Mathematics

3:00 PM

You can definitely estimate a size. The point is by mean value theorem $\phi$ is Lipschitz: $|\phi(x) - \phi(y)| \leq C |x - y|$ where $C = \max_{x \in I} |\phi'(x)|$, $I$ being the interval where $|\phi'(-)| < 1$.

So let's see what Banach fixed point theorem says explicitly.

Secret

[Day 3: Trial number 12] Evaluate the integral such that the evaluation highlights as many of its symmetries as possible
$$\int x^n \sin x e^x dx$$
I hope I can stop making careless mistakes!

Balarka Sen

@Alessandro Wait, no, I didn't read what you said correctly. I don't understand the question: if $I$ is the nbhd of $\alpha$ in which $|\phi'| < 1$, it's precisely where the convergence is guaranteed...?

Why's that not a "relation"

Riccardo

I've a quick question about Steenrod Squares. I want to compute explicit generators of H_5(K(Z_2,2);Z_2), and according to UCT, H_5(K(Z_2,2);Z_2)=Z_2+Z_2. According to Serre's Theorem about the cohomology rings of K(Z_2,n), it should be generated by image of the fundamental class i in H^2(K(Z_2,2);Z_2) via iterated Steenrod Squares. In particular, I should be able to find 2 non zero composition of iterated Steenrod Squares evaluated in i. I'm only able to find one of them, Sq^2Sq^1(i)

Balarka Sen

I see, the theorem you quoted does not mention that. Strange.

Alessandro Codenotti

@Balarka I'm not sure, it's phrased in an horrible way in my textbook, I know that there is a nbhd $I_1$ of $\alpha$ in which convergence is guaranteed and that $|\phi'(\alpha)|<1$ (so by continuity there is a nbhd $I_2$ of $\alpha$ in which $|\phi'(\alpha)|<1$), what I'm trying to understand is whether $I_1=I_2$

I'm quite certain that $I_2\subseteq I_1$

Riccardo

3:06 PM

the other sequences are zero evaluated on i, (1,1,1) or (1,2) or (3). Can't find another one sadly

Balarka Sen

$I_2$ can be any closed neighborhood of $\alpha$ contained in $I_1$.

Alessandro Codenotti

Hmm, that's weird. The theorem before that is basically a specific instance of the contraction theorem: if $\phi\in C^0([a,b])$, $\phi([a,b])\subseteq [a,b]$ and $\exists L<1$ such that $|\phi(x_1)-\phi(x_2)|\le L|x_1-x_2|$ then there is a unique fixed point $\alpha$ of $\phi$ in $[a,b]$ and the sequence $x_{k+1}=\phi(x_{k})$ converges to $\alpha$ for all $x_0\in[a,b]$

Balarka Sen

"Contained" was not strict. If $I_1$ is closed then $I_2 = I_1$ is totally fine.

Alessandro Codenotti

yeah so in general the nbhd with guaranteed convergence is at least as big as the one in which the derivative is bounded by $1$

Balarka Sen

right

Mary Star

3:17 PM

Why is $\mathbb{Q}(\sqrt{2})\cup \mathbb{Q}(\sqrt{3})$ not a field?

Alessandro Codenotti

it doesn't contain $\sqrt{2}+\sqrt{3}$

Mary Star

@AlessandroCodenotti Ah ok. Thanks!!

Pichi Wuana

3:31 PM

Can any right triangle be inscribed in a circle? I'm thinking the fact that a inscribed angle = 90° is pointing to the diameter.

DHMO

@PichiWuana yes, and you've just explained why.

Balarka Sen

ANY triangle can be inscribed in a circle

Pichi Wuana

Oh hahaha

DHMO

@BalarkaSen I was just about to say that

Alessandro Codenotti

for a right triangle a semicircle is enough

DHMO

3:33 PM

Can any triangle be inscribed in a semicircle?

Balarka Sen

no

DHMO

for example?

Balarka Sen

literally anything. one of the interior angles have to be 180/2

SteamyRoot

Equilateral triangle?

