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# Problem: Calculating Square Area

The first exercise from this topic is the following: write a console program that inputs an integer a and calculates the area of a square with side a. The task is trivial and easy: input a number from the console, multiply it by itself and print the obtained result on the console.

## Hints and Guidelines

We create a new project in the existing Visual Studio solution. In the Solution Explorer right-click on Solution 'Simple-Calculations'. Choose [Add] -> [New Project…]:

A dialogue window is going to be opened for choosing the project type for creation. We choose C# console application with name “Square-Area”:

We already have a solution with one console application in it. What remains is to write the code for solving this problem. For this purpose, we go to the main method's body Main(string[] args) and write the following code:

The code inputs an integer through a = int.Parse(Console.ReadLine()), afterwards it calculates area = a * a and finally prints the value of the variable area. We start the program with [Ctrl+F5] and test it with different input values:

## Testing in the Judge System

Test your solution here: https://judge.softuni.org/Contests/Practice/Index/504#0. You have to get 100 points (completely correct solution):


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# Calling Conventions for Libraries on Mac and UNIX

This topic is 2786 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Hi all,

Here's another question that burns me while I'm trying to write a cross-platform game engine:

What calling convention(s) are usually used on POSIX (MaxOSX and UNIX) for functions to be exported from libraries (DLLs)?

Is it tradition to use C Declaration, or Standard (Windows) Call?

I know Windows traditionally prefers stdcall (Standard Call), such as "CALLBACK" etc. for all DLL functions.
So if I write cross-platform libraries, should I create a cross-platform macro that maps to "_stdcall" on Windows and "_cdecl" on POSIX and stick that macro before every function to export?

Mark

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I personally don't use any on my application of shared objects (dll's on unix)

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POSIX using the C calling convention, I believe, but I don't know if that's actually called out in the POSIX standards or just at the lever of the POSIX C API. Apple has a document on the various calling conventions appropriate for their OS on various platforms (some of which, like the 68k and PPC chips, aren't relevant any more): click here.

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Sometimes I think I should just FORGET about cross-platform compatibility entirely. Linux and games... Not really, Apple? Oh well... But then I keep hearing how "easy" it supposedly is if you just keep a few important things in mind... And that there might be even other advantages of separating OS-specific code from non-OS-specific code.

Hmmm...

I guess I'll create a cross platform macro "LIBRARYFUNCTIONCALLINGCONVENTION" that maps to whatever convention (or noting at all) that I will use before every DLL function prototype / body.

Regards,
Mark

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Quote:
 Original post by MarkieHi, thanks for the answers!Sometimes I think I should just FORGET about cross-platform compatibility entirely. Linux and games... Not really, Apple? Oh well... But then I keep hearing how "easy" it supposedly is if you just keep a few important things in mind... And that there might be even other advantages of separating OS-specific code from non-OS-specific code.Hmmm...I guess I'll create a cross platform macro "LIBRARYFUNCTIONCALLINGCONVENTION" that maps to whatever convention (or noting at all) that I will use before every DLL function prototype / body.Regards,Mark

The "trick" to making crossplatform development easy is to use existing cross platform middleware, when writing a game engine the calling conventions really aren't a big issue, both gcc and msvc use __cdecl by default.

The drawback of __cdecl is that it results in slightly larger code for non inlined functions, its advantage is that it allows for functions with a variable number of arguments and won't cause your program to crash catastrophically if you call a function using the wrong number of arguments. (If you use c or c++ the compiler will handle that anyway but for asm applications it can cause some problems)

There is absolutely nothing wrong with using __stdcall on Linux/MacOS or __cdecl on Windows, in fact, most Windows software use __cdecl for the majority of their functions, choose calling convention based on how the function in question is used, not what OS its running on.

The one thing you should avoid however is __fastcall since its implementation varies between compilers (Thus if you use it to compile your engine any user of your engine will have to compile their code using the same compiler (or one that happens to be compatible with it (Ofcourse, if there is a significant performance advantage by using this it might be worth the restriction it brings))

[Edited by - SimonForsman on December 4, 2010 11:02:31 PM]

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Quote:
 The one thing you should avoid however is __fastcall since its implementation varies between compilers

With 64 bit this might be necessary non the less. On 64 bit, gcc (on linux, didn't test mingw though) uses registers for the first 6 parameters (the rest goes on the stack as in 32 bit c call) where as the MS compiler only uses 4 registers.
It might be better to stick with one compiler and calling convention for engine and application but use whatever is default on the platform instead of trying to use the same calling convention on all platforms.

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Quote:
 Original post by MarkieI guess I'll create a cross platform macro "LIBRARYFUNCTIONCALLINGCONVENTION" that maps to whatever convention (or noting at all) that I will use before every DLL function prototype / body.
That's actually not as crazy as you make it sound. I personally have all my commonly used functionality arranged as modules in a hierarchy with the one at the very bottom being dedicated to platform isolation.

For example, you could define a macro like this in a shared library of yours
#ifdef __WINDOWS__#   ifdef __cplusplus#       define MYEXPORT extern "C" __declspec(dllexport)#   else#       define MYEXPORT __declspec(dllexport)#   endif#elif defined(__APPLE__)#   define MYEXPORT ...#elif defined(__LINUX__)#   define MYEXPORT ......
The major benefit of keeping macros like this in a separate module/file is that you only need to write them once; afterward you can conveniently import them into each of your projects
#include <mymacros.h>MYEXPORT int MyExportedFunction ( void );
And the party doesn't stop here, because re-using the module like this all the projects using it are affected immediately everytime you update it; all you need to do is re-compile them!

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Hi SimonForsman!

Awesome post! Thanks!!! :-)

Quote:
 Original post by SimonForsmanThe "trick" to making crossplatform development easy is to use existing cross platform middleware, when writing a game engine the calling conventions really aren't a big issue, both gcc and msvc use __cdecl by default.
I think Petzold writes that in Windows programming, it is a tradition (that is expected by other Windows programmers as well), that functions exported from DLLs use the _stdcall (STanDard) Windows calling convention.
The argumentation is that a DLL can be seen as a part of the operating system (Windows) rather than the application, because multiple apps can use one and the same dll, hence it's part of the OS, seen from a certain perspective.
And... Windows internally, meaning Windows System functions use _stdcall, so, so should functions in DLLs.
That's the whole point of Petzold (and other Windows programmers?) using _stdcall (or CALLBACK, etc.) for DLL functions.
Obviously this tradition is not in place on POSIX.

Quote:
 Original post by SimonForsmanThe drawback of __cdecl is that it results in slightly larger code for non inlined functions, its advantage is that it allows for functions with a variable number of arguments and won't cause your program to crash catastrophically if you call a function using the wrong number of arguments. (If you use c or c++ the compiler will handle that anyway but for asm applications it can cause some problems)There is absolutely nothing wrong with using __stdcall on Linux/MacOS or __cdecl on Windows, in fact, most Windows software use __cdecl for the majority of their functions, choose calling convention based on how the function in question is used, not what OS its running on.
Yes, thanks, I was aware of the drawbacks and advantages of _std and _cdecl. Thanks. Most Windows software indeed uses __cdelc for the majority of their functions, but that goes only for function not in DLLs, right?
At least Petzold seems to be saying so.

You're absolutely right I think that one should choose the correct calling convention based on what one needs, (multiple arguments OR smaller code). So this means I really shouldn't mind about either Windows or POSIX traditions?
If so, I guess I'd specifically use _stdcall for all library functions, both on Windows and POSIX, except for those functions with variable arguments, where I'd be using _cdecl on both platforms.
Is that really smart / correct? Should I thus really abandon the Windows tradition as Petzold shows and encourages?

Quote:
 Original post by DragonionThe major benefit of keeping macros like this in a separate module/file is that you only need to write them once; afterward you can conveniently import them into each of your projects#include MYEXPORT int MyExportedFunction ( void );And the party doesn't stop here, because re-using the module like this all the projects using it are affected immediately everytime you update it; all you need to do is re-compile them!
Hey, thanks to you too for your help!
Yes, I got a similar example from "Programming Applications for MS Windows" by Jeffrey Richter. I even took it to another level.
I'm actually creating a library-specific "IMPORT" / "EXPORT" macro (the one you call "MYEXPORT". It's defined in the header of each library as "IMPORT" (if it was not defined yet - #ifndef) because the header is included in the library-using source code, so the functions are correctly imported by the app. Then in the library source code it is defined as "EXPORT" BEFORE the library header is included (which would define it as "IMPORT"). Thus in the compiled library header (which would define it as "IMPORT" only if it's not defined yet - #ifndef) and source code it is correctly "EXPORT".
- Actually that's an old trick I assume.
I also created a header to define different import and export specifiers, depending on the type of library, you know, static, run-time dll or load-time dll, as they have different import / export specifiers.
The "IMPORT" / "EXPORT" specifier of each library actually maps / expands to the correct one of those, so I can use one common cross-platform import / export macro for all functions from libraries.

Anyway, all this is totally independent of the calling convention which my original question actually was about, as the calling convention must go between the function's return type and the function name, not BEFORE the function's return type as the "IMPORT" / "EXPORT" specifiers. But thanks for helping! :-)

Mark

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Quote:
 Original post by MarkieThanks. Most Windows software indeed uses __cdelc for the majority of their functions, but that goes only for function not in DLLs, right?At least Petzold seems to be saying so.

In general you want __stdcall for dynamic libraries yes since those functions won't be inlined anyway allthough the only real benefit is marginally smaller code and compatibility with languages that doesn't support __cdecl (such as VB6), i was wrong in my earlier post though, gcc and msvc doesn't handle __stdcall in the same way. for gcc you want:
__attribute__((stdcall)) rather than __stdcall

#ifdef __GNUC__
#define __stdcall __attribute__((stdcall))
#endif

or better yet i guess:

#ifndef WIN32
#define __stdcall
#endif

then simply ignore all non Microsoft compilers for the Windows platform and let other platforms use their default convention.

