help-49

[hallow meteor]

Hmm

Yea mb

True

But

Including prim recursion

Will give you more than presburger?

[glacial slate]

if you work in lambda calculus then I guess you can do it without the usual notion of recursion

[glacial slate]

yes, in the sense that multiplication is definable

[hallow meteor]

Oh how so? (I'm yet to get to the lambda calculus)

[glacial slate]

defining naturals as church numerals allows you do simulate a form of iteration

which allows addition

(and multiplication)

[hallow meteor]

I suppose I'll havr to read up on it first

Thnx a lot!

.close

[midnight plank]

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[woeful turret]

could someone tell what a,b,c, delta are?

[raven zenith]

a, b, c, are side lengths, delta is the area

[full leaf]

a b c are the lengths

yea delta is area

[woeful turret]

is it an important formula to remember?

cause i dont recall learning this

[raven zenith]

if you're doing comp math, potentially. for school math, I'm not sure

[woeful turret]

oh wait

i learnt a formula R=delta/S where S = semiperimeter

is that the same thing?

[raven zenith]

that's for the incenter, no?

inradius

[woeful turret]

oh

really

[raven zenith]

yeah

[woeful turret]

oh ok

ill just remember it dosent seem too hard

thank u!

.close

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[hallow meteor]

Kinda cooked trying to understand unbounded search

[glacial slate]

this probably isn't the best starting example to understand it

try define like, the indicator function of equality or something simple first

[hallow meteor]

Wait

No but

How exactly does unbounded minimisation even work

[glacial slate]

what do you mean by "work"?

[hallow meteor]

I mean what exactly is the function "searching" for

It's iterating over y's that's for sure

[glacial slate]

yeah

[hallow meteor]

It's finding the first y for which ? is 0?

[glacial slate]

the first y such that the function is defined for all values below y, and the function is 0 at y

[hallow meteor]

What function

[glacial slate]

whatever function you're doing unbounded search on

it's like primitive recursion, you need to put some functions in first

[hallow meteor]

Okay so theres this computable function g

Which takes in n+1 args

Ah so it's n args and one y arg

And we iterate over all ys

Checking at what point g becomes 0

[glacial slate]

what are you stuck on

[hallow meteor]

Wtf is f

[glacial slate]

like, is it what the resulting function does?

f is the resulting function

that you obtain by applying unbounded search on g

[hallow meteor]

What exactly is the resulting function

[glacial slate]

f(x1,x2,...,xn) returns the least y such that g(y,x1,x2,...,xn) = 0 and g(z,x1,...,xn) is defined for all z<y

we say that f is defined by unbounded search on g

[hallow meteor]

So g is taking in x1,...,xn,f(x1,...xn)?

[glacial slate]

no

[hallow meteor]

y is essentially f?

[glacial slate]

g is an n+1 arg function, yes?

it takes arguments (y, x1,...,xn)

we define a new function f, on n arguments

such that when we pass in f(x1,...,xn), it tells us the smallest y makes g(y,x1,...,xn) = 0

x1,...,xn to f

[hallow meteor]

So that function simply points to the first y which makes g = 0

[glacial slate]

yes

[hallow meteor]

I'm just not able to fit this into my understanding of semidecidability/RE

[glacial slate]

example: say $g(y,x) = |x-y|$ (where - is truncated subtraction). Then, unbounded search on $g$ gives a function $f$ such that $f(x) =$ the least y such that $g(y,x) = 0$.

[grand pond]

Desync

[hallow meteor]

We are enumerating over ys

[glacial slate]

if $y=x$, then $g(y,x) = |x-y| = 0$, and if $y$ is any smaller, then $g$ is not $0$, so $f(x) = x$ is the identity function

[grand pond]

Desync

[glacial slate]

it's RE because the result is generally partial

if you want to think about it more in terms of computability, we can just plug in each value of y starting from 0

since g is computable, we can find whatever the result is

and move on to the next y if it's not 0

since we only demand that it's semidecidable, it's ok if we never find a y; it just runs forever

[hallow meteor]

Yeah

Where in this is the guarantee of Turing recognisability

[glacial slate]

what's your definition for that?

isn't that the same as semidecidable (and RE)

[hallow meteor]

Yeah, I'm asking if this doesn't accidentally end up defining an uncomputable function

Whats the guarantee on that

[glacial slate]

well, think about how that could happen

we start with a computable function g

and all we're doing is computing it at input 0

(the other inputs are fixed)

[hallow meteor]

Yeah if a minimising input exists well eventually find it..

[glacial slate]

then computing it at 1, then 2,...

[glacial slate]

yes, and that's all that's required for RE

[hallow meteor]

Hmmm

[glacial slate]

if it's not in the set the machine does not have to halt

that's the point of semidecidability

[hallow meteor]

So for a given function g

Oh ok yeah sorry

Dumb question

I'm trying to learn functional programming

I keep getting diverted into things

Thanks tho

I think I kind aget it

Uh hope this is relevant to lambda calculus?.

[glacial slate]

yes, moreso than turing machines

PR functions/lambda calculus/FP are much more closely related

[hallow meteor]

Yeah..

[glacial slate]

turing machines are more related to procedural/imperative programming

[hallow meteor]

Yeah

❤️

y is nothiing special ig

It's just one of the n+1 inputs of g

Which we search over

And f, takes in the other n inputs and spits the first value of that remaining input to g,

Whivh makes g0

[glacial slate]

yup

it takes some inputs x_1,...,x_n and asks "what is the smallest y that makes g(y,x_1,..,x_n?) = 0?"

[hallow meteor]

So in terms of halting problem, the function g could be taking in M and x (machine and input tape) along with some y which is it's expected number of steps in which it halts

And iterate over y

If it never finds such a y it means the machine never halts

And otherwise it finds a y and M halts on x

[glacial slate]

you need to be a bit clearer with your encoding, and what g is actually doing here

you can indeed construct a halting-semidecider

Let $\varphi_e(x)$ denote the $e$-th partial computable function, and define
\begin{equation*}
h(e,x,y) = \begin{cases}
0 & \text{if the computation of } \varphi_e(x) \text{ halts in exactly } y \text{ steps},\
1 & \text{otherwise}.
\end{cases}
\end{equation*}
and define by unbounded search:
\begin{equation*}
H(e,x) = \mu y [ h(e,x,y) = 0 ].
\end{equation*}
then, $H(e,x)$ is the number of steps until the computation halts, and if $\varphi_e(x)$ does not halt, then $H(e,x)$ is undefined.