DHMO

not really

If one angle is obtuse, then it is certainly possible

because the inscribed circle would have the center outside of the triangle

so the triangle would occupy less than a semicircle

Balarka Sen

3:35 PM

oh i'm dumb yes

DHMO

@SteamyRoot the semicircle has its diameter touching the ground

make one tip of the equilateral triangle touch the center

never mind i'd just draw a diagram

Balarka Sen

for some reason i was assuming center of the circle is supposed to lie on a side but of course that's nonsense

@DHMO that is not inscribed. one of the vertices is not on the circle

SteamyRoot

^

DHMO

@BalarkaSen semicircle.

SteamyRoot

Semicircle is half a circle

DHMO

3:37 PM

yes

Balarka Sen

semicircle doesn't contain that diameter

DHMO

what the hell

SteamyRoot

Not the boundary of half a disk

DHMO

that's new to me

the diagram on wiki includes the diameter

but the one on mathworld does not

Balarka Sen

read the definition, not the diagram

a semicircle is in particular an arc

DHMO

3:39 PM

Euclid's definition includes it though

> A semicircle is the figure contained by the diameter and the circumference cut off by it. And the center of the semicircle is the same as that of the circle.

(The Elements: Book I: Definition 18)

Balarka Sen

that seems to indicate it's half a disk, not a one dimensional object we think of it as. nobody actually uses that defn

SteamyRoot

Euclid defines things differently

In particular, his circle is our disk

DHMO

that's interesting

SteamyRoot

and our circle is something he always describes as the "circumference of a circle"

DHMO

i learnt something new today

how do you describe the wrong semicircle then

I'm interested if every triangle can be inscribed in a wrong semicircle

SteamyRoot

3:41 PM

Boundary of half a disk?

DHMO

let's just call it the wrong semicircle

@BalarkaSen thoughts?

SteamyRoot

Wouldn't that be kind of trivial?

Pichi Wuana

But, are all 90° angles that point to the diameter inscribed in the circle?

What I mean is that I have an ellipse

DHMO

@SteamyRoot so you mean yes?

why?

SteamyRoot

Look at the diagram you drew before

Pichi Wuana

3:44 PM

SteamyRoot

Choose one angle to touch the diameter

make the opposite edge parallel to the diameter

and let the center of that edge be "right above" the center of the semicircle

Hmmm... actually, that might not work

You'll have to pick the right angle first.

Balarka Sen

@DHMO Well, equilateral triangle is a ctrexample

DHMO

@BalarkaSen it can be inscribed in the wrong semicircle

just as demonstrated above

Balarka Sen

@DHMO oh you want to do it with the wrong one. shrug.

SteamyRoot

Hmmm. My idea works if you pick the largest angle to be the one touching the diameter.

Pichi Wuana

3:51 PM

In my ellipse D and E are the focal points. F is a point in the first quadrant and on the ellipse. DFE = 90°. Can I take DE as a diameter, $(0,0)$ as the center of a circle, and F as a point in the circle?

DHMO

@SteamyRoot nice. I tried it in geogebra

Pichi Wuana

And then FO = EO = DO

DHMO

@PichiWuana what is your question?

Pichi Wuana

If I can take that as a fact.

DHMO

you don't need the ellipse at all

Pichi Wuana

3:54 PM

hmm

DHMO

@PichiWuana let O be mid point of BC

construct foot of perpendicular from O to AB as D, to AC as F

by definition, ODAF is a rectangle

OA is a diagonal of the rectangle, so ΔOAD = ΔAOF

i'm probably making everything complicated

by intercept theorem, CF=FA

by SAS, ΔAOF = ΔCOF

similarly, ΔOAD=ΔOBD

so ΔOBD = ΔOAD = ΔAOF = ΔCOF

so OB = OA = AO = CO

Secret

Finally
THE
****
IS
DONE
!

[Integral symmetries] Cauchy Riemann theorem like analogues in reals:

$$\frac{d^n}{dx^n}f=\lambda\int^{(m)}fdx$$

(where $f=\sin, \cos,e^x$ (possibly more?),$\lambda \in \mathbb{R}$)

For example

$$\int x^n \sin x e^x dx$$
Consider

$(uv)''=u''v+2u'v'+uv''$

Plug $u=\sin x$, $v=e^x$. Then

$(uv)''=-uv+2u'v+uv$

Note how the antisymmetry of $\sin x$ wrt the functional $()''$ results in cancellation. Hence

$(uv)''=2u'v$

Now it is easy to check that $\cos x$ is also antisymmetric wrt $()''$ thus we could have start with $u' =\sin x\implies u = -\cos x$. Hence

DHMO

@Secret what would fit in the asterisks... "fuck" doesn't fit there.

oh, "shit" fits there.

