Mathematics - 2019-03-28

Rithaniel

00:34

Alright, here's a puzzle: given an nxn matrix A with coefficients randomly selected from {-m,...,m}, what is the probability that det(A) is in {-m,...,m}?

Rithaniel

00:55

(Though, "puzzle" might not be the best word. I don't know if the answer is known by anyone.)

Rumplestillskin

01:07

Can somebody help a tired old brain out. I have an integral of the form $\int^{a}_{0}\int^{x}_{0} f(y)dy dx$ and apparently I can change it to $\int^{a}_{0} \int^{a}_{y} f(y) dy dx$?? This is probably very trivial but I just can't seem to get my head in to it!!

user76284

@Rumplestillskin y is both the integration variable and one of the endpoints? Doesn't seem right.

Rumplestillskin

O my dizzy days!! That is what I thought as well and the notes I am looking at just have it written down wrong! I think the person has just reversed the order of integration except forgot to actually change the order. Like they have changed the limits but not changed the order of integration. Does that sound righT? sorry I am having a slow day

user76284

Perhaps what they meant is y goes from 0 to a and then x goes from 0 to y? The region of integration forms a triangle between (0,0), (0,a), and (a,a).

Sorry, x goes from y to a.

Yeah, just switch dx and dy.

Rumplestillskin

Yep that's precisely what it is! @user76284

user76284

@Rithaniel Not sure what the answer is, but the empirical distribution looks like this

Secret

01:34

looks surprisingly tidy

Leaky Nun

07:15

let $f : \Bbb R \to \Bbb R$ continuous

is $C[a,b] \to C[a,b] : g \mapsto f \circ g$ continuous with the sup norm?

Silent

07:55

T/F: The polynomial $x^6-27x^4-13x^3+11x^2+24x$ has exactly one real root.

The sol i have says: polynomial is degree 6, so it may have 6 roots at most. since complex roots come in pairs, it can't have exactly one real root.

But, how do we know that multiple roots do not occur?

Tobias Kildetoft

@Silent They might, but that would not count as a single real root

Silent

oh :( so this question was kind of open to interpretation

Tobias Kildetoft

@Silent Well, does it have a multiple root?

Silent

yes, i saw in desmos

Tobias Kildetoft

so no interpretation needed

You could also have checked this by hand if you needed to

Silent

08:04

how?

Tobias Kildetoft

by finding gcd with its derivative

Silent

i am sorry. but gcd of derivative? i have heard about that first time

Tobias Kildetoft

gcd with its derivative, i.e. $gcd(f,f')$

Silent

ok, i googled it, found some articles, reading. thank u

Larry Eppes

m.wolframalpha.com/input/…

i just test wolfram alpha

Tobias Kildetoft

08:09

You don't need to. It has been thoroughly tested already :)

Larry Eppes

the ans is two complex root of your equation

Silent

@TobiasKildetoft, can you please suggest reading for that gcd (f,f')? The ones i googled already assume some familiarity with material.

Tobias Kildetoft

@Silent It does require some familiarity with the abstract theory of polynomials

but the proof is fairly straightforward, and the result simply says that multiple roots are the same as shared roots between a polynomial and its derivative (and to find whether they have shared roots, you can find their gcd)

Silent

ok

nbro

10:20

> Fathers of the Deep Learning Revolution Receive ACM A.M. Turing Award: Bengio, Hinton and LeCun Ushered in Major Breakthroughs in Artificial Intelligence

Akiva Weinberger

@Silent Prove that if $(x-a)^2$ divides $f(x)$ then $(x-a)$ divides $f'(x)$

By the way, note that we don't need calculus to define derivatives in $k[x]$. We just need to define the derivative of $\sum a_nx^n$ to be $\sum a_nnx^{n-1}$.

In fact, if we're working in a finite field like $\Bbb Z/p\Bbb Z$, it doesn't even make sense to define it with calculus and limits and things like that.

You should check that this definition still follows the product rule no matter what $k$ is (i.e. $(fg)'=f'g+fg'$).

(Also: if $k=\Bbb Z/p\Bbb Z$, then $(x^p)'=px^{p-1}=0$. So there can be nonconstant things with zero derivative.)