IIRC this applies to gcc on all platforms, even Windows so its necessary if you want to support more than one compiler.

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Quote:
 Original post by SimonForsmanIn general you want __stdcall for dynamic libraries yes since those functions won't be inlined anyway allthough the only real benefit is marginally smaller code and compatibility with languages that doesn't support __cdecl (such as VB6), i was wrong in my earlier post though, gcc and msvc doesn't handle __stdcall in the same way. for gcc you want:__attribute__((stdcall)) rather than __stdcallso you should probably add:#ifdef __GNUC__#define __stdcall __attribute__((stdcall))#endifor better yet i guess:#ifndef WIN32#define __stdcall#endifthen simply ignore all non Microsoft compilers for the Windows platform and let other platforms use their default convention.IIRC this applies to gcc on all platforms, even Windows so its necessary if you want to support more than one compiler.

Hey yea, thanks!
That was exactly my next question, I was gonna add it by edit, but you've already addressed it: :-)
The implementations of invoking the std and cdecl calling conventions are not the same in Windows and in POSIX, are they?
If they are not, as your posts suggests (on Linux, std calling convention is invoked by "__attribute__((stdcall))" instead of "__stdcall", it might be wise to create cross-platform (cross-compiler as well?) macros such as "cp_CALLCONV_STD" and "cp_CALLCONV_CDECL" which map to std and cdecl with the correct implementation on every platform / compiler.
For instance:
#ifdef __GNUC__#define cp_CALLCONV_STD      __attribute__((stdcall))#define cp_CALLCONV_CDECL  ??__attribute__((cdeclcall))??#endif#ifdef __MACOSX__#define cp_CALLCONV_STD    ??whatever??#define cp_CALLCONV_CDECL  ??whatever??#endif#ifdef WIN32#define cp_CALLCONV_STD     __stdcall#define cp_CALLCONV_CDECL   __cdecl#endif

I prefer a "complicated" setup like this instead of a "nifty" #ifndef on one platform only. This way I can explicitly SET any calling convention I want for any function on each platform with one single macro. And if I omit the macro, I simply get the native convention.
(That would probably be in line with your probably good suggestion of using the calling convention which best suits you personally for every function call, regardless of platform or any possible traditions, etc. :-)

For you Linux gurus out there, I'd be curious what the "attribute" stands for in __attribute__((stdcall))?
Does it have to be replaced with anything (if so, what?) or is that literally how std calling convention is invoked? If so what's the meaning / background of "attribute"??
:-)

Mark

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once beyond the monthly The amount is bound will g

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# How do you express 45 degrees in radian in term of pi?

$\frac{\pi}{4}$
Consider that $\pi$ rad corresponds to 180°.
So 45° is (180°)/4 or $\frac{\pi}{4}$.


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# Assigning proper powers of $i$ to vertices of Feynman diagram

I'm reading the chapter 9 "the path integral for interacting field theory" of the Srednicki's QFT book. The lagrangian we are dealing with here is given by $$\begin{gather} \mathcal{L} = \mathcal{L}_0 + \mathcal{L}_1\\ \mathcal{L}_0 = -\frac 12 \partial^\mu\phi\partial_\mu\phi - \frac 12 m^2\phi^2 \,\,\mbox{(free lagrangian)}\tag{9.8}\\ \mathcal{L}_1 = \frac16 Z_g g\phi^3 + Y\phi -\frac 12 (Z_\phi-1)\partial^\mu\phi\partial_\mu\phi - \frac 12 (Z_m-1) m^2\phi^2.\tag{9.9} \end{gather}$$ Considering only the $$\phi^3$$ term, the generating functional is \begin{align} Z_1(J) \propto &~ \exp{\left(\frac i6 Z_g \,g \int d^4x \left(\frac{\delta}{i\delta J(x)}\right)^3 \right)} \,Z_0(J)\\ =& \sum_{V=0}^\infty \frac{1}{V!}\left[\frac i6 Z_g \,g \int d^4x\left(\frac{\delta}{i\delta J(x)}\right)^3 \right]^V \\ &\times \sum_{P=0}^\infty \frac{1}{P!}\left[\frac i2 \int d^4y\,d^4z J(y)\Delta(y-z)J(z) \right]^P \tag{9.11} \end{align} where $$Z_0(J)$$ is the generating functional for $$\mathcal{L}_0$$ and $$\Delta(y-z)$$ is the Feynman propagator. Here $$V$$, $$P$$, and $$E :=2P-3V$$ are the number of the vertices, propagators (edges), and the sources (external lines) of each Feynman diagram, respectively. As a Feynman rule, Srednicki assigns $$iZ_g g \int d^4x$$ for each vertex, $$\frac 1i \Delta(y-z)$$ for each propagator, and $$i\int d^4x J(x)$$ for each source. My question is how the powers of $$i$$ of these assigned values are determined? First I guessed that since each term of $$(1)$$ has $$i^V(\frac 1i)^{3V} i^P = i^{V-P+E}$$ as its prefactor, it is natural to assign $$i$$ for vertex, $$\frac 1i$$ for propagator, and $$i$$ for source as mentioned.

However, my guess turns out to be wrong when considering the next term, $$Y\phi$$. The generating functional is then $$Z_Y (J) \propto \exp{\left(iY \int d^4x \left(\frac{\delta}{i\delta J(x)}\right) \right)} \,Z_1(J). \tag{*}$$ According to my guess, a new kind of vertex introduced by $$Y\phi$$ should stand for $$Y\int d^4y$$ because two $$i's$$ cancel off in $$(*)$$ and a prefactor for each term is still $$i^{V-P+E}$$. But in eq. (9.19) on p. 66 (of the 1st edition), Srednicki assigns $$iY\int d^4y$$ instead and I cannot figure out how $$i$$ appears. Can anyone help me understand this? I appreciate any help.

1. By the way, one can restore the dependence of Planck's constant $$\hbar$$ in eq. (9.11) by replacing $$i\to \frac{i}{\hbar}$$, so the counting of $$i$$s is related to the counting of $$\hbar$$s.
2. OP is discussing Feynman diagrams in the $$J$$-picture, i.e. diagrams of the partition function $$Z[J]$$. (In particular there are no amputated diagrams.) Then the simplest weight assignment goes as follows: All propagators (internal and external) have weight $$\frac{\hbar}{i}$$, and all vertices and sources have weight $$\frac{i}{\hbar}$$. (Sources may be viewed as 1-vertices.)
4. Finally let us address OP's last question. Since $$Y$$ is a 1-vertex, it has weight $$\frac{i}{\hbar}$$.


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# Two sided binomial hypothesis test interpretation, what is the meaning of the two sided test?

I am on the Binomial Test entry in Wikipedia and in it, there is an example where a die is rolled 235 times, and 6 comes up 51 times. They claim there are two ways to construct a two-sided hypothesis test to compute a p-value for this.

There are two methods to define the two-tailed p-value. One method is to sum the probability that the total deviation in numbers of events in either direction from the expected value is either more than or less than the expected value. The probability of that occurring in our example is 0.0437. The second method involves computing the probability that the deviation from the expected value is as unlikely or more unlikely than the observed value, i.e. from a comparison of the probability density functions.

I understand what the second method is, and that is the only method I am familiar with. That is, define the p-value to be

$$\displaystyle P_{val} = \sum_{p(y) \leq p(51)}p(y)$$

This means summing up the probabilities smaller or equal to that of the event of 6 coming up 51 times.

However, I am lost what the first method is. What does it mean, "sum the probability that the total deviation in numbers of events in either direction from the expected value is either more than or less than the expected value."?

This seems like it would imply summing probabilities from 51 to 235, then summing from 1 to 50, which is obviously incorrect. What are they trying to say?

• "... sum the probability that the total deviation in numbers of events in either direction from the expected value is either more than or less than the expected value" is awful English. Given an observation $x$ (treated now as a number, not a random variable) and an expected value of $\mu$ under the null hypothesis, it's trying to refer to the probability of the event $|Y-\mu|\ge|x-\mu|$ when $Y$ follows the null distribution.
– whuber
Aug 6 at 13:19


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## Good Latex Guide

Hello,
for those of you who are new to latex, there is a really good guide that can be found here:

ftp://ftp.ams.org/pub/tex/doc/amsmat...math-guide.pdf

Pretty much has everything I need.
Hope this helps


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# Measure has density with respect to Lebesgue measure

When we assume that a probability measure $$G$$ has density $$g$$ with respect to the Lebesgue measure $$\lambda$$. Do we refer to Radon-Nikodym theorem, where we have that

$$\frac{dG}{d\lambda}=g$$ and $$G$$ is absolutely continues to $$\lambda$$ ??

• Yes, that is exactly what a density function is. Oct 1 '20 at 11:41

For a measure $$\mu$$ and a non-negative function $$g$$ we can always define a measure $$\nu$$ as $$$$\nu(A) = \int_A g(x) \mu(dx) \tag{1}$$$$ and it is immediately obvious that $$\mu(A)= 0 \Rightarrow \nu(A)=0$$, so $$\nu$$ is absolute continuous with respect to $$\mu$$.
The Radon-Nikodym theorem however states the reverse implikation, namely that if $$\mu,\nu$$ are $$\sigma$$-finite measures with $$\nu << \mu$$, then there exist a non-negative function $$g$$ such that $$(1)$$ holds. The fact that such a density is unique $$\mu$$-almost everywhere allows us to speak of the density of $$\nu$$ with respect to $$\mu$$, which we call the Radon-Nikodym derivative and we often use the notation $$g(x) = \frac{d\nu}{d\mu} (x)$$ to motivate intuitive formulas, such as $$\int_A d\nu = \int_A \frac{d\nu}{d\mu} d\mu.$$ So just to recap. You do not need to use the Radon-Nikodym theorem to assume that $$G$$ has a density with respect to $$\mu$$, you can simply assume that $$G$$ is given as in $$(1)$$ (with $$G=\nu$$). However the Radon-Nikodym theorem states that it is equivalent to assume absolute continuity with respect to $$\mu$$ (when we are dealing with $$\sigma$$-finite measures).


