Mathematics - 2017-01-13 (page 5 of 5)

Balarka Sen

20:00

I prefer your basis-wise definition with $dx_I$'s

Ted Shifrin

Right, @Socrates. You're given $n$ real numbers.

@Balarka: There are good reasons to do everything with tensors, but not at the level of my course.

Semiclassical

About the only time I'd use reciprocal as notation for matrix inverse is if I'm doing something like $(A-\lambda I)^{-1}$.

Abcd

Could someone help me with a question ?

Ted Shifrin

I'm still flunking you, @Semiclassic. Next thing you know you'll be dividing by a vector and thinking I won't notice.

Astyx

Depends what the question is @Abcd

Ted Shifrin

20:01

Just ask it, @Abcd.

Tobias Kildetoft

@TedShifrin Well, how else could he define the derivative of a function between vector spaces? :)

Socrates

well, certainly the rank of the vandermonde matrix with all $x_i$ being 1, is 1. but more obvious is impossible^^

Balarka Sen

@TedShifrin I don't immediately know why one would do that, except for understanding that they eat k-tensors.

Semiclassical

So that I can default to Taylor series mode and write $\frac{1}{A-\lambda I} = A^{-1}\frac{1}{I-\lambda A^{-1}}=A^{-1}+\lambda A^{-2}+\cdots$

Abcd

This one ^

Ted Shifrin

20:02

The interesting case is distinct values, @Socrates.

@Semiclassic. Please erase that, lest some unsuspecting reader think it is correct mathematics.

Semiclassical

The more hilarious version of this is when people have $\frac{1}{1-X}$ with $X$ a Feynman diagram.

Socrates

@TedShifrin so one example might be $x_n=\frac{1}{2^n}$

Ted Shifrin

It's any finite set of $n$ real numbers, @Socrates. The $n$ is fixed.

Semiclassical

I'm pretty sure your disclaimer is enough to forestall that :P

Balarka Sen

@Semiclassical heavens what new garbage is that

Ted Shifrin

20:03

Better to try integers. This has nothing to do with series.

@Abcd: First of all, that diagram is not drawn to scale or anything, so be careful. OK, where are you stuck?

Abcd

See How I am going about it:

I found the equation of AB

Mike Miller

@Balarka: Only way to redo campaign missions is to start again. No way I keep up this S-rank streak.

Ted Shifrin

puts Semiclassic on permanent ignore for blasphemy

Abcd

got y = x+7

Then

Mike Miller

I'm not going to replay missions for S, either.

Abcd

20:05

Using distance formula

Ted Shifrin

Whoa, slow down, @Abcd.

Abcd

Tried to calculate

@TedShifrin Okay

Ted Shifrin

What is that equation for?

Abcd

AB

Line seqment AB

Ted Shifrin

You mean the line, not the line segment.

Semiclassical

20:06

I'm almost positive I've seen that aforementioned garbage, but I don't see a quick reference to it. (If it does, it's under the rubric of the Dyson series.)

Ted Shifrin

But that cannot be right. The line has negative slope.@Abcd

Tobias Kildetoft

@MikeMiller What are you playing?

Ted Shifrin

Can't you find B without doing that?

Abcd

@TedShifrin I found it using 2 point form

Socrates

@TedShifrin why is the leading 1 at all lines essential in vandermond matrices?

Ted Shifrin

20:07

What's the slope of the line?

Socrates

$x^0$?

Abcd

!

1

Ted Shifrin

Right @Socrates :)

Balarka Sen

@MikeMiller Aww, that's a shame.

Abcd

@TedShifrin 1

Ted Shifrin

20:07

Compute again, @Abcd.

Abcd

Okay

Balarka Sen

Alright, I gotta wake up early tomorrow, so I'm gonna head off to bed. See ya.

Ted Shifrin

Night, @Balarka. Don't let those factorials bite.

Abcd

@TedShifrin its 1 for AC

Calculated again

Ted Shifrin

No, it isn't.

