex}}\text{m}$ from a proton (the average distance between the proton and electron in a hydrogen atom)?

(a) A sphere has a surface uniformly charged with 1.00 C. At what distance from its center is the potential 5.00 MV? (b) What does your answer imply about the practical aspect of isolating such a large charge?

(a) 1.80 km

(b) A charge of 1 C is a very large amount of charge; a sphere of radius 1.80 km is not practical.

How far from a $1\text{.}\text{00 µC}$ point charge will the potential be 100 V? At what distance will it be $\text{2.00}×{\text{10}}^{2}\phantom{\rule{0.25em}{0ex}}\text{V}?$

What are the sign and magnitude of a point charge that produces a potential of $\text{–2.00 V}$ at a distance of 1.00 mm?

$–2\text{.}\text{22}×{\text{10}}^{–13}\phantom{\rule{0.25em}{0ex}}\text{C}$

If the potential due to a point charge is $5\text{.}\text{00}×{\text{10}}^{2}\phantom{\rule{0.25em}{0ex}}\text{V}$ at a distance of 15.0 m, what are the sign and magnitude of the charge?

In nuclear fission, a nucleus splits roughly in half. (a) What is the potential $2\text{.}\text{00}×{\text{10}}^{–14}\phantom{\rule{0.25em}{0ex}}\text{m}$ from a fragment that has 46 protons in it? (b) What is the potential energy in MeV of a similarly charged fragment at this distance?

(a) $3\text{.}\text{31}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{V}$

(b) 152 MeV

A research Van de Graaff generator has a 2.00-m-diameter metal sphere with a charge of 5.00 mC on it. (a) What is the potential near its surface? (b) At what distance from its center is the potential 1.00 MV? (c) An oxygen atom with three missing electrons is released near the Van de Graaff generator. What is its energy in MeV at this distance?

An electrostatic paint sprayer has a 0.200-m-diameter metal sphere at a potential of 25.0 kV that repels paint droplets onto a grounded object. (a) What charge is on the sphere? (b) What charge must a 0.100-mg drop of paint have to arrive at the object with a speed of 10.0 m/s?

(a) $2\text{.}\text{78}×{\text{10}}^{-7}\phantom{\rule{0.25em}{0ex}}\text{C}$

(b) $2\text{.}\text{00}×{\text{10}}^{-10}\phantom{\rule{0.25em}{0ex}}\text{C}$

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward a gold nucleus, and its path was substantially deflected by the Coulomb interaction. If the energy of the doubly charged alpha nucleus was 5.00 MeV, how close to the gold nucleus (79 protons) could it come before being deflected?

(a) What is the potential between two points situated 10 cm and 20 cm from a $3\text{.}0 µC$ point charge? (b) To what location should the point at 20 cm be moved to increase this potential difference by a factor of two?

Unreasonable Results

(a) What is the final speed of an electron accelerated from rest through a voltage of 25.0 MV by a negatively charged Van de Graaff terminal?

(c) Which assumptions are responsible?

(a) $2.96×{10}^{9}\phantom{\rule{0.25em}{0ex}}\text{m/s}$

(b) This velocity is far too great. It is faster than the speed of light.

(c) The assumption that the speed of the electron is far less than that of light and that the problem does not require a relativistic treatment produces an answer greater than the speed of light.

how can chip be made from sand
is this allso about nanoscale material
Almas
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where is the latest information on a no technology how can I find it
William
currently
William
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Got questions? Join the online conversation and get instant answers!


text:
Is there a way to convince Mathematica to perform quantifier elimination on a semialgebraic set and yield a resulting semialgebraic set whose representation does not involve any roots or radicals, i.e., is given as a Boolean combination of polynomial inequalities? Here is a simple example:

org = Reduce[Exists[{z}, x^2 + 2 x y <= z^3 && x y + z <= 5], {x, y}, Reals]

yields

(x < 0 && y >= Root[-125 + x^2 + 77 x #1 - 15 x^2 #1^2 + x^3 #1^3 &, 1]) || x == 0 ||
(x > 0 && y <= Root[-125 + x^2 + 77 x #1 - 15 x^2 #1^2 + x^3 #1^3 &, 1])

Update: In this specific example, we can use ToRadicals and manually introduce two variables w1, w2 for which the following holds:

cond = w1^3 == Sqrt[3] w2 - 90 x^6 - 9 x^8 &&
w2^2 == 2732 x^12 + 540 x^14 + 27 x^16 &&
w2 >= 0

Substituting these variables for the respective radical expressions yields the following Boolean combination of polynomial inequalities:

new =
((x < 0 && ((6 w1 x^3 y >=
2^(2/3) 3^(1/3) w1^2 + 30 w1 x^2 - 4 2^(1/3) 3^(2/3) x^4 &&
6 w1 x^3 > 0) || (6 w1 x^3 y <=
2^(2/3) 3^(1/3) w1^2 + 30 w1 x^2 - 4 2^(1/3) 3^(2/3) x^4 &&
6 w1 x^3 < 0))) ||
x == 0 ||
(x > 0 && ((6 w1 x^3 y <=
2^(2/3) 3^(1/3) w1^2 + 30 w1 x^2 - 4 2^(1/3) 3^(2/3) x^4 &&
6 w1 x^3 > 0) || (6 w1 x^3 y >=
2^(2/3) 3^(1/3) w1^2 + 30 w1 x^2 - 4 2^(1/3) 3^(2/3) x^4 &&
6 w1 x^3 < 0)))) &&
w1^3 == Sqrt[3] w2 - 90 x^6 - 9 x^8 &&
w2^2 == 2732 x^12 + 540 x^14 + 27 x^16 &&
w2 >= 0

Reduce/Resolve are able to prove the equivalence:

Reduce[ForAll[{x, y, w1, w2}, Implies[cond, Equivalent[org, new]]], {}, Reals]
True

Resolve[ForAll[{x, y}, Exists[{w1, w2}, Equivalent[org, new]]], Reals]
True

However, I wonder about a general solution which should in principle exist.