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# qutip.sesolve and qutip.optimize_pulse_unitary produce different results

I'm currently trying out some things with qutip in the field of optimal control (state-to-state transfer) and have some trouble to reproduce my results from the optimization process with sesolve.

My code for the optimization is:

# define the main hamiltonian of our system
H_0 = Qobj(w[0:dimension]*np.identity(dimension))
# define the drive-hamiltonian
H_d = Qobj(matrix)
# define the control-list for later optimization
H_c = [H_d]

# define the initial state of our system
psi_0 = basis(dimension,0)
# define the target state of our system
psi_target = basis(dimension,1)

# define the projector onto the inital state (|psi_0><psi_0|)
proj0 = ket2dm(psi_0)
# define the projector onto the target state (|psi_target><psi_target|)
proj1 = ket2dm(psi_target)
.
.
.
result = cpo.optimize_pulse_unitary(H_0, H_c, psi_0, psi_target, n_ts, evo_time,
max_iter=max_iter, max_wall_time=max_wall_time,
init_pulse_type=p_type,
log_level=log_level, gen_stats=True)

Where I get a final fidelity error of 2.0313561943652303e-10.

I wanted to verify this result by taking the optimized pulse and plugging it into the qutip.sesolve function but doing so gives me fidelity errors which are not compatible with the result from the optimization.

My code for the sesolve function is:

# define an evenly spaced list with the number of intervals being equal to the the time steps of our control pulse
tlist = np.linspace(0, evo_time, n_ts)
# define the optimized pulse
optim_pulse = result.final_amps[:, 0]
# define our "Hamiltonian"
H = [H_0, [H_d, optim_pulse]]
# define the e_ops list
e_ops = [proj0, proj1]
# define the intial state
init_state = psi_0

options = Options()
options.store_states = True

# compute the time-evolution of the given Hamiltonian (including the drive term) from an inital state to a final state
evolved_State = qutip.sesolve(
H=H,
psi0=init_state,
tlist=tlist,
e_ops=e_ops,
options=options)

Which produces a fidelity error of 0.0011998990068043947.

I wondered if I'm doing something wrong with the application of the sesolve function or if its something else which produces this difference?

Using other inital pulses I even sometimes get final populations of the target state lower then the inital state while the optimization result says it converged in the fidelity error.

Would be awesome if anyone could share some input on this idea, thank you!


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## Notes to Properties

1. On viewing Plato’s forms as universals, cf. Harte (2008). As Harte makes it clear, this interpretation is common, but not unanimous. For a recent challenge to it, see Marmodoro (forthcoming), according to whom, it is Aristotle who “‘discovers’ universals”. The universal/particular distinction, which we are taking for granted here, though typical and widespread, is not uncontroversial; MacBride (2018) questions it, on the basis of a reconstruction of its manifestations in early analytic philosophy.

2. As is well-known, Quine (1948) rejects quantification over predicates and this is why he views predicates as devoid of ontological commitment. Castañeda (1976: §3) argues against this.

3. In its more radical versions nominalism involves a sweeping rejection of abstract entities, including predicates understood as types rather than tokens, and sets. Predicate nominalists usually view singular terms for properties with suspicion. For, following Quine, they take singular terms to carry an ontological commitment that predicates lack. Moreover, they must somehow capture the similarity in the classifying jobs made by predicates of different languages such as the English “red” and the French “rouge”. Finally, if they take predicates to be types, then the predicates are abstract and can even be conceived of as properties (see §4.2). To confront such issues, nominalists typically appeal to quotation devices and paraphrases. A locus classicus of this strategy is Sellars 1963, which many nominalists still find compelling. Many problems for approaches of this sort have been pointed out (Castañeda 1976; Armstrong 1978a: ch. 2; Bealer & Mönnich 1989: §7; Loux 1998: ch. 2; Szabó 2003). Alternatively, in so-called Class nominalism, properties are proxied by predicate extensions. There are severe problems here as well (Armstrong 1978a: ch. 4): most notoriously the fact that co-extensive predicates cannot be appropriately discriminated (see §3.1; Lewis’ conception of properties as sets of possibilia discussed therein may be viewed as a sort of class nominalism). Perhaps, the most radical version is Ostrich nominalism (Armstrong 1978a: 16; Devitt 1980; Armstrong 1980). This sees a brute fact where the realist finds an explanandum requiring mind-independent universals. For a detailed taxonomy of nominalist positions, see Allen 2016: ch. 4, and Koons & Pickavance 2017: ch. 8. Recently, Schulte (2019) has proposed “grounding nominalism”, which takes properties, possibly understood as universals, to be non-fundamental. Still, their existence is granted, and this concession makes it dubious that this is really nominalism. For sustained criticisms of conceptualism, see Armstrong 1978a: ch.3, and Bealer & Mönnich 1989: §8.

4. Leibniz is usually viewed as a philosopher who stands out for his attempts to effectively show how relations can be reduced to monadic properties (Mugnai 1992; in contrast, Orilia 2000a defends the claim that Leibniz acknowledged the existence of irreducibly relational facts). Some have remained attached to the old idea that there are no external relations (e.g., Fisk 1972), but this position remains largely minoritarian nowadays. The view that there are relations but no monadic properties, or at least that the former have ontological priority over the latter, has also been considered. It is defended in different forms by Dipert (1997) and by various authors in the context of ontic structural realism (see, e.g., French & Ladyman 2003; Esfeld 2003; entry on structural realism, §4). This view is also far from having gained some consensus (Ainsworth 2010; Sider 2020: §3.4). In sum, most philosophers would agree that there are both genuine monadic properties and genuine relations (for recent dissenting voices see, however, Marmodoro & Yates 2016).

5. Benovsky (2014) argues that choosing between universalism and tropism may be an insurmountable problem.

6. The disagreement between the universalist and the tropist is operative at a very basic ontological level. One may wonder however whether divergences at this foundational layer have some impact on more specific philosophical issues and indeed it has been claimed that this is the case in philosophy of mind, in particular as regards mental causation and the tenability of reductive physicalism (Robb 1997; entry on mental causation, §6.5; Gozzano & Orilia 2008).

7. “Property” is being used here as elsewhere in this entry in a most generic sense, possibly not in line with Lowe’s more restricted usage of it. In general, in Lowe’s account, kinds and attributes are instantiated by objects and modes, respectively; kinds and objects are characterized by attributes and modes, respectively; attributes are exemplified by objects, whether dispositionally or occurrently. Exemplification is dispositional to the extent that the exemplifying objects instantiate kinds characterized by the attributes in question, while it is occurrent when the exemplifying objects are characterized by modes that instantiate the attributes in question.

8. There is an analogous tropist bundle theory in which the elements of the bundles are tropes rather than universals (see entry on tropes, §3.2).

9. Going very much beyond what Armstrong explicitly say, we can perhaps take the elements of these collections to be tropes, so that the first collection is a class of perfectly resembling tropes (playing the role of a universal); and the second collection is a class of compresent tropes as in the tropist bundle theory of objects.

10. Contrary to what Allen (2016: §2.4.1) emphasizes, the main problem here is not so much that the state of affairs in question would have an infinite complexity, as it would have an infinite number of exemplification relations as constituents. As Allen notes, this could be swallowed, after granting Cantorian actual infinity. The problem is rather that no relation ends up occurring as attributed in this state of affairs (recall the distinction of §1.3 and §1.4 between occurring as argument and occurring as attributed).

11. Lowe vacillates between taking these formal relations to be internal and declaring them non-existent (Lowe 2006: 111, 167). Perhaps, they are considered non-existent precisely because internal, but separating internality from non-existence may be a better choice, since after all formal relations are in Lowe’s ontological inventory. Moreover, it is really their internality, rather than lack of existence, that is appealed to in his attempt to avoid Bradley’s regress, as we shall now see.

12. See, e.g., Peacock 2012 for an approach resorting to Russell’s notion of a relating relation, as opposed to a relation that is not doing its relating work; Tegtmeier 2013, for a view according to which a fact itself grounds its unity; Schneider 2013 for an appeal to “unrepeatable nexuses” in the context of an Aristotelian framework in the spirit of the one defended in Lowe 2006; Meinertsen 2018 for a view of exemplification as relating itself to its relata.

13. Quine (1957 [1969: 21]) contrasts the “murky intensionality of attributes” to the “limpid extensionality of classes” and speaks of all intensional entities as “creatures of darkness” (1956 [1976: 186]).

14. Here, however, one could more generally invoke causal roles rather than causative ones and claim that epiphenomenal properties are at least caused by certain other properties.

15. Formal systems of property theory are often provided with an algebraic semantics that associates primitive predicative terms of the language with “basic” properties and the lambda abstracts with complex properties obtained from the basic ones by means of operations that generate new properties from given ones (Bealer 1973; 1982; McMichael and Zalta 1981; Leeds 1978; Menzel 1986; Swoyer 1998; Zalta 1983). Thus, for example, one assumes that there is an operation, $$\amp$$, that maps each pair of properties, $$\bP$$ and $$\bQ$$, to the conjunctive property $$\bP \amp \bQ$$. If “$$P$$” and “$$Q$$” stand for $$\bP$$ and $$\bQ$$, respectively, then “[$$\lambda x(P(x) \amp Q(x))$$]” will stand for $$\bP \amp \bQ$$. For another example, it is typically assumed that there is an operation, $$\PLUG_1$$, that, given a two-place relation $$\bR$$ and an object $$\bd$$, generates the monadic property $$\PLUG_1(\bR, \bd)$$. If “$$R$$” and “$$d$$” denote $$\bR$$ and $$\bd$$, respectively, then the property $$\PLUG_1(\bR, \bd)$$ will be denoted by the lambda term “[$$\lambda x R(d, x)$$]”. The property in question is the one that something has when $$d$$ bears the relation $$R$$ to it.


