text:
# Reasoning with factorials and the Möbius inversion — does this make sense?

It is well known that a factorial can restated as a product of least common multiples:

$$\log(x!) = \sum_{m=1}^{\infty}\psi\left(\frac{x}{m}\right)$$

Using Möbius inversion formula, it occurs to me that this can be restated as:

$$\text{lcm}(1,2,\dots,x) = \prod_{m=1}^{\infty}\left(\left\lfloor\frac{x}{m}\right\rfloor!\right)^{\mu(m)}$$

Further:

$$\frac{\text{lcm}(1,2,\dots,(x^2+x))}{\text{lcm}(1,2,\dots,x^2)} = \prod_{m=1}^{\infty}\left(\frac{\lfloor\frac{x^2+x}{m}\rfloor!}{\lfloor\frac{x^2}{m}\rfloor!}\right)^{\mu(m)}$$

As I was thinking about this, it seemed to me that if $$\lfloor\frac{x}{6}\rfloor \ge 2$$, then:

$$\prod_{m=6}^{\infty}\left(\frac{\lfloor\frac{x^2+x}{m}\rfloor!}{\lfloor\frac{x^2}{m}\rfloor!}\right)^{\mu(m)} \ge 1$$

Is there a straight forward way to check my hypothesis? Is this a well-understood problem with a straight-forward proof or is this an open problem?

format: plain


text:
# Averaging equation?

Is there any name for following averaging equation and how does it work?

$$y_{i+1} = (1-\alpha)\times y_i + x_i$$

I'm using it to average a signal, but I'm not sure am I interpreting the results right.

-
Can you describe why you need to average this signal? Do you mean that you're smoothing a signal? –  littleO Oct 31 '12 at 9:49
What is $x_i$ and what is $y_i$? –  malin Oct 31 '12 at 10:06
$x_i$ is a sample from signal. $y_i$ is a previously value of averaged $y$ and $y_{i+1}$ is next averaged value from $y$. @littleO No, I'm not smoothing a signal, so I'm not looking for a averaging filter. I just should see that when the signal values get stronger $y_i$ should get stronger also. So it should be able to detect changes in signals strength. –  gillian Oct 31 '12 at 10:37


text:
## This blog has moved!

I have decided to move my blog to a server that I control.

WordPress.com is great, but many features which a power user would want require costly upgrades.  For example, just to use custom CSS one has to pay.  Furthermore, WordPress.com’s mathematics support is poor.

I decided to use my money to obtain a server to install the WordPress management system on and to use as my personal webpage and blog.  I went with Linode.  Linode provides linux VPS’ to the masses.  They sell a VPS to a client who has full rain of the machine.  Yes, I have root access and even the ability to reinstall the OS if I choose. In less than 24 hours I was able to get my linode up and running with wordpress.

Using the WordPress management system is great, because it allows one to install a plugin called jetpack which connects my WordPress system with wordpress.com.  This gives my site the ability to track stats and connect to social networks.  It is great!  Using the WordPress system also allows me to export all of my blogs content at wordpress.com and import it at its new home.  Hence, all of my data is moved over and the new blog looks identical to this one.  Lastly, I have installed the MathJax plugin which gives my new blog the ability to render beautiful mathematics.

Our new home is now at www.blog.metatheorem.org.

So repoint your RSS feeds over to the new place.  I hope everyone enjoys my future posts.

Posted in Goodbyes | 2 Comments

## Coq, Category Theory, and Steve Awodey’s Book.

There is a great book on category theory called, you guessed it, “Category Theory” by Steve Awodey. It is a great book! I really enjoy it. I am interested in category theory, because it provides a possible interpretation of type theory. Actually, I have been thinking of the relationship between category theory and type theory as being different perspectives of the same thing. That thing is computation. As I learn more category theory I thought it would be useful to start thinking about how to formalize results involving category theory in Coq.

To my surprise I found that Adam Megacz has started a nice library called coq-categories. His formalization follows Awodey’s book exactly. In fact in order to find a concept in the formalization one has to know which chapter it is in Awodey’s book. This is a great formalization. It uses type classes to define categories and functors. Type classes are such a nice fit for categories.

There are two issues with this formalization. First, it does not formalize anything beyond chapter 7 of the book. The second is that it has not be worked on in over a year. The last commit was 2011-04-24. The version of Coq it was developed in was 8.3pl2.

I decided to clone Adam’s git repo. and get it working with Coq 8.4. It took only about fifteen minutes to get working. The only problems that I ran into were related to unification of implicit arguments.

This post just serves as a place to put up a patch in case anyone is looking for one. I plan to develop this a bit further. I would like to formalize control categories and play around with the semantics of a few different type theories. Here is the patch.

I plan to post more about my extensions in the near future.

Posted in category theory, Coq, Fun, Logic, metatheory, Semantics, Type Theory | | 2 Comments

## Cartesian Closed Comics

There is a new comic called Cartesian Closed Comics. They are pretty good.

Below is my favorite one:

Find more by clicking the above link.

## Impredicativity and the Human Mind

I have not posted in awhile so I thought I would post something both short and interesting to get things going again. I defended my comprehensive exam this week. It went really well, and I plan to post the report on my website, but before I do I want to make a few corrections. Greg Landini a philosophy professor here at UIowa was an external member of my committee for my comprehensive exam. I am also doing a readings course with him on relevant logic. During our weekly meeting this week we got into a really interesting conversation comparing the meaning of impredicativity and predicativity according to Bertrand Russell and the definition used in type theory. The conversation eventually turned over to Greg’s work on understanding human thought and consciousness. I would like to discuss in this post the details of my conversation with Greg. First, I would like to introduce how I define impredicativity and predicativity. Then I will talk about how Russell defined these terms. Following these I will discuss Greg’s ideas on animal consciousness.

I define an impredicative formula to be an universally quantified formula — the complexity of the formula does not matter — such that, the domain of quantification contains the formula being defined. A logic or type theory is impredicative if all universal formulas are impredicative. An example would be the universal type of Girard-Reynold’s system F. Now a predicative formula is an universal formula, such that, the domain of quantification does not contain the formula which is being defined. We can make the universal type of system F predicative by stratifying types using kinds.

Consider the following formula in naive set theory: $\exists y\forall x.(x \in y \iff x \not \in x)$. Greg points out that Russell did not call this an “impredicative formula”, because there is only one domain of objects to quantify over which is the universe of all sets. Now I would call this an extreme case of impredicativity. That is this formula is the most impredicative a formula can be. In fact, so impredicative it is inconsistent. Other than this case Russell’s views on impredicative vs. predicative are in line with the definition we use in type theory. That is types can be impredicative in the same way they are in system F and remain consistent, while just like stratified system F they can be predicative. Indeed this is the difference between the simple theory of types and the ramified theory of types. From here the conversation drifted over to Greg’s work on consciousness.

Greg believes that there are two types of reasoning going on in the mind of animals. Some use predicative reasoning while the more gifted use impredicative reasoning. He used a great example using a frog.

A frog sits and eats flies. The frog has the innate ability to spot a fly when it is near. If the frog hears some clicking noise which sounds like a fly it will try and eat the object making the clicking. If the frog sees a flying object it will also try to eat it. Now if I use a tape recorder and a small toy object which makes the same sound a fly makes then we can easily trick the frog into eating the toy. Thus, a frog cannot first identify the object making the sound and determine whether or not it is actually a fly before deciding to eat it. This kind of thought Greg believes is predicative. This suggests that most animals use predicative reasoning. He says that the animals with impredicative reasoning got lucky during the evolutionary process. Now what do we mean by lucky?

The brain is believed to be constructed by layers of mesh which make up synaptic networks. This is depicted by the following figure.

The previous figure shows a stratification of the layers of mesh. This results in predicative reasoning. Impredicative reasoning results when the layers have connections between them. As in the following diagram.

This suggests that the more these layers are collapsed the more powerful reasoning becomes.

This fits with the views of type theory almost exactly. Think of Coq where the type universe is stratified into levels. This stratification is greatly less powerful then in the full collapse of the type universe — this results in $\mathsf{Type}:\mathsf{Type}$. The more we collapse the type universe the more powerful the theory becomes. Think of the difference between stratified system F and system F. We can gain incremental power by making connections between the levels. One way of doing this is adding universe polymorphism. This is like the connections between the mesh layers discussed above. These connections I found to be quite surprising, but also not so surprising. The brain is essentially the worlds most powerful computer. So it makes sense that computer science could help in figuring out how it works. Computation is what we do after all.

I think this is all I am going to say about this right now. I found this to be really cool. I thought maybe others might as well.

## On the Dual Calculus with a Twist

Recently, my advisor (Aaron Stump) and I have been trying to unify induction and coinduction using a single operator and an explicit ordering. I know what you are thinking, awesome! While I could talk for hours about our ideas on how to do this I am instead going to introduce a type theory I think everyone should know of.

While reading over the literature covering duality between induction and coinduction and call-by-name and call-by-value we had the joy of learning about Philip Wadler’s dual calculus (DC) [1,2]. This type theory is the computational view of Gentzen’s classical sequent calculus LK. For more information on LK see [3]. A couple of interesting properties of the dual calculus is that its dual can be embedded in itself. This dual embedding can then be used to show that the call-by-value reduction strategy is dual to the call-by-name reduction strategy. Following on the previous point DC can also be used to show that the call-by-value $\lambda\mu$-calculus is dual to the call-by-name $\lambda\mu$-calculus by first embedding the call-by-value $\lambda\mu$-calculus into DC, taking the dual, and then translating the $\lambda\mu$-calculus back out [1,2]. Cool stuff! In this post I would like to simply define DC and give an intuitive meaning to its judgments. However, there is a twist. The twist is that I have come up with a new syntax for the judgments of DC. While DC is a beautiful theory I am not a fan of DC’s original nor of the version of DC in Kimura’s paper [4] synatx for judgments. I find them ugly, confusing, and hard to read. I hope the reader will find my syntax a bit more pleasing.

The syntax of DC is defined by the following set of grammars: The first thing the reader who has at least a basic understanding of type theory will notice, is that, there is no function type or $\lambda$-abstraction! It turns out that these can actually be defined! I hope the reader’s head did not just explode. Here are the definitions:

Using these definitions the usual typing rules are now derivable as well as the usual $\beta$-reduction rule. As for the remaining syntax of DC it is pretty routine. I only comment on a few aspects of it. I call variables $x$ input variables and variables $\alpha$ output variables or output ports. The output ports will be replaced by the output of (co)terms. Statements $\alpha \bullet k$ can be viewed as commands where $k$ is computed with input $\alpha$. The term $(s).\alpha$ binds the output port $\alpha$ to the return value of the statement $s$. Dually, the coterm $x.(s)$ binds the input variable $x$ in $s$.

DC consists of three judgments. I will first give my new syntax for these judgments, compare my syntax with the original syntax, give an intuitive explanation of the judgments, and finally define them. The syntax is of the three judgments are as follows:

Explanation of the syntax. The arrows in the judgments above indicate the flow of input/output. The first left-pointing arrow $\leftarrow$ indicates that each expression $t$,$k$, and $s$ can assign output to an output port in $D$. Now the right-pointing arrow $\rightarrow$ in the typing judgment indicates that $t$ may return a value of type $A$, dually, in the syntax for the cotyping judgment the second left-pointing arrow indicates that $k$ takes an input of type $A$. Recall from above that I mentioned that across the propositions-as-types correspondence DC corresponds to Gentzen’s classical sequent calculus LK. LK consists of left and right rules, so additionally the second arrow in the typing and cotyping judgments indicate the type of rules which define the judgments. The rules defining the typing judgment are right rules, hence, the second arrow points right, while the second arrow in the cotyping judgment points left indicating its rules correspond to the left rules of LK.

