help-19

[leaden karma]

$\lim_{n\to\infty}\frac{\sqrt[n^2]{1!2!\dots n!}}{n^\frac1{2}}$

[clever fjord]

kheerii

[leaden karma]

can I use Sterling's approximation for the factorials here?

I'm assuming no

basically is $\lim_{n\to\infty}\prod_{r=1}^n\frac{r!}{\sqrt{2\pi r}\left(\frac{r}{e}\right)^r}$ finite?

[clever fjord]

kheerii

[odd edge]

@leaden karma Has your question been resolved?

[leaden karma]

<@&286206848099549185>

[noble solstice]

are you

lol

micheal penn moment

[leaden karma]

yes lol 😭

[noble solstice]

💀

[leaden karma]

I'm trying to find an alternative to Stolz-Cesaro theorem

[noble solstice]

mh

goodluck!

[leaden karma]

if I can manage to prove this then I'd be done

but this looks... bad

[noble solstice]

how about

logarithm

turning product into sum

check if converge using various tests

maybe that can give some insight into how to solve the product

[leaden karma]

we don't really need to solve it, since this product is inside of an n^2'th root

also solving it sounds like hell

if I can prove it's finite then that ratio^(1/n^2) just goes to 1

[noble solstice]

https://en.wikipedia.org/wiki/Infinite_product#Convergence_criteria

[leaden karma]

well I tried the ratio test and that gave me 1

[cerulean vortex]

some basic bounds on 1!2!...n! don't work
like 1!2!...n! <= n^(n^2) and
1!2!...n! >= n^n
so i guess you will need better bounds and maybe sterling's approximation

[leaden karma]

I'm trying to prove that Sterling's approximation does work for products of the type 1!2!...n!

[leaden karma]

$p_r=\log r! -\frac1{2}\log(2\pi) - \left(r+\frac1{2}\right)\log r + r - 1$

[clever fjord]

kheerii

[leaden karma]

this looks bleak

[low locust]

do you have any reason to believe it does? all the errors would have to cancel out. and especially early on the errors are probably significant

[leaden karma]

yeah, but I think the product as n tends to infinity may be finite

and if the product I defined above is a finite positive number then this would be true

[clever fjord]

kheerii

[leaden karma]

also I happen to know the original answer and it does match

[clever fjord]

kheerii

[leaden karma]

oh wait, I can just use the higher order terms of the Sterling approximation

[cerulean vortex]

$\forall \epsilon > 0,\lim_{n\to\infty} \frac{\sqrt[n^2]{n^{n^{2-\epsilon}}}}{n^{\frac{1}{2}}}=0$

[clever fjord]

Axe

[leaden karma]

interesting

this is just $n^{n^{-\epsilon}-\frac1{2}}$

[clever fjord]

kheerii

[cerulean vortex]

yeah

this might help if you can show that 1!2!...n! < n^n^(2-eps)

[vernal pilot]

k ! <= n^k

so the product <= n^(n(n+1)/2)

[cerulean vortex]

unfortunately the exponent is bigger than n^2

[vernal pilot]

is it?

[cerulean vortex]

uh

well you do have a factor of 1/2

[vernal pilot]

n^2/2 + n/2 isnt bigger than n^2 if n is sufficiently large

i mean it will obviously be a better bound than n^(n^2) because this bound is derived by writing k!<n^n which is horrible

so writing k!<n^k will obviously give you a better bound

[cerulean vortex]

but n^2/2 + n/2 will dominate n^(2-epsilon)

[vernal pilot]

indeed

this is still O(n^2)

we can better the bound by writing k!<k^k

but im not aware of the asymptotics of products of this type

,w \sum_{k=1}^{n}klog_n(k)

not too helpful

[vernal pilot]

this should surely give an improvement but not sure if its enough to take it below O(n^2)

[odd edge]

@leaden karma Has your question been resolved?

[odd edge]

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[raw palm]

Prove that if $K\subset\mathbb{R}^t$ is compact and convex then $K=\bigcap_{\alpha}B_{\alpha}$ where ${B_{\alpha}}$ is a collection of balls

[raw palm]

so first if K is compact then K is closed and bounded (K⊂R^t) so K is contained in some ball of radius r>0 which proves the first direction

but i am having trouble with the second direction

so after i take K to be the intersection of some balls

ik that K is bounded

[vernal pilot]

im not sure how that gives the first direction

[raw palm]

i also know that K is convex since if x,y∈K then x,y∈_B_α for all B_α that satisfy this

[raw palm]

if it is contained in a ball then it is contained in the intersection of some balls right ?

[vernal pilot]

well yeah but we need to show it is equal to the intersection of some balls

[raw palm]

ah wait i was trying to prove something wrong

what i proved is a part of the first direction

i still need to prove the inclusion B_α⊂K

or i should try something else

[vernal pilot]

just think about why this wont be true necessarily

[raw palm]

yes but then K being convex will make this necessarily true

[vernal pilot]

so yes you should try something different

[vernal pilot]

well check R^2 now

take a closed square

it will look something like this

[raw palm]

ohh yes you are right

[vernal pilot]

moreover a K cannot be equal to a single ball since K is closed and a single open ball is open

infact the intersection of any finite number of open balls is open

[raw palm]

yes that's right too i didnt think about this

so if it is the intersection of open balls then it must be the intersection of infinite number of balls

[raw palm]

but doesnt this say balls

so if i prove that K is the intersection of closed balls

[vernal pilot]

actually yeah i think it needs to be closed balls

[raw palm]

oh wait maybe there are compact and convex subset of R^t that are the intersection of an infinite collection of open balls

[raw palm]

ohh

for the second direction, if $K=\bigcap_{\alpha}B_{\alpha}$ where ${B_{\alpha}}$ is a collection of closed balls then $K$ is closed and bounded and thus compact. Moreover, balls are convex so if $x,y\in K$ then $x,y\in B_{\alpha}$ for all $B_{\alpha}$ in the above collection then $bx+(1-y)b\in B_{\alpha}$ whenever $0<b<1$ and thus $bx+(1-y)b\in K$ so that $K$ is convex

is this correct

for the second direction

but now i am assuming that the balls are closed

[raw palm]

following this

[vernal pilot]

im not sure about that one tho. i couldnt come up with examples that must need closed balls

its 2 am so i have to sleep but you will get help asking in #real-complex-analysis

[raw palm]

ohh ok

[raw palm]

tysm and good night

i will ask there

[odd edge]

@raw palm Has your question been resolved?

[raw palm]

<@&286206848099549185>

[feral lark]

Where does this question come from? I don't really believe the statement... If you have a line segment in R^2, every closed ball that contains this line segment also contains the ball whose diameter is this line segment, so the intersection of such a collection cannot equal the line segment itself.

[cerulean vortex]

You can take the intersection of circles that have the segment as a chord

[odd edge]

@raw palm Has your question been resolved?

[feral lark]

Ah, good point 😅

Okay, I do believe the statement now

So you can take ${B_\alpha}$ to be the set of all closed balls containing K. Then clearly $K \subseteq \bigcap_{\alpha} B_\alpha$. For the other direction, suppose $x \not \in K$. We must then show there is some closed ball $B$ containing $K$ such that $x \not \in B$. You can do this by finding a hyperplane separating $x$ and $K$, then choosing $B$ to be, say, tangent to this hyperplane and have radius large enough to contain $K$ (you probably need to use some tools to formally establish this).