[grand pond]

Desync

[glacial slate]

you can't just say "put in a turing machine", because they're not really what we're working with; you have to phrase it in terms of PR functions

[hallow meteor]

I suppose all possible partial functions are countable

Since TMs have finite description

So you are attaching number e to this particular partial function

[glacial slate]

it's more than countable; it's effectively enumerable

you can do this via godel numberings

[hallow meteor]

whats the difference

[glacial slate]

countable just means there exists a bijection to N

doesn't mean that it's constructive, or if we do know it, that it's computable

effectively enumerable means there's a computable bijection

[hallow meteor]

Ahh

So not only bijection exists, we can compute that bijection

[hallow meteor]

So this function h is the equivalent partial function of halting problem, takes in equivalent partial function of some turing machine, and its equivalent input, and iterates over ys

[glacial slate]

for an example of the difference, take the set of godel numbers of turing machines that do not halt on input 0

this is countable, since it is a subset of N, but is not effectively enumerable, because computing the bijection would give a halting oracle

[glacial slate]

yes

[hallow meteor]

Computing the bijection means listing out all turing machines that do not halt on Input 0

[glacial slate]

yup

[hallow meteor]

How is this a halting oracle tho

It's just for one particular input no?

[glacial slate]

oh, right yeah

just define similar sets for each input and diagonalise

[hallow meteor]

Yeah and all possible inputs is also a enumerable set

Hmm

Uh

NGL still kinda confused but

I'll choose to move on

Thanks a lot for the help

.close

[midnight plank]

Channel closed

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[glacial slate]

oh, actually, you can do this even more directly

[hallow meteor]

.reopen

[midnight plank]

✅

[glacial slate]

just add a wrapper around the programs and make it take 0 as an input

so given $\langle P,x\rangle$ (program and input), we can construct a program $Q_{P,x}$ that ignores its input, and simulates $P$ on $x$.

[grand pond]

Desync

[glacial slate]

then run the 0-oracle on Q_P,x

Q_P,x(0) halts if and only if Q_P,x halts for any input, if and only if P,x halts

[hallow meteor]

[glacial slate]

basically, on any input $y$, $Q_{P,x}$ ignores $y$ and simulates $P$ on $x$ such that if $P$ halts, then it also halts

[grand pond]

Desync

[glacial slate]

since it doesn't depend on the input y, it'll halt on any input if and only if it halts on 0

[hallow meteor]

Ah

[glacial slate]

and it halts if and only if P halts on x

so we can decide if P halts on x by checking if Q_{P,x} halts on 0

and the mapping from (P,x) to Q_{P,x} is computable, because it's just simulating another machine

[hallow meteor]

kinda not understanding

So now we are dealing with this subset of Turing machines which all take 0 as input and ignore it and then do what the turing machines would do on inout x

[glacial slate]

ok, we want a halting oracle

so given a program P and input x, we want to decide if P halts on x

given P and x, construct a new program Q_{P,x}

that, when given an input y, ignores y, and simulates P on x; if P halts on x, then Q_{P,x} also halts

[hallow meteor]

Ahh I get it now

[glacial slate]

now check if Q_{P,x} halts on 0 with our oracle

and this decides if P halts on x

[hallow meteor]

So if you have a way to list all turing machines which don't halt on some input, then you include those which ignore that input and simulate all possible turing machines, and don't halt

Giving halting oracle

[glacial slate]

yes; the point is that, to check any given program and input, we can construct a new program that can be checked by the specific-input-oracle, since it ignores its input

this is a pretty common trick for computability reductions

[hallow meteor]

Hmmm

Where else have I seen this

I guess in some sort of "emptiness is undecidable" result

Anywhoo

Thax a lor

[glacial slate]

here's another if you want to practice the trick btw

and here's a similar proof

[midnight plank]

@hallow meteor Has your question been resolved?

[midnight plank]

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[viral dagger]

let $a,b$ be positive reals such that
$$\frac{1}{a}+\frac{1}{b}\leq 2\sqrt{\frac{2}{5}} \text{ and } (a-b)^2=\frac{4}{25}(ab)^3$$
find the maximum of $a^2+b^2$

[grand pond]

ihave<skissue>

[viral dagger]

xy just in case (ignore what i answered)

ok, so i squared the first inequality, since lhs and rhs are positive then there wouldnt be any problems

$$\frac{1}{a}+\frac{1}{b}\leq 2\sqrt{\frac{2}{5}}$$
$$\frac{a+b}{ab}\leq 2\sqrt{\frac{2}{5}}$$
$$(\frac{a+b}{ab})^2\leq (2\sqrt{\frac{2}{5}})^2$$
$$\frac{(a-b)^2+4ab}{(ab)^2}\leq \frac{8}{5}$$
$$\frac{\frac{4}{5}(ab)^3+4(ab)}{(ab)^2}\leq \frac{8}{5}$$
$$\frac{4}{5}(ab)^3-\frac{8}{5}(ab)^2+4(ab)\leq 0$$
$$\frac{4}{5}(ab)((ab)^2-2(ab)+5)\leq 0$$

[grand pond]

ihave<skissue>

[viral dagger]

x^2-10x+25 is always >=0, so the bottom inequality is <=0 when ab<=0, but we have that a and b are positive reals, which would be impossible for ab<=0

why did i get a contradiction?

squaring should be fine given both sides are positive, and sending stuff over (like the (ab)^2) is fine given positive

[civic lynx]

$(a-b)^2 = \frac{4}{25}(ab)^3$, you put $\frac{4}{5}$

[grand pond]

haseeb

[civic lynx]

hmm but that would give $(ab)^2 - 10(ab) + 25$ which is nonnegative

[grand pond]

haseeb

[midnight plank]

@viral dagger Has your question been resolved?