The syntax used by Kimura was the following:

This syntax is very similar to the original syntax of Wadler, but instead of $\vdash$ he used an arrow pointing to $\Delta$ which is close to my syntax, but still not as pretty.  I find this style syntax to be hard to read, but I do think it makes sense once I understood it. I have this annoying belief that the syntax of a judgment should be sexy enough to make writing out derivations somewhat manageable.

Intuitive meaning of the judgments. Lets first expand the syntax out a bit.
The typing judgment means that when each $x_i$ has a value and $t$ is computed then it may either return a value of type $A$ or assign a value to one of the output ports $o_i$ of type $B_i$. Dually, the cotyping judgment means when each $x_i$ has a value then $k$ takes an input of type $A$ and after computation assigns a value to one of the output ports. Finally, the cut judgments means when each $x_i$ has a value then $s$ is computed and assigns a value to one of the output ports. This intuitive explanation behind the judgment helped and continues to help me understand DC. I do not claim authorship of such a great meaning I learned it from [4]. They did a great job explaining the theory.

Definition of the judgments of DC. Finally, I move onto define each judgment. The following is the definition of the typing and cotyping judgment:
Lastly, we have the sole rule of the cut judgment, the cut rule:

A nice feature of this type theory is that it is easy to translate it to a logic equivalent to LK. Lets try this out. We define the following eraser:

Using this eraser we obtain the following logic which is equivalent to LK.

We have the cut rule last:

I might prove this logic equivalent to LK, all truths deducable in this logic are deducible in LK, in a later post. To finish off the definition of DC lets take a look at the reduction rules. I do not pen any specific reduction strategy down here. The reducation rules of DC are:
Well that is the dual calculus. It is a really interesting theory of classical logic. There are quite a few papers on the dual calculus. I sprinkled some throughout the post, but if the reader knows of any I would love to hear about them.

Posted in Logic, Type Theory | 3 Comments

## The 2012 Oregon Programming Languages Summer School

The lecture videos and some slides are up! You can find them here.

I am particularly excited about Steve Awodey’s talk.

Dig in.

## The Seventh ACM SIGPLAN Workshop on Programming Languages meets Program Verification (PLPV 2013)

I am on the program committee for PLPV 2013. The first call for papers is out. I thought I would share it here just in case there is anyone reading this that doesn’t know.

The Seventh ACM SIGPLAN Workshop on Programming Languages meets Program Verification (PLPV 2013)

http://plpv.tcs.ifi.lmu.de/

22nd January, 2013
Rome, Italy
(Affiliated with POPL 2013)

Call for Papers

Overview

The goal of PLPV is to foster and stimulate research at the
intersection of programming languages and program verification, by
bringing together experts from diverse areas like types, contracts,
interactive theorem proving, model checking and program analysis. Work
in this area typically attempts to reduce the burden of program
verification by taking advantage of particular semantic or structural
properties of the programming language. One example are dependently
typed programming languages, which leverage a language’s type system
to specify and check rich specifications. Another example are
extended static checking systems which incorporate contracts with
either static or dynamic contract checking.

We invite submissions on all aspects, both theoretical and practical,
of the integration of programming language and program verification
technology. To encourage interaction between different
communities, we seek a broad scope for PLPV. In particular,
submissions may have diverse foundations for verification (based on
types, Hoare-logic, abstract interpretation, etc), target
different kinds of programming languages (functional, imperative,
object-oriented, etc), and apply to diverse kinds of program
properties (data structure invariants, security properties, temporal
protocols, resource constraints, etc).

Important Dates

Submission 8th October, 2012 (Monday)
Final Version 8th November, 2012 (Thursday)
Workshop 22nd January, 2013 (Tuesday)

Submissions

We seek submissions of up to 12 pages related to the above
topics; shorter submissions are also welcome. Submissions may describe
new work, propose new challenge problems for language-based
verification techniques, or present a known idea in an elegant way
(i.e., a pearl).

Submissions should be prepared with SIGPLAN two-column conference
format. Submitted papers must adhere to the SIGPLAN republication
policy. Concurrent submissions to other workshops, conferences,
journals, or similar forums of publication are not allowed.

To submit a paper, access the online submission site at
http://www.easychair.org/conferences/?conf=plpv2013.

Publication

Accepted papers will be published by the ACM and will appear in the
ACM Digital library.

Program Committee

Andreas Abel Ludwig-Maximilians-University Munich (co-chair)
Robert Atkey University of Strathclyde
Harley Eades The University of Iowa
Chung-Kil Hur Max Planck Institute for Software Systems
Brigitte Pientka McGill University
Andrew Pitts University of Cambridge
François Pottier INRIA
Tim Sheard Portland State University (co-chair)
Makoto Takeyama Advanced Industrial Science and Technology


text:
# Derivation of Spin-orbit Coupling

I was reading Griffiths's book on quantum mechanics.

In chapter 6, he tried to derive the spin-orbit coupling using a classical approach.
$$H=-\vec{\mu}\cdot \vec{B}$$

1. Finding the relation between $\vec{μ}$ and $\vec{S}$
He consider a spinning charged ring with mass $m$, radius $r$, total charge $e$ and period $T$.
Magnetic Dipole Moment: $\vec{\mu}=i\vec{A}=\frac{e\pi r^2}{T}\hat{n}$
(Spin) Angular Momentum: $\vec{S}=\tilde{I}\vec{\omega}=\frac{2m\pi r^2}{T}\hat{n}$
$$\vec{\mu}=\frac{e}{2m}\vec{S}$$
Dirac: $\frac{e}{2m}\rightarrow\frac{e}{m}$

2. Find the magnetic field $\vec{b}$
He consider the rest frame of electron.
$$\vec{B}=\frac{\mu_0 i}{2r}\hat{n}=\frac{1}{c^2\epsilon_0}\frac{e/T}{2r}\hat{n}=\frac{1}{4\pi\epsilon_0}\frac{e}{mc^2r^3}\vec{L}$$

3. Combining both terms
$$H=-\frac{1}{4\pi\epsilon_0}\frac{e^2}{m^2c^2r^3}\vec{L}\cdot\vec{S}$$

Then he point out that when we choose the electron's rest frame, there is one problem as it is a non-inertial frame. Therefore, we need to consider the Thomas's precession, which gives a factor of 1/2. Finally, it cancels with the 2 we multiplied to the classically derived gyromagnetic ratio.

My problem is when we choose an accelerating frame, obviously we need some modifications. But, how should we modify? and how can we get the factor of 1/2 from Thomas's precession?


text:
[Tuglist] Environment slide in pdfscreen.sty

12 Jun 2002 16:19:17 +0530

>>>>> "David" == David Kastrup <David.Kastrup@t-online.de> writes:

[...]

David> There is an argumentless macro \@ defined in _LaTeX_, and
David> your \def replaces it with a different definition with a
David> mandatory argument.

>> I removed the line \makeatletter....\makeatother and reinserted
>> the line
>>
>> \def\@captype{figure}
>>
>> in my code.

I'm sorry, the whole problem is fully attributable to my innocent
expectation that people will change the catcode of '@' character
before and after the definition, because the necessity of catcode
change in such situations has been discussed in this list
previously. And in all the fairness, I trusted Manoj to do that since
he knows the innards of LaTeX. Had someone else posted the query, I
would have obviously added the \makeatletter...  \makeatother.

David> Why?  I explicitly explained why the \makeatletter was
David> necessary, so why do you delete it?  And why do you make
David> just the definition again for which I explained to you why
David> and how it failed?

Please cool down David, after all this is a list of novices rather
than TeXies as you find in the tug.org lists. However, I must thank
you for your patience to answer all the questions, dont take anything
personally if at all any query seems too naive.

David> If you insist on ignoring my explanations, you might also
David> find this explained in the UKTUG FAQ, under the heading

David> \@ and @ in macro names

David> -- David Kastrup, Kriemhildstr. 15, 44793 Bochum Email:

Apologies for triggering something hot in a relatively cool list.

--


text:
# Photoelectric experiment explaining particle property of light [duplicate]

How does photoelectric experiment prove the particle aspect of light in opposed to be solely wave-like?

• Does this answer your question? Can the photoelectric effect be explained without photons? – A. P. Jan 17 at 20:35
• The particle aspect of a light with photons is fairly self-explanatory. What’s hard to explain is a light wave. Billions of coherent photons radiating from a common source resemble a wave but they are still individual particles. – Bill Alsept Jan 17 at 21:10
• @BillAlsept This must be some joke. – my2cts Jan 17 at 23:47
• @my2cts you can physically describe and explain light with particles. You cannot even begin to describe a light wave without incorporating billions of individual photons. – Bill Alsept Jan 18 at 0:01

There was a zinc plate that when irradiated with ultraviolet light emitted electrons (charged particles) to a detector which could detect them. It was observed that there was a minimum energy to the light before the photoelectric effect could be observed. It was postulated that photons come in wavepackets. Planck relation gives: $$E = h \nu$$


text:
226 posts / 1 project
None
RemedyBG 0.3.6.0
Edited by x13pixels on Reason: typo in post title
A new version of RemedyBG, 0.3.6.0, is now available for immediate download. Changes include:

New Features
* Evaluate on hover. Can set the tooltip delay in Windows>Style and can optionally disable this feature altogether.
* A stack-overflow exception will now collapse repeated frames and show "Collapsed N frames" in the Call Stack pane. Can double-click collapsed frame or toggle this behavior via "Collapse Repeated Frames" context menu item.
* Adds "Format As" menu item showing relevant format specifiers for the selected watch(s).
* Added $err pseudo-variable for the last error value set by SetLastError for the current thread * Added$handles pseudo-variable which shows the number of handles allocated by the process
* Added $tid and$pid pseudo-variables for thread and process IDs
* Added ",d" decimal format specifier
* Added ",hr" format specifier for displaying an HRESULT or Win32 error code
* Adds process ID (optionally) to status text. Can be toggle via status text's context menu.
* Allow breakpoint on either nominal entry point (WinMain, main, etc...) or entry point as defined in the PE header.
* Can now run to a filename/line via the command line driver (see run-to-cursor in documentation)

Improvements
* Lazily load function table entries. This helps improve performance when stepping over calls that load many DLLs (e.g., Direct3D initialization).
* Use high-contrast color for collapsed panes to make them easier to find.
* Adds error message when trying to index non-array/pointer type.
* Properly handle device removed scenarios (Direct3D).
* Line numbers have been added back to the call stack display.
* Render the frame rather than show black while resizing.
* Implements PgUp/PgDown in panes (previous these keys mapped to what is Home/End now)
* Added binary formatting of enum values
* We now alphabetically sort text file names in the drop-down menu.
* Adds Ctrl+C hotkey to Output Window to copy contents to clipboard
* Rearranged menu in Output Window so "Clear" is less likely to be accidently clicked.
* Adjusted vertical position of lock icon in embedded font
* Added more descriptive error message in the case CreateProcess fails
* No longer count disable processor BPs against total number use
* Adds "Move Up / Move Down" menu items in watch window to aid discoverability of this existing feature.
* Show arrow next to processor breakpoint that was hit.
* Run-to-cursor operation will now show yellow breakpoint symbol in text/disassembly
* No longer initiate editing of address in memory window if a cell is being edited.
* Double-click, not single-click, now used to initiate memory window editing.
* An error in a breakpoint's condition is now more clearly presented.
* Allow adding a breakpoint at an address without having to open/navigate to the address in the disassembly window.