Secret

I have been working for 3 days on this ever since simpleart gave me that challenge. It should not took that long but I keep making careless mistakes after careless mistakes

DHMO

what is the challenge?

Secret

4:08 PM

in Simply Beautiful Art's room, 2 days ago, by Simple Art

@Secret can you take this integral without IBP?$$\int x^2\sin(x)e^x\ dx$$

(Please let me type the nxt msg before responding)

For simpleart's challenge, I do managed to find a IBP free way to evaluate it, and then some careless mistake or something result in the integral to get wrong ans, but simpleart said it is fine as my ans agrees with the form of the ans of that integral

When doing his challenge, I got lazy (and at the same time I want to test my hypothesis on integrand symmetries) thus I generalise his integral to the one above

I then found (as you saw in that long wall of text), the n power version of that integral has a close form which is captured by the symmetry of the integrands

(The IBP free method to do simpleart's integral is the starred message of today)

Secret

NB whenever I got lazy, I went into generalisation mode to derive a universal formula for all types of problems similar to the problem I am working in. However because generalisation proofs are in general hard, paradoxically, it result in a lot more work than simply blowing through the problem directly

DHMO

@Secret I really like your passion.

Secret

Alternately, I have a weird metric on measuring effort. To me the more I need to repeat the exact same thing, the more effort it takes

Sophie

4:32 PM

can someone prove $\int_{-1}^0 \frac{e^x+e^{1/x}-1}{x}dx=\gamma$?

DHMO

@Sophie trapezoidal rule?

Sophie

what?

DHMO

Riemann sum

In mathematics, a Riemann sum is an approximation that takes the form ∑ f ( x ) Δ x {\displaystyle \sum f(x)\Delta x} . It is named after German mathematician Bernhard Riemann. One very common application is approximating the area of functions or lines on a graph, but also the length of curves and other approximations. The sum is calculated by dividing the region up into shapes (rectangles, trapezoids, parabolas, or cubics) that together form a region that is similar to the region being measured, then ...

basically the definition of definite integral

Sophie

I know that $\int_{-1}^0 \frac{e^x-1}{x}dx=\sum_{h=1}^\infty \frac{(-1)^{h+1}}{h\cdot h!}$

@DHMO I know this, but how do you use it to prove the integral is exactly $\gamma$? I already know it is pretty close numerically, which is why I'm wondering

DHMO

@Sophie just an idea; I haven't tried it

Secret

4:38 PM

Got a very simialr looking expression in this page

Exponential integral

In mathematics, the exponential integral Ei is a special function on the complex plane. It is defined as one particular definite integral of the ratio between an exponential function and its argument. == Definitions == For real non zero values of x, the exponential integral Ei(x) is defined as Ei ⁡ ( x ) = − ∫ − x ∞ e − t ...

In fact the $\int \frac{e^{x^{-1}}}{x}dx$ bit migh be related to the exponential integral

Sophie

this came up because I think $\int_0^x\frac{e^x-1}{x}dx\sim \gamma-\ln(-x)$ as $x\to -\infty$

Secret

Sophie

that's what I'm looking for, I think

Sophie

4:59 PM

this guy says there's no perfect cuboid arxiv.org/abs/1506.02215

DHMO

@Sophie isn't this one of the million dollar questions?

Sophie

nope

DHMO

no?

right

Secret

So if there is no perfect cuboid, then it should not be possible to have a perfect hypercuboid since at least one of its (n-1)D faces will need to be a perfect cuboid

DHMO

@Secret there you are, generalizing again :p

Mike Miller

5:05 PM

There's no particular reason to believe that paper.

Alessandro Codenotti

There are papers on the arvix claiming all sort of outrageous things from P=NP to the Goldbach conjecture...

Secret

Well, I am not even sure what's the significance of (non)existence of parallepippeds where the diagonals are all integers

DHMO

cuboids @Secret not parallelpipeds

Secret

that paper define 'cuboids' as some kind of parallelpipped

Goldbach's conjecture

Goldbach's conjecture is one of the oldest and best-known unsolved problems in number theory and all of mathematics. It states: Every even integer greater than 2 can be expressed as the sum of two primes. The conjecture has been shown to hold up through 4 × 1018, but remains unproven despite considerable effort. == Goldbach numberEdit == A Goldbach number is a positive integer that can be expressed as the sum of two odd primes. Since four is the only even number greater than two that requires the even prime 2 in order to be written as the sum of two primes, another form of the statement ...

