text:
# Consistency strength: If Con($T+A$) implies Con($T+B$), can we infer anything about $A$ and $B$?

To be more specific, let $T$ be a first order theory and let $A$ and $B$ be two different first-order sentences, both in the same language as $T$ but independent of $T$. Additionally, suppose we have (working in some meta-theory) that Con($T+A$) implies Con($T+B$) (but not vice-versa).

$\textbf{Question}$: What, if anything, can we say about the sentences $A$ and $B$?

That is, can we only speak of the difference between $A$ and $B$ in terms of $T$? Are we limited to saying that adding $A$ to $T$ results in a theory which is "more likely" to derive a contradiction than adding $B$ to $T$? Or can we infer anything about $A$ and $B$ on their own? (Does one prove or refute the other? Might it be that $A$ has more first-order consequences than $B$ does?)

$\textbf{Follow-up question}$: Could it ever be the case that replacing $T$ with a different theory $S$ (also in the same language and unable to decide $A$ or $B$) leads to Con($S+B$) being equivalent to, or stronger than, Con($S+A$)?

Here is an example to better illustrate my questions:

It is known that the consistency strength of ZF together with the Axiom of Choice is exactly that of ZF:

$$\text{Con(ZF+AC)} \iff \text{Con(ZF)}$$

However, the consitency strength of ZF together with the Axiom of Determinacy is very much greater:

$$\text{Con(ZF+AD)} \iff \text{Con(ZF}+\ \psi )$$

where $\psi$ is the statement "there are infinitely many Woodin cardinals".

Do these relations of consitency strength tell us about AC and AD as sentences set apart from ZF? Is AD in some regard more powerful than AC (even though it's refutable in ZFC)? Could, for some new theory of sets $T$ in the same language as ZF, it be possible that $T$+AC is equiconsistent with $T$+AD?

The answers to these questions may turn out to be trivial, but I'm unable to settle on a satisfactory conclusion myself. Any responses offering a greater intuition about consistency strength would be very much appreciated.

-
For the last part, take $T=\mathrm{ZF}+\psi$. –  Asaf Karagila Jul 25 '12 at 21:38

Sadly, there isn't anything we can glean from this. For example, working in ZF, we find that the principle of dependent choice--DC--is equivalent to AC for countable sets of non-0 finite sets--CC(fin)--together with the principle of dependent multiple choice--DMC. It is also equivalent to DMC together with the assumption that Cantor cubes $\{0,1\}^X$ are Baire spaces. Thus, taking T=ZF+DMC, we have Con(T+CC(fin))$\Leftrightarrow$Con(T+Cantor cubes $\{0,1\}^X$ are Baire). Moreover, we know that if Cantor cubes $\{0,1\}^X$ are Baire spaces, then CC(fin) holds. However, the converse does not hold.

There are other, similar examples, where T+A, T+B are equiconsistent, but where A, B aren't logically related to each other outside of the context of T.

-
Thanks for your answer, Cameron. I wasn't aware of the example you provide and will read further into it. Can I take from what you say that the consistency strength never has an influence on logical strength? –  AJCR Jun 11 '12 at 19:59
That is a good question, to which I don't know the answer. :-( We can certainly conclude that consistency strength doesn't always have an influence on logical strength. Let me sum up. In answer to your first question, we can't necessarily infer anything about A and B outside the context of T, solely by observing the relative consistency strengths of T+A and T+B. As to your follow-up question, I suppose it may be possible in some circumstances, but I can't imagine how one might (dis)prove it for general A, B...or even specific A, B, for that matter. –  Cameron Buie Jun 11 '12 at 20:23
(apologies for my month-late reply!) Thanks again for your help with the question. I've left things unaccepted (for now) in case a wise logician stumbles upon the question and can offer any other clues towards a solution... –  AJCR Jul 25 '12 at 20:58
That's fine. I'm glad I could help. –  Cameron Buie Jul 25 '12 at 21:10

Let us agree to work in $\ZF$ as our theory and $\ZFC$ as the meta-theory. First we consider the seemingly strange equiconsistency of $\ZF$ and $\ZFC$ alongside with many weak choice principles in between.

What does that tell us? It means that if $\ZF$ is our meta-theory, then whenever we happen to run into something which is a model of $\ZFC$ then we are guaranteed to run into a model of $\ZF+\lnot\AC$ somewhere in the universe. Often proofs of equiconsistency also tell us how to find such model, for example by forcing arguments or inner model arguments, etc.

Philosophically this means that if $\ZFC$ is true in some structure then $\ZF+\lnot\AC$ is true in another. Since $\ZF$ is a reasonable theory to consider, it tells us that there is no "leap of faith" if we assume $\ZFC$ instead. In particular it means that we are free of the worry that $\AC$ introduced some inconsistency to the system, that of course if $\ZF$ did not have one already.

From this we can deduce that Lebesgue and the school of people leading a mathematical life that $\AC$ is false are actually rejecting $\ZF$ as a whole, since from a proof of contradiction in $\ZFC$ we know how to generate a contradiction in $\ZF$. Then again, these proofs were given quite later than the early days of $\AC$ where people raged over its unthinkable consequences.

On the other hand, we have $\ZF+\AD$ which is infinitely stronger than $\ZF$ or $\ZFC$ in the consistency strength. This tells us that in contrast to the situation above even if we believe that $\ZF$ itself is consistent we still have to believe that there are large cardinals, and that there are many of them. So many that anything less than $\omega$ Woodin cardinals is not even close to enough.

This is quite the leap of faith, specifically since we know that anything that proves the consistency of $\ZF$ is quite the strong theory, and coming from outside set theory it may seem suspicious and borderline contradictory. However there are still arguments in favour of large cardinals, and either way these are interesting.

### Coda:

It seems nowadays like a reasonable idea to believe that the axioms of $\ZFC$ are consistent. I personally believe that if no shocking contradictions are found, in several decades most people will assume $\ZFC+\varphi$ where $\varphi$ is some large cardinal axiom which is strong enough to support category theory related constructions.

Until that happens, people may be wary of large cardinals, or indifferent to them.

-
An interesting read, thanks. I find justifications (both philosophical and mathematical) for accepting large cardinal axioms intriguing. I've read a little about the programs to find reasonable-sounding (even arithmetical) mathematical statements that are independent of ZFC + large cardinal axioms and look forward to learning more about the discoveries that have been made through this research. –  AJCR Jul 25 '12 at 21:59
@AJCR: It's an interesting question. As a strongly agnostic person, I refuse to decide any question on the existence or consistency of mathematical theories (or any other theory which is non-mathemtical). My main justification for large cardinal axioms is that they generate a lot of interesting questions and results. The surprising thing is that they are tied with "low cardinality" mathematics like determinacy of projective sets and measurability. –  Asaf Karagila Jul 25 '12 at 22:03


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# Am I calculating my Kelly Criterion correctly?

I'm taking a look at my trading history over a particular time period and have 500 trades on with an win rate of 82%. My average win is $W$. My average loss is $L$. So am I correct in assuming the Kelly Criterion is:

$$\frac{0.82 \times (W / L + 1) - 1}{W / L}$$

$$f=\frac{pW-qL}{WL}=p/L-q/W$$
where $q=1-p$ and $p$ is the probability of a win.
When the wins and losses are random, with average $W$ and $L$ respectively, I am not sure this formula is completely justified. But it might be a good approximation
• But is it fair to use my existing track record to justify the p, W and L? – Shamoon Oct 17 '17 at 20:18

format: plain


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## The Annals of Probability

### An annihilating–branching particle model for the heat equation with average temperature zero

#### Abstract

We consider two species of particles performing random walks in a domain in ℝd with reflecting boundary conditions, which annihilate on contact. In addition, there is a conservation law so that the total number of particles of each type is preserved: When the two particles of different species annihilate each other, particles of each species, chosen at random, give birth. We assume initially equal numbers of each species and show that the system has a diffusive scaling limit in which the densities of the two species are well approximated by the positive and negative parts of the solution of the heat equation normalized to have constant L1 norm. In particular, the higher Neumann eigenfunctions appear as asymptotically stable states at the diffusive time scale.

#### Article information

Source
Ann. Probab., Volume 34, Number 6 (2006), 2382-2405.