• What different form might one obtain? Jun 18, 2021 at 14:10
• A Boolean combination of polynomial inequalities p(x) >= 0 (x being a vector here). Jun 18, 2021 at 14:12
• I meant for this specific example. When everything is an equation this is standard variable elimination. But for inequalities it is not clear to me what the result above might be. Jun 18, 2021 at 14:16
• @DanielLichtblau: I extended my post accordingly. However, the approach is not insightful for the general case. Jun 18, 2021 at 15:56
• @MathGaudium: Could you indicate the edited places in your question? Such behavior is standard. Jun 18, 2021 at 16:13

Yes, there is such away:

ToRadicals[Root[-125 + x^2 + 77 x #1 - 15 x^2 #1^2 + x^3 #1^3 &, 1], Assumptions -> x < 0]

5/x - (2 (2/3)^( 1/3) x)/(-90 x^6 - 9 x^8 + Sqrt[3] Sqrt[2732 x^12 + 540 x^14 + 27 x^16])^( 1/3) + (-90 x^6 - 9 x^8 + Sqrt[3] Sqrt[2732 x^12 + 540 x^14 + 27 x^16])^(1/3)/( 2^(1/3) 3^(2/3) x^3)

ToRadicals[Root[-125 + x^2 + 77 x #1 - 15 x^2 #1^2 + x^3 #1^3 &, 1], Assumptions -> x > 0]

5/x - (2 (2/3)^( 1/3) x)/(-90 x^6 - 9 x^8 + Sqrt[3] Sqrt[2732 x^12 + 540 x^14 + 27 x^16])^( 1/3) + (-90 x^6 - 9 x^8 + Sqrt[3] Sqrt[2732 x^12 + 540 x^14 + 27 x^16])^(1/3)/( 2^(1/3) 3^(2/3) x^3)

• Thank you. I also extended my post with a manually constructed solution that involves two additional variables. However, I wonder how this can be done in general with Mathematica without relying on ToRadicals which cannot be used for in general. It should be possible though because a semialgebraic set can always be represented as a Boolean combination of polynomial inequalities. Jun 18, 2021 at 15:54


text:
How does AUC of ROC equal concordance probability? [duplicate]

If this answer on Quora is correct then I think I understand what concordance probability is. However, I also find that this answer on StackExchange provides the formula of concordance probability that is a little bit different from the one on Quora: It include the counts of tied pairs with weight $0.5$. Which one should I trust?

I also understand AUC stands for Area Under a Curve and how to compute it... visually.

What I do not understand is how AUC equal concordance probability?

I have just found my question is the same as this answered question on StackExchange.

marked as duplicate by Sycorax, gung♦Sep 16 '17 at 1:36

• What do you mean by computing it visually? You can just sweep your threshold across the range to get the points on the curve. Then compute the area underneath like you would compute an integral. – Aaron Apr 6 '17 at 18:25
• @Aaron That is what I meant by saying "compute it visually"... I do not know of a shorter formula to calculate AUC except concordance probability. I'm sorry for my incomprehensible words. – ntvy95 Apr 7 '17 at 0:38
• The sklearn library computes it using the trapezoidal rule for integration scikit-learn.org/stable/modules/generated/… – Aaron Apr 7 '17 at 0:41
• @Aaron Thank you very much for your suggestion! :) – ntvy95 Apr 7 '17 at 1:08

I finally have just figured it out. I'm not sure if my reasoning is correct so I will leave it here for anyone to point out the errors.

The discrete formula of AUC should be:

$$\sum_{i} \frac{TP_{i}}{P}\frac{FP_{i+1}-FP_{i}}{N}$$

(borrowed from http://mlwiki.org/index.php/ROC_Analysis)

That is, I also happen to calculate the (sometimes maybe just approximate) number of concordance pairs in the numerator ($TP_{i}(FP_{i+1}-FP_{i})$) and the number of total pairs in the denominator ($PN$).

For an illustration:

• When $i = 2$, $TP_{2} = 2$ and $FP_{3} - FP_{2} = 1$, that is we have found $2$ possible concordance pairs at $i = 2$ (i.e. 1-3, 2-3).

• When $i = 6$, $TP_{6} = 5$ and $FP_{7} - FP_{6} = 1$, that is we have found more $6$ possible concordance pairs at $i = 6$ (i.e. 1-7, 2-7, 4-7, 5-7, 6-7). The iterative process keeps going on.

• ...

However, this formula of AUC does not count tied pairs with 0.5 weight according to this answer on StackExchange.

• I have found that this answer on StackExchange with some modifications can cover the case counting tied pairs with weight $0.5$. – ntvy95 Apr 7 '17 at 1:48

I use basic integration to prove that Area Under ROC and % of Concordant pairs are the same. I am adding a description below each illustration to make the understanding clear.

Figure 1 contains the distribution of 0's and 1's , $x(c), y(c)$ respectively. The cutoff $c$ is represented on the X-axis.

Figure 2 contains the distribution of ROC curve. The sensitivity is $Y$ and the (1-specificity) is $X$

In the above picture, we begin by counting the total number of concordant pairs. The number of concordant pairs, for a given $x(c)dc$ number of 1's, is given by $[x(c)dc]*[\int_0^c y(c)]$.

So, the total number of concordant pairs is found by integrating the above expression from 0 to 1, i.e. $\int_0^1[x(c)*[\int_0^c y(c)]]dc$

Proportion of concordant pairs is found by ${(\int_0^1[x(c)*[\int_0^c y(c)]]dc) \over ((\int_0^1 x(c)dc)*(\int_0^1 y(c)dc))} \tag{1}\label{1}$

Now we turn to the ROC plot. For a given cutoff $c$, we obtain the specificity ($1-X$) by ${\int_0^c y(c)dc \over \int_0^1 y(c)dc}$.

