text:
• #### Extending the Internet Explorer context menu

• 46 Comments

In a comment, Darrell Norton asked for a "View in Mozilla" option for Internet Explorer.

You can already do this.

Internet Explorer's context menu extension mechanism has been in MSDN for years. Let me show you how you can create this extension yourself.

First, create the following registry key:

[HKEY_CURRENT_USER\Software\Microsoft\Internet Explorer\MenuExt\View in Mozilla]
@=REG_SZ:"C:\some\path\to\ViewInMozilla.htm"
Contexts=REG_DWORD:1

Of course, you need to change C:\some\path\to to an actual path.

How did I know to do this? Because steps 1, 2 and 3 in the "Implementation Steps" section tell me (1) what key to create, (2) what to set the default value to, and (3) what to set Contexts to. I chose a Context value of 1, which means "Default".

Okay, now to write the script ViewInMozilla.htm. Well, the documentation says that I can access context from the menuArguments property of the external object. So let's start with that.

<SCRIPT>
alert(external.menuArguments);
</SCRIPT>

Okay, let's run this puppy. Launch IE, right-click on a blank space in the web page, select "View in Mozilla", and you get...

    [object]

Woo-hoo! This is a major accomplishment: Something happened at all. Doing things in small steps makes it easy to identify where a problem is. If we had run full steam ahead to completion and then it didn't work, we wouldn't have known whether it was due to a bug in the script, a bad registration, a bad filename...

Now that I have the menu arguments, I can use that to suck out information about the item that the context menu applies to. Let's try this:

<SCRIPT>
alert(external.menuArguments.document.URL);
</SCRIPT>

Woo-hoo, now it gives me the URL. Almost there. All that's left to do is to run a program with that URL as the command line parameter.

<SCRIPT>
var shell = new ActiveXObject("WScript.Shell");
shell.run("mozilla \"" + external.menuArguments.document.URL + "\"");
</SCRIPT>

Mission accomplished.

Now you too can create Internet Explorer context menu extensions.

In fact, go ahead and do it, since Darrell asked for it: Create an Internet Explorer context menu extension that operates on anchors and opens the linked-to page in Mozilla.

(Bonus: Tony Schreiner cooked up something similar for zooming.)

• #### Callback, the safety newsletter for the aviation community

• 9 Comments

Yes the "S" in NASA stands for "Space", but don't forget that the "A" stands for "Aeronautics". One of the programs that I (amateur aviation wannabe) find fascinating is the Aviation Safety Reporting System, where pilots can submit anonymous reports of "stupid things I have done" in order to teach other pilots "Don't do what I did."

• #### Do you know when your destructors run? Part 2.

• 33 Comments

Continuing from yesterday, here's another case where you have to watch your destructors. Yesterday's theme was destructors that run at the wrong time. Today, we're going to see destructors that don't run at all!

Assume there's an ObjectLock class which takes a lock in its constructor and releases it in its destructor.

DWORD ThreadProc(LPVOID p)
{
... do stuff ...
ObjectLock lock(p);
... do stuff ...
return 0;
}

Pretty standard stuff. The first batch of stuff is done without the lock, and the second batch is done inside the lock. When the function returns, the lock is automatically released.

But suppose somebody adds a little code to this function like this:

DWORD ThreadProc(LPVOID p)
{
... do stuff ...
ObjectLock lock(p);
...
if (p->cancelled) ExitThread(1);
...
return 0;
}

The code change was just to add an early exit if the object was cancelled.

But when does that ObjectLock destructor run?

It runs at the return statement, since that's when the lock goes out of scope. In particular, it is not run before you call ExitThread.

Result: You left an object locked permanently.

You can imagine how variations on this code could lead to resource leaks or other problems.

• #### You can do anything at zombo.com, anything at all

• 15 Comments

Zombo has been around for many years, and yet I still find it hilarious (requires Flash).

I just went back to check, and alas the introduction actually ends. But fortunately, they made it even cooler by having a text-only version. (Still requires sound.)

• #### Do you know when your destructors run? Part 1.

• 10 Comments

Larry Osterman discussed the importance of knowing when your global destructors run, but this problem is not exclusive to global objects. You need to take care even with local objects. Consider:

void Sample()
{
if (SUCCEEDED(CoInitialize(NULL))) {
CComPtr<IXMLDOMDocument> p;
if (SUCCEEDED(p.CoCreateInstance(CLSID_IXMLDOMDocument))) {
...
}
CoUninitialize();
}
}

Easy as pie. And there's a bug here.

When does the destructor for that smart-pointer run?

Answer: When the object goes out of scope, which is at the closing brace of the outer if statement, after the CoUninitialize call.

So you shut down COM, and then try to access a pointer to a COM object. This is not good. (Or as Larry describes it, "Blam!")

To fix this problem, you have to release all your COM pointers before the CoUninitialize. One way would be to insert a p.Release() at the end of the inner if. (But of course, if you're going to do that, then why bother using a smart pointer?)

Another fix would be to introduce a seemingly unnecessary scope:

void Sample()
{
if (SUCCEEDED(CoInitialize(NULL))) {
{
CComPtr<IXMLDOMDocument> p;
if (SUCCEEDED(p.CoCreateInstance(CLSID_IXMLDOMDocument))) {
...
}
} // ensure p is destructed before the CoUninit
CoUninitialize();
}
}

Make sure you leave that comment there or the next person to come across this code is going to "clean it up" by removing the "redundant" braces.

Of course, this is still too subtle. Here's another solution: Put the CoUninitialize inside a destructor of its own!

class CCoInitialize {
public:
CCoInitialize() : m_hr(CoInitialize(NULL)) { }
~CCoInitialize() { if (SUCCEEDED(m_hr)) CoUninitialize(); }
operator HRESULT() const { return m_hr; }
HRESULT m_hr;
};

void Sample()
{
CCoInitialize init;
if (SUCCEEDED(init)) {
CComPtr<IXMLDOMDocument> p;
if (SUCCEEDED(p.CoCreateInstance(CLSID_IXMLDOMDocument))) {
...
}
}
} // CoUninitialize happens here

This works even if you put the smart pointer at the same scope, as long as you put it after the CCoInitialize object:

void Sample()
{
CCoInitialize init;
CComPtr<IXMLDOMDocument> p;
if (SUCCEEDED(init) &&
SUCCEEDED(p.CoCreateInstance(CLSID_IXMLDOMDocument))) {
...
}
}

This works because objects with automatic storage duration are destructed in reverse order of declaration, so the object p wil be destructed first, then the object init.

Mind you, this is basically subtle no matter now you slice it. Nobody said programming was easy.

Tomorrow, part 2.

• #### Coffee machine or assault weapon?

• 12 Comments

Coffee Machine Assault Weapon
AK-47   X
M-90 Automatic X
KF550-BK X
XM15-E2S   X
• #### Beware the hash reset attack

• 18 Comments

There are a variety of message digest algorithms out there, MD5 being a particularly popular one. These generate a "message digest" (essentially, a hash) so you can detect whether somebody has tampered with a file, the theory being that it's hard to tamper with a file without changing its hash.

But make sure you record the file size as well as the digest.

Not that collisions are necessarily easy to create by mistake. (I've heard a rumor that the deployment team has seen an MD5 collision, but it's just a rumor. I have no evidence. Heck, maybe what really happened was that somebody on the deployment got their MR2 into a car accident...)

Anyway, the possibility of a "reset attack" makes collisions trivial to create.

Hash generators typically operate on a stream. The hash engine maintains some state. The file to be hashed is broken up into chunks, and each chunk is combined with the engine's state variables in some complex way. When you have passed all the data through the engine, you push a button on the engine and out pops the hash value (which is typically a copy of the state variables, or possibly a subset of them).

Now suppose somebody came up with a way of "resetting" the engine; that is, returning it to the initial state. Here's how they can make any document match your digest:

First, create an alternate message and send it through the hash engine.

Next, generate the bytes necessary to "reset" the engine.

Finally, append the original message.

In other words, the fake file looks like this:

[alternate message][garbage][original message]

where "garbage" is the reset.

This fake file has the same hash as the original message, since the "garbage" resets the hash engine to the initial state, at which point the replay of the original message regenerates the hash.

Result: A file with the same hash as the original, but with different content!

In a proper attack, of course, the "alternate message" would be crafted so the garbage and original mesage would be ignored. You might end it with a marker that means "Ignore everything after this point." (For HTML, you can just say <NOFRAMES> and everything after that point will be largely ignored by all modern browsers.) Many other file types encode the expected file length in the header, in which case you can append whatever garbage you want without having any effect.

But if you also store the file size with the hash, then the reset attack fails, because a reset attack always generates a file bigger than the original. To create a collision, they would have to create a shorter alternate message than the original, and then fiddle with the extra bytes to get the desired target hash to come out. This is significantly harder than just resetting.

(I'm not aware of anybody who has successfully been able to reset MD5, mind you. This is a protective measure: If somebody figures out how to reset MD5, a small bit of work on your side will prevent you from falling victim.)

• #### Letterboxing: Geocaching without the GPS

• 4 Comments

Geocaching is sort of like global hide-and-seek (with hints) for technology geeks.

I stumbled upon a non-GPS version of this diversion, known as Letterboxing (not to be confused with letterboxed movies, another subject entirely).

By the way, if geocaching isn't enough "fun with a GPS" for you, you can try The Degree Confluence Project. See the world with a GPS, experience other cultures, learn a few words of the local language:

The Thais kept asking how many farang (foreigners) did this and how much they got paid. I must have heard the phrase "farang Baar", (foreigners are crazy) more than a couple of times that evening.
• #### String sorting is not done by ASCII code any more.

• 31 Comments

Just because you have the ASCII table memorized doesn't mean you know how sorting works.

I remember a bug filed where somebody said that the "sort" command was sorting underscores incorrectly:

AAA__
AAAAA

this was claimed to be wrong "because underscore character follow uppercase letters and precedes lowercase letters".

Well perhaps it does if you think ASCII sorting is the way sorting should be.

ASCII sorting is so last century.

The CompareString function describes two types of sorting, "word sort" and "string sort". And both of them sort punctuation before letters.

• #### The coolest thing you can do with a Gameboy is not "play a game on it"

• 5 Comments

From Public Radio International, The Next Big Thing covers The Gameboy Music Match.

These days, a lot of electronic music is performed directly off of a laptop, which most people complain looks a lot like someone up there on stage checking their email. But Gameboy music isn't just for listening. It's also a performance. Nullsleep plays with both thumbs tapping, both feet moving, and head swaying. Okay, yeah, it looks like he's playing a video game. But a really great video game.