Are we doing the same problem?

Abcd

20:09

Yes

Ted Shifrin

I get -4/3.

Abcd

How ?

Ted Shifrin

$\Delta y/\Delta x$

Abcd

SLope is y2-y1 / x2-x1 right ?

Ted Shifrin

$\Delta y = 8$, $\Delta x = -6$.

Yes. You're making arithmetic errors.

Abcd

20:10

@TedShifrin I cant see what you are typing it comes like dollar sign backslash delta y dollar sign backslash delta x

Kasmir Khaan

Ted can you please tell me how to get better at tripple integrals and visualization of 3d ?

Ted Shifrin

I'm using MathJax. See the link over there >>>>^^^^

It's hard to type math without it.

Anyhow, change in y divided by change in x is correct.

What's the change in y ?

Kasmir Khaan

you have some lectures of that ?@TedShifrin

Ted Shifrin

Yes, Kasmir. Several of them. Start with 2-d, then 3-d. I even did a few later on in 4-d.

Kasmir Khaan

2 d i can handle boundry set up

but in 3 d i get lost

Astyx

20:12

What is so interresting about dual spaces ?

Ted Shifrin

Still, @Kasmir, listen to my explanation for 2, because you use the same algorithm in higher dimensions.

Tobias Kildetoft

@Astyx What is so interesting about Hom spaces?

Ted Shifrin

@Astyx: Linear maps are important.

Abcd

@TedShifrin I just cant understand you :(

Ted Shifrin

Most of this is not for you, @Abcd. What is the change in y? What is the change in x? Tell me your numbers.

Kasmir Khaan

20:13

@TedShifrin youtube.com/watch?v=fnjKUVJHRxE

Abcd

I got it like 5-3 / -2+4

user243301

Good friday all users, bye.

Kasmir Khaan

i start with that ?

Abcd

This is the slope right?

Ted Shifrin

NOOOOO, @Abcd. Subtraction!!

5-(-3) = 5+3 = 8

@Kasmir: The labels should make it obvious. You can play around. I'm busy now.

Kasmir Khaan

20:14

okay thanks :=)

Astyx

Meh, I guess my question was a bit too general

Abcd

What a blunder

!

Ted Shifrin

LOL @Abcd :)

Studentmath

G'night all!

Ted Shifrin

@Abcd, but I don't see how the equation of that line will help you find B, anyhow.

Night, @Studentmath :)

Astyx

20:15

Bye @Studentmath

Abcd

@TedShifrin Then what to do?

Ted Shifrin

You know C is halfway from A to B. So can't you change x and y by the same amounts and take one more step?

Abcd

@TedShifrin Can I use distance formula?

Ted Shifrin

No, the distance formula will not help.

Abcd

Why not AB = Bc ?

right?

Ted Shifrin

20:16

When you go from A to C, x goes from 4 to -2, so it goes down by 6. Going from C to B it must go down by 6 again.

You need x and y coordinates, not distance.

Abcd

@TedShifrin Thats what in the formula I will replace the value of the euation of line >

equation*

Ted Shifrin

I have no idea what you're thinking.

Abcd

like I will substitute y = x+1

In the formula

Ted Shifrin

That won't help.

Abcd

WHy?

Ted Shifrin

20:18

Didn't your teacher show you how to calculate the midpoint of a line segment?

Astyx

I guess my question was more like : What is so special about dual spaces to motivate their study as dual spaces ? What changes from the study of a generic space ?

Abcd

OHHHH ! Midpoint formula is what I neeeeeeeded !!

Semiclassical

haha, found an example

Abcd

Just went out of my head !!

Ted Shifrin

Uh huh, @Abcd.

Tobias Kildetoft

20:19

@Astyx Nothing intrinsically

Abcd

@TedShifrin Thanks a lot !