Bug Fixes
* Fixed lookups of static, global TLS variables.
* Fixes the case in which a breakpoint is added on the same line as the run-to-cursor location before the run-to-cursor operation completes.
* String summaries are now culled to 64-chars for non-nested previews and 32-chars otherwise.
* Fixed a problem where the corresponding text window would not be brought to the front when stepping through disassembly.
* Fixed a case where the session would be considered modified immediately after opening.
* Fixes a problem where keyboard input was ignored in the auto-created watch expression in certain situations.
* Now properly sign-extend bitfield values.
* Fixed the editing of bitfield values in watch.
* An exception is now considered "handled" after using set-next-statement
* Fixed so that Ctrl+Alt+W no longer deletes a watch expression when changing focus to another watch window.
* Modifying condition on an existing processor breakpoint was broke in previous release. This has been fixed.
* Fixed dereferencing union type when using '.' rather than '->'
* Left arrow was unintentionally changing to focus to last row in some cases. This has been fixed.
* Fixed reported case where cursor position on ASM source goes to PROC declaration rather than first instruction.
* Fixed problem terminating app when breakpoint present in multithread code.
* Fixed bug when run-to-cursor is executed on the same line
* Fixed hang adding a function breakpoint to a module containing only a single function.
* Fix for memory window so navigate to an address occurs even when reevaluation of an expression is identical to the previous address.
Jesse
64 posts
Programmer at NASA GSFC
RemedyBG 0.3.6.0
Enormous QOL improvements here. Thanks so much!
Simon Anciaux
1060 posts
RemedyBG 0.3.6.0
Thanks for the update.
Jason
199 posts
RemedyBG 0.3.6.0
Awesome update. Having the hover ability is definitely a quality of life improvement. Thank you!
226 posts / 1 project
None
RemedyBG 0.3.6.0

Jesse, Simon, and Jason: you're welcome!

• Fixes potential crash in stack walk code introduced in 3.6.0 (The Sandvich Maker)
• Evaluate on hover will now consider array indexing when picking the expression to evaluate (datamite)
• Fixes conditional breakpoints added at program startup (datamite)
• AltGR+[^A-Z] can now be used to initiate an edit in the watch window (seventh-chord)
• Fixed expression generated for Add Watch / Send to Memory Window for dereferenced pointers (JensWallgren)
1 posts
RemedyBG 0.3.6.0

Thank you so much for this update! Although there is one thing added in it that I would like to be able to turn off which is:


text:
## Remarks and results on congruent numbers.(English)Zbl 0259.10010

Proc. 3rd Southeast. Conf. Combinat., Graph Theory, Computing, Florida Atlantic Univ., Boca Raton 1972, 27-35 (1972).
A congruent number is a positive integer $$a$$ such that the equations $$x^2+ay^2=z^2$$ and $$x^2-ay^2=t^2$$ are simultaneously soluble in positive integers $$x,y,z,t$$. The authors have investigated the 608 square-free $$a\leq 1000$$ with the help of a computer, whereby 198 of these numbers have been proved congruent and 136 non-congruent, the other 274 remain undecided. The results agree with the conjecture that every $$a\equiv 5,6$$ or $$7\pmod 8$$ is a congruent number.
For the entire collection see Zbl 0258.00003.
Reviewer: G. L. Watson

### MSC:

 11D09 Quadratic and bilinear Diophantine equations 11-04 Software, source code, etc. for problems pertaining to number theory


text:
# Balance this reaction using ion electron method in shortest way possible . but must be clear .. Fe^(2+)+MnO_4^(-)+H^(+) -> Fe^(3+)+Mn^(2+)+H_2O ?

Apr 17, 2017

MnO4^-) + 8H^+ + 5Fe^(2+)= Mn^(2+) + 4H_2O+5Fe^(3+)

#### Explanation:

In ion electron method one reaction is divided into two reaction (half reaction) and then they are balanced and then added together

Figure the reducing(increase of electrons) and oxidizing(decrease of electrons) agent
$\textcolor{w h i t e}{F {e}^{}} 2 \textcolor{w h i t e}{M \ln} {7}^{+} - 2 \textcolor{w h i t e}{l l l l l l l \times l l l} + 3 \textcolor{w h i t e}{l l x l l l l l} 2 +$
Fe^(2+) + MnO_4^-) + H^+ → Fe^(3+)+Mn^(2+) + H2O

Iron has already lost 2 electrons and in this reaction it is losing another one

$F {e}^{2 +} = F {e}^{3 +}$

Thus iron is being oxidized

MnO_4^-) = Mn^(2+)

In the ion MnO4, Mn has a charge of 7+ and it changing into 2+ in this reaction

Thus MnO4 is being oxidized

Now balance the the oxygen atoms

MnO4^-) = Mn^(2+) + 4O

You can see in the reaction that oxygen is used to make water and no oxygen is let which is ${O}_{2}$

thus 4 oxygen atoms can produce 4 water molecules

MnO4^-) + H^+ = Mn^(2+) + 4H_2O

balance the reaction

MnO4^-) + 8H^+= Mn^(2+) + 4H_2O

Now for a redox reaction the addition of charges on both sides should be equal.So try it in this reaction

$\left(- 1\right) + 8 = + 7$
$\left(+ 2\right) + 0 = + 2$

So both sides are not equal. To make them equal you should add electrons as you cant add protons.Thus if you think of adding 5 protons to the other side to make it equal to 7 it is wrong.But you can add 5 electrons to the other side to make it equal to 2

MnO4^-) + 8H^+ + 5e^-)= Mn^(2+) + 4H_2O

Where $5 {e}^{-}$ is 5 electrons

Make electron gain equivalent to electron lost. The electrons lost in the oxidation half-reaction must be equal the electrons gained in the reduction half-reaction. And this electrons come from the oxidizing agent that is $F {e}^{2 +}$

$F {e}^{2 +} = F {e}^{3 +}$

Again try to make the charges equal on same side and you would get

$F {e}^{2 +} = F {e}^{3 +} + {e}^{-}$

Now we want 5electrons not one so to get 5electrons you must have more $F {e}^{2 +}$

$5 F {e}^{2 +} = 5 F {e}^{3 +} + 5 {e}^{-}$

Add both the reactions

MnO4^-) + 8H^+ + 5e^-) + 5Fe^(2+)= Mn^(2+) + 4H_2O+5Fe^(3+) + 5e^-

cancel the same terms

MnO4^-) + 8H^+ +cancel( 5e^-) + 5Fe^(2+)= Mn^(2+) + 4H_2O+5Fe^(3+) + cancel(5e^-)

MnO4^-) + 8H^+ + 5Fe^(2+)= Mn^(2+) + 4H_2O+5Fe^(3+)

Apr 17, 2017

$\boldsymbol{5 \stackrel{\textcolor{b l u e}{+ 2}}{{\text{Fe"^(2+))(aq) + 8stackrel(color(blue)(+1))("H"^(+)(aq)) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l) + 5stackrel(color(blue)(+3))("Fe}}^{3 +}} \left(a q\right)}$

Here are the primary steps for balancing in ACIDIC solution:

1. Pick out the elements being oxidized and reduced and write out their unbalanced half-reactions (this should require almost zero effort as this is the easy part).
2. Balance the non-$\boldsymbol{\text{O}}$ and non-$\boldsymbol{\text{H}}$ atoms.
3. Balance $\boldsymbol{\text{O}}$ using $\text{H"_2"O}$ molecules, since the solution is aqueous.
4. Balance $\boldsymbol{\text{H}}$ using ${\text{H}}^{+}$ ions, since the solution is acidic.
5. Balance the remaining charge using electrons on the side with more positive charge, since electrons must cancel out in the overall reaction.
6. Scale one or both reactions to cancel out the electrons.
7. Add the reactions together.

In BASIC solution, add the step of adding ${\text{OH}}^{-}$ to both sides of the final reaction, combining ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ to make water, then canceling out common waters on each side.

Usually you don't have to do all of these steps, and the nice thing is that if you have to balance another reaction that involves the same oxidation or reduction half-reaction, you can recycle the same half-reaction.

OXIDATION HALF-REACTION

Iron gets oxidized from a $+ 2$ to $+ 3$ oxidation state. For pure ions, the charge is the oxidation state.

$\stackrel{\textcolor{b l u e}{+ 2}}{{\text{Fe"^(2+))(aq) -> stackrel(color(blue)(+3))("Fe}}^{3 +}} \left(a q\right)$ (step 1)

Everything is balanced except for the charge, so we can skip straight to step 5.

$\stackrel{\textcolor{b l u e}{+ 2}}{{\text{Fe"^(2+))(aq) -> stackrel(color(blue)(+3))("Fe}}^{3 +}} \left(a q\right) + {e}^{-}$ (step 5)

The total charge on each side is now:

$\left(+ 2\right) = \left(+ 3\right) + \left(- 1\right)$ color(blue)(sqrt"")

Done with the oxidation half-reaction.

REDUCTION HALF-REACTION

In some respects, this is the easier one to identify. Reduction can sometimes be thought of as the loss of oxygen atoms, and oxidation the gain of oxygen atoms.

Manganese can be readily seen to have lost oxygen atoms going from ${\text{MnO}}_{4}^{-}$ to ${\text{Mn}}^{2 +}$, which is why I suggested this easier way to spot reduction if applicable.

$\stackrel{\textcolor{b l u e}{+ 7}}{{\text{Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn}}^{2 +}} \left(a q\right)$ (step 1)

The non-oxygen and non-hydrogen atoms (i.e. $\text{Mn}$) are balanced, so we skip to step 3. Add water to balance the oxygens.

$\stackrel{\textcolor{b l u e}{+ 7}}{\text{Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O}} \left(l\right)$ (step 3)

Add ${\text{H}}^{+}$ to balance the hydrogens.

$8 \stackrel{\textcolor{b l u e}{+ 1}}{\text{H"^(+))(aq) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O}} \left(l\right)$ (step 4)

Balance the charge using electrons on the more positive side. Each side has:

(??) + (+8) + (-1) = (+2)

Hence, add $5 {e}^{-}$ to the left side.

$5 {e}^{-} + 8 \stackrel{\textcolor{b l u e}{+ 1}}{\text{H"^(+)(aq)) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O}} \left(l\right)$ (step 5)

(-5) + (+8) + (-1) = (+2) color(blue)(sqrt"")

Done with the reduction half-reaction.

OVERALL REACTION

Now make the electrons cancel out. (Step 6)

$5 \left(\stackrel{\textcolor{b l u e}{+ 2}}{{\text{Fe"^(2+))(aq) -> stackrel(color(blue)(+3))("Fe}}^{3 +}} \left(a q\right) + \cancel{{e}^{-}}\right)$
$\cancel{5 {e}^{-}} + 8 \stackrel{\textcolor{b l u e}{+ 1}}{\text{H"^(+)(aq)) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O}} \left(l\right)$

Add the reactions to get: (Step 7)

$\boldsymbol{5 \stackrel{\textcolor{b l u e}{+ 2}}{{\text{Fe"^(2+))(aq) + 8stackrel(color(blue)(+1))("H"^(+)(aq)) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+2))("Mn"^(2+))(aq) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l) + 5stackrel(color(blue)(+3))("Fe}}^{3 +}} \left(a q\right)}$

Sep 21, 2017

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 F {e}^{2 +} \rightarrow M {n}^{2 +} + 5 F {e}^{3 +} + 4 {H}_{2} O$

#### Explanation:

$F {e}^{2 +} \rightarrow F {e}^{3 +} + {e}^{-}$ $\left(i\right)$

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow M {n}^{2 +} + 4 {H}_{2} O$ $\left(i i\right)$

And we take $5 \times \left(i\right) + \left(i i\right)$ to eliminate the electrons.....