Read the episode rundown or listen directly to the fourth story, Hot GameBoy Music Club. (The story was re-run a year later under the much catchier title New Old Sounds.)

Page 377 of 415 (4,146 items) «375376377378379»


text:
# Height and Distance Question & Answers

## Heights and Distances

Quantitative aptitude questions are asked in many competitive exams and placement exam.'Height and Distance' is a category in Quantitative Aptitude. Quantitative aptitude questions given here are extremely useful for all kind of competitive exams like Common Aptitude Test (CAT), MAT, GMAT, IBPS Exam, CSAT, CLAT, Bank Competitive Exams, ICET, UPSC Competitive Exams, CLAT, SSC Competitive Exams, SNAP Test, KPSC, XAT, GRE, Defense Competitive Exams, L.I.C/ G. I.C Competitive Exams, Railway Competitive Exam, TNPSC, University Grants Commission (UGC), Career Aptitude Test (IT Companies) and etc., Government Exams etc.

We have a large database of problems on "Height and Distance "answered with explanation. These will help students who are preparing for all types of competitive examinations.

If an object travels at five feet per second, how many feet does it travel in one hour?

 A) 30 B) 3000 C) 18 D) 1800

Explanation:

If an object travels at 5 feet per second it covers 5x60 feet in one minute, and 5x60x60 feet in one hour.

$\inline \fn_cm \therefore$Answer = 1800

Subject: Height and Distance - Quantitative Aptitude - Arithmetic Ability
Exam Prep: GRE

6

A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30º with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60º. What is the distance between the base of the tower and the point P?

 A) Data inadequate B) 8 units C) 12 units D) None of these

Explanation:

One of AB, AD and CD must have given.

Subject: Height and Distance - Quantitative Aptitude - Arithmetic Ability

5

An observer 1.6 m tall is $\inline \fn_cm 20\sqrt{3}$ away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The heights of the tower is:

 A) 21.6 m B) 23.2 m C) 24.72 m D) None of these

Explanation:

Draw BE ${\color{Black}&space;\bot}$ CD

Then, CE = AB = 1.6 m,

BE = AC = ${\color{Black}&space;20\sqrt{3}m}$

${\color{Black}\frac{DE}{BE}=tan30^{0}=\frac{1}{\sqrt{3}}}$

${\color{Black}\Rightarrow&space;DE=\frac{20\sqrt{3}}{\sqrt{3}}m=20m}$

${\color{Black}\therefore&space;}$ CD = CE + DE = (1.6 + 20) m = 21.6 m.

Subject: Height and Distance - Quantitative Aptitude - Arithmetic Ability

6

Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is:

 A) 173 m B) 200 m C) 273 m D) 300 m

Explanation:

Let AB be the lighthouse and C and D be the positions of the ships.

${\color{Black}&space;Then,AB=100m,&space;\angle&space;ACB=30^{\circ}&space;and&space;\angle&space;ADB=45^{\circ}}$

${\color{Black}&space;\frac{AB}{AC}=\tan&space;30^{\circ}=\frac{1}{\sqrt{3}}\Rightarrow&space;AC=AB\times&space;\sqrt{3}=100\sqrt{3}m}$

${\color{Black}&space;\frac{AB}{AD}=\tan&space;45^{\circ}=1\Rightarrow&space;AD=AB=100&space;m}$

${\color{Black}&space;\therefore&space;CD=(AC+AD)=(100\sqrt{3}+100)m=100(\sqrt{3}+1)=100\times&space;2.73=273&space;m}$

Subject: Height and Distance - Quantitative Aptitude - Arithmetic Ability

3

Jack takes 20 minutes to jog around the race course one time, and 25 minutes to jog around a second time. What is his average speed in miles per hour for the whole jog if the course is 3 miles long?

 A) 6 B) 8 C) 9 D) 10

Explanation:

Average speed = total distance / total time

Total distance covered = 6 miles; total time = 45 minutes = 0.75 hours

Average speed = 6/ 0.75 = 8 miles/hour

format: plain


text:
## Characterize the elements of ZZ/(n)

Prove that the distinct equivalence classes in $\mathbb{Z}/(n)$ are precisely $\overline{0}, \overline{1}, \overline{2}, \ldots, \overline{n-1}$. Use the Division Algorithm.

Clearly $\overline{k} \in \mathbb{Z}/(n)$ for each $0 \leq k < n$. Now let $\overline{a} \in \mathbb{Z}/(n)$. By the Division Algorithm, we have $a = qn + r$ for some $q,r$ with $0 \leq |r| < n$. Thus $\overline{a} = \overline{qn+r} = \overline{qn} + \overline{r} = \overline{r}$.

Finally, note that the $\overline{0}, \overline{1}, \overline{2}, \ldots, \overline{n-1}$ are distinct as follows. Suppose $0 \leq r_1, r_2 < n$ with $r_1 \neq r_2$ and $\overline{r_1} = \overline{r_2}$. Then $n|r_2 - r_1$, so $nk = r_2 - r_1$ for some $k$, giving $r_2 = kn + r_1$. But we also have $r_2 = 0n + r_2$, and both $|r_1|$ and $|r_2|$ are less than $n$. So by the uniqueness part of the division algorithm we have $(k,r_1) = (0,r_2)$, hence $r_1 = r_2$, a contradiction.

Thus the distinct equivalence classes in $\mathbb{Z}/(n)$ are precisely $\overline{0}, \overline{1}, \overline{2}, \ldots, \overline{n-1}$. $\blacksquare$


text:
How to understand quadratic finance or practice of Value-at -Risk(VaR)

We define the following notions for a jointly normally distributed random vector $P=(P_1,...,P_n)$ with f the density function.

$$\mu=\int_{-\infty}^{\infty}(x_i-\mu_i)f_i(x_i)dx_i$$

$$\sigma^2_{ij}=\int_{-\infty}^{\infty}(x_i-\mu_i)(x_j-\mu_j)f_{ij}(x_i,x_j)dx_idx_j$$ where the density function of n-dimensional jointly normally distributed random vector is given by

$$f(x)=\frac{1}{(2\pi)^\frac{n}{2}\sqrt{\det{V}}}\exp\left(-\frac12 \left\langle x-\mu,V^{-1}(x-\mu)\right\rangle\right)$$

and where $V={(\sigma^2_{ij})}_{1\leq{i,j}\leq{n}}$ is the variance-covariance matrix.

• I don't understand the density function of a jointly jointly normally distributed random vector.
• To understand all of these above terms, is the knowledge of multi-variate normal distribution necessary?

If any quant-finance expert could explain me the above terms, it would be helpful to me in understanding quadratic finance.

• What is the meaning of $\langle{x-\mu,V^{-1}(x-\mu)\rangle}$ – Dhamnekar Winod Jun 6 '16 at 14:51
• Short answer is yes, at least some basic knowledge of multivariate statistics would be necessary to understand the notion of the $n$-dimensional normal distribution. By the way, it seems to me there are some minor typos in your question, e.g. the $\mu_i$ on the rhs of the first equation. – BerndH Jun 6 '16 at 15:25


text:
# How would you have answered Pauls question, "why do you think I oughtn't to?"

its a real aloud question.


text:
# The total head to be developed by a centrifugal pump is expected to be up to 50 m. The normal ratio of radii of impeller rim and impeller eye of 2 is maintained. The design is for a speed of 1300 rpm. What is the nominal diameter of impeller? Take √g = 3.13 and $$\frac{1}{\pi } = 0.318$$

Free Practice With Testbook Mock Tests

## Options:

1. 53 cm

2. 57 cm

3. 60 cm

4. 64 cm

### Correct Answer: Option 1 (Solution Below)

This question was previously asked in

ESE Civil 2016 Paper 1: Official Paper

## Solution:

Given,

$$\frac{{Radii\;of\;rim}}{{Radii\;of\;eye}} = 2\;or\;\frac{{{D_{rim}}}}{{{D_{eye}}}} = 2$$

For starting of the pump,

And Hm is given by

$${H_m} = \left( {\frac{{u_{rim}^2 - u_{eye}^2}}{{2g}}} \right)$$

Where urim and ueye are speeds of impeller rim and eye:

Now,

Min available Head = 50 m (given)

$$50 \ge {\left( {\frac{{\pi \;{D_{rim}}\;N}}{{60}}} \right)^2} - {\left( {\frac{{\pi \;{D_{eye}}\;N}}{{60}}} \right)^2}$$

$$50 \times 2 \times 9.81 \ge \frac{{{\pi ^2}{N^2}}}{{{{\left( {60} \right)}^2}}}\;\left[ {{{\left( {2\;{D_{eye}}} \right)}^2} - {{\left( {{D_{eye}}} \right)}^2}} \right]$$

$$\frac{{50 \times 2 \times 9.81 \times {{\left( {60} \right)}^2}}}{{{\pi ^2} \times {{\left( {1300} \right)}^2}}} \ge 3\;D_{eye}^2$$

⇒ Deye = 0.265 m

$$\frac{{{D_{rim}}}}{{{D_{eye}}}} = 2 \Rightarrow {D_{rim}} = 0.53\;m\;or\;53\;cm$$

∴ Nominal diameter = 53 cm


text:
## Banach Journal of Mathematical Analysis

### Disjointness preserving linear operators between Banach algebras of vector-valued functions

#### Abstract

We present vector-valued versions of two theorems due to A. Jimenez-Vargas, by showing that, if $B(X,E)$ and $B(Y,F)$ are certain vector-valued Banach algebras of continuous functions and $T:B(X,E)\to B(Y,F)$ is a separating linear operator, then $\widehat{T}:\widehat{B(X,E)}\to \widehat{B(Y,F)}$, defined by $\widehat{T}\hat{f}=\widehat{Tf}$, is a weighted composition operator, where $\widehat{Tf}$ is the Gelfand transform of $Tf$. Furthermore, it is shown that, under some conditions, every bijective separating map $T:B(X,E)\to B(Y,F)$ is biseparating and induces a homeomorphism between the character spaces $M(B(X,E))$ and $M(B(Y,F))$. In particular, a complete description of all biseparating, or disjointness preserving linear operators between certain vector-valued Lipschitz algebras is provided. In fact, under certain conditions, if the bijections $T:Lip^{\alpha}(X,E)\to Lip^{\alpha}(Y,F)$ and $T^{-1}$ are both disjointness preserving, then $T$ is a weighted composition operator in the form $Tf(y)=h(y)(f(\phi(y))),$ where $\phi$ is a homeomorphism from $Y$ onto $X$ and $h$ is a map from $Y$ into the set of all linear bijections from $E$ onto $F$. Moreover, if $T$ is multiplicative then $M(E)$ and $M(F)$ are homeomorphic.