We obtain $dX = {y(c)dc \over \int_0^1 y(c)dc}$. As $X$ varies from 0 to 1, $c$ varies from 1 to 0.

Similarly, we obtain the expression for Sensitivity. And we go on to get the area under ROC = $\frac{[\int_0^1[\int_0^1 x dc] y dc]}{(\int_0^1 x(c)dc)(\int_0^1 y(c)dc)} \tag{2}\label{2}$

Now we have to show that the numerators of (1) and (2) are the same. This can be easily shown by using integration by parts.


text:
# help me with inverse z transform

• February 10th 2011, 07:48 PM
TechnicianEngineer
help me with inverse z transform
Find the causal signal x(n) for the ff. z transform
$X(z) = \frac{0.5z}{z^2-z+0.5}$
• February 11th 2011, 06:41 AM
Ackbeet
Dividing top and bottom by $z^{2}$ yields

$X(z)=\dfrac{0.5z^{-1}}{1-z^{-1}+0.5z^{-2}}.$

Comparing this expression to # 20 in this list, we see that we must have, if possible,

$0.5=a\sin(\omega_{0}),$

$1=2a\cos(\omega_{0}),$ and

$0.5=a^{2}.$

You can make this work out. What do you get for $a$ and $\sin(\omega_{0})$ and $\cos(\omega_{0})?$ What is your radius of convergence?
• February 11th 2011, 10:42 AM
chisigma
Quote:

Originally Posted by TechnicianEngineer
Find the causal signal x(n) for the ff. z transform
$X(z) = \frac{0.5z}{z^2-z+0.5}$

The 'rexcursive relation' corresponding the the z-transform...

$\displaystyle X(z)= \frac{1}{2}\ \frac{z^{-1}}{1-z^{-1} + \frac{z^{-2}}{2}}$ (1)

... is...

$\displaystyle x(n)= x(n-1)- \frac{x(n-2)}{2} + \frac{\delta(n-1)}{2}\\,\\ x(-1)=0\\,\\ x(-2)=0$ (2)

... that has to be solved...

Kind regards

$\chi$ $\sigma$
• February 11th 2011, 04:40 PM
TechnicianEngineer
the region of convergence is |z| > 0.707106

solving for a
$\sqrt(0.5) = a$
$a = 0.707106$

solving for $\omega_o$
$\sin^{-1}(\frac{1}{2*0.707106}}) = \omega_o$
$\omega_o = 45 degrees$

the inverse z transform is
$x[n] = (0.707107)^n \sin(\frac{\pi}{4}n)u[n]$

is this correct?
• February 11th 2011, 04:57 PM
Ackbeet
You should check that the values you obtained check out with the third equation (three equations and two unknowns renders your system overdetermined, but the third equation must be satisfied). If it does, then I'd say you have the correct answer.


text:
# Boundary value problem, numerical methods for partial differential equations

Approximate methods of solution which yield the solution to the problem in the form of a numerical table. An exact construction of a solution (in terms of explicit formulas, series, etc.) to a boundary value problem is only rarely possible. The most prevalent of the methods of approximate solution are difference methods (see [1]); they are applied in the most general problems and are readily computerized. The essential idea of difference methods is to replace the original domain of variation of the independent variables by a discrete set of points, a grid, and to approximate the derivatives figuring in the equation and the boundary conditions by difference quotients at the grid points. As a result of this procedure the original problem is replaced by a system of finitely many algebraic equations (linear or non-linear); this system is known as a difference scheme. The solution of the difference scheme is taken to be an approximate solution of the original problem. The accuracy of the approximation depends on the method of approximation and on the fineness of the grid, i.e. on how densely the grid points fill the original domain. In the rest of this article attention will be restricted to linear boundary value problems for partial differential equations, and it will be assumed that the problem in question is well-posed. In order to justify difference methods one must investigate the well-posedness of the difference problem and its convergence as the grid is refined. A difference problem is said to be well-posed if it has a unique stable solution, whatever the terms on the right-hand sides of the equations. By stability of a difference scheme one means that its solution depends continuously on the right-hand side, uniformly with respect to the grid spacing.

For example, suppose one wishes to solve the Dirichlet problem for the Poisson equation in the square $G = \{ 0 < x _ {1} , x _ {2} < 1 \}$ with boundary $\Gamma$:

$$\frac{\partial ^ {2} u }{\partial x _ {1} ^ {2} } + \frac{\partial ^ {2} u }{\partial x _ {2} ^ {2} } = \ - f (x _ {1} , x _ {2} ),\ \ (x _ {1} , x _ {2} ) \in G,$$

$$u (x _ {1} , x _ {2} ) = \mu (x _ {1} ,\ x _ {2} ),\ (x _ {1} , x _ {2} ) \in \Gamma .$$

The domain $G$ is replaced by a square grid $G _ {h}$ with spacing $h$, i.e. by the set of points

$$G _ {h} = \ \{ ( x _ {1} ^ {(i)} , x _ {2} ^ {(j)} ) : x _ {1} ^ {(i)} = ih,\ x _ {2} ^ {(j)} = jh;$$

$${} i, j = 1 \dots N - 1 \} ,\ hN = 1,$$

with boundary

$$\Gamma _ {h} = \ \cup _ {i, j = 1 } ^ { {N } - 1 } \{ ( 0, x _ {2} ^ {(j)} ) \cup ( 1, x _ {2} ^ {(j)} ) \cup ( x _ {1} ^ {(i)} , 0 ) \cup ( x _ {1} ^ {(i)} , 1 ) \} ,$$

and the derivatives figuring in the equation by difference quotients

$$( \Lambda _ {1} u) _ {i, j } = \ u _ {\overline{x}\; _ {1} x _ {1} , i, j } = \ \frac{(u _ {i + 1, j } - 2u _ {i, j } + u _ {i - 1, j } ) }{h ^ {2} } ,$$

$$( \Lambda _ {2} u) _ {i, j } = u _ {\overline{x}\; _ {2} x _ {2} , i, j } = \frac{(u _ {i, j + 1 } - 2u _ {i, j } + u _ {i, j - 1 } ) }{h ^ {2} } ,$$

where $u _ {i, j } = u (x _ {1} ^ {(i)} , x _ {2} ^ {(j)} )$. The difference scheme is

$$\tag{1 } ( \Lambda _ {h} y) _ {i, j } = \ y _ {\overline{x}\; _ {1} x _ {1} , i, j } + y _ {\overline{x}\; _ {2} x _ {2} , i, j } = \ - f _ {i, j } ,$$

$$i, j = 1 \dots N - 1; \ \left . y _ {i, j } \right | _ {\Gamma _ {h} } = \mu _ {i, j } ,$$

where $y _ {i, j }$ is the solution.

