FRAC

Empiece por romper los espejos de su casa, deje caer los brazos, mire vagamente la pared, olvídese. Cante una sola nota, escuche por dentro. Si oye algo como un paisaje sumido en el miedo, con hogueras entre las piedras, con siluetas semidesnudas en cuclillas, creo que estará bien encaminado, y lo mismo si oye un río por donde pasan barcas pintadas de amarillo y negro, si oye un sabor de pan, un tacto de dedos, una sombra de caballo. Después compre solfeos y un frac, y por favor no cante por la nariz y deje en paz a Schumann.
— 

Julio Córtazar

Historia de Cronopios y de Famas, Instrucciones para cantar

kubleeka  asked:

I think you can define integrals without formally appealing to area (or any geometry ). Like yeah, the definition is inspired by rectangles, but if you just define \displaystyle\int_a^b f(x)=\lim_{n\to\infty}\sum_{i=1}^n f(a+\frac{i(b-a)}{n})(\frac{b-a}{n}) then couldn't you just define the area of a rectangle as an integral? There's nothing inherently geometric about the limit definition

I beg to disagree… what you more or less have is as below:

but with f( x_i ) = f( a + i(b-a)/n) ). 

What the limit definition is doing is dividing the area under f into n subintervals, the width of each being given by (b-a)/n. The graph above is more or less what I’m talking about; in that case there are 8 subintervals. 

The height of the ith subinterval, often written as f( x_i ), is  f( a + i(b-a)/n) ). (The term a+i(b-a)/n) is an x value within the ith subinterval; plus i widths.) 

So to find the area under the curve, the limit definition is taking the limit of sum of (height)(width) as width gets infinitely narrow. This is consistent with how I usually describe integrals: as the sum of infinitely many, infinitely skinny rectangles. 

So that definition, at least, is inherently geometric. The formula for the area of a rectangle is foundational to the limit definition. (Though that’s something some calc classes spend about 20 minutes on, as someone added to this discussion earlier.) There may be a way to define integrals that doesn’t use rectangles, but I don’t know it. 

anonymous asked:

Do you know of any interesting buildings post 2000 designed with new tech insulation skin or have found good solutions to such problems?

Not sure what you are asking. The design of insulated skins is always progressing. Here are some recent examples:

Media-TIC Enric Ruiz Geli

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How I Attempted to Mathematically Model the Spread of Memes on the Internet, pt 3 (The Final Chapter)

Welcome back! If you have not read the earlier parts of this series, you can find them here:

  • Part 1 (Intro to Markov Chains)
  • Part 2 (Setting Up The Model)

This will be the last part of this series and will primarily involve ~actually~ using the model. Finally, amirite?

Things I’m Going To Touch On:

  1. More About I(t)
  2. Fitting I(t) To Data
  3. Predicting Probabilities
  4. For the Future

Here we go!

More About I(t)

First, let’s recall the equations we obtained in the last post. 

We can figure out the long term behavior of this I(t) by taking the limit as t approaches $\infty$. Let’s note that since $\alpha$, $\beta$ and $\gamma$ are less than 1, the exponential terms will go to zero when taking the limit. So, 

What does this mean in context?

This $\frac{\gamma}{\beta+\gamma}$ tells us the probability of someone being in the infected state in the long term (that is, after the initial “outbreak” of the meme). We can also verify this from our steady state distribution, but I’ll leave that as an exercise. 

Additionally, I think it’s interesting to note that this long term value does not depend whatsoever on $\alpha$. This means that the probability for moving from susceptible to infected has no effect on the long term probability of being in the infected state. 

Fitting I(t) to Data

I obtained my data from Google Trends. Here’s what I did:

First, I looked up a particular meme as a search term on Google Trends and got a spreadsheet of popularity values. Google plots the data so that the search term’s peak popularity is given a value of 100 and then the rest is scaled in relation to that value. Since I’m concerned with probabilities, I can’t use the values given by Google. So, I scaled the values down so that they were between 0 and 1.

Now that I have my adjusted values, I want to try and fit my I(t) equation to them. I did this by performing regression analysis using SageMath to find the $\alpha$, $\beta$, and $\gamma$ values that best fit I(t) to the data (Will provide my code upon request!). 

Below is the Google Trends graph for “dat boi”. I do want to note that each data point obtained from this graph is for a one week time span. 

