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[–]default bindings, org, magit, ox-hugo 2 points3 points  (5 children)

I have been working on getting the ox-hugo package out. It's an Org exporter backend to export to Hugo markdown posts. See Why ox-hugo? for more.

I am stuck at this one Org/elisp problem (more info here and here) and trying to solve it.

[–] 0 points1 point  (4 children)

I'm working on something similar, but instead of markdown, just pure org-mode since Hugo now natively supports that. There are a few things I find awkward, mainly supporting a series of posts. Looks to me that to implement a lot of the cool stuff in Hugo, I have to add many annotations to the org file, which makes the package more for self-consumption than for general use.

[–]default bindings, org, magit, ox-hugo 0 points1 point  (3 children)

I'm working on something similar, but instead of markdown, just pure org-mode since Hugo now natively supports that.

So you use Org to Org exporter and then let Hugo parse the Org? I export from Org to Markdown because Hugo has a much more mature Markdown parser (Blackfriday).

mainly supporting a series of posts.

How do you determine a post's 'series-ness'? If it is just by setting a front matter variable, you can do this in ox-hugo.

* Series ABC line
:PROPERTIES:
:EXPORT_HUGO_CUSTOM_FRONT_MATTER: :series "How to do ABC"
** Post 1 in series ABC
:PROPERTIES:
:EXPORT_FILE_NAME: post-1-abc
:EXPORT_DATE: 2017-07-30
:END:
As level-1 subtree is setting the custom
'series' front matter to some value, all
the subtrees under that (like this one)
will inherit that value. So all series ABC
posts need to be just under that level, and
it will get that series property by Org
property inheritance.
** Post 2 in series ABC
:PROPERTIES:
:EXPORT_FILE_NAME: post-2-abc
:EXPORT_DATE: 2017-07-24
:END:
Here's another post in the same series.

I have a lot more examples here. The exported markdown files from that one all-posts.org are here.

Looks to me that to implement a lot of the cool stuff in Hugo, I have to add many annotations to the org file

Can you give few examples of that? I am working on this package so that we can write normal Org files (referencing to the same example above) that can easily export to standard formats like HTML and PDF. ox-hugo is designed to do the right thing when exporting that same file to Hugo.

My idea is to have to do as less explicit Org property settings for Hugo as possible, and derive the Hugo front matter from implicit Org attributes. For example, at the moment, ox-hugo needs just #+HUGO_BASE_DIR Org keyword, and EXPORT_FILE_NAME and EXPORT_DATE properties in the post subtree.

• The latter 2 can be prefilled using a helper function, Org capture template, or any other mechanism like yasanippet.
• To set Hugo tags, categories, you just use Org tagging mechanism. Also you need to set the Org tags for a new post only if they are not inherited from a parent subtree.

[–] 0 points1 point  (2 children)

First off, I'd like to thank you for your contributions to the Emacs community. Your comments here have been enlightening, and I've also looked through your emacs config for some inspiration (nice hydras :P)

So you use Org to Org exporter and then let Hugo parse the Org? I export from Org to Markdown because Hugo has a much more mature Markdown parser (Blackfriday).

I simply copy the org subtree, demote until it's not possible to further demote, remove the top level header and the properties drawer, then demote each heading one more level. How does LaTeX formulae play out with the BlackFriday parser? I'm hoping someday Org will overtake markdown in terms of features via Goorgeous.

How do you determine a post's 'series-ness'?

I considered having a definition like the following: A header denoting a series has sub-headings with property drawers containing Hugo properties. I was hoping this wouldn't be computationally expensive, since I don't tend to write really long articles. The reason why I want this is so I don't get unexpected output if I accidentally forget to annotate it with the series property.

I haven't implemented this yet, still thinking things through.

Can you give few examples of that?

Clearly I haven't thought things out more than you have. I litter my properties drawer with hugo properties like slug, and just used the slug as the export file name, but I realize that this leads to a lot of inflexibility. Your :EXPORT_HUGO_CUSTOM_FRONT_MATTER: looks to be a solution I might adopt.

The latter 2 can be prefilled using a helper function, Org capture template, or any other mechanism like yasanippet.

Atm I use a derived major mode from Org, binding keybindings for exporting subtrees and inserting property drawers. I think this has worked out really well for me.

It wasn't my intention to duplicate efforts. I just wanted something simple that worked for me, which can be generalized to something more useful. When I first started out, I wanted it to be pure elisp, but I'm now of the opinion that Hugo can do a lot of the heavy lifting. I'd definitely move over to ox-hugo when it's ready, but for now the little I have works well enough for me. Oh, did I mention I'm terrible at elisp? :P

[–]default bindings, org, magit, ox-hugo 0 points1 point  (1 child)

(nice hydras :P)

They are awesome, thanks to /u/abo-abo for hydra.el :)

I simply copy the org subtree, demote until it's not possible to further demote, remove the top level header and the properties drawer, then demote each heading one more level.

Cool. I do the same effectively by exporting only the post subtree. In addition, the post does not need to be always on level 1; it could be on any nested level. Any subtree having the EXPORT_FILE_NAME property is considered a valid Hugo post subtree and that entire subtree will be exported as a post. So you could have your point within a sub-heading at any level within a post, but when you export the subtree, it figures out what subtree level to export by looking for the presence of that property.

How does LaTeX formulae play out with the BlackFriday parser?

I just added few commits to support native Org supported latex fragments to be exported to Blackfriday compatible Markdown (commits: here and here).

So now the Equations -> Inline equations subtree in Org exports to this in Markdown. That looks like this in Hugo-rendered HTML using the bare_min theme I use for ox-hugo testing.

I'm hoping someday Org will overtake markdown in terms of features via Goorgeous.

It would be a high-ambition project.. imagine re-writing the whole of Org (plus parts of Emacs) in Go.. because Goorgeous would need to support more than syntax.. it's supporting stuff like tags, tag groups, property drawers, inheritance of tags and properties, recognition of all the ox export options used in #+OPTIONS, recognition of file local variables, .dir-locals.el, etc. If one is using any of these features, it might be best to just manage the blog in Emacs+Org.