Semiclassical

@BalarkaSen equation 10.34 here: books.google.com/…

Ted Shifrin

Then to find the equation of AT, you'll need the correct slope and then figure out what the perpendicular line slope is. @Abcd

Tobias Kildetoft

@Astyx But for example if you have a space of operators on some other space, then the eigenvalues live in the dual of the space of operators

Abcd

Yes Thats simple

Ted Shifrin

20:19

Only simple if you do arithmetic right to do the slope, @ABcd. Pay attention :)

Abcd

@TedShifrin I had copied the question wrongly in my notebook

Ted Shifrin

Ahhhh.

Astyx

I see (kinda)

Abcd

Thats why I made a mistake

Yes

Ted Shifrin

That's why I asked if we were doing the same question, @Abcd :P

OK, cool. You're all set now.

Abcd

20:20

Yes Thanks :)

Ted Shifrin

You're welcome.

Tobias Kildetoft

@Astyx Assuming the operators are simultaneously diagonalizable that is

Ted Shifrin

@Astyx: You learn things about vector spaces by studying linear maps on them. And then multilinear maps ...

@Tobias: I wondered why you were going down that road.

OK, lunchtime for Ted. Bye, all.

Astyx

Such as ? @Ted

Bon appétit !

Liad

Hi, i got a question :
Im proving that any $G $ , simple group with order 60 is isomorphic to $ A_5$.
I assumed it is not.
I have proved that :
1)there is no subgroup of G with index less then or equal to 5(and greater then 1).
2)$n_5 = 6 $ where $n_5 $ is the number of 5-sylow subgroups of G
3) $n_2$ = 15.
now i want to show that given $P,Q $ , two different 2-sylow subgroups of G, then $<P,Q> = G $. i know that $|<P,Q>| \in \{5,6,10,60\}$ but im not sure how to eliminate $\{5,6,10\} $
any ideas?

Astyx

20:24

What is $A_5$ ?

Tobias Kildetoft

@Liad Well, what is the order of a $2$-Sylow subgroup?

@Astyx The alternating group of degree $5$

Astyx

Right

Socrates

20:37

What should I read to grasp change of basis?

Astyx

Have you studied linear algebra ?

Socrates

well, I am at it, but this one is hard for me

Astyx

What approach did you have ?

Socrates

none, the only thing I know is, that it is useful(or even neccessary) expressing a matrix in basis B, in basis C

Astyx

So you don't know about vector spaces ?

Socrates

20:41

yes I do!

Akiva Weinberger

@TedShifrin I have no idea how to prove that the two definitions of ellipse are equivalent

Mike Miller

@Balarka A-rank on Eagle. My 8-day finish screwed me. I took a little too much damage to the T-copter.

Socrates

a lot of things can be vectorspaces, to the point of a vectorspace of vectorspaces

Tobias Kildetoft

@MikeMiller What game?

Mike Miller

Advance Wars

Astyx

20:43

And do you know about dimension theory then ?

Socrates

No.

Astyx

Do you know what a basis is ?

Socrates

a set of independand vectors, whose span is the space

Astyx

Right

Socrates

and change of basis would be, we have two different sets

Astyx

20:45

So if you have a matrix $A$ in some basis $B = (e_1, ..., e_n)$

Simplex

Is there a term for the total volume of an object's path as it moves through space? (Specifically in 3d)

Astyx

Then the first collumn of $A$ would be the decomposition of the image of $e_1$ in $B$ (which is unique since the vectors are independant)

Socrates

@Simplex a similar thing would be rotating volume (the space it takes, if you rotate an object)

Semiclassical

@Simplex An example of what you have in mind would help.

Astyx

The same goes for the second collumn and $e_2$, and more genrally the i-th collumn and $e_i$

Now changing basis only means taking another basis $B' = (e_1', ... e_n')$

And you get a new matrix $N$ (my notations suck btw) that satifies the properties described above

Does that help ? @Socrates

Socrates

20:49

@Astyx ok, that clarified what we talk about.

Astyx

And does it answer your question ?

