## 0055: "Useless"

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pinkgothic
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### Re: "Useless" Discussion

I tend to look at ASK and think... yes, of course you should ask, that's always the best way to find a possible mate.
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### Re: "Useless" Discussion

I only get an ASK of 1.2 (out of ten) with the girl I'm currently kissing (we're not dating, but that's because of distance). Also, the numbers seem to leave a small window for people to actually ask, they're either astronomically high, or sub 1. Also, what if she's in no relationship, and actively looking for a potential partner; would that make her R = 0 istead of 1?
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AvalonXQ
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### Re: "Useless" Discussion

pinkgothic wrote:

I tend to look at ASK and think... yes, of course you should ask, that's always the best way to find a possible mate.

Precisely. When dealing with the possibility of a relationship, you're talking about ONE human being -- not an average or representative member of a pool of human beings. Individual people are unpredictable enough that equations dealing with populations or averages are pretty much useless.
When I chose to move such that the average education of the population dropped dramatically, putting me in a situation where the population is less desirable for me in terms of finding a mate, I shared my concerns with my father. He replied, "Sure, son, there are fewer women out here that you might be compatible with. Remember, though, that you're not trying to maximize your number of hits. You don't have to have a great relationship with every woman you date. You only have to find ONE that really works."
... he's a smart man.

pwnb0t
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### Re: "Useless" Discussion

This is the groom's (my) cake from my wedding that was December 29, 2007

unfortunately the cake maker didn't put xkcd.com at the bottom like I had requested (but I told everyone where it was from, no I didn't pass it off as my idea)
and apparently it's difficult to make actual infinity symbols on the fourier transform
All the engineers(/math/phys/etc) that came liked it and explained it to the laypeople

I was going to email this picture to Randall Munroe but I didn't exactly find a "Hey Buddy!" email address (only "business" ones)
I imagine he's busy, but I'd like it if he got to see it (since I was honoring (and advertising!) his awesome comic in a way).

The cake was good. (And this isn't a lie)

Teelabrown
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### Re: "Useless" Discussion

I think you are all missing the obvious. Take the inverse of cos and you get that cos of some unknown is love. I think that's about as deep as it gets, man.

natturner
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### Re: "Useless" Discussion

anyone notice that there should actually be a negative in the exponential of the fourier transform?

ConMan
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### Re: "Useless" Discussion

natturner wrote:anyone notice that there should actually be a negative in the exponential of the fourier transform?

There are several equivalent formations of the Fourier transform, each with a corresponding inverse transform. Some move negative signs around, some use a 1/sqrt(2*pi) term in both while others use a single 1/(2*pi), and so forth. They all have different "nice" properties depending on the application.
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afireinside13t
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### Re: "Useless" Discussion

ConMan wrote:
natturner wrote:anyone notice that there should actually be a negative in the exponential of the fourier transform?

There are several equivalent formations of the Fourier transform, each with a corresponding inverse transform. Some move negative signs around, some use a 1/sqrt(2*pi) term in both while others use a single 1/(2*pi), and so forth. They all have different "nice" properties depending on the application.

This has always bothered me. Can't we all just say that the Fourier transform has a negative exponent and that the 2 Pi is in the exponent? I'm studying for my qualifying exam, and every professor uses a different convention.

TheNige
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### "Useless" Discussion

In the spirit of the "Useless" comic, I offer the following:

∫ex = f (u)n

which, of course, reads that sex ... is fun. I hope I haven't violated any rules here with my first posting.

Fledermen64
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### Re: "Useless" Discussion

Men its just a cardioid how hard could it be to solve for.
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bluegenetic
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### Re: "Useless" Discussion

Love is blind. So if you perform any operations on it the operator can't see it...

joedaka
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### Re: "Useless" Discussion

Hey all,

here comes a complete and well structured solution by my friend and me. As it's a quite mathematical answer - and we felt forced to make it a nice LaTeX-made paper.
Attachments
xkcd55-2.pdf
Complete solution for XKCD 55

ejleon
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### Re: "Useless" Discussion

joedaka wrote:Hey all,

here comes a complete and well structured solution by my friend and me. As it's a quite mathematical answer - and we felt forced to make it a nice LaTeX-made paper.

Could you possibly post the LaTeX source?

joedaka
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### Re: "Useless" Discussion

ejleon wrote:Could you possibly post the LaTeX source?

Yea ... well - why soever - for what do you need it?

Here it is (made with LaTeX again ... )
Attachments
xkcd55-2_print.pdf
source code of xkcd55-2.pdf

chesterbr
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### Re: "Useless" Discussion

Well, here is my take on the problem - if I didn't screw up with the calculations (which I typically do). Adapting the cardioid formula to a more heart-like shape led to a weird, but credible result. Tried the cosine, but I guess it could be done for the other formulas:

http://chester.blog.br/archives/2009/04 ... heart.html

NCY.Jay
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### Re: "Useless" Discussion

Hehe
I say:

Sin <3= ostensible/happenstance
Cos <3 = angler/hopeless
Tan <3 = officially/aerate

let me explain
sin= stated to be true when it may not be the case/ chance esp. when it results in something good.
cos= angler: a person who catches fish => poor hopeless fish
tan= officially/ make it possible to mix air with soil or water

xyko
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### Re: "Useless" Discussion

Hum, the Fourier transform of love seams to be wrong.

If heart is

$\text{heart}(t)$

so:

$\mathcal{F}\left\{\text{heart}(t)\right\} = \text{HEART}(\omega) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty }^{+\infty} \! \text{Heart}(t) e^{i \omega t} dt$

And the plus infinity formally should be placed with the plus signal.