[clever fjord]

zkzach

[raw palm]

what are hyperplanes

and what tools do i need

[feral lark]

like you need to be able to say that if $K$ is closed and compact and $x \not\in K$, then there exists a closed ball $B$ such that $K \subseteq B$ and $x \not\in B$

[clever fjord]

zkzach

[feral lark]

this is some type of separation theorem. have you encountered any results similar to this?

[raw palm]

no i havent encountered this before

i know a bit of topology from studying analysis

[feral lark]

🤔 where does this problem come from?

[raw palm]

the book contains a chapter about topology

and someone i know asked me about this problem

then i got curious and started trying to solve it

so thats where this problem came from

[feral lark]

I see. you don't really need topology here

[raw palm]

everything related to topology that i know is only what is found in rudin's book

[raw palm]

wait let me think about this for a bit

[feral lark]

I think a typical thing one establishes quickly in convex analysis is that if you have a closed convex set C and a point x outside of C, then there is a hyperplane separating x from C

[feral lark]

a hyperplane is a generalization of a line in R^2 or a plane in R^3

so in R^2, it says you can cut space into two pieces with a line, one piece containing the point x and one piece containing the closed convex set C

you need something similar in this problem, but basically you will approximate the line with an enormous circle that encloses the convex set C (hence C must be bounded, which is why the problem is for compact rather than just closed sets)

[raw palm]

is this necessary for the proof

so is there a way that avoids the use of this

[feral lark]

I don't think you need to use this result exactly. but at least in the solution I outlined above, you do need to prove that for any compact convex set K and point x outside of K, there is a closed ball that contains K and does not contain x (i.e., separates x from K and while swallowing K entirely)

this seems like the most direct approach

but it is stronger than the separating hyperplane theorem because it's easy to show that there is a hyperplane separating a closed ball from a point

[raw palm]

and this is enough to prove the original statement

because this means that no point x outside K is contained in every closed ball containing K

[feral lark]

yes

[raw palm]

thus the points contained in the intersection of these balls are exactly the points of K

[raw palm]

hmmm i will try to prove that

[vernal pilot]

Nice, the convexity is being used implicitly for the existence of said plane...

[raw palm]

so since x is not in K then for any 0<a<1 and any y in K , ax+(1-a)y is not in K

so the set {ax+(1-a)y|0<a<1 and y in K} separates K from x no ?

[feral lark]

I don't think what you wrote is true

you're saying that if you connect x and a point y \in K with a line segment, the entire intierior of the line segment is not in K?

[raw palm]

how does this cover the entire line segment ?

[vernal pilot]

It is not a dichotomy

Half the segment can be in the set and half outside

The negation of A subset B isnt A intersection B = empty

[feral lark]

btw I think you are also using "separates" incorrectly. a separating hyperplane looks like this (in 2 dimensions):

it is not a line segment joining x to a point in C, but rather an (infinite) line such that x is on one side and C is on the other

(image from https://www.princeton.edu/~aaa/Public/Teaching/ORF523/S16/ORF523_S16_Lec5_gh.pdf)

I think a proof could go like this: fine a separating hyperplane and fix a point z on this hyperplane. now consider all balls that are tangent to (supported by) the hyperplane at z. argue that there exists one that contains C (must use that C is bounded)

this part does not need convexity or closed-ness; just boundedness

[vernal pilot]

I think they are proving the existence of separating hyperplane part in their attempted proof

[feral lark]

ah ok

[raw palm]

yes i was trying to do this

[feral lark]

I don't follow; there is no problem with a hyperplane intersecting a convex set in more than two points. consider a line going through a disc

[vernal pilot]

indeed, i have edited the message

[feral lark]

I think you can just do something like this: take y \in K to minimize dist(x, y) (such a point exists because K is compact). now argue that the perpendicular bisector the line segment between x and y separates x from K

[vernal pilot]

Hmm i had thought of this briefly but for some reason assume the plane may intersect K at some point. But it looks like this is it

Btw what subject is this? If i wanted to look further into this

[raw palm]

the original question ?

[vernal pilot]

Yes

[raw palm]

i am not sure myself

someone asked me about this after someone else gave him the question

[vernal pilot]

@feral lark you must know

[raw palm]

now imagine this chain going further than that

[feral lark]

convex analysis?

[vernal pilot]

I see, any prerequisites?

[feral lark]

I'm not sure—I've never properly studied the subject myself. I know there is a well-regarded (but somewhat old) book by Rockafellar. maybe that lists some prereqs

I have a cs background and the basics (like separating hyperplane theorems and Farkas lemmas) come up in convex optimization

[feral lark]

I believe if K intersects the plane you should be able to find a point z in K closer to x than y (but don't see the details rn)

[raw palm]

the perpendicular bisector separates the set A of all the points on the line segments into 2 sets, a set containing all points of A that satisfy d(y,p)<r/2 and a set containing all points in A that satisfy r/2<d(x,q)<r where r=min d(x,y)

these 2 new sets are separated since the intersection of the closure of any of them with the other is empty

[vernal pilot]

I think that point would be on the line joining y and z, namely the foot of the perpendicular from x to the line joining y and z

And that foot must be in the set K since K is convex

But that leaves the question of what if that line is parallel to the plane

[iron bear]

hilbert projection theorem says there is a unique y in K such that dist(x, y) = dist(x, K)
consider the plane P perpendicular to the line joining x and y passing through y
suppose that a point z in K lies on the side of P containing x, then all of the points on the line joining z and y lie in K: these are the points tz + (1 - t)y
now compare the distance of tz + (1 - t)y to x and the boundary of the ball centred on x with radius dist(x, y)
since tz + (1 - t)y is linear and the boundary of the ball is quadratic, one arrives at a contradiction since there must be some t such that dist(x, tz + (1 - t)y) < dist(x, y)

[feral lark]

nice. I feel like "hilbert projection theorem" is overkill since we are just talking about R^d tho

[iron bear]

well its just a name to throw around

one just needs that y exists

maybe uniqueness isnt used here

[feral lark]

if you don't need uniqueness, y exists because K is compact and dist(x, y) (as a fn of y) is continuous

[iron bear]

i was worried about it being non-unique at some point but maybe its fine

[raw palm]

ohhh

[feral lark]

you can rule out two distinct minima using a similar argument

[raw palm]

so this proves that there is a hyperplane that separates x from K

[vernal pilot]

I think the non unique part would mean that the two points lie on a ball and the plane segment joining them has to be a chord so there is a point on the (hyper)chord with more minimal distance

[feral lark]

yes

[iron bear]

you just assumed the euclidean norm 🫵

anyway that wasnt really the point

[vernal pilot]

This is nice

[vernal pilot]

Logically i feel like this should work too

[iron bear]

to fill in the details, tz + (1 - t)y intersects the sphere at t = 0, and isnt tangent to the sphere, so theres another intersection

[feral lark]

the boundary of the ball is quadratic

[iron bear]

like its the usual expand out ||a + tb||^2 thing and since you have a minima you force the linear term to be 0 kinda deal

[feral lark]

I was trying to suggest that you also assumed the euclidean norm

[iron bear]

i did

[vernal pilot]

Yeah i was gonna say lol

[iron bear]

that was not a serious comment

[feral lark]

it was an interesting point though lol

[raw palm]

now since there is a hyperplane that separates x and K then there exists B that contains K with x not in B whenever x not in K

you can take the ball to be of radius max dist(z,y) with z in K

and center z

this covers K

and still doesnt cross to the other side of the hyperplane

[feral lark]

why is it bounded by the hyperplane?