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# Determining possible minimal polynomials for a rank one linear operator

I have come across a question about determining possible minimal polynomials for a rank one linear operator and I am wondering if I am using the correct proof method. I think that the facts needed to solve this problem come from the section on Nilpotent operators from Hoffman and Kunze's "Linear Algebra".

Question: Let $V$ be a vector space of dimension $n$ over the field $F$ and consider a linear operator $T$ on $V$ such that $\mathrm{rank}(T) = 1$. List all possible minimal polynomials for $T$.

Sketch of Proof: If $T$ is nilpotent then the minimal polynomial of $T$ is $x^k$ for some $k\leq n$. So suppose $T$ is not nilpotent, then we can argue that $T$ is diagonalizable based on the fact that $T$ must have one nonzero eigenvalue otherwise it would be nilpotent (I am leaving details of the proof of diagonalization but it is the observation that the characteristic space of the nonzero eigenvalue is the range of T and has dimension $1$). Thus the minimal polynomial of $T$ is just a linear term $x-c$.

Did I make a mistake assuming that $T$ can have only one nonzero eigenvalue?

Thanks for your help

• Assume that $n\ge2$ and pick $v\ne0$ in the image of $T$, then $Tv=av$ for a given $a$ and the minimal polynomial is $x^2-ax$. – Did Jul 19 '11 at 13:14
• If $T$ has more than one non-zero eigenvalue then it has more than one eigenvector, hence rank exceeding 1. – Gerry Myerson Jul 19 '11 at 13:18
• Thank you Didier and Gerry so am I correct in saying that the only possible minimal polynomials are of the form $x^k$, $x-a$, and $x^2 -ax$ – user7980 Jul 19 '11 at 14:16
• As I said: if $n\ge2$, the only possible minimal polynomial is $x^2-ax$ (for $a=0$, this includes $x^2$). – Did Jul 19 '11 at 17:02

If $n=1$ then $T$ is multiplication by some $a\in F$, the minimal polynomial$\mu_T$ of $T$ is $X-a$, and $\def\rk{\operatorname{rk}}\rk T=1$ means that $a\neq0$. Now assume $n>1$; then $T$ cannot be any scalar multiplication, so one certainly has $\deg(\mu_T)>1$.
Since $T$ has rank$~1$, it can be factored through (the $1$-dimensional vector space) $F$, that is $T=f\circ\alpha$ where $\alpha\in\mathcal L(V,F)=V^*$ and $f\in\mathcal L(F,V)$ is just scalar multiplication of a fixed vector $v\in V$. One has $T^2=f\circ\alpha\circ f\circ\alpha$, and since $\alpha\circ f\in\mathcal L(F,F)$ is scalar multiplication by $\def\tr{\operatorname{tr}}c=\alpha(v)=\tr T$, one gets $T^2=c(f\circ\alpha)=cT$, so that $\mu_T=X^2-cX=X(X-c)$. Here $c\in F$ can be anything; if $c=0$ then $T$ is nilpotent of order$~2$, hence not diagonalisable, otherwise $T$ is diagonalisable with distinct eigenvalues $0$ and $c$.
An alternative argument without using an explicit decomposition of $T$ goes as follows. By rank-nullity $T$ has an eigenvalue$~0$, so $\mu_T$ contains a factor$~X$. The remaining factor $\mu'=\mu_T/X$ of$~\mu_T$ is equal to the minimal polynomial of the restriction of$~T$ to its image. That image has dimension $\rk T=1$ on which $T$ therefore acts by some scalar$~c$, so $\mu'=X-c$, and $\mu_T=X(X-c)$. There are many ways to see that $c$ is the trace of$~T$, one of which it that the trace of any operator equals that of its restriction to its image.


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Home > linux > Multiple Kerberos realms on single host

Multiple Kerberos realms on single host

We are using Kerberos for central authentication at our university department; recently, we have set up a second realm for authentication to other services than UNIX logins (e.g. webmail) – we did not like the idea of reusing the same passwords.

Setting up a second realm has been pretty straightforward, just add the other realm to /etc/krb5.conf, write explicit port number to admin_server and kadmin_port options, add another init script for secondary admin server and… it almost works.

But suddenly, users find themselves unable to change their password within the original realm using passwd or kpasswd, with less than helpful Authentication error: Failed reading application request. And that’s the reason I’m writing up this, since you explicitly need to set the totally undocumented option kpasswd_port to a different value than 464! This is because kadmind provides the kadmin service on TCP port (749 by default), but also chpw service on UDP port (464 by default).

Categories: linux Tags:


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# Revision history of "Antiscalar"

• curprev 03:30, 18 May 2022  574 bytes +574 Created page with "An ''antiscalar'' in a geometric algebra is an element having antigrade 0. Antiscalars are multiples of the volume element given by the wedge product of all basis vectors. The basis element representing the unit antiscalar is denoted by $$\large\unicode{x1D7D9}$$, a double-struck number one. The unit antiscalar $$\large\unicode{x1D7D9}$$ is the multiplicative identity of the geometric antiproduct. For a general element $$\mathbf a$$, the notation a_{\larg..."


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Change bold font

I want to use Avenir Next Medium as the bold font for my document whenever I use \textbf. Is there a way I can use \textbf and specify a font to be used. After that, can I also customize the size for any font rendered using \textbf?

• I assume you're using LuaLaTeX or XeLaTeX? If so it's not hard to do. – Alan Munn Nov 18 '13 at 22:47

Since the Avenir Next font is a commercial font, I'm assuming you are using either XeLaTeX or LuaLaTeX. It's easy to specify a different font for bold text. Here's a simple example:

% !TEX TS-program = XeLaTeX
\documentclass[]{article}
\usepackage{fontspec}
\setmainfont[Ligatures=TeX,BoldFont={Linux Biolinum O},Scale=MatchLowercase]{Linux Libertine O}
\begin{document}

This is some text. \textbf{This is some bold'' text}
\end{document}

Assuming you use the LuaLaTeX format, you could achieve your objective by including the following instructions in the preamble of your document:

% !TEX TS-program = lualatex
\usepackage{fontspec}
\setmainfont[BoldFont = {Avenir Next Pro Medium},
BoldItalicFont = {Avenir Next Pro Medium Italic}]
{Avenir Next}

This setup obviously requires that fonts of the Avenir Next Pro family, distributed by Linotype, are available on your system.

If you use XeLaTeX instead of LuaLaTeX, you will probably have to use the actual file names of the fonts rather than "logical" font names such as "Avenir Next Pro Medium".

• I had no (well, only a few) problems with XeLaTeX logical font names. – marczellm Nov 19 '13 at 7:30


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# In a Pentagon, Two Angles Are 40° and 60°, and the Rest Are in the Ratio 1 : 3 : 7. Find the Biggest Angle of the Pentagon. - Mathematics

Sum

In a pentagon, two angles are 40° and 60°, and the rest are in the ratio 1 : 3 : 7. Find the biggest angle of the pentagon.

#### Solution

In a pentagon, two angles are 40° and 60° Sum of remaining 3 angles = 3 x 180°

= 540° – 40° – 60° = 540° – 100° = 440°

Ratio in these 3 angles =1 : 3 : 7

Sum of ratios =1 + 3 + 7 = 11

Biggest angle =(440 xx 7)/11= 280°

Concept: Sum of Angles of a Polynomial
Is there an error in this question or solution?

#### APPEARS IN

Selina Concise Mathematics Class 8 ICSE
Chapter 16 Understanding Shapes
Exercise 16 (A) | Q 14 | Page 181


text: $100$ people are at a party, and each is friends with at most $10$ of the other guests. Determine the greatest $n$ for which there must exist a set of $n$ guests such that no two of them are friends.


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# Can massive ammounts of 0603 SMD 1uF ceramics displace power caps?

Edit:Warning: there's an error in one of my assumptions in this question, which more or less invalidates it. See Tom Carpenters answer.

I'1m designing a switched mode power supply, and of course need various capacitors in the design.

For instance, to spare the mains from all the noise from the switching action, I'm thinking about building an LC filter with a 3uF capacitor.

So now I've been trying to source 3uF capacitors. Of course it's not difficult at all to source a 3uF 450V capacitor. Polypropylene caps seem to be suitable.

But the cost of a capacitor capable of handling at worst 1A of ripple current, is signficiant (say 5 euro). See for example.

However, 1uF ceramics can be had almost for free, see for instance. 2000 caps for ~13 euro.

If I put the 1uF 100V caps 15 in series, 3 in parallell, I need 45 caps. This will cost approximately 0.3 euro.

Of course I'll need some resistors to balance the caps. Placing the caps at a reasonable distance I calculate that they would all fit within a 12.5mm x 12.5mm box, which means they take up a fraction of the space of the Epcos one.

One possible problem could be that DC bias reduces capacitance of ceramics. But for my application I need approximately 400V, so with 15 in series the 100V 0603 caps are loaded only marginally, and should retain most of their 1uF capacitance.

I'm not really considering doing this, but would it work? Or will the 0.3A+ ripple current kill the little 0603:s?

• If you put capacitors in series, the capacitance goes down - so if you put 15 of the 1uF caps in series you would essentially have a 67nF capacitor. Also, unless the capacitors are identical, the voltage drop of each will not split evenly which may or may not cause issues with the voltage across a given capacitor being higher than its rating. – Tom Carpenter Aug 10 '15 at 13:29
• Darn! Hadn't thought about that. That invalidates the entire idea. Many thanks!! – avl_sweden Aug 10 '15 at 13:31
• Generally the same idea is employed for stuff like tesla coil capacitors (google mmc capacitor) but you need bleeder/equalization resistors. You can design it, do the math, and calculate the costs (not only in components, but also in pcb space and power lost). Then it is very likely the dedicated cap doesn't look so bad anymore. – PlasmaHH Aug 10 '15 at 13:34
• "the cost of a capacitor capable of handling at worst 1A of ripple current is signficiant" -- Then how about paralleling smaller caps? 3x 1µF reduce the ESR to 1/3 and provide 3x the current of a single 1µF; scales to 6x 500nF, 30x 100nF, or whatever provides best cost/benefit ratio for you. – JimmyB Aug 10 '15 at 15:14
• Possibly useful: Temperature and Voltage Variation of Ceramic Capacitors (Maxim IC) and COMPARISON OF MULTILAYER CERAMIC AND TANTALUM CAPACITORS (AVX). Note the AVX doc is significantly more optimistic than the Maxim one (based on Murata data). – The Photon Aug 10 '15 at 16:03

When placing capacitors in series, the overall voltage rating does go up as you have identified. You have to be careful though as if the capacitance is not equal some will end up with a larger voltage drop than others which may push them outside their rated operating voltage.