I considered having a definition like the following: A header denoting a series has sub-headings with property drawers containing Hugo properties. I was hoping this wouldn't be computationally expensive, since I don't tend to write really long articles. The reason why I want this is so I don't get unexpected output if I accidentally forget to annotate it with the series property. I haven't implemented this yet, still thinking things through.

Cool.

I litter my properties drawer with hugo properties like slug, and just used the slug as the export file name, but I realize that this leads to a lot of inflexibility.

Though ox-hugo supports setting the Hugo slug, as I worked more on ox-hugo, I realized that slug is not needed at all.. I just need to update EXPORT_FILE_NAME to what I want. Once this issue gets resolved, that flow will be seamless (Right now, one would need to delete the old md if EXPORT_FILE_NAME is changed after the first export. Though, git diff would tell me if such file renaming happened by mistake.)

Cool, thanks.

Atm I use a derived major mode from Org, binding keybindings for exporting subtrees and inserting property drawers. I think this has worked out really well for me.

It wasn't my intention to duplicate efforts. I just wanted something simple that worked for me, which can be generalized to something more useful. When I first started out, I wanted it to be pure elisp, but I'm now of the opinion that Hugo can do a lot of the heavy lifting. I'd definitely move over to ox-hugo when it's ready, but for now the little I have works well enough for me.

No problem. Thanks for sharing your config! I have taken inspirations for this project from whyarethingsthewaytheyare.com (as well) and www.holgerschurig.de :)

I am planning on working on a minor mode instead of a major mode that would use a hydra to do stuff like add a new post subtree, update export date, update export file name based on title, etc.

Oh, did I mention I'm terrible at elisp? :P

I feel the same about myself.. I only know as much elisp as I need. If I need something from elisp but I am terrible at that, I will gradually become less terrible in that :).

[–] 0 points1 point  (0 children)

Thanks for working on ox-hugo! I'll try it out and give you my feedback.


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NOAO >   Observing Info >   Approved Programs >   2010A-0381

# Proposal Information for 2010A-0381

PI: Jeffrey L. Carlin, University of Virginia, jc4qn@mail.astro.virginia.edu
Address: Department of Astronomy, P. O. Box 400325, Charlottesville, VA 22904-4325, USA

CoI: Steven R. Majewski, University of Virginia
CoI: Ricardo R. Munoz, Yale University

Title: Fixing the Extragalactic Reference Frame For the Proper Motion of the Carina Dwarf Spheroidal

Abstract: \vskip -0.1in The orbital shapes of Milky Way dwarf spheroidal (dSph) galaxies provide specific constraints on CDM models of hierarchical structure formation on small scales and for late infalling subhalos. Furthermore, many physical properties of these dSph systems are shaped by tidal impulses, the magnitude of which depends on the orbit of the dSph. Unfortunately, the tangential (i.e., \it proper) motions of distant Milky Way dSphs are extremely difficult to measure, and thus remain poorly constrained (or in most cases unknown). We propose to remedy this for the Carina dSph, for which we have used a large, unique set of photographic plates spanning a 14 year baseline and covering large enough area to allow for a precise proper motion measurement. We have digitized these plates and derived precise (1-2 mas/yr per star) \it relative proper motions for 565 confirmed Carina members, and >1000 more likely members; these lead to a bulk relative proper motion for the system defined to 0.04 mas/yr. Despite this extreme precision, the \it absolute (relative to the fixed'' extragalactic background) proper motion of Carina is still poorly constrained, due to the paucity of identified QSOs (only 7) in the field of view to fix the reference frame. With our proposed search for QSOs among the ~2000 objects in our proper-motion catalog, we expect to discover ~50 QSOs, which will improve the accuracy of our proper motion zero point by a factor of 3 in each dimension, and likely more. This will allow for a precise determination of the orbit of the Carina dwarf galaxy, with an expected accuracy 2\times better than the recent proper motion for this object, which disagrees with the proper motion expected from the direction of Carina's tidal tails.

National Optical Astronomy Observatory, 950 North Cherry Avenue, P.O. Box 26732, Tucson, Arizona 85726, Phone: (520) 318-8000, Fax: (520) 318-8360

 NOAO >   Observing Info >   Approved Programs >   2010A-0381

format: plain


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# Read return value as string by Mathlink.h

I'm using "mathlink.h" to call Mathematica from a C program. I send a computation to Mathematica and its return result is a very big number (100 digits). Hence, I want to read that number as a string.

How can I do it using MathLink?

Thanks,

-
add comment

## 1 Answer

You could simply wrap your computation in

MLPutFunction(link, "ToString", 1);

and convert your result to a string. After that you can use MLGetString.

-
One problem is if the output is an expression, say $x^2 + x + 1$, then the ToString method will return in un-readable format. What should I do if I want to read arbitrary output as a string? Like for the expression above, I expect to get output as exactly $x^2 + x + 1$. –  Loi.Luu Oct 21 '13 at 6:37
and I see there is no other option for ToString method. –  Loi.Luu Oct 21 '13 at 6:37
@Loi.Luu You question was how you can send very big numbers as string through MathLink. Why do you now want to send expressions? Addtionally, ToString[x^2 + x + 1, InputForm] gives a string of valid Mathematica syntax. –  halirutan Oct 21 '13 at 7:36
Sorry to make you confused, I thought I can apply the same technique to expression. You are correct in both answers, thank you for your help :) –  Loi.Luu Oct 21 '13 at 7:48
add comment


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## Precalculus (6th Edition) Blitzer

The formula for $_{n}{{C}_{r}}$ has the same numerator as the formula for $_{n}{{P}_{r}}$ but contains an extra factor of $r!$ in the denominator.
We know that permutations of $n$ things taken $r$ at a time can be defined by the formula: ${}_{n}{{P}_{r}}=\frac{n!}{\left( n-r \right)!}$ And combinations of $n$ things taken $r$ at a time can be defined by the formula: ${}_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$ And by comparing both formulas it can be seen that r! is an extra factor. Therefore, there are r! times as many permutations of n things taken r at a time as combinations of n things taken r at a time.


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## Friday, September 27, 2002

### Experience Mathematics # 15 : OR and AND

Suppose that your mom says that you can have either an Apple or a Banana. Can you have both? One of the most fundamental rules of logic says that the expression either $A$ or $B$ is true only if one of $A$ or $B$ is true. That is to say, you cannot have both the Apple and the Banana (assuming you wish to obey your mom.)