[raw palm]

because it stops at y

[feral lark]

sure, but z may be closer to the hyperplane than y

[raw palm]

if it is then dist(z,x)<dist(y,x)

but we assumed that d(x,y)=d(x,K) so it is the minimum distance between x and K right ?

[feral lark]

i'm worried about this type of thing:

the ball centered at z will not be good

it doesn't even need to be closer to the hyperplane than y

ah wait, maybe it is good

[vernal pilot]

I think you just need to change the center up a bit

[feral lark]

I can't really tell. but I had another approach earlier that I think should work

just fix any point q on the hyperplane H, and consider all balls that are supported by H at q and contained in the K-side

or for concreteness, take the balls B_k of radius k

for integer k

I claim that there is some k such that B_k contains K

[vernal pilot]

Ah actually no

That lower bound is surely necessary but not sufficient

[raw palm]

wdym by supported by H at q

so the balls only intersect H at q ?

[vernal pilot]

I think they mean tangent

[feral lark]

yes

ok so why does some B_k work? well we just need to show that for any point t on the K-side of H, there is some B_k that contains t

then we take max k over all t in K. this works because the {B_k} are nested

[vernal pilot]

That construction doesnt work either

[feral lark]

you mean mine, or the one you just posted and deleted?

[vernal pilot]

The one i posted and deleted

[vernal pilot]

This should definitely work

[feral lark]

yeah, you can try to be clever and come up with a precise center and radius, but we don't need to be that clever 😅

[vernal pilot]

Wait a minute, why must the balls be nested?

[feral lark]

maybe it requires a proof. but the point of the construction is precisely have a sequence of nested balls that are increasing in size

[vernal pilot]

I forget this is the picture

[feral lark]

then they will necessarily eventually engulf anything

^ yes this is precisely the picture I had in mind

[raw palm]

man i didnt do anything in this question so far i still need training

[vernal pilot]

No i feel this is not standard for analysis because my course hasnt covered anything like this

(Real analysis)

[raw palm]

no this isnt standard for real analysis i am sure about that

[feral lark]

ah sorry I hope I didn't ruin the problem for you too much

[raw palm]

but this was not meant as a real analysis question to begin with

so yes

[raw palm]

no i am not saying this

in fact its the opposite

i am just blaming myself a bit

this will remind me that i always need to work more hahaha

[pulsar elbow]

are we still finding a separating hyperplane

sorry i’m kind of lost as to where y’all are

[vernal pilot]

We did that a while ago

[feral lark]

I think we are basically done (?)

[raw palm]

no that is done

[pulsar elbow]

oh good

[vernal pilot]

Most of tbe proof has veen done yes

Or all if you count the "obvious" separating ball existence

[raw palm]

tbh i am having trouble proving that proving that some B_k exists

but dont mind me

[feral lark]

ah ok. do you understand the idea of the {B_k} ?

[raw palm]

yes

finding a B_k for each t in K and then taking the maximum k

which does the job since B_k are nested

[feral lark]

right

so you just need to show that for a fixed point t (on the same side as K), there is some k for which t \in B_k

[raw palm]

yes

so i need to show that given t in K , there exists k in N such that d(z,t)<k

where z is the center of B_k

d(z,t)=<d(q,t)+d(z,q)<d(q,t)+k

but i dont think this is what i should do

[feral lark]

ok so d(z, t) <= d(z, q) + d(q, t)

and you want to choose k so that d(z, t) < k

[raw palm]

ah

[feral lark]

ah sry, but this is a bit weird because we are fixing z before k

[raw palm]

ah yes we cant do that

because q is fixed so once z fixed then k will also be fixed

[feral lark]

finding a nice expression for k probably involves some trigonometry, as it depends on dist(q, t) and the angle between the line segment from t to q and the hyperplane

[raw palm]

does that mean that there is a way to prove its existance without actually finding it

[feral lark]

there should be, but I don't quite see it yet 😅

[raw palm]

np

[feral lark]

intuitively, the balls are nested and growing symmetrically, so they should eventually swallow up any point. but not exactly clear how to formalize

[raw palm]

and its not good to have the center in terms of t

is it ?

[feral lark]

maybe we can consider all balls (not just of integer radius) tangent to H at q

[raw palm]

because the same problem of taking t to be the center will arise

[feral lark]

we also have to argue that max { k_t : t \in K} exists, which requires compactness 🤔

my approach is starting to get a bit messy

[raw palm]

yes it probably exists since there is a finite subcover of K

[vernal pilot]

Or maybe i dont get what you are saying

[raw palm]

but we need max {k_t : k in K} to exist

it exists because of the finiteness of the numbers of balls that form a subcover of K

if left like this then what guarantees that there is a radius where a ball of this radius cover K

[vernal pilot]

What i said came out completely wrong. What i mean is i dont see how any open cover having a finite subcover will help

[raw palm]

it helps to show that if each t is contained in a ball of radius k_t , then there is a radius k where B_k covers K

[vernal pilot]

Okay so this is the same as what we are trying to do

[raw palm]

what we are trying to do is to prove that for each t there exists k_t

[raw palm]

if we dont prove this then the above statement doesnt do anything

[raw palm]

by the above statement i mean this

[raw palm]

ohh wait

K is closed

so each limit point is a center of ball which contains another point t in K

in particular if you take the ball of center t and radius d(q,t) then you obtain a ball that has q on its boundary and contains a point t in K

but then these balls may not cover every point in K

[feral lark]

yeah, then you can't just take the max

the center of B_k is just q + k * a, where a is the unit vector defining the hyperplane (and pointing in the direction of K)

so we need to argue that there is some k for which dist(q + k * a, t) <= k

[vernal pilot]

Tbh i am convinced after the separation part

[raw palm]

yes it is convincing

because the separation proved that there is a ball which only contains K

so K is the ball

is this wrong

i dont think that it is

[vernal pilot]

Why does K need to be a ball

K is just any convex compact set

[raw palm]

the separation proved that if x is not in K then x is not in B

so if x is in B then x is in K

[vernal pilot]

I think you mean K is the intersection of all such balls

[raw palm]

oh wait yes

the separation was made for all x not in K

so that gives all such balls and there intersection is contained in K

[raw palm]

but arent we forming balls that contain K

and at the same time dont contain points outside K

[feral lark]

you just need to argue that there is a ball that contains K and is separated from x by H

I agree it is "obvious" but formalizing it has been somewhat painful

[raw palm]

for each x not in K there is a ball that separates K from x

[feral lark]

yeah, x is arbitrary

an arbitrary point outside of K

then we need to find a ball B that contains K and not x

[raw palm]

is it this or is it one ball that excludes all x not in K

[feral lark]

no, for each x you can find a ball

[raw palm]

each ball excludes an x outside K

so that the intersection of all these balls is K

[feral lark]

right, from the top, this is the structure of the argument:

Let ${B_\alpha}$ be the set of all closed balls containing $K$.