[viral dagger]

oh oops

$$\frac{1}{a}+\frac{1}{b}\leq 2\sqrt{\frac{2}{5}}$$
$$\frac{a+b}{ab}\leq 2\sqrt{\frac{2}{5}}$$
$$(\frac{a+b}{ab})^2\leq (2\sqrt{\frac{2}{5}})^2$$
$$\frac{(a-b)^2+4ab}{(ab)^2}\leq \frac{8}{5}$$
$$\frac{\frac{4}{25}(ab)^3+4(ab)}{(ab)^2}\leq \frac{8}{5}$$
$$\frac{4}{25}(ab)^3-\frac{8}{5}(ab)^2+4(ab)\leq 0$$
$$\frac{4}{25}(ab)((ab)^2-10(ab)+25)\leq 0$$

[civic lynx]

the last step should also change, because different factoring

unfortunately, the quadratic you get is still positive or zero

[grand pond]

ihave<skissue>

[civic lynx]

that's all i see for now, unfortunately :/

[viral dagger]

hm by cheating the graph looks like this

[civic lynx]

lol we pulled up the same thing

so there are definitely solutions, and the max is proabably that cusp

i wonder if you can do this via calc 🤔 or are you supposed to just do manipulations?

[viral dagger]

i think calc is allowed, just discouraged

also wouldnt you need like laplace multipliers or smth i havent learnt that yet

[civic lynx]

possibly, i might try that later

sorry i cant be fully helpful :<

[small jasper]

Let $s=a+b$ and $p=ab$. Then, $$s^2-4p=\frac{4}{25}p^3$$ $$\frac{s}{p} \leq 2 \sqrt{\frac{2}{5}} \implies s^2 \leq \frac{8}{5} p^2$$

Now, consider ||substituting the second inequality into the first equation||

As a check, the sum of the digits of the answer is 3.

[viral dagger]

is it not $\frac{4}{25}p^3+4p\leq \frac{8}{5}p^2$

[grand pond]

ihave<skissue>

[viral dagger]

oh wait wrong one

$$s^2-4p-\frac{4}{25}p^3\leq -\frac{4}{25}p^3+\frac{8}{5}p^2-4p$$

[grand pond]

ihave<skissue>

[viral dagger]

actually wait what why do you have s here

[grand pond]

Civil Service Pigeon

[viral dagger]

second inequality to first equality?

oh wait is it 30

[small jasper]

Yes

[viral dagger]

i just realized ab=5 works here

it really was the 4/5

i swear i did this problem like thrice now and i think every time i used 4/5

alr ty!!

.solved

[midnight plank]

Channel closed

Closed by @viral dagger

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[midnight plank]

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• Show your work, and if possible, explain where you are stuck.
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[nocturne harness]

Ok so i understand this video https://www.youtube.com/watch?v=0VBQnR2h8XM

[nocturne harness]

Right but Its hard to kinda imagine whats happening under the hoot

So we start at a point that is on the unit circle

we can form a triangle of x and y components and that x and y are the cos and sin of say alpha

cos a and sin a right

ok so lets image that we want to rotate once more by B

To do that wed want to stack another angle on top of a

and to do that then it means that the hypotenuse of the first triangle would become cos B that would be its base

So like Im kinda trying to visualize how x and y (cos a and sin a) get scaled as its rotated by B

So yea then the height of the triangle B sorta creates a slant

[fallow scarab]

https://www.cut-the-knot.org/triangle/SinCosAddition.gif

that one?

[nocturne harness]

Yes

It looks kinda weird but im trying to visualize how the point cos(a), sin(a) gets scaled

[fallow scarab]

what part of this diagram are you having trouble with

[nocturne harness]

Ok so its kind of weird to imagine how the hypotenuse would be cos B

[fallow scarab]

why

[nocturne harness]

does it mean that we rotated a then we technically have a new starting point on the circle and then rotated b more on it

hold on

let me make it clearer

@fallow scarab

Ok this is the first rotation

So now we can think of cos a and sin a as the starting point

[fallow scarab]

what is the question

[drowsy ferry]

Maybe this helps answer why it's cos B?

[nocturne harness]

I said I'm trying to learn how cos a sna sin a scale when rotated by b

[fallow scarab]

that's still not a question

[fallow scarab]

if this diagram didn't answer your question, ask more specifically

[nocturne harness]

Yes it is

Huh??

How do cos a and sin a scale when rotated by b?

That's a question

[nocturne harness]

So is the base of the red triangle cos B?

[drowsy ferry]

Yes, because it's a right-angled triangle with hypotenuse 1 and angle B.

[fallow scarab]

these values aren't scaled, they're rotated

[drowsy ferry]

(Oh, the labels didn't come through)

[nocturne harness]

so cos a and sin a dont reach the unit circle?

then what is that bigger traingle?

Thats has the same ratio

[drowsy ferry]

The larger triangle's apex is at (cos A, sin A)

It's on the unit circle.

[nocturne harness]

I know but can you describe what rotation does? Like how when you scale you multiply each point by a magnitude

I meant how it transforms cos a and sin a

[nocturne harness]

so then how would its hypotenuse*** be cos B

[fallow scarab]

https://external-content.duckduckgo.com/iu/?u=https%3A%2F%2Fi.pinimg.com%2Foriginals%2F60%2F04%2Fd5%2F6004d59c301fe5bf051cb1135e5ebbb6.jpg&f=1&nofb=1&ipt=59b79b71e015e2f6d2fa857f4fbed70f3d803ed56cb0e22bc9f39eafb7759c73

you multiply the starting vector (x, y) by the rotation matrix R(theta)

e.g. take (x,y) = (cos(a), sin(a))

[nocturne harness]

Sorry @drowsy ferry I meant hypotenuse

[fallow scarab]

theta is the angle from vector a to vector b in the diagram

[nocturne harness]

So how is that rotation matrix derived

Why is it the way it is

[fallow scarab]

https://www.sunshine2k.de/articles/RotationDerivation.pdf

there's also this proof which may be simpler than using matrices

[midnight plank]

@nocturne harness Has your question been resolved?

[nocturne harness]

This doesnt help me really

[nocturne harness]

again this is not showing why you need to multiply cosb and sin a or cosa and sin b

[fallow scarab]

the steps are given in reverse

[fallow scarab]

why not?

did you read that part?

[nocturne harness]

Yes I did

[fallow scarab]

so what more do you need?