Socrates

well, I didn't understand how we now calculate the new matrix.

Astyx

One way to do that would be to take $e_1'$, compute its image, decompose this image in $B'$ and write the collumns accordingly

Socrates

and that's why I asked, what should I read, watch, search

Astyx

Then with $e_2'$, ..., $e_n'$

Socrates

20:52

so, we write the change from $e_1$ to $e_1'$ down first?

Astyx

What do you mean ?

Socrates

or do we plug in $e_1'$ in the starting matrix, and what we get is the column of the second matrix?

Astyx

Plug $e_1'$ in the first matrix, what you get is the coordinates of the image of $e_1'$ in $B$ (and you want them in $B'$)

Socrates

to get it in $B'$ we write down the coefficients such that $M(e_1')=a_1e_1'+...a_ne_n'$?

Liad

@TobiasKildetoft the order is 4

Astyx

20:57

@Socrates Right

Socrates

and those coefficients are the first column

Tobias Kildetoft

@Liad Right, and it will be contained in the subgroup generated by two of them

Liad

@Astyx $A_5$ is is the Alternating group

@TobiasKildetoft of course, so the order of this subgroup is at least 5

Tobias Kildetoft

@Liad Even better, use Lagrange

Socrates

mmh, but I must admit that I don't grasp why this works ;)

but that's ok, I guess it will come

Astyx

20:59

If you haven't studied vector spaces that much, it might be hard to get

Liad

@TobiasKildetoft how? i just know from lagrange that $|<P,Q>| | 60 $ , and i already eliminated the numbers above using that

Tobias Kildetoft

@Liad You also know that it has a subgroup of order 4

Socrates

@Astyx I grasp the general concept. We maybe have a wierd vectorspace, like a quotientspace. We express it first in our field, do something to the matrix. then translate it back into the wierd space. So we don't have to compute with wierd rules.

like 2+2=0 in $F_3$

Astyx

It's not so much about the space being weird, it's about the matrix being inappropriate for computation and stuff

Socrates

ah, ok, that explained my prof. Because we don't know a priori wether a certain vector that goes through the matrix is maybe 0(or maybe not)

like a quotientspace about planes, we dont know if (a,b,c) is already in the plane

and to make such errorprone calculations easier, we first translate into a better one?

Astyx

21:06

For instance

Socrates

21:22

but does this only make something easier, or does it make something possible?

i mean, is something impossible without change of basis

Astyx

21:34

I doubt so

Anyway I gotta go

Godd day/night to you

Danu

o/ Ted

Ted Shifrin

@AkivaWeinberger Which two definitions? The standard algebraic one and the sum of distances? That's just algebra. The more interesting one is the conic section one (#9).

Bonne nuit @Astyx

So, @Danu, did you learn the secrets?

Danu

Nope---he also didn't remember. Sigh. But the idea is to add a bunch of terms which are actually zero due to Leray-Hirsch to simplify the expression.

Ted Shifrin

Hmm ...

Maybe I'll work on this more when I have time.

Socrates

@Astyx cu

Danu

21:41

In other news, I have a small question

robjohn

@Danu I'll alert the media!

Danu

so if I compute the cohomology of $\Bbb P(TS^6)$, I find that it's generated by $x\in H^6(\Bbb P(TS^6))$ and $y\in H^2(\Bbb P(TS^6))$ which satisfy the relations $x^2=0$ and $y^3=-2 x$. Here $x$ is the pullback of the orientation class of $S^6$ while $y$ is $c_1(H)$.

@robjohn hehe

Ted Shifrin

@robjohn: Make sure it's not fake news.

Danu

Now I want to do the same thing for $T^*S^6$

Then everything is the same except the (Leray-Hirsch) relation $y^3=2x$.

Ted Shifrin

Right.

Danu

21:43

In the $TS^6$ case, $xy^2$ is a positive generator of top cohomology right, because $x$ is positive generator on base and $y^2$ on fiber

In the $T^*S^6$ case... is $-xy^2$ a positive generator now?!