Dates
First available in Project Euclid: 13 February 2007

https://projecteuclid.org/euclid.aop/1171377448

Digital Object Identifier
doi:10.1214/009117906000000511

Mathematical Reviews number (MathSciNet)
MR2294987

Zentralblatt MATH identifier
1122.60085

#### Citation

Burdzy, Krzysztof; Quastel, Jeremy. An annihilating–branching particle model for the heat equation with average temperature zero. Ann. Probab. 34 (2006), no. 6, 2382--2405. doi:10.1214/009117906000000511. https://projecteuclid.org/euclid.aop/1171377448

#### References

• Bañuelos, R. and Burdzy, K. (1999). On the hot spots'' conjecture of J. Rauch. J. Func. Anal. 164 1--33.
• Bañuelos, R. and Pang, M. (2006). Level sets of Neumann eigenfunctions. Indiana Univ. Math. J. 55 923--939.
• Burdzy, K. and Chen, Z. (2002). Coalescence of synchronous couplings. Probab. Theory Related Fields 123 553--578.
• Burdzy, K., Hołyst, R., Ingerman, D. and March, P. (1996). Configurational transition in a Fleming--Viot-type model and probabilistic interpretation of Laplacian eigenfunctions. J. Phys. A 29 2633--2642.
• Burdzy, K., Hołyst, R. and March, P. (2000). A Fleming--Viot particle representation of Dirichlet Laplacian. Comm. Math. Phys. 214 679--703.
• Conti, M., Terracini, S. and Verzini, G. (2005). Asymptotic estimates for the spatial segregation of competitive systems. Adv. Math. 195 524--560.
• Conti, M., Terracini, S. and Verzini, G. (2005). A variational problem for the spatial segregation of reaction--diffusion systems. Indiana Univ. Math. J. 54 779--815.
• Courant, R. and Hilbert, D. (1953). Methods of Mathematical Physics. Interscience Publishers, New York.
• Cybulski, O., Babin, V. and Hołyst, R. (2004). Minimization of the Renyi entropy production in the stationary states of the Brownian process with matched death and birth rates. Phys. Rev. E 69 016110.
• Cybulski, O., Babin, V. and Hołyst, R. (2005). Minimization of the Renyi entropy production in the space partitioning process. Phys. Rev. E 71 046130.
• Cybulski, O., Matysiak, D., Babin, V. and Hołyst, R. (2005). Pattern formation in nonextensive thermodynamics: Selection criterion based on the Renyi entropy production. J. Chem. Phys. 122 174105.
• Dawson, D. A. (1993). Measure-valued Markov processes. École d'Été de Probabilités de Saint-Flour XXI---1991 1--260. Lecture Notes in Math. 1541. Springer, Berlin.
• Ethier, S. N. and Kurtz, T. G. (1986). Markov Processes. Characterization and Convergence. Wiley, New York.
• Grieser, D. (2002). Uniform bounds for eigenfunctions of the Laplacian on manifolds with boundary Comm. Partial Diff. Eq. 27 1283--1299.
• Grigorescu, I. and Kang, M. (2004). Hydrodynamic limit for a Fleming--Viot type system. Stochastic Process. Appl. 110 111--143.
• Grigorescu, I. and Kang, M. (2006). Tagged particle limit for a Fleming--Viot type system. Electron. J. Probab. 11 311--331.
• Hempel, R., Seco, L. A. and Simon, B. (1991). The essential spectrum of Neumann Laplacians on some bounded singular domains. J. Func. Anal. 102 448--483.
• Morrey, C. (1966). Multiple Integrals in the Calculus of Variations. Springer, New York.
• Netrusov, Yu. and Safarov, Yu. (2005). Weyl asymptotic formula for the Laplacian on domains with rough boundaries. Comm. Math. Phys. 253 481--509.
• Protter, Ph. (1990). Stochastic Integration and Differential Equations. A New Approach. Springer, Berlin.
• Stroock, D. and Varadhan, S. R. S. (1971). Diffusion processes with boundary conditions. Comm. Pure Appl. Math. 24 147--225.


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# Arbitrarily Complex Fish

148 views

Written a vector based editor for creating the environment for my game:

Here are a collection of spheres bouncing off and interacting physically accurately with an arbitrarily complex fish.

I'd like to think that that sentence has never been posted before.

It will all look a lot better when textures are dropped in. I've got texturing working in the editor and it looks a lot better.

The overall idea though is that you construct the environment out of as few physics polygons as possible, but since the quads are forced to multiples of the texture base size (64x64), they are just textured with wrap mode and look like a grid of smaller quads.

This reduces the number of polygon intersection tests.

The editor generates the polygon vertex info and will soon generate the texture co-ordinates as well. I'm planning to atlas the map textures so the map can hopefully be drawn with one DrawPrimitive call.

Oh, further to my revamped graphics wrapper, I've got atlased fonts working. I was suprised to find I can fit Tahoma, size 8, bold A-Z, 0-9, a-z and some punctuation onto a 128x128 texture.

I did try to write a new font file generator with C#, but it seems that GDI+ returns some very weird numbers from MeasureString(). According to various bods on the internet, this is all to do with how clever it is at rendering text in a device-independant manner or something. Just a bit of a bastard as far as I'm concerned.

My current, working font file generator was written ages ago with Borland Builder and since I don't have Builder installed anymore (it won't coexist nicely with Borland's command line compiler, which I still use) I can't alter it.

Might have to rewrite it in straight Win32 to avoid this GDI+ problem.

That is indeed, a very arbitrarily complex fish. Looks cool though.

Looking good, EC. I'm looking forward to the point where we can swap editors and compare our work. Mine runs pretty much on homebrewn GUI, since there's no Win32/WinForms API around my code, so I'm guessing you'll win on the user-friendliness front. (Since we both seem to be writing polygonal map editors. [grin])

Best of luck!

Quote:
 Original post by HopeDagger Looking good, EC. I'm looking forward to the point where we can swap editors and compare our work. Mine runs pretty much on homebrewn GUI, since there's no Win32/WinForms API around my code, so I'm guessing you'll win on the user-friendliness front. (Since we both seem to be writing polygonal map editors. [grin]) Best of luck!

That's really weird. Just had a look at your screenshots. I caught your comment in Sir Sapo's journal the other day. I genuinely started this project before I read that he had started on this polygon intersection stuff, and have literally just looked at your screenshots and heard Twilight Zone music in my brain.

You seem to have cracked texturing more complex shapes than me though. My physics code can cope with collisions with any convex polygons, but working out how to texture them is eluding me at the moment.

Regarding the fonts, it doesn't use Win32 to render them. A seperate program uses Win32 to generate a file containing a list of square textures representing each letter, along with an unsigned char Index[256]; array that gives the index in the list of each letter.

The images are then atalased onto a Direct3D texture when the font is loaded, and the TextureFont class works out the tex co-ords for two little triangles when rendering each letter into a vertex buffer.

So in theory, you could use my font files with pretty much any 2D or 3D graphics API. I thought that having the one file was a bit tidier than that Bitmap Font Generator program that everyone seems to be using. It's just an extension of my .spr format for storing multiple images with alpha information in a single file.

Nothing against your wonderful fonts, but I wrote 'front', not 'font'. [grin]

Regarding the texturing:

It's far easier than you think, good friend. The texture coordinates simply equivalent (or with some arbitrary scaling) to the world coordinates of the vertex. Since texture coordinates repeat (or rather, they can repeat if the right texture states are set), the texture will simply tile seamlessly over the polygon. Since your world rotates around you, remember to use the non-transformed world coordinates instead of the screen coordinates.

Quote:
 Original post by HopeDagger Nothing against your wonderful fonts, but I wrote 'front', not 'font'. [grin]

Doh! [smile]

Quote:
 Original post by HopeDagger Regarding the texturing: It's far easier than you think, good friend. The texture coordinates simply equivalent (or with some arbitrary scaling) to the world coordinates of the vertex. Since texture coordinates repeat (or rather, they can repeat if the right texture states are set), the texture will simply tile seamlessly over the polygon. Since your world rotates around you, remember to use the non-transformed world coordinates instead of the screen coordinates.

Guess that makes sense. I'll have to have a play around with it.

I was actually quite inspired by your level editor screenshots so I'm rewriting mine.

Quote:
 Original post by EasilyConfused I was actually quite inspired by your level editor screenshots so I'm rewriting mine.

I can send you a copy of what I have so far if you think you would be able to sheer off some/any ideas for your own. Heck, you're welcome to build off of what I have so far, if you're so inclined.

Just tried your suggestion with the texture co-ords=world.co-ords and of course it works.

I'm dividing the world co-ord by the texture width, and all systems go.

I guess though that I can't atlas textures that I want to wrap.