#### Article information

Source
Banach J. Math. Anal., Volume 8, Number 2 (2014), 93-106.

Dates
First available in Project Euclid: 4 April 2014

https://projecteuclid.org/euclid.bjma/1396640054

Digital Object Identifier
doi:10.15352/bjma/1396640054

Mathematical Reviews number (MathSciNet)
MR3189541

Zentralblatt MATH identifier
1308.47047

#### Citation

Ghasemi Honary, Taher; Nikou, Azadeh; Sanatpour, Amir Hossein. Disjointness preserving linear operators between Banach algebras of vector-valued functions. Banach J. Math. Anal. 8 (2014), no. 2, 93--106. doi:10.15352/bjma/1396640054. https://projecteuclid.org/euclid.bjma/1396640054

#### References

• Y.A. Abramovich and A.K. Kitover, Inverses of disjointness preserving operators, Mem. Amer. Math. Soc., 143 (2000), no. 679, 1–162.
• Y.A. Abramovich, A. I. Veksler and A.V. Koldunov, Operators preserving disjointness, Dokl. Akad. Nauk USSR, 248 (1979), 1033–1036.
• J. Araujo and K. Jarosz, Separating maps on spaces of continuous functions, Contemp. Math. 232 (1999), 33–37.
• W. Arendt, Spectral properties of Lamperti operators, Indiana Univ. Math. J. 32 (1983), 199–215.
• E. Beckenstein and L. Narici, A nonarchimedean Stone-Banach theorem, Proc. Amer. Math. Soc. 100 (1987), 242–246.
• E. Beckenstein and L. Narici, Automatic continuity of certain linear isomorphisms, Acad. Roy. Belg. Bull. Cl. Sci. 73 (1987), no. 5, 191–200.
• E. Beckenstein and L. Narici, Automatic continuity of linear maps of spaces of continuous functions, Manuscr. Math. 62 (1988), 257–275.
• H.X. Cao, J. H. Zhang and Z.B. Xu, Characterizations and extensions of Lipschitz-$\alpha$ operators, Acta Math. Sin. (Engl. Ser.) 22 (2006), no. 3, 671–678.
• H.G. Dales, Banach Algebras and Automatic Continuity, LMS Monographs 24, Clarendon Press, Oxford, 2000.
• K. Esmaeili and H. Mahyar, Weighted composition operators between vector-valued Lipschitz function spaces, Banach J. Math. Anal. 7 (2013), no. 1, 59-72.
• J.J. Font, Automatic continuity of certain isomorphisms between regular Banach function algebras, Glasg. Math. J. 39 (1997), 333–343.
• J.J. Font and S. Hernandez, On separating maps between locally compact spaces, Arch. Math. (Basel) 63 (1994), 158–165.
• H-L. Gau, J-S. Jeang and N-C. Wong, Biseparating linear maps between continuous vector-valued function spaces, J. Aust. Math. Soc. 74 (2003), 101–109.
• M. Hosseini and F. Sady, Banach function algebras and certain polynomially norm-preserving maps, Banach J. Math. Anal. 6 (2012), no. 2, 1–18.
• K. Jarosz, Automatic continuity of separating linear isomorphisms, Canad. Math. Bull. 33 (1990), 139–144.
• A. Jimenez–Vargas, Disjointness preserving operators between little Lipschitz algebras, J. Math. Anal. Appl. 337 (2008), 984–993.
• A. Jimenez–Vargas and Ya-Shu Wang, Linear biseparating maps between vector-valued little Lipschitz function spaces, Acta Math. Sin. (Engl. Ser.) 26 (2010), no. 6, 1005–1018.
• E. Kaniuth, A Course in Commutative Banach Algebras, Springer, Graduate Texts in Mathematics 246, 2009.
• J.S. Manhas, Weighted composition operators and dynamical systems on weighted spaces of holomorphic functions on Banach spaces, Ann. Funct. Anal. 4 (2013), no. 2, 58–71.
• A. Nikou and A.G. O'Farrell, Banach algebras of vector-valued functions, Glasgow Math. J., to appear, arXiv:1305.2751.
• D.R. Sherbert, The Structure of ideals and point derivations in Banach algebras of Lipschitz functions, Trans. Amer. Math. Soc. 111 (1964), 240–272.
• B.Z. Vulkh, On linear multiplicative operations, Dokl. Akad. Nauk USSR 41 (1943), 148–151.
• B.Z. Vulkh, Multiplication in linear semi-ordered spaces and its application to the theory of operations, Mat. Sbornik 22 (1948), 267–317.


text:
# Accelerating patient recruitment: The growing role of clinical trial educators

Keywords:

Stewart H. Rosen
Vice President, Medical Affairs, QuintilesIMS

Nicola Smith
Senior Director, Head of Field-based Medical Operations, QuintilesIMS

Life sciences companies spend an average of $300,000 to recruit each participant for a Phase III clinical trial. (1) However, despite this investment, 80% of trials fail to meet their initial patient enrolment quotas. Challenges associated with the identification of subjects to participate in a study can cause serious bottlenecks and delay study completion, slowing development, delaying patient access to important therapies, and causing financial losses to sponsors as high as$8 million a day. (2)

This challenge is compounded when it comes to specialty medicines where the patient pool is small and often geographically dispersed. In cancer clinical trials, for example, it costs approximately \$50,000 to initiate a study site, yet 20% to 30% of study sites never enroll a single patient. (3) In addition, attrition rates for Phase III clinical trials, which have reached 30% to 40%, are often attributed to issues such as convenience and efficacy rather than safety.

Surprisingly, the biggest stumbling block that prevents clinical trials from reaching recruitment goals is a lack of awareness. A Harris Interactive Survey found that 85% of patients surveyed were either u ...


text:
# Trying to find $\sum\limits_{k=0}^n k \binom{n}{k}$ [duplicate]

\begin{align} &\sum_{k=0}^n k \binom{n}{k} =\\ &\sum_{k=0}^n k \frac{n!}{k!(n-k)!} =\\ &\sum_{k=0}^n k \frac{n(n-1)!}{(k-1)!((n-1)-(k-1))!} = \\ &n\sum_{k=0}^n \binom{n-1}{k-1} =\\ &n\sum_{k=0}^{n-1} \binom{n}{k} + n \binom{n-1}{-1} =\\ &n2^{n-1} + n \binom{n-1}{-1} \end{align}

1. Do I have any mistake?
2. How can I handle the last term?

(Presumptive) Source: Theoretical Exercise 1.12(a), P18, A First Course in Pr, 8th Ed, by S Ross

-

## marked as duplicate by Mike Spivey, Henry T. Horton, TMM, hardmath, Stefan HansenJan 14 '13 at 19:12

$\binom nr=0$ if $r<0$ or $r<n$ –  lab bhattacharjee Jan 14 '13 at 15:21
By convention $\binom{n}k=0$ if $k$ is a negative integer, so your last line is simply $$n\sum_{k=0}^n\binom{n-1}{k-1}=n\sum_{k=0}^{n-1}\binom{n}k=n2^{n-1}\;.$$ Everything else is fine. –  Brian M. Scott Jan 14 '13 at 15:22
Thank you, you should have put that as an answer –  Bojan Serafimov Jan 14 '13 at 15:23
Okay; I’ve made it an answer and added a longer comment that I didn’t make before. –  Brian M. Scott Jan 14 '13 at 15:31
Alternatively you could drop $k=0$ from the initial summation since its contribution is zero. –  hardmath Jan 14 '13 at 15:42

By convention $\binom{n}k=0$ if $k$ is a negative integer, so your last line is simply $$n\sum_{k=0}^n\binom{n-1}{k-1}=n\sum_{k=0}^{n-1}\binom{n}k=n2^{n-1}\;.$$ Everything else is fine.

By the way, there is also a combinatorial way to see that $k\binom{n}k=n\binom{n-1}{k-1}$: the lefthand side counts the ways to choose a $k$-person committee from a group of $n$ people and then choose one of the $k$ to be chairman; the righthand side counts the number of ways to select a chairman ($n$) and then the other $k-1$ members of the committee.

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If you know some elementary calculus, there's a simple way to do this.

By the Binomial theorem we have $$(x+1)^n=\sum_{k=0}^n \binom{n}{k}x^k$$ Differentiate both sides with respect to $x$: $$n(x+1)^{n-1}=\sum_{k=0}^n\binom{n}{k}kx^{k-1}$$ Set $x=1$ $$n2^{n-1}=\sum_{k=0}^n k\binom{n}{k}$$

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lol genious :DD –  Bojan Serafimov Jan 14 '13 at 15:49

The annoying $\binom{n-1}{-1}$ at the end can be most easily avoided by starting with the observation that $$\sum_0^nk\binom{n}{k}=\sum_1^nk\binom{n}{k}.$$

Remark: For another way of summing the series, let $X$ be the number of tosses when a fair coin is tossed $n$ times. By symmetry we have $E(X)=n/2$. But $$E(X)=\sum_0^n k\binom{n}{k}\frac{1}{2^n}.$$ The result follows immediately. A mean proof!

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+1. André, I've seen you give or refer to mean proofs a few times. I really like them. Do you know of any references or collections of mean proofs? –  Mike Spivey Jan 14 '13 at 20:46
@MikeSpivey: It is a term I may have invented, in imitation of Proofs that Really Count. Do not know of any collection. –  André Nicolas Jan 14 '13 at 20:52

As an completely different method to compute the sum (though that's not what was asked for): Note that the derivative of $$\sum_{k=0}^n {n\choose k}x^k=(1+x)^n$$ is $$\sum_{k=0}^n k {n\choose k}x^{k-1}=n(1+x)^{n-1}$$ and we want the value at $x=1$.

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What is the maximum speed of the ejected electrons? good times!

Hello everyone, i'm alittle confused on why this isn't right.
What is the maximum speed of the ejected electrons?

They give you the following:
With the help of others i found:

If the work function for a certain metal is 1.5 eV, what is its stopping potential for electrons ejected from the metal when light of wavelength 337 nm shines on the metal?
2.19 V

Andrew Mason said it best:
The stopping potential is the potential (energy / unit charge) measured in volts (joules/coulomb) that must be applied to stop the electrons from being ejected from the surface when the light is shone on it.

If the energy of the incident photon is greater than the work required to remove the electron from the surface plus the applied (-) potential, electrons will leave the surface with some kinetic energy. The stopping potential is the applied potential that makes this KE = 0.