I don't understand why it would... but if it's not the Euler characteristic flips sign.

Ted Shifrin

Wait. I was just thinking about this. $\chi(E^*)$ versus $\chi(E)$.

Danu

Those are equal

Ted Shifrin

Are you sure?

Danu

Oh, wait.

Well, the Euler characteristic has to be the same

Ted Shifrin

Huh?

Danu

21:45

because both of these manifold $\Bbb P(...)$ are diffeomorphic to $G_2/U(2)$.

Ted Shifrin

We're talking Euler class of a bundle, not of a topological space.

Danu

I'm talking Euler class of $T\Bbb P(TS^6)$.

and T^*S^6$

Ted Shifrin

BTW, unlike our discussion of complex geometry, I'm not sure that there's a canonical notion of "positive" here.

Danu

Something something almost complex structure on $S^6$

Ted Shifrin

But back up. The Leray-Hirsch relation has the Euler characteristic of the bundle in there.

Danu

21:47

I'm sketchy on the background, sorry about that.

Ted Shifrin

Ah, ok, something something in that case.

Danu

@TedShifrin It has the Euler class of $TS^6$/$T^*S^6$ in there.

Ted Shifrin

So we need to settle the sign if you dualize. For a surface, it changes sign. OK, when $E$ is a bundle of rank $2k$, you get a sign of $(-1)^k$, agreed?

Danu

In what, the Euler class? Yes.

But the Euler characteristic of course doesn't change sign.

Ted Shifrin

We're talking different bundles. Euler characteristic is tangent bundle only.

Danu

21:49

Yeah, but I'm always talking about tangent bundles of the projectivized bundles.

Ted Shifrin

But the tautological bundles on those are different if you use $T$ and $T^*$.

Danu

There are two projectivized bundles. I want the Chern numbers (including Euler characteristic) of those total spaces

@TedShifrin They are? The taut. and hyperplane switch?

Ted Shifrin

Even though you're claiming the base spaces are diffeo, the bundles are off by a duality, aren't they?

You certainly have a nondegenerate pairing.

Danu

So maybe I'll start making claims, and you tell me where I go wrong.

Ted Shifrin

Nah ... You just go sort this out :)

Danu

21:51

But I don't understand what you're saying :P

Ted Shifrin

I'm actually working on my taxes.

I'm saying that the tautological line bundles on $\Bbb P^n$ and $\Bbb P^{n*}$ are different.

The latter acts on the former.

Danu

@TedShifrin Meaning... that they're dual?

Semiclassical

hi chat

Ted Shifrin

Uh huh.

I guess you have to sort out how the diffeomorphism of base spaces works, though.

This stuff takes attention.

Danu

@TedShifrin So you're saying when I change $T\mapsto T^*$, in addition to $c_3(TS^6)=-c_3(T^*S^6)$ I have to use $c_1(H\to \Bbb P(TS^6))=-c_1(H\to \Bbb P(T^*S^6))$?

where I denoted the hyperplane bundles over the respective total spaces by the same letter

Ted Shifrin

21:55

Offhand that's what I was claiming.

Semiclassical

Something worse than not knowing the answer to a question: not being sure you know what the question is :/

Danu

I'm pretty confused

Simple Art

Busy chat room today.

Danu

It's a shame that Kotschick wrote "proof omitted because it is exactly the same as [proposition for the tangent bundle case]" for the cotangent bundle case...

Semiclassical

A solid bit of that is me rambling about stuff that doesn't make sense. @SimpleArt

Simple Art

21:57

@Semiclassical Ah, I see then. Good luck! :-)

Ted Shifrin

Well, it is the same, once you sort out signs.

Danu

@TedShifrin The signs make me sad

Ted Shifrin

Signs (and factors of i) are always something one needs to pay attention to. I warned you of this months ago.

