Mathematics - 2016-10-30 (page 3 of 8)

@DHMO I get the feeling that you are trying to stack subjects I don't understand on top of this arbitrarily

Josué

Let $X=Y=\mathbb R$.

you are restricting the operation to only look at the values of the two inputs point by point, which is precisely what composition means

when all I am saying is that it is a set of functions such that if you take their unevaluated images and place them into a multivariate function, the resulting new function is a function in that srt.

@DHMO but composition with any function regardless of inside or outside the set

Also, I didn't know "composition" applied to multivariate functions/

@TheGreatDuck i'm saying, you are composing the function f(x,y)=ln(x+y) with A(x) and B(x)

6:03 AM

I thought that was only with single variate.

@DHMO yes

and technically f can be any number of variables

but I think it should work the same regardless of how many

since f(x,y) can be composed into g(z,w) to imitate a 3 variable function composition

@DHMO Basically what I found is that any set of functions that fulfill that qualification can be used to replace the notion of "constant" in integration/differentiation while still preserving most forms we are familiar with.

it might serve as a useful technique for differential equations

and composition is not the only operation

I know piecewise constant seems to work well with it

@DHMO Well, and I was meaning operation upon NUMBERS.

@TheGreatDuck that means composition

but unfortunately my writing is bad unless I really take a lot of time to think about it.

@DHMO Well, I have yet to learn those things formally. So I just kinda picked up based on what people here say.

I'm just in a differential equations class

my only experience with "abstract" like things is discrete math

and right after predicate logic we switched gears to computer science stuff cause it was a CS version of discrete math

so most of my terminology is just whatever habits I;ve seen from others discussing things

tbh, I know I sometimes study things I really shouldn't know yet.

granted, that's probably a good thing.

XD

anyway

now that I cleared that up

I have to go

it's late here

and I need sleep

@TheGreatDuck a little knowledge is a dangerous thing

it's good to explore things beyond your syllabus

but do so thoroughly

6:31 AM

@DHMO by "knowing things I shouldn't" I mean that I know subjects completely unrelated and the subjects that I do know I know correctly.

rehi @TedShifrin

For instance I have a healthy knowledge of coordinate systems

It honestly surprises me that they teach polar coordinates (something that uses the complicated expressions of sin and cos to define their relation to Cartesian) before parabolas coordinates which is literally converted to cartesian via arithmetic

granted parabolic is a niche subject

parabola coordinate?

Polar coordinates is more useful.

@DHMO auto correct struck me

XD

parabolic

6:35 AM

mathworld.wolfram.com/ParabolicCoordinates.html

that's new to me

Ted Shifrin

Hi @Balarka. So you've totally messed up your sleep cycle. Mazltov!

But now it's bedtime for me.

Oh well.

Sleep well.

Talking about 2d parabolic

Ted Shifrin

Night.

en.m.wikipedia.org/wiki/Parabolic_coordinates

6:37 AM

I like trilinear coordinate en.wikipedia.org/wiki/Trilinear_coordinates

Orthogonal coordinates

In mathematics, orthogonal coordinates are defined as a set of d coordinates q = (q1, q2, ..., qd) in which the coordinate surfaces all meet at right angles (note: superscripts are indices, not exponents). A coordinate surface for a particular coordinate qk is the curve, surface, or hypersurface on which qk is a constant. For example, the three-dimensional Cartesian coordinates (x, y, z) is an orthogonal coordinate system, since its coordinate surfaces x = constant, y = constant, and z = constant are planes that meet at right angles to one another, i.e., are perpendicular. Orthogonal coordinates...

Hopefully it takes you to the table of formulai

formulae

Beautiful reference material

yeah formulae

Haven't seen trilinear yet

i intend to understand all of those eventually

y'know it's funny

most people think of a mulitivariate function returning multiple values as a vector field

i dont

i think of it as a coordinate transformation

granted, either interpretation is valid

But the former feels more natural to me

@DHMO I also know quite a bit regarding the floor function (more than one would think to think about it from just seeing it in a book).

@TheGreatDuck for example?