So the stopping potential is given by:

$$q_eV_s = E_{photon} - q_e\phi$$
$$V_s = h\nu/q_e - \phi$$

where $V_s$ is the stopping potential and $\phi$ is the work function (Joules/coulomb)./QUOTE]

So i figured this formula would work,
Stopping Potential = 2.19 V
.5*mv^2 = Stopping Potential;
solve for V to find the speed.
V = sqrt(StoppingPotential/(.5*m));
V = 2.5E15 m/s

but it was wrong. ANy ideas what i'm missing? Thanks!

Last edited:

dav2008
Gold Member
eV is a measure of energy, not voltage, so in your equation $$V_s = h\nu/q_e - \phi$$ you are equating potential to potential plus energy.

Also you set voltage equal to kinetic energy. Again, voltage is energy/charge.

Last edited:
Your right, my units don't work out at all. I end up getting volts/kg hm...
I don't see how this is going to work out, are you saying I should convert my 2.9 V into somthing else?

dav2008
Gold Member
You have to realize what is going on.

Once the electron has been shot out, it takes a potential of 2.19 Joules/coulomb(if you did your calculations correctly) to stop them.

If it takes 2.19 Joules/Coulomb to stop the electron, what does that tell you about its initial kinetic energy?

dav2008
Gold Member
I guess the key thing to realize is that all this is is conservation of energy. You don't need to memorize any formulas if you just understand the situation and write a conservation of energy statement for it.

In the first case you have the energy of the photon converted into the work function energy plus the energy of the electron. Then you have the energy of the electron converted into electric potential (qV)

Oooo! i got you so it should be:
.5*mv^2 = (charge of an electron)(2.19 J/C)?


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2

First of all, some unsolicited advice: your derivation for (b) need not be so messy. It greatly simplifies things to observe that $$f(n,m,k) = f(n-1,m,k) + f(n-1,m-1,k) - f(n-k-1,m-k,k) \tag{1}$$ holds not just when $0 \le k \le m \le n$, $k < n$. In fact, it holds for any $n \ge 0, m \ge 0, k \ge 1$, except in the two cases $m = n = k$, and $m = n = ... 0 One solution is:$G(z) = k z^{\displaystyle{-(1+\ln(z)/\ln(16))}}$I discovered this by considering how the logs of both sides of the function scale. 2 $$f(x)=\frac{2}{1-2x}=2\sum_{k=0}^\infty 2^k x^k=2+2^2x+2^3x^2+\cdots$$ 2 If the contestants are all distinct, you can take account of that by changing your coefficients: Of the$5$entrants from each state, there are$\binom{5}{0}$ways to choose 0 of them,$\binom{5}{1}$ways to choose$1$,$\binom{5}{2}$ways to choose$2, and so on. So, the generating function for contestants from a single state is $$... 1 The generating function for the Central Binomial Coefficients is$$ \sum_{k=0}^\infty\binom{2k}{k}x^k=(1-4x)^{-1/2}\tag{1} $$Integrating (1) and dividing by x yields$$ \sum_{k=0}^\infty\binom{2k}{k}\frac{x^k}{k+1}=\frac{1-\sqrt{1-4x}}{2x}\tag{2} $$Sum the formula against x^m:$$ \begin{align} ... 1 Whenn=1$it's easy to evaluate for all$m\ge1$. Then it's easy to prove the$n+1$by induction, given the$n$case (for all$m$): just write $$\binom{n+k}{m+2k} = \binom{n-1+k}{m+2k} + \binom{n-1+k}{m-1+2k}.$$ 0 I don't know whats going on but the question's title is provoking laymen. Stirling numbers(1st kind)$ 1, 1+1/2, 1+1/2+1/3, 1+1/2+1/3+1/4, ..... $after computing lcm gives Stirlings in the numerators and factorial in the denominators. 1$F_0 = 0$,$F_1 = 1$, and$F_{n+2} = F_n + F_{n+1}$.$F_{n+3} = F_n + 2F_{n+1}F_{n+6} = 5F_n + 8F_{n+1} = F_n + 4F_{n+3}$Let G be a series where$G_0 = 0$,$G_1 = 2$, and$G_{n+2} = G_n + 4G_{n+1}$. In other words,$G_i = F_{3i}$. Its generating function is$g(x) = \dfrac{2x}{1 - 4x - x^2}$.$\sum_{n \ge 0}F_{3n}2^{-3n} = \sum_{n \ge 0} ...

1

Here's an approach via generating functions. As the Fibonacci recurrence is defined by $F_{n+2} = F_{n+1} + F_n$, we have $$\sum_{n \ge 0} F_{n+2}z^{n+2} = \sum_{n \ge 0} F_{n+1}z^{n+1}z + \sum_{n \ge 0}F_nz^nz^2$$ which with the generating function $G(z) = \sum_{n\ge0} F_n z^n$ gives $$G(z) - F_0 - F_1z = zG(z) - zF_0 + z^2G(z)$$ and therefore (using $F_0 = ... 3 The middle expression is the result of iterated use of the power rule. If you take the$n^\text{th}$(formal) derivative of$\frac1{1-x}$, you end up with$\frac{1}{(1-x)^{n+1}}$times some polynomial in$n$. If you then take the fifth derivative of the geometric series, then you can work out the coefficient to be$\binom{n+5}{5}$but missing some terms; ... 2 As suggested, the simplest would consist in a use of Binet formula for Fibonacci number. It write $$F_n=\frac{\varphi ^n-\psi ^n}{\sqrt{5}}$$ with$\varphi=\frac{1}{2} \left(1+\sqrt{5}\right)$and$\psi=\frac{1}{2} \left(1-\sqrt{5}\right)$. So, if we consider each term of the sum, we can rewrite it as $$F_{3n}2^{-3n}=\frac{\Phi ^n-\Psi ^n}{\sqrt{5}}$$ ... 1 For someone who may need this. \begin{equation*} \begin{split} d(x) &= (4-2x+x^2)^{-1} \\ &= \frac{1}{(\alpha -x)(\beta -x)}\\ &= \frac{A}{\alpha -x} + \frac{B}{\beta -x} \end{split} \end{equation*} We can determine that : \begin{equation*} A(\beta -x) + B(\alpha -x) = 1 \end{equation*} So, we have: \begin{equation*} A = \frac{1}{\beta - ... 1 Alf van der Poorten, wrote about this, calling it the "Hadamard product" of the two generating series. It's not easy stuff, but you can read about it in section 8 of this paper. See the example in Section 11, where $$\sum{2h\choose h}x^h=(1-4x)^{-1/2}$$ is an elementary and algebraic function, but $$\sum{2h\choose h}^2x^h$$ is a complete elliptic integral ... 1 The strategy is to look for the generating function $$G(x,y)=\sum_{n,m}F(n,m)x^ny^m.$$ To do so: Multiply by$x^ny^m$each term in the identity, getting $$F(n,m)x^ny^m = n\cdot F(n-1,m) x^ny^m+ (n-m)\cdot F(n-1,m-1)x^ny^m.$$ The LHS terms sum to the generating function$G(x,y)$. For the first terms on the RHS, use the identity $$... 1 Determining the asymptotic growth of the coefficients of a generating function is essentially the entire subject of Analytic Combinatorics, on which see the 810-page book by Flajolet and Sedgewick available online. The basic setup is this: for a large class of generating functions G(z) = g_0 + g_1z + g_2z^2 + \dots, we can show that the coefficients ... 1 By the Cauchy-Hadamard theorem, \limsup\sqrt[n]{|a_n|}=1/R. If the lim sup is in reality a lim, and even better, if the limit of the quotient formula exists, then the asymptotic growth rate is about the reciprocal of the distance of the origin to the closest singularity of f. 1 I presume that in$$G(z) = a_0 + a_1 z + a_2 z^2 + \dots,$$a_k represents the probability that a certain discrete random variable (call it X) takes the value k, for each k. For the mean E[X] defined as E[X] = a_0(0) + a_1(1) + a_2(2) + a_3(3) + \dots, we can differentiate to get$$G'(z) = a_1 + 2a_2z + 3a_3z^2 + 4a_4z^3 + \dots,$$whose value ... 1 I'm guessing your recurrence is: a_n -7a_{n-1} +10a_{n-2}=0 \longleftrightarrow a_n = 7a_{n-1}-10a_{n-2} \longleftrightarrow a_{n+2} = 7a_{n+1} -10a_n . Let's say a_0=0, a_1=1 for sake of example. Set F(x) = a_0 + a_1x + a_2x^2+\ldots = \sum_{n=0}^{\infty} a_nx^n So \frac{F(x) - a_0}{x} = a_1 + a_2x + a_3x^2+\ldots =\sum_{n=0}^{\infty} ... 2 Solution I. Introduce the sequences a_{n,k} and b_{n,k} that count the number of k element-subsets of n that contain no consecutive integers where the a_{n,k} count such sequences that do not use n and the b_{n,k} the ones that do use n. Then we clearly have$$a_{n,1} = n-1 \quad\text{and}\quad b_{n,1} = 1.$$Furthermore we have the ... 2 The combinatorial species here seems to be$$\mathcal{T} = \mathcal{Z} + \mathcal{Z}(\mathfrak{M}_1(\mathcal{T}) + \mathfrak{M}_2(\mathcal{T})).$$This is one of several interpretations of the question. The one we chose here is that the trees are not labelled and two trees that differ only in the left and right children being exchanged are considered the ... 0 The Coefficient of x^9 in \displaystyle(1+x^3+x^8)^{10}=[(1+x^3)+x^8]^{10}=(1+x^3)^{10}+\binom{10}1(1+x^3)^9x^8+\cdots+(x^8)^{10} = the Coefficient of x^9 in \displaystyle (1+x^3)^{10}+10x^8(1+x^3)^9 Now, the Coefficient of x^9 in \displaystyle(1+x^3)^{10} is \displaystyle\binom{10}3 and the Coefficient of x^9 in ... 1 Every term in the expansion of (1+x^3+x^8)^{10} is of the form 1^a(x^3)^b(x^8)^c with a+b+c = 10. It is clear that the only way to get x^9 as a term is when a = 7,b=3,c=0. And the coefficient corresponding to this term is the number of ways you can choose three x^3's from 10 i.e \binom{10}{3} 0 Hint: Solve for integer solutions to 9 = 8a + 3b + 0c subject to a+b+c = 10. 0 For sure, what lab bhattacharjee gave is the answer. However, before starting, you could have simplified your problem since$$\frac{\left(1-x^6\right)^4}{(1-x)^4} = \left(x^5+x^4+x^3+x^2+x+1\right)^4$$and the problem becomes simpler. 1 So, the coefficient of x^{12} = the coefficient of x^0 in (1-x^6)^4 \cdot the coefficient of x^{12} in (1-x)^{-4} +the coefficient of x^6 in (1-x^6)^4 \cdot the coefficient of x^{12-6} in (1-x)^{-4} +the coefficient of x^{12} in (1-x^6)^4 \cdot the coefficient of x^{12-12} in (1-x)^{-4} ... 0 You can determine a function that simultaneously "generates" both sin(nt) and cos(nt). If you assume x to be real, then you're looking for the imaginary part of the coefficient of \frac{x^n}{n!} in the following expression.$$ \sum_{n=0}^\infty \left(cos(nt) + i.sin(nt)\right)\frac{x^n}{n!} \\ = \sum_{n=0}^\infty e^{int}\frac{x^n}{n!} \\ = ... 0 Here's a way. The number of$l$-step walks from$(0,\cdots,0)$to$(u_1,\cdots,u_n)$will be $$[x_1^{u_1}\cdots x_n^{v_n}](x_1+\cdots+x_n)^l \tag{\cdot}$$ where the polynomial is considered an element of$\Bbb Z[x_1,\cdots,x_n]/(x_1^2-1,\cdots,x_n^2-1)$. We can decompose$[x_1^{u_1}\cdots,x_n^{u_n}]=[x_1^{u_1}]\cdots[x_n^{u_n}]$informally and consider ... 0 I'll start writing down my approach. I don't know yet whether it will work out or not, but at least you can see it and maybe continue it. There are$n$dimensions. For each move, we choose a direction$1\leq i\leq n$(north/south, east/west, up/down and so on for higher dimensions) and move in that direction. In vector notation:$a_0=(0,0,\dots,0)$, ... 0 I'm taking a Discrete Mathematics course right now, using Applied combinatorics by Roberts and Tesman. It does have some good 'real life' examples and applications. 0 Just Cauchy's product followed by some cleanup. 2 I've been thinking of this problem on and off for more than a day, and at the end the answer I've reached is very trivial and already implicit in your calculations. Specifically, it's just this simple fact: Lemma: If a number$A$has$n$digits, then for$m > n$, the number$A(10^m - 1)$has a run of$m-n$nines, namely in the positions having place ... 1 Not a generating function approach, but I would just calculate how many numbers there are. The digits will appear in their proper proportion ($\frac 1{45} 1's, \dots \frac 9{45} 9's$), so if there are$N$numbers the sum will be$\frac N{45}(1^2+2^2+\dots 9^2)(\frac {10^{45}-1}9)$where the last factor is the number with$45\ 1's$To evaluate$N$, we can ... 0 Using generating functions: Let$P$be the generating function for one Bernoulli trial, i.e. for one$X_i$: we have $$P(s) = q + ps,$$ as the outcome is$X_i = 0$with probability$q$, and$1$with probability$p$. Then the generating function for the sum of$n$trials is $$P(s)^n = (q + ps)^n = \sum_{r=0}^{n}\binom{n}{r}(ps)^r q^{n-r},$$ in which the ... 4 Starting with index$0$this is $${(-1)}^{1+n}\left(\binom{14}{n}-\binom{14}{n-2}\right)$$ (using the fact that$\binom{N}m=0$if$m<0$or$m>N$. 1 The usual way to do this would be to sum each series before you multiply, getting$\frac 1{(1-x)(1-x^2)(1-x^5)}$and expand that, getting$1+x+2x^2+2x^3+3x^4+4x^5+\dots$In your case, I believe the upper limit of the$l$sum should be$k$, not$n$, to get the proper convolution. 1 You say you are looking for an easier way - so set x2=0,1,2,3 and you will get x1=2,4,6,8 respectively (x2>3 doesn't work). The remainder to 12 (10,7,4,1) is to be split between x3 and x4 such that x3≤x4. This yields 14 solutions in total (6+4+3+1). Certainly not the mathematically most beautiful, but the easiest way to solve this question. 1 One way to simplify the computation would be to take out the$\frac{1}{1-x^3}$, so that you only have to do partial fractions on$\frac{1}{(1-x^2)(1-x)}. Then you get \begin{align*} \frac{1}{(1-x^2)(1-x)} &= \frac{1}{4(1+x)} + \frac{1}{4(1-x)} + \frac{1}{2(1-x)^2} \\ &= \frac{1}{4} \sum_{i=0}^\infty (-x)^i + \frac{1}{4} \sum_{i=0}^\infty x^i + ... 2 If you calculateF(0.01)$in Windows calculator you'll get$0.010102030508132134559046368320032$. Look at pairs of digits and you'll see you get the Fibonacci sequence right up until it gets into the 3-digit range. So a simplistic way to do it is to calcuate$F(10^{-5})$to something like 105 digits of accuracy.$F(10^{-5}) \approx 0.00001 00001 00002 ...