What if your mom says you can have an Apple or a Banana? In this case, you can have both.

Suppose your mom asks you if you have had an Apple or a Banana. Can you honestly say yes if you have had an Apple and an Orange? The answer is yes. If she asks you if you have had an Apple and a Banana, you can answer yes only if you have had both.

Suppose your mom insists that you should not have an Apple. Is it OK to have a Banana? How about Baked Beans? It depends. The alternatives to an Apple allowed by your mom depend on the context. For example, if the alternatives allowed consist of the other fruits in the house, you cannot have baked beans instead of the Apple, but you could have a Banana. However, if the context of discussion is the five servings of fruits and vegetables that you must have every day, then Baked Beans are allowed. In Mathematics, when we refer to a set $A$, then we must specify the universal set $U$ from where the elements of $A$ are picked. Then the complement of $A$ is the set of all the elements that are in $U$ but not in $A$. Then there is no confusion when we claim: $a$ is not an element of $A$. By this statement we mean that $a$ is an element of the complement of $A$.

## Tuesday, September 17, 2002

### Experience Mathematics #14 - Uncountable sets

Many of the infinite sets we have encountered are countable. Even numbers, the set of prime numbers, integers and rational numbers, all have the same number of elements as $N$, the set of natural numbers. Are there any infinite sets that have more elements than the natural numbers?

This question was answered by Cantor, who showed that the real numbers outnumber the natural numbers. All the real numbers between $0$ and $1$ have a decimal expansion such as $x=0.13212987\dots$. Cantor showed that all numbers of this form cannot be put into one-to-one correspondence with the set of natural numbers. To be able to understand his proof, find a number that differs from $x$ in the first decimal place. Take any number $y$ with $2$ in the first decimal place. Since $2$ is different from $1$, $y$ differs from $x$ in the first decimal place.

To return to Cantor’s proof, suppose that you are able to find a one-to-one correspondence between the natural numbers and all the real numbers in the interval $(0,1)$. Let us denote by $x_1$ the number corresponding to $1$; $x_2$, the number corresponding to $2$, and so on. Now consider a number $y$ (between $0$ and $1$) that is different from $x_1$ in the first place after the decimal; different from $x_2$ in the second place after the decimal; and so on. Clearly, $y$ cannot appear in the list, since it is different from all the $x$’s. Thus we have found a real number between $0$ and $1$ that is not in the above correspondence. This contradiction shows that no such correspondence is possible. In other words, the real numbers are uncountable in number.

Are there any infinite sets that have more elements than $N$ but less elements than the set of real numbers?

## Thursday, September 12, 2002

### Experience Mathematics #13 - The cartesian society in Hilbert Hotel

A set S is countable if it can be put in one-to-one correspondence with $N$. Suppose that you are the manager of the Hilbert Hotel, a hotel with a countable number of rooms. Now, even though the hotel is full, when $200$ new guests arrive after lunch at the Restaurant at the End of the Universe, you can accommodate them. All you have to do is to move the guests in Room $1$ to Room $201$, the guests in Room $2$ to Room $202$, etc. In other words, you will move the guests in Room $n$ to Room $200+n$.

Now, suppose the hotel is empty. The members of the Cartesian Society (Motto: We Think, Therefore We Exist!) decide to have a convention. Each member of the Cartesian Society has an identification mark of the form $(a, b)$, where $a$ and $b$ are natural numbers. The chairman of the society is $(1,1)$ and for any two numbers $a$ and $b$, there is a member corresponding to the ordered pair $(a, b)$. Don’t confuse $(2,3)$ with $(3,2)$: they are quite different people. There are many, many members in this society. But all of them can be accommodated in the Hilbert Hotel. Can you find a one-to-one correspondence of the members of the Cartesian Society with the natural numbers?

First write down the ID numbers of all the members of the Cartesian Society in the form of a table. For example, put $(3,5)$ in the third row and the fifth column of the table. Now find a way to “count” them. In other words, assign a natural number to each of them in a systematic fashion.

## Friday, September 06, 2002

### Experience Mathematics #12 -- A part can be equal to the whole!

A set S is countable if it can be put in one-to-one correspondence with $N$. For example, if we take the set of even numbers, we can establish a one-to-one correspondence as follows. $1$ corresponds to $2$, $2$ to $4$, $3$ to $6$, and so on. This shows that the number of even numbers is equal to the number of natural numbers.

This contradictory idea—that a part of an object is equal to the whole—troubled many philosophers. But they got over it, and began to compare the concept of infinity with the concept of God. However, when Biologists made it possible to clone human beings, they have stopped approving of the idea.

The famous mathematician Hilbert told the story of a hotel with an infinite number of rooms. Suppose the Hilbert Hotel is full, but the hotel manager wants to accommodate a guest who arrives suddenly. How does he manage that? Well, he asks the guest in room number $1$ to move to Room $2$, the guest in Room $2$ to move to Room $3$, and so on. Room $1$ becomes empty and is readied for the new guest. Can you figure out how to accommodate $30$ guests, even if the hotel is full?

What if a travel agent calls the manager, and says she is sending groups of tourists to the hotel. The numbers of people in the groups are: $3$, $7$, $11$, $15$, and so on. Help the manager come up with the required one-to-one correspondence in these cases, so that he can accommodate all the guests.


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# Cícero

Joined 4 December 2005

### Cícero

Hi! I'm an undergraduate student in mathematics at UFRGS and would like to collaborate in mathematical topics. I read English very well, but don't write so (maybe so bad that you would think I learned English with the awesome manual English as she is spoke, but it's not true!). Please be condenscendent about language mistakes.

 ${\displaystyle e^{i\pi \ }}$ This user is a mathematician.


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# Posts Tagged ‘ PROC SCORE ’

## Obtain Trace of the Projection Matrix in a Linear Regression

October 6, 2011
By

Recently, I am working on coding in SAS for a set of regularized regressions and need to compute trace of the projection matrix: S=X(X'X + \lambda I)^{-1}X'. Trace of the project matrix S is a key concept in modern regression analysis. For example, ...