Appreciate the offer of your code, but I reckon it would just confuse me. I never seem to be able to make sense of all but the most trivial of other people's code.

Believe me, your suggestion above has solved the vast majority of the problems I was having.

Quote:
 Original post by EasilyConfused Just tried your suggestion with the texture co-ords=world.co-ords and of course it works. I'm dividing the world co-ord by the texture width, and all systems go. I guess though that I can't atlas textures that I want to wrap.

Not at all. The auto-wrapping is a nice feature to take advantage of, but it's not like it's the *only* way to repeat your texture over the 0..1 range. Right now, your UV coordinates are generated something like this:

float u = vertex->x / polygon->texture->width;
float v = vertex->y / polygon->texture->height;

Which will give you the right values over an infinite plane filled with the wanted texture repeated. However, you can simply divide out the whole-number portion of each coordinate to keep the number between 0 and 1:

float u = vertex->x / polygon->texture->width;
float v = vertex->y / polygon->texture->height;

u /= int(u);
v /= int(v);

/* Note: If maintaining the sign is important (above gives you positive always), then "u /= abs(int(u));" preserves it. */

I do recommend precalculating these though, since divisions and integer castings will eventually catch up with you on bigger maps.

Quote:
 Appreciate the offer of your code, but I reckon it would just confuse me. I never seem to be able to make sense of all but the most trivial of other people's code.

It's always difficult to try and make sense of someone else's code -- it's not just you. Everyone has their own unique style and methods to doing things, so it's tricky to make sense of it. However, it is an excellent skill to learn (at least if you plan on getting a job writing code), as it can oftentimes prevent you from reinventing the wheel for the 100th time, and even learn new tricks.

Gripes aside, hopefully that helps. Continued good luck on your development.

Cheers. I was wondering about making maps bigger and bigger and buggering up the texture co-ords. Looks like the above will solve it.

## Create an account

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text:
# Partitioning and sorting even and odd numbers within an array

I solved a problem where it was told to sort an array of numbers such that the even numbers are kept first and the odd numbers should follow. Moreover, the even numbers must be sorted in ascending order among them and the odd numbers as well. Only arrays can be used. My code is as follows , which works fine but it seems bit long to me. I wanted to know if there is anything I can do to shorten it and make it more readable.

#include <stdio.h>

int main(void)
{
int n,i,k,p,j,temp;

scanf("%d",&n);

int arr[n],arr_even[n],arr_odd[n];

k=0,p=0;
for(i=0;i<n;i++){
scanf("%d",&arr[i]);
if(arr[i]%2==0) arr_even[k++]=arr[i];
else arr_odd[p++]=arr[i];
}

for(i=0;i<k;i++){
for(j=0;j<=i;j++){
if(arr_even[j]>arr_even[i]){
temp=arr_even[i],arr_even[i]=arr_even[j],arr_even[j]=temp;
}
}
}

for(i=0;i<p;i++){
for(j=0;j<=i;j++){
if(arr_odd[j]>arr_odd[i]){
temp=arr_odd[i],arr_odd[i]=arr_odd[j],arr_odd[j]=temp;
}
}
}

p=0;
for(i=0;i<n;i++)
{
if(k) arr[i]=arr_even[i],k--;
else arr[i]=arr_odd[p++];
}

for(i=0;i<n;i++)
{
printf("%d ",arr[i]);
}

return 0;
}

## Input:

10
0 5 1 2 3 4 6 12 10 9

## Output:

0 2 4 6 10 12 1 3 5 9

Your code is quite complicated. The C standard library provides all the ingredients you need to sort an array. You just need to define a comparison function. All the rest is done by the qsort function from stdlib.h.

The comparison function should look like:

static int even_first(const void *a, const void *b) {
int left = *(const int *)a;
int right = *(const int *)b;

int res = (left % 2 != 0) - (right % 2 != 0);
if (res == 0)
res = (left > right) - (left < right);
return res;
}

The expressions of the form cond1 - cond2 may look strange at first, but they are commonly used in C code in comparison functions like this one.

The benefit over a naïve left - right is that no integer overflow can happen. Integer overflow is a common source of undefined behavior.

To make the code more readable, it's also possible to extract the basic integer comparison into a separate function:

static int compare(int a, int b) {
return a < b ? -1 : a > b ? +1 : 0;
// alternatively: return (a > b) - (a < b);
// alternatively: return a < b ? -1 : a > b;
}

Then, the comparison function becomes:

static int even_first(const void *a, const void *b) {
int left = *(const int *)a;
int right = *(const int *)b;

int res = compare(left % 2 != 0, right % 2 != 0);
if (res == 0)
res = compare(left, right);
return res;
}

This form is much less of a brain twister than the above variant.

Firstly, we should separate the sorting from the reading of inputs and writing of outputs. We can create a function sort_evens_first() - that's the foundation of writing re-usable code. An advantage (even in this small program) is that a separable function can more easily be tested - no need for an external script to run many instances of the program with different inputs, making tests run much faster; also it makes it easier to distinguish bugs in the I/O from bugs in the algorithm

Secondly, the Standard Library provides us with qsort() to save us having to re-implement sort every time (and it's usually more efficient than the bubble sort implemented here). We need to give it a comparator function as follows:

• if one element is even and the other is odd, the even number sorts before the odd one,
• else, the numerically smaller one is lower.

That looks like:

int compare_evens_first(const void *va, const void *vb)
{
const int *a = va;
const int *b = vb;

if (*a % 2 != *b % 2)
return (*a % 2) - (*b % 2);

// else both odd, or both even
// return *a - *b might overflow, so avoid that
return (*a > *b) - (*a < *b);
}

Then we can simply use it:

#include <stdlib.h>

void sort_evens_first(int *array, size_t count)
{
qsort(array, count, sizeof *array, compare_evens_first);
}

N.B. I've not had time to test this code; bugs may be lurking.

## Additional problems with the supporting code

When reading input with scanf() and family, it is essential to confirm the return value before using any of the results.

It's less important to check the result of printf() as failure there is less likely to lead to bad outcomes, but it's still worth considering so that we can return EXIT_FAILURE if the output wasn't successfully written (it might not be obvious to the user if directed to a file or pipeline, for example).

• I'd tend to write int const * const a = va; (note the two consts). I'd also be tempted to write const int parity = (*a % 2) - (*b % 2); if (parity != 0) return parity; – Martin Bonner supports Monica Aug 16 at 8:59
• The answer from Roland actually extracts the int from the void pointer once, and then uses that - I prefer that too. – Martin Bonner supports Monica Aug 16 at 9:01
• @Martin - I feel that the conversion by assignment is slightly safer than casting the pointers, so that one is just two different preferences. There's arguments for both, and they could even be combined if you don't mind two more temporaries. – Toby Speight Aug 16 at 9:21


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# What is the second value in \fontsize for?

I know that it is possible to change the size of the font of the section titles in a document using \usepackage{sectsty} and the command \sectionfont; for example, to set the section titles' font to 14pt, I would use

\sectionfont{\fontsize{14}{15}\selectfont}

But what does the second number (15) stand for? Nothing seems to change when I modify this value.

• The 15 is the \baselineskip in points. The real distance is computed by the product of \baselineskip and \baselinestretch – user31729 Aug 3 '16 at 21:03
• generally it's better to use one of the defined sizes such as \large (and then if necessary define \large to be \fontsize{14}{15}\selectfont that way you get consistend fonts across the document structures. Note that this is \fontsize syntax, unrelated to \sectionfont – David Carlisle Aug 3 '16 at 21:07

The two arguments for \fontsize{<size>}{<bskip>} sets respectively the font size and the baseline skip. After setting, you have to \selectfont for the parameters to become active, making them available in \f@size and \f@baselineskip.

Some considerations:

• Not all font sizes are available, so your choice for <size> should be made to avoid font substitution, or choose a package that provides support for your font choice (like lmodern).

• The choice of font size and baseline skip is usually paired to avoid descenders/ascenders from sticking into the line below/above it. As such, it is advised to use the predefined or default font switches. Another motivation for using these is that they vary depending on the default document font size selected. That is, \large (say) has a different font size under 10pt, 11pt and 12pt document class option.