Claim: $K = \bigcap_\alpha B_\alpha$.

=> If $x \in K$, then $x \in B_\alpha$ for every $\alpha$. Since the $B_\alpha$ are closed, it follows that $x \in \bigcap_{\alpha} B_\alpha$.

<= Suppose $x \not\in K$. We must show that $x \not\in B_\alpha$ for some $\alpha$. In other words, there exists a closed ball containing $K$ but not containing $x$.

[clever fjord]

zkzach

[feral lark]

So in the <= direction, we first choose x \not\in K, then we find the ball.

[raw palm]

isnt the B_α are closed not necessary here

for the => direction

since if x is in B_α for every α then x is in their intersection

[feral lark]

ah, yeah, sorry

[raw palm]

i believe this is more than enough as a proof

including the separation by a hyperplane for sure

the last part about the existance of a Ball in the nested balls isnt necessary

tysm for your huge help everyone

[feral lark]

I am not completely happy with our proof, but I am definitely convinced of the statement

[raw palm]

you are not happy from which side

so what is bothering you

[feral lark]

I don't feel like we have formally established that there for every x, there exists a closed ball B such that K \subseteq B and x \not\in B

I do feel like the separating hyperplane and the balls B_k we defined should suffice. but I don't think we have a bulletproof argument for why there is some k such that B_k contains K

[raw palm]

ahh yes i get what you mean

so you are bothered by the last thing that we tried to prove but couldnt

[feral lark]

yeah. like it's pretty intuitive, at least in dimensions 2 and 3

[feral lark]

I suspect this should work (if I set it up correctly) but didn't follow it through

[pulsar elbow]

can B_k just any closed ball?

[raw palm]

now that you mention it where exactly did we require only a closed ball

and not just any ball

[feral lark]

this is the argument so far:

• There exists a hyperplane H separating x from K.
• Let q be a point on this hyperplane.
• Define B_k (for integer k, or maybe for real k) to be the ball tangent to H at q (and on the same side as K) with radius k

we now want to say that there is some B_k that contains K. since the balls are nested, it suffices to show that for any point t, we can find a k such that B_k contains t

this seems completely obvious

[raw palm]

it seems completely obvious except that it isnt

[feral lark]

like it should follow from the fact that locally, a ball looks like a hyperplane. so if you make the ball big enough, you should be approximating the hyperplane locally. (and any point t defines a sufficiently local region)

[pulsar elbow]

okay, i think i’ve got something, it’ll use two hyperplanes, suppose K is a compact subset of H = R^{n-1}x[0,infty) and suppose further that no point of K contains a point with final coordinate 0, let z* be that coordinate (later i call this the “top hyperplane”). if you take some point (0,…,0,N), where N > sup{z: z is the last coordinate of some point in K}, then there is some cone whose base is on the top hyperplane which contains K. then, the ball will just be determined by the intersection of the cone with the top hyperplane and the point (0,…,0) (i think this’ll determine a ball in R^n? and then this will be tangent to the hyperplane with last coordinate 0)

i never used some of the things that i constructed so it’s very sloppy sorry

i’m just stressed about other things and wanted to throw something out and hope it sticks

but you don’t need a ball now, only a cone, which is good because K is convex and cones are straight-liney objects

[feral lark]

sorry, I didn't quite follow. a couple questions:

• what is z* and where did you use it?
• isn't the intersection of a cone and hyperplane going to have dimension at most n - 1? (since hyperplane has dimension n - 1)

[raw palm]

but in R^3 for example, isnt this just a circle in the plane determined by z*?

also what is z*

[pulsar elbow]

yes so, i want a ball whose boundary contains the intersection of the (boundary of the) cone with the hyperplane determined by having last coordinate z*, and also that contains (0,…,0)

[raw palm]

and one more thing, what guaratness that such a cone exists . I mean if we claim this then how is it different from claiming that a ball containing K on the K-side and tangent to the hyperplane exists

[pulsar elbow]

and this determines a unique ball in R^n

[pulsar elbow]

yeah so this is something i didn’t show at all

but this seems easier than finding a ball outright

[raw palm]

are N and z* the same thing ?

[feral lark]

fyi I found this question on math.stackexchange.com

[pulsar elbow]

no they are not

z* is supposed to be small. i wanna draw a picture rq

wtf

[raw palm]

nani

[feral lark]

I think the accepted answer is doing something similar to our approach, but they just say "since in the limit this ball will contain the entire half-space" lol

[pulsar elbow]

here’s a crude drawing in R3

the ball won’t in general have any relation to the cone like in the drawing, except for the fact that the boundary contains the n-2-sphere (here, the circle) and the origin, i just ran out of space. thinking about it more, the existence of such a cone C isn’t even going to be that hard, so as long as the rest of the pieces work then this is good.

[raw palm]

but does the statement that i want to prove only true for closed balls?

i didnt see anywhere were it was necessary for the balls to be closed

[raw palm]

what does this imply

so what does the fact that the boundary of the ball contain the n-2-sphere mean

[pulsar elbow]

it’s just something i’m using to define a ball

what will happen is that the cone is going to lie entirely within the ball, and since K is a subset of the (filled in) cone, then the ball will contain K

[raw palm]

but wont this ball contain points not in K ?

[pulsar elbow]

yes

the only thing that you needed was a ball that was tangent to the plane at (0,...,0) and also contained K

[raw palm]

i want a collection of balls which have an intersection contained in K

[pulsar elbow]

yes, so if you have some x not in K you're gonna do this construction to show the existence of a ball that contains K and not x

to guarantee that the family that you wanted to take the intersection of has at least one ball not containing x

[raw palm]

ah this follows due to the choice of z*?

[pulsar elbow]

well, it follows because there has to be at least two parallel hyperplanes separating x and K, so yes you'll use some isometry of R^n to take the plane that's closer to x to 0 and then the plane that is closer to K will have coordinate z*

[raw palm]

honestly i still dont fully get what z* is , all that is said about z* is that it is the last coordinate of points in K

but there are no more details that specify what it is

[pulsar elbow]

well, the moral is really the existence of two parallel hyperplanes which separate x and K, and taking an isometry of Rn will make them both hyperplanes that are determined by having last coordinate 0 and... something else (which i happened to call z*)

[raw palm]

ohh ok now i get it

[raw palm]

one last thing which is this

or does it work for any ball regardless of whether it is closed or open

[pulsar elbow]

i have a suspicion that if you take the collection that zkzach mentioned earlier, removed all the balls in which K \cap boundary(B_alpha) were nonempty, and then took the interiors of all those guys, that would get you a suitable collection of open balls.

[raw palm]

isnt that the same as just removing the boundaries of the balls

[pulsar elbow]

no

you have to be careful not to accidentally take some ball which doesn't contain K

but you're removing the boundaries of most of the balls and removing the ones where doing that causes a problem

[raw palm]

but every taken B_α contains K

[pulsar elbow]

yes, but if the intersection of B_alpha and K is nonempty, then removing the boundary of B_alpha requires the big intersection to not contain some point of K

[raw palm]

oh so you mean that maybe some point of K is on the boundary of one of the balls ?