Akiva Weinberger

12:07

"Abel prize is certainly more difficult to get than Fields Medal." - Cedric Villani

Rumplestillskin

Does any one know where and why the absolute value shows up in the accepted answer here

4

Q: Simplifying a double integral to a single integral

In a recent cross-validated post a comment was left by Dilip Sarwate that stated that the following double integral: $$\int_0^L \int_0^L \rho(t-u)dudt$$ Could be simplified to the following single integral: $$\int_{-L}^{L} \rho(s)(L-|s|)\,\mathrm ds$$ Where $\rho(t-u)$ is the correlation func...

Akiva Weinberger

12:33

It's a bit like rotating a square 45 degrees and cutting it down the diagonal @Rumplestillskin

geometrically

Rumplestillskin

12:47

@AkivaWeinberger I’m not sure I follow?

Akiva Weinberger

Think of $\int_0^L\int_0^L$ as integrating over a square

Draw a plane where one axis is the $u$ axis and the other axis is the $t$ axis. You're integrating over the region where $t$ and $u$ both range from $0$ to $L$

Since you're integrating $\rho(t-u)$, it doesn't matter quite where on the square you are as much as how far away you are from the diagonal $u=t$ line

For any point in the square, we can draw the line of slope $1$ (parallel to the $u=t$ line) and see how much of that line lies in the square

and that's gonna depend on the absolute value of your distance away from the diagonal $u=t$ line

@Rumplestillskin You know what would be a more algebraic (but less geometric) way of doing this

Define $r=t+u$ and $s=t-u$

and do the change of variables

noting that $dudt=\frac12drds$ (using the Jacobian)

Silent

14:13

@AkivaWeinberger Thank you!

Semiclassical

hi chat

another question from the AMM which I like

Suppose $f(x)=\prod_{k=1}^n (x-a_k)$ with all roots distinct. Show that $\displaystyle \sum_{k=1}^n \frac{a_k^{n-1}}{f'(a_k)} = 1.$

James Groon

14:31

When I put $f(z)=\mathopen|z\mathclose|^2$ into wolfram alpha, it says that the function is nowhere differentiable in the complex plane. As far as I'm awared of, this function is differentiable at 0. What seems to be the problem here?

Semiclassical

@JamesGroon how did you get that output from wolfram alpha, to be clear?

My hunch would be that WA isn't smart enough to see the differentiability at zero

James Groon

@Semiclassical I typed in: f(z)=|z|^2 and scrolled down

Semiclassical

thought so

when I do that, I see two things on the "derivative" section

James Groon

Yup

Semiclassical

there's "nowhere differentiable in the complex plane" as you said, but also "assuming a function from reals to reals"

those don't exactly seem compatible :P

(especially since, on the real line, |z|^2=z^2 is differentiable.)

so my knee jerk reaction is to assume WA is being dumb

James Groon

14:41

but

I believe

it outputs the derivative and then (assuming from reals to reals ) . In addition , nowhere differentiable in the complex plane. I believe it separates those two brackets

Semiclassical

maybe but

James Groon

i mean, it says complex plane right ?

Semiclassical

we're not exactly in a position to interrogate wolfram alpha about what it's exactly saying there

nor about what method it's using to get to that conclusion

James Groon

You're absolutely right.

Semiclassical

I mean, WA is definitely wrong about it being nowhere differentiable. it's differentiable almost nowhere, but the C-R equations are satisfied at z=0

James Groon

14:44

But still , I believe your first hunch makes the most sense

Yup

Semiclassical

so it's definitely being dumb -somewhere-. figuring out where, though, is not so obvious

Leaky Nun

@Semiclassical it's not complex differentiable

(except at z=0)

Semiclassical

...yes, which is exactly what we've been saying

James Groon

xd

Leaky Nun

@Semiclassical do you like Poincare-Bendixson theorem?

Semiclassical

14:49

I don't touch that part of ODEs enough for it to be of much interest to me.

I mean, limit cycles are neat

Leaky Nun

how about the absolute counter-example in the 3D analogue: Lorenz system

Lorenz system

The Lorenz system is a system of ordinary differential equations first studied by Edward Lorenz. It is notable for having chaotic solutions for certain parameter values and initial conditions. In particular, the Lorenz attractor is a set of chaotic solutions of the Lorenz system which, when plotted, resemble a butterfly or figure eight. == Overview == In 1963, Edward Lorenz developed a simplified mathematical model for atmospheric convection. The model is a system of three ordinary differential equations now known as the Lorenz equations: ...

Semiclassical

strange attractors are weird

Leaky Nun

you mean they're strange

Semiclassical

lol

I don't have a great feel for the Lorentz attractor tho

my main knowledge of chaotic dynamical systems is in 1D

Leaky Nun

Logistics?