Some examples highlighting the effect of the second argument:

\documentclass{article}

\usepackage{lmodern}
\usepackage[nopar]{lipsum}
\usepackage[margin=1in]{geometry}

\begin{document}

% https://tex.stackexchange.com/q/24599/5764
{\tiny tiny\par}% 5pt font / 6pt baseline skip
{\scriptsize scriptsize\par}% 7pt font / 8pt baseline skip
{\footnotesize footnotesize\par}% 8pt font / 9.5pt baseline skip
{\small small\par}% 9pt font / 11pt baseline skip
{\normalsize normalsize\par}% 10pt font / 12pt baseline skip
{\large large\par}% 12pt font / 14pt baseline skip
{\Large Large\par}% 14pt font / 18pt baseline skip
{\LARGE LARGE\par}% 17pt font / 22pt baseline skip
{\huge huge\par}% 20pt font / 25pt baseline skip
{\Huge Huge\par}% 25pt font / 30pt baseline skip

\clearpage

{\fontsize{15}{15}\selectfont \lipsum[2]\par}

{\fontsize{15}{18}\selectfont \lipsum[2]\par}

{\fontsize{15}{20}\selectfont \lipsum[2]\par}

\end{document}

The default font switches provide a good balance between font size and baseline skip to make the font readable:

Too small a baseline skip value makes the text look bunched together:


text:
# Code-Golf: Farey sequence (I)

## Challenge

In this task you would be given an integer N (less than 10^5), output the Farey sequence of order N.

The input N is given in a single line,the inputs are terminated by EOF.

Input

4
3
1
2

Output

F4 = {0/1, 1/4, 1/3, 1/2, 2/3, 3/4, 1/1}
F3 = {0/1, 1/3, 1/2, 2/3, 1/1}
F1 = {0/1, 1/1}
F2 = {0/1, 1/2, 1/1}

Constraints

• The number of inputs would not exceed 10^6 values
• You can use any language of your choice
• Shortest solution wins!
• This will get loooong.....the output i mean. – st0le Mar 26 '11 at 8:49
• Is N=0 permitted? – Eelvex Mar 26 '11 at 18:50
• What's with the »(I)« in the title? – Joey Mar 28 '11 at 16:49
• @Joey: Hmm. there's a Farey Sequence (II) now. Must be first edition! :-) – mellamokb Mar 30 '11 at 18:08
• @mellamokb: Well, that one's a code challenge, though, so no title clash in any case. But yes, that sort of answers my question. – Joey Mar 30 '11 at 19:03

## J, 96

('F',],' = {0/1',', 1/1}',~('r';'/')rplc~', ',"1":"0@(3 :'}./:~~.,(%~}:\)i.1x+y')&".);._2(1!:1)3

( /:~~.,(%~}:\)i.>:x:y gives the list; the rest is I/O and formatting (with bad style))

E.g:

4
3
1
2
F4 = {0/1, 1/4, 1/3, 1/2, 2/3, 3/4, 1/1}
F3 = {0/1, 1/3, 1/2, 2/3, 1/1}
F1 = {0/1, 1/1}
F2 = {0/1, 1/2, 1/1}

### Edits

• (114 → 106) Clearer appending ,
• (106 → 105) Cap [: to At @
• (105 → 101) Delete superfluous ": conversion
• (101 → 99) Use infix \ for the list
• (99 → 96)
• I get |value error: rplc. Are you sure you didn't load 'strings' earlier in the session and forget about it? – Jesse Millikan Mar 29 '11 at 4:33
• @Jesse: absolutely. I (almost) never use 'strings'. I just use the default linux-j-7.01 environment. – Eelvex Mar 29 '11 at 4:51
• Ugh... I switched to j602 for wd and now I may need to switch back. :) – Jesse Millikan Mar 29 '11 at 5:04

# Common Lisp, 156

(do((l()()))((not(set'n(read()()))))(dotimes(j n)(dotimes(i(1+ j))(push(/(1+ i
)(1+ j))l)))(format t"~&F~D = {0/1~{, ~A~}/1}"n(sort(delete-duplicates l)'<)))

(newlines not necessary)

Very brutal, but languages with native rationals are an invitation to that.

                                        ; at each iteration:
(do ((l()()))                           ; - reset l to nil
((not (set 'n (read()()))))         ; - read a term (nil for eof)
;   assign it to n
;   stop looping if nil
(dotimes (j n)                        ; for j in 0..n-1
(dotimes (i (1+ j))                 ;   for i in 0..j
(push (/ (1+ i) (1+ j)) l)))      ;     prepend i+1/j+1 to l
(format t "~&F~D = {0/1~{, ~A~}/1}"   ; on a new line, including 0/1,
; forcing the format for 1
n                             ; print sequence index, and
(sort                         ; sorted sequence of
(delete-duplicates l)        ;   unique fractions
'<)))                        ; (in ascending order)

# Python, 186 Chars

import sys
p=sys.stdout.write
while 1:
a=0;b=c=x=1;d=y=N=input();p("F%d = {%d/%d, %d/%d"%(d,a,b,c,d))
while y-1:x=(b+N)/d*c-a;y=(b+N)/d*d-b;p(", %d/%d"%(x,y));a=c;c=x;b=d;d=y
p("}\n")

• +1,but are you sure this will be fast for 10^6 number of inputs? – Quixotic Mar 26 '11 at 15:11
• @Debanjan No. It would be really slow for 10^6 inputs. It is linear in complexity (in terms of the number of terms), though. – fR0DDY Mar 26 '11 at 15:27

# J, 156135117 112

d=:3 :0
wd;'F';(":y);' = {';(}.,(', ';2|.'/';|.)"1(<@":)"0(2)x:/:~~.,(-.@>*%)"0/~i.x:>:y),<'}'
)
d@".;._2(1!:1)3

j602 or similar (wd). Input on stdin, output on stdout.

Still puzzling over how to golf the output code, which is 100 characters or so.

Edit: (156->135) Tacit->explicit for long monadic verb chains, less braindead list generation

Edit: (135->117) Found raze. Took me long enough. Switched string handling around.

Edit: (117->112) Slightly less braindead way to exclude fractions above 1. Unnecessary open.

• Maybe you can omit one of your two x:s? – Eelvex Mar 30 '11 at 1:56
• @Eelvex: The left one is 2&x:, e.g. split a rational number into numerator and denominator. – Jesse Millikan Mar 30 '11 at 3:24
• oic. Pity ... :( – Eelvex Mar 30 '11 at 3:31

~:c[,{){.}c(*}%.c/zip{+}*]zip{~{.@\%.}do;1=},{~<},{~\10c?*\/}${'/'*}%', '*'F'c+' = {0/1, '+\', 1/1}'  # Ruby, 11010810297949291 89 #!ruby -lp$_="F#$_ = {#{a=[];1.upto(eval$_){|d|a|=(0..d).map{|n|n.quo d}};a.sort*', '}}"

• I think you should output "0/1" and "1/1" instead of "0" and "1" respectively. Also, does this only work for ruby 1.9? – Eelvex Mar 27 '11 at 16:51
• @Eelvex: It does output 0/1 and 1/1 on my system. And yes, it requires 1.9 (because of the character literals). – Lowjacker Mar 27 '11 at 18:54

f n="F"++show n++" = {"++(intercalate", ".("0/1":).map(\(i:%d)->show i++"/"++show d).sort.nub$[i%d|d<-[1..n],i<-[1..d-1]])++"}" main=interact$f.read`


text:
# Machine Learning for Hackers, Chapter 7

The title of this chapter, “Optimization: Breaking Codes”, sounds more exciting than it really is. The code breaking is attempted via an optimization algorithm called the Metropolis method. The authors show you how to encrypt text using a simple Caesar cipher and then how to implement the Metropolis method to attempt to decipher the message. Their test message is “here is some sample text”. They encrypt it using the Caesar cipher, so it looks like “ifsf jt tpnf tbnqmf ufyu “. The encryption rule is simply replacing every letter with the following letter in the alphabet. The letter a is replaced with b, b with c, and so on until you replace z with a. They manage to decipher their test text through the course of 50,000 Metropolis steps (on the 45,609th attempt). I suppose this is a fun demonstration of optimization. It just seemed a little outdated. I guess I was expecting to see them crack an actual message from World War II or something.