The trouble is a more fundamental one. Placing capacitors in series causes the capacitance of the chain to be less than each individual one. Essentially the following equation becomes valid:

$$\frac{1}{C_{series}} = \frac{1}{C_1} +\frac{1}{C_2} +\frac{1}{C_3} + \space...$$

If all of the capacitors are the same value, this simplifies down to:

$$C_{series} = \frac{C_{single}}{N}$$

Where in this case $N$ is the number of capacitors in the series chain. So if you were to place $N=15$ of the $1\mathrm{\mu F}$ capacitors in series, this would mean the overall capacitance reduces to $~67\mathrm{nF}$, so you would actually have to place $45$ of these chains in parallel to get the $3\mathrm{\mu F}$ that you require (or to put it another way $675$ capacitors in total).

As PlasmaHH points out, for 450V rating, you only need 5 ideal capacitors in the series chain to get the voltage rating required rather than the 15 you mentioned in the question. The feasibility of this would depend on how the capacitance derates with voltage, and would also be contingent on ensuring the capacitors are correctly balanced with bleed resistors to ensure that none of the chain go out of spec. Adding more would increase the total count, but also increase the resilience of the circuit to voltage fluctuations and capacitor tolerance.

So lets repeat the calculations. If you were to place $N=5$ of the $1\mathrm{\mu F}$ capacitors in series, this would mean the overall capacitance reduces to $~200\mathrm{nF}$, so you would have to place $15$ of these chains in parallel to get the $3\mathrm{\mu F}$ that you require which brings to total requirement down to a minimum of $75$ capacitors in total. You would have to determine if this is cost effective in terms of space and component cost.

• I am not sure where (here and in the question) the 15 comes from, isn't he trying to up from 100V to 450V ? – PlasmaHH Aug 10 '15 at 13:45
• @PlasmaHH Here it comes from it what was asked in the question. You would theoretically only need 5 if they were balanced, so that would actually be considerably less required. – Tom Carpenter Aug 10 '15 at 13:48
• Yes, but ceramic caps lose their capacitance when operated near their rated voltage (!). Also, as the voltage sharing won't perfectly even, it's good to have some margin. But the number 15 was just arbitrary. – avl_sweden Aug 10 '15 at 13:48
• 15 might be a high number. But I think even with 5 in series you would need a huge number in parallell. 5 to offset the series reduction, 3 to get from 1uf to 3 uf and maybe 3 again because of ceramic cap reduction at high voltage. – avl_sweden Aug 10 '15 at 13:52
• And 225 caps would be a bit impractical... :-) – avl_sweden Aug 10 '15 at 13:55


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## Valuation for investors

Twice in the past week I have had pre-product entrepreneurs tell me that they were raising seed rounds at an approximately $10mm pre-money. In both cases I had to pass, despite the merit of the management teams. Both companies told me that they have other early-stage investors ready to fill their rounds, and I’m glad. I am generally of the opinion that if an entrepreneur can get a better valuation while still getting value-add investors, then they should. Plenty has been written recently by venture capitalists about venture capital to help entrepreneurs. Not much has been written to help newer venture capitalists. I think, in a way, this is because VCs don’t care so much if those new to the industry succeed or not. If a new angel loses his shirt, well, one less competitor for me down the line. I don’t believe that, though. I think we need more investors. But smart ones, investors who make some money on their investments and so feel confident reinvesting it in a new round of entrepreneurs. Investors, like everyone else, get smarter and more helpful the more experience they have. Smarter investors is better for the ecosystem. So, valuation. Valuation in venture capital is tough. The amount of uncertainty between investment and exit is immense. But that doesn’t mean that you shouldn’t try to pay the right price. Valuing a startup correctly means estimating risk, not contemplating the unknowable. The idea that was briefly tossed around that valuation doesn’t matter because a startup either goes big or dies is wrong. Ludicrous, in fact. There’s a range of outcomes for every fund. The cliche that out of each ten investments, two are winners, three are failures and five go sideways shows this. The two winners determine whether the fund is an overall winner or loser, but how sideways the five go determines whether it’s a good return or a great return. What happens to the second tier of investments matters. And, for the math challenged, even if startup returns were binary, when you invest in many of them you get a binomial distribution, so expected value matters. I look at valuation this way. For every company, • I think about what the expected exit would be if the entrepreneurs were right: if they are right about the problem, about what their customers want, about their ability to execute, right about everything. • I figure out how much dilution I expect before the exit. • I decide how likely it is they are right and multiply by the expected exit to get an expected value. • I divide by three, because I’d like my investments as a whole to return 3x*. This is the post-money valuation. An example. Company X is an amazing data-driven adtech company. It’s going to disrupt some existing companies and if it does it should sell for$400mm in five years. I think, given the risks, that there’s a one in ten chance they will succeed.

But I know they will need to raise a Series A to commercialize the product once it’s ready, and a Series B to ramp sales once they have product-market fit, and a Series C to expand the product line. The Series A will be 33% of the company, the Series B will be 25%, and the Series C 20%. My stake will be diluted down to 40% of my original ownership.

So my post-money expected value of the company is $400mm * 10% * 40% =$16mm. I would be looking for a post-money of $5mm. If the company is raising$1mm in the seed round, the pre-money valuation would be $4mm. You can see the difficulty in the$10mm pre-money. If the company is raising, say, $2mm at a$10mm pre, then the expected exit value would have to be $12mm/(10% * 40%) * 3 =$900mm.

Billion dollar exits are the sine qua non of the venture business. But they are rare. Rarer than you think.

I made a list off the top of my head of some 125 business-to-business advertising exits. I may be missing some obvious ones, but there were only a handful of $500mm plus exits in the last ten years, even fewer billion dollar ones (M&A exits, I didn’t count IPOs, so I probably undercounted by one or two.) Here are the$500 million and up exits I have.

 Company Acquiror Price ($mm) Date aQuantive Microsoft$5,900 May-07 Doubleclick Google $3,100 Apr-07 Omniture Adobe$1,800 Sep-09 Overture Yahoo! $1,630 Jul-03 Digitas Publicis$1,300 Dec-06 NetRatings Nielsen $817 Feb-07 AdMob Google$750 Nov-09 Right Media Yahoo! $680 Jul-07 Lending Tree IAC$675 Aug-03 24/7 Real Media WPP $650 May-07 Rosetta Publicis$575 May-11 Razorfish Publicis $530 Aug-09 Of the 125 exits, five were more than a billion, seven were between$500 million and a billion, 20 were between $200 million and$500 million, and 15 were between $100 million and$200 million. The rest were sub-$100 million. Remember, these were all exits–companies that didn’t make it weren’t counted. There’s also a bias in the list because I am more aware of the large exits; I would be surprised if I missed too many billion dollar exits but I am sure I missed many$10mm exits. Also note that only a couple of the billion dollar exits here were as straightforward as my model: aQuantive was built through acquisition (and thus had substantially more dilution), Doubleclick had gone through several owners including the public markets, etc.

In fact, all else being equal (a priori, that is) billion dollar exits returned less overall than $500mm-$1bn exits, because there were fewer of them. Exits between $200mm and$500mm probably returned slightly more than $500mm-$1bn exits (also because there were more of them). The $100mm-$200mm range and less than $100mm range each return less than the$200mm-$500mm range**. Here’s my estimate of what each of these ranges returned.  Exit Range Companies Total Est. Value$1bn + 2 $3,000$500mm – $1bn 7$5,000 $200mm –$500mm 15 $6,000$100mm – $200mm 25$4,000 $50mm –$100mm 50 $3,750 <$50mm 100 $2,500 The sweet spot in adtech, the “average” expected exit value, seems to be around$400mm.

Every industry is different, you need to know yours. Make a list of exits over the last ten years, all exits not just the good ones. Then try and figure out how many companies were funded in your industry. This will inform your expected exit values in the success case as well as help you decide what percentage of funded firms get to an exit. Conditions change all the time, of course, but looking at the last ten years will probably keep you reasonably conservative.

—–
* I asked an engineer friend of mine how comfortable he is being in the buildings he helped design. “Pretty comfortable”, he said. “You never worry?” I asked. “Look”, he said, “There’s a lot of math and experience behind choosing exactly how much steel and concrete the building needs to bear its load. I do the work carefully, run the calculations twice and make absolutely sure the answer I am getting is the right one. Then I multiply by three.” [Edit: For those for whom anecdote is not analysis, I’ll point out that a 25% IRR compounded annually for five years is 3x].

** The list I made is up here. Click on the headers to sort. I think all of the #N/As are sub $50mm exits except the two bolded ones, which I think are ~$200mm exits. [Edit: Wow, the sorting on the linked table was all screwy. Sorry. Fixed it.]

1. Jerry, your thinking is most helpful to the entire ecosystem. Thank you.

2. Wow, very insightful, thank you

3. Is it a coincidence that the AdMeld acquisition was $400m? http://selnd.com/jcsOlZ Also, FWIW, counting the aQuantive deal as well as Publicis acquisition of Razorfish double-counts Razorfish. Only the aQuantive shareholders had an exit. 4. I’m sure Microsoft upped their bid to$400m after getting cover from my post. Nobody ever got fired for taking ReactionWheel advice. (If you have been, I will refund you your ReactionWheel subscription payment.)