[nocturne harness]

It all comes down to this

This is probably better

Let me get one i had before

[fallow scarab]

yes i shared a similar picture earlier

[nocturne harness]

Ok so what i dont understand is

[fallow scarab]

yes i agree that one is better

[nocturne harness]

basically how cosa*cosB,sina x cosB wouldnt be touching the unit circle

[fallow scarab]

define "touching"

[nocturne harness]

Nor would its hypotenuse

[nocturne harness]

On the unit circle

[fallow scarab]

well it does if beta = 0

[nocturne harness]

Huh

[fallow scarab]

B = 0 implies cos(B) = 1

[nocturne harness]

Because cosine of 0 degrees is 1 right?

[fallow scarab]

yes

if 0 < B < 90 degrees, can cos(B) = 1?

[nocturne harness]

No

[fallow scarab]

do you know the length of the hypotenuse formed by these coordinates

[nocturne harness]

Wdym

We could use phythagoreas to solve for it but if ur talking about it it shows its CosB

[fallow scarab]

yes

do you agree that if 0 < B < 90 then cos(B) < 1

[nocturne harness]

Yes

[fallow scarab]

does that answer this then

[nocturne harness]

Ok wait so that proves that

when you take a rotation a

[fallow scarab]

?

[nocturne harness]

Like this

Then we can Plot another rotation like this

right

[fallow scarab]

i don't know what you're asking. anytime you have a hypotenuse of 1, you can rotate it to align with the x-axis

[nocturne harness]

So what happens if we take rotation a cos and sin output and use that as the starting position

[nocturne harness]

Ok this is what i made

Wait wrong labels

it would technically be cos(a+b)

and sin(a+b)

@fallow scarab

So how do i go further than this

[fallow scarab]

go further to do what?

[fallow scarab]

if you're still trying to prove the sum formulas, this image has all you need.

[nocturne harness]

Help me with this

[nocturne harness]

^

Help me prove this one

With

[fallow scarab]

just follow the diagram you shared earlier

[fallow scarab]

.

[nimble leaf]

remember the hypotenuse of both triangles is 1

[nocturne harness]

I'm saying i dont understand it

[nimble leaf]

so when finding the sides of one triangle you can basically ignore the other

[fallow scarab]

which part of the diagram

[nocturne harness]

Im saying i dont under stand the way its formatted

[nimble leaf]

wdym?

[nocturne harness]

So we are looking for the opposite and adjacent side of the a+b triangle right

[nocturne harness]

In this graph

[nimble leaf]

yes

i assume we're finding sin and cos a+b

[nocturne harness]

Yes im trying to derive how the formulas for them come

[fallow scarab]

which part exactly?

[nimble leaf]

i think we're both confused as to what specifically you don't understand

why don't you try explaining your understanding of the graph from the beginning until the point you get stuck

[nocturne harness]

Ok I dont understand why the blue triangle is scaled by cos B i mean it looks weird to imagine it in a box for me id rather use a unit circle

[nimble leaf]

cause the hypotenuse is cos B

but yes we can use the circle too

[fallow scarab]

do you understand that cos(B) is a number between 0 and 1?

[nocturne harness]

So we split it at where cosB stops

[nocturne harness]

Yes i know but the graph makes it seem like cos B is 1 so it looks weird

[nimble leaf]

yes

in the box representation

[fallow scarab]

i don't see why you think so

[nocturne harness]

Oh i see where that comes from

[fallow scarab]

cos(B) is the side of a right triangle whose hypotenuse is 1 so cos(B) is < 1

[fallow scarab]

specifically the red right triangle

[nimble leaf]

yeah

[nocturne harness]

Ok so im less confused its overall from the fact of how sina * sinb gets subtracted from cosa*cosb i can see it but its weird

How would you desribe that why does sina*sinb affect the x

[fallow scarab]

what

"affect the x"

[nocturne harness]

Like how would you describe that

[nimble leaf]

draw a lilne from the top of the red triangle straight down to the bottom of the rectangle

[nimble leaf]

why are we subtracting sinasinb from cosa cosb

[fallow scarab]

the length of the top side of the rectangle is equal to the bottom side of the rectangle

[nocturne harness]

Let me just use the unit circle its hard to visualize that

So this is the box method on the unit circle

@fallow scarab @nimble leaf

Alright and so

[fallow scarab]

yes?

[nocturne harness]

Oh thats what hardy showed me

Anyways

[nimble leaf]

yes

[nocturne harness]

To get what sin(a+b) would be we need to find those 2 purple lines at the right end

the one on the top and bottom

[nimble leaf]

yes

[nocturne harness]

So we scale our original traingle by cos B

[nimble leaf]

which one is the original triangle?

green?

[nocturne harness]

Green one

YEs

Yes

[nimble leaf]

well i guess you can say that, we're using the same box setup

[nocturne harness]

Yes its just more easy for me to see now

Ok so to get the y value of a rotation

and the x value

we scale down our original points by cosB

Then lengths of our new triangle would be

cos(a)cos(B) and sin(a)cos(B)

So we can say the first step is to scale the original values by the cosine of the angle your adding

Ok then

We now know the first portion of sin(a+B) will be sin(a)cos(B)

Ok and they

then*

[nimble leaf]

now you need to find the lengths of that top-right purple trianlge

[nocturne harness]

Yes

Let me draw some right angles

@nimble leaf

Wow that took a bit

Anyways the next step is to figure out that other purple portion so we know that we first scale by cosB which give us the first piece, sin(a)(cos(B) then we find the other piece by understanding that we still have to add on the y from B so instead of only scaling by cosB we also add on a piece that takes the cos(a) and scales it by the sin(B) So basically the y part on A is scaled by the x part on B and the y part on B is scaled by the x part on a then these two Y pieces add up to give you the sum of sin(a+B)

@nimble leaf

Is this right?