They start the chapter by demonstrating how to use the R optim function. For some reason this function has always confused me. Probably because I don’t use it that often. But I have to say I feel much better about it now. They give clear demonstrations of its use and visually show you how it works. Their example involves the height and weight dataset from chapter 2. First we do linear regression and find coefficients for the slope and intercept. Then we find the same coefficients using optim. The function minimized in the call to optim is a hand-built sum of squared errors function. I thought this was a great way to demonstrate how optim works. But after this, they do something puzzling: they continue to demonstrate how to use optim with Ridge Regression. Now Ridge Regression is a type of regression that helps remedy multicollinearity. If you have predictor variables that are highly correlated, you may get some weird coefficients in your linear modelling results. I’m talking about getting a negative value where you would expect a positive value, that sort of thing. Ridge regression helps address this by “modifying the method of least squares to allow biased estimators of the regression coefficients” (Kutner, et al., p. 432). Knowing that, it doesn’t make much sense to demonstrate Ridge regression on height and weight data. There’s only one predictor. I thought maybe they were going to do more with Ridge regression in this chapter, but they don’t. They move on to code breaking. It just seemed out of place to bring up Ridge regression in the context of single predictor regression.

I won’t go into the code breaking program. The code is available on the book’s git hub page. What I wanted to review for my own benefit is their function for encrypting text using the Caesar cipher. I thought it was a good example of using R’s indexing capabilities with list objects. First we need a vector with the English alphabet and two empty list objects for the ciphers: one for encrypting text and another for deciphering text:

english.letters <- letters
caesar.cipher <- list()
inverse.caesar.cipher <- list()

Now we create the ciphers:

# need to use index %% 26 + 1 to handle letter "z" at end
# 26 %% 26 + 1 = 0 + 1 = 1
for (index in 1:length(english.letters))
{
caesar.cipher[[english.letters[index]]] <- english.letters[index %% 26 + 1]
inverse.caesar.cipher[[english.letters[index %% 26 + 1]]] <- english.letters[index]
}

The "index %% 26 + 1" confused me for a couple of reasons. First off, the %% operator means modulus and I'm used to see something like 12 %% 5, where the first number is bigger than the second. The answer to that is of course 2. If you divide 12 by 5 you get remainder 2, and that's what the %% operator returns. But in the code above, you'll notice the "index" value is less than 26 on every iteration except the last. How do you calculate something like 2 %% 26? I wasn't sure so I had to google around and figure it out. It turns out the modulus operation a %% n is defined as follows:

$a \, mod \, n = a - \lfloor \frac{a}{n} \rfloor \times n$

So, 2 %% 26 is calculated like so:

$2 \, mod \, 26 = 2 - \lfloor \frac{2}{26} \rfloor \times 26 = 2 - \lfloor 0.0769 \rfloor \times 26 = 2 - (0 \times 26) = 2$

Once I was clear on that, I was confused why we needed to use the %% operator at all. Again if you look at the code above, you'll see "index %% 26 + 1". When index = 2, "index %% 26" gives 2. When index = 3, "index %% 26" gives 3. And so on. Why not just use "index + 1"? Well, when we get to the letter "z", we need to substitute "a". The index number for "a" is 1. But if we do "index + 1", then we would get 27 at the letter "z" since its index number is 26. That would not index anything in the 26 element "english.letters" vector and produce an error. Therefore we need to use the %% operator to allows us to recycle through the numbers 1 - 26. When index = 26 (i.e., when we get to the letter "z"), we do 26 %% 26 + 1, which equals 1 since 26/26 has remainder equal 0.

Now that I understand how the ciphers are generated, I can print them and have a look if I like:

print(caesar.cipher) # encrypt text
print(inverse.caesar.cipher) # decipher text

Since these are lists, we investigate with indices using double brackets. For example, to see what we substitute for letter "h", we do the following:

caesar.cipher[["h"]]

That returns "i". It seems like a fun extension to this would be to create a general function that allowed you to specify the substitution index. In the Caesar cipher, this is 1. It would be neat to specify, say, 13 so "a" becomes "n", "b" becomes "o", and so on. Anyway, moving on...

Now we use the ciphers we just made to encrypt text. To do this the authors build two functions. The first encrypts text letter-by-letter using the cipher you give it. The second function encrypts words using the first function. Let's have a look at these:

apply.cipher.to.string <- function(string, cipher)
{
output <- ''
for (i in 1:nchar(string))
{
output <- paste(output, cipher[[substr(string, i, i)]], sep='')
}
return(output)
}
apply.cipher.to.text <- function(text, cipher)
{
output <- c()
for (string in text)
{
output <- c(output, apply.cipher.to.string(string, cipher))
}
return(output)
}

The first function loops through each letter of a word (or "string"). Using the substr function, we extract each letter, use that letter to look up the substitution in the cipher, and "paste" the substituted letter to the previously substituted letter. At the conclusion of the loop we have encrypted the word using our cipher. The second function uses this first function to encrypt whole phrases. Here is how the authors encrypt the message "here is some sample text" using the Caesar cipher:

apply.cipher.to.text(c('here', 'is', 'some', 'sample','text'), caesar.cipher)

That gives us "ifsf" "jt" "tpnf" "tbnqmf" "ufyu" . If we wanted to decipher the text, we would use the inverse cipher as follows:

apply.cipher.to.text(c('ifsf', 'jt', 'tpnf', 'tbnqmf', 'ufyu'), inverse.caesar.cipher)

Again it would be a good exercise to generalize these functions to accept a whole text file and return an encrypted text file. Not that I'm going to do that, but it sounds like it would make a good project assignment in some programming class.

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text:
rowcontains¶

Purpose¶

Checks whether any element in the row of a matrix or string array matches any element from a separate symbol.

Format¶

mask = rowcontains(haystack, needles)
Parameters
• haystack (matrix) – multi-dimensional array or string array in which to search.

• needles (vector or string array) – the elements to search for in haystack.

Returns

mask (vector) – with the same number of rows as the input haystack, containing only 1’s and 0’s. Each element of the output, mask, will contain a 1 if any element in the corresponding row of haystack matches any element in needles.

Examples¶

Find observations matching one or more criteria¶

// A string array with categorical variables
string haystack = { "married" "unemployed"      "king",
"single" "unemployed" "snohomish",
"single"   "employed"      "king",
"married"   "employed"      "king",
"single"   "employed"    "pierce" };

needles = "single" $| "unemployed"; // Find any observations in which the participant // is either single or unemployed mask = rowcontains(haystack, needles);  After the above code, mask will equal: 1 1 1 0 1  Find rows with specified invalid values¶ data = { 26 121 4, 999 139 2, 45 145 2, 51 . 3 }; // Search for 999 or a GAUSS missing value '.' bad_vals = 999 | miss(0,0); // Search 'names' for instances of any of // the strings in 'missing' mask = rowcontains(data, bad_vals);  After the code above, mask will equal: 0 1 0 1  Remarks¶ rowcontains() is similar to the dot operators .== and .$==. The differences are that rowcontains() allows for the comparison of more than one needle in each element of the haystack and rowcontains() reports one value for each row in haystack, rather than one value for each element in haystack.

To return a mask of ones and zeros the same size as haystack, use ismiss().


text:
Chapter 68

• The current practice of transfusion in the ICU is conservative, reflecting a balance between the benefits and adverse consequences of transfused blood.
• Anemia developing in the ICU should not be attributed solely to phlebotomy; red cell loss, destruction, or underproduction should also be considered.
• Evaluation of the basis for anemia, beginning with analysis of the blood smear, reticulocyte count, and red cell indices, may lead to easier management of the anemia and may prove useful in overall treatment.
• For most critically ill patients, a transfusion target for hemoglobin of 7 to 9 g/dL is at least as safe as a higher threshold of 10 to 12 g/dL; the lower value may be superior.
• Transfusion is rarely associated with life-threatening complications such as transfusion-related acute lung injury, major hemolysis, or lethal infection.
• Erythropoietin and oxygen-carrying blood substitutes remain unproven for critically ill patients.
• Massively transfused patients are at risk for hypothermia and dilutional coagulopathy.

Red blood cells are the most efficient vehicle by which oxygen is delivered to the tissues of whole organisms. The evolution of red cells and the circulatory system allowed for the development of larger, more complex organisms with higher levels of activity. Deficiency in the number and in the local delivery of red blood cells causes relative oxygen deficiency in tissues. Oxygen deficit causes metabolic insufficiency, with inadequate energy production to sustain activity, or even life. Hemoglobin, the complex molecule contained within human red blood cells, has unique properties in and of itself, allowing for the efficient uptake of oxygen in an oxygen-rich environment, and the efficient divestiture of oxygen in an oxygen-poor environment. The whole organism also has evolved to attempt to regularly recycle and reuse the important portions of the hemoglobin molecule, thus reducing the need for ongoing sources of iron. Anemia, the lack of red blood cells in normal volumes, is therefore a disorder that threatens the homeostasis of the entire organism, and when it occurs in the setting of already life-threatening illness, anemia is a potential contributor to mortality.