And here’s a graph of the best fit I(t) (*drumroll*):

This is awesome! Look how closely I(t) matches the actual data. In this case, the $\alpha$, $\beta$, and $\gamma$ that best fit I(t) are:

  • $\alpha \approx$ 0.5354
  • $\beta \approx$ = 0.3308
  • $\gamma \approx$ = 0.03285

We can also see that $\frac{\gamma}{\beta+\gamma} = 0.09$.

I performed this analysis with 10 different memes/search terms. I’ll present the results in the table below, just listing the $\alpha$, $\beta$, and $\gamma$ values. 

Quick note: those terms with an asterisk had data points that were taken daily, those with a cross were monthly, and those with nothing were taken weekly.

I included “Flint, Michigan” because although it isn’t a meme, it does follow the same viral pattern that the memes do. This is due to the increased media coverage due to the Flint Water Crisis. I do want to stress that I am in no way trying to diminish the significance of the Flint Water Crisis or the struggle that those involved are facing, but rather I want to show that this model can be applied to more things than just memes. 

It’s neat that we can find these probabilities with memes that have already died down from their “outbreaks,” but is there a way we can try and predict the $\alpha$, $\beta$, and $\gamma$ values for up and coming memes? This brings us to the next section of this post. 

Predicting Probabilities

I want to try to predict the $\alpha$, $\beta$, and $\gamma$ values for a particular meme given an initial set of data points. Let’s assume that we have the data points up to and including the peak popularity. We’ll talk about why we need to assume this in a little bit. 

Let’s start by attempting to predict $\alpha$. Briefly recall that $\alpha$ is the probability of moving from the susceptible state to the infected state. I know that $\alpha$ has to do with how quickly the meme rises in popularity. Because of this, I approximate $\alpha$ using slopes in the following process:

  1. Take the set of points up to and including the peak point.
  2. Find all of the slopes between consecutive pairs of points.
  3. Take the maximum over all these slopes. This is the estimated $\alpha$. 

This process is why we have to assume the peak is included in our data set. If we didn’t have the peak, we wouldn’t know if there was a slope that would be higher than the maximum we found over the points we have. 

Using this method over all 10 examples gives the following results for $\alpha$.

This seems promising at first, but with a closer look, the differences for some of the $\alpha$ values are pretty large. Is there a way to make it an even better approximation?

Spoiler alert: yes. 

Using SPSS, I performed various regressions with this estimated $\alpha$ as the independent variable and the actual $\alpha$ value obtained from the best fit I(t) as the dependent variable. See the table below:

Let’s quickly note that for every regression performed, the p-value is < 0.05, and so the relationship between $\alpha$ and our approximation of $\alpha$ is statistically significant (via F-Test of Overall Significance). We want to pick the equation that gives us the highest R-Square value. The quadratic and the cubic equations have the same R-Square value, so I’ll pick the quadratic just so I have one less term to deal with. 

Thus,

where $\alpha_{est}$ is that largest slope value. 

Using this new way to estimate $\alpha$, we find our new approximate $\alpha$ values by plugging in the maximum slope the equation above:

We can see that the differences between the estimated and the actual $\alpha$ values are much smaller than in the previous estimate! The bold values are those that have decreased significantly.

Now we have a pretty nice approximation for $\alpha$!

Can we do the same for $\beta$?

Let’s try! Since $\beta$ is the probability of moving from the infected state to the recovered state, we can’t really use the slopes as an approximation. Instead, we’ll perform a regression analysis with $\alpha$ as the independent variable and $\beta$ as the dependent variable to see if there’s a connection. 

In this case, we will pick the cubic equation. Thus,

Let’s look at how this approximation compares to the actual $\beta$ values obtained from this model. 

The average difference between the expected and the actual $\beta$ with this approximation is 0.06214. I believe it’s safe to say that provided we have an accurate $\alpha$ value, we can obtain an estimated $\beta$ value that is somewhat close to the actual $\beta$. 

Of course, there is probably a method that will better predict $\beta$, but for now, this is the best we’ve got.

What about $\gamma$?

Well, here’s where we have problems. Unfortunately, there is no significant relationship between $\alpha$ and $\gamma$ or $\beta$ and $\gamma$, as determined by performing various regressions. Because of this, we can’t use the same methods of approximation that we used for $\alpha$ and $\beta$. This brings me to the final section of this post. 