[pulsar elbow]

yeah

but barring that there's nothing bad that's gonna happen

now, all you have to do is show that this new intersection is equivalent to the old one

[raw palm]

ah yes thats exactly what you said

mb

[pulsar elbow]

which is fine, because $\bigcap_{\epsilon > \epsilon_0} B_0(\epsilon) = \overline{B_0(\epsilon_0)}$

[clever fjord]

spire shield

[pulsar elbow]

where B_0 is the open ball of radius epsilon, and overline is the closure in R^n

[pulsar elbow]

now just apply this moral to every closed ball that we removed and note that we have all the open balls of bigger radius

pretend B_0(eps_0) is the "problem ball" that we removed

[raw palm]

yes

also K is in every B_α so removing some of these sets wont affect the intersection

[pulsar elbow]

yeah, in some sense this is literally the same intersection, since we take $\bigcap_\alpha B_\alpha = \bigcap_\alpha\bigcap_{\beta > 0} B_{\alpha,\beta}^\circ$ where $B_{\alpha,\beta}^\circ$ is an open ball with the same center as $B_\alpha$ but radius $R + \beta$ where $R$ was the radius of $B_\alpha$

[clever fjord]

spire shield

[pulsar elbow]

sorry for all the edits

no more edits

[raw palm]

dont worry about that almost all of my messages are editted in this channel xD

[raw palm]

finally something i understand directly

[pulsar elbow]

so that's good, and then we'd be done, right? because the right hand side is a collection of open balls

[raw palm]

idk maybe because i am not that used to this stuff

[raw palm]

exactly

[pulsar elbow]

you develop a sense for it over time. this is a great problem to work through

where did you find this?

[raw palm]

exactly

[pulsar elbow]

the problem

[raw palm]

if i dont experience this then i will never learn

[raw palm]

someone showed it to a friend of mine and that friend sent it to me

[pulsar elbow]

nice

[pulsar elbow]

genuinely one of the best mindsets to have

[raw palm]

tysm for all your explanations, insights and hardwork everyone

[raw palm]

tysm

it also feels great when you get the ideas and or reach the solution after having a hard time

[steel spire]

oh i see a slight problem you missed

the first two words of your name are flipped 😳

ur welcome 😌

[pulsar elbow]

that is true. idk if yall get this but analysis always feels like i'm in some alternate reality where i literally have the power to grow and shrink as i please, and the goal is literally to morph into the right shape to take care of the task i'm given

like

i can't really describe this properly

[raw palm]

this correction was my username originally

but when discord kept asking me change username i changed it to this

because i was lazy to come up with something

[raw palm]

yes i get what you mean

[pulsar elbow]

there is genuinely a small gap in one of the arguments that i made but idt it's actually worth going through atp

[pulsar elbow]

here, it's possible that the ball doesn't contain the cone, so you have to take two balls and take whichever one contains the other. this can be fleshed out in some detail if you really want, but the ball will either be the one that i described originally, or the one whose center is the cone point and radius is the last coord of the cone point (then i think that should work)

[raw palm]

and although it is not on the side of super difficult areas ( real analysis) it is like a major stepping stone

[pulsar elbow]

real analysis can be made as difficult as you want

[raw palm]

i will experience that as i delve deeper into the subject

[raw palm]

yes the ball should contain the cone if the center and radius are taken the same as the ones at the base of cone

[pulsar elbow]

no that's not true

[raw palm]

how so

ah wait

maybe the height exceeds the diameter ?

is that the reason ?

[feral lark]

I wonder if there's a cute way to do this problem where you can transform between hyperplane and ball.

like if you had a ball B that separated K from x (i.e., K is inside B and x is outside), then you can invert the circle. this transformation would turn B into a hyperplane separating a convex compact set K' and a point x'. we basically want to do this in reverse

(sorry to interject)

[raw palm]

isnt that possible

just like how you transform from cartesian to spherical coordinates for example

something like that

[pulsar elbow]

sorry, what did you mean by inverting the circle?

[pulsar elbow]

yes, something whose moral is analogous to this. you can have wide cones and skinny cones, and for the wide cones you use the ball that i originally described, and for skinny cones you use the new ball in my last message

[raw palm]

ohhh ok i got that

[pulsar elbow]

i am trying to think of some sphere that you can do some inversion over

or like

[feral lark]

I don't know how to formalize what I'm saying, but I want the "obvious" transform between a sphere and hyperplane

like all points but one of the sphere will be mapped to the hyperplane

[pulsar elbow]

there are a few good ones

[feral lark]

I see. well I think anything should be fine if it preserves convexity

[pulsar elbow]

if it takes a hyperplane to a sphere it won't preserve convexity

[feral lark]

I was hoping it would preserve convexity of the side it's "wrapping over"

that's where K will appear

[pulsar elbow]

i think i would need a picture to help me visualize it. i have a few ideas for what i think you want. is there an easy transformation of R2 that you had as an example?

[odd edge]

@raw palm Has your question been resolved?

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[mint canyon]

I need to prove wether there is a function g that maps from the whole numbers to the power set of the natural numbers, that the function F cannot return
F = \lambda f \in \mathbb{N} \to \mathcal{P}(\mathbb{Z}). \lambda z \in \mathbb{Z}. {n \in \mathbb{N} \mid z \in f(n)}

[mint canyon]

Sorry how do I make it organized with all the symbols

[desert marlin]

What do you mean

that the function F cannot return?

[mint canyon]

I need a function g that the function F cannot have as an output, as it returns functions

Nevermind, I got it. It was the function that takes a whole number and returns a natural number if the natural number doesn't belong to the output of F(f(z))

.close

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[modest widget]

Btw what does * mean in xi*?

[modest widget]

And would be the difference between taking the $lim_{n\to\infty}$ and simply, $\sum_{i=0}^\infty$ ?

😭

[upper onyx]

i guess its somewhere defined before the formula.

[modest widget]

No?

wait

[upper onyx]

ok, then i dont know. i can mean anything.

[frigid grove]

i means sqrt(-1)

i also means me

i also means integer

[modest widget]

[modest widget]

huh??

[frigid grove]

im just kidding lol

[clever fjord]

Jake

[frigid grove]

https://www.youtube.com/watch?v=bLz5N_6fkZ8

[modest widget]

alr

Would be nice if the source had mentioned what the hell the snowflake meant 💀

[frigid grove]

yeah

[upper onyx]

read the previous section.

[modest widget]

oooooh

tysm

.close

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[topaz tangle]

need help understanding this

how LHS is equal to RHS isn't clear to me

(k,n) = gcd(k,n)

mobius function etc etc, but i think that it doesn't matter

[odd edge]

@topaz tangle Has your question been resolved?