Blood transfusion has long been recognized as a potential remedy for blood loss, although the first efforts to transfuse blood from animals to human beings or from one human to another were marred by the lack of knowledge of the unique antigens on the surface of red blood cells and the immune response to them. Once a systematic approach to the collection, storage, cross-matching, and transfusion of blood was developed, enthusiasm for transfusion reached its peak. Tempered by the recognition of new transmissible illnesses, by the increasing limitations of the blood supply,1 and by the awareness of the limitations of aggressive transfusion, modern critical care practice has evolved to a more conservative approach to the use of transfusion as a remedy for anemia in critically ill patients.

The frequency of anemia in critically ill patients has been studied in a variety of settings, all with the conclusion that it is a ...

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text:
### Math Notes

Subjects

#### Algebra Solutions

##### Topics || Problems

A rectangular plot of ground is 120 yards long and 80 yards wide. To double the area of the plot, while retaining the rectangular shape, strips of equal width will be added at one end and along one side. Find the width of the strip.

Let $$x$$ be the width of the strip. Thus the new length of the rectangle is $$120 + x$$ and the new width is $$80+x$$.

The area of the old plot: $$A_o = 120x80 = 9600 yd^3$$

The area of the new plot is twice the rea of the old plot. Thus, $$A_n = 2A_o$$

$$A_n = (120 +x)(80+x)$$

$$A_n =9600 + 200x + 80x^2$$

p>$$9600 + 200x + 80x^2 = 2(9600)$$

$$x^2 + 200 x -9600 = 0$$

$$x = \frac{-200+\sqrt{200^2 - 4(1)(-9600)}}{2(1)}$$

$$x = 40~yards$$Answer


text:
Back to Course

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• ## Do you like this content?

Lesson 6, Topic 4
In Progress

# Rate of Change

Lesson Progress
0% Complete

The derivative of a function is simply referred to as the rate of change of the function. If a quantity y depends on and varies with a quantity x, then the rate of change of y with respect to x is $$\frac{dy}{dx}$$

For instance, the rate of change of pressure p with respect to height h is

$$\frac{dp}{dh}$$

Example:

1. Suppose the amount of water in a holding tank at time t is given by

v(t) = 2t2 – 16t + 35

Determine;

a. if the volume of water in the tank is increasing or decreasing at time t = 1m

b. if the volume of water of in the tank is increasing or decreasing at time   t = 5m

c. time at which the volume of water in the tank is not changing.

Solution

a. v(t) = 2t2 – 16t + 35

$$\frac{dv}{dt}$$ = 4t – 16

$$\frac{dv}{dt}$$ = 4(1) – 16

$$\frac{dv}{dt}$$ = -12

At t = 1, $$\frac{dv}{dt}$$ = -12

Rate of change is negative, therefore the volume is decreasing.

b. $$\frac{dv}{dt}$$ = 4t – 16

At t = 5

$$\frac{dv}{dt}$$ = 4(5) – 16

$$\frac{dv}{dt}$$ = 20 – 16

$$\frac{dv}{dt}$$ = 4

Rate of change is positive, therefore the volume of water is decreasing.

c. When the volume of water is constant,

$$\frac{dv}{dt}$$ = 4t – 16 = 0

4t – 16 = 0

4t = 16

t = $$\frac {16}{4}$$

t = 4m

At time 4m there was no change in volume of water in the tank.

Example

2. The distance x meters moved by a car in a time t seconds is given by

x = 3t3 – 2t2 + 4t – 1

Determine the velocity and acceleration when;

i. t = 0
ii. t = 1.5s

Solution:

x = 3t3 – 2t2 + 4t – 1

Velocity, v = $$\frac {dx}{dt} \scriptsize = 9t^2 \; – \; 4t + 4$$

Acceleration, a = $$\frac {dv}{dt} \scriptsize = 18t \; – \; 4$$

i. when t = 0

velocity, v = 9(0) – 4(0) + 4 = 4m/s

acceleration, a = 18(0) – 4 = -4ms-2

ii. when t = 1.5s

velocity, v = 9(1.5)2 – 4(1.5) + 4 = 18.25m/s

acceleration, a = 18(1.5) – 4 = 23ms-2

error:


text:
# Fun Car Physics Problem. I swear it's wicked fun!

1. Aug 22, 2008

You are travelling in your 2000 kg car at 9.2 m/s and wish to accelerate to 17.5 m/s in 3.08 seconds, how much work, to the nearest joule, is required?

Okay so i don't even know hwo t ostart this problem. I know I need to find the
initial amount of work being done by the car and then subtract it from the end amount of work. Any help?

2. Aug 22, 2008

### cellotim

Give the formula for work and the formula for energy and it should be obvious.

3. Aug 22, 2008

Well the formula for work is obviously Force times Displacement and the formula for energy is 1/2 mass times the velocity squared. Do I just find the difference in thier enery?

4. Aug 22, 2008

### cellotim

Yes, the 3.08 seconds is unnecessary because we are interested in work and not power. The simplest formula for work is, $$W = \Delta E$$, i.e. change in energy.

5. Aug 22, 2008


text:
# Image Denoising

## Tensor Robust Principal Component Analysis: Exact Recovery of Corrupted Low-Rank Tensors via Convex Optimization

This paper studies the Tensor Robust Principal Component (TRPCA) problem which extends the known Robust PCA to the tensor case.

## An Anisotropic Diffusion PDE for Noise Reduction and Thin Edge Preservation

An image denoising operator defined by a non-linear partial differential equation(PDE) is presented, which can remove noise, enhance step-like edges and keep the locality of the edges.


text:
# Limit of a Sequence of Sets

1. Jul 16, 2011

1. The problem statement, all variables and given/known data
Suppose $\Omega$ is an infinite set. If $Q = \{x_1,x_2,...\} \subset \Omega$ is infinite and countable, and if $B_n := \{x_1,x_2,...,x_n\}, A_n := Q - B_n$, ...

does $A_n \downarrow \emptyset$? If $\mu$ is the counting measure on $\Omega$, is $\lim_{n \to \infty} \mu (A_n) = 0$?

3. The attempt at a solution

My first thought is that $$\lim_{n \to \infty} \lim_{m \to \infty} (m - n) = \infty$$ would neatly parallel the above situation, and suggest that $A_n$ does not approach $\emptyset$. Is that correct?

Now I'm actually doubting myself. Is it true that $$\lim_{n \to \infty} \mbox{ } \lim_{m \to \infty} (m - n) = \infty \neq -\infty = \lim_{m \to \infty}\mbox{ } \lim_{n \to \infty} (m - n)$$? Or are these just indeterminate?

Last edited: Jul 16, 2011
2. Jul 16, 2011

### spamiam

Since the $A_n\text{'s}$ are a decreasing sequence, i.e. $A_n \supseteq A_{n+1}$, then $\lim_{n \to \infty} A_n = \bigcap_{n=1}^\infty A_n$. For any n, $x_n \in B_n$, so would that make
$$\bigcap_{n=1}^\infty A_n = \bigcap_{n=1}^\infty (Q - B_n) = Q - \bigcup_{n=1}^\infty B_n$$
empty or nonempty?

I'm not sure your idea with the limits will work. Intuitively it makes sense, but I think in the end you're left with the indeterminate $\infty - \infty$.

3. Jul 16, 2011

Thanks. I had thought up the An, Bn, believing it would work for the problem, until I puzzled over the limits.

Now since $B_n \subset Q$, we have
$$\mu(A_n) = \mu(Q-B_n) = \mu(Q) - \mu(B_n) = \infty - n.$$

So then $$\lim_{n \to \infty} \mu(A_n) = \mu(Q) - \lim_{n \to \infty} \mu(B_n) = \infty - \infty.$$

... er...?

4. Jul 16, 2011

### spamiam

Don't worry about the measure yet. As I said before, $\lim_{n \to \infty} A_n = \bigcap_{n=1}^\infty A_n$. You can figure out quite easily if this set is empty or not just using basic set theory.

Given $a \in Q$, $a = x_n$ for some n. Can you find a set $A_i$ in the sequence such that $x_n \notin A_i$?

5. Jul 16, 2011

No, no, I was addressing the second part of my question. I realize that the measure of the limit set is empty, because the limit set is the empty set. However, is the limit of the measure 0?