For The Future

We’ve done a lot with this already, but there is always more to be done! There are several things that my model does not account for:

  1. If a meme has two peaks instead of just one (i.e. it died and then became popular again)
  2. Extremely sharp increase followed by an extremely sharp decrease (say, in the matter of a couple days)
  3. If $\beta + \gamma = \alpha$ or if $\beta + \gamma > 1$. This is because of issues with our I(t) equation. $\beta + \gamma  = \alpha$ gives a denominator of zero and $\beta + \gamma > 1$ makes $1-\beta-\gamma$ negative, which causes issues when raising it to t

Problem 3 is an issue because initially our Markov chain has no restrictions on what $\alpha$, $\beta$, or $\gamma$ could be, yet when I come up with the I(t) equation, we somehow have these problems. It’d be nice to have a way to circumvent this issue! 

Additionally, we still cannot predict what $\gamma$ will be, and until we can do so, we can’t create an equation given just an initial data set. Finding a way to $\gamma$ is the first thing on the list for future work. 

These are all things to consider more in the future, and I’d (of course) love to hear any ideas on how to combat these issues. 

Thanks for reading!

As always, if there are any questions/comments/concerns or even suggestions for another post, please feel free to send an ask!

Hope all is well and (as always) stay positive!

heyyo! I recently reached 7k awhile back and since it’s Valentine’s Day, I thought it might be a good day to spread my appreciation and luv to my lovely followers / followings. Thank you so much for all you’ve done and dealing with my semi-hiatus a$$ these past years. I’ve honestly been a passive aggressive member of tumblr and yet here we are!!! I’ve received 7k because of all of you. To my followings, mutual or not, thank you for also filling my dashboard with an unimaginable quantity of feed & quality at that. As for my followers, I wouldn’t be here without you guys, thank you so much for always adding notes to my posts and even my writing!! Awe shucks I’m getting emo. Thank you, so so much!! 

note: bolded are long time mutuals / most feed on my dashboard. 

@diomodias @kumeichan @kaizune @myuumin @not1ce-m3-senpa11 @usagiwaltz @sad-empty @mikoooh @emptyvoice @flwrtea @okaykimchi @sasak-i @ao-kun @orangepies @thelastdeadsurvivor @rainyscenes @ante-i-ku @monoblur @murderered @imiiko @00m @kawaiioyori @sui-u @kuro-ichi @bakaletters @chunchuun @usotsuki-san @kuzuun @ikasle @112yk @un-flipped @yugen-ai @kokyu @itoshigoyo @steamed-bun @furawa-su @cielillumine @buruberry @massushiro @yuureitachi @daisyalyssum @seijochan @effetnuit @thorneater @shikudo @tubschan @konayukii @endlessnoctis @gothjerk @fukai-danieru @atsu-i @sadasuna @haneurii @yuki-zora @ochiru-unmei @fushimii @evanqelic @hoshiko-san @ddeku @yumetsugu @ohmilklu @yechiru @almightyone @errichi @transcendently @hiiyorrin @norunir @kuuka-i @isntoolate @8zip @nataku-s @nobuo-senpai @omphhh @yce-cream-i @chihiruu @muryokuna @frac-t-ures @shoujoromance @shojonotes @voicelessbaka @bakaland @oyasuumi @kuronanashi @c0verted @nijieen @fuy-u @monokuronoai @shinjukuu @nemuicchi @fightforyourideals @y-u-u-n-a-r-i @someothermangacaps @wearylimbs @instaliu @tor-u @ne-zushii @sireai @baham-t @sola-nin @serelyne @torihaizu @hopelessbaka @raikee-hime @kurai-reii @hui-kun @anything-shoujo @eulaylia @e-s-u-r-a @otsukai @nakiyurei @aichirou @addictivities @h-xrder @s-eramic @shinigamichii @chidoki @f-uwa @paralian-s @10kou @uruka @oikawawas @lluvina @habuti @d-reamt @tykikakusei @prince-vegeta @yumierro @watashi-no-hikari @theabearr @invisible-depression 

The Simple Brilliance of David Aja

As usual, I am late to the party: Fraction/Aja’s run on Hawkeye is one of the best things Marvel has done in the past few years.  I had already owned a few issues that I quite like, but sometime between the weekend Matty Frac-fracs was matching donations from his Welovefine shop to Futures Without Violence (wherein I purchased a sweet Hawkguy messenger bag) and a sale Comixology had on Hawkeye issues, I finally partook of the series as a whole.