4

\begin{eqnarray*} {z\over 1-z-z^2}&=&z\sum_{k=0}^\infty (z+z^2)^k\\ &=&z\sum_{k=0}^\infty z^k(1+z)^k\\ &=&z\sum_{k=0}^\infty z^k \sum_{j=0}^k {k\choose j} z^j\\ &=&\sum_{k=0}^\infty \sum_{j=0}^k {k\choose j} z^{1+j+k} \end{eqnarray*} From this we see that the coefficient of $z^N$ is $$\sum_{j=0}^{N-1}{N-1-j\choose j},$$ in ...

1

First, find the roots of your functions denominator $1-z-z^2$ which are $$z_\pm=-\frac{1\pm\sqrt{5}}{2}$$ Then find the partial fraction decomposition $$\frac{z}{1-z-z^2}=\frac{z}{(z-z_-)(z-z_+)}=\frac{a}{z-z_+}+\frac{b}{z-z_-}$$ which turns out to be $a=\tfrac{1}{2}(1-\tfrac{1}{\sqrt{5}})$ and $b=\tfrac{1}{2}(1+\tfrac{1}{\sqrt{5}})$ Now you can easily ...

3

You can do the following: Solve the equation $1-x-x^2=0$. The roots are $\frac{-1-\sqrt{5}}{2}$ and $\frac{-1+\sqrt{5}}{2}$. Then you have $1-x-x^2=-(x-\frac{-1-\sqrt{5}}{2})(x-\frac{-1+\sqrt{5}}{2})=-(x+\frac{1+\sqrt{5}}{2})(x+\frac{1-\sqrt{5}}{2})$ ...

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Hint : Subtract the sequence $x^3+x^6+x^9+...$ from $x+x^2+x^3+...$ Both sequences are geometric.

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Quantitative Aptitude Questions – Ratio and Proportion 1.     Find the fourth proportional to the numbers 8,12 and 16 Explanation: ...

# Aptitude Questions for Bank Tests - Ratio and Proportion Solved Problems 1

## Quantitative Aptitude Questions – Ratio and Proportion

1.    Find the fourth proportional to the numbers 8,12 and 16
Explanation: The fourth proportion of a,b and c is
bc / a
Therefore, fourth proportional of 8,12 and 16 =
12 x 16 / 8
= 24

2.    What is the third proportional to 18, 12?
Explnation: Third proportion of a and b is
b2 / a
Third proportion of 18 and 12 =
122 / 18
= 8

3.    What is the mean proportional between 16 and 25 ?
Explnation: The mean proportional of a and b is  a x b
mean proportion of 16 and 25 is =  16 x 25  =  400 = 20

4.    If A:B = 3:1 and B:C = 2:3 . What is the value of A:B:C?
Explnation:   In the given 2 ratios, the common term is B. To write the two ratios into a single ratio, the values of the common term must be equal.
A : B : C                 A : B : C
3  : 1       x  2        6 : 2
2: 3                      2 : 3
6 : 2 : 3

5.    If A:B=2:5   , B:C = 5:4  and C:D = 6:1. Then what is the value of A : D ?
Explanation:
A / B
x
B / C
x
C / D
=
A / D
A:D =
2 / 5
x
5 / 4
x
6 / 1
=
3 / 1
= 3:1

6.    Rs 3798 is divided into parts in the ratio of  5:4. What is the value of the 1st part?
Explanation:: Divided in the ratio= 5:4
Sum of the parts= 5+4= 9
1st part = Rs 3798 = $\frac{5}{9}$ = Rs 2110

7.    A sum of money is divided between A and B in the ratio of 11:3 . The difference between their shares is Rs 6400. Find the total amount?
Explanation:  : Let A and B got 11x and 3x .
The difference between their shares 11x – 3x = 8x =6400 =>x = 800
Total amount= Sum of the shares = 11x + 3x
=> 14x = 14 x 800 =Rs 11200

8.    The sum of the two numbers is 72. Which of the following cannot be the ratio of those two numbers ?
a)8:1       b ) 5:3          c)11:7           d)7:3
Explanation: The sum of the numbers is 72.
The sum of the terms of the ratio must divide the sum of the numbers.
Taking option 1 : Sum of the terms = 8+1 = 9 which divides 72.
Taking option 2 : Sum of the terams = 5 +3 = 8 which divides 72.
Taking option 3: Sum of the terms = 11 +7 = 18 which divides 72
Taking option 4: Sum of the terms = 7+3 = 10 which is not factor of 72.
So this cannot the  required ratio.

9.    The price of a TV and a Radio are in the ratio of 7:2. If a T.V. costs 25000 more than a Radio, find the price of the T.V.?
Explanation : When TV costs more than the radio by Rs 25000 means , the difference between the price of T.V. and Radio is Rs 25000
Prices of TV and Radio are in the ratio 7:2. Let their prices be 7x and 2x.
The difference of their prices => 7x-2x => 5x = Rs 25000=> x= Rs 5000
The price of the TV => 7x = 7 x 5000 = Rs 35000

10. Rs 2400 is divided among A, B and C . Half of the A’s share , $\frac{1}{3}$ of the B’s share and $\frac{1}{5}$ th of the C’s share are equal. Then find the share of C?
Explanation:
1 / 2
A =
1 / 3
B =
1 / 5
C are equal.