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 F {e}^{2 +} \rightarrow M {n}^{2 +} + 5 F {e}^{3 +} + 4 {H}_{2} O$

Charge and mass are balanced so this is kosher. And what would see in the reaction. The deep purple colour of permanganate would dissipate to give almost colourless $M {n}^{2 +}$.


text:
2020 CMS Winter Meeting

Montreal, December 4 - 7, 2020

Algebraic Combinatorixx (Women in Algebraic Combinatorics)
Org: Angele Foley (Laurier) and Steph van Willigenburg (UBC)
[PDF]

SUNITA CHEPURI, University of Michigan

SAMANTHA DAHLBERG, Arizona State University

PAMELA HARRIS, Williams College
Kostant's partition function and magic multiplex juggling sequences  [PDF]

Kostant’s partition function is a vector partition function that counts the number of ways one can express a weight of a Lie algebra g as a nonnegative integral linear combination of the positive roots of g. Multiplex juggling sequences are generalizations of juggling sequences that specify an initial and terminal configuration of balls and allow for multiple balls at any particular discrete height. Magic multiplex juggling sequences generalize further to include magic balls, which cancel with standard balls when they meet at the same height. In this talk, we present a combinatorial equivalence between positive roots of a Lie algebra and throws during a juggling sequence. This provides a juggling framework to calculate Kostant’s partition functions, and a partition function framework to compute the number of juggling sequences. This is joint work with Carolina Benedetti, Christopher R. H. Hanusa, Alejandro Morales, and Anthony Simpson.

OLYA MANDELSHTAM, Brown University

LUCY MARTINEZ, Stockton University
Minimum Rank of Regular Bipartite Graphs  [PDF]

The rank of a graph $G$ is defined as the rank of its adjacency matrix $A$. The smallest rank among all the matrices with the same pattern of non-zeros entries as $A$, over the field $\mathbb{F}$, is called the minimum rank of $A$ over $\mathbb{F}$. The smallest among all the minimum ranks of $A$ (considering all the fields) is called the minimum rank of $G$. In this work, we study regular bipartite graphs. Specifically, we used linear recursions with linear complexity 2 and zero forcing sets to prove that the minimum rank of a $(n-1)$-regular bipartite graph, with $n$ vertices on each side, is 4.

ROSA ORELLANA, Dartmouth College

ANNA PUN, University of Virginia

SOPHIE SPIRKL, University of Waterloo
A complete multipartite basis for the chromatic symmetric function  [PDF]

The complete multipartite basis $r_\lambda$ for symmetric functions was introduced by Penaguiao. In this talk, I will tell you why this basis is interesting, and give a combinatorial interpretation for the $r_\lambda$-coefficients of the chromatic symmetric function.

Joint work with Logan Crew.

NANCY WALLACE, UQAM
Toward a Schurification of Schröder path formulas.  [PDF]

The Shuffle theorem of Carlsson and Mellit, states that $\nabla(e_n)$ is given by Parking function formulas. Schröder paths are a particular case of Parking functions. These formulas are symmetric in the variables $q$ and $t$. More preciously, for all $n$, $\nabla(e_n)$ can be seen as a $GL_2\times \mathbb{S}_n$-module. In this talk we will put forth a partial formula for the irreducible bicharacters of these modules. Namely we will write subsets of the Schröder paths formulas as products of Schur functions in the variables $q$ and $t$ and the usual Schur functions in the variables $X=\{x_1, x_2, \ldots\}$.


text:
# statsmodels.stats.diagnostic.linear_rainbow¶

statsmodels.stats.diagnostic.linear_rainbow(res, frac=0.5)[source]

Rainbow test for linearity

The Null hypothesis is that the regression is correctly modelled as linear. The alternative for which the power might be large are convex, check

Parameters: res (Result instance) – fstat (float) – test statistic based of F test pvalue (float) – pvalue of the test


text:
# Degree of field extention $\mathbb{F}_{p^q} / \mathbb{F}_p$

Let $p, q$ be prime numbers.

What is the degree of field extension of $\mathbb{F}_{p\ ^q} / \mathbb{F}_p$

I have no idea on this but if I had to guess I would say $q$ ?

Any help or insight is deeply appreciated.

• Iteratively construct a basis to show that $\mathbb{F}_{p^q}$ is a $q$-dimensional $\mathbb{F}_p$-vector space. – reuns Sep 7 '17 at 5:05
• How many elements are there in an $n$-dimensional vector space over $\Bbb{F}_p$? – Jyrki Lahtonen Sep 7 '17 at 5:55


text:
# Representation of simple groups

Let $G$ be a finite simple group, prove that $G$ does not have a non-trivial represention of degree $1$.

Remark: I intend to prove by contradiction. Let $V$ be a non-trivial $\mathbb{C}G$-module of $G$ with dimension $1$, let $H:=\{x\in G: vx=v, \forall v\in V \}$. Observe that $H$ is normal in $G$ and $H \neq G$ due to $V$ is not trivial, so $H=\{1\}$ which implies $V$ is faithful. But how does the existence of a faithful representation of degree $1$ lead to a contradiction?

Note: This problem arose when I tackle the other problem :"Let $G$ be a group of order 20. Prove that $G$ has a non-trivial representation of degree 1, and hence deduce that $G$ can not be simple." In which to prove $G$ is not simple I think it is equivalent to prove the above assertion.

-
The problem should say "non-Abelian simple group". Hint: Show that a group with a faithful 1-dimensional representation of degree 1 is Abelian. –  Geoff Robinson Oct 24 '12 at 6:24
I see now, thanks! –  user31899 Oct 24 '12 at 6:45

Let $G$ be a finite simple group. Let $f\colon G \rightarrow \mathbb{C}^\times$ be a non-trivial homomorphism. Since $G$ is simple, $Ker(f) = 1$. Hence $G$ is isomorphic to a finite subgroup of $\mathbb{C}^\times$. Hence $G$ is a finite cyclic group of a prime order.


text:
Find the Lcm and Hcf of the Following Pairs of Integers and Verify that Lcm × Hcf = Product of the Integers: 510 and 92 - Mathematics

Numerical

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = Product of the integers:

510 and 92

Solution

(ii) 510 and 92

Let us first find the factors of 510 and 92

510=2 x 3 x 5 x 17

92 = 2^2xx23

L.C.M of 510 and 92=2^2xx3xx5xx23xx17

L.C.M of 510 and 92 = 23460

H.C.F of 510 and 92 = 2

We know that,

L.C.M x H.C.F = First number x Second Number

23460 x 2 = 510 x 92

46920 = 46920

Hence verified

Concept: Fundamental Theorem of Arithmetic
Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 10 Maths
Chapter 1 Real Numbers
Exercise 1.4 | Q 1.2 | Page 39
Share


text:
# Where am i going wrong here?

• Oct 10th 2008, 11:33 AM
Flexible
Where am i going wrong here?
Where am i going wrong? I'm trying to find x using the cosine rule from:

http://www.mathhelpforum.com/math-he...ccbeee97-1.gif

so i've got:

$7^2 = (x+2)^2 + 4^2 - 2\times{(x+2)}\times{4}\times(\frac{\sqrt2}{2})$

$49 = x^2 + 4x + 4 + 16 - 8(x+2)\times(\frac{\sqrt2}{2})$

$49 = x^2 + 4x + 20 - 4(x+2)\times(\sqrt2)$

$x^2 + 4x - 29 - 4(\sqrt2)x - 8(\sqrt2)$

$x^2 - 4(\sqrt2-1)x - (29+8\sqrt2) = 0$


text:
# 2645. Extended RSA

RSA is a well-known algorithm for public-key encryption. It uses a number which is the product of two primes as a public-key. Usually the two primes are selected randomly, but Dr. Imo found that the number can contain more information by choosing particular primes. Of course, Dr. Imo is intelligent enough to demonstrate that the public-key of the new method is as marvelously strong as that of the older method.

The Imo method is simple. Seek the pair of primes if you are given one prime and two number.

The two primes and the two number meet the following requirements.

• There are two primes a, b and two number c, d (c < 2d).
• ab ≡ c (mod 2d). ((a * b) mod (2 ** d) = c)

Thus, you can generate an RSA public-key ab which has information about c. For example, if a is 2, b is 13, c is 10 and d is 4, the remainder of division of a key 26(= ab) by 24(= 2d) includes information 10(= c).

One prime b and two numbers c, d are given. You need to find the minimum prime a to assist him.

### 输入格式

The input is a sequence of datasets followed by a line containing three zeroes separated by a space.

A dataset is a line containing three integers b, c and d separated by a space. You may assume that b is a prime, 2 ≤ b < 221, 0 < c < 2d ≤ 220.

CAUTION!

Although the sample input contains only small datasets, note that a × b is larger than 231.

### 输出格式

The output should be composed of lines, each corresponding to an input dataset. An output line should contain an integer indicating a prime a that meets ab ≡ c (mod 2d). You may assume that it is less than 221.

The output must be written to standard output.

### 样例

Input
13 10 4
19 7 5
0 0 0

Output
2
29

38 人解决，45 人已尝试。

61 份提交通过，共有 228 份提交。

4.1 EMB 奖励。


text:
linalg(deprecated)/issimilar - Help

linalg(deprecated)

 issimilar
 determine similarity of matrices

 Calling Sequence issimilar(A, B) issimilar(A, B, 'P')

Parameters

 A - square matrix B - square matrix 'P' - (optional) assigned a transformation matrix

Description

 • Important: The linalg package has been deprecated. Use the superseding command, LinearAlgebra[IsSimilar], instead.
 - For information on migrating linalg code to the new packages, see examples/LinearAlgebraMigration.
 • The function issimilar(A, B) returns true if A is similar to B and false otherwise.
 • If called in the form issimilar(A, B,'P'), then P will be assigned a transformation matrix such that $A=\mathrm{inverse}\left(P\right)BP$.
 • This function is part of the linalg package, and so can be used in the form issimilar(..) only after performing the command with(linalg), or with(linalg,issimilar).  This function can always be accessed in the long form linalg[issimilar](..).

Examples

Important: The linalg package has been deprecated. Use the superseding command, LinearAlgebra[IsSimilar], instead.

 > $\mathrm{with}\left(\mathrm{linalg}\right):$
 > $A≔\mathrm{matrix}\left(3,3,\left[1,2,3,4,5,6,7,8,9\right]\right)$
 ${A}{≔}\left[\begin{array}{ccc}{1}& {2}& {3}\\ {4}& {5}& {6}\\ {7}& {8}& {9}\end{array}\right]$ (1)
 > $B≔\mathrm{diag}\left(\mathrm{eigenvalues}\left(A\right)\right)$
 ${B}{≔}\left[\begin{array}{ccc}{0}& {0}& {0}\\ {0}& \frac{{15}}{{2}}{+}\frac{{3}{}\sqrt{{33}}}{{2}}& {0}\\ {0}& {0}& \frac{{15}}{{2}}{-}\frac{{3}{}\sqrt{{33}}}{{2}}\end{array}\right]$ (2)
 > $\mathrm{issimilar}\left(A,B,P\right)$
 ${\mathrm{true}}$ (3)
 > $\mathrm{print}\left(P\right)$
 $\left[\begin{array}{ccc}\frac{{1}}{{6}}& {-}\frac{{1}}{{3}}& \frac{{1}}{{6}}\\ {-}\frac{{5}}{{12}}{+}\frac{{7}{}\sqrt{{33}}}{{132}}& {-}\frac{{1}}{{6}}{+}\frac{\sqrt{{33}}}{{198}}& \frac{{1}}{{12}}{-}\frac{{17}{}\sqrt{{33}}}{{396}}\\ \frac{{5}}{{12}}{+}\frac{{7}{}\sqrt{{33}}}{{132}}& \frac{{1}}{{6}}{+}\frac{\sqrt{{33}}}{{198}}& {-}\frac{{1}}{{12}}{-}\frac{{17}{}\sqrt{{33}}}{{396}}\end{array}\right]$ (4)
 > $\mathrm{map}\left(\mathrm{normal},\mathrm{evalm}\left(\left({P}^{-1}&*B\right)&*P\right)\right)$
 $\left[\begin{array}{ccc}{1}& {2}& {3}\\ {4}& {5}& {6}\\ {7}& {8}& {9}\end{array}\right]$ (5)


text:
open pdf to specific page via macro

Hi, I am attempting to convert a VBA macro from Excel to Libreoffice. The function of the macro is fairly simple- it picks up data from the current cell and, using that information, opens a PDF file then advances to a specific page number. However, the code to open the PDF and move to specific page uses windows features to walk through open windows and the code doesn't work for Libreoffice. I've done a bit of research and experimentation but have yet to find any good references or examples to open the PDF to a specific page. Below is the VBA code I am attempting to migrate - any suggested references or examples appreciated.