[fickle silo]

on the left we sum over the set A of all pairs (k, d) where 1 <= k <= n and d | gcd(k, n). Now let's construct this set in a different way, we define B to be the set of all (dl, d) with d | n and dl is a multiple of d with dl <= n. We need to prove A = B now. If (k, d) is in A, then first of all, d | n and d | k need to hold in order for d | gcd(k, n) to hold. Thus k is a multiple of d, say k = dl, and (k, d) is in B. Then let (dl, d) be in B, then say k = dl and we have d | k. We also have that d | n and thus d | gcd(k, n) and (dl, d) is in A

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[topaz tangle]

.reopen

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✅

@topaz tangle Has your question been resolved?

[topaz tangle]

why the order is reversed to the set B?

why it isn't (d,dl)

[fickle silo]

so that k matches with dl and d matches with d

ofc you can take B' with reversed order and then sum over that

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[topaz tangle]

.reopen

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✅

[topaz tangle]

i think i got it now, thank you

.close

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[mystic saffron]

Hello

[mystic saffron]

Helllo?

Nobody is online

I am alone

Gotta go eat some cone

Talking all day to a bone

[sand horizon]

Ask your question

[mystic saffron]

Till the bright sun ever shone

[mystic saffron]

W motivation

Let me fail

I will try again

Ty

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[quick hamlet]

Hey guys I have a question, I^ve been solving some problems recently about finding the limits of a sequence and I stumbled upon some problems where the result is an unmarked symbol like infinity - infinity etc and I was wondering what techniques can help me solve these equations for example I had a problem like this: 3n - cube root of(27n^3 - n +1) and unfortunately I can’t solve it with timing it by itself but it’s 3n+…/3n+…

[quick hamlet]

How do I solve these problems?

[narrow crypt]

compare the degree of denominator and numerator

[narrow crypt]

${\lim_{n \to \infty} \sqrt[3n]{27n^3 - n + 1}}$?

[clever fjord]

k

[odd edge]

@quick hamlet Has your question been resolved?

[quick hamlet]

I did this and the result was something divided by 0…

[quick hamlet]

3n - this

[narrow crypt]

how?

${\lim_{n \to \infty} \sqrt[3n]{27n^3 - n + 1}}$?

[clever fjord]

k

[oblique echo]

One thing you can do to turn this into a fraction is this frac (____)^(1/3) is equivalent to (__)^(1/3) / 1

So multiply the numerator and denominator by the root.

Let me try it out and see where I get. I'll try something else if it doesn't work.

[narrow crypt]

another interesitng approach would be to take ln of the limit

[oblique echo]

Yep you could say y = ()^1/(3n)

[quick hamlet]

it’s 3n- the whole thing

and the whole thing is cubed

I mean the root of cube

And it was +2 inside instead of +1 my bad dry

Sry

[narrow crypt]

${\lim_{n \to \infty} 3n - \sqrt[3]{27n^3 - n + 2}}$?

[clever fjord]

k

[quick hamlet]

Yes

[quick hamlet]

Exactly

[narrow crypt]

well then

this should go to infinity

3n term doesnt converge

by eyeballing 3\sqrt{27n^3 - n + 2} does converge

[oblique echo]

bro.....

why didn't u say it was 3n - cbrt(thingy) god gotta be specific. Was wasting time.

[scarlet pawn]

!done

[odd edge]

If you are done with this channel, please mark your problem as solved by typing .close

[oblique echo]

answer I believe is infinity. Here is my reasoning.

[quick hamlet]

Sorry bro

My bad it’s my first time here

[oblique echo]

Hold up made an error/

wait nvm.

[quick hamlet]

I tried to do it the denominator way u guys said before but the cube didn’t cancel out like it would with sqrt haha

[oblique echo]

Mk, so there is a slight error. Basically when I multiplied numerator and denimoniator with conjugate. denominator should be 3n + a, but thankfully 3n - a didn't change the answer.

But know that the denominator should be (3n +a), because we aren't changing the function, we are just algebraically manipulating it.

if you take the limits separately of just the first line using limit laws then you would see that 3n goes to infinity and the cbrt function since the power converges to 0, that would converge towards 1.

But 1 is simply negligible so we can just negate that.

Now this is assuming that limit laws are applicable, but we can't do that since we don't know if there is a real number or not. In this case its not a real number as we found since the expression tends to infinity as n approaches infinity.

[quick hamlet]

Okay thank you

Can I screenshot this?

your explanation

[oblique echo]

do as you wish.

[quick hamlet]

Thank you

[oblique echo]

hold up

got another way

to explain it to you as well. This is important.

There we go.

So what I did was I rewrote the initial expression as a fraction. What I then did was multiply the numerator and denominator by (1/n^(2)). What I did was just observed the leading power and then algebraically manipulated the expression into this format where it proves that as the limit tends to infinity we see that the term on the right of the expression for both the numerator and denominator become insignificant as we approach infinity.

So we hen just consider those to be 0, we are then left with 9/3/n = 3n. As the limit approaches infinity of the expression the expression itself also approaches infinity.

[quick hamlet]

Wait shouldn’t it be 0?

Ohh never mind

I mistaken something

Thank you 🙏

[quick hamlet]

.close

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[quick hamlet]

.reopen

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✅

[quick hamlet]

!done

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[quick hamlet]

.close

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[wanton bison]

maybe you can substitute x-1/x and use ibp

[sweet needle]

i tried

but it didnt rly help

[quasi sparrow]

Show

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[wanton bison]

Determine all units and null divisors of the polynomial ring $\mathbb{Z}_4[x]$ of degree 1. $\$
Can somebody check my solution?
$\ \textbf{Solution.}$
[ \mathcal{U}(\mathbb{Z}_4[x]^1) = {2x+1,2x+3} ]
and
[ \mathcal{N}(\mathbb{Z}_4[x]^1) = {2x, 2x+2}. ]

[clever fjord]

𝔸dωn𝓲²s

[weary pelican]

yep correct for me

[wanton bison]

Thank

.solved

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[naive night]

Hello! Can u pls help solving this:
p, q - prime numbers, n1, n2 - natural numbers. Find all pairs of p and q that meet the condition:
p+q=(n1)²
p+4q=(n2)²

Okay, I got pairs 11,5 and 13,3 but how to find other pairs of prove that there is no more pairs?

[celest acorn]

hm

so you have

3q = (n2-n1)^2

3q=(n2+n1)(n2-n1)

@naive night

[naive night]

Yes

[celest acorn]

from here it is pretty straight forward

[naive night]

Oh yeah

Ty

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[celest acorn]

welcome

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[polar matrix]

how can someone give that value for du after saying u= arctan ???

[wicked kestrel]

Differentiate arctan

[polar matrix]

omg im stupid i was taking the integral of arctan

sorry ty 😭

.close

[scarlet pawn]

its ok

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[shy patio]

hello, i must find a side of an equilateral triangle

[shy patio]

what exactly am i meant to exploit here?

no further info given

abc is equilateral

[rich iris]

even tho i have my own question, ill try to help you out

we can see that the origin is at the center of the triangle

[shy patio]

does that allow us to do anything?