Semiclassical

14:52

yeah, that's the big one

or whatever one gives you the mandelbrot set

Leaky Nun

that's not 1D...

Semiclassical

Sarkovskii's theorem is pretty cool

pfft, one complex dimension

Leaky Nun

I don't get it

surely some sine is a counter-example?

Semiclassical

To what, Sarkovskii?

Leaky Nun

yes

Semiclassical

14:56

Sarkovskii isn't about periodicity of functions in the sense of f(x+L)=f(x)

Leaky Nun

oh

in the sense of iteration

Semiclassical

it's about periodicity in the sense of f(f(...f(x)...)) = x

Leaky Nun

that's interesting

Semiclassical

right

yeah

in particular there's the whole "period-3 implies chaos" business

Leaky Nun

I do not expect this at all

Semiclassical

14:58

yeah

i don't remember if we proved it in the course I had on chaotic dynamical systems

i'm guessing not

the wikipedia page includes this external link for a proof: math.arizona.edu/~dwang/BurnsHasselblattRevised-1.pdf

i make no claim for its authority tho

Leaky Nun

could you demonstrate by means of an example?

(it's quite hard to cook one up)

Semiclassical

Probably something to do with the logistic map would suffice.

So let's see...

Take $f(x)=r x(1-x)$.

Then the condition to have a period-3 point is that $f(x)\neq x$ but $f(f(f(x))) = x$

So that's $$x=r^3 x (1 - x) (1 - r x + r x^2) (1 - r^2 x + r^2 x^2 + r^3 x^2 - 2 r^3 x^3 + r^3 x^4)$$

hrm. This seems ill-advised.

Leaky Nun

15:15

yeah exactly

vzn

@LeakyNun did someone say dynamical system? :D ps counterexample to what?

Leaky Nun

@vzn Poincare-Bendixson

anakhro

@LeakyNun how are you?

Semiclassical

one thing that may help: the equation I just wrote should have $x=r x(1-x)$ as a trivial solution

Leaky Nun

@anakhro great

anakhro

15:19

I was looking at Poincare-Bendixon recently.

Foliations on S^2.

Leaky Nun

everyone is looking at PB these days

vzn

@LeakyNun not familiar but skimming wikipedia its for 2d systems which Lorenz attractor doesnt fit (3d).

anakhro

@LeakyNun why is that

Leaky Nun

how do I know

anakhro

I AM NOT SURE, LEAKY YOU JUST SAID IT WAS THE CASE.

hi

vzn

15:21

btw just found this on reddit & am curious/ interested

in theory salon, 6 mins ago, by vzn

Proof Finds That All Change Is a Mix of Order and Randomness — All descriptions of change are a unique blend of chance and determinism, according to the sweeping mathematical proof of the “weak Pinsker conjecture.” / quanta

Semiclassical

okay, slight refinement @LeakyNun

anakhro

>combinatorics
>distant

Semiclassical

you'll get solutions with least period 3 when $$ 1 + r + r^2 - r (1 + 2 r + 2 r^2 + r^3) x + r^2 (1 + 3 r + 3 r^2 + 2 r^3) x^2 \\- r^3 (1 + 3 r + 5 r^2 + r^3) x^3 +
r^4 (1 + 4 r + 3 r^2) x^4 - r^5 (1 + 3 r) x^5 + r^6 x^6=0$$ has real solutions

oof

anakhro

@vzn when will they learn

Semiclassical

I mean, that's still not great but

for any particular choice of r, you can plot that in mathematica and see if there's solutions :/

@LeakyNun playing around with mathematica, it looks like that polynomial has no real solutions when $r$ is smaller than roughly 3.82842

Leaky Nun

15:31

oof

Semiclassical

after that it does have real solutions, and therefore period 3 solutions

that might seem esoteric but

anakhro

Is mathematica good for graphing paths in R^3?

Semiclassical

looking at wikipedia, it seems that that's $1+\sqrt{8}$

which does play some significance in the logistic map

see the long discussion here: en.wikipedia.org/wiki/Logistic_map

I think the logistic map is maybe the wrong one to look at tho

$f(x)=x^2+c$ may have been better

@LeakyNun i guess one thing we can do

If $r=4$ in the logistic map, you can show that the logistic map has 3-periodic solutions

which means that, at that point, you've definitely got solutions of arbitrary periodicity

the tricky thing is that, as $r$ goes down, you eventually lose the 3-cycles but you still have 5-cycles

and after that you've still got 7-cycles etc

they don't disappear all at once

so what happens to the periods as r goes down is Complicated(TM).