Because it seems like the y part on a is scaled by the x part on b and the y part on b is scaled by the x part on a then its added

[nimble leaf]

so in the end you get sin a+b = sina cosb + cosa sinb

[nimble leaf]

yes

[twilit jetty]

its not every day you see the roblox font in a diagram

[nocturne harness]

omg it finally clicked

Multiplying sin(a) by cos(b) basically gives you how much the original Y when rotated stays vertical

So for example

[nimble leaf]

i think you're getting towards intuition that will help a lot when understanding matrix rotaion

you should check out 3b1b's essence of linear algebra series

it's not too advanced and i think you can solidify some of what you're saying mathemetically

[nocturne harness]

OH I SEE IT OH MY GOSH

Wait i can visualize it omg

as cos(B) gets smaller since that is what we first scale by we can see that that first part will get smaller so year sin(a)cos(b) basically gives how much the original y stays vertical when being rotated

and sliding across cos b you can see it

and then cos(a)*sin(b) well that gives how much the x gets affected and turns vertical

so

First scale down the deep blue dot

We get our first value, sin(a)*cos(B)

since we know that the scaled down triangle is a right triangle then its other acute angle is 90-a

which means 90-a+90=180

180-a=180

So the missing angle to add is a

Ok now we know after scaling down by cos(B) that we now move up by cos(a)*sin(B)

[nimble leaf]

yeah i think you've got it

yay

[midnight plank]

@nocturne harness Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

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[summer venture]

i have a geometry worksheet i need help with

[austere blade]

sure, could you send an image

[scenic wyvern]

please show the worksheet!

[summer venture]

it's to find the gradient using the first derivative

[graceful drum]

,rccw

[scenic wyvern]

this is calculus btw

not geometry

[summer venture]

oh sh sorry i didnt know

[scenic wyvern]

npnp

just letting you know

[summer venture]

thanks

[scenic wyvern]

um you seem to have done the first one

so what would be the issue with the rest?

[summer venture]

it's from the help with the teacher but i actually didn't understand anything

[scenic wyvern]

i see

do you know how to take a derivative?

[austere blade]

well they applied the power rule for the first one

[summer venture]

no

[scenic wyvern]

i see

[summer venture]

i dont know calculus in general anything at all

[scenic wyvern]

i guess we should start with a crash course on how to take derivatives using the power rule

[summer venture]

yeah sure

[austere blade]

do you know what a derivative is though at least?

[summer venture]

no😥

[scenic wyvern]

the simple explanation of a derivative is that it is the tangent to the curve at a particular point

[austere blade]

https://www.youtube.com/watch?v=N2PpRnFqnqY&t=246s&pp=ygUUZGVyaXZhdGl2ZSBleHBsYWluZWQ%3D. I think it's best you go first learn what a derivative is from this video, to understand it.

[scenic wyvern]

the curve of a function*

but i agree that you should probably understand what a derivative is first

[summer venture]

ok ill watch the video to understand even a little

[graceful drum]

bruh how does your teacher expect you to do this sheet

[summer venture]

bro i skipped half the grade now im Back again and don't understand a thing

[austere blade]

yeah that's a problem

[scenic wyvern]

i'm not gonna ask how you managed to skip half a grade

but let's work with what we have rn

[summer venture]

lowest in my grade fml 😭

[scenic wyvern]

anyway take a look at the video first

[graceful drum]

time to grind

[summer venture]

ok i watched the video

[scenic wyvern]

welcome back

do you get what derivatives are now?

[summer venture]

im sorry but no i didnt understand anything

[scenic wyvern]

oh dear 😅

[merry moat]

do you know the formula?

[summer venture]

nothing at all, the only thing in maths i understand is adding subtracting and dividing

[scenic wyvern]

hm.

[austere blade]

What about stuff like pre-calculus

[summer venture]

nope

[austere blade]

algebra and geometry stuff

[scenic wyvern]

do you at least understand algebra?

[austere blade]

you shouldn't really be doing calculus then

[summer venture]

i used to but forgot everything

[scenic wyvern]

sorry to ask, but how did that happen

did you quit math for some time or?

[summer venture]

dont really have a choice it's my school work and i have to take an exam like next year

[austere blade]

It is best you go visit back to those basics of algebra, geometry and trigonmetry before you do calculus

[summer venture]

nope i quit everything for some time and came back

[austere blade]

You're going to struggle a lot though

[merry moat]

and you jumped to calculus without the basics?

[scenic wyvern]

i'm surprised they let you jump grades then

[summer venture]

I know I know

[scenic wyvern]

because you seem to need a crash course in algebra

[summer venture]

i didnt the teachers dont know that i have 0 knowledge of this stuff

[scenic wyvern]

did you tell them?

[summer venture]

yeah but they dont really take it seriouslu

[scenic wyvern]

actually, do they not make you take some sort of assessment test when coming back

[summer venture]

nope just jumped straight into calculus without any idea of how i can do it

[scenic wyvern]

this is... not good honestly.

to understand what a derivative is, you need to know what limits are

and to understand what limits are, you might wanna understand what functions are

[modern sapphire]

well, whats done is done, but since you got time in hand rn, you should focus on learning algebra/pre-calc stuff for now before starting these exercises

[scenic wyvern]

and to understand what functions are, that's algebra

so you'll need to quickly brush up on algebra

[summer venture]

learned algebra in like 6th grade but forgot every single thing

[summer venture]

i have to have this worksheet done by this week

[scenic wyvern]

yea you might wanna revise that then

[mystic condor]

if you want to learn calculus you need to at least do: (they are involved in calculus)

• algebra
• functions
• coordinate geometry
• indicies and logarithms
• sequences and series (probably)
• trigonometry (probably)
• limits

probably some other stuff

[scenic wyvern]

exponents are more or less a must

otherwise you can't even use the power rule

[summer venture]

i don't even know what the power rule is😭🙏🏻

[scenic wyvern]

it's the simplest rule of taking derivatives

[austere blade]

and I assume you also don't know what exponent rules are as well?

[summer venture]

your safe to assume i dont know anything because i don't

[mystic condor]

$\frac{d}{dx}x^n=nx^{n-1}$

[grand pond]

ImOakley

[scenic wyvern]

sorry OP, but this... might be a lot of trouble if you try to tackle it right now

no offense to you meant, but you really ought to quickly brush up on algebra and ask if you have questions

[mystic condor]

but again you first need to do the other stuff

[scenic wyvern]

https://math.libretexts.org/Bookshelves
you can refer to this site for open-source texts

[summer venture]

i mean i think i know how to find dy/ dx

[scenic wyvern]

if you do and you insist on trying without the basics, do question 8

i think that's the simplest of the remaining 3

there's only one term

[sudden yacht]

How's that possible??