What happens if we can show there exists integers that are not sum of two primes?

Astyx

The Goldbach Conjecture is disproved ?

Sophie

5:15 PM

@MikeMiller why not?

PVAL-inactive

That's a 45 page paper with essentially no explanation, no sketch of the proof just random likely meaningless equations

Lucas Henrique

Hello there

Secret

No I mean: Suppose there exists integers that are not sum of two primes, what does that mean?

Alessandro Codenotti

5:33 PM

Probably nothing. The interesting part will be the techniques used to prove that there is (or there isn't) such an integer

Secret

I see

Lucas Henrique

Are any of you guys acquainted with graphs? (graph theory)

Sophie

does $\int_\pi \frac{1}{t}dt=\ln(x)$ for a real positive number $x$ for any contour $\pi$ that goes from 1 to $x$ in the complex plane?

DHMO

@Sophie good luck passing through the origin

if $\pi$ doesn't pass through the origin, then yes.

the origin is the only pole of $f(z) = \dfrac 1z$.

Sophie

@LucasHenrique this is like asking someone if they're familiar with number theory. Maybe they'll say yes and then you'll ask something bizarre about inter-universal Teichmuller theory. Maybe they'll say no and you'll ask an introductory question they did know how to answer. So if you state the question you're more likely to get an answer

arctic tern

5:45 PM

@Sophie no

it will be ln(x) + 2pi i times the number of times the contour goes around 0 counterclockwise

Lucas Henrique

Ok then :p

My problem is the following: I have, for each vertex of a graph, the set of vertex it's connected to

I want to build a graph that is closed given this information

How can I do this?

s.harp

What do you mean with "closed"?

Lucas Henrique

Eh... I want to build "a circle"

(I have neither vocabulary nor theorical basis on graphs, the book just solves problem using graphs without an introduction)

A cyclic graph

s.harp

If your graph is not directed you and for $v$ a vertex $N(v)$ is the set of vertexes connected to it you get the set of edges to be $\{ v\to w \mid w\in N(v) \}$

Sophie

if you have the set of vertices any vertex is connected to then your graph is already determined

s.harp

5:50 PM

Knowing what vertices are connected to what other vertices completely describes the graph

Sophie

that's how you define a graph. The dots and lines interpretation is "just" a representation

Lucas Henrique

I don't know how can I represent my graph tho

I need to prove that, given the set of vertices, it's possible to create a cyclic graph

(I know it's a cyclic graph)

It's given by the table (each number is corresponding to a different graph)

Secret

[Random Rambles]
Suppose the Goldbach conjecture is true. Then for all even integers $2n \in \mathbb{Z}^+$, there exists $m,r \in \mathbb{Z}^+$ such that it can be wrote as $2n=(2m-1)+(2r-1)=2(m+r-1)$, where $2m-1$,$2r-1$ are primes.

Rewriting the equation, we get

$n=m+r-1$

$m-r=-(n+1)------(*)$

which is an equation of a 3D linear subspace in $\mathbb{R}^4$

By taking cross sections $n \in \mathbb{Z}^+$, the problem becomes the above equation always have a solution in terms of integers $m,r$ such that they corresponding to the aforementioned primes

s.harp

If you just have vertices take the graph so that all vertices are connected to each other

Lucas Henrique

$1: 6, 7, 9\\
2: 3, 9, 10\\
3: 2, 9, 11\\
4: 10, 12\\
5: 7, 11\\
6: 1, 8, 12\\
7: 1, 5, 12\\
8: 2, 6\\
9: 1, 3\\
10: 2, 4, 11\\
11: 3, 5, 10\\
12: 4, 6, 7$

I want to prove that it's possible to find a path that it's possible to walk at every vertex exactly one time and end in the first vertex

Sophie

5:59 PM

draw the graph on a piece of paper and it shouldn't be hard

Astyx

Hi @Ted

Lucas Henrique

@Sophie I have been trying... but it is hard :|

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$Mathematics$ Mathematics