Tags: , , , , ,
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## Obtain Trace of the Projection Matrix in a Linear Regression

October 6, 2011
By

Recently, I am working on coding in SAS for a set of regularized regressions and need to compute trace of the projection matrix: S=X(X'X + \lambda I)^{-1}X'. Wikipedia has a well written introduction to Trace @ here. Trace of the project matrix S i...

Tags: , , , , ,
Posted in SAS | Comments Off on Obtain Trace of the Projection Matrix in a Linear Regression

## Obtain Trace of the Projection Matrix in a Linear Regression

October 6, 2011
By

Recently, I am working on coding in SAS for a set of regularized regressions and need to compute trace of the projection matrix: $$S=X(X'X + \lambda I)^{-1}X'$$. Wikipedia has a well written introduction to Trace @ here. To obtain the inverse of ...

Tags: , , , , ,
Posted in SAS | Comments Off on Obtain Trace of the Projection Matrix in a Linear Regression

## Low Rank Radial Smoothing using GLIMMIX and its Scoring

September 30, 2010
By

Low Rank Radial Smoothing using GLIMMIX , a semiparametric approach to smooth curves . Specifying TYPE=RSMOOTH option in RANDOM statement, we can implement this spline smooth approach. The bast thing is that for future scoring, data preparation...

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Posted in SAS | Comments Off on Low Rank Radial Smoothing using GLIMMIX and its Scoring

## Regularized Discriminant Analysis

September 30, 2010
By

Using SAS to implement Regularized Discriminant Analysis .

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Posted in SAS | Comments Off on Regularized Discriminant Analysis

August 9, 2010
By

Partial Least Square is one of several supervised dimension reduction techniques and attracts attention in recent years. In the one hand, PLS is able to generate a series of scores that maximize linear correlation between dependent variables and inde...

Tags: , ,

## Implement Randomized SVD in SAS

July 13, 2010
By

In the 2010 SASware Ballot®, a dedicated PROC for Randomized SVD was among the options. While an official SAS PROC will not be available in the immediate future as well as in older SAS releases, it is fairly simple to implement this algorithm using ...

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## SAS implementation of Kernel PCA

February 6, 2010
By

Kernel method is a very useful technique in data mining that is applicable to any algorithms relying on inner product . The key is applying appropriate kernel function to the inner product of original data space. I show here SAS/STAT+BASE ex...

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## Welcome!

SAS-X.com offers news and tutorials about the various SAS® software packages, contributed by bloggers. You are welcome to subscribe to e-mail updates, or add your SAS-blog to the site.


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# Math Help - Integral of a floor function

1. ## Integral of a floor function

Greetings.

A bit of help would be much appreciated on the following integral:

$\int_{0}^{\infty} \lfloor x \rfloor e^{-x} dx$

Edit:

Ah this is just a geometric series: $\int_{n-1}^{n} n e^{-x} dx$

2. Note $\int_{j}^{j+1}{e^{-x}\,dx}=(e-1)e^{-j-1},$ so $\int_{0}^{\infty }{\left\lfloor x \right\rfloor e^{-x}\,dx}=\sum\limits_{j=0}^{\infty }{\int_{j}^{j+1}{\left\lfloor x \right\rfloor e^{-x}\,dx}}=(e-1)\sum\limits_{j=0}^{\infty }{je^{-j-1}}.$

Computation of that series will give ya the answer.

3. Good catch it looks like we cross posted.


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# Math Help - Derivative of e.

1. ## Derivative of e.

Please check if i have done these correctly.

$f(x) = e^{-x^2}$

$f'(x) = -2xe^{-x^2}$

$f''(x)= -4x^2e^{-x^2} - 2e^{-x^2}$

Can i also write f''(x) as $-4e^{-x^2}x^2-2e^{-x^2}$ ??

2. Originally Posted by el123
Please check if i have done these correctly.

$f(x) = e^{-x^2}$

$f'(x) = -2xe^{-x^2}$

$f''(x)= -4x^2e^{-x^2} - 2e^{-x^2}$

Can i also write f''(x) as $-4e^{-x^2}x^2-2e^{-x^2}$ ??
Your $f''(x)$ is incorrect as it should be $4x^2e^{-x^2} - 2e^{-x^2}$ since we started with a factor of $-2x$ and multiplied it by $-2x$ again when we use the chain rule to get $-2x*-2x = 4x^2$. And as for your re-writing of $f''(x)$, you just moved the $x^2$ term around...that's not different in any meaningful way ie. just like $ax^2$ and $x^2a$ are obviously the same since multiplication in $\mathbb{R}$ is commutative. If you wish to factor $f''(x)$ you can rewrite it as $f''(x) = (4x - 2)e^{-x^2}$


text:
Class 9 Maths Worksheet for Quadrilaterals

Notes Ncert Solutions Assignments Revision sheet

Given below are the Class 9 Maths Worksheet for Quadrilaterals
a. Concepts questions
b. Calculation problems
c. Multiple choice questions
e. Fill in the blanks
Question 1
ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that :
(i) $\Delta APB \cong \Delta CQD$
(ii) AP = CQ

Question 2
The angles of a quadrilateral are in the ratio 3 : 5 : 7 : 9. Find the angles of the quadrilateral.
Question 3
In a quadrilateral ABCD, AO and BO are the bisectors of $\angle A \; and \; \angle B$ respectively. Prove that $\angle AOB = \frac {1}{ 2} (\angle C + \angle D)$
Question 4
In the figure, ABCD is a parallelogram. E is the mid-point of BC. DE and AB when produced meet at F. Prove that AF = 2AB

Question 5
ABCD is a square and on the side DC, an equilateral triangle is constructed. Prove that
(i) AE = BE
(ii) $\angle DAE = 15^0$

Question 6
ABCD is a rectangle in which diagonal BD bisects ∠B. Show that ABCD is a square.
Question 7
If ABCD is a trapezium in which AB || CD and AD = BC, prove that ∠A = ∠B
Question 8
In the figure, diagonal BD of parallelogram ABCD bisects ∠B. Show that it bisects ∠D also

Question 9
E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. Show that BFDE is a parallelogram.
Question 10
The two opposite angles of a parallelogram are $(3y - 10)^0$ and $(2y + 35)^0$. Find the measure of all the four angles of the parallelogram

Question 11
ABCD is a rhombus in which AC = 16 cm and BC = 10 cm. Find the length of the diagonal BD.
Question 12
Diagonals of a quadrilateral ABCD bisect each other. If ∠A = 35°, then ∠B = 145°. Is it true ? Justify your answer.
Question 13
Can $\angle 95^0 \; ,\; 70^0 \; ,\; 110^0 \; and \; 80^0$ be the angles of a quadrilateral ? Why or why not ?
Question 14
Prove that quadrilateral formed by bisectors of the angles of a parallelogram is a rectangle.