Leaky Nun

15:49

I thought there is no real number greater than 4 @Semiclassical

anakhro

@LeakyNun that was complex number.

For example, 2.6>4 in the reals.

Semiclassical

lol

nbro

How would Bayes theorem apply in the following case P(s | a, o, b)?

Leaky Nun

it woudln't

nbro

It can be applied, because I have seen it applied

Leaky Nun

15:55

you were hallucinating

nbro

Rather than attacking me, why don't you explain and argue your claims?

Leaky Nun

lol i'm just joking

nbro

Jokes are of the form "why did the chicken cross the road". I think you wouldn't make a life out of a comedy career

Semiclassical

to quote something from a video game

"A joke has structure. I believe that was 'messing with you'."

Leaky Nun

nah real jokes look like this

it's funny cuz it's true

nbro

16:00

👍

Semiclassical

wel

nbro

Ok, but can someone help me with Mr. Bayes?

Semiclassical

it's funny because it's true without telling you anything

So bayes when you've got multiple conditions?

nbro

Yeah, not just two conditions, but 3

I have already seen the case of 2 conditions, but I can't generalise to 3

Semiclassical

well, you can start by noting that "A and B and C" = " "A and B" and C"

in other words, start by thinking of your three conditions as two: that conditions a,o both be true, and that condition b be true

You should then be able to use Bayes' to get conditional probabilities involving stuff like P(x|a,o) and P(x|b)

nbro

16:04

I see. I thought about doing something like that, but I didn't come up with anything

Semiclassical

at which point you can do bayes' once more

well, let's suppose we want to compute something like P(A|B1,B2,B3)

nbro

We can then try to calculate P(A|(B1, B2), (B3))

But isn't the result depend on how we aggregate the conditions? It shouldn't

Semiclassical

well, it'd better not if this is going to make sense

P(A|B1,B2,B3) = P(A,B1,B2,B3)/P(B1,B2,B3) at minimum

bleh, i should remember this better than I actually do

I mean, you'll be able to do stuff like P(B1,B2,B3)=P(B1|B2,B3) P(B2,B3)

and if some of those conditions are independent then you can factorize

Leaky Nun

and if they aren't?

Semiclassical

shrug

what's here may be useful: math.stackexchange.com/questions/408774/…

Leaky Nun

16:16

what's the probability that it is useful?

Secret

Define the event "useful"

[Random]

Amorphous bread

No matter how you cut it, you always get some loaf, and finite number of pieces of bread

Dedekind bread

You always get a loaf when you cut it, except the loaf is smaller

Aleph bread

You always have a loaf, and you can magically cut as many pieces of bread and loafs you want

Nonmeasurable bread

You cannot even illustrate the loaf because it is too complicated

Also the loaf can get bigger or smaller depending on its orientation

Semiclassical

set theory terminology question

Suppose I've got a set. Then the k-element subsets of S are those subsets with exactly k elements.

What if I want to consider subsets with at most k elements?

I'm trying to figure out how to word something without it becoming a mess

and all I can think of is "all subsets of S with size at most k"

Secret

How about $x \in \mathscr{P}(S)$ such that $|x| \leq k$?

or you can also say m-elements subsets of S such that $m \leq k$

Leaky Nun

16:38

@Semiclassical $[S]^{\le k}$

Sir Cumference

proofwiki.org/wiki/…

I don't suppose this can be proven without assuming the axiom of choice?

Karl Kronenfeld

16:53

@SirCumference Yeah, you need the axiom of (countable) choice.

Semiclassical

I'm not entirely happy with what I've put together but here's a question I just put on the main site:

0

Q: Symmetric subsets of random variables

I came up with the following question while formulating the properties for a certain statistical model I'm dealing with. I will refer to a random variable as being symmetric (about zero) if $X$ and $-X$ are identically distributed. I will moreover extend this to sets of random variables, i.e., a ...

probability probability-distributions random-variables conjectures

Sir Cumference

@KarlKronenfeld :(

Welp thanks lol

Karl Kronenfeld

Without the axiom of choice, you actually do get sets with equivalence relations that have more equivalence classes than elements, (admittedly from what I've just read).

Sir Cumference

@KarlKronenfeld Woah what

Karl Kronenfeld

Oh, that is not exactly what I was looking for, but still remarkable indeed.