[summer venture]

a friend explained a bit to me

[scenic wyvern]

if you don't understand that... yeah, please start from the basics

[summer venture]

dont even know the times table..

[sudden yacht]

That's really serious!!

[austere blade]

like how to multiply and divide?

[mystic condor]

yeah definitely go back

[summer venture]

mind you im in grade 10

[mystic condor]

maths is a ladder you need to understand one thing in order to understand the next

[summer venture]

i know how to divide and maybe do long divison

[austere blade]

what about multiplication though

[sudden yacht]

And you don't know the times table??

[mystic condor]

how do you know division but not multiplication

[summer venture]

wait i forgot what that is

[sudden yacht]

.....

[mystic condor]

3x2=6

[summer venture]

i use my fingers i dont know it at the top of my head

[scenic wyvern]

oh dear.

[sudden yacht]

You can't do anything then...

[summer venture]

fr🙏🏻

[scenic wyvern]

oh so you can multiply, i guess? you just don't know the name

ok one more question

[summer venture]

ok sure

[scenic wyvern]

have you heard of and used negative numbers?

[summer venture]

what

[scenic wyvern]

oh.

[mystic condor]

numbers can go below zero

[austere blade]

Is english your native language?

[summer venture]

😭🙏🏻

[austere blade]

There might be a language barrier maybe

[summer venture]

nope maybe that's why i didnt understand the derivative explanation video

[scenic wyvern]

i see

[mystic condor]

negative numbers arent that bad, if you had a car with a negative speed like -10 it just means its going backwards

[summer venture]

oh yeah that

[scenic wyvern]

so have you heard of them? have you ever used them in math?

[summer venture]

ive heard of it but never understood ts

i did question 8 in the way i understoo

[scenic wyvern]

but question 8 involves a negative...

ok, since you said you did question 8

could you please show your working?

[mystic condor]

if youre worried about not completing work in the required time you can just tell people about this

[summer venture]

i did tell my teacher but im too embarrassed to say i dont understand anything at all

[scenic wyvern]

and you want to get stuck in this cycle of torture where you get tossed increasingly difficult stuff?

[austere blade]

You did say your teacher didn't take you seriously

[summer venture]

kill me

[mystic condor]

i mean i know people who spent ages on simple 3 digit addition problems even though they had been doing maths for years

[scenic wyvern]

,rccw

[scenic wyvern]

valiant attempt

but unfortunately incorrect

[austere blade]

it should be -6x^2 for dy/dx

[scenic wyvern]

and then you'd substitute x = -2

[summer venture]

wait what

[scenic wyvern]

yeah, that's how derivatives work

[mystic condor]

you multiply the equation by the exponent and then reduce the exponent by 1

[summer venture]

1.) dy/dx = -6 xand a small 2
2.) x= -2?

[sudden yacht]

A small 2??

[mystic condor]

small 2 = to the power of 2

[austere blade]

the small 2 just means raised to the power of 2

[sudden yacht]

Don't you even know what exponent means?

[summer venture]

...yeah😥

[scenic wyvern]

oh, dear. oh my yellow fox.

[austere blade]

he said before he that he didn't know exponent rules

[mystic condor]

do you know how powers work?

[scenic wyvern]

do you know what powers are?

[summer venture]

it's like this √?

[scenic wyvern]

that's a root

[austere blade]

that's a square root

[mystic condor]

close, thats the opposite

[scenic wyvern]

but that's the opposite of what we're looking for

[summer venture]

isnt that how Power works

[sudden yacht]

Yeah I remember, but this was the confirmation

[scenic wyvern]

ok, what IS a power?

[mystic condor]

$x^3=(x)(x)(x)$

[grand pond]

ImOakley

[summer venture]

is when someone has authority over something or idk

[austere blade]

you can think of powers as repeated multiplication

[scenic wyvern]

in the mathematical sense

[austere blade]

we meant in maths

[summer venture]

in mathermeyical sense i don't know..😭

[scenic wyvern]

ok

[sudden yacht]

2³ = 2•2•2

This is a power @summer venture

[scenic wyvern]

so do you know how multiplication is repeated addition?

[summer venture]

oh yeah wait it's 2 but 3 times?

[sudden yacht]

Sure

It's 2 multiplied by 2 multiplied by 2

[summer venture]

it says 8

[sudden yacht]

Three times as you can clearly see

[scenic wyvern]

it is 8

[sudden yacht]

What is 2•2?

[summer venture]

4

[sudden yacht]

And 4•2?

[summer venture]

uh

8

[sudden yacht]

There you go

[summer venture]

so 2and a small 3 means 8?

[austere blade]

yes 2 raised to the power of 3 is 8

[sudden yacht]

Yes.

[summer venture]

ok im smart

[sudden yacht]

Small number = EXPONENT

[summer venture]

ohhh

[scenic wyvern]

just in case you need it

[austere blade]

I do think this may be some language barrier problem though

[sudden yacht]

And number with an exponent above to the right is a POWER

[austere blade]

since you said english isn't your native and you struggle

[sudden yacht]

Maybe, but ...

[summer venture]

nono im fluent

[austere blade]

hmm ic

[scenic wyvern]

either way we're kind of at an impasse atm

[summer venture]

yeah true

[scenic wyvern]

having you jump straight to derivatives seems like a very long stretch

we'd have to cover god knows how many years of material in one help channel

[summer venture]

no it's ok i won't bother you guys

[austere blade]

plus it looks like you might have to self-teach it all

[sudden yacht]

Yeah it's like not knowing the whole alphabet and writing a dissertation

[summer venture]

atleast now i know what a exponsinet is

[austere blade]

Or I guess hire a private tutor

[scenic wyvern]

i think right, OP

[summer venture]

am i op

[scenic wyvern]

hit up that site i gave you and khan academy

you are the OP, yes

[sudden yacht]

Exponent, though

[austere blade]

yes OP just means original poster

[summer venture]

oh yeas

[scenic wyvern]

anyway, hit up that site i gave you, plus khan academy, plus other resources you can find online

[summer venture]

ok exponent is the little number at the top

[summer venture]

yes i will look into it

[scenic wyvern]

then, practice with those resources, and if you have questions about those, ask in this server

i think i'd rather have you scaffold back to calculus

rather than doing this task right now

[summer venture]

thank you for your help tho ill just ask my teacher and embarrass myself

[mystic condor]

you cant be embarrassed by a teacher

[scenic wyvern]

if a teacher embarrasses you

i don't think that's a teacher worthy of being called one

source: i'm a teacher

[summer venture]

trust she WILL embarrass me Infront of the whole class

oh wow your a teacher

[scenic wyvern]

anyway

[mystic condor]

a teacher is literally paid to help you

[scenic wyvern]

see what you can do about it

you need scaffolding and assistance

no shame in signalling for help

[summer venture]

i think the first thing i need to do is learn the times table

[scenic wyvern]

hold.

do you know how to multiply?