Question 15
Show that the line segments joining the mid-points of opposite sides of quadrilateral bisect each other.
Question 16
ABCD is a parallelogram in which X and Y are the mid-points of AB and CD. AY and DX are joined which intersect each other at P. BY and CX are also joined which intersect each other at Q. Show that PXQY is a parallelogram
Question 17
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus
Question 18
If one angle of a parallelogram is a right angle, it is a rectangle. Prove.

Reference Books for class 9 Math

Given below are the links of some of the reference books for class 9 Math.

1. Mathematics for Class 9 by R D Sharma One of the best book for studying class 9 level mathematics. It has lot of problems to be solved.
2. Secondary School Mathematics for Class 9 by R S Aggarwal This is also as good as R.D. Sharma. Either this or the book by R.D. Sharma will do. I find book R.S. Aggarwal little bit more challenging than the one by R.D. Sharma.
3. Pearson IIT Foundation Series - Maths - Class 9 Buy this book if you want to challenge yourself further and want to prepare for JEE foundation.
4. Pearson IIT Foundation Physics, Chemistry & Maths combo for Class 9 Only buy if you are prepared to study extra topics and want to take your studies a step further. You might need help to understand topics in these books.

You can use above books for extra knowledge and practicing different questions.

Note to our visitors :-

Thanks for visiting our website.


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# Testbook test series

64 views

edited
1

R' ∩ C' = ( R U C )' = C' ===> is it CFL closed under complementation? No ===> Is it CSL closed under Complementation? yes

therefore it must be CSL but it may or may not CFL

When CFL is DCFL then it is closed under complementation
0
R is regular so complement(R) is also regular as regular language are closed under complementation .

C is cfl so complement (C) is necessarily not cfl as cfl are not closed under complementation bt its is necessarily csl becoz cfl is subset of csl.

so intersection of csl and regular is definitely csl .(becoz regular are also csl and csl ∩ csl = csl as csl are closed under intersection)....

L1=regular so L1(complement) is also regular as well as CFL.

L2=CFL so its complement will be CSL .now $L1\bigcap L2$ will be CSL not CFL

selected
0
it may be CFL

## Related questions

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Consider the infinite two-dimensional grid G={(m,n)| m and n are integers} Every point in G has 4 neighbors, North, South, East, and West, obtained by varying m or n by 1. Starting at the origin (0,0), a string of command letters N, S, E, W generates a ... a closed path. Which of the following statements is TRUE? i) L is Regular. ii) L is context free. iii) L complement is context free. Thanks!
$Que-$ The minimum number of states in the $NFA$ for the regular expression $(a + a(b + aa)*b)* a(b + aa)*a$ is ______. Approach ?
$\overline{L(M)}$ is Regular DCFL but not regular CFL but not DCFL Recursive but not CFL


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Inelastic Collision Derivation Help

1. Nov 2, 2006

GreenSabbath

How wxactly has this equation be derived?

Can someone help me with the derivation. I cant trace the orgin of the second part od the systems of equation.

Help needed urgently Thanks

Last edited by a moderator: Apr 22, 2017 at 1:44 PM
2. Nov 2, 2006

Staff: Mentor

(1) Looks like conservation of momentum (OK)
(2) Looks like "c" is the coefficient of restitution (not quite right)

You'll have to fix the second equation. It should read:
$$V_{1f} - V_{2f} = c(V_2 - V_1)$$

Then, to combine those equations to get the equation in your first link: multiply the 2nd equation by M2 and add it to the first equation. Isolate V1f.

3. Nov 2, 2006

arildno

You derive it by perseverance.
Solve the linear system of equations you are supposed to solve.

If you don't have the skill to solve a system of linear equations, that is what you should focus on.

If you can't be bothered to perform the operations yourself, there is no reason for us to help you out, is there?


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Math Help - Prove

1. Prove

Prove that every line normal to the cone with equation z=sqrt(x^2+y^2) intersects the z-axis.

2. Originally Posted by esum0209
Prove that every line normal to the cone with equation z=sqrt(x^2+y^2) intersects the z-axis.

Define $F(x,y,z):=\sqrt{x^2+y^2}-z\Longrightarrow$ the normal line to this surface at any given point $(x_0,y_0,z_0)$ is given by the parametric equations

$x(t):=x_0+\frac{x_0}{\sqrt{x_0^2+y_0^2}}t\,,\,\,y( t)=y_0+\frac{y_0}{\sqrt{x_0^2+y_0^2}}t\,,\,\,z(t)= z_0-t\,,\,\,t\in\mathbb{R}$ .

Now prove that there exists $t_1\in\mathbb{R}\,\,s.t.\,\, x(t_1)=y(t_1)=0$ ...

Tonio

3. Originally Posted by tonio
Now prove that there exists $t_1\in\mathbb{R}\,\,s.t.\,\, x(t_1)=y(t_1)=0$ ...

Tonio
Sorry, but I don't know to prove this... Can u help me a bit more?

4. What happens if $t = \dfrac{{ - 1}}
{{\sqrt {x_0 ^2 + y_0 ^2 } }}?$

5. Originally Posted by esum0209
Sorry, but I don't know to prove this... Can u help me a bit more?

No. If you're dealing with partial derivatives you must be able to cope with this: show that the solution to $x(t_1)=0$ is the same as the solution to $y(t_1)=0$ . This is HS stuff.