Anyway, did you find some differential equation of love on "t" that you want to solve by using Fourier? Could you share it with us? Isn't a numerical method applicable? Or another kind of computer-based brute force solution? We may setup a cluster to work on it. The project may be called love@home.

Anyway[2], this topic name is denigrating the serious discussions we're trying to have here.

Cheese!
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### Re: "Useless" Discussion

tradiuz wrote:Here is an article that might propose the equation for finding a mate. It's not the equation for love, but it is a good step in the right direction.

I don't like this equation at all, because it implies that the effect of being witty diminishes when the girl is hotter. I think it should be more like
This would make more sense, though the limits of the function would have to be moved. (You would need an even higher number.)

xyko
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### Re: "Useless" Discussion

So, the Fourier transform is wrong. You guys are ok with that?

Algrokoz
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### Re: "Useless" Discussion

Lol you are all wrong. cosine <3 = sine <3 + pi. =)

xyko
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### Re: "Useless" Discussion

Kinda... nope. [imath]cos \left( Love \right) = sin \left( Love - \frac{\pi}{2} \right)[/imath].

I was thinking about the discrete transform of the heart... I think there's a hope, if the we respect the Nyquist sampling principia of for love.

scarletmanuka
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### Re: "Useless" Discussion

xyko wrote:Kinda... nope. [imath]cos \left( Love \right) = sin \left( Love - \frac{\pi}{2} \right)[/imath].

Another fail; you're out by a minus sign. [imath]cos \left(x\right) = sin \left(\frac{\pi}{2}-x\right)= -sin \left(x-\frac{\pi}{2}\right).[/imath]

xyko
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### Re: "Useless" Discussion

You're quite right. Sorry.

callili
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### Re: "Useless" Discussion

RebeccaRGB
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### Re: "Useless" Discussion

When I learned about the equations for gravity (F_g = G m_1 m_2 / r^2) and electromagnetism (F_em = K q_1 q_2 / r^2) in AP Physics in high school, I came up with one for love: F_love = A ♥_1 ♥_2 / r^2.

Soon after, I found this quote from Albert Einstein: "Gravitation cannot be held responsible for people falling in love."
Stephen Hawking: Great. The entire universe was destroyed.
Fry: Destroyed? Then where are we now?
Al Gore: I don't know. But I can darn well tell you where we're not—the universe!

sirKris
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### Re: 0055: "Useless"

What about using a cardoid function, a set of parametric functions, or an implicit curve to represent the heart?

scarletmanuka
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### Re: 0055: "Useless"

sirKris wrote:What about using a cardoid function, a set of parametric functions, or an implicit curve to represent the heart?

Congratulations on totally missing the point.

RoeCocoa
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### Re: 0055: "Useless"

That wedding cake callili linked last September showed up on Cake Wrecks today: http://cakewrecks.blogspot.com/2011/06/sunday-sweets-geek-wedding-cakes.html

Here's the bakery's official page for the cake (with commentary and a two-tier version): http://blog.pinkcakebox.com/xkcd-comic-wedding-cake-2010-09-05.htm

SirMustapha
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### Re: 0055: "Useless"

Fortunately they chose a good xkcd comic to make a cake about. Imagine a cake referencing the "you can't milk me right now. I'd have to be lactating" one.

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### Re: lets give this a go...

chiggs wrote:We start by noting that hollywood would teach us that love is the only true constant in the Universe (hey - a thousand heroes can' be wrong!). Using this we get immediately:

d/dx (love) = 0, and

[ [ 1 0 ] [ 0 1 ] ](love) = [ [ love 0 ] [ 0 love ] ]

Next, my mother always told me that "friendship is the root of love" and who am I to argue with my mother? So, although I can't pretend to understand how she derived it, I do know that:

sqrt(love) = friendship

For cos(love) I think we have to start with that old "sine" of love "it's in his kiss" as expounded by Cher. Next we note that to us men, women seem to go off on a tangent at random, while I'm sure women think the same of us. Using

cos(love) = sin(love)/tan(love) we get

cos(love) = kiss/women if you're a man and,

cos(love) = kiss/men if you're a woman

Which leaves is with the Fourier transform of a function f(t) into the love domain (which I won't try to write mathematically here). Clearly we can't really answer this unless given a particular function to transform, but whatever the answer is, being "transformed into the love domain" sounds like it must be good.

Colin.

Your work is good. I do not have the skill to do an assessment on the Math that got you there; But, I recognize the right answer when I see it.
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cdstelly
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### I've figured it out.

http://www.wolframalpha.com/input/?i=%28x^2+%2B+y^2+-+1%29^3+-+%28x^2+*+y^3%29

$(x^2 + y^2 - 1)^3 - (x^2 * y^3)$

UniqueScreenname
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### Re: 0055: "Useless"

Well, everyone knows cos = acute/hypotenuse. So you need to divide a cute guy by bondage (hypote-noose).

Spoiler:
This is not my personal theory, just so you know. I have this shirt. That is my personal theory. No answer.
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Monox D. I-Fly
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### Re: 0055: "Useless"

Shouldn't the derivative equal zero?
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Copper Bezel
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### Re: 0055: "Useless"

It's really sweet that you think love is a constant, but I also kinda feel bad.
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somitomi
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### Re: 0055: "Useless"

Copper Bezel wrote:It's really sweet that you think love is a constant, but I also kinda feel bad.

Wait, so love is a function of location? That's odd.
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