Problem (1) has a — unique — solution for any inhomogeneous term $f$ and any boundary conditions $\mu$( see [2]). Moreover, the solution of the difference scheme (1) converges as $h \rightarrow 0$ to the solution of the original problem in such a way that the scheme is a second-order approximation in the maximum norm, i.e.

$$\max _ {(x _ {1} , x _ {2} ) \in G _ {h} } | y (x _ {1} , x _ {2} ) - u (x _ {1} , x _ {2} ) | \leq Mh ^ {2} ,$$

where $M$ is a constant independent of $h$.

The difference scheme (1) is a system of linear algebraic equations, characteristically with a large number of equations (to be precise: $(N - 1) ^ {2}$ equations, with $N \rightarrow \infty$ as $h \rightarrow 0$) and a large number of zeros in the matrix of the system; moreover, the system is generally ill-conditioned (the ratio of the minimum eigen value to the maximum one is of the same order of magnitude as $h ^ {2}$, $h \rightarrow 0$). Such systems of equations, which arise when differential equations are replaced by difference equations, may be solved using various efficient methods, both direct and iterative. Direct methods yield an exact solution of the difference equation after a finite number of arithmetic operations. This category includes various versions of the double-sweep method, including matrix double-sweep, the decomposition method, the fast Fourier transform, and the method of representations by sums (see [1], [2], [3], [6]). A measure of the efficiency of direct methods is the order of magnitude of the number of operations as $h \rightarrow 0$. Thus, solving problem (1) by matrix double-sweep requires $O (h ^ {-4} )$ operations, whereas the fast Fourier transform method needs — for the same problem — $O (h ^ {-2} \mathop{\rm log} h ^ {-1} )$ operations. The most widely used iterative methods for the solution of difference problems are Richardson's method with Chebyshev interpolation points, the alternating-triangular iterative method, and various alternating-direction methods (see [2]). The efficiency of iterative methods is measured by the order of magnitude of the minimum number of iterations, $n _ {0} ( \epsilon )$, necessary to decrease the error of the initial approximation by a factor $1/ \epsilon$. For example, in solving problem (1) by Richardson's method, $n _ {0} ( \epsilon )$ is of the order $h ^ {-1} \mathop{\rm log} (1/ \epsilon )$, while the alternating-direction method with optimally-selected iteration parameters has $n _ {0} ( \epsilon ) = O ( \mathop{\rm log} h ^ {-1} \mathop{\rm log} ( 1/ \epsilon ))$. Iterative methods are more universal and simpler to implement than direct methods, and consequently have been widely utilized to solve difference problems.

For example, consider solving the first boundary value problem for the heat equation:

$$\tag{2 } \left . \begin{array}{ll} \frac{\partial u }{\partial t } = \ \frac{\partial ^ {2} u }{\partial x _ {1} ^ {2} } + \frac{\partial ^ {2} u }{\partial x _ {2} ^ {2} } , & t > 0,\ (x _ {1} , x _ {2} ) \in G \\ u (x _ {1} , x _ {2} , 0) = \ u _ {0} (x _ {1} , x _ {2} ); &{} \\ u (x _ {1} , x _ {2} , t) = 0, & (x _ {1} , x _ {2} ) \in \Gamma . \\ \end{array} \ \right \}$$

To solve this problem, one prescribes a time grid with spacing $\tau > 0$:

$$\omega _ \tau = \ \{ {t _ {n} = n \tau } : { n = 0, 1 ,\dots } \}$$

and the grid $G _ {h}$( see above) for the space variables. Let $u _ {i, j } ^ {n} = u (x _ {1} ^ {(i)} , x _ {2} ^ {(j)} , t _ {n} )$. The derivative $u _ {t} ^ \prime$ is approximated by the quotient

$$u _ {t, i, j } = \ \frac{u _ {i, j } ^ {n + 1 } - u _ {i, j } ^ {n} } \tau$$

and the Laplacian $\Delta$ by the difference operator $\Delta _ {h}$. The original equation (2) is replaced by a corresponding difference scheme

$$\tag{3 } \left . \begin{array}{l} y _ {t, i, j } = \ \sigma ( \Delta _ {h} y) _ {i, j } ^ {n + 1 } + (1 - \sigma ) ( \Delta _ {h} y) _ {i, j } ^ {n} ; \\ y _ {i, j } ^ {0} = \ u _ {0} ( x _ {1} ^ {(i)} , x _ {2} ^ {(j)} ) ,\ \ i, j = 1 \dots N - 1; \\ \left . y _ {i,j} ^ {n} \right | _ {\Gamma _ {h} \times \omega _ \tau } = 0. \\ \end{array} \ \right \}$$