We say they are equal to k

1 / 2
A =
1 / 3
B =
1 / 5
C= k
Therefore A=2k B= 3k and C=5k.
The ratio of the shares of A,B and C = 2k :3k: 5K= 2:3:5
The share of C = Rs 2400 x
5 / 10
= Rs 1200

11. Rs 3720 is divided into three parts in proportional to the fractions $\frac{1}{2}$ : $\frac{1}{3}$: $\frac{1}{5}$ . What is the second part?
Explanation:
1 / 2
:
1 / 3
:
1 / 5
=
15 / 30
:
10 / 30
:
6 / 30
The required ratio = 15: 10:6
second part = Rs 3720 x
10 / 31
=Rs 1200

12. Two numbers are in the ratio of 4:5 and their product is 500.Find the bigger number?
Explanation: Let two numbers be 4x and 5x.
Product of the numbers =>4x x 5x => 20x2 = 500
=> x2 = 25 =>x=5
Bigger number = 5x= 5 x 5 = 25

13. Rs 3120 is divided among A,B and C so that 3 times the A’s share, and twice the B’s share and 4 times of the C’s share are equal. Find the share of A ?
Explanation:
3 times the A’s share, and twice the B’s share and 4 times of the C’s share are equal 3A = 2B= 4C = k
A =
k / 3
, B=
k / 2
C=
k / 4
=> A:B:C =
k / 3
:
k / 2
:
k / 4
A: B : C =
4:6:3 / 12

The required ratio = 4:6:3
Share of A = Rs 3120 x
4 / 13
= Rs 960

14. Rs 7820 are divided among A, B and C so that A receives $\frac{5}{12}$ as much as B and C together receive and B receives $\frac{1}{3}$ as much as A and C together receive. Find the difference between the shares of A and B ?
4 / 13
as much as B and C together receive.
A =
5 / 12
(B+C)=> A : (B+C)= 5:12
A’s share = Rs 7820 x
5 / 17
=Rs 2300
4 / 13
as much as A and C together receive
B =
4 / 13
(A+C)=> B: (A+C)= 4:13
B’s share = Rs 7820 x
4 / 17
= Rs 1840
Difference between A and B’s share = Rs 2300 –Rs 1840 =Rs 46013.
15.
Three numbers are in the ratio 5:4:3 such that the sum of their squares is 0. What is the smallest number?
Explanation: The numbers are in the ratio of 5:4:3.
Let numbers be 5x,4x and 3x. Given that sum of their squares is 450
Sum of their squares = (5x)2+(4x)2+(3x)2
ð  25x2 + 16 x2 + 9 x2 = 450
ð  50 x2 =450
ð  x2=9
ð  x=3
Smallest number = 3x = 3 x 3 =9


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• 算法设计与分析 •

### 融合三重注意力和评论评分的深度推荐算法

1. （江苏科技大学计算机学院，江苏镇江212100）
• 出版日期:2022-06-08 发布日期:2022-06-08
• 作者简介:张宗海 (1995—)，男，山东临沂人，硕士研究生，研究方向：数据挖掘，深度学习，E-mail： 15192921420@126.com；於跃成(1971—)，男，副教授，博士，研究方向：数据挖掘，模式识别，E-mail： 857454249@qq.com； 冯申(1994—)，男，硕士研究生，研究方向：数据挖掘，深度学习，E-mail： 1193411772@qq.com。
• 基金资助:
国家自然科学基金资助项目（61806087）； 江苏省研究生创新项目（SJCX20_1475）

### Deep Recommendation Algorithm Integrating Triple Attention Mechanism and Review Score

1. (College of Computer, Jiangsu University of Science and Technology, Zhenjiang 212100, China)
• Online:2022-06-08 Published:2022-06-08

Abstract: Many shopping websites have a large amount of review information written by users. Although most recommendation systems use review information, there is still much room for improvement. On the one hand, the information in the comments is uneven, mixed with a lot of useless information; on the other hand, most existing recommendation systems assume that a user’s attention to a certain product feature is the same for all products and cannot accurately reflect user preferences. This paper proposes an aspect-aware depth recommendation model ANAP that integrates triple attention and review score. Starting from the two levels of words and features, the important information in the review text is extracted by constructing two different attention networks to reduce the impact of useless information; in order to accurately reflect user preferences, the attention interaction network is constructed to capture the user’s different attention to various aspects of different items, and to achieve fine-grained modeling of aspect perception. This paper conducts experiments on 6 real data sets and designs an attention mechanism comparison experiment. The results show that the ANAP model effectively improves the score prediction accuracy, and the mean absolute error (MAE) is lower than the existing best algorithm by 4.86 percentage points.


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# Photons and waves

1. Dec 8, 2015

### Prannoy Mehta

Moderator's note: this post follows from the Is light a wave or a particle FAQ

Nothing to add, rather just a query.

1) Do Photons move in a wavelike manner, as given by De Broglie's Equation?
2) If so, will the wavelength obtained over there be different from that of the electromagnetic wave which the photon consists of.

- A Laymen who wants to reason.

Cleared a few of my doubts, thank you for posting an answer to the FAQ.

Last edited by a moderator: Dec 8, 2015
2. Dec 8, 2015

### Staff: Mentor

What does "moving in a wavelike manner" mean? They are not going in curved lines if you imagine something like that.
That question does not make sense.

3. Jan 1, 2016

### Prannoy Mehta

I am sorry for the late reply. mfb and bhooba. I am sorry as I am not very fluent in English.

By Wave like I meant Sinusoidal form.
Do photons move up and down through space as shown in various animations? Thank you for that answer, I always they thought they moved in Sinusoidal manner.

The second question is wrong. As per the first answer. Thank you.

4. Jan 1, 2016

### Staff: Mentor

No, photons do not move "up and down". They do not even have a well-defined position.

5. Jan 2, 2016

### A. Neumaier

Photons do not have a well-defined position, so one cannot describe realistically how they move.

But each photon has a shape, and these shapes take the form of a wave (more precisely of a nonzero solution of the vacuum Maxwell equations). These shapes oscillate, and their frequencies define the color of the photon. Thus, you can have photons of different colors, and in the shape of for example beams or radial waves, etc..

6. Jan 2, 2016

### ZapperZ

Staff Emeritus
I think this has the risk of someone interpreting it the wrong way. Each photon does not have a "shape", and certainly they don't have the shape of "beams or radial waves". The "nonzero solution" to the Maxwell equation is the classical wave picture of "a lot of photons", not "each photon". And what oscillate are the electric and magnetic field components, to be precise. If this is what people call "shape", then that's fine, but it needs to be explicitly stated, because someone not familiar with physics, such as the OP's layman, has a different concept of what a "shape" is (i.e. a DEFINITE physical boundary of the object in real space).

Zz.

7. Jan 2, 2016

### A. Neumaier

To be completely precise: A photon (even more precisely a pure 1-photon state) is mathematically specified by its wave function, which I called its shape to make it intuitive, and which (in a bounded region, cavity) is an arbitrary normalized solution of the vacuum Maxwell equations. That the shape has a fuzzy boundary is nothing exceptional - the same holds for the shape of a city, a meadow, a cloud, a star, a sun ray, or any material object if its surface is looked at in enough detail.

This picture of shape is the same as that used for simple quantum mechanical systems like a free nonrelativistic particle, which is a point in a Dirac delta state and a fuzzy ball in a radial state. The electron density of a single hydrogen atom in the ground state determines the spherical shape of the atom and its radius, although the boundary is also fuzzy.

A single photon cannot be localized in a point since there are no Dirac delta solutions of the Maxwell equations - but it can be localized in a beam or a spherical wave.

8. Jan 2, 2016

### Prannoy Mehta

I have understood what mfb, ZapperZ and A. Neumaier tried to convey. I guess I should have specific and mentioned photon rather than photons. Though it would take me more time to grasp the concept of photons I believe this would be the answer I needed. Thank you for the answer(s).

9. Jan 2, 2016

### ZapperZ

Staff Emeritus
You are mixing up the spread in its POSITION and the SIZE of the object itself! I'm not arguing about the various probability density of the location. I'm arguing that when we talk about "size", we are defining a definite shape and volume of the object, NOT the spread it its position!

A ping-pong ball has a volume (if you include the empty inside as part of the ball). It has a spherical shape. Yet, it's location, when it is stationary, is defined by the position of its center of mass. An electron is a point object. It's location may be defined to within the HUP, but that is its location. It's probability density may be spread out, signifying where it can be found at some instant, but its VOLUME is still zero, i.e. it has no or undefined shape!

A photon has no size and shape. It may have a range of position, and we may use some properties such as its wavelength to define some characteristics length, but there has been no physics that defines the same and volume of a single photon.

Zz.

10. Jan 2, 2016

### drphysica

Photons travel as waves and particles due to wave particle duality and it also applies to electrons and all bosons and fermions i.e. all particles. The only difference between photon an electron is its energy relations for electron particle energy relates to its mass is E^2=(mp)^2+(mc^2)^2 and for photon E = hf where f ~ is frequency and h is Plank Constant 6.62607004 × 10-34 m2 kg / s Thus you can calculate energy of your AM frequency favorite radio buy just plugging in a number between 535 to 1700 kHz into equations and getting the energy of your photon. Remember photons don't have mass and electrons and other particles do.

11. Jan 2, 2016

### ZapperZ

Staff Emeritus
We have had bad consequences when people state this. So you should explain yourself.

What exactly do you mean when you say "photons travel as waves and particles"? Start by what you mean by "photons travel as waves"? Note that laymen will likely think that these "photons" are going through space bouncing up and down like the water waves they are familiar with. If you don't believe me, browse through many posts on this forum, and you'll come across such a misunderstanding. All this because they have a picture of what a "wave" is, and someone simply and callously wrote something similar to what you've written.

We ALL need to be careful in looking at what we write. What you write, and what you MEAN, may not translate exactly to what the other person understands! This is especially true if you use similar terms and words, but having different meaning when applied to the context of physics. If you do not spend the effort to clarify and elaborate, and if you do not anticipate the misunderstanding, then you will be perpetuating the mistake!

Zz.

12. Jan 2, 2016

### Prannoy Mehta

Okay, I guess I have messed it up again. Anyhow, the reason I found it reasonable because I read this. (It is the text we are referring now) Please go to Page 19 or Pg. 432. There is a big blue box regarding LASERS in that the third paragraph has the lines below. (Please see the figure also)

www.ncert.nic.in/ncerts/l/leph204.pdf

These are the lines I am trying to refer to.

" The wavelength of each packet is almost the same. Also, the average length of the packet of waves is much larger. This means that there is better
phase correlation over a longer duration of time. This results in reducing the divergence of a laser beam substantially."

As per what you have mentioned it has no size and shape. (By Shape we refer to the oscillating electric and magnetic fields.) What does length stand for here? The length of the packet seemed more reasonable if the conclusion I came up to earlier was true.

Anyhow, just another curiosity question. As Zapper Z earlier mentioned,

" The "nonzero solution" to the Maxwell equation is the classical wave picture of "a lot of photons", not "each photon". "

So for each photon, it can be zero?

Also, the picture I requested you to see, (In the lasers box.)

What is the thing which is depicted as a wave? Is it the motion of a photon through space or is it with respect time.