' Input Variables: ' ~~~~~~~~~~~~~~~~ ' sFile : Fully qualified path, filename & extension of the PDF file ' lPage : Page number to view ' ' Usage: ' ~~~~~~ ' Call ChangePDFPage("C:\Users\Daniel\Documents\AccessTechniques.pdf", 65) ' ' Revision History: ' Rev Date(yyyy/mm/dd) Description ' **************************** ' 1 2018-11-25 Initial Release '--------------------------------------------------------------------------------------- Public Function ChangePDFPage(sFile As String, lPage As Long)

Dim hwndParent&
Dim hwndChild&
Dim sFileName             As String

'Dim sTempWindowText As String
'Dim sTempRetVal As Long

On Error GoTo Error_Handler

sFileName = Right(sFile, Len(sFile) - InStrRev(sFile, "\")) 'Get just the filename from the full path/filename/ext

hwndParent = FnFindWindowLike("*" & sFileName & "*")

If hwndParent = 0 Then
'This means that we can't find an instance of Adobe with the specified File, so we could use the traditional approach for the first time
Call OpenPDF(sFile, lPage)

Else
PostMessage hwndParent, WM_CLOSE, 0, ByVal 0&
Call OpenPDF(sFile, lPage)
GoTo skipchild

hwndChild = GetWindow(hwndParent, GW_CHILD)        'Find the 1st Child window handle - Zoom %
hwndChild = GetWindow(hwndChild, GW_HWNDNEXT)        'Find the next window handle - Page No.

'an use PostMessage instead if wanted
Call SendMessage(hwndChild, WM_SETTEXT, 0&, lPage)        'Enter our page number into the Adobe Reader Page number textbox
Call SendMessage(hwndChild, WM_KEYDOWN, VK_RETURN, 0) 'Submit an Enter key to actually switch to the input page
Call SendMessage(hwndChild, WM_KEYUP, VK_RETURN, 0)

skipchild:

End If

Error_Handler_Exit: On Error Resume Next Exit Function

Error_Handler: MsgBox "The following error has occured" & vbCrLf & vbCrLf & _ "Error Number: " & Err.Number & vbCrLf & _ "Error Source: ChangePDFPage" & vbCrLf & _ "Error Description: " & Err.Description & _ Switch(Erl = 0, "", Erl <> 0, vbCrLf & "Line No: " & Erl) _ , vbOKOnly + vbCritical, "An Error has Occured!" Resume Error_Handler_Exit End Function

'Purpose : Enumerates all the child windows and returns the handle of a child window with ' a matching caption or class name. 'Inputs : lParentHwnd The handle to a parent window ' [sClassName] The class name to search for. ' [sCaption] The caption to search for. ' [bSearchChildWindows] Searchs within each child window for a matching window 'Outputs : Returns the handle of child window or zero if not found 'Source : https://visualbasic.happycodings.com/... Public Function FindChildWindow(ByVal lParentHwnd As Long, Optional ByVal sClassName As String, Optional ByVal sCaption As String, Optional bSearchChildWindows As Boolean = False) As Long Dim bClassMatches As Boolean, bCaptionMatches As Boolean Const clMaxName As Long = 255 Const GW_CHILD = 5 Const GW_HWNDNEXT ...

edit retag close merge delete

Just try open Reader with key /A "page=<pagenum>"

( 2021-01-26 08:45:52 +0200 )edit

JohnSUN - thanks. I went to the referenced link and it appears to be how to use a command line for Acrobat. My need is to automate starting acrobat from a CALC sheet. Have you seen any references on how to run a command line such a acrobat from a CALC macro?

( 2021-01-26 13:46:00 +0200 )edit

Sort by » oldest newest most voted

This is written a little shorter than you wrote in your question - just use Shell:

Sub getParamsAndRunAcrobat(oEvent As Variant)
Dim oData As Variant
oData = ThisComponent.getCurrentController().getActiveSheet().getCellRangeByName("B1:B2").getDataArray()
Call OpenPDF(oData(0)(0), oData(1)(0))
End Sub

Sub OpenPDF(sFile As String, lPage As Long)
Shell (ConvertToURL(PATH_TO_READER), 3, " /A ""page=" + lPage + """ """ + sFile + """")
End Sub

Demo file - C:\fakepath\ShellAcrobat.ods

more

JohnSUN - PERFECT! Now the fun starts to get the rest of the VBA macro converted. THANKS!

( 2021-01-26 23:30:11 +0200 )edit

The solution for Windows using acrobat reader proposed by johnSun, whom I thank for the idea, can be extended to linux-ubuntu with the standard document viewer "evince"

Sub Main
xEvince ("/ media /...../ test.pdf", 2)
End Sub

sub xEvince (sFileFullPath, iPageNumber)
Shell ("evince --page-label =" + cstr (iPageNumber) + "  " + sFileFullPath)
end sub

The sub xEvince opens the pdf file requested on the iPageNumber page passed as a parameter

I used as a test a pdf file created with libreoffice writer containing TOC, bookmarks and single hyperlinks to local bookmarks by activating "tagged pdf" and "export bookmark" in the export form. The pdf exported from writer maintains the TOC with relative automatic links to the headings of the original document and preserves both the bookmarks and the hyperlink functionality

While with 'evince' it is possible to open a pdf in correspondence of a page, I have not found any command to open the pdf in correspondence of a bookmark

However, I believe that it is possible to bypass the problem, in the case of pdf exported by writer By scanning the writer document with a macro to search for bookmarks, it is possible obtain the list of pairs (bookmark-page number) to be used to open the pdf each time in correspondence with the page containing the bookmark

Even when the TOC and hyperlinks imported from libreoffice work in the pdf, and therefore the imported bookmarks work, the bookmarks are not always also visible from evince in the side panel

When it appears, using the command from the terminal:

gvfs-info -a "metadata :: evince :: bookmarks" fileFullPath.ext

all pairs (page number-bookmark name) detected by 'evince' are displayed with the format:

metadata :: evince :: bookmarks: [(0, 'bookName0'), (2, 'bookName1')]

Page numbering in this list is 0 based (page 1 corresponds to 0)

more


text:
# cv::SolvePnP with CV_ITERATIVE under OpenCV 3.3

Hi all, I have a problem with the function cv::SolvePnP. The same function works PERFECTLY under OpenCV 3.1 with three points (markers). This is the call:

cv::solvePnP(Markers3DCoord_world,ImagePoints,Mat(IntrinsicMatx),Mat(Distortionparams),rot_vec_l,vec_tras_l,true,cv::CV_ITERATIVE); Markes3DCoord_world are std::vector of Point3f, whereas ImagePoints are std::vector of Point2f

I know that the problem per se would be ambigous with three points (P3P problem), but using the Iterative optimization of the Levenberg-Marquardt optimization algorithm (flags set to true and method: CV_ITERATIVE) if I use a good initial guess I obtain the solution that I am seeking under OpenCV 3.1. Now the problem is that the same method on OpenCV 3.3 seems to crash. Why is that? If i use the same method under OpenCV 3.3 but with a higher n° of points = 12 ( as in the case of a chessboard tracking) it works perfectly. In this case the call is: flag_err=solvePnP(Markers3DCoord_world,ImagePoints,Mat(IntrinsicMatx),Mat(Distortionparams),rot_vec_l,vec_tras_l,false,cv::CV_ITERATIVE);

What's the matter with solvePnP under OpenCV 3.3?

Thanks

edit retag close merge delete

In OpenCV 3.3, the minimum number of points is 4 for solvePnP (it has been changed, there is an assert).

With an initial guess with the iterative method, what is the minimum number of points? With a good initial pose, does it make sense mathematically to use one point / two points / three points?

( 2017-11-07 07:52:53 -0500 )edit

@Eduardo Thanks for your reply. With the iterative method 3 points are enough to converge to a local minimum. As for the second question, if you have more than one camera you might be in the condition of properly solving a standard absolute orientation problem ( link text which can provides you a good initial guess for the subsequent exterior orientation problem or PnP problem link text

( 2017-11-08 05:05:25 -0500 )edit

@fab_cut If you have built OpenCV from source, it should be easy to change the assert and add the case 3 points with an initial guess with the iterative method (the file is in modules/calib3d/src/solvepnp.cpp).

I think the following changes can be made for the next OpenCV release: accept 3 points if there is an initial guess and the iterative method is used, otherwise 4 points are required for solvePnP.

Thanks for the report.

( 2017-11-09 07:46:23 -0500 )edit


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## Background

Suppose we are given $L_\infty$-algebras $(g,Q)$ and $(g',Q')$ and an $L_\infty$-morphism $F$ from $(g,Q)$ to $(g',Q')$. Furthermore, we have a Maurer-Cartan element $\pi$ of $(g,Q)$. One can twist $(g,Q)$ with the Maurer-Cartan element $\pi$ and obtains a new $L_\infty$-algebra that we call $(g,Q_\pi)$. Furthermore, we can construct a Maurer-Cartan element $\pi'$ of $(g',Q')$ by the formula

$\pi' = \sum_{n=1}^\infty \frac{1}{n!} F_n(\pi, \ldots , \pi)$,

where $F_n$ is the $\bigwedge^n g \rightarrow g'$-part of $F$. I don't know whether there is a (better) term, so I call $\pi$ and $\pi'$ associated Maurer-Cartan elements

One can twist the morphism $F$ with the Maurer-Cartan elements $\pi$ and $\pi'$ and obtain an $L_\infty$-morphism $F_\pi$ from $(g,Q_\pi)$ to $(g',Q'_{\pi'})$. The references I found are Dolgushev: A Proof of Tsygan's Formality Conjecture for an Arbitrary Smooth Manifold (section 2.4) and Yekutieli: Continuous and Twisted L_infinity Morphisms (section 3).

## Question

Given $L_\infty$-algebras $(g,Q)$ and $(g',Q')$, an $L_\infty$-morphism $F$ from $(g,Q)$ to $(g',Q')$, a Maurer-Cartan elements $\pi$ of $(g,Q)$ and a Maurer-Cartan element $\omega$, $\omega\neq \pi'$, of $(g',Q')$. Can one construct an $L_\infty$-morphisms between $(g,Q)$ twisted with the Maurer-Cartan element $\pi$ and $(g',Q')$ twisted with the Maurer-Cartan element $\omega$, where the Maurer-Cartan elements are not "associated"? I.e. can one construct an $L_\infty$-morphism between $(g,Q_\pi)$ to $(g',Q'_\omega)$?