The parameter $\sigma$ appearing in these equations determines the stability and accuracy of the scheme. If $\sigma \geq 0.5$, the scheme is stable for any grid spacings $\tau$ and $h$( an absolutely-stable difference scheme). But if $\sigma < 0.5$, the scheme is stable only if some restriction is imposed on $\gamma = \tau /h ^ {2}$( a conditionally-stable difference scheme); for example, the explicit scheme ( $\sigma = 0$) is stable if $\gamma \leq 1/4$. If $\sigma = 0.5$, the scheme is second-order accurate with respect to $\tau$ and $h$; for other values of $\sigma$ it is first-order accurate in $\tau$ and second-order accurate in $h$. The difference problem (3) is solved "in layers" . The $n$- th layer is the set of all points of the grid $G _ {h} \times \omega _ \tau$ for some fixed $n$. The values of $y _ {i, j } ^ {0}$ on the zero-th layer ( $n = 0$) are known from the initial conditions. If the values on the $n$- th layer are known, then the values on the next layer $y _ {i, j } = y _ {i, j } ^ {n + 1 }$ are determined from the system of equations

$$\tag{4 } \left . \begin{array}{l} y _ {i, j } - \sigma \tau ( \Delta _ {h} y) _ {i, j } = \ f _ {i, j } ^ {n} ,\ \ i, j = 1 \dots N - 1, \\ \left . y _ {i, j } \right | _ {\Gamma _ {h} } = 0, \\ \end{array} \ \right \}$$

where

$$f _ {i, j } ^ {n} = \ y _ {i, j } ^ {n} + (1 - \sigma ) \tau ( \Delta _ {h} y) _ {i, j } ^ {n} .$$

The solution of problem (4) may be found by any of the methods for solving the stationary problem (1). However, there are more economical algorithms for the solution of multi-dimensional non-stationary boundary value problems, namely alternating-direction methods (see [1][5]), which enables one to reduce the solution of a multi-dimensional problem to that of a sequence of one-dimensional problems. Thus, the heat equation may be solved using the following alternating-direction scheme:

$$\left . \begin{array}{l} \frac{y _ {i, j } ^ {n + 1/2 } - y _ {i, j } ^ {n} }{0.5 \tau } = \ \Lambda _ {1} y _ {i, j } ^ {n + 1/2 } + \Lambda _ {2} y _ {i, j } ^ {n} , \\ \frac{y _ {i, j } ^ {n + 1 } - y _ {i, j } ^ {n + 1/2 } }{0.5 \tau } = \ \Lambda _ {1} y _ {i, j } ^ {n + 1/2 } + \Lambda _ {2} y _ {i, j } ^ {n + 1 } . \\ \end{array} \ \right \}$$

This scheme is absolutely stable, second-order accurate and is solved by successive inversion of one-dimensional difference operators.

Even when the difference problem is solved exactly, the solution may well be different — not only quantitatively but also qualitatively — from the solution of the original differential problem. The discrepancy is particularly prominent in dealing with equations the coefficients — or solutions — of which involve singularities (thus, in computing discontinuous solutions of the equations of gas dynamics one usually encounters zones in which the discontinuities are strongly "spread out" ). Thus, direct approximation of a differential problem by a difference problem, when the derivatives are replaced — with a considerable degree of arbitrariness — by difference quotients, does not always yield a good difference scheme. Various principles have been developed for the construction of difference schemes of good quality. Among the successful methods are the balance method and the method of variational-functional approximation (see [1][3]). The difference schemes obtained in these methods correctly reflect the integral laws of conservation which are valid for the original equations, and ensure that the relevant difference operators have fixed signs. In the theory of homogeneous difference schemes (see [7]) one considers questions of constructing difference schemes for equations with variable (including discontinuous) coefficients and investigating their convergence.

#### References

 [1] A.N. Tikhonov, A.A. Samarskii, "Equations of mathematical physics" , Pergamon (1963) (Translated from Russian) MR0165209 Zbl 0111.29008 [2] A.A. Samarskii, G.I. Marchuk, "The theory of difference schemes" , Moscow (1983) (In Russian) MR1818323 MR0514784 MR0378439 MR0347102 MR0438725 MR0207234 MR0203968 Zbl 0971.65076 Zbl 1090.65500 Zbl 0368.65032 Zbl 0359.65083 Zbl 0155.43603 [3] G.I. Marchuk, "Methods of numerical mathematics" , Springer (1982) (Translated from Russian) MR0661258 Zbl 0485.65003 [4] N.N. Yanenko, "The method of fractional steps: solution of problems of mathematical physics in several variables" , Springer (1971) (Translated from Russian) MR307493 [5] E.G. D'yakonov, "Difference methods for solving boundary value problems" , 1–2 , Moscow (1971–1972) (In Russian) Zbl 0434.65078 Zbl 0308.65051 [6] G.N. Polozhii, "Numerical solution of two- and three-dimensional boundary value problems of mathematical physics and functions of a discrete argument" , Kiev (1962) (In Russian) [7] A.N. Tikhonov, A.A. Samarskii, "On homogeneous difference schemes" Zh. Vychisl. Mat. i Mat. Fiz. , 1 : 1 (1961) pp. 5–63 (In Russian) MR168127 Zbl 0125.07303


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## nincompoop 2 years ago use the definition of derivative to find f'(x)

1. nincompoop

$f(x)=\sqrt{(x-4)}$

2. myininaya

Do you need help pluggin in or multiply by the the top's conjugate?

3. anonymous

hmmm do you mean $\frac{f(a+h)-f(a)}{(a+h)-a}$

4. nincompoop

that is the definition of derivative @swissgirl YES :D

5. nincompoop

so you're left with $f'(x)=\frac{ \sqrt }{ h }$

6. anonymous

$f'(x)=\frac{ \sqrt{(x+h)-4}-\sqrt{x-4}}{h}$

7. nincompoop

$f'(x)=\lim_{h \rightarrow 0}\frac{ \sqrt }{ ? }$

8. nincompoop

WTFFFFFF

9. shamil98

LOL

10. shamil98

That latex fail

11. nincompoop

why is not going through?