To be fair though

@DHMO $\lfloor \frac x{sqrt(x^2+1)} \rfloor$

if you are at all familiar with C++ or C

6:43 AM

$\left\lfloor\dfrac x{\sqrt{x^2+1}}\right\rfloor$

you are welcome

@TheGreatDuck i know a bit

then you'll understand when I say that function is equivalent to (x >= 0)

its a mathematical version of the digital operation

yes

No, that is NOT how c does it

of course not

(trust me I had to write it once. It's bitwise set operations)

(We think of bytes as 8 element sets, or at least I do)

maximum of 8 that is

6:46 AM

@TheGreatDuck I think it uses flags instead...

@DHMO no. Flags are an OS thing.

they are Boolean parameters in essence

@TheGreatDuck but C compiles to assembly

which uses zero flags

oooh

but those flags are actually bitwise operations in the hardware

alright

I had to implement a bunch of operations using bitwise operators and assignment

and we were told that ultimately pretty much all assembly operations work using them

except the control flow ones

6:49 AM

I see

those just set the appropriate code pointers

in fact, computers run almost entirely on Boolean algebra.

well actually entirely

@TheGreatDuck boolean algebra?

yes

and or and not specifically

alright

Computers don't run on 0's and 1's

they run on true and false

well unless you look at circuitry

then they run on quarks and the fabric of the universe

XD

6:52 AM

sure

Well actually

quantum computers run on qbits

normal computers use voltage for true and false

CPU engineers use Nand Nor and Not

but that's just cause they can save some wiring that way

apparemtly

@DHMO well you know you're router? Or modem?

@TheGreatDuck i don't

they actually don't use bits. They use 32 different signals to supercompress the data

or was it 16?

just a random fact I know

6:54 AM

interesting

They get away with that because routers rarely if ever do data manipulation

I don't conceptually understand geodesics well enough I guess. I understand what they are, and the geodesic equation, but I would have to solve the tedious PDE to find geodesics on even the simplest surfaces.

all they do is pack and unpack the data

I can't guess, by looking at the surface, what would the geodesics be.

@DHMO in fact, computers way back when used multivalued bits in the form of gears.

6:56 AM

I see

it wasnt until electricity forced us to use two values that we realized Boolean algebra was the key to doing everything easily

and then it all just kind of fell into place that way

Do you know what a graphics card is?

its basically a giant linear algebra calculator

for manipulating spacial geometric figures

AKA 3D models

@BalarkaSen ODE. Play with examples. Explicitly calculate the geodesics on a torus, for instance.

Sorry, ODE, right. System of ODE's actually, but yeah.

well and of course talking to the monitor

Geodesics are both a good way of understanding manifolds (that's the first way I understood any of Riemannian geometry) and also a mysterious source (question: does every Riemannian manifold have infinitely many closed geodesics? what Riemannian manifolds have all of their geodesics closed? if there is a non-closed geodesic, is there a dense geodesic?)

6:59 AM

but that's just a case of literally taking an array of pixel values and converting the values to be read by the monitor

mathoverflow.net/questions/127530/… is a nice question

@DHMO technically one could write a renderer without a graphics card but the matrix multiplication would be severely expensive

Nice questions.

GPUs have a special set of registers purely for computing the dot product

ok

7:02 AM

@DHMO I also know implied calculus

which is a subject I made up XD

I was about to ask you what that is

it basically uses piecewiss constant in place of constant

@MikeMiller Thanks for them. I'll also calculate what you suggested.

and the primary purpose is to relate the solution of an implied differential equation to a regular differential equation

because the former is simpler to calculate

there's no argument about it not being easier

The implied integral of floor equaling floor*x is far easier than the regular integral

@DHMO there's also a nice geometric property

ok

7:05 AM

the portion of a piecewise constant function that consists purely of jumps

in other the piecewise constant solution C(x) to the functional equation f(x) = g(x) + C(x) where g is the given function and f is an arbitrary continuous function. I.E. f is merely known to be continuous

it's also the function that represents the error between the implied and regular integral

as such c can only be found up to an arbitrary constant

which makes sense seeing as how antiderivitives are the same

If you go to my profile there is a (old version mind you) link to the paper I started writing to keep all my notes regarding it down.

interestingly enougj

robjohn

@MikeMiller what are the poles of a surface of revolution "with poles"?

the additive relation formula is really all I need

docs.google.com/document/d/…

@DHMO I have yet to see a C(x) that I can not separate from x terms in a differential equation solution through algebra

That is it

@robjohn It's where the axis of revolution hits the surface.