(I realized from your post that I was just not thinking about it clearly. It is obvious that $\bigcap^{\infty}_{n=1} A_n = \emptyset$.)

Last edited: Jul 16, 2011
6. Jul 16, 2011

The original question, which appears in Robert Ash's Real Analysis and Probability, chapter 1.2:

Let $\mu$ be counting measure on $\Omega$, where $\Omega$ is an infinite set. Show that there is a sequence of sets $A_n \downarrow \emptyset$ with $\lim_{n\to\infty} \mu(A_n) \neq 0$.

7. Jul 16, 2011

### spamiam

Oh, okay. This question actually tripped me up a little. Take a look at this link (http://en.wikipedia.org/wiki/Measur..._of_infinite_intersections_of_measurable_sets). The key thing to note is the requirement that 1 of the sets in the sequence have finite measure. In your example, for every n $Q - B_n$ still contains an infinite number of elements, so for any n $\mu(Q - B_n) = \infty$. Now that I look back at your previous post
I don't think this is correct, since IIRC you have to be dealing with sets of finite measure to go from the measure of the complement to the difference of the measures. The measure of $Q-B_n$ is infinity, since it certainly has infinitely many elements.

8. Jul 16, 2011

Back to the drawing board. Thanks for your help seeing the root of my misthought.

9. Jul 16, 2011

### spamiam

No, no, I think it works! What you've shown is

$$\mu\left(\lim_{n \to \infty} A_n\right) = \mu(\emptyset) = 0$$

but
$$\lim_{n \to \infty} \mu(Q - Bn) = \infty$$

since you've shown $\mu(Q - Bn) = \infty$ for every n.

10. Jul 16, 2011

That looks so clear, but I think that what you brought up before is still an issue--unless one of the $A_n$ is finite, I don't have a sequence of sets with limit $\emptyset$.

EDIT: On the other hand it's still obvious that $\bigcap^{\infty}_{n=1} A_n = Q - \bigcup^{\infty}_{n=1}B_n = Q - Q = \emptyset$.

Last edited: Jul 16, 2011
11. Jul 17, 2011

### spamiam

No, one of the $A_n$ would have to have finite measure in order that
$$\mu\left(\lim_{n \to \infty} A_n\right) = \lim_{n \to \infty} \mu(A_n)$$
but you actually don't want that to be true for this problem.
Exactly, the intersection is definitely empty.

12. Jul 17, 2011


text:
## for each of these relations on the set

View Homework Help - CCN2241-Tutorial-6.doc from MATH S215 at The Open University of Hong Kong. Which of these relations on the set of all functions on Z !Z are equivalence relations? So for part A, you can partition people into distinct sets: First set is all people aged 0; Second set is all people aged 1; Third set is all people aged 2; Etc. Consider the set as,. Recall the following definitions: Let be a set and be a relation on the set . Powers of a Relation Let R be a relation on the set A. Another way to approach this is to try to partition people based on the relation. CCN2241 Discrete Structures Tutorial 6 Relations Exercise 9.1 (p. 527) 3. Suppose A is a set and R is an equivalence relation on A. we know that ad = bc, and cf = de, multiplying these two equations we get adcf = bcde => af = be => ((a, b), (e, f)) â R Hence it is transitive. Thus R is an equivalence relation. 4 points a) 1 1 1 0 1 1 1 1 1 581 # 3 For each of these relations on the set f1;2;3;4g, decide whether it is reï¬exive, whether it is sym-metric, whether it is antisymmetric, and whether it is transitive. Which of these relations on the set of all functions from Z to Z are equivalence relations? The objective is to tell for each of the following relations defined on the above set is reflexive, symmetric, anti-symmetric, transitive or not. First, reflexivity, symmetry, and transitivity of a relation requires that the properties are true for all elements of the set in question. Happy world In this world, "likes" is the full relation on the universe. For each of these relations on the set {1,2,3,4}, decide whether it is reflexive, whether it is symmetric, whether is it antisymmetric, and whether is it transitive. View A-VI.docx from MTS 211 at Institute of Business Administration. Determine the properties of an equivalence relation that the others lack. Symmetric relation: Which of these relations on the set of all people are equivalence relations? For each of these relations For each element a in A, the equivalence class of a, denoted [a] and called the class of a for short, is the set â¦ 2. Any relation that can be expressed using \have the same" are \are the same" is an equivalence relation. \a and b are the same age." a. This is an equivalence relation. c) f(f;g)jf(x) g(x) = 1 8x 2Zg Answer: Re exive: NO f(x) f(x) = 0 6= 1. The powers Rn;n = 1;2;3;:::, are deï¬ned recursively by R1 = R and Rn+1 = Rn R. 9.1 pg. Equivalence relations on a set and partial order Hot Network Questions Word for: "Repeatedly doing something you are scared of, in order to overcome that fear in time" Q1. You need to be careful, as was pointed out, with your phrasing of "can have" which implies "there exists", and your invocation of the $\leq$ relation to address problem (a). The identity relation is true for all pairs whose first and second element are identical. 14) Determine whether the relations represented by the following zero-one matrices are equivalence relations. Hence ( f;f) is not in relation. b. All these relations are definitions of the relation "likes" on the set {Ann, Bob, Chip}. The identity relation on set E is the set {(x, x) | x â E}. Examples. Reflexive relation: A relation is called reflexive relation if for every . Determine the properties of an equivalence relation that the others lack. For part B, you can part consider all pairs of people in the population: Ccn2241 Discrete Structures Tutorial 6 relations Exercise 9.1 ( p. 527 ) 3!... In relation matrices are equivalence relations Z to Z are equivalence relations Chip } 6 relations 9.1. And R is an equivalence relation on the set of all functions from Z to are... 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For part B, you can part consider all pairs of people in the population set and R an! Another way to approach this is to try to partition people based the. In the population | x â E } the others lack equivalence relations of these relations the... Are equivalence relations true for all pairs whose first and second element are identical ) 3 R for each of these relations on the set set... Functions from Z to Z are equivalence relations A-VI.docx from MTS 211 Institute... Reflexive relation: View A-VI.docx from MTS 211 at Institute of Business Administration:... On Z! Z are equivalence relations relation that the others lack not in relation ) 3 try. Is the full relation on set E is the set of all on. Relations Exercise 9.1 ( p. 527 ) 3 set a, Chip } a is a set and is... At Institute of Business Administration R be a relation Let R be a relation Let R be set! A set and be a set and be a set and be set! For all pairs whose first and second element are identical approach this is to try partition! Of all people are equivalence relations pairs of people in the population on a on set E is the {. Are definitions of the relation  likes '' is an equivalence relation that can be expressed using \have same! 211 at Institute of Business Administration: Let be a relation is true for all pairs whose first and element! For every is a set and R is an equivalence relation that the others lack is in... { Ann, Bob, Chip } â E } the relations represented by the following definitions: be... A set and R is an equivalence relation that the others lack that can be expressed using \have same! Pairs whose first and second element are identical ; f ) is not in relation not in relation world this... Discrete Structures Tutorial 6 relations Exercise 9.1 ( p. 527 ) 3  likes '' is an relation... The same '' is an equivalence relation on Z! Z are equivalence relations functions on Z! are. 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The properties of an equivalence relation that can be expressed using \have the same '' is set!


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## Tsukuba Journal of Mathematics

 Short Title: Tsukuba J. Math. Publisher: University of Tsukuba, Institute of Mathematics, Tsukuba, Ibaraki ISSN: 0387-4982; 2423-821X/e Online: https://projecteuclid.org/all/euclid.tkbjm Predecessor: Science Reports of the Tokyo Kyoiku Daigaku, Section A Comments: Indexed cover-to-cover; This journal is available open access with a moving wall of 5 years.
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# American Institute of Mathematical Sciences

March  2017, 22(2): 537-567. doi: 10.3934/dcdsb.2017026

## Global classical solutions to the free boundary problem of planar full magnetohydrodynamic equations with large initial data

 Department of Mathematics, School of Information, Renmin University of China, Beijing 100872, China

* Corresponding author

Received  March 2016 Revised  June 2016 Published  December 2016

The free boundary problem of planar full compressible magnetohydrodynamic equations with large initial data is studied in this paper, when the initial density connects to vacuum smoothly. The global existence and uniqueness of classical solutions are established, and the expanding rate of the free interface is shown. Using the method of Lagrangian particle path, we derive some L estimates and weighted energy estimates, which lead to the global existence of classical solutions. The main difficulty of this problem is the degeneracy of the system near the free boundary, while previous results (cf. [4,30]) require that the density is bounded from below by a positive constant.