12. shamil98

oh gratz on 90 ss nin

13. anonymous

Just btw my definition is not complete. This is the correct definition $\lim_{h \to 0}\frac{f(a+h)-f(a)}{(a+h)-a}$

14. nincompoop

$f'(x)=\lim_{h \rightarrow 0} \frac{ \sqrt{(x+h)-4} -(\sqrt{x-4})} { h }$

15. nincompoop

finally!!!!!!!!

16. shamil98

f(x) = (x-4)^1/2 $f'(x)= (\frac{ x-4 }{ 2 })^{\frac{ -1 }{ 2 }}$

17. nincompoop

I didn't want to pick on you, @swissgirl LAUGHING OUT LOUD but yes. without the limit would be the MVT or the mean value theorem

18. shamil98

o.o

19. nincompoop

no @shamil98 you've changed the entire thing by changing h = 2

20. nincompoop

h -> 0

21. nincompoop

you must be using a Plutonian Rule

22. anonymous

Ok so basically the next step would be to multiply by the conjugate. Gonna write it out and scan it in cuz it will take me forever to type it in latex

23. nincompoop

I tried, omg that thing takes forever using the equation tool bar

24. nincompoop

you know at a glance, I thought it is not differentiable because it involves imaginary number and when that is the case, there's no limit with real number, hence no derivatives.

25. anonymous

no there is a derivative just rewriting it.

26. anonymous

27. nincompoop

i know that >.<

28. nincompoop

was showing @shamil98

29. nincompoop

how'd you scan that so quickly?

30. anonymous

ummm it takes no longer than a half a second to scan

31. nincompoop

omg my scanner takes like a light year to scan a page

32. anonymous

I have an issue I cant draw straight lines -_-

33. nincompoop

ruler usually solves that problem

34. anonymous

lol thats what I usually hafta us

35. nincompoop

it is less of a technology than a scanner

36. anonymous

hahahaha

37. nincompoop

@shamil98 yo review this LAUGHING OUT LOUD so you can skip 2 years of college in math

38. anonymous

Ya I am great at bargaining and I managed to still get warranty coverage after the warranty was over for a year and I complained that the scanning was slow so they sent me a newer model printer which works awesome

39. nincompoop

can you send me an extra scanner then? I want the fujitsu desktop scanner. the brother machine I have is uber slow

40. anonymous

ohhh I kept it and gave it to my bros ato destroy

41. nincompoop

bought it for 99 bucks just to have something to scan all my lab and experiment draft crap

42. anonymous

ummm I have hp

43. anonymous

mine was \$150 I think

44. nincompoop

yo I saw the HP one, it's "touch" technology, but it's bulky and it gets smudgie

45. anonymous

But I gotta scan in all my work every week so it was a necessity

46. anonymous

ya the touch aint bothersome But have not noticed it smudgy but then again I rarely print I usually scan

47. nincompoop

so do you think you can get the fujitsu desktop scanner? although it's not for high-paper scanning, it's fast and very small

48. anonymous

for you or for me???

49. anonymous

Not gonna waste my bargaining skills on you :P

50. nincompoop
51. nincompoop

you're mean ... just mean mean mean cheese

52. anonymous

OMG its expensive -_-

53. anonymous

Totally not on my budget

54. nincompoop

it's actually cheaper now. it used to be around 900

55. anonymous

ya still dont excite me

56. nincompoop

I've been waiting for that crap to be like 200 bucks so I can buy it, but they kept discontinuing the older models

57. anonymous

Smart guys

58. nincompoop

59. anonymous

Did ya hear ma mayor is a crack head :D

60. nincompoop

it's fujitfrichinsu, it should be Japan

61. nincompoop

your mayor? or the re-elected Christie of NJ?

62. nincompoop

wait he's a governor LAUGHING OUT LOUD

63. anonymous

Hey I dont live in NJ remembah

64. nincompoop

I would have bashed him for being unhealthy fat slob and use that against him he neglects his own health and that is why he doesn't care about other people's health, which is the reason why he keeps cutting the university hospital's budgets

65. anonymous

lol ohhh I forgot u live in NJ

66. nincompoop

I am a New Yorker... LAUGHING OUT LOUD

67. anonymous

But u do live in NJ or at least at one point u did

68. anonymous

idk I only have one place of residence

69. nincompoop

don't you live in like NJ, NY and Canada?

70. anonymous

I visit those places but I only live in canada. I may be coming into NJ in 2 weeks -_-

71. nincompoop

laughing out loud

72. anonymous

Its not funny!!! idk i may just stay at home alone. Seriously dont wanna head out there


text:
# Construction of $C^1$/$H^2$-conforming finite element basis for triangular or tetrahedral mesh

In the paper Hierarchical Conforming Finite Element Methods for the Biharmonic Equation, P. Oswald claimed Clough-Tocher type elements has $C^1$-continuity while being a cubic polynomial on each triangle. He didn't give a set of explicit basis functions just the standard degrees of freedom on the quadrature points.

Similarly, in the book The Mathematical Theory of Finite Element Methods Chapter 3, the authors give us the construction of cubic Hermite finite elements, but they didn't mention the continuity of the cubic Hermite elements.

However, in the paper Differential complexes and numerical stability, Doulgas Arnold proposed that for $C^1$/$H^2$-conforming discrete space we should use the Hermite quintic(or rather Argyris) finite elements, which is very complicated to express explicitly.

So here are my questions:

(1) Is there any paper that comes up with an explicit formula for the $C^1$/$H^2$-conforming finite elements on triangular or tetrahedral mesh?

(2) Should piecewise cubic be the minimal degree of polynomials requirement for $C^1$-continuity?

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The cubic Hermite elements have a continuous normal derivative but not full $C^1$ continuity. In particular, the normal derivatives may not match at the boundary of two elements, away from the vertices. If you want full $C^1$ continuity you will have to use the Argyris element or Hsieh-Clough-Tucker or something. I recommend the discussion in chapter 6 of Ciarlet's finite element book.