Semiclassical

Phases don't stop being a pain just because you move from physics to math :)

Ted Shifrin

I infer that was not directed at moi.

Semiclassical

22:07

Nah.

I'm still entirely unsure of what exactly I'm trying to find with this polynomial.

Best I can come up with is the very vague: Consider the family of polynomials $y^2=(x-c)(x^2-1)(x^3+ax+b-c(x^2-1))$ with moduli space $a,b,c$.

In what sense is the Riemann surface different when $c=0$ vs. $c\neq 0$?

Painfully, painfully vague. Ugh.

Ted Shifrin

No, that's not vague. But it is surely varying with $a$ and $b$ as well.

Semiclassical

That's the problem. I want something which doesn't change when $a,b$ are varied. :/

Ted Shifrin

Certainly for the cubic those alter the complex structure.

Semiclassical

At the level of the underlying model, I know that $c\neq 0$ represents a breaking of symmetry in a way that isn't true for $a,b$. But I can't see how on earth that could be apparent for the above polynomial.

$c\neq 0$ does not seem like a special condition in any obvious way.

It moves the root at $x=c$, and it changes the locations of the roots inside the cubic. But that's all.

I feel like I'm asking the wrong question, but that doesn't move me any closer to the right one :/

Ted Shifrin

@Semiclassic: I am no expert on curves. I presume that somewhere it is written down how to compute things analogous to the j-invariant for cubics for hyperelliptics of higher degree.

Semiclassical

22:23

Yeah.

Ted Shifrin

You might even look at the webpage for the math department at UM and see if there are any people working on algebraic geometry/curves.

Semiclassical

Good point.

I know one prof at least who I could ask.

Ted Shifrin

I would help more if I could ...

Semiclassical

Looks like it might be the Igusa invariants

Ted Shifrin

Looks like it, indeed. Well done.

Semiclassical

22:26

Alas, it then says that to compute those you need the Igusa-Clebsch invariants...and the link for that doesn't work

Ted Shifrin

LOL

Google separately. Or find a book in the library (I know — so old-fashioned).

Hey, the link worked fine for me.

Maybe it's a popup and you have 'em blocked.

@Semiclassic: You saw what I just typed?

OneRaynyDay

@Semiclassical Hey Semi

Ted told me last night that you might be familiar with entropy. Could I confirm something with you?

Couldn't find such information online

Kirill

Hello! some time ago somebody posted a link to the list of algebraic structures like magma, monoid, groups etc. Do you know the link or an alternative source where I can find almost all the structures?

Alessandro Codenotti

is wikipedia enough?

Kirill

no

it could be enough, but I cannot be sure if I have seen all the structures that are more or less common in definitions @AlessandroCodenotti

making an alphabet, want to cover as much as possible

Ted Shifrin

22:37

rehi @Alessandro

Alessandro Codenotti

Hi again @Ted

Kirill

or, do you mean, they are almost all on the link you sent, @AlessandroCodenotti?

Alessandro Codenotti

I don't know what you want to do with such a list @Kirill, that's just a starting point if you want to get an extremely detailed one, I think at most 10 of the things listed there where mentioned in the courses I attended so far,

Semiclassical

@TedShifrin Yeah, will check it out

@OneRaynyDay You can ask the question, not sure I can confirm right now.

Ted Shifrin

Cool, @Semiclassic.

Kirill

22:42

@AlessandroCodenotti as I said, I am making an alphabet of structures, according to the axioms they have. Something like that.

OneRaynyDay

@Semiclassical Sure, thanks :) It's basically that

my professor wanted to calculate the entropy of a multinomial distribution

it was later found that he wanted the entropy of a multiNOULLI distribution which trivializes it

but I was still interested in what the actual answer was for a multinomial distribution with k classes, in terms of maximizing entropy.

I don't have a good intuition on entropy because it's not necesasrily required for this class, but my solution was:

Semiclassical

Ah. That's definitely beyond my technical expertise.