[rich iris]

gimme one moment

[shy patio]

ok

[rich iris]

ok i found a website that should teach you!

https://www.khanacademy.org/math/geometry-home/cc-geometry-circles/area-inscribed-triangle/v/area-of-inscribed-equilateral-triangle-some-basic-trig-used#:~:text=You can draw an equilateral,* sqrt(3).)

hope this helps :3

[shy patio]

maybe.. i'll have to see

[ebon harness]

!status

[odd edge]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[shy patio]

2

[shy patio]

kind of opened my eyes to something that might work

will see

[rich iris]

oki dokie

[shy patio]

still unsure, i dont think that ended up being relevant unfortunately

[rich iris]

ok then lemme grab some paper and help you out

i think i gave you a wrong article that i thought was correct oops

/mb

[shy patio]

its not wrong, just not relevant

[rich iris]

i think i might have it give me a moment

[shy patio]

ok

[rich iris]

OOPS DIDNT MEAN TO SEND YOU THE OTHER PHOTO

[shy patio]

HE DID GET IT THO...

[rich iris]

But uhh yeah does this help

[shy patio]

uhhh wait

[rich iris]

?

[shy patio]

you got it wrong btw

sqrt of 5 isnt an alternative

[rich iris]

R u sea

Man

[shy patio]

yeah its multiple choice

[rich iris]

I tried 😭

[shy patio]

thanks

i appreciate it tho

[rich iris]

Let’s try again if

Ig*

[shy patio]

so

[shy patio]

ur trying to use

law of cosines for a non right triangle

you dont have an angle in it unfortunately

so that doesnt work

[rich iris]

So let’s assume that the pic isn’t to scale

WAIT

ITS EQALATERAL

[shy patio]

abc is

[rich iris]

But not and

Abd

Oh

I’m dum

why is this problem so hard lol

[shy patio]

yeah hahahaha

you just cant get 3 things on either of those triangles

[rich iris]

yeah

what are the options

to the question

[shy patio]

id prefer not to say

they dont really help

[rich iris]

can i just see?

please

[shy patio]

why?

if you think you have an answer shoot

ill verify

[rich iris]

ok then

OMG I MIGHT HAVE SOMETHING

hopefully i get this

[shy patio]

nice?!

[rich iris]

honestly i think im going in the wrong rabbit hole

[shy patio]

what are u doing

[rich iris]

here lemme show u

[shy patio]

im not sure a+b+c+d = 360

[rich iris]

its a quadrilateral

[shy patio]

oh hang on

[shy patio]

i...

think C is visibly under 180 degrees

i dont see how the quadri would help

we go from knowing 3 overall things to 2

[rich iris]

idk what im doing rn 😭 im only 13 doing trig

[shy patio]

hahaha ur doing good

dont worry

this stumped me too

[rich iris]

most progress is just finding the first step

[rich iris]

but why does my work show that angle C is 180

should i pull up a ai to help us?

!status

[odd edge]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[rich iris]

i think ima give up since i cant find out what to do 😭

[odd edge]

@shy patio Has your question been resolved?

[odd edge]

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[velvet estuary]

Does anyone have a paper with all the trig identities

[cold sage]

you mean just a summary?

[grave roost]

Define all of them

There's a lot

[dense moth]

I had

[velvet estuary]

Yes

[velvet estuary]

The main ones

[strange glacier]

google images

[low locust]

print out the wikipedia article

[cold sage]

this is a number of them for pre calc

[velvet estuary]

Where is Sin(x)×Cos(x)

[cold sage]

?

[velvet estuary]

It's not on the image

[cold sage]

are you talking about the double angle for sin?

[low locust]

product to sum, middle left

choose x=y

[velvet estuary]

I don't got a calc

The ones for a exam with no calculator

Like basic basic ones

They are about 20 if I remeber

Like 2sin(x)cos(x)=sin(2x)

[low locust]

bottom left

[velvet estuary]

Just basic knes

[low locust]

like just look properly

[velvet estuary]

.close

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[quiet mulch]

I need help, but is easy.

[swift lake]

which one

[odd edge]

@quiet mulch Has your question been resolved?

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[robust ledge]

say i have a function that applies f(1)=3, lim(x->inf)f(x)=-3

you can easily see that the function crosses y=x because it starts above it and finishes below it
but how can i prove this formally

[robust ledge]

the function is continuous

[stone cipher]

intermediate value theorem

[grave roost]

IVT with f(x)-x tho

[robust ledge]

can you elaborate

i know the theorem

[stone cipher]

ok then tell me the theorem

[robust ledge]

let f be a continuous function, let a be a number where f(a)<a<f(b), there exists a point c where a<c<b s.t f(c)=a

something like this

[robust ledge]

good idea

thanks that was so silly of me it was too easy 😭

.close

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[boreal crag]

are there any polynomials p(x,y) and q(x,y) such that (xy+1)p(x,y) + (1+x+y)q(x,y) has all even coefficients except for an odd constant term?

[odd edge]

@boreal crag Has your question been resolved?

[hasty dome]

Do you know modular arithmetic?

[boreal crag]

yes

[hasty dome]

OK, so do the coefficients mod 2.

We know the constant term has to be odd.

So, the constant coefficient on p can be 0 or 1. The constant coefficient on q can be 0 or 1.

[boreal crag]

right

and they should be different

[hasty dome]

Right.

So, there are two cases.

p has an odd constant and q has an even constant or vice versa.

[boreal crag]

right

[hasty dome]

OK, so what do you get with case 1?

[boreal crag]

other than 1+0 for constant terms?

idk

you get an xy in there though that could be canceled if p(x,y) also has an xy

[hasty dome]

Right.

What happens if p(x, y) has an xy?

[boreal crag]

you'd get x^2y^2 on the left

[hasty dome]

Right.

[boreal crag]

unless p also has x^2y^2 etc

so you get an uncancelled x^ny^n

[hasty dome]

Right.

[boreal crag]

for some n >= 1

[hasty dome]

OK, so what does that mean?

[boreal crag]

(1+x+y)q(x,y) needs to also have an x^ny^n

[hasty dome]

Why does it need to have that?

[boreal crag]

since the left side has one

and they need to cancel to make 1

(as in the polynomial 1 mod 2)

[hasty dome]

OK, so how can you get x^n y^n on the right?

[boreal crag]

well q(x,y) could have a x^ny^n, or an x^(n-1)y^n or an x^ny^(n-1)

each of which would also produce other extra terms on the right side

[hasty dome]

OK, let's say we choose x^n y^n on the right.

[boreal crag]

sure so q(x,y) has an x^ny^n

which produces an additional x^(n+1)y^n and x^ny^(n+1) on the right which have to be cancelled out either by something on the left or by another different combination on the right

[hasty dome]

Hmm.

[boreal crag]

ok if we look mod x+y+1

xy+1 becomes x(x+1)+1 = x^2+x+1

[hasty dome]

How did you get that?

[boreal crag]

mod x+y+1, x+y+1 is 0

x+y+1=0

x+1=y

(by adding y to both sides)

then I plugged that into xy+1

so we see that mod x+y+1, we get (x^2+x+1)p(x,y) = 1

[hasty dome]

OK, what does that lead to?

I found something else.

If we take the x exponent to be a and the y exponent to be b, we can write x^a y^b as (a, b).

With xy + 1, we can produce either (a + 1, b + 1) + (a, b) or (a, b) + (a - 1, b - 1) depending on whether we multiply xy + 1 by (a, b) or (a - 1, b - 1).