The degree of polynomial required for $C^1$ continuity will depend on your spatial dimension, but in 2D or 3D I don't think you can get away with less than cubic polynomials. You could consider some kind of nonconforming method which may allow a simpler finite element space.

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Err, if a function is continuous across the interface between two cells, and if the function on each cell is in $C^\infty$ as it must be if it's a polynomial, then how can the tangential derivative be discontinuous on a cell interface? Or did you mean that the tangential derivative can be discontinuous at the vertices, i.e. the end-points of each interface? –  Wolfgang Bangerth May 3 '12 at 0:18
You're absolutely right, I edited the answer. –  Andrew T. Barker Jun 8 '12 at 14:51

I refer you to the book Splines on Triangulations. I cannot locate my copy at the moment to give you a better answer, but I recall a discussion/theorems on the polynomial order required for $C^1$ spaces. If I recall correctly, Lai proves that under certain conditions $p=3$ is OK, but $p=5$ is always sufficient.

Unfortunately, I also remember that Lai does not then show how to construct $C^1$ spaces, only prove they exist given a triangulation and a spline space. Once he has this proof, he solves his application with additional linear constraint equations to enforce the $C^1$ condition.

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welcome to scicomp Mr. Collier :) –  Aron Ahmadia Apr 25 '12 at 6:25

You can refer to the following pages for a full listing of the basis functions for Argyris: FEMList.pdf Wikipedia entry (French)

Also, you can use the VT-ICAM ArgyrisPack that a colleague and I developed.

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Can't share more than two links, so the third link was suppose to be github.com/VT-ICAM/ArgyrisPack –  lvleph Nov 16 '12 at 12:50


text:
# Table Tension

I came across a cool design principle known as "tensegrity" today. Here is a example. It's looks unreal but it's actually true! So how did it works?

If you look the two chains in the middle actually gives an upward force to the upper part of the table. This is because the lower part of the table is fixed means all the joints with the lower part of the tables are also fixed. Now due to gravity the upper part applies a force on the chain downward and in return the chain pull it upward.

Now we got the upward force but it's not enough to keep the table there. We also need to counteract the torque applied due to gravity and other forces which does not pass through the center of mass of the body (due to which the body can rotate about it's center of mass, which is something you don't want to happen when you keep your laptop on it).

That's why the four corners are connected with chains to the bottom part which prevents rotation by cancelling the torque produced.

For example if we place a object on one side of the table. Due to the weight of the object there will be a downward force on that side and an upward force on the opposite side. Now as all the corner are connected with chains. The opposite sides chain will experience an upward force which,in turn, will give a downward force that will cancel force thus produced that prevent rotation.

It may not seem that stable but it actually is! Just don't jump on it.

Note by Sanjay Swain
5 months, 3 weeks ago

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Sorry for bad quality photos. I wanted to create some 3D designs with blender but in order to create it I must know how to use blender. Also if you found any problem with explanation please say it. Thanks!

- 5 months, 3 weeks ago

That is some good stuff. Looks almost like its floating. Blender is pretty easy to use. You just have to use a few shapes to make this. Just watch some tutorials and you'll be fine.

- 5 months, 3 weeks ago

Yes

- 5 months, 3 weeks ago

Thanks I am already learning blender. I can now make a burned donut!

- 5 months, 3 weeks ago

lol

- 5 months, 3 weeks ago


text:
# 1.3.5.3: Grouping Numeric Data

Another way to organize raw data is to group them into class intervals, and to then create a frequency distribution of these class intervals.

There are many methods of creating class intervals, so we will simply focus on creating intervals of equal width.

How to create class intervals of equal width and a frequency distribution

1. Choose how many intervals you want. Best is between 5 and 15 intervals.
2. Determine the interval width using the formula and rounding UP to a convenient value:

$\text { IW }=\text { Interval Width }=\dfrac{\text { Maximum Value - Minimum Value+ } 1}{\text { Number of Intervals }} \nonumber$

1. Create the class intervals starting with the minimum value:

Min to under Min + IW,

Min +IW to under Min +2(IW), ...

1. Calculate the frequency of each class interval by counting the values in each class interval. Values that are on an endpoint should be put in the lower class interval. This result is called a frequency distribution.

Example: Students browsing the web

Let's return to the data that represents how much time 30 students spent on a web browser in a 24 hour period. Data is rounded to the nearest minute.

First we choose how many class intervals. In this example, we will create 5 class intervals.

Next Determine the Class Interval Width and round up to a convenient value.

$\mathrm{IW}=\frac{125-67+1}{5}=11.8 \rightarrow 12 \nonumber$

Now create class intervals of width 12, starting with the lowest value, 67.

$\begin{array}{lllll} (67 \text { to } 79) & (79 \text { to } 91) & (91 \text { to } 103) & (103 \text { to } 115) & (115 \text { to } 127) \end{array} \nonumber$

Now, create a frequency distribution, by counting how many are in each interval. Values that are on an endpoint should be put in the higher class interval. For example, 103 should be counted in the interval (103 to 115):

As we did with categorical data, we can define Relative Frequency as the proportion or percentage of values in any Class Interval.

n = sample size ‐ The number of observations in your sample size.

Frequency ‐ the number of times a particular value is observed in a class interval.

Relative  frequency ‐  The proportion or percentage of times a particular value is observed in a class interval.

Relative Frequency = Frequency / n

Note that the value for the (91 to 103) class interval was deliberately rounded down to make the totals add up to exactly 100%

From the frequency distribution, we can see that 30% of the students are on the internet between 103 and 115 minutes per day, while only 10% of students are on the internet between 67 and 79 minutes.

Example: Comparing weights of apples and oranges

A Chilean agricultural researcher collected a sample of 100 Royal Gala apples and 100 navel oranges and measured their weights in grams (see previous example on dot plots).