The Razorfish shareholders had many exits! I was one of the original Razorfish shareholders, from back in the mid-’90s. We had an awesome exit to the public markets in 1998. Then Razorfish got acquired by SBI (the public markets did not have such an awesome exit), who had an awesome exit to aQuantive, who had an awesome exit to Microsoft, who had an awesome (well, in a sense, not to Microsoft) exit to Publicis. When Publicis changes its name to publicisRazorfish, they will have an awesome exit too, I am sure, probably to the government of France, who will rename themselves the République Razorfish.

But yes, you’re right, the exit to Publicis is of a company that can no longer be called venture backed. Regardless, it’s the price that someone would pay for an internet advertising business, so it’s definitely a data point. If someone buys something and then resells it, that’s two data points.The logical inconsistency isn’t that I included Razorfish twice, it’s that I didn’t include the first (IPO) and the second (going private) exits. Same with Doubleclick. I would love to add more data points to the list. If you have any, suggest them!

5. What a great guarantee!

Didn’t mean to short-change the exits that Razorfish went through pre-aQuantive; many of my co-workers (i worked at RF from 06 to 09) were from the Scient/iXL days (Some are still there!) and I know the company changed its fair “share” of times.

I suppose the Microsoft acquisition always bothered me. For a company with so many products, and that now admits it costs about $8 of services per$1 of license fee to get their enterprise products working, they had no appetite for the Razorfish services portfolio; a shortsighted position. Still, a digression from our learning objective about ad tech exits. I would add the Franchisegator deal @$21.5m (later spun out by M$FT for “around $20m!) I was looking for a price for WPP’s minority stake in iconmobile, or more recently Blue State Digital, just as examples, but no prices were disclosed. 6. Anonymous says: This is great stuff. Only thing you need to keep in mind (I believe) is the annual cash these investments are throwing off to their shareholders. For some of these businesses that are high margin, those$s can add up.

7. Really good post…I ran my own startup through your formulas to see what would happen. :-)

One more ad tech company that isn’t on your list is AdMission, which was (sort of) acquired by Cobalt Group in May 2008: http://blog.kelseygroup.com/index.php/2008/05/06/most-of-admission-corp-sold-to-cobalt-group/

But I couldn’t find any pricing info. Maybe someone else has more data on the deal.

I’m not even sure whether AdMission was venture-backed. But they did something similar to what my company does: automate the process of display ad design and creation.

Thanks!


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# Examples and further results about the order of the product of two elements in a group

Let $G$ be a group and let $a,b$ be two elements of $G$. What can we say about the order of their product $ab$?

Wikipedia says "not much":

There is no general formula relating the order of a product $ab$ to the orders of $a$ and $b$. In fact, it is possible that both $a$ and $b$ have finite order while $ab$ has infinite order, or that both $a$ and $b$ have infinite order while $ab$ has finite order.

On the other hand no examples are provided. $(\mathbb{Z},+), 1$ and $-1$ give an example of elements of infinite order with product of finite order. I can't think of any example of the other kind! So:

• What's an example of a group $G$ and two elements $a,b$ both of finite order such that their product has infinite order?

Wikipedia then states:

If $ab = ba$, we can at least say that $\mathrm{ord}(ab)$ divides $\mathrm{lcm}(\mathrm{ord}(a), \mathrm{ord}(b))$

which is easy to prove, but not very effective. So:

• What are some similar results about the order of a product, perhaps with some additional hypotheses?
• The product of two planar symmetries with respect to some lines has an infinite order if (and only if) the angle between these lines is an irrational part of the full angle, although each of these symmetries has order two. – Did May 25 '11 at 16:32
• I think that this example shows why in general we cannot say too much about the order of the product: In $S_n$ any permutation is a product of transpositions.... – N. S. May 25 '11 at 16:37
• Yes, Coxeter groups are groups which are generated by elements of order two with arbitrary conditions on the orders of $xy$ where $x$ and $y$ are two of the generators. I've always been fascinated by equivalence of the fact: "Finite simple groups are either cyclic of prime order or are of even order" can be restated as "finite simple groups are either rotation groups of prime order or images of a Coxeter group." – Thomas Andrews May 25 '11 at 17:12
• My definition of Coxeter group was imprecise, see en.wikipedia.org/wiki/Coxeter_group. But my statement about the relationship between Coxeter groups and finite simple groups stands. – Thomas Andrews May 25 '11 at 17:21

Simple examples are given by free products, assuming you know the normal form for a free product; otherwise, I'm not really saying much.

If you want a more concrete example, take the $2\times 2$ matrices with coefficients in $\mathbb{Q}$, and $$a = \left(\begin{array}{cc}0&1\\1&0\end{array}\right),\qquad b=\left(\begin{array}{cc}0 & 2\\\frac{1}{2}& 0\end{array}\right).$$ Then $a^2=b^2=1$, but $$ab = \left(\begin{array}{cc}\frac{1}{2} & 0 \\0 & 2\end{array}\right)$$ has infinite order.

If $a$ and $b$ commute, with $\mathrm{ord}(a)=m$ and $\mathrm{ord}(b)=n$, then you can do better than Wikipedia. We have that: $$\frac{\mathrm{lcm}(m,n)}{\mathrm{gcd}(m,n)}\quad\text{divides}\quad \frac{\mathrm{lcm}(m,n)}{|\langle x\rangle\cap\langle y\rangle|}\quad\text{divides}\quad \mathrm{ord}(ab)\quad\text{divides}\quad \mathrm{lcm}(m,n).$$ For example, if for every prime $p$ that divides $\mathrm{lcm}(m,n)$, the highest power of $p$ that divides $m$ is different from the highest power of $p$ that divides $n$, then $\mathrm{ord}(ab)=\mathrm{lcm}(m,n)$.

If you are willing to impose global conditions (conditions on $G$), then for example Easterfield proved in 1940 that if $G$ is a $p$-group of class $c$, $a$ has order $p^{\alpha}$, and $b$ has order $p^{\beta}$, then the order of $ab$ is at most $p^m$, where $$m = \max\left\{\alpha,\beta+\left\lfloor\frac{c-1}{p-1}\right\rfloor\right\}.$$ From this you can get similar results for a finite nilpotent group (which is necessarily a product of $p$-groups), and hence for the product of two elements of finite order in any nilpotent group (or in fact, in any locally nilpotent group, or even more strongly, in any group in which the 2-generated subgroups are nilpotent).

• The first example is just a rephrasing of the question: it is true that a product of an element of order $2$ and an element of order $3$ need not have finite order if and only if this is true of the free product, which is the universal example. So there still remains something to be proved: why doesn't a relation of the form $(ab)^n$ hold in the free product? – Qiaochu Yuan May 25 '11 at 16:43
• @Qiaochu: Fair enough; I tend to think of free groups and free products in terms of their normal form (as sets of words, with operation "concatenate and cancel") rather than as defined by their universal property, so to me the fact that the free product of finite groups has elements of infinite order is clear (since there is no cancellation if we repeat a word that "starts" in one group and "ends" in the other). If you think of free products mainly in terms of their universal properties, then it's of course not clear at all why that's the case. – Arturo Magidin May 25 '11 at 16:54
• Ah, yes, matrices! $G$ had to be non commutative and had to contain elements of both finite and infinite order... why didn't I think of matrices? I'm accepting this answer because it's the most complete. I will read the cited paper by Easterfield once Cambridges fixes their website. Thank you for your time! – Jacopo Notarstefano May 25 '11 at 22:09

Here is a rephrasing of the question which I think aptly illustrates the value of thinking in terms of universal properties. Let $n, m$ be two positive integers. There is a group $C_n * C_m$, the free product of the cyclic groups $C_n, C_m$, with presentation $\langle a, b | a^n, b^m \rangle$. It satisfies the following universal property:

Let $G$ be a group with elements $g, h$ such that $g^n = h^m = 1$. Then there is a unique homomorphism $\phi : C_n * C_m \to G$ such that $\phi(a) = g, \phi(b) = h$.

Depending on how comfortable you are with presenting a group by generators and relations, this might be obvious. Nevertheless, the existence of the free product $C_n * C_m$ is helpful because it allows us to recast a question about an entire class of groups (the class of groups with an element of order $n$ and an element of order $m$) into a question about a single group:

Is it true that $(ab)^k = 1$ for some $k$ in $C_n * C_m$?

Note that there are only two possibilities: either $ab$ has order $k$ for some $k$, which means that $gh$ has order (dividing) $k$ in any group $G$ such that $g^n = 1, h^m = 1$, or $ab$ has infinite order. In particular, it is impossible for $gh$ to always have some finite but arbitrarily high order (since that would necessarily imply that $ab$ has infinite order). This is a special case of a very general theorem in first-order logic called the compactness theorem.

But now it should be intuitively obvious that $ab$ has infinite order. More explicitly, it should be intuitively obvious that the elements of $C_n * C_m$ are elements of the free group $F_2 = \langle a, b \rangle$ on $a, b$ such that $a, a^{-1}$ don't appear $n$ times in a row and $b, b^{-1}$ don't appear $m$ times in a row, with the obvious multiplication (cancel all of the obvious relations), and in this group there are no obvious relations to cancel in the word $(ab)^k$.

Of course, this is not a proof. (I believe it follows from standard properties of free groups, but what I've written is not a proof.) But what I'm trying to get at here is that you shouldn't need to see explicit examples to be convinced that this is intuitively true. Undoubtedly writing down explicit examples (nice quotients of $C_n * C_m$) is probably the easiest way to answer this particular question, but for more general questions about the group generated by two elements of finite order (for example, does $a b a^{-1} b a$ have infinite order in $C_n * C_m$?) it's good to be aware that there is a goal you should be aiming for: to understand the specific group $C_n * C_m$.

As far as writing down explicit examples, here now is a chance to illustrate the value of thinking in terms of representation theory in the general sense. Take two random matrices of orders $n$ and $m$. Their product should have infinite order with probability $1$; in fact, they should generate $C_n * C_m$ with probability $1$. An easy way to generate such matrices is just to conjugate a matrix which obviously has order $n$ or $m$ (for example an appropriate permutation) by a random matrix.