7:11 AM

im gonna head out

So you have a sort of singularity, near which the surface looks like a cone.

robjohn

@BalarkaSen Oh, so a torus does not have a pole.

it's gets late and by now I'm usually just vegging on YouTube.

Nah

cya

8:18 AM

@AndrewThompson It all works.

Danu

8:48 AM

@MikeMiller Congratulations

user228700

@DHMO: Up for a quick calculus-tsy problem?

@Kaumudi yes

user228700

Well, it's not very calculus-tsy, so I hope u don't get too excited. The problem is actually very boring, but I have a very small question. My textbook is asking me to find the domain of the function given by

user228700

$f(x)=1/log(2-x) + \sqrt{x+1}$

that's the worst clickbait i ever seen

$f(x) = \dfrac1{\log(2-x)} + \sqrt{x+1}$

user228700

8:55 AM

@DHMO Ikr :-P

for $\log(2-x)$ to be defined, we have $2-x>0$

user228700

Yeah, that.

user228700

Wait, wait!

for $\dfrac1{\log(2-x)}$ to be defined, we have $\log(2-x)\ne0$

for $\sqrt{x+1}$ to be defined, we have $x+1\ge0$

user228700

You always assume that my question is exactly what my textbook is asking me to do!

8:55 AM

??

user228700

Pls wait for me to finish asking the question :-P

ok

user228700

Yes, so after we find the domain for the individual functions involved, we have $x<2$ and $x≥-1$.

user228700

So, we have the domain of $f(x)$ to be $[-1,2]$

user228700

So far, am I correct?

8:59 AM

you forgot $\log(2-x)\ne0$

alright, that is irrelavent

@Kaumudi no, it should not include $2$

user228700

To check for the log function to be defined, we need to check for two things; the base and the number inside should both be greater than zero.

and the base should not be $1$

user228700

Since the base is e in this case (Yeah, my textbook uses $log$ for base e also), that condition holds true.

yes

user228700

@DHMO Yeah, anything above zero, but not 1.

user228700

9:02 AM

Then I only gotta check for $(2-x)>0$.

user228700

Which gives me $x<2$.

user228700

And I have $x≥-1$ from the thing where I don't want to deal with complex numbers, since my function is only defined in $R$.

I said $x\ge-1$

$\sqrt0$ is not complex

user228700

Yeah, sorry. Typo.

yes, so $-1\le x<2$

user228700

9:05 AM

Yeah, so everything makes sense right, so far?

yes

user228700

Ik that writing $[-1,1)$ U $(1,2)$ is the same as $[-1,2)$.

user228700

But why would u wanna write that?!

no, they are not the same

user228700

Essentially, they are...

9:06 AM

also, $\log(2-x)\ne0$

so $2-x\ne1$

so $x\ne1$

@Kaumudi No, they are not

user228700

@DHMO Why not?

@Kaumudi because the former does not include $1$ while the latter does

user228700

@DHMO Neither includes $1$.

@Kaumudi no, $[-1,2)$ includes $1$

user228700

Ah, damn.

user228700

9:09 AM

@DHMO Why do we have to specify this explicitly?

user228700

Ah...

user228700

'Cause.

user228700

OK, I understand why.

user228700

Damn, I missed that subtlety. OK, this was my question. Thanks :-)

user228700

And I get why I missed it too.

user228700

9:11 AM

I changed $1/log(2-x)$ to $-log(2-x)$

which is wrong

user228700

So, as a rule of thumb, no changing stuff unless absolutely necessary and then check the original function again.

no, it is wrong

the rule is $\log\dfrac1x=-\log x$

user228700

What dyou mean "it is wrong"?

user227867

Hello @robjohn Happy Halloween. I pray that you are well.

9:13 AM

not $\dfrac1{\log x}=-\log x$

user228700

@DHMO Yeah ._.

user228700

Sigh. OK, thanks.

user228700

@DHMO: God, WikiHow has just about everything: (Seriously, pictures included!)

user228700

m.wikihow.com/Find-the-Range-of-a-Function-in-Math