We will start with a value of 100 and make the interval width equal to 30. Using the tally feature of Minitab, we can create a frequency distribution for the two fruits. Minitab uses “Count” for “Frequency” and reports “Percent” for “Relative Frequency”

The most frequently occurring interval for apples is 220 to 250 grams while the most frequently occurring interval for oranges is 280 to 310 grams. Notice that there are some intervals with 0 observations, showing a potential high outlier for apples and a low outlier for oranges.

1.3.5.3: Grouping Numeric Data is shared under a CC BY-SA license and was authored, remixed, and/or curated by LibreTexts.


text:
#### Volume 4, issue 2 (2004)

 Recent Issues
 The Journal About the Journal Editorial Board Editorial Interests Editorial Procedure Submission Guidelines Submission Page Subscriptions Author Index To Appear Contacts ISSN (electronic): 1472-2739 ISSN (print): 1472-2747
Non-triviality of the $A$–polynomial for knots in $S^3$

### Nathan M Dunfield and Stavros Garoufalidis

Algebraic & Geometric Topology 4 (2004) 1145–1153
 arXiv: math.GT/0405353
##### Abstract

The $A$–polynomial of a knot in ${S}^{3}$ defines a complex plane curve associated to the set of representations of the fundamental group of the knot exterior into ${SL}_{2}ℂ$. Here, we show that a non-trivial knot in ${S}^{3}$ has a non-trivial $A$-polynomial. We deduce this from the gauge-theoretic work of Kronheimer and Mrowka on ${SU}_{2}$–representations of Dehn surgeries on knots in ${S}^{3}$. As a corollary, we show that if a conjecture connecting the colored Jones polynomials to the $A$–polynomial holds, then the colored Jones polynomials distinguish the unknot.

##### Keywords
knot, $A$–polynomial, character variety, Jones polynomial
##### Mathematical Subject Classification 2000
Primary: 57M25, 57M27
Secondary: 57M50


text:
##### Exercise7

Determine whether the following statements are true or false and provide a justification for your response.

1. The eigenvalues of a diagonal matrix are equal to the entries on the diagonal.

2. If $$A\vvec=\lambda\vvec\text{,}$$ then $$A^2\vvec=\lambda\vvec$$ as well.

3. Every vector is an eigenvector of the identity matrix.

4. If $$\lambda=0$$ is an eigenvalue of $$A\text{,}$$ then $$A$$ is invertible.

5. For every $$n\times n$$ matrix $$A\text{,}$$ it is possible to find a basis of $$\real^n$$ consisting of eigenvectors of $$A\text{.}$$

in-context


text:
# Separability of Banach Spaces

A homework problem from Folland Chapter 5, problem 5.25.

If $\mathcal{X}$ is a Banach space and $\mathcal{X}^{\star}$ is separable, then $\mathcal{X}$ is separable.

I tried the following approach: For every $\epsilon >0$ I wanted to show the existence of a linear map from $x_{1},\ldots,x_{n}$ such that for any $x\in\mathcal{X}$ $\| x-L(x_{1},\ldots,x_{n})\|\leq \epsilon$.

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 the sentence in which you describe your approach does not seem to be complete. – Mariano Suárez-Alvarez♦ Feb 8 '12 at 6:16 The book provides a hint for this problem: Let $\{f_n\}_1^\infty$ be a countable dense subset of $\mathcal{X}^*$. For each $n$ choose $x_n\in\mathcal{X}$ with $\|x_n\|=1$ and $|f_n(x_n)|\geq \frac{1}{2}\|f_n\|$. Then the linear combinations of $\{x_n\}_1^\infty$ are dense in $\mathcal{X}$. – john w. Feb 8 '12 at 6:24 sorry I forgot to mention the hint. – user17523 Feb 8 '12 at 6:34

Use the Hahn-Banach Theorem:

Taking $f_n$ and $x_n$ as in your hint.

Let $Y$ be the set of all linear combinations of the $x_i$ with rational coefficients.

Suppose $Y$ were not dense in $X$. Then the closure of $Y$ is a proper subspace of $X$, and thus, there is an $f\in X^*$ of norm 1 with $f(Y)=\{0\}$. Then $${1\over 2}\Vert f_n\Vert\le|f_n(x_n)| =|f_n(x_n) - f(x_n)| \le \Vert f_n-f\Vert \Vert x_n\Vert =\Vert f_n-f\Vert$$

Take $\Vert f_{n_i}-f\Vert\rightarrow 0$. Then from the above, $\Vert f\Vert=0$, a contradiction.

You could also use Riesz' lemma:

Let $Y$ be a proper closed subspace of the normed space $X$ and $0<\theta<1$. Then there is an $x_\theta$ of norm 1 for which $\Vert x_\theta-y\Vert>\theta$ for all $y\in Y$.

If $X$ were not seperable, you could use Hahn Banach to construct uncountably many functionals $f_\alpha\in X^*$ with $\Vert f_\alpha-f_\beta\Vert\ge \theta$ whenever $\alpha\ne\beta$.

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 And, of course, Hahn-Banach (or some other consequence of the Axiom of Choice) is essential. The non-separable Banach space $l_\infty / c_0$ has dual $\{0\}$, as far as we can tell without AC... – GEdgar Feb 8 '12 at 15:26


text:
What is the mass of 7.9 * 10^23 of O_2 at STP?

Jan 30, 2016

The mass of $7.9 \times {10}^{23}$ ${O}_{2}$ molecules is INDEPENDENT of temperature and pressure. It is approx. $40$ $g$.

Explanation:

Avogadro's number of ${O}_{2}$ molecules has a mass of $32.00$ $g$.

We have $7.9 \times {10}^{23}$ ${O}_{2}$ particles; i.e.

$\left(7.9 \times {10}^{23} \text{ particles")/(6.022xx10^23" particles } m o {l}^{-} 1\right)$ $= 1.31$ $m o l$.

So we take this molar quantity, and multiply this by the molecular weight:

$1.31 \cdot m o l \times 32.00 \cdot g \cdot m o {l}^{-} 1$ $=$ ??? g


text:
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## Arguments

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