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 If I understand your question correctly, my guess is that the answer is yes. Maurer-Cartan elements are to be thought of as deformations of the "underlying formal manifold". $Q$ is a "vector field" on this formal manifold. "Twisting" on either side should just correspond to deforming the vector field... But I'm not sure. Again this is my guess. – Kevin Lin Apr 26 2010 at 14:33 Unfortunately, I know too little about the formal manifold language of L-infinity algebras, so I did not get any intuition from there. Do you or anyone know a good introductory reference for this point of view? I only know the Kontsevich letter. (Maybe I should make this a separate question?) – C. Jost Apr 26 2010 at 16:35

Here's an idea for simplifying the problem a bit. I don't know the technical details or whether it will work, but it wouldn't fit in the comment box, so I'll put it as an answer, even though it's incomplete.

First, answer this simplified question: If $\alpha$ is an MC-element of $(h,P)$, is there a morphism $(h,P_\alpha) \to (h,P)$ and a morphism $(h,P)\to (h,P_\alpha)$? (This corresponds to $\pi$ or $\omega$ being zero and $g=g'$, $Q=Q'$ in the original question.) Then compose the morphisms

$(g,Q_\pi) \to (g,Q) \to (g',Q') \to (g',Q'_\omega)$,

where the first arrow comes from $h=g$, $P=Q_\pi$, and $\alpha = -\pi$, the second is the given morphism, and the third comes from $h=g'$, $P = Q'$, and $\alpha = \omega$.

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 Hm, it has not helped me yet. But thanks anyway, Peter! – C. Jost Apr 27 2010 at 8:43


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# Topological Quantum Field Theory Seminar

### Poincaré Duality as Duality of Categories

We give a construction that associates a small category $\mathcal{C}(X)$ to a CW-decomposition $X$ of a manifold. We obtain interesting families of finite categories from spheres and projective spaces as examples. Under some conditions this category $\mathcal{C}(X)$ seems to represent the homotopy type of $X$. Interestingly, for finite dimensional $X$ the Poincaré dual $\hat{X}$ has associated to it the opposite category $(\mathcal{C}(X))^{\rm{op}}=\mathcal{C}(\hat{X})$.

This is part of a joint project with Benjamin Heredia.

Please note the new date of 19th October - this seminar was originally scheduled for 12th October.


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Next, use the information you know to work out the average acceleration. It is denoted by symbol a and is articulated as-The S.I unit for acceleration is meter per second square or m/s 2. You do have the initial angular velocity; it is given as 32 rad/s. For example a car traveling at 50 km/hr starts to accelerate, 10 seconds after, its speed changes to 100 km/hr then the acceleration of the car during the time can be calculated as below: initial speed = 50 km/hr final speed = 100 km/hr Velocity is not exactly the same as speed. T ( f) is the final time and t ( i) is the initial time. The acceleration due to gravity at the surface of Earth is represented as "g" and has a standard value of 9.80665 m/s2. I would appreciate it if somebody could help me show the average acceleration of the flea during take off. Physics calculator to solve for velocity given initial, constant acceleration and time with constant acceleration This allows you to measure how fast velocity changes in meters per second squared (m/s^2). where a is acceleration, v is the final velocity of the object, u is the initial velocity of the object and t is the time that has elapsed. v 2 = u 2 + 2 a s for a particle undergoing constant acceleration. Change in time mostly means, the difference in time from time 0 to the final time recorded. Acceleration is the rate of change of velocity of an object. Hence it is denoted by â a. Want to improve this question? Under acceleration (a) is defined as a physical quantity that characterizes the change in velocity of a body in the period of time during which a body changes its location in space. After that, you will get the acceleration value. 15.04.2020BillRecommendations. The first method involves the Speed Difference. The average angular acceleration is the change in the angular velocity, divided by the change in time. Where, Dividing distance by time twice is the same as dividing distance by the square of time. Acceleration is change in velocity. If t (time taken), v (final velocity) and u (initial velocity) are provided. Median response time is 34 minutes and may be longer for new subjects. The corresponding graph of acceleration versus time is found from the slope of velocity and is shown in Figure $$\PageIndex{6}$$(b). Subtract the final angular velocity from the initial angular velocity to get the change in angular velocity. If s represents the displacement of the object, then: s = â¦ How do you find angular acceleration? Rate of change in position, or speed, is equal to distance traveled divided by time. Calculation of Distance from Acceleration –. An object is thrown up, released from a height of 1m, and reaches a maximum height of 1.5m. The unit of angular acceleration is radians/s 2. $= 1025.57 ms^{-2}$, For a particle moving linearly, in three dimensions in a straight line, with constant acceleration, you can use the following equation, In case you are wondering where the answer comes from: $$a=\frac{dv}{dt}=\frac{ds}{dt}\frac{dv}{ds}=v\frac{dv}{ds}$$ which gives $$ads=vdv$$ integrating gives $$v^2=v_0^2 + 2\int_{s_0}^{s}ads$$ using the fundamental theorem of calculus then $$v^2=v_0^2+2\bar{a}(s-s_0)$$, 2021 Stack Exchange, Inc. user contributions under cc by-sa, Have you tried $v^2 = v_0^2 + 2a(r_0 - r)$. The only reason I've accepted Chris's answer is that it is more complete, https://physics.stackexchange.com/questions/79645/calculate-average-acceleration-without-time/82971#82971, Calculate average acceleration without time? You can calculate the acceleration of an object from its change in velocity and the time taken. On the other hand A = 2S/T^2 is the acceleration when S is the distance and T is the time to cover that distance. Based on the above definition, this value is easy enough to find. Science Forums is a good one If. Velocity has a direction as well as a speed. t is the time is the time of travel. Then the acceleration is given by the formula. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Formula of Acceleration. No that it makes any significant difference in the value of $a$... @User58220, yes you're right, I put the wrong number in the formula, i'll edit now. Givens: Vi (initial velocity) = 1200 m/s. No, the acceleration due to gravity is constant. Note initial vel. $v^2 = u^2 + 2as$ for a particle undergoing constant acceleration. https://physics.stackexchange.com/questions/79645/calculate-average-acceleration-without-time/79693#79693. I know that to calculate average acceleration you can use $\frac{dv}{dt}$, however I only have the following information: At the moment the flea's leg leave the surface its body is raised 0.44 Find the initial and final time for the period being considered. Deceleration also is known as negative acceleration. Acceleration Due To Gravity Formula: g = G*M/R2 Follow the below tutorial which guides on how to calculate acceleration due to gravity. $= \frac{0.95^2}{2\times0.00044}$ Your acceleration is 26.6 meters per second 2, and your final speed is 146.3 meters per second. Then it asks me to find acceleration and velocity at the maximum height. How do you find time without acceleration? Velocity is distance travelled divided by time. $a = \frac{v^2-u^2}{2s}$ mm and it is moving at a speed of 0.95$ms^{-1}$. You can write this as a formula like this: a av = (Îv/Ît), where delta represents change. v is the final velocity. To solve for time, divide the distance traveled by the rate. How To Find Acceleration Without Time? Physics: That Equation, finding final velocity without time –. For example, if Cole drives his car 45 km per hour and travels a total of 225 km, then he traveled for 225/45 = 5 hours. v ( f) â v ( i) t ( f) â t ( i) In this acceleration equation, v ( f) is the final velocity while is the v ( i) initial velocity. This equation can be rearranged to give: v = u + at. Great. $u$ represents initial velocity - in this case 0 Shouldn't $u$ stay at $u=0$? Acceleration is the rate of change in velocity to the change in time. This graph is depicted in Figure $$\PageIndex{6}$$(a), which is a straight line. The definition of acceleration Acceleration is the rate of change of velocity as a function of time. Time-velocity graph of a particle is shown in figure. v u = â(v i 2 + (2 x Acceleration x Distance(m))) Where, v u = Final Velocity without Time v i = Initial Velocity It is the final velocity minus the initial velocity, with a negative sign in the result because the velocity is decreasing. Plugging in the numbers, you get the following: Okay, the acceleration is approximately 27 meters per second 2. average acceleration formula without time: find the magnitude of acceleration: acceleration formula using mass and force: speed time acceleration formula: instant acceleration formula: newtonâs second law of motion calculator: equation for velocity time graph: initial angular acceleration formula: Find the initial and final angular velocity in radians/s. The constant acceleration equations use the following symbols: a stands for acceleration, v means final velocity, u means starting velocity, s means displacement (i.e. Acceleration is also a vector quantity, so it includes both magnitude and direction. Given how far it travels in a given time, calculate the acceleration of a car undergoing constant acceleration. You can rearrange this equation with a little algebra to solve for acceleration; just divide both sides by t 2 and multiply by 2 to get. Example: Loise just bought a new car which goes from 0 to 50 m/s in just 5 seconds. In a physics equation, given a constant acceleration and the change in velocity of an object, you can figure out both the time involved and the distance traveled. @ProgrammingEnthusiast Oh my gosh, nope. I think that you need to find a better website, that one was rather elementary. Acceleration can be positive (for example, start the train from the platform) and negative (train starts braking at the destination). To find average acceleration, start by remembering that acceleration means how quickly something is speeding up or slowing down. In this case pf a varying acceleration, this formula can be used to calculate the "average" acceleration, which represents the total change in velocity over the total change in time. We have a question that is confusing me. We cannot find value of acceleration without knowing time. $s$ represents displacement - in this case 0.44mm, or in SI units, 0.00044m. In my case, I know the vehicle is going at 20 mph and it comes to a full stop after 12 meters. I have a question about a flea jumping, for which I need to show that the average acceleration is around 1000 $ms^{-2}$. Time can be measured with clocks. For instance, imagine youâre a drag racer. Example 4. Thus the SI unit of acceleration is the meter per second squared . Acceleration (a) is the change in velocity (Îv) over the change in time (Ît), represented by the equation a = Îv/Ît. Now find the total distance traveled. usually to find acceleration you need time, but its not in this equation. The angular acceleration is a vector that points in a direction along the rotation axis. Calculating acceleration involves dividing velocity by time â or in terms of SI units, dividing the meter per second [m/s] by the second [s]. The answer hasn't changed however. Providing the object does not build up too much speed and air resistance is not a factor the acceleration measured should be the same. The angular acceleration is â 0.7 rad/ s 2, it is negative because the gyro is slowing. What is the formula for acceleration? Would you like to add that as an answer for me, so I can accept it? In this case pf a varying acceleration, this formula can be used to calculate the "average" acceleration, which represents the total change in velocity over the total change in time. Don't I need a time component (how long it takes to reach maximum height) to find acceleration â¦ Find its instantaneous acceleration at following intervals (i) at t = 3s (ii) at t = 6s (iii) at t = 9s Solution: (i) Instantaneous acceleration at t = 3s, is given by a = slope of line AB = zero (ii) Instantaneous acceleration at t = 6 s, is given by a = slope of line BC The formula for acceleration expressed in terms of the initial velocity (speed), final velocity and the acceleration duration (time) is: where a is the acceleration, v0 is the starting velocity, v1 is the final velocity, and t is the time (acceleration duration or t 1 - t 0). https://physics.stackexchange.com/questions/79645/calculate-average-acceleration-without-time/79649#79649, +1 Thank you very much for your answer. The second method involves the Distance Traveled. [closed]. Acceleration due to gravity can be used as an example of constant acceleration, but problems often specify when the acceleration continues at a constant rate. where Î v is the change in velocity and Î t is the change in time. Vf (final velocity) = 0 m/s. You can also write the acceleration equation like this: a =. If starting velocity, final velocity and time taken are given, Deceleration Formula is given by. *Response times vary by subject and question complexity. Calculate the acceleration of a bullet if it travels at a speed of 1200 m/s and stops within a bulletproof vest that is 1.0 cm thick. Subtract â¦ Whatâs that in more understandable terms? First, a simple example is shown using Figure 3.3.4(b), the velocityversus-time graph of Example 3.3, to find acceleration graphically. u is the initial velocity. Average Acceleration is a vector quantity that measures the rate at which an object changes with respect to velocity. I knew it would be something simple like that. Update the question so it's on-topic for Physics Stack Exchange. So to find the stopping time you have to solve 32 â 0.7 t â¦ Can I calculate the acceleration? The magnitude of the angular acceleration is given by the formula below. Acceleration (a) is the change in velocity (Îv) over the change in time (Ît), represented by the equation a = Îv/Ît. $v$ represents final velocity - in this case 0.95m/s Formula to calculate average acceleration. Calculate the acceleration of the car. Here, enter the values of the Initial Speed, Final Speed, and Time then choose the unit of measurement from the drop-down menu. s is the distance travelled. Change in velocity is the difference between the initial velocity and the final velocity. For the period being considered have the initial velocity ) = 1200.... Time mostly means, the acceleration measured should be the same be longer for subjects... Write this as a speed: v = u + at a undergoing! Can calculate the acceleration when s is the rate of change of velocity an! \ ( \PageIndex { 6 } \ ) ( a ), (. It would be something simple like that speeding up or slowing down ( m/s^2 ) not in this.. Starting velocity, with a negative sign in the numbers, you get the acceleration due to gravity constant. Information you know to work out the average acceleration, start by remembering that acceleration means quickly! Stack Exchange g = g * M/R2 how do you find time without acceleration t is the time.... And Î t is the distance and t is the change in velocity and Î t is the angular. Taken ), v ( final velocity minus the initial velocity, with a sign! Time twice is the rate of change in velocity is the rate result because the gyro is slowing be to., which is a straight line v^2 = u^2 + 2as $for a particle constant! You do have the initial angular velocity from the initial and final time for period! T is the rate of change in time from time 0 to 50 m/s in 5. Like to add that as an answer for me, so i can accept it not in this equation be... Your acceleration is given by 146.3 meters per second squared change of velocity as a speed ), where represents! Speed and air resistance is not a factor the acceleration value the SI unit of acceleration acceleration is the of! From time 0 to 50 m/s in just 5 seconds also a vector that points in a given time calculate. A particle undergoing constant acceleration gravity at the maximum height articulated as-The S.I unit acceleration! Guides on how to calculate acceleration due to how to find acceleration without time Formula: g = g * M/R2 how you... A full stop after 12 meters ( initial velocity ) are provided not. Average acceleration of an object from its change in angular velocity in radians/s difference between the initial.... Velocity of an object is thrown up, released from a height of 1.5m angular velocity in.! Going at 20 mph and it comes to a full stop after meters. Show the average acceleration a direction as well as a function of time ) and u ( initial and... Up or slowing down change in angular velocity in radians/s unit of without!, i know the vehicle is going at 20 mph and it comes to full... Acceleration you need time, but its not in this equation can be rearranged to give: v u... Velocity ; it is negative because the velocity is decreasing up, released from a height of 1.5m both and... Too much speed and air resistance is not a factor the acceleration value as a function of.. Direction as well as how to find acceleration without time speed also write the acceleration of an object is thrown up, released from height. A ), v ( final velocity minus the initial velocity ) are provided undergoing constant acceleration height of,! = u 2 + 2 a s for a particle undergoing constant acceleration,:! Final time recorded it would be something simple like that the numbers you... And time taken ), v ( final velocity ) = 1200 m/s is... 9.80665 m/s2 2 = u 2 + 2 a s for a particle constant. How fast velocity changes in meters per second 2, it is given by the displacement of angular! Acceleration, start by remembering that acceleration means how quickly something is speeding up slowing... Is decreasing m/s 2 a negative sign in the result because the velocity is decreasing stay$! Me, so i can accept it ) are provided easy enough to find acceleration you need time, its. The definition of acceleration without knowing time up too much speed and air resistance is not factor! A maximum height the magnitude of the angular acceleration is approximately 27 meters per second ) are provided know... Response times vary by subject and question complexity released from a height 1m! Calculate acceleration due to gravity thus the SI unit of acceleration acceleration is the time is rate... Something is speeding up or slowing down of 9.80665 m/s2 if s represents the displacement of flea! And your final speed is 146.3 meters per second 2, and reaches a maximum height 1m! Velocity, with a negative sign in the result because the gyro is.... And it comes to a full stop after 12 meters acceleration measured should be the same full! The angular acceleration is given by the square of time being considered a av = ( ). Other hand a = 2S/T^2 is the change in velocity and Î is! Straight line Îv/Ît ), where delta represents change } \ ) ( ). Standard value of 9.80665 m/s2 delta represents change could help me show the average acceleration the... Includes both magnitude and direction # 79649, +1 Thank you very much for your answer the! Î t is the change in time vehicle is going at 20 mph and it comes to a full after. Sign in the numbers, you get the following: Okay, the acceleration of an object from its in... Going at 20 mph and it comes to a full stop after 12 meters is! Earth is represented as  g '' and has a standard value acceleration... By subject and question complexity height of 1m, and your final speed is 146.3 meters second! That distance, calculate the acceleration measured should be the same your final speed is 146.3 meters second! Final speed is 146.3 meters per second along the rotation axis u=0 \$ without knowing time as an answer me. Build up too much speed and air resistance is not a factor acceleration! To add that as an answer for me, so it 's on-topic for physics Exchange! From a height of 1m, and your final speed is 146.3 meters per second and (. 20 mph and it comes to a full stop after 12 meters u at. Equation like this: a av = ( Îv/Ît ), v ( final velocity of time acceleration velocity... Or slowing down it is denoted by symbol a and is articulated as-The S.I unit for acceleration â.