OneRaynyDay

oh nevermind then

thanks anyways though :)

Semiclassical

There's some references re: entropy of a multinomial distribution in this MO question: mathoverflow.net/questions/251467/…

might be useful

Kirill

ok, thanks a lot!

Semiclassical

22:45

@TedShifrin Ah, I see it now. I didn't realize it was opening up below the text shown.

OneRaynyDay

hmm interesting

I think I got something similar but instead of 1/k I got xi/k

I'll look into his references, thanks :)

Alessandro Codenotti

Reviewing for next week topology exam I finally understood why we defined a projective transformation $f$ as having an associated isomorphism of the underlying vector space $\varphi$ such that $f([v])=[\varphi(v)]$

Danu

Gotta commute those diagrams :)

JessyunBourne

Hello everyone

Semiclassical

new name/icon, I see

JessyunBourne

22:59

Yup. It's Cat Damon.

Hence the Jason Bourne reference.

Finally done with pdes

now I'm working on algebra.

OneRaynyDay

oh my gah

I never knew it was this hard @Semiclassical

I spent half a day on it yesterday and came to incorrect conclusions because my summation was too specific and wrong

Semiclassical

23:20

That's probably why your prof had you do the multinoulli version, so that it'd be tractable

OneRaynyDay

Yep yep. I still can't understand what the papers were about - too complicated for me at this time haha

Semiclassical

That's what makes research both difficult and interesting: you have to figure out what problems are both intriguing but doable

Adeek

hey

Socrates

what does $_B[\vec{v}]$ mean?

Stella Biderman

Socrates That's hard to answer without context

Socrates

23:26

ok, I thought that where some widly common notation

OneRaynyDay

Isn't that just v in a basis of B

omg i solved it

the multinoulli one

so unsatisfying when you already know the answer

Mary Star

How could we check the pointwise and uniform convergence for $\displaystyle{\sum_{n=0}^{\infty}\frac{\sin (x)}{(1+x^4)^n}}$ ? Do we use the Weierstrass criterium or do we use the definition of the convergence?

Stella Biderman

23:43

Uniform convergence on can be shown by the Weierstrauss-M test and letting M_n = 1/x^{4n}

Sylent Nyte

Hey guys, whats a way of going about finding a factor of a number thats 911 digits long?

TheGreatDuck

@SylentNyte brute force

Sylent Nyte

@TheGreatDuck im doing that currently

TheGreatDuck

and time

good luck

Sylent Nyte

yea im pretty impatient

i have a simplistic method, whats the quickets

TheGreatDuck

23:46

if you have a solution in general that actually finds the factor (like a simple equation) then you have rendered cryptography worthless and win some massive prize or something.

so be patient

:-)

Sylent Nyte

okay XD

then i shall set up two systems

Daminark

Hey everyone!

*Also hey Stella!

Mary Star

@StellaBiderman I see. So, we have to check if $\sum_{n=1}^{\infty}\frac{1}{x^{4n}}$ converges. Checking it with the ratio test or with the root test we get an expression with $x$ and so it is not always $<1$, is it?

TheGreatDuck

no you know what you do?

start with some odd number

and then test divisibility whilst incrementing by 2

Stella Biderman

@MaryStar Indeed. That tells you that it is uniformly convergent on (-1,1), though it's possible for it to be pointwise or uniformly convergent outside that interval.

Mary Star

23:51

@StellaBiderman Can we not check for $x\in \mathbb{R}$ ?

Stella Biderman

@MaryStar Actually, the shift makes the M-test I mentioned not work. However, you can just directly use the geometric series test. I had thought the 1

The 1+ was outside the exponent, not inside

The series converges for |1+x|>1 by the geom series test

Mary Star

Which exactly is the geometric series test? I got stuck right now... @StellaBiderman

Stella Biderman

A series of the form $\sum x^n converges on the interval |x|<1

here x is your 1/(1+x)^4

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$Mathematics$ Mathematics