With 1 + x + y, we can produce (a, b) + (a + 1, b) + (a, b + 1) or (a - 1, b) + (a, b) + (a - 1, b + 1) or (a, b - 1) + (a + 1, b - 1) + (a, b).

So, we can get all combinations of a + {-1, 0, 1} and b + {-1, 0, 1}.

[boreal crag]

I have to go atm I'll be back

[hasty dome]

OK.

We started off by using the (a, b) -> (a + 1, b + 1) + (a, b) rule to get to x^n y^n.

[iron bear]

if we mod (x + y + 1, xy + 1), then we get that 0 = 1

but clearly x is not 1 mod (x + y + 1, xy + 1)

[quasi sparrow]

A wild Blake has appeared

[odd edge]

@boreal crag Has your question been resolved?

[odd edge]

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[lost fog]

<@&286206848099549185>

.close

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[unkempt fable]

how they introduced area of AOC and BOC

[meager juniper]

Not area, angle

Introduced because PON and CON' are similar triangles, as well as POM and COM'

So they have the same angles.

[mental lotus]

Nahh, its area. [AOC] = 0.5 * OA * height, and the height is CM'. So, you can substitute CM' as 2 * [AOC] / OA. Same for CN'. They just skipped a step and multiplied both by 2, so that 2 is not there.

They converted the ratio and expressed it in terms of quantities which are known to be constant

[meager juniper]

How is OA part of any triangle on that image?

[mental lotus]

You can construct AC, and create the triangle AOC

OA, OB and OC are finite length segments

[meager juniper]

Fair enough, I believe you I just don't see it right now. Probably owing to the fact that it's late

[mental lotus]

probably you need to sleep now

[odd edge]

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[lost vortex]

$y''x^2-2y=\sin(\ln x) \
y''-\frac{2y}{x^2}=\frac{\sin(\ln x)}{x^2} \
y''-\frac{2y}{x^2}=0 \
y=x^r, \quad y'=rx^{r-1}, \quad y''=r(r-1)x^{r-2} \
r(r-1)x^{r-2}-2x^{r-2}=0 \
r^2-r-2=0 \
r_1=2, \quad r_2=-1 \
y_g=C_1x^{r_1}+C_2x^{r_2}=C_1x^2+\frac{C_2}{x} \
y_p=\phi_1(x)x^2+\phi_2(x)x^{-1} \
\begin{cases}
\phi_1'x^2+\phi_2'x^{-1}=0 \
2\phi_1'x-\phi_2'x^{-2}=\frac{\sin(\ln x)}{x^2}
\end{cases} \
\phi_2'=-\phi_1'x^3 \
2\phi_1'x+\phi_1'x=\frac{\sin(\ln x)}{x^2} \
\phi_1'=\frac{\sin(\ln x)}{3x^3} \
\phi_1=\frac{1}{3} \int \frac{\sin(\ln x)}{x^3} , dx$

[clever fjord]

uguu~ba

[lost vortex]

Did I make a mistake somewhere? Because such integral cannot be taken (it can be taken lol)

maybe i don't see something, tell me

,w 1/3 Integral sin(lnx)/x^3 dx

[clever fjord]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by lambdacuc

[lost vortex]

but how to solve this?

i suck

[grave roost]

I think you can solve the integral w a u sub

[lost vortex]

yeah but what after sub?

[grave roost]

It should end up like e^-2u sin u, right?

[lost vortex]

integration by parts i dont think so

maybe a recurrent formula?

no idea

[grave roost]

It's integration by parts

[lost vortex]

but it doesn't work

[grave roost]

It's a classical integration by parts problem

You do integration by parts twice and end up with e^-2u sin u again

[lost vortex]

okay

it is

thanks

im just blind

[grave roost]

Np

Happens to all of us sometimes

[lost vortex]

but wait

its cycle

isnt it?

[odd edge]

@lost vortex Has your question been resolved?

[grave roost]

Yeah

Not your problem but you have to do what they do here

[lost vortex]

wait

[odd edge]

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✅

[lost vortex]

,w 1/3 Integral sin(lnx)/x^3 dx

[clever fjord]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by lambdacuc

[tulip wing]

do yall need help on anything

[lost vortex]

I also wanted to ask how to write this more correct way

wait a sec

i lost the minus

[odd edge]

@lost vortex Has your question been resolved?

[odd edge]

@lost vortex Has your question been resolved?

[boreal crag]

solution ended up being: let y = x+1, then reduces to (x^2+x+1)p(x,y)=1 which is clearly impossible from a degree argument

[odd edge]

@lost vortex Has your question been resolved?

[lost vortex]

i think i wrote it not quite clearly, where the integrals

[lost vortex]

.close

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[mossy zealot]

how to put this in the calculator to solve it

[meager juniper]

You can input it as two lines

-4 - 1/(-32.25) * (26 * -13.5 + 15.5 * 18.5)
-3 - 1/(-32.25) * (15.5 * -13.5 + 8 * 18.5)

@mossy zealot ^

[mossy zealot]

i wanna use the matrix in the calculator

[meager juniper]

Then that depends on the calculator

[mossy zealot]

991es plus

[meager juniper]

https://support.casio.com/global/en/calc/manual/fx-115ESPLUS_991ESPLUSC_en/using_calculation_modes/matrix.html

[mossy zealot]

i can't use this button while i have matrix stored

[meager juniper]

Make your matrix entries 26/-32.25 etc

[mossy zealot]

can you please elaborate

[meager juniper]

Instead of using 1/-35.25 just use scalar multiplication to modify each of the 4 entries

So you work around having to use that button

[odd edge]

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[odd edge]

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[warm hare]

hello i need some help about calculating if this improper integral is converging or diverging ?

[odd edge]

@warm hare Has your question been resolved?

[odd edge]

@warm hare Has your question been resolved?

[dusky cobalt]

Write regular expressions for the following languages over the alpha-
bet

Σ = {a, b, c, d}.
(a) All strings that do not end with aa.
(b) All strings that contain an even number of b’s.
(c) All strings which do not contain the substring ba.

1. Strings Not Ending with “aa”
   The language cannot end with aa, but can have any starting combination.
   L = {ε, a, b, c, d, ab, ac, ad, . . . }
   Regular Expression:

R.E = ((a + b + c + d)* (b + c + d))* + a

2. Strings containing even number of b’s
   The regular expression for the language with even no of b’s can be formulated
   as follows;
   L = {ε, a, c, d, bb, abb, bba, ac, ad......}
   Regular Expression:

R.E = ((a + c + d) * b(a + c + d)* b(a + c + d))

3. Strings not containing “ba”

The regular expression for the language with no substring of ba can be formu-
lated as follows;

L = {ε, a, b, c, d, aa, ab, ac, ad, bb, bc, bd, ca, cb, cc, cd......}
Regular Expression:

R.E = ((a + c + d) * (b* (c + d))* )* + b

is it correct?
can anyone please help with the first one.

[odd edge]

@warm hare Has your question been resolved?

[noble solstice]

@dusky cobalt please open a new channel #❓how-to-get-help

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[amber schooner]

how many minutes is 0.8 hours

[warped glacier]

1 hour = 60 minutes
0.8 hours = .... minutes

cool!

[clever fjord]

Result:

48