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# Ticket #3356: sage-3356-part9.patch

File sage-3356-part9.patch, 13.9 KB (added by was, 14 years ago)

this patch gets doctest coverage to 100%

• ## sage/finance/markov_multifractal.py

# HG changeset patch
# User William Stein <wstein@gmail.com>
# Date 1214893576 25200
# Node ID 62116b6e86e394cf5610481d41def0bc0e03a857
trac #3356 -- get doctest coverage up to 100% and fix some minor issues with hurst exponent.

diff -r c7f6d59ded41 -r 62116b6e86e3 sage/finance/markov_multifractal.py
 a class MarkovSwitchingMultifractal: n -- a positive integer EXAMPLES: sage: msm = finance.MarkovSwitchingMultifractal(8,1.4,1.0,0.95,3) sage: msm.simulation(5) [0.0059, -0.0097, -0.0101, -0.0110, -0.0067] sage: msm.simulation(3) [0.0055, -0.0084, 0.0141] """ return self.simulations(n, 1)[0]
• ## sage/finance/markov_multifractal_cython.pyx

diff -r c7f6d59ded41 -r 62116b6e86e3 sage/finance/markov_multifractal_cython.pyx
 a def simulations(Py_ssize_t n, Py_ssize_t def simulations(Py_ssize_t n, Py_ssize_t k, double m0, double sigma, int kbar, gamma): """ Return k simulations of length n using the Markov switching multifractal model. INPUT: n, k -- positive integers m0, sigma -- floats kbar -- integer gamma -- list of floats OUTPUT: list of lists EXAMPLES: sage: set_random_seed(0) sage: msm = finance.MarkovSwitchingMultifractal(8,1.4,1.0,0.95,3) sage: sage.finance.markov_multifractal_cython.simulations(5,2,1.278,0.262,8,msm.gamma()) [[0.0014, -0.0023, -0.0028, -0.0030, -0.0019], [0.0020, -0.0020, 0.0034, -0.0010, -0.0004]] """ cdef double m1 = 2 - m0 cdef Py_ssize_t i, j, a, c cdef TimeSeries t, eps
• ## sage/finance/stock.py

diff -r c7f6d59ded41 -r 62116b6e86e3 sage/finance/stock.py
 a class Stock: def get_data(exchange=''): """ This function is used internally. EXAMPLES: This indirectly tests the use of get_data. sage: finance.Stock('aapl').historical()[:2]    # optional -- requires internet [ 2-Jan-90 8.81 9.38 8.75 9.31    6542800, 3-Jan-90 9.50 9.50 9.38 9.38    7428400 ] """ url = 'http://finance.google.com/finance/historical?q=%s%s&output=csv&startdate=%s'%( exchange, symbol.upper(),startdate)
• ## sage/finance/time_series.pyx