Last edited: Jan 2, 2016
13. Jan 2, 2016

### Prannoy Mehta

I don't think it is mp. I think pc was intended.

14. Jan 2, 2016

### drphysica

I have no idea what you talking about! I simply state the fact that photons are waves and particles and it also applies to any other particles. Now if you don't agree with that you shouldn't be in physics forums. If you want to talk philosophy you should be at philosophy forums. This place needs to be cleaned up a bit.

15. Jan 2, 2016

### drphysica

Yes you right I made a mistake there.

16. Jan 2, 2016

### blue_leaf77

I must acknowledge that the physics community is divided into (at least) two groups with regard to how they picture the word "photon" in their mind. Physicists working closer towards the experimental side or theoriticians in the scope of applied physics (note that I'm not by any means referring to anyone involved in this thread, this is just evident) tend to depict photons as the grains making up light. The other groups, the genuine theoretical physicists (when I call "genuine", I don't mean to disregard the other types of theoretical physicists), i.e. those which go down the path of physics rigor, will frown upon hearing that photons are ball-like entities. The correct view of photons are that depicted by the second group. In spite of this, I would refrain from blaming the first group for their simplified view of photons because to them, that is the most convenient way of explaining the phenomena they are dealing with, phenomena in which the correct view of photon are not urgent to be incorporated. For example, in the field of diffraction imaging, many people will probably say something like "photons being scattered" or "photons hitting the camera pixel chip", such wording will easily be interpreted as photons having dynamics in space. While that kind of depiction is not justified in the quantum theory of light, the correct one will not get them anywhere either, because most of the times diffraction physicists are working with high intensity light, in which the classical nature heavily overshadows the quantum one.

So, @Prannoy Mehta, to layman like you which is probably not yet certain to which group of the two physicists I talk about above you are going to, photon is better to be regarded as a quantum object, not "worldly" object. One photon is described by the so-called Fock state, which is basically the same state as you encounter in quantum harmonic oscillator. One photon state is written as $|n=1\rangle$, two photons state as $|n=2\rangle$ and so on. This is what ZapperZ was trying to tell you. It obeys the same quantum properties as the harmonic oscillator eigenstate, e.g. if you apply raising operator to $|n=2\rangle$ photon state, you will get $|n=3\rangle$ photon state.
That passage does not talk about the quantum nature of light, instead it tries to explain the difference between laser light and light bulb in terms of classical electrodynamics. What the author refers to when he gave an example of a crowd walking in the same direction vs soldiers marching synchronously is the phase relationship at any point in the light beam, again, the phase, not the photons.
ZapperZ didn't try to refute your answer, your answer is not wrong in the loose sense. He asked you to elaborate on what you said that photons travel in space, because this has the potential of misleading the OP given his limited theoretical knowledge of photons. Probably, had this thread been in general physics section, you may not get such response.

17. Jan 2, 2016

### Prannoy Mehta

This is not the first time you have replied to my questions, so I will take a blind word for that one (The different group of physicists). The fact that photons are not 'worldly' makes it very counter intuitive (Can't blame them. Can we? ). We generally like the relate to things to what we see every day. We do not get to play/ do the major experiments we have just read about them. Though Zapper Z did state that imagination without knowledge is dangerous. We still tend to test it out every day it helps us deepen our understanding regarding the subject.

Anyhow, back to the topic. What do you mean by 'Fock state'? What does the length of a packet mean? Could you please answer the question added earlier on what the image depicts. That is one of the main questions from the beginning of this forum.

PS: I have a very small idea of what those Quantum Harmonic Oscillators are. For now, I will just assume they are energy. If it is not enough for answering my queries. I think it would be better I keep this question aside and come back to it when I have a better understanding of light.

Thank you

Last edited: Jan 2, 2016
18. Jan 2, 2016

### ZapperZ

Staff Emeritus
I shouldn't be in physics forums? How long have you been here again?

Note that I asked you to ELABORATE on what you mean by "photons are waves" and not just to leave it at that! I painstakingly explained why this can easily mislead a lot of people who do not understand the physics. Photons are not "waves", and certainly not in the conventional, physical sense. Tell that to someone off the street, and the picture that the person will get is a bunch of objects bouncing up and down as it moves through space. This is not even true for electrons, etc... etc!

This isn't philosophy. It is a plea to pay attention to what you say, because often times, especially on here, what you mean is not what they understand!

Understand?!

Zz.

19. Jan 2, 2016

### blue_leaf77

I come from the first group, probably all of my laboratory colleagues will tend to depict photons as ball-like entities. But as I have also taken quantum optics course, I realize that photons are actually not as simple as it would have been if it really were like particles in the classical sense.
Fock state is the number state a photon can have, it is therefore the eigenstate of the number operator $a^\dagger a$. More on this :http://www.qms.uni-rostock.de/filea...heel/Quantenoptik/Quantenoptik-Vorlesung3.pdf.

20. Jan 2, 2016

### A. Neumaier

This is not true. Since there is no position operator for the photon, one cannot speak of a photon of having a range of positions (or a probability distribution of being in some small region of space) - this is impossible for conceptual reasons.

But a single photon has an energy density, which very well describes its shape. This is evident in the design and analysis of experiments that create single photons on demand. See my lecture Classical and quantum field aspects of light.


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# Have we managed to make a perfect vacuum?

Have we managed to make a device without any atom inside it on earth? I was reading about vacuum here, and I found in the examples part here that even on the best man made vacuum devices, there are still millions of "molecules per $cm^3$". So my other question here is, what are these molecules and what is the mass of them?

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No. Every vacuum is measured with the scale given en.wikipedia.org/wiki/Free_space#Measurement, not being "Perfect Vacuum". – Rafael Reiter Jan 25 '13 at 23:59
In most UHV (ultra high vacuum) systems, including mine, the main component is by far hydrogen molecules (H2). They're light and very difficult to pump out. For that reason, many pumps in the UHV range focus solely on pumping hydrogen (e.g., titanium sublimation or non-evaporable getters). – emarti Jan 26 '13 at 23:29

As zonk said, there is no perfect vacuum. Even the 'vacuum' of space contains a few atoms per cubic meter on average.

In the lab, the lack of a high vacuum usually results from not having a pump that can effectively extract enough of the particles inside the chamber you're trying to evacuate.

There are several different kinds of pumps used, depending on how 'good' of a vacuum you want: mechanical roughing pumps, ion pumps, turbo-molecular pumps among others. Each works either by transferring gas particles or by capturing them. Since transfer works by preferentially forcing gas molecules in a certain direction, it cannot remove ALL the particles (as in the electrons inside a wire: on average, you can force the current preferentially in the direction of the applied voltage but each individual electron will zip around in all directions). A capture pump is better at removing particles, but it is still limited by how many materials they have already captured and can saturate.

Even if we had a perfect pump, you would still need to be careful about what is inside the chamber or what the chamber is made of. Fingerprint grease, zinc (and brass), and plastics actually out-gas under high enough vacuums. This means that the particles normally trapped under atmospheric pressures can break loose and rattle around inside your vacuum if you're not careful.

Also, pumps such as mechanical pumps use lubricating oil to function (though some fancier ones such as the ion pumps do not) and that can get into the chamber as well and increase the number of particles.

That's probably more than you wanted to know, but there you have it.

I do not think the tag 'particle physics' is accurate. Particle physics generally refers to subatomic physics and encompasses things like quarks, QED, and the Higgs boson. Perhaps 'experimental physics' would be better?

Glad it helped. Richard Terrett is correct, (charged) anti-matter is confined in a magnetic trap in as high of a vacuum as we can get. Uncharged anti-matter must be contained using laser traps ('optical tweezers' is something to look into if you're curious).

There will still be some particles in the antimatter container but, contrary to the DaVinci Code and popular perception, when one particle of anti-hydrogen meets a particle of hydrogen, the energy released (according to E=mc^2, which I believe is correct) is 3*10^-10 Joules or 0.3 nanojoules.

To give this a sense of scale, Wikipedia says a nanojoule (10^-9 Joules) is equal to 'about 1/160 of the kinetic energy of a flying mosquito'. So an antihydrogen annihilating with a stray hydrogen in the vessel would be have less of an effect than a mosquito bumping the outside of the container, energetically speaking.

Now, we can also determine how many particles would be in a container. Let's take the highest vacuum rating; ultra-high vacuum, which is classified as having a pressure under 10^-7 pascal, with 10^-10 pascal (10^-12 torr) being the golden standard among people who care about excellent vacuum vessels. A lot of UHV systems operate at very low temperatures, but let us take the temperature of the inside of the vessel to be room-temperature, T=293 Kelvin. We can use the ideal gas equation if we know the volume of the container, PV=nRT. Plugging in the pressure, room temperature, and assuming a container size of a rubik's cube [V=(5.5*10^-2 meters)^3=1.66*10^-4 meters^3], we arrive at a number of particles of n=6.83*10^-18 moles or 4.113*10^6 molecules. To put this in perspective, at standard pressure and temperature in the same size vessel (STP, 101325 pascal, 293K), we would have 4.155×10^21 molecules. So UHV is a pretty damn good vacuum.

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Great answer, tag edited too. Now one last question since you seem to know much about this. I have asked earlier about the mass of these particles still in there. Wikipedia mentioned molecules but I don't know molecules of what. Anyway to clarify this, I asked this question to know if there is 100% safe place to store antimatter when we somehow get to know how. So now the important thing is about the mass of these "molecules" or particles, have you got any idea about how many are left in these different vacuum methods and what the mass of them is ? – Hurricane Jan 26 '13 at 8:58
@Hurricane - Antimatter can be stored in a Penning trap. – Richard Terrett Jan 26 '13 at 9:11
@RichardTerrett, Wikipedia says it is used to store charged particles like antiprotons, so does this also apply for complete atoms like antihydrogen ones since atoms are not charged ? – Hurricane Jan 26 '13 at 23:05
@Hurricane You should ask that as a new question (assuming it doesn't already exist on this site). – Chris White Jan 26 '13 at 23:14

I'd like to add another dimension to this discussion. Even if we were to hypothetically evacuate all atoms from a certain region of space, quantum field theory tells us that there would still be something there, namely the "zero point" energy corresponding to the ground state of fields. So in this sense, at least as far as I am aware, our current model of fundamental physics renders the existence of a region of space with "nothing" there an impossibility, even in principle. http://en.wikipedia.org/wiki/Zero-point_energy

Cheers!