The value of doing this is that there are lots of easy-to-verify conditions which imply that a matrix has infinite order: for example, if it is a $d$-dimensional matrix with trace greater than $d$ in absolute value, then it must have infinite order (exercise). It is also surprisingly easy to guarantee that a $2 \times 2$ matrix has order $n > 2$: you just have to make sure that its characteristic polynomial is $x^2 - 2 \cos \frac{2\pi}{n} x + 1$, or equivalently that its determinant is $1$ and that its trace is $2 \cos \frac{2\pi}{n}$.

Let me illustrate with $n = m = 3$: we just need to find two matrices of determinant $1$ with trace $-1$ whose product has trace greater than $2$ in absolute value. In fact it's easy to generate families of such matrices: just take

$$A = \left[ \begin{array}{cc} a-1 & -(a^2-a+1) \\ 1 & -a \end{array} \right], B = \left[ \begin{array}{cc} b-1 & b^2-b+1 \\ -1 & -b \end{array} \right].$$

We compute that the trace of the product is $(a-1)(b-1) + (a^2 - a + 1) + (b^2 - b + 1) + ab$, which in particular is a large positive number as soon as $a, b$ themselves are large and positive. In fact all I had to do was guarantee that this wasn't a constant polynomial: any nonconstant polynomial takes on arbitrarily large values (exercise).

This same trick allows us to construct $2 \times 2$ matrices $A, B$ of orders $n, m$ such that their product has order $k$ for any triple $(n, m, k)$. The corresponding universal groups are called the von Dyck groups and are related to some interesting geometry.

• I didn't understand the argument based on the compactness theorem: I'm not acquainted with the concepts of model and first-order sentence. But, if I did understand correctly the intuitive argument, I claim that if $m=n=2$ then $aba^{-1}ba$ has finite order (it's two), in all other cases it has infinite order. Writing two copies of $aba^{-1}ba$ with the relations $a^2=1$ and $b^2=1$ I see that this product telescopes; the same does not happen with different relations. – Jacopo Notarstefano May 26 '11 at 11:19
• The trace of a matrix is also the sum of its eigenvalues. A matrix of dimension $d$ has $d$ eigenvalues. If $A$ is a matrix of dimension $d$ with $\abs{\trace{A}} > d$ then one eigenvalue must be larger than $1$ in modulo. Since the eigenvalues of $A^{k}$ are the $k$-th powers of the eigenvalues of $A$ this implies that for any $k$ the matrix $A^{k}$ hasn't all eigenvalues equal to one, a necessary (but not sufficient) condition to be the identity. – Jacopo Notarstefano May 26 '11 at 11:19
• @Jacopo: The argument based on the compactness theorem is as follows: suppose there are groups $G_i$ with elements of orders $n$ and $m$ whose product has order at least $i$. Then by the compactness theorem, there is a group $G$ (say the ultraproduct of the groups above) which has elements of orders $n$ and $m$ whose product has infinite order. It's easier to see with ultraproducts, but in more detail: add a 0-ary function $x$ to the language and axioms dictating that $x$ is the product of two elements of orders $n$ and $m$. Then any finite – Qiaochu Yuan May 26 '11 at 12:26
• ... collection of the axioms "$x$ has order at least $i$" is consistent, so the entire collection is consistent. – Qiaochu Yuan May 26 '11 at 12:26

Product of two elements of infinite order = element of finite order?

Let $G$ be any group containing an element $a$ of infinite order. Then $a$ is clearly not the identity, $e$ of G, since $ord(e) = 1$, which is finite. Nor can $a$ be its own inverse, since that would mean $ord(a) = 2$, also finite.

So there must exist an element $b$ such that $b$ = $a^{-1},\; b \neq a,\;b \neq e$, since for every element $x \in G,\; x^{-1} \in G$, and it follows that $ord(a) = ord(a^{-1}) = ord(b)$. Hence $b$, like $a$, is of infinite order Now clearly, $a * b = a*a^{-1} = e \in G$, and, of course, the order of $e$ is finite. So we have if a set has an element of infinite order, it also has an element of finite order which is the product of two elements of infinite order.

Product of elements of finite order = element of infinite order?

Another even more concrete example of two elements of finite order, the product of which is of infinite order:

Take $G=GL_2(\mathbb{Z})$, and two elements $A$ and $B$ such that $$A = \left(\begin{array}{cc}-1&1\\0&1\end{array}\right),\qquad B=\left(\begin{array}{cc}-1 & 0\\ 0 & 1\end{array}\right).$$ $A$ and $B$ each have order 2, yet there product $AB$ is of infinite order: $$AB = \left(\begin{array}{cc}1&1\\0&1\end{array}\right)$$ To see this, note that $$(AB)^n = \left(\begin{array}{cc}1&n\\0&1\end{array}\right)$$ which is the identity matrix only for $n = 0$.

Also, the product $$BA = \left(\begin{array}{cc}1&-1\\0&1\end{array}\right) = (AB)^{-1}$$ has infinite order. So here again, you have $(AB), (AB)^{-1} \in GL_2(\mathbb{Z})$, both of infinite order, whose product is $I_{2x2}$, of finite order!

• Not every infinite group has elements of infinite order (for example, a product of countably many copies of a finite group). – Qiaochu Yuan May 25 '11 at 17:07
• @Qiaochu: Thanks for catching that. I meant to write that every infinite group with elements of infinite order...I'll edit my answer. – Namaste May 25 '11 at 17:18
• Your matrix example is the "affine group" version of the infinite dihedral group I mention: (-1,1;0,1) is the affine transformation x→-x+1, "b", and (-1,0;0,1) is the affine transformation x→-x+0, "a". – Jack Schmidt May 25 '11 at 17:44
• @Jack: Yes, I noticed that after I posted and read your answer. I found your post quite interesting; I'm really fascinated with dihedral groups. BTW: I'm not fast, in terms of constructing a post...I started my response (on paper) when no one had yet responded...I get buried in composing answers, and by the time I post, it always seems others manage to post before me :-S – Namaste May 25 '11 at 18:13
• I like it. I should probably be more clear in my comment. "(+1) Your matrix example is...". When people are trying to understand something simple like this, it is hard to predict what sort of example will click. Your answer looks very different, and so stands a good chance of helping where my answer will not. – Jack Schmidt May 25 '11 at 18:55

For product of two elements of finite order having infinite order, simple examples can be built as follows.

Take two lines, say through the origin, making an angle $\theta$ with each other. Let $a$ be reflection in one line, $b$ reflection in the other. Both $a$ and $b$ have order $2$. Their product is rotation about the origin through $2\theta$. If $\theta$ is not commensurable with $\pi$, then $ab$ has infinite order.

Added comment: That takes care of infinite order. By taking $\theta$ to be $\pi/n$, we can produce $a$ and $b$ of order $2$ with product of any desired finite order.

• Great example, because it also shows that such products can have any order, which supports the claim that there is no general formula. – Jacopo Notarstefano May 25 '11 at 22:12

Dihedral groups are generated by two elements of order two. The order of their product is arbitrary, and defines which dihedral group you get. For instance, the symmetries of a regular n-gon, for n ≥ 3 a positive integer, is a dihedral group of order 2n generated by two reflections a, b whose product is a rotation of order n.

In the symmetric group $$\operatorname{Sym}(E)$$, where $$E$$ is a finite or infinite set, every element can be expressed as the product of two involutions. For example, in $$\operatorname{Sym}(\mathbb Z)$$, the permutation $$f(n)=n+1$$ can be expressed as $$f(n)=g(h(n))$$ where $$g(n)=1-n$$ and $$h(n)=-n$$.


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## Latex with libreoffice (figures and tables)

Previous post here.

The issue with tables is that, unlike the normal flow of the latex document, they are completely foreign to someone who is not a latex user. My proposal, though still a bit unfriendly on the eyes (because a text table is not as pretty as a pdf one) allows my collaborator to get and edit the text.

My approach assumes that you have your tables and figures in separate files and you import them to your main latex with the input command. The main idea is to transform the figure and table files into text and then automatically insert them in the main latex document where they are inputed. The result is the same latex document except for the figures and tables which are in text format.

I use catdvi to do the latex->text formatting. I also use sed, echo, cat and latex. The following is the Makefile which I am using to do stuff. You should change the first three variable to the names of your main document, figures and tables files. Then just run make -f NameOfFile. You might also need to add different targets depending on what you need to compile your latex figure and table documents.

maindoc = paper.tex
figdoc = paperfig.tex
tabdoc = papertab.tex
TXTs = $(figdoc).txt$(tabdoc).txt
$(maindoc).txt:$(TXTs)
cat $(maindoc) >$(maindoc).txt
for tfile in $(subst .txt,,$(TXTs)); do \
sed -i -e "/input{$$tfile}/{r$$tfile.txt" -e "d}" $(maindoc).txt; \ done rm -rf$(TXTs)
$(TXTs): echo '\\documentclass{article}' >$@.tex
cat $(maindoc) | grep usepackage >>$@.tex
echo '\\begin{document}' >> $@.tex cat$(subst .txt,,$@) >>$@.tex
echo '\\end{document}' >> $@.tex latex$@.tex
catdvi $@.dvi >$@

Notice that this code will not include the actual figures into the txt document. I guess I’ll be doing that by hand. But with 5 to 10 figures should not be too time consuming. Also, the script does not have tab spaces, so Make might complain (I blame wordpress).


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# What is the physical meaning of reactive power in non-sinusoidal steady-state (with harmonics)?

In sinusoidal steady-state (linear loads, no harmonics), I understand what is reactive power $$\Q = V_{\text{rms}} I_{\text{rms}} \sin{(\theta)}\$$ where $$\\theta = \theta_v - \theta_i\$$: its absolute value is the maximum rate of transfer of the energy that's oscillating between the device and the external circuit. This conceptual interpretation is in accordance with the mathematical definition above. I haven't seen/read this interpretation anywhere, but I proved it.