[summer venture]

is that like 2x6 and stuff like that

[scenic wyvern]

yes

[summer venture]

i need to count with my fingers

[scenic wyvern]

if i give you two random numbers and tell you to multiply them

can you?

[summer venture]

maybe i think

[scenic wyvern]

i don't care what you use

if you are able to multiply two numbers, and you understand the concept of multiplication

imo, the multiplication table is the least of your worries right now

[summer venture]

i dont even know the full 2 times table

[austere blade]

you don't really need to though

[mystic condor]

if you are allowed a calculator then definitely prioritise algebra and stuff

[scenic wyvern]

maybe you don't. but you can multiply.

you don't need to

[summer venture]

oh yes i can use calculator

[scenic wyvern]

right now, i advise you to learn the following

[summer venture]

ok

[scenic wyvern]

fractions, decimals, exponents, radicals (including how to work with them, and their laws/rules)

if you already know these, you can go to algebra. if not, you might wanna master these first.

[summer venture]

i think i did a bit of fractions and decimals can you give an example

[scenic wyvern]

what is 2/5 + 1/5?

these are both fractions

[summer venture]

3/5?

[scenic wyvern]

then, what is 2/5 + 1/6?

[summer venture]

17/30

[scenic wyvern]

good

what is 4/3 times 2/3?

[summer venture]

8/9

[scenic wyvern]

4/3 divided by 2/3?

[summer venture]

2

[scenic wyvern]

i think you can skip fractions for now

possibly decimals too

[summer venture]

yeah i remember fractions actually

[scenic wyvern]

but exponents - that one you need to work on

[summer venture]

exponents are just the small number on top of X or any number?

[scenic wyvern]

to the top right, but yes

[mystic condor]

this channel is about to cover an entire high school maths curriculum

world record

[lyric charm]

wait didnt we start with a derivative problem

and now op says he struggles with arithmetic? or sth?

[scenic wyvern]

we did, but turns out OP skipped ten thousand miles

[summer venture]

im dumb

[scenic wyvern]

can't say that when you quit for some time

and not really productive to say that right now

time to start the grind sir

i think for the time being, grind your basics

then come back to this later

[summer venture]

I'll try my best 🫡

[mystic condor]

i mean im pretty sure 90% of the world wouldnt be able to add and multiply fractions first try after not doing anything for ages

[scenic wyvern]

i'm that 90%
anyway, i think you can close this channel first, then reopen another one if you get stuck anywhere in your grind

[summer venture]

no no i did a lot of division and multiplying and fractions before so i kinda remember all of that

ok sure i dont even know where ill get the help ill try

[scenic wyvern]

remember the two sites i mentioned?

https://math.libretexts.org/Bookshelves
and Khan Academy

[summer venture]

khan academy?

oh yeah ill check

thank you everyone who tried to help my slow ah

[scenic wyvern]

godspeed sir

do remember to close this channel and claim a new one when you have a question

see you around if you have nothing else to ask

[summer venture]

cu

close

[scenic wyvern]

.close

[summer venture]

.close

[midnight plank]

Channel closed

Closed by @summer venture

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dry ore]

,rotate

[nimble leaf]

it was fine the first time

[dry ore]

Oh 😭

Nah ik how to do js a qn

[mystic condor]

id switch axes first because when one function has top and bottom its confusing

actually thats dumb

[dry ore]

So if I int with respect to X from O to C I always take the + x axis right ? Like the 2nd quad here?
If let's say curve was not symmetrical is there any way to int such that I wld want it in the 3rd quadrant

[mystic condor]

the first thing to do is find the points a c and b

[dry ore]

Because curve is symmetrical we are able to take bottom area=top area from O to C right but technically when we integrate we are getting the area in 2nd quad not 3rd

[dry ore]

I alreadly did it

[mystic condor]

ok good

[dry ore]

Js some conceptual clarifying yea

[mystic condor]

i think just take a regular integral from C to O, then the area of that top triangle

then one from O to B and subtract the empty triangle

[dry ore]

That's not my qn bro 😭

[worldly canopy]

If it were me I’d actually find A first and transform the whole graph such that A lies on the x axis

Like move it all down

Then integrate the bottom line equation in the relevant bounds to find the area under the graph, take the mod of this

Then subtract the right triangle with hypotenuse AB

[midnight plank]

@dry ore Has your question been resolved?

Channel closed

Closed by @dry ore

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[timber rampart]

@visual summit

[timber rampart]

Bro I cant talk

In chat

Whyd u make me do that bru

[visual summit]

Ok close this and go to #bots

[scenic wyvern]

??

[past jay]

???

Okay?

[austere blade]

?

[visual summit]

Type .close

[mortal mirage]

why bots??

[visual summit]

Gave him the studying role

[past jay]

And why did you make him claim a help channel?

[visual summit]

U get out of here

[visual summit]

I did not

[austere blade]

@timber rampart you can type ,iamnot studying in bots to remove the study role

[timber rampart]

Oh m

[visual summit]

Close this these are for math help

[scenic wyvern]

!done

[midnight plank]

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

[past jay]

.close

[midnight plank]

Channel closed

Closed by @past jay

Use .reopen if this was a mistake.

[timber rampart]

.close

[past jay]

Please don't do that again

People need these channels

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[jade magnet]

1.i

[lyric charm]

make a table of n vs. T(n)

with n = 1, 2, 3, ...

[jade magnet]

oh ok

ill try

[vivid yoke]

Trace back value of T(k+1)=T(k)+k+1=T(k-1)+k+(k-1)+1+1=...