diff -r c7f6d59ded41 -r 62116b6e86e3 sage/finance/time_series.pyx
 a cdef class TimeSeries: return v def linear_filter(self, M): """ Return a linear filter using the first M autocovariance values of self.  This is useful in forcasting by taking a weighted average of previous values of a time series. WARNING: The input sequence is assumed to be stationary, which means that the autocovariance $\langle y_j y_k \rangle$ depends only on the difference $|j-k|$. INPUT: M -- integer OUTPUT: TimeSeries -- the weights in the linear filter. EXAMPLES: sage: set_random_seed(0) sage: v = finance.TimeSeries(10^4).randomize('normal').sums() sage: F = v.linear_filter(100) sage: v [0.6767, 0.2756, 0.6332, 0.0469, -0.8897 ... 87.6759, 87.6825, 87.4120, 87.6639, 86.3194] sage: v.linear_forecast(F) 86.017728504280015 sage: F [1.0148, -0.0029, -0.0105, 0.0067, -0.0232 ... -0.0106, -0.0068, 0.0085, -0.0131, 0.0092] """ acvs = [self.autocovariance(i) for i in range(M+1)] return linear_filter(acvs) def linear_forecast(self, filter): """ Given a linear filter as output by the linear_filter command, compute the forecast for the next term in the series. INPUT: filter -- a time series output by the linear filter command. EXAMPLES: sage: set_random_seed(0) sage: v = finance.TimeSeries(100).randomize('normal').sums() sage: F = v[:-1].linear_filter(5); F [1.0019, -0.0524, -0.0643, 0.1323, -0.0539] sage: v.linear_forecast(F) 11.782029861181114 sage: v [0.6767, 0.2756, 0.6332, 0.0469, -0.8897 ... 9.2447, 9.6709, 10.4037, 10.4836, 12.1960] """ cdef TimeSeries filt if isinstance(filter, TimeSeries): filt = filter cdef class TimeSeries: return v def show(self, *args, **kwds): """ Calls plot and passes all arguments onto the plot function.  This is thus just an alias for plot. EXAMPLES: Draw a plot of a time series: sage: finance.TimeSeries([1..10]).show() """ return self.plot(*args, **kwds) def plot(self, Py_ssize_t plot_points=1000, points=False, **kwds): cdef class TimeSeries: t._values[i] = s/k return t def exponential_moving_average(self, double alpha, double t0 = 0): def exponential_moving_average(self, double alpha): """ Return the exponential moving average time series over the last .  Assumes the input time series was constant with its starting value for negative time.  The t-th step of the output is the sum of the previous k-1 steps of self and the kth step divided by k. Return the exponential moving average time series.  Assumes the input time series was constant with its starting value for negative time.  The t-th step of the output is the sum of the previous k-1 steps of self and the kth step divided by k. The 0th term is formally undefined, so we define it to be 0, and we define the first term to be self[0]. INPUT: alpha -- float; a smoothing factor with 0 <= alpha <= 1. alpha -- float; a smoothing factor with 0 <= alpha <= 1 OUTPUT: a time series with the same number of steps as self. cdef class TimeSeries: EXAMPLES: sage: v = finance.TimeSeries([1,1,1,2,3]); v [1.0000, 1.0000, 1.0000, 2.0000, 3.0000] sage: v.exponential_moving_average(0,0) sage: v.exponential_moving_average(0) [0.0000, 1.0000, 1.0000, 1.0000, 1.0000] sage: v.exponential_moving_average(1,0) sage: v.exponential_moving_average(1) [0.0000, 1.0000, 1.0000, 1.0000, 2.0000] sage: v.exponential_moving_average(0.5,0) sage: v.exponential_moving_average(0.5) [0.0000, 1.0000, 1.0000, 1.0000, 1.5000] Some more examples: sage: v = finance.TimeSeries([1,2,3,4,5]) sage: v.exponential_moving_average(1) [0.0000, 1.0000, 2.0000, 3.0000, 4.0000] sage: v.exponential_moving_average(0) [0.0000, 1.0000, 1.0000, 1.0000, 1.0000] """ if alpha < 0 or alpha > 1: raise ValueError, "alpha must be between 0 and 1" cdef class TimeSeries: cdef TimeSeries t = new_time_series(self._length) if self._length == 0: return t t._values[0] = t0 t._values[0] = 0 if self._length == 1: return t t._values[1] = self._values[0] cdef class TimeSeries: double EXAMPLES: sage: v = finance.TimeSeries([1,2,3]) sage: v.central_moment(2) 0.66666666666666663 Note that the central moment is different than the moment here, since the mean is not $0$: sage: v.moment(2)     # doesn't subtract off mean 4.666666666666667 We compute the central moment directly: sage: mu = v.mean(); mu 2.0 sage: ((1-mu)^2 + (2-mu)^2 + (3-mu)^2) / 3 0.66666666666666663 """ if k == 1: return float(0) cdef class TimeSeries: return sqrt(self.variance(bias=bias)) def range_statistic(self): sums = self.sums() return (sums.max() - sums.min())/self.standard_deviation() """ Return the range statistic of self, which is the difference between the maximum and minimum values of this time series, divided by the standard deviation of the series of differences. OUTPUT: float EXAMPLES: Notice that if we make a Brownian motion random walk, there is no difference if we change the standard deviation. sage: set_random_seed(0); finance.TimeSeries(10^6).randomize('normal').sums().range_statistic() 1873.9206979719115 sage: set_random_seed(0); finance.TimeSeries(10^6).randomize('normal',0,100).sums().range_statistic() 1873.920697971955 """ return (self.max() - self.min())/self.diffs().standard_deviation() def hurst_exponent(self): """ Returns a very simple naive estimate of the Hurst exponent of this time series.  There are many possible ways to estimate this number, and this is perhaps the most naive.  The estimate we take here is the log of the range statistic divided by the length of this time series. We define the Hurst exponent of a constant time series to be 1. NOTE: This is only a very rough estimator.  There are supposed to be better ones that use wavelets. EXAMPLES: The Hurst exponent of Brownian motion is 1/2.  We approximate it with some specific samples: sage: set_random_seed(0) sage: bm = finance.TimeSeries(10^5).randomize('normal').sums(); bm [0.6767, 0.2756, 0.6332, 0.0469, -0.8897 ... 152.2437, 151.5327, 152.7629, 152.9169, 152.9084] sage: bm.hurst_exponent() 0.5174890556918027 We compute the Hurst exponent of a simulated fractional Brownian motion with Hurst parameter 0.7.  This function estimates the Hurst exponent as 0.6678... sage: set_random_seed(0) sage: fbm = finance.fractional_brownian_motion_simulation(0.7,0.1,10^5,1)[0].sums() sage: fbm.hurst_exponent() 0.66787027921443409 Another example with small Hurst exponent (notice how bad the prediction is...): sage: fbm = finance.fractional_brownian_motion_simulation(0.2,0.1,10^5,1)[0].sums() sage: fbm.hurst_exponent() 0.30450273560706259 The above example illustrate that this is not a very good estimate of the Hurst exponent. """ cdef double r = self.range_statistic() if r == 0: return 0 return float(1) return log(r)/log(self._length) def min(self, bint index=False): cdef class TimeSeries: else: return s def clip(self, min=None, max=None): def clip_remove(self, min=None, max=None): """ Return new time series obtained from self by removing all values <= a certain maximum value, >= a certain minimum, or both. Return new time series obtained from self by removing all values that are less than or equal to a certain miminum value or greater than or equal to a certain maximum. INPUT: min -- None or double max -- None or double OUTPUT: time seriessx time series EXAMPLES: sage: v = finance.TimeSeries([1..10]) sage: v.clip_remove(3,7) [3.0000, 4.0000, 5.0000, 6.0000, 7.0000] sage: v.clip_remove(3) [3.0000, 4.0000, 5.0000, 6.0000, 7.0000, 8.0000, 9.0000, 10.0000] sage: v.clip_remove(max=7) [1.0000, 2.0000, 3.0000, 4.0000, 5.0000, 6.0000, 7.0000] """ cdef Py_ssize_t i, j cdef TimeSeries t cdef class TimeSeries: We take the above values, and compute the proportion that lie within 1, 2, 3, 5, and 6 standard deviations of the mean (0): sage: s = a.standard_deviation() sage: len(a.clip(-s,s))/float(len(a)) sage: len(a.clip_remove(-s,s))/float(len(a)) 0.68309399999999998 sage: len(a.clip(-2*s,2*s))/float(len(a)) sage: len(a.clip_remove(-2*s,2*s))/float(len(a)) 0.95455900000000005 sage: len(a.clip(-3*s,3*s))/float(len(a)) sage: len(a.clip_remove(-3*s,3*s))/float(len(a)) 0.997228 sage: len(a.clip(-5*s,5*s))/float(len(a)) sage: len(a.clip_remove(-5*s,5*s))/float(len(a)) 0.99999800000000005 There were no "six sigma events": sage: len(a.clip(-6*s,6*s))/float(len(a)) sage: len(a.clip_remove(-6*s,6*s))/float(len(a)) 1.0 """ if distribution == 'uniform': def unpickle_time_series_v1(v, Py_ssize_ def linear_filter(acvs): """ Create a linear filter with given autococvariance sequence. Create a linear filter with given autocovariance sequence. EXAMPLES:


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Section4.3Varignon’s Theorem

Varignon’s Theorem is a method to calculate moments developed in 1687 by French mathematician Pierre Varignon (1654 – 1722). It states that sum of the moments of several concurrent forces about a point is equal to the moment of the resultant of those forces, or alternately, the moment of a force about a point equals the sum of the moments of its components.

This means you can find the moment of a force by first breaking it into components, evaluating the scalar moments of the individual components, and finally summing them to find the net moment about the point. The scalar moment of a component is the magnitude of the component times the perpendicular distance to the moment center by the definition of a moment, with a positive or negative sign assigned to indicate its direction.