When there're harmonics due to non-linear loads, there're various mathematical definitions of reactive power. This article shows some. One definition I've seen quite often is the following (in the article, it's called Budeanu reactive power in section 3.1):

$$\ Q = \displaystyle \sum_{n=1}^{\infty} V_{n,\text{rms}} I_{n,\text{rms}} \sin{(\theta_n)} \tag*{}\$$

This mathematical definition is really an "extrapolation" of the definition in sinusoidal SS. But unlike the latter case, here I can't understand what this infinite sum with units of watts represents. (Yes, I know reactive power is measured in VAR, but that unit is dimensioanlly the same as watts and joules per second and volt-amperes.)

Do you know what this means? Can you explain why your interpretation is correct? I'm not asking for analogies, since they at some point break.

I read a textbook in which the author used the same mathematical definition as above, but he said the significance of that was unclear.

• Are you familiar with function orthogonality? sin() of different frequencies (e.g. harmonics) can form an infinite orthogonal basis, so the extrapolation to infinite-sum makes sense. Oct 19 '20 at 9:30
• @Vicente Unfortunately the only time where I was taught about orthogonal functions was in a communications class, but not in an ODE class or any other. I somewhat understand that the integral of the functions would be zero; and this fact simplifies the expression for average/active power in non-sinusoidal steady-state, for example. But how does this apply to my question? Would you explain it, please? Oct 19 '20 at 15:51
• Just saying that defining a power as an infinite sum of harmonic components (as the article you linked says Budeanu postulated) is not an unnatural thought and simetrically fits the active power definition. Have you read the reference mentioned [3] ? It explictely says "The usefulness of QB (Budeanu reactive power) for quantifying the flow of harmonic nonactive power has been questioned by many engineers". I think this was just an intuitive mathemathical definition to decompose apparent power. Oct 19 '20 at 16:59
• "Have you read the reference mentioned [3] ?" Nope but I will, thanks. // "I think this was just an intuitive mathemathical definition to decompose apparent power" I'm trying to understand what Q means in the equation (non-sinusoidal SS) without needing to talk about apparent power. For example, look at my interpretation of Q in sinusoidal SS; I didn't mention apparent power. I'll try to use the orthogonality as you pointed earlier to see if I come up with something. Oct 20 '20 at 4:44

The only thing that I can contribute that might be helpful is that the reactive component of harmonic power (harmonic VARs) is orthogonal to both the fundamental real power and the fundamental reactive power. The power triangle becomes a three dimensional diagram.

Any harmonic power in the system would presumably add a 4th dimension to the diagram since that would seem to be orthogonal to the other quantities. Considering the harmonic power might require considering each harmonic individually with components orthogonal to those of the fundamental and all of the other harmonics

The harmonic VARs can not be compensated by pf compensation capacitors. Harmonic compensation capacitor values must be determined using only the fundamental VARs. It is possible to build harmonic filters that mitigate harmonic distortion while also compensating fundamental VARs.

Harmonic distortion in a power system leads to high harmonic VARs, but not a lot of harmonic power. If the source voltage is undistorted, no real harmonic power is transmitted. All of the real harmonic power is the result of harmonic I squared R dissipated in the power system conductors, transformers, motor windings and the ESR of the pf compensation capacitors. That is certainly important, but it needs to be minimized by filtering or by preventing it at the source.

Look at IEEE Standard 1459-2010.

• I already knew the geometrical interpretation in your first and second paragraph, though another person may not; thanks anyways Charles. // Regarding the third paragraph, does "harmonic VAR" mean the second and higher-order terms in the equation of my question? I agree with you in the filtering, I've read caps can even amplify harmonics and lead to resonance, but I don't see it relevant to my question. // Regarding the fourth paragraph, I found that more helpful; what's the difference between "harmonic VARs" and "harmonic power"? Does the latter refer to distortion power? Oct 20 '20 at 4:36
• I will read the standard you and Vicente suggested me. Oct 20 '20 at 4:36

$$\n\$$ denotes the various harmonics of the fundamental sinewave. For example $$\n=3\$$ is the third harmonic.

The sum $$\ \displaystyle \sum_{n=1}^{\infty}\$$ is saying that the total power is found by adding the powers for each and every individual harmonic.

• Which integral? Oct 20 '20 at 0:06
• $\displaystyle \sum_{n=1}^{\infty}$. Sorry, I should have said sum not integral, I'll make that change. Oct 20 '20 at 6:44


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Re: Hiya :)
Author: Kae Verens
Date:   06-07-02 10:26

no period.

balancing on the forearm is easy, but it takes practice.

place a ball in the middle of the forearm, and try to balance it. If it moves left, move the arm further to the left (quickly). the same goes for every direction.

try moving around with the ball balanced there.

it will take practice, but after a while, you will be able to catch the ball there (I think I uploaded a move with something like that (one of the Mesh moves, I think) that uses that).

Kae

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text:
# car descending at Constant speed

I understand now, that the static friction opposes the tendency of a surface to slip over another. When a wheel rotates, static friction pushes the wheel forward to keep the contact surface stationary with respect to the ground.

When a car is going at constant speed down a hill, static friction must push the car down the hill where you also have gravity helping too. The only way for this friction (static friction) to oppose gravity is if the wheels are locked, where the wheels are not rotating and the car is stationary, or for the car to go up the hill, where the static friction is again pushing the car forward, countering the slipping of the tires that would result due to the rotation of the tires.

How can you get a constant velocity when going down hill, when the static friction and gravity are both creating a net force down the hill (if you ignore rolling friction)?

A question asks me to find the steepest hill a car can descend at constant speed given the static friction coefficient, but I think what is needed is the rolling friction coefficient. I don't think its possible to answer the question without rolling friction.

• "When a car is going at constant speed down a hill, static friction must push the car down the hill where you also have gravity helping too." This is incorrect. When rolling downhill at constant speed, Newton's 2nd law says that the forces pulling downwards must be balanced by some force pulling upwards. Static friction is in fact pulling upwards along the hill in order for the wheel not to skid and to make this balance. – Steeven May 4 '17 at 9:00
• but what about the rotation of the wheel? – Think May 4 '17 at 9:01
• "but I think what is needed is the rolling friction coefficient." In general, many question use an idealized world-model, so a rolling friction is often not present (surfaces and tires assumed ideal/non-deforming). So I would not start thinking into rolling friction, unless the surface is clearly non-ideal (like a sand beach). @Think, what is your comment about the rotation of the wheel referring to? – Steeven May 4 '17 at 9:03
• surely the static friction is such that is stops the sliding of the wheel that would result by its rotation? – Think May 4 '17 at 9:05
• What sliding results from its rotation? The only thing that can cause sliding is a force that pulls (in the point of the wheel that touches the ground). Such force must be opposed by static friction. But rotation in itself (rotation without forces present for example) says nothing about sliding. – Steeven May 4 '17 at 9:08

I understand now, that the static friction opposes the tendency of a surface to slip over another.

Exactly! Let's remember this point in the next sentences.

When a wheel rotates, static friction pushes the wheel forward to keep the contact surface stationary with respect to the ground.

Not necessarily forward. It can also be backwards. It pulls in whatever direction necessary to keep the contact point stationary. And only if necessary (e.g. there is no static friction when a wheel rolls at constant speed over a horizontal surface. There are no forces at all present, so nothing for the static friction to oppose. The rolling motion just continues effortless until stopped.)

When a car is going at constant speed down a hill, static friction must push the car down the hill where you also have gravity helping too.

No. It pulls uphill.

Think for a moment of a star rolling down. It's legs touches the ground one at a time. While a leg touches the ground, it must not slide. We can think of it as a stationary object in that moment while it is touching. Downwards forces (gravity) must be opposed by static friction upwards, just like for a stationary object resting on the hill.

Then the next leg takes over, and the same thing is the case. Static friction must hold back upwards to avoid sliding of the leg.

Now add more legs to the star. Many, many more. Soon you almost have a round circle, where the legs are the "points" of the circle that touch the ground for only a split-second. Nevertheless, the same is the case; while touching, static friction holds on to the touching point upwards to avoid it from sliding because of gravity.

[...] for the car to go up the hill, where the static friction is again pushing the car forward, countering the slipping of the tires that would result due to the rotation of the tires.

Yes. Again static friction pulls uphill to prevent sliding because gravity pulls down. The direction of static friction does not depend on the rolling direction; it doesn't care if you roll up or down the hill.

If the wheel accelerates at the same time, the static friction might point differently. This is again regardless of the rolling direction but only the acceleration direction comes into play.

How can you get a constant velocity when going down hill, when the static friction and gravity are both creating a net force down the hill (if you ignore rolling friction)?

You are exactly on point here. It can't have constant speed, if all forces pull the same way. Such thinking will lead to the understanding that static friction in fact must point the other way.

A question asks me to find the steepest hill a car can descend at constant speed given the static friction coefficient, but I think what is needed is the rolling friction coefficient. I don't think its possible to answer the question without rolling friction.

As mentioned in a comment, most questions would assume ideal world-models. No deformation of surfaces e.g. So rolling friction will be assumed 0 most often, unless you drive on a clearly non-ideal surface, like a sandy beach or a flexible trampoline.

The idea to solve this question is to remember the formula for maximum static friction:

$$f_s\leq \mu_s n$$

If you have a certain friction coefficient $\mu_s$, you can do your Newtons 2nd law calculations on the car and put this formula in with an $=$ instead of $\leq$, because you are looking for the maximum limit. Then solve it for the slope angle (the angle will be a part of the force components).

It is possible without rolling resistance. The question is not about rolling resistance. The question is about steepest hill.

When you are braking static fiction is not pushing you down the hill. The brakes are opposing gravity. You do not have to lock up the wheels to brake. In fact when you lock up the wheels it is no longer static friction.

Take a block of wood and put it on board. Lift one end to create an angle. The angle at which the block slips the pull from gravity is greater than the static friction. Use the sine of the angle to get the pull from gravity.

You can think of zero as a constant speed.