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# Source code for cameo.exceptions

# Copyright 2014 Novo Nordisk Foundation Center for Biosustainability, DTU.
#
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
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# Unless required by applicable law or agreed to in writing, software


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Integral elements of a non-archimedean completion of a number field

Let $k$ be a number field, $v$ a discrete non-archimedean valuation on $k$. Let $k_v$ be the completion of $k$ wrt $v$ and $\mathcal O_v$ its valuation ring.

My question is:

If $x\in k_v$ then is there an integer $m\in \mathbb{Z}$ (not merely $x\in \mathbb{Z}_p$) such that $mx\in \mathcal{O}_v?$

Hint: $v$ corresponds to a prime ${\frak p}$ of $k$. Then ${\frak p}$ lies above $(p)$ for some rational prime $p$. Thus we have an integer with positive valuation...
• I see, so for a large enough positive integer $n$, $v_\mathfrak{p}(p^nx) \ge 0$, hence $p^nx\in \mathcal{O}_v$. Why must a prime ideal of $\mathbb{Z}_p$ be of the form $p\mathbb{Z}_p$ for $p$ a rational prime number? – Mark Rodriguez Aug 18 '13 at 16:57
• @MarkRodriguez While that's true, I didn't say that in my answer. Is that a separate question you wanted to ask? | At any rate, here's a hint: argue that the minimal (normalized) valuation of an element in a prime ideal of ${\Bbb Z}_p$ must be $1$. – anon Aug 18 '13 at 17:04
• Sorry I momentarily forgot that $\mathfrak{p}$ was a prime ideal of the ring of integers of $k$ (and not of $\mathcal{O}_v$). Please ignore my question. Regards – Mark Rodriguez Aug 18 '13 at 17:11


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## Sunday, March 18, 2012

### Smoothing splines and interest rate curves

Yield curves are important in Economics and used by finance professionals to analyze bonds and look for trading opportunities and by economists, to try to understand economic conditions.

The yield is the amount in cash that returns to the owners of a security, for example, if a sovereign bond with maturity at time $T$ is bought at time $t$ for a price $P(t,T)$, the yield would be high if $P(t,T)$ is much less than the maturity price $P(T,T)$ (which would obviously be the risk-free amount received by the lender). A higher yield allows the owner to make more profit from the investment, but it may also mean that the security is being sold at a discount as a result of some risk related to the security. For a recent example of this, Greek bonds, which were sold at such discounts because the market anticipated a default, this made yields soar, with risk-averting investors unloading bonds at the lower prices that risk-taking investors were comfortable buying in. Ultimately everybody took a haircut.

For fixed income mathematics, a set of axioms must be assumed, namely:
1. The market trades continuously over its trading horizon: it extends from the current time to some distant future time such that the maturities of all the instruments to be valued fall between now and the trading horizon..
2. The market is efficient: information is available to all traders simultaneously, and every trader makes use of all the available information, also related to the axiom that
3. There are no arbitrage opportunities: the price of a portfolio is the sum of its constituent parts.
4. The market is complete: any desired cash flow can be obtained from a suitable self-financing strategy based on a portfolio of discount bonds.

It is assumed further that there are no legal barriers to trading and that the market is rational (traders try to maximize their profits).

Now, define the zero yield in terms of the bond price above. The zero yield, as seen at time $t$, of a bond that matures at time $T$, is denoted by $y(t,T)$ and is defined by
$$P(t,T)=e^{-(T-t) y(t,T) }$$
for every $t<T$, which means that the current price of the bond is equal to discounted price at maturity (using compound interest).

Suppose that at time $t$ we enter into a forward contract to deliver at time $T_1$ a bond that will mature at time $T_2$ (obviously $T_2>T_1$). Let the forward price of the bond be denoted by $P(t,T_1,T_2)$. At the same time, a bond that matures at time $T_1$ is purchased at a price of $P(t,T_1)$. Again at time $t$, a bond that matures at time $T_2$ is bought with a price of $P(t,T_2)$. Notice that the axiom specifying that there are no arbitrage opportunities implies that the price of the bond maturing at time $T_2$ must be equal to the product of the price of the bond maturing at time $T_1$ and the forward price $P(t,T_1) P(t,T_1,T_2)$. Let the implied forward rate $f(t,T_1,T_2)$, valued at time $t$, for the period $(T_1,T_2)$ as
$$P(t,T_1,T_2)=e^{-(T_2-T_1) f(t,T_1,T_2) }$$
Notice that $P(t,T_2) =P(t,T_1) P(t,T_1,T_2)=e^{-(T_1-t) y(t,T_1) } e^{-(T_2-T_1) f(t,T_1,T_2) }$ and, summing the exponents, we get
$$f(t,T_1,T_2)= \frac{(T_2-t) y(t,T_2) - (T_1 - t) y(t,T_1)}{T_2 - T_1}$$
which is the period forward rate. However, the instantaneous forward rate is of much greater importance in the theory of the term structure. The instantaneous forward rate for time $T$, as seen at time $t$, is denoted by $f(t,T)$ and is the continuously compounded rate defined by
$$f(t,T)= \lim_{h \rightarrow 0} {f(t,T,T+h) } = y(t,T) + (T-t) y_T (t,T)$$
where $y_T = \frac{\partial}{\partial T} y(t,T)$. We can interpret the previous equation as the current  yield value plus the instantaneous change of the yield with the maturity time.

To interpolate curves, and thus have values for all the points within the interval and not only the data points that are made available to us by the problem, we can use a number of different methods. One needs some complexity to be able to capture what the nature is saying, but at the same time this complexity might be cause by some "noise", or perturbations that we want to avoid (market panics, manipulations or poorly registered data). We can choose a parametric family, such as polynomials, fix the order and fit the coefficients. However, this appears as somehow arbitrary. One can choose to interpolate with a non-parametric method such as splines. When using some functional space, one must restrict himself to functions that meet some criteria, not only that approximate well (or maybe the best approximation in that space). This is to avoid overfitting the said noise.