This may sound like more work than just finding the moment of the original force, but in practice it is often easier. Consider the interactive to the right. If we break the force into components along the wrench handle and perpendicular to it, the sum of the moments of these component is

$$M = F_\perp d \text{,}\label{moment-fperp}\tag{4.3.1}$$

where $$d$$ is the length of the handle, and $$F_\perp$$ is the component of $$F$$ perpendicular to the handle. Here, the contribution of the parallel component to the sum is zero, since its line of action passes through the moment center $$A\text{.}$$

This result agrees with our intuitive understanding of how a wrench works; the greatest torque is developed when the force is applied at a right angle to the handle.

Equations (4.2.1) and (4.3.1) not only produce the same result, they are completely identical. If the length of the handle is $$d$$ and the angle between the force $$\vec{F}$$ and the handle is $$\theta\text{,}$$ then $$d_\perp = d \sin \theta$$ and $$F_\perp = F \sin \theta\text{.}$$ Using either equation to calculate the moment gives

$$M = F\ d\ \sin \theta\text{.}\label{moment-general}\tag{4.3.2}$$

Subsection4.3.1Rectangular Components

Verignon's theorem is particularly convenient to use the diagram provides horizontal and vertical dimensions, which is often the case. If you decompose forces into horizontal and vertical components you can find the scalar moments of the components without difficulty.

The moment of a force is the sum of the moments of the components, so

$$M = \pm F_x d_y \pm F_y d_x\text{.}\label{verignon}\tag{4.3.3}$$

Take care to assign the correct sign to the individual moment terms to indicate direction; positive moment tend to rotate the object counter-clockwise and negative moment tend to rotate it clockwise according to the standard right hand rule convention.

Example4.3.3.Verignon's theorem.

A $$\lb{750}$$ force is applied to the frame as shown. Determine the moment this force makes about point $$A\text{.}$$

\begin{equation*} \vec{M}_A = \ftlb{174} \text{ Clockwise}. \end{equation*}
Solution.

Force $$\vec{F}$$ acts $$\ang{60}$$ from the vertical with a $$\lb{750}$$ magnitude, so its horizontal and vertical components are

\begin{align*} F_x \amp = F \sin \ang{60} = \lb{649.5} \\ F_y \amp = F \cos \ang{60} = \lb{375.0} \end{align*}

For component $$F_x\text{,}$$ the perpendicular distance from point $$A$$ is $$\ft{2}$$ so the moment of this component is

\begin{equation*} M_1 = 2 F_x = \ftlb{1299} \text{ Clockwise}\text{.} \end{equation*}

For component $$F_x\text{,}$$ the perpendicular distance from point $$A$$ is $$\ft{3}$$ so the moment of this component is

\begin{equation*} M_2 = 3 F_y = \ftlb{1125} \text{ Counter-clockwise}\text{.} \end{equation*}

Assigning a negative sign to $$M_1$$ and a positive sign to $$M_2$$ to account for their directions and summing, gives the moment of $$\vec{F}$$ about $$A\text{.}$$

\begin{align*} M_A \amp = - M_1 + M_2\\ \amp = - 1299 + 1125\\ \amp = -\ftlb{174} \end{align*}

The negative sign indicates that the resultant moment is clockwise, with a magnitude of $$\ftlb{174}\text{.}$$

\begin{equation*} \vec{M}_A = \ftlb{174} \text{ Clockwise}. \end{equation*}

The interactive diagram below will help you visualize the different tools to find moments that were covered in this section.


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# Coupling an electric charge to a gauge field. How is it done in this setup?

In page 9 of Tachikawa's N=2 susy dynamics for pedestrians it says that an electric particle with charge $n$ in the first quantised setup (in what sense first quantised?), Wick rotated to Euclidean signature, couples to a gauge field via $$S = i n \int_L dx^{\mu} A_{\mu} = in \int_L A.$$

where $L$ is the worldline of the particle. Now, why is this true? Can some further justification be given.

Also, in the following step he uses Maxwell's action and the equation I just wrote above and adds them together, and uses the EOM for $A_{\mu}$ to derive the fact that $n$ must be an integer. Specifically he writes

$$\int_{S^2} \frac{4\pi}{e^2} \vec{E} \cdot d\vec{n} = \int_{S^2}\frac{4\pi}{e^2} \star F = 2\pi n$$

Where does the $S^2$ come from and where does more generally the above equation come from? I cannot understand how he gets it.

The "first-quantized setup" is the setup of quantum mechanics, where single particles are considered quantum objects, but fields like the electromagnetic field are still treated classically. However, the derivation of the action in the following is wholly classical (and the first-quantized setup arises when considering $x^\mu$ as quantum fields in the sense of QM as a 0+1 dimensional QFT).
A particle with charge $n$ couples to the gauge field by $n\int_L A$ along its worldline because gauge fields, as a general principle, couple minimally to their four-current $j^\mu$ by $\int j_\mu A^\mu$. The four-current is simply the charge of the particle flowing along the worldline $x^\mu(\tau)$ of the particle, i.e. $j^\mu(y) = n \frac{\partial x^\mu}{\partial \tau}\int \delta(x(\tau) - y)\mathrm{d}\tau$. Inserting this into $\int j^\mu A_\mu$ gives $$\int j^\mu(y)A_\mu(y)\mathrm{d}^4 y = \int \left(n \frac{\partial x^\mu}{\partial \tau}\int \delta(x(\tau) - y)\mathrm{d}\tau\right)A_\mu(y)\mathrm{d}y$$ and carrying out the $y$-integration yields $$\int n \frac{\partial x^\mu}{\partial\tau}(\tau)A_\mu(x(\tau))\mathrm{d}\tau$$ and the chain rule yields the claimed expression $$\int n A_\mu \mathrm{d}x^\mu$$
For your second question, the "sphere at infinity" $S^2$ appears from considering Stokes' theorem $$\int_\Sigma \mathrm{d}\omega = \int_{\partial\Sigma}\omega$$ and formally writing $\partial\mathbb{R}^n = S^n$. A better consideration would be restricting the integral to a ball of radius R in $\mathbb{R}^n$ and then letting $R$ tend to infinity.