Sir Cumference

16:56

Welp I'll just assume the axiom of choice like most people. Makes things convenient, and the above theorem is pretty useful

Karl Kronenfeld

Jech's book on the axiom of choice should have this and related phenomena.

Semiclassical

some discussion in that vein here: math.stackexchange.com/a/397938/137524

nbro

No bread, no mathematics 👍

L.K.

17:14

I'm facing a problem with the density plot, for e.g
`ListDensityPlot[
 Table[Sin[j^2 + i], {i, 0, Pi, 0.02}, {j, 0, Pi, 0.02}],
 PlotLegends -> Automatic]`
Is there away to get large axes labels, with large ticks label?

Semiclassical

You may have more luck if you ask this over in the Mathematica chat room: chat.stackexchange.com/rooms/2234/wolfram-mathematica

L.K.

@Semiclassical Thanks, I first posted it on tex stackexchange than here but not at the right place. I'm fully disoriented lol

Semiclassical

np

that chat room isn't as active as this one, unfrotunately, so you may need to wait for a bit

Sha Vuklia

17:34

Guys, I have a question. So we have irreducible polynomials $X^3+X+1,X^3+X^2+1\in\mathbb Z_2[X]$. Let $\alpha$ be a root of $X^3+X+1$, and $\beta$ a root of $X^3+X^2+1$. It’s easy to show that $\alpha^{-1}$ is a root of $X^3+X^2+1$, but I’m a bit confused on how to construct the isomorphism between $\mathbb Z_2(\alpha)$ and $\mathbb Z_2(\beta)$.

I would think that we would have to send $\alpha$ to $\alpha^{-1}$, since $\mathbb F_2(\alpha^{-1})$ and $\mathbb F_2(\beta)$ are isomorphic, because $\alpha^{-1}$ and $\beta$ are both roots of the same minimumpolynomial.

But I’m a bit stuck to finish this off. I know that $\{1,\alpha,\alpha^2\}$ is a basis for $\mathbb F_2(\alpha)$, and $\{1,\beta,\beta^2\}$ is a basis for $\mathbb F_2(\beta)$, and if we were to have an isomorphism, we would at least need a one-to-one correspondence between the basis elements. But yea, I’m not sure how to proceed exactly.

Oh, actually it makes more sense to send $\alpha$ to $\beta^{-1}$, because $\beta^{-1}$ in a way behaves the same as $\alpha$, since $a^{-1}$ is a root of $X^3+X^2+1$. Still a bit confused about the details of this, but alright

Oh, never mind, I was confused because I was confused about the fact that $\mathbb F_2(\alpha^{-1})$ is isomorphic to $\mathbb F_2(\beta)$, and not equal, but I get it now.

Danu

o/

Semiclassical

17:53

whenever i see someone answer their own question, i always think of this

a line of thinking comes along, spontaneously creates a question, and then this question annihilates itself, thus returning to the original line of thinking

Daminark

@Semi what's that actually a diagram of?

Semiclassical

it's a feynman diagram

Daminark

Or I mean, which specific process?

Semiclassical

a photon comes along and spontaneously decays into an electron-positron pair. this pair has equal and opposite charges, so it's possible for them to come back together and collide, annihilating each other to give a photon back

...or, at least, that's how you might understand that diagram. the tricky thing with feynman diagrams is that you don't actually think of electrons/positrons/photons as having a well-defined trajectory in QM

Daminark

I see

Semiclassical

18:07

the interpretation of feynman diagrams is a weird thing, frankly

visually it lets you tell a cute story about one possible chain of events for a photon

Daminark

Yeah I've heard that they apparently have a good bit of computational power? Which is kinda wow

Semiclassical

yeah, what justifies them isn't the visual story you get from them (which is good, because in a lot of cases that visual story wouldn't make much sense)

rather, it's that you can actually use those pictures to calculate things

for a flavor of it, suppose you expanded out the product $(x_1+x_2)^2=x_1^2+x_2^2+x_2 x_1+x_1 x_2$

you can 'visualize' those four terms in the following way: Put down two dots on the left, labelled as 1 and 2, and two dots on the right, labelled as 1 and 2 again

then draw a line from one dot on the left to one on the right

there's four ways to do that: 11, 12, 21,22

so that's a pictorial way of expressing that, and it'd work if I had $(x_1+x_2+....+x_n)^2$ instead

that's obviously a toy example, though. actual feynman diagrams are more complicated than that

Daminark

Interesting

So what exactly do they calculate?