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Plus a bath of photons from the walls. – dmckee Jan 26 '13 at 3:59
-1: The "zero-point energy" is referring to virtual particles, not real particles. – Chris Gerig Jan 26 '13 at 4:36
@ChrisGerig: true, but the Casimir effect shows there is something there so I think your downvote is unkind. – John Rennie Jan 26 '13 at 10:11
@JohnRennie: First sentence of question: "a device with no ANY atom inside it" ... atom... not fake particle. – Chris Gerig Jan 26 '13 at 18:15
@ChrisGerig I think you make a good point that my answer isn't immediately relevant to the body of the question, but that's why I added the phrase "add another dimension to this discussion;" I simply thought certain people might find this, admittedly somewhat tangential, answer interesting given the broader scope of the question title. Oh btw I was a physics/math undergrad at Cal; I'm not surprised you left physics for math. Cheers. – joshphysics Jan 26 '13 at 18:25

The current record vacuum, made at CERN, is around 1000atoms/cubic cm

For comparison intersteller space (well away from the solar system) is around 1 atom/cc and deep intergalactic, really empty, space is around 0.001 atom/cc

The bits of space we can reach, in Earth orbit, are only about the sort of vacuum you can get in an ordinary lab vac pump.

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in Earth orbit Could you please say whether that would be low-earth orbit (LEO) or geostationary orbit (GEO), thanks. – Eugene Seidel Jan 26 '13 at 21:19
Upto about 600km (LEO) it's the atmosphere reducing in density with height. Above that (MEO and GSO) it's mostly ions from the sun - don't remember the figures – Martin Beckett Jan 26 '13 at 21:29


text:
## Univn Shx .rar

Univn Shx .rar

But when I download it to my lap, it is a.rar file which cannot be downloaded. So is there any way to add the.rar extension? The Unzipped File is at C:\ProgramData\FileRepository\UserData\Univn\Shx\Ph.zip A: Your file is not actually a zip file. If you right click on the file and select properties, you should see the file extension is « rar » If you click properties and select the content tab there is a button to select the « associated program ». The associated program is RAR. If you go to control panel and find the control Rar and click « uninstall » the RAR program will be uninstalled from your computer.  » As Monarchs maneuver around their changing stage and learn to dance their way to the royal wedding, how will they react when they eventually realize that they’re trapped in the middle of a performance? Will their perfect worlds crumble to the dance floor? This time, fate keeps the Empress locked into a pregnancy that she isn’t ready to handle. As punishment, the Empress is transported to a nearby colony planet. When her son is born, the Empress escapes. Will she find her way home before the prince turns 21? If she misses her son’s 21st birthday by even a moment, she must face the consequences. Will she become the King’s right hand counselor and chief advisor? Or will she fall to pieces, never to see her son?  » Z is for Emperor. It’s time to let the world know that this is NOT a story about politics – unless you like a story told with a bit of nuance and a lot of suspense. « Q: Proving the analytic continuation of a complex function. I don’t know how to prove the equality between two complex functions $z_1$ and $z_2$ given below. How do I prove it? $\begin{cases} z_1(z) = z + \frac{1}{z} \\\\ z_2(z) = z – \frac{1}{z} \end{cases}$ A: The following is probably the easiest way (but unfortunately only works when


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1. ## Question(s) about complex numbers.

• I understand that $\displaystyle |z| = \sqrt{x^2+y^2}$ (from the Argand Diagram). Is there an algebraic way to show that $\displaystyle |z| = \sqrt{x^2+y^2}$?

• Why is the principal value of the argument taken to be between $\displaystyle -\pi$ and $\displaystyle \pi$? I was asked to find the modulus and argument of $\displaystyle -1-i$. Here is what I did:

$\displaystyle |z| = \sqrt{\left(-1\right)^2+\left(-1\right)^2} = \sqrt{1+1} = \sqrt{2}$. Right.

$\displaystyle \arg{z} = \arctan\left(\frac{-1}{-1}\right)+\frac{\pi}{2} = \arctan(1)+\frac{\pi}{2} = \frac{\pi}{4}+2\pi = \frac{2\pi+\pi}{4} = \frac{3\pi}{4}$. Obviously that isn't between $\displaystyle -\pi$ & $\displaystyle \pi$. What am I supposed to do then?

PS. For the last part (as noticed by Archie Meade), I meant: $\displaystyle \arg{z} = \arctan\left(\frac{-1}{-1}\right)+\pi = \arctan(1)+\pi = \frac{\pi}{4}+\pi = \frac{4\pi+\pi}{4} = \frac{5\pi}{4}$.

2. $\displaystyle There\ are\ 2\ main\ geometric\ ways\ to\ show\ |z|=\sqrt{x^2+y^2}\ with\ x\ perpendicular\ to\ y.$

$\displaystyle The\ angle\ is\ taken\ as\ 0\ to\ \pi\ radians\ if\ z\ is\ on\ or\ above\ the\ horizontal\ axis.$

$\displaystyle It's\ taken\ as\ 0\ to\ -\pi\ for\ z\ below\ the\ horizontal\ axis.$
$\displaystyle It's\ the\ convention.$

$\displaystyle 0\ to\ -\pi\ are\ the\ angles\ from\ 360\ to\ 180\ degrees.$

$\displaystyle Arctan\ doesn't\ distinguish\ between\ opposite\ ends\ of\ the\ same\ line,$
$\displaystyle so\ you\ should\ ask\ yourself\ why\ you\ are\ adding\ 90\ degrees.$

$\displaystyle Adding\ \frac{\pi}{2}\ radians\ rotates\ a\ line\ through\ 90\ degrees.$

$\displaystyle From\ the\ real\ and\ imaginary\ parts\ of\ the\ complex\ number,$
$\displaystyle you\ can\ identify\ it\ as\ being\ between\ 180\ and\ 270\ degrees.$

$\displaystyle You\ must\ make\ this\ evaluation\ first,\ then\ calculate$
$\displaystyle the\ acute\ angle\ z\ makes\ with\ either\ the\ real\ or\ imaginary\ axis.$

3. Here is the way to find the principal argument.
$\displaystyle \arg (x + yi) = \left\{ {\begin{array}{rl} {\arctan (y/x)} & {,0 < x} \\ {\arctan (y/x) + \pi } & {,x < 0\,\& \,y > 0} \\ {\arctan (y/x) - \pi } & {,x < 0\,\& \,y < 0} \\ \end{array} } \right.$

4. Originally Posted by Archie Meade
$\displaystyle There\ are\ 2\ main\ geometric\ ways\ to\ show\ |z|=\sqrt{x^2+y^2}\ with\ x\ perpendicular\ to\ y.$
I understand one (using pythagoras' theorem). What is the other? And is there any other way to show it apart from plotting the Argand Diagram?
$\displaystyle The\ angle\ is\ taken\ as\ 0\ to\ \pi\ radians\ if\ z\ is\ on\ or\ above\ the\ horizontal\ axis.$
$\displaystyle It's\ taken\ as\ 0\ to\ -\pi\ for\ z\ below\ the\ horizontal\ axis.$
$\displaystyle It's\ the\ convention.$$\displaystyle 0\ to\ -\pi\ are\ the\ angles\ from\ 360\ to\ 180\ degrees. Thank you. That is what I have been missing. \displaystyle Arctan\ doesn't\ distinguish\ between\ opposite\ ends\ of\ the\ same\ line, \displaystyle so\ you\ should\ ask\ yourself\ why\ you\ are\ adding\ 90\ degrees. \displaystyle Adding\ \frac{\pi}{2}\ radians\ rotates\ a\ line\ through\ 90\ degrees. That was a mistake. What I meant was \displaystyle \arg{z} = \arctan\left(\frac{-1}{-1}\right)+\pi = \arctan(1)+\pi = \frac{\pi}{4}+\pi = \frac{4\pi+\pi}{4} = \frac{5\pi}{4} \displaystyle From\ the\ real\ and\ imaginary\ parts\ of\ the\ complex\ number, \displaystyle you\ can\ identify\ it\ as\ being\ between\ 180\ and\ 270\ degrees. I could see that. But I was measuring anticlockwise from zero. I thought that the angle I needed was the acute angle I found plus 180 (measuring anticlockwise from zero). \displaystyle You\ must\ make\ this\ evaluation\ first,\ then\ calculate$$\displaystyle the\ acute\ angle\ z\ makes\ with\ either\ the\ real\ or\ imaginary\ axis.$
Alright. I will do that. Little me first take a little grasp of what I've so far gathered.

5. Originally Posted by Plato
Here is the way to find the principal argument.
$\displaystyle \arg (x + yi) = \left\{ {\begin{array}{rl} {\arctan (y/x)} & {,0 < x} \\ {\arctan (y/x) + \pi } & {,x < 0\,\& \,y > 0} \\ {\arctan (y/x) - \pi } & {,x < 0\,\& \,y < 0} \\ \end{array} } \right.$
So for $\displaystyle -1-i$, $\displaystyle \arg(z) = \arctan\left(\frac{-1}{-1}\right)-\pi = \arctan\left(1\right)-\pi = \frac{\pi}{4}-\pi = \frac{\pi-4\pi}{4} = -\frac{3\pi}{4}$. Brilliant. Thanks.

6. $\displaystyle Hi\ Captcha,$

$\displaystyle The\ proof\ is\ just\ 2\ alternative\ ways\ to\ derive\ Pythagoras'\ Theorem.$

$\displaystyle One\ way\ is\ to\ split\ a\ right-angled\ triangle\ into\ two$
$\displaystyle smaller\ right-angled\ triangles,\ then\ line\ these\ 2\ and\ the\$
$\displaystyle original\ one\ up, then\ write\ the\ side\ ratios\ in\ order\ to\ get\ x^2\ and\ y^2.$

$\displaystyle The\ 2nd\ way\ is\ to\ draw\ a\ square,\ draw\ 4\ interlocking\ circles$
$\displaystyle whose\ diameters\ are\ the\ sides\ of\ the\ square$
$\displaystyle and\ whose\ centres\ are\ the\ midpoints\ of\ the\ sides.$
$\displaystyle Inside\ the\ square\ are\ now\ 4\ semicircles.$
$\displaystyle You\ can\ now\ draw\ lines\ such\ that\ you\ end\ up\ with$
$\displaystyle 4\ identical\ right-angled\ triangles\ and\ a\ central\ square.$
$\displaystyle the\ hypotenuse\ of\ the\ triangles\ are\ the\ sides\ of\ the\ main\ square.$
$\displaystyle You\ can\ then\ quickly\ write\ the\ square\ area\ in\ terms\ of$
$\displaystyle the\ 4\ triangles\ and\ smaller\ central\ square.$

$\displaystyle The\ formula\ is\ the\ expression\ of\ the\ geometry.$

$\displaystyle Plato\ nicely\ summarised\ the\ angle\ calculation.$