And holy sweet shit Aja is brilliant.  I had read Fraction doting on the guy, and obviously I read #11 and knew that anybody who could throw together that issue was special, but my goodness he’s clever.  There are just too many examples, too many small things to focus on, so let me just take you through a few pages in Hawkeye #6.

Here’s page 9, not counting the cover and the credits:

This is as close to a perfect comic page as you are going to see: writer, artist, and colorist, all firing on all cylinders.  I mean, we’re talking about a page covered in the word “bro,” where Clint gets called “Hulk-Guy” but dismisses it with a sentence containing the term “joyous-ass Kwanzaa.”  Delightful.  Not to mention Hollingsworth, who is always solid, making that panel where the arrow hits the bat just purple enough so that you know what’s up.  Ugh, so good.

But Aja is on another level here.  Literally.  Because what do you think this page is about?  The arrow hitting the bat?  Nah.  The punk girl coming out and letting him know about the bros?  Nope.  This page is about Clint seeing what’s downstairs.  That’s it.  It’s not about him being a hero (at least not directly, not yet), it’s not about nobody knowing how to say “Hawkeye,” and it’s not about him being ready to pew-pew some bros: it’s about seeing the bros, and then acting accordingly by going downstairs.  

It’s a page about… well, about downstairs.  And what does Aja do?  He gives you that fifth and, especially, that sixth panel.  That glorious fucking comics move where an artist uses a gutter to divide time but not space.  Specifically because he does that, we get the exaggerated effect of really feeling like Clint and Co. are running to the side of that building and then looking over: it gives the reader a little bit of extra vertigo, and it completely determines how you ought to read this page.  As the sixth panel descends to ground level, we get the seventh panel of the bro shouting “HEY, BROOOO.”  And then, as the sixth panel cuts off before the ground, the eight panel puts us at ground level, looking at the bros, just before Clint’s arrival and the extra dose of juxtaposed silhouetted genius.

Do you know how boring this page could have been?  That sixth panel giving all that gravity to the page is a very specific and effective artistic choice, one that 9/10 artists don’t make, even if a couple of them make pretty good choices.  Aja does this constantly: he takes something simple and he makes it wonderful in a way that only comics could make something wonderful.  Here’s the page immediately after:


Okay, what is this page about?  A confrontation?  That happened last page.  An ass-beating?  That happens on the next page.

This is a page of a story that is about people getting out of fucking vans.  That’s it!!!  Imagine reading something akin to this page in a textual narrative: “Two vans pulled up.  The doors opened.  A bunch of reinforcement bros got out of the vans and charged at Clint, bats at the ready.”  Who the fuck cares.  Sure, it builds supsense: obviously setting up a bunch of people about to kick the shit out of someone and then ending it before the beatdown will build suspense.  But that doesn’t really make a bunch of dudes getting out of a van interesting.

But Aja does.  Aja actually makes a page about men climbing out of a fucking vehicle interesting.  How?  Symmetry, bro.  Mirroring, bro.  Clint comes out, across from the bros.  Next beat, we see him talking with a bro, juxtaposed next to the bro.  So we get distance, and then closeness.  Then we see the vans pull up across from each other (distance), then we see that mirrored, flipped around (closeness) yet still divided by white space in order to give Clint’s words prominence (which creates distance), and then end the page by seeing the bros close the distance between them and Clint.

The best part of this symmetrical mirroring juxtaposition stuff that Aja’s got going on here?  Clint is in the center of the page the entire time.  The page is about dude’s getting out of vans, but obviously we ought to be considering the fact that the object of their vodka-fueled ass-kickery is one Clint Barton.  The only time he’s not right in the middle of the page is when his words are the only thing in the middle of the page, which allows for an extra beat of suspense.

It would be one thing if Aja only turned in these two pages, but the dude is consistently making simple actions and interactions seem like magic on the page.  I just hope that every writer, like Fraction, recognizes how special this guy is and hands him Marvel-style scripts.  AJA IS BETTER THAN YOU, WRITERS.  HE IS BETTER THAN YOU.