Alessandro Codenotti

@SirCumference It is consistent with $\mathsf{ZF}$ that $\Bbb R$ is a countable union of countable sets! It is also consistent with $\mathsf{ZF}$ that $\mathrm{cof}(\omega_1)=\omega$

Rithaniel

18:24

cof?

Alessandro Codenotti

cofinality

Rithaniel

Ah (still not familiar with that, but now I know what I should google if I want to know more :P )

Alessandro Codenotti

If you have a poset $(P,\leq)$ a cofinal subset is just a subset $A$ such that for all $p\in P$ there is $a\in A$ with $p\leq a$. The cofinality of $P$ is the least cardinality of a cofinal subset

$\mathsf{ZFC}$ proves that $\mathrm{cof}(\aleph_\alpha)=\aleph_\alpha$ for all successor ordinals $\alpha$ (or in other words that successor cardinals are regular)

Semiclassical

18:42

substantial edit of my question is up now:

0

Q: Symmetry conditions for symmetric random vectors

While formulating the properties for a certain statistical model I'm dealing with, I came up with the following question (with credit going to MikeEarnest in comments for the proper formulation). A random variable $X$ is symmetric (about zero) if $X$ and $-X$ are identically distributed. This in ...

probability probability-distributions random-variables conjectures

Sir Cumference

@AlessandroCodenotti Aw man

Well meh, either way I don't think most mathematicians these days reject the axiom of choice

Daminark

@SirCumference ~~none with taste at least~~

Alessandro Codenotti

18:58

You need AC everywhere in modern math, there's not reason not to use it

But I also think that ZF+AD is a very interesting theory and I'd like to know more about it

Daminark

n e r d

:P

Sir Cumference

@AlessandroCodenotti Yeah but there are people out there who get triggered

Akiva Weinberger

19:38

Hielo

Alessandro Codenotti

19:51

Hi @Akiva

MatheinBoulomenos

Hi @Alessandro

I get triggered by people getting triggered over AC

The Coding Wombat

Is this room for asking questions about mathematics?

Alessandro Codenotti

Hi @Mathei

MatheinBoulomenos

among other things, yeah

Alessandro Codenotti

I'm sorry for using AC, but I had no choice

Érico Melo Silva

19:54

hah

The Coding Wombat

Okay quick question: If I have three lines, L,M,N, in projective P^4 space, do I have that <<L,M>,N>=<L,M,N>? It seems to make sense, but my normal intuition has not yet been of use in projective geometry

Akiva Weinberger

What's <>?

The Coding Wombat

I was told it's the span

So if L and M are lines, <L, M> is the set of all lines intersecting L and M I think

doppelgreener

20:23

0

Q: Add a horizontal scrollbar to MathJax formula blocks

MathJax formula blocks (i.e. $$these$$, not single-dollar-sign $inline$ mathjax) sometimes get wide, especially when they're arrays. Wide formulae don't play nicely with the responsive design though and instead overflow into the sidebar if the window is narrow enough. Could MathJax formula...

bug mathjax responsive-design

I don't know if this has come up before on Math Meta, but I've just posted this as a bug on main Meta and suggested a lightweight fix.

Alessandro Codenotti

21:02

Oh, I just discovered that the Rudin in set theory/set theoretic topology was the wife of the other Rudin

Skies burn

21:41

Why is this closed

3

Q: Automorphism group of a finite group

For which finite groups $G$ is $\operatorname{Aut}(G)$ isomorphic to a (non necessarily strict) subgroup of $G$?

finite-groups automorphism-group

Alessandro Codenotti

I don't know but there are some reopen votes

BAYMAX

22:02

How parameterizaytion of $y = a + 2ax - 8ax^2$ leads to $x = a(1-2t)(1+2t-8t^2), y = a - 12at^2$

*parameterization

nevermind, it wont be!

Akiva Weinberger

23:25

Calculus proof of the Pythagorean theorem

The tangent to a circle is perpendicular to its radius. This means that, for a circle,
dy/dx = −y/x
Rearranging, we get:
x dx + y dy = 0
Integrating and multiplying by 2, we get:
x^2 + y^2 = C
Since the point (r,0) in on a circle of radius r, we see that C=r^(2)+0^(2), so:
x^2 + y^2 = r^2

QED

Rumplestillskin

23:54

@AkivaWeinberger I appreciate the response and the information!! I finally got it. Safe to say my brain is not firing on all cylinders!! Cheers

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$Mathematics$ Mathematics