Huh

What's wrong with this

[jade magnet]

i was trying to do this originally and uh

yeah i struggled

ok im noticing it's a series formula

[vivid yoke]

1+2+...+k=(k+1)(k+2)/2

[jade magnet]

it's 1, 3, 6, 10, 15, 21, 28 etc

1+2+3+4+... n times

[vivid yoke]

!noans

[midnight plank]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[quartz heart]

Oh

Sorry

[vivid yoke]

Not that serious I'm just joking

Yeah I also got 12

[jade magnet]

(n^2 + n)/2 ?

so it's just (n^2 + n)/2 > 100 which is a quadratic?

[vivid yoke]

Yeah sum from 1 to n is (n+1)(n+2)/2 if iirc

[jade magnet]

i mean you can test it

(1+1)(1+2)/2

(2)(3)/2

3?

[nova pike]

It’s n(n+1)/2

[quartz heart]

Isn't it n(n+1)/2

[jade magnet]

T(1) is 1

not 3

so yeah

[vivid yoke]

Oh shot

Damn

[jade magnet]

why didnt you just check if it worked bruh 💀

[vivid yoke]

I'm rusting

[nova pike]

You can prove this via mathematical induction btw

[jade magnet]

yeah you can

but im not going to rn because i'd like to save time

for the later questions

[nova pike]

Yes, but you don’t need to solve the quadratic. Just approximate values of n given the inequality. Even by the problem statement, you can tabulate values of T(n) and find when it’s greater than 100.

[nova pike]

that’s good

[vivid yoke]

T(k)=n(n+1)/2+n, no?

[nova pike]

How? And also you mean T(n) I’m pretty sure

[jade magnet]

bruh

we already did the question

[vivid yoke]

T(k+1)=T(k)+k+1
=T(k-1)+k+(k-1)+2
=T(k-2)+k+(k-1)+(k-2)+3
.....
=T(1)+[k+(k-1)+....+1]+k

T(k+1)=T(1)+k(k+1)/2+k

[jade magnet]

im pretty sure there's no +k in the end

it's just (k)(k+1)/2

[nova pike]

That’s correct. However previously you wrote a formula for T(k) in terms of n which is completely different

[vivid yoke]

It's a typo lol

[nova pike]

Okay then you’re good

[nova pike]

+k comes from the k +1’s. He basically backtracked the formula all the way in terms of T(1)

[midnight plank]

@jade magnet Has your question been resolved?

[midnight plank]

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[woeful turret]

im not able to start

i only took parametric coordinates for P and Q

[fallow scarab]

Did you draw a picture

[woeful turret]

i tried

but it wasnt coming out wel

ill send in a few mins

[jade magnet]

try plotting it in a graph?

thats usually what i'd do

[last slate]

If a triangle is inscribed in a circle with one side as diameter, then the angle opposite to the diameter is a right angle

put the points

of origin

and another ones equal to area

[woeful turret]

oh what

i didnt know this property

this is what my sketch was looking like

[last slate]

u got like any graph with ques

[woeful turret]

huh?

[last slate]

its right now name the points

and

[woeful turret]

yeah i named them

[last slate]

put them equal to area

[woeful turret]

oh u mean t1,t2

[last slate]

and solve using section fromula

[last slate]

yh like dat

[woeful turret]

wait what

section formula

[last slate]

😭 udk like

nvm i cant explain them

[woeful turret]

ik it but how to use it here

[last slate]

its also u can find zeroes by making it a polynimal

@woeful turret

[woeful turret]

yeah

[last slate]

do it -

[woeful turret]

bro i cant understand what ur saying 😭

its also u can find zeroes??

hindi m baat kro

[last slate]

ehh zeroes = alpha, beta

alpha beta aata hn?

[woeful turret]

h par konsa function ka zeroes alpha,beta?

[dreamy lichen]

u can also find the area using thales

[woeful turret]

what is thales :(

[last slate]

..

[dreamy lichen]

this thing

triangles above diameter are right

sth like that

[woeful turret]

first could u confirm that POQ will be a right angle? is it a property of parabola?

[dreamy lichen]

that tells you that POQ is right

[woeful turret]

oh shit lol

my bad

[dreamy lichen]

not a property of parabola, rather a property of circle

[woeful turret]

yea yeah got it

was just being dumb

ok now that its a right triangle

how to find P,Q?

oh ok m1m2 = -1

[dreamy lichen]

i'd try to introduce some variables

[woeful turret]

that is one condition

and another condition i can get from area in the determinant formula

that should do it right

[dreamy lichen]

i mean sure, if you wanna work with coordinates....

working with side lengths might be easier though

you can very very easily get the side lengths

[woeful turret]

oh yeah right

[dreamy lichen]

and then computing the coordinates is easy, though a bit tedious

[woeful turret]

yeah ur right

sorry im at the end of my solving for the day

so im making mistakes like this

i think ill get it now

thank you!

@last slate thanks bhai

[last slate]

:))

[midnight plank]

@woeful turret Has your question been resolved?

[midnight plank]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[half tangle]

Hi, I need to prove C (a 2D ring) is an open set, but can't find a suiting radius of the neighborhood for all points inside the region, so I thought of dividing the ring into two rings (A and B), and then proving the points P0 (in A) and P1 (in B)are in C. Is this valid? is there an easier way to prove this?

[runic hamlet]

how would you do it for the interval (2,4)

there are also other options like considering it as an intersection of two open sets

or as the preimage of an open set under a continuous function

[midnight plank]

@half tangle Has your question been resolved?

[mortal falcon]

This was my first thought too

You can look at the closed disk with radius sqrt(2) and take its complement, and intersect that with the open disk of radius 2

If you wanna do the basis constructoon you could just make the radius of each ball < the min of the two bounds you found

instead of having to do any cases

[glacial slate]

if you wanted to directly prove that it's open from the definition and find an open ball around any point in the set, you could also look at the ray from the origin through the point, and look at the distance from your point to the edge of the sets

then you can find the radius of the open ball from there and proceed via the triangle-inequality

[half tangle]

You mean something like this?

[mortal falcon]

mhm yup

[half tangle]

But the problem is that there are 2 distances, you could measure the distance to the inner or outer circumference depending on which one is the smallest

[glacial slate]

yes, just take the minimum