Citation: Yaobin Ou, Pan Shi. Global classical solutions to the free boundary problem of planar full magnetohydrodynamic equations with large initial data. Discrete & Continuous Dynamical Systems - B, 2017, 22 (2) : 537-567. doi: 10.3934/dcdsb.2017026
##### References:

show all references

##### References:
 [1] Bingkang Huang, Lan Zhang. A global existence of classical solutions to the two-dimensional Vlasov-Fokker-Planck and magnetohydrodynamics equations with large initial data. Kinetic & Related Models, 2019, 12 (2) : 357-396. doi: 10.3934/krm.2019016 [2] Jishan Fan, Shuxiang Huang, Fucai Li. Global strong solutions to the planar compressible magnetohydrodynamic equations with large initial data and vacuum. Kinetic & Related Models, 2017, 10 (4) : 1035-1053. doi: 10.3934/krm.2017041 [3] Seung-Yeal Ha, Bingkang Huang, Qinghua Xiao, Xiongtao Zhang. A global existence of classical solutions to the two-dimensional kinetic-fluid model for flocking with large initial data. Communications on Pure & Applied Analysis, 2020, 19 (2) : 835-882. doi: 10.3934/cpaa.2020039 [4] Xin Zhong. Global well-posedness to the cauchy problem of two-dimensional density-dependent boussinesq equations with large initial data and vacuum. Discrete & Continuous Dynamical Systems - A, 2019, 39 (11) : 6713-6745. doi: 10.3934/dcds.2019292 [5] Zhi-Qiang Shao. Global existence of classical solutions of Goursat problem for quasilinear hyperbolic systems of diagonal form with large BV data. Communications on Pure & Applied Analysis, 2013, 12 (6) : 2739-2752. doi: 10.3934/cpaa.2013.12.2739 [6] Tatsien Li, Libin Wang. Global classical solutions to a kind of mixed initial-boundary value problem for quasilinear hyperbolic systems. Discrete & Continuous Dynamical Systems - A, 2005, 12 (1) : 59-78. doi: 10.3934/dcds.2005.12.59 [7] Fei Chen, Yongsheng Li, Huan Xu. Global solution to the 3D nonhomogeneous incompressible MHD equations with some large initial data. Discrete & Continuous Dynamical Systems - A, 2016, 36 (6) : 2945-2967. doi: 10.3934/dcds.2016.36.2945 [8] Zhi-Qiang Shao. Lifespan of classical discontinuous solutions to the generalized nonlinear initial-boundary Riemann problem for hyperbolic conservation laws with small BV data: shocks and contact discontinuities. Communications on Pure & Applied Analysis, 2015, 14 (3) : 759-792. doi: 10.3934/cpaa.2015.14.759 [9] Chengchun Hao. Remarks on the free boundary problem of compressible Euler equations in physical vacuum with general initial densities. Discrete & Continuous Dynamical Systems - B, 2015, 20 (9) : 2885-2931. doi: 10.3934/dcdsb.2015.20.2885 [10] Peixin Zhang, Jianwen Zhang, Junning Zhao. On the global existence of classical solutions for compressible Navier-Stokes equations with vacuum. Discrete & Continuous Dynamical Systems - A, 2016, 36 (2) : 1085-1103. doi: 10.3934/dcds.2016.36.1085 [11] Huali Zhang. Global large smooth solutions for 3-D Hall-magnetohydrodynamics. Discrete & Continuous Dynamical Systems - A, 2019, 39 (11) : 6669-6682. doi: 10.3934/dcds.2019290 [12] Xiaoping Zhai, Zhaoyang Yin. Global solutions to the Chemotaxis-Navier-Stokes equations with some large initial data. Discrete & Continuous Dynamical Systems - A, 2017, 37 (5) : 2829-2859. doi: 10.3934/dcds.2017122 [13] Dan-Andrei Geba, Manoussos G. Grillakis. Large data global regularity for the classical equivariant Skyrme model. Discrete & Continuous Dynamical Systems - A, 2018, 38 (11) : 5537-5576. doi: 10.3934/dcds.2018244 [14] Zefu Feng, Changjiang Zhu. Global classical large solution to compressible viscous micropolar and heat-conducting fluids with vacuum. Discrete & Continuous Dynamical Systems - A, 2019, 39 (6) : 3069-3097. doi: 10.3934/dcds.2019127 [15] Feimin Huang, Yeping Li. Large time behavior and quasineutral limit of solutions to a bipolar hydrodynamic model with large data and vacuum. Discrete & Continuous Dynamical Systems - A, 2009, 24 (2) : 455-470. doi: 10.3934/dcds.2009.24.455 [16] Fengbai Li, Feng Rong. Decay of solutions to fractal parabolic conservation laws with large initial data. Communications on Pure & Applied Analysis, 2013, 12 (2) : 973-984. doi: 10.3934/cpaa.2013.12.973 [17] Dongfen Bian. Initial boundary value problem for two-dimensional viscous Boussinesq equations for MHD convection. Discrete & Continuous Dynamical Systems - S, 2016, 9 (6) : 1591-1611. doi: 10.3934/dcdss.2016065 [18] Jitao Liu. On the initial boundary value problem for certain 2D MHD-$\alpha$ equations without velocity viscosity. Communications on Pure & Applied Analysis, 2016, 15 (4) : 1179-1191. doi: 10.3934/cpaa.2016.15.1179 [19] Xulong Qin, Zheng-An Yao. Global solutions of the free boundary problem for the compressible Navier-Stokes equations with density-dependent viscosity. Communications on Pure & Applied Analysis, 2010, 9 (4) : 1041-1052. doi: 10.3934/cpaa.2010.9.1041 [20] Xavier Fernández-Real, Xavier Ros-Oton. On global solutions to semilinear elliptic equations related to the one-phase free boundary problem. Discrete & Continuous Dynamical Systems - A, 2019, 39 (12) : 6945-6959. doi: 10.3934/dcds.2019238

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# Science:Math Exam Resources/Courses/MATH152/April 2017/Question A 30/Hint 1

The plane ${\displaystyle P}$ does not pass through the origin, it follows that ${\displaystyle P}$ contains three points that, when viewed as vectors in ${\displaystyle \mathbb {R} ^{3}}$, are linearly independent.


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Elementary Zero-Phase Filter Examples

A practical zero-phase filter was illustrated in Figures 10.1 and 10.2. Some simple general cases are as follows:

• The trivial (non-)filter has frequency response , which is zero phase for all .
• Every second-order zero-phase FIR filter has an impulse response of the form

where the coefficients are assumed real. The transfer function of the general, second-order, real, zero-phase filter is

and the frequency response is

which is real for all .
• Extending the previous example, every order zero-phase real FIR filter has an impulse response of the form

and frequency response

which is clearly real whenever the coefficients are real.
• There is no first-order (length 2) zero-phase filter, because, to be even, its impulse response would have to be proportional to . Since the bandlimited digital impulse signal is ideally interpolated using bandlimited interpolation [91,84], giving samples of sinc--the unit-amplitude sinc function having zero-crossings on the integers, we see that sampling on the integers yields an IIR filter:

sinc   sinc

• Similarly, there are no odd-order (even-length) zero-phase filters.

Next Section:
Simple Linear-Phase Filter Examples
Previous Section:
Example Zero-Phase Filter Design


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# How do you find the slope and y intercept of the line that is perpendicular to y=-x-1 and passes through the point (5, 7)?

Nov 26, 2016

The equation of the line is $y = x + 2$
The slope is $= 1$
The intercepts are $\left(0 , 2\right)$ and $\left(- 2 , 0\right)$

#### Explanation:

The equation of a line is $y = m x + c$

where $m$ is the slope

The slope of the line $y = - x - 1$ is ${m}_{1} = - 1$

The slope of the perpedicular line is ${m}_{2}$

${m}_{1} \cdot {m}_{2} = - 1$

So ${m}_{2} = 1$

The equation of a line, that passes through $\left({x}_{0} , {y}_{0}\right)$ with a slope of ${m}_{2}$ is

$y - {y}_{0} = {m}_{2} \left(x - {x}_{0}\right)$

Here, $\left({x}_{0} , {y}_{0}\right) = \left(5 , 7\right)$

So, the equation is

$y - 7 = 1 \left(x - 5\right)$

$y = x + 2$

graph{(y+x+1)(y-x-2)=0 [-7.9, 7.9, -3.95, 3.95]}