Now we can choose to fit an approximating yield or the forward curve that approximates the true yield or forward curves, with the information given by actual bond prices as of time $t$, by minimizing the following functional (assuming we fit the yield curve $\varphi$):
$$R(\varphi,t) = \sum_{i=1}^n (P(t,T_i) - P(t,T_i,\varphi))^2 + \lambda \int \left( \frac{\partial^2}{\partial s^2} \varphi(s) \right)^2 ds.$$
Where $\lambda$ is a regularization parameter, and $P(t,T_i,\varphi)$ is a functional that prices a bond of maturity $T_i$ at time $t$ with the yield curve $\varphi$. We change formulation, for practical derivation purposes (we now compute the observed yield $y(t,T_i)$ from the observed bond price)
$$R(\varphi,t) = \sum_{i=1}^n (y(t,T_i) - \varphi(T_i))^2 + \lambda \int \left( \frac{\partial^2}{\partial s^2} \varphi(s) \right)^2 ds.$$
And then we apply the standard smoothing splines theory to compute the solution, which is:
$$\mathbf{w} = (I+\lambda K)^{-1} \mathbf{y}$$
Where $\mathbf{y}$ is a vector with elements $y(t,T_i)$, $K$ is a matrix whose elements are of the form $\int \frac{\partial^2}{\partial s^2} \phi_i (s) \frac{\partial^2}{\partial s^2} \phi_j (s) ds$, with $\phi$ is a choosen spline basis (notice the dot-product structure) and $\mathbf{w}$ is a vector with the coefficients on that basis, so that the estimated function is
$$\varphi = \sum_i w_i \phi_i$$


text:
# Question

Formatted question description: https://leetcode.ca/all/269.html

 269	Alien Dictionary

There is a new alien language which uses the latin alphabet.
However, the order among letters are unknown to you.

You receive a list of words from the dictionary, where words are sorted lexicographically
by the rules of this new language.

Derive the order of letters in this language.

For example,

Given the following words in dictionary,
[
"wrt",
"wrf",
"er",
"ett",
"rftt"
]

The correct order is: "wertf".

Input:
[
"z",
"x",
"z"
]

Output: ""

Explanation: The order is invalid, so return "".

Note:
You may assume all letters are in lowercase.
If the order is invalid, return an empty string.
There may be multiple valid order of letters, return any one of them is fine.

@tag-graph

# Algorithm

It is actually a problem of directed graph traversal.

For each node in the directed graph, calculate its in-degree, and then start BFS traversing the directed graph from the node with an in-degree of 0, and then save the traversal path and return.

Java


text:
# How to find isomorphic group of the quotient $\mathbb{Z}\times \mathbb{Z} / \langle (m,n) \rangle$?

It's easy when $$\langle(m,n)\rangle=\langle(1,2)\rangle$$

Just define the homomorphism $$f\colon \mathbb{Z}\times \mathbb{Z} \to \mathbb{Z}$$ by $$f(a,b)=2a-b$$

Then $$f$$ onto

And, $$kerf=\langle (1,2) \rangle$$

So, $$\mathbb{Z}\times \mathbb{Z} / \langle (1,2) \rangle \cong \mathbb{Z}$$

if $$\langle (m,n) \rangle$$( where $$m,n$$ prime to each other)

Then , $$\mathbb{Z}\times \mathbb{Z} / \langle (m,n) \rangle \cong \mathbb{Z}$$

What if $$\langle (m,n) \rangle = \langle (2,2) \rangle$$

What if $$\langle (m,n) \rangle = \langle (2,4) \rangle$$

And others arbitrarily....

• In general this problem can be approached with Smith normal forms. A manual way to find eg $\Bbb Z \times \Bbb Z / \langle (2, 2) \rangle$ is to define $\phi: \Bbb Z \times \Bbb Z \to \Bbb Z \times \Bbb Z$ by $\phi(a, b) = (a - b, a)$. Show that $\phi$ is an isomorphism and calculate the image of this subgroup under $\phi$. You should find that this makes the quotient group easier to identify! Mar 18 at 4:04
• Here in your case,,you have shown $\mathbb{Z}\times \mathbb{Z}$ is isomorphic to itself,,,but ask for $\mathbb{Z}\times \mathbb{Z} / \langle(2,2)\rangle$ Mar 18 at 4:14
• The idea is that if $\phi$ is an isomorphism, then $G/N \equiv \phi(G) / \phi(N)$, and in this case $\phi(N)$ is much easier to work with. This is exactly the same as what reuns' lovely answer below is doing, except they are representing the isomorphism by a matrix. Mar 18 at 4:26
• @Izaak van Dongen Ohh, thanks. Just assure me here ,, $\mathbb{Z}\times \mathbb{Z} / \langle (2,2) \rangle$ is isomorphic to $\mathbb{Z}×\mathbb{Z}_{2}$.......and hence $\mathbb{Z}\times \mathbb{Z} / \langle (m,n) \rangle$ is isomorphic to $\mathbb{Z}×\mathbb{Z}_{GCD(m,n)}$.... Mar 18 at 4:47

$$(m,n)=l(a,b)$$ with $$\gcd(a,b)=1$$. Then $$ad-bc=1$$ and $$\pmatrix{a&b\\c&d}$$ has an integer matrix inverse $$U$$ (can you find it?).
$$\Bbb{Z^2}/\Bbb{Z}(m,n)\cong \Bbb{Z^2}U/\Bbb{Z}(m,n)U=\Bbb{Z}^2/\Bbb{Z}(l,0)$$
• Which part? ${}{}$ Mar 18 at 4:44
• Just assure me here ,, $\mathbb{Z}\times \mathbb{Z} / \langle (2,2) \rangle$ is isomorphic to $\mathbb{Z}×\mathbb{Z}_{2}$.......and hence $\mathbb{Z}\times \mathbb{Z} / \langle (m,n) \rangle$ is isomorphic to $\mathbb{Z}×\mathbb{Z}_{GCD(m,n)}$. Mar 18 at 4:48


text:
Elementary math that professors cant solve

Njorl
StonedPanda said:

[(9-sqrt(9))!]/[(sqrt(9)!)^2]

the square root and the square kind of mess it up, but it's still pretty damn sweet

That equals 20.

sqrt(9)=3
9-sqrt(9)=6
6!=720

6^2=36

720/36=20

Njorl

Njorl
1 1 1 = 6................(1+1+1)!
2 2 2 = 6................2+2+2
3 3 3 = 6................3x3-3
4 4 4 = 6................(4!/4)x40
5 5 5 = 6................5+5/5
6 6 6 = 6................6+6-6
7 7 7 = 6................7-7/7
8 8 8 = 6................(8-81/3)x80
9 9 9 = 6................9-9/(91/2)

Njorl

How come I never thought of those answers that Njorl did. Hmmm, something is wrong with my brain. lol!

Njorl said:
1 1 1 = 6................(1+1+1)!
2 2 2 = 6................2+2+2
3 3 3 = 6................3x3-3
4 4 4 = 6................(4!/4)x40
5 5 5 = 6................5+5/5
6 6 6 = 6................6+6-6
7 7 7 = 6................7-7/7
8 8 8 = 6................(8-81/3)x80
9 9 9 = 6................9-9/(91/2)

Njorl

what does 8^0 mean?
what does is equal?

and where do you get 9-9/9^1/2
doesnt 9-9=0, then 0/9^1/2=0?

x^0 is equal to 1 for all real (and complex) x.

9 - 9/9^(1/2) is interpreted as $$9 - \frac{9}{9^{1/2}}$$. If he had meant $$\frac{9 - 9}{9^{1/2}}$$, he would have written (9 - 9)/9^(1/2). The parantheses are important :P

ExecNight
How bout this..

Get -1 using 0,0,0

Oh and this is mathematically possible without using any tricks..

Last edited by a moderator:
plover
Homework Helper
ExecNight said:
Get -1 using 0,0,0

0 + 0 - 0!

Or to get 6:

(0! + 0! + 0!)!

futb0l
$$0 - 0^0$$
is this qualified?

futb0l
$$-\cos{0} - 0 + 0$$

and mm...

$$- ( \sin ^2 0 + \cos ^2 0 ) + 0$$

Zurtex
Homework Helper
futb0l said:
$$0 - 0^0$$
is this qualified?
Strictly speaking $0^0$ is not defined. As:

$$x^0 = \left( x^1 \right) \left( x^{-1} \right)$$

Therefore:

$$x^0 = \frac{x}{x}$$

Which means $x^0 = 1$ when $x \neq 0$

plover
Homework Helper
Or for those whose tastes run to notation sadism:

$$- \lim_{0\rightarrow0} 0!$$

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futb0l
Zurtex said:
Strictly speaking $0^0$ is not defined. As:

$$x^0 = \left( x^1 \right) \left( x^{-1} \right)$$

Therefore:

$$x^0 = \frac{x}{x}$$

Which means $x^0 = 1$ when $x \neq 0$

mmm.. if you do 0^1 in google, it will come up as 1.
and ... http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

so i don't think
$x^0 = 1$ when $x \neq 0$
is true.

futb0l
there should be a rule that says when any number is to the power of 0 it will be equal to 1.

ExecNight
Most of em are true solutions..

Now the funny thing here is we are getting something from noting..

How come we can get 1 from 0 by using only 0? That always makes my head iching...

futb0l
NoNose
Hi
First: Sorry if my english isn´t correct or couldn´t be understand, but i´m trying to.

Without to resume the discussion, if factorial and bases are elemental math, i´m thinking i´ve found a solution for the problem down this text for all positive and negative real numbers and the 0:
$$\left( \left( x^2 \right) ^0 + \left( x^2 \right) ^0 + \left( x^2 \right) ^0 \right) ! =6$$

Gunni said:
There's another fun variation on this theme where you line up all the numbers from one to nine in threes and are supposed to make them add up to six by adding only plus, minus, division, multiplication, root and power signs (whole powers and roots, no logs!). You can also use ( and ) (forgot what they're called).

Like this:
Code:
1   1   1 = 6
2   2   2 = 6
3   3   3 = 6
4   4   4 = 6
5   5   5 = 6
6   6   6 = 6
7   7   7 = 6
8   8   8 = 6
9   9   9 = 6

For example (I hope I'm not ruining anything for anyone here ):
6 + 6 - 6 = 6

Have fun.

Zurtex
Homework Helper
NoNose said:
Hi
First: Sorry if my english isn´t correct or couldn´t be understand, but i´m trying to.

Without to resume the discussion, if factorial and bases are elemental math, i´m thinking i´ve found a solution for the problem down this text for all positive and negative real numbers and the 0:
$$\left( \left( x^2 \right) ^0 + \left( x^2 \right) ^0 + \left( x^2 \right) ^0 \right) ! =6$$

I like that , but using the square function is kind of using a 2 really (where as the square root actually has a symbol). So perhaps before anyone complains about this it could be easily fixed as:

$$\left( |x|^0 + |x|^0 + |x|^0 \right) ! = 6$$

For $x \neq 0$

Njorl said:
10 (44-4)/4

I had to use one "44". Is there a way to get 10 without resorting to this?

Njorl

4 * 4 - 4! / 4 works

arildno said:
$$9+\frac{9}{9}=20_{(base 5)}$$

I rarely do math for fun so my so I dont know much math indepth. From what I know base 2 or binary numbers are like this

01 this means that 0*(2^1)+1*(2^0)=1

So in base 5 it would be

0,1,2,3,4 = 0*(5^4)+1*(5^3)+2*(5^2)+3*(5^1)+4*(5^0)=194

how do you get 9 in base 5? Is it a different base system or something?

edit--------------

Oh I think I know

9 =14
14/14=1
14+1=20
right?

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