Study With Me: Line Integrals

Hey guys! I’m currently studying for the Mathematics Subject Test of the GRE, which I plan on taking in the fall. One of the ways I like to study is by explaining the material to someone else. I currently have weekends off from research, and since Saturdays are for the boys, it leaves Sundays for GRE preparation. 

Because of this, every Sunday, I’ll explore a different undergraduate topic that could appear on the Mathematics Subject Test. This week: Line Integrals.

I’ll talk about the following:

  1. What is a line integral?
  2. How do you calculate a line integral?
  3. An Example

As a brief note, this post contains LaTeX code and will be much easier to read when viewed directly on my blog, where the code will compile!

What is a line integral?

Let’s first recall what we already know about integrals. We’re used to integrating functions of one variable over an interval [a,b]. We can think of this as integrating over the path on the x-axis from a to b, and the value of the integral as giving the area bounded by the curve y=f(x) over the path [a,b]. 

But! We can also integrate over paths that aren’t just straight lines along the x-axis. The resulting integral is called a curve, contour, or path integral. Most commonly, it is known as a line integral. 

In this post, I’ll be talking about line integrals with respect to arc length. 

Before we get into it, I’d like to start by defining what it means for a curve to be smooth. A curve, C, with parameterization r(t) = <x(t), y(t)> is smooth if the derivative r’(t) is continuous and nonzero. Additionally, we can say C is piecewise smooth if it is composed of a finite number of smooth curves joined at consecutive endpoints. Basically, this means I have a bunch of curves $C_1, C_2, …, C_n $ that are all individually smooth and Each $C_i$ has its endpoints connected to $C_{i-1}$ and $C_{i+1}$. 

Back to line integrals. Suppose we have a function f(x,y) and a smooth curve, C, in the x-y plane. We want to think about breaking C into n tiny pieces of arc length $\Delta s_i$. For each of the tiny pieces of C, choose any point $P_i = (x_i, y_i)$ and then multiply $f(P_i) = f(x_i, y_i)$ by the length $\Delta s_i$. This process is fairly similar to how we define integration for the case where the path is a line on the x-axis. We want to sum up these multiplied terms for all n terms. If the value of that sum approaches a finite, limiting value as $n \rightarrow \infty$, then the result is the line integral of f along C with respect to arc length. Below is a comparison of the single variable case integrating over a path [a,b] on the x-axis (left) and the line integral with respect to arc length over the curve C (right). 

Note the notation used for the line integral. If we’re integrating over a path C, we write C at the bottom of the integral. 

What does this mean geometrically?

The value of this integral is the area of the region whose base is C and whose height above each (x,y) point is given by f(x,y)

How do we actually calculate the line integral?

First, parameterize C. That is, for a parameter t, find the equations x=x(t) and y=y(t)  for $a \leq t \leq b$. We consider C to be directed, which means we’re saying that we trace C in a definite direction, which is called the positive direction. Basically, we’re saying that t runs from a to b, so A = (x(a), y(a)) is the initial point and B = (x(b), y(b)) is the final point. 

Since we have $(ds)^2 = (dx)^2 + (dy)^2$ (think Pythagorean theorem), we can write:

$\frac{ds}{dt} = \pm \sqrt{( \frac{dx}{dt})^2+(\frac{dy}{dt})^2}$

which can then be rewritten as:

$\pm \sqrt{(x’(t))^2+(y’(t))^2}$.

We use the + sign if the parameter t increases in the positive direction on C and the - sign if t decreases in the positive direction on C

So, we have:

$\int_C f \,ds = \int_{a}^{b} f(x(t), y(t))\frac{ds}{dt} \,dt$.

An Example

Determine the value of the line integral of the function f(x,y) = x + y^2 over the quarter-circle x^2 + y^2 = 4 in the first quadrant, from (2,0) to (0,2).

Solution below.

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image isn't anything

I am the smoke
you are the mirror

I cloud your hope
you project unclear

You will never be
anything at all

except an image
too skinny, fat or tall

the damage I cause
will haunt you for years

I am the source
whispers in your ears

you are the mirror

you are a sky without clouds
over a cornfield

you are on the yellow frac-
tured
road to fucking awesome

you have a beautiful heart and beauty
lives in the smile of the Soul

those aren’t scars on your heart
they are only scratches

The Soul gives no shits
about smoke

bg-4/27/17