help-27

[simple grotto]

Note projected x and y are not on plane x and y and I will need on plane x and y. X E.

You can ignore dot. X E.

A not lying thing, read bio no links. X E.

Thanks. X E.

Channel 3^3. X E.

Where is doctor strange the -1st above?

maybeJosiah, the 0th. X E.

I have tried cosine*distance in various ways for this, not working, how do I with distances find how far to a right triangle?

[craggy dagger]

i'm still figuring out what the goal is, and trying to figure out what the code is doing

[simple grotto]

The goal is to make x, y, and t known to relate somehow more than just scaling. X E.

[craggy dagger]

is the idea of the code that the output should be a 0 vector?

[simple grotto]

Yes, to prove that we can combine vectors to get there. X E.

Also, the scaling and how it exists is important. X E.

let t=[cosine(x,point,plane)*distance(point,plane),cosine(y,point,plane)*distance(point,plane)];
console.log(t,setx,sety);
//X E

That even gives wrong results. X E.

[craggy dagger]

you may want to use the fact that cosine(a, b, c) should be the same as dot(minus(c, b), minus(a, b))/(distance(b, c)*distance(b, a))

so, in other words:
con[0] is the signed length of projection of the vector from point to view onto the vector from point to x
con[1] is the signed length of projection of the vector from point to view onto the vector from point to y

wait it's the other way around

[simple grotto]

Like this?

con[0] is x of view onto plane with point and con[1] is y. X E.

[craggy dagger]

[note: i'm probably reading your code much more than your words]

[simple grotto]

Either way, neither is finding the value of line closest to point. X E.

Of the dot or distance cosine. X E.

[craggy dagger]

i'm not sure what "value of line closest to point" is exactly

[simple grotto]

That is not a correct thing anyway. X E.

Not what I am finding. X E.

[craggy dagger]

yeah okay now I'm pretty sure what's going on with the code

now the problem for me would be figuring out what you are asking about

in fact, it really didn't matter what values you put into con

@simple grotto i'm still trying to figure out what you are asking

[simple grotto]

This is an insane truth. X E.

VC does not matter. X E.

[craggy dagger]

yeah, since the amount you added in the cent variable is later completely subtracted out by:

con[0] over here
scale(vx,s*(setx-con[0]))

and con[1] over here
scale(vy,s*(sety-con[1]))

inside this expression
console.log(minus(add(view,add(add(scale(minus(cent,view),s),scale(vx,s*(setx-con[0]))),scale(vy,s*(sety-con[1])))),proj));

[simple grotto]

The point is that any 3 ways will do. X E.

[craggy dagger]

not sure what "any 3 ways means"

like it works with any 3 points?

[simple grotto]

A vector x+vector y+vector of any point from P to plane with scaling can work. X E.

[craggy dagger]

let me try to figure out what you are asking: do you think the code is showing a nontrivial relation between some variables, if so, which variables in the code?

if not, then I must have guessed wrongly. I would be quite confused on what you are exactly asking.

[simple grotto]

Between view, proj, vx, and vy and some trivial point on plane. X E.

[craggy dagger]

hmm, now that you know that con[0] and con[1] don't matter, perhaps you can set them to 0 and simplify the code to understand what's going on?

[simple grotto]

The scaling factors do matter to an extent though. X E.

[craggy dagger]

The idea is that no matter what they are, the amount these two values add to cent is eventually subtracted in the console.log line. I mentioned it up here as well.

I'm thinking that these two factors are simply complicating what is happening in the code

[simple grotto]

This works. X E.

You are correct. X E.

Would a measure of how off a way is from another way help?

[craggy dagger]

do you see what the code is doing now? Do you see why it's just recomputing proj with another approach?

[simple grotto]

3d like?

[craggy dagger]

im not sure how such a measure helps, and with what exactly that it helps

[simple grotto]

Given a distance to one point and a distance to another, the off amount could indicate how much it should go in the one way to form a right angle. X E.

Another geometric way to compute how much in one way a thing travels. X E.

a way to find t. X E.

Basically define a way as one dimension straight forward. X E.

Maybe not though. X E.

One thing I learned, cosine does not work like this. X E.

[craggy dagger]

Hmm are you trying to find the projection of a point on a plane

[simple grotto]

Yes. X E.

By another point. X E.

[craggy dagger]

Firstly you should consider finding a pair of perpendicular vectors that lie in the plane

not just any vx vy

and then normalizing these to have length 1

[simple grotto]

Not like I can without cosine. X E.

Technically anything is perpendicular nearly. X E.

[craggy dagger]

Conceptually you may want to use the idea of dot products and projections

[simple grotto]

Like what?

[craggy dagger]

First you start with 2 vectors in the plane, and then you consider the length of the projection of one vector on the other

To do so, you use dot product

Then you want to subtract the projection from a vector to make the two vectors perpendicular

so if the two vectors are a, b

replace b with b - a * dot(a, b) / dot(a, a)

[simple grotto]

And that will stay true to the plane that was?

[craggy dagger]

That would stay within the plane because it is a linear combination of a and b

[devout snow]

@simple grotto Has your question been resolved?

[simple grotto]

Well, Gaussian elimination?

Does Gaussian elimination work for inequalities?

Bed time. X E.

.close

[devout snow]

Channel closed

Closed by @simple grotto

Use .reopen if this was a mistake.

[devout snow]

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[sand quarry]

In a certain town consisting of $n$ citizens, there are $m$ distinct clubs. The clubs are formed according to the following two strict rules:
\e{itemize}{
\ii Every club has an \textbf{odd} number of members
\ii Every pair of distinct club shares an even number of members
}
Prove that the number of clubs cannot exceed the number of citizens ($m \le n$).

[sand quarry]

so uh

dabbled with inclusion-exclusion

didnt work

tried generating functions

got nowhere

pigeonhole stuff didnt work either

im lost

blud thinks he is him after ing me

[river solar]

i dont even know what this means

what is a distinct club

(btw i will be of no help im just excited by the problem)

[devout snow]

@sand quarry Has your question been resolved?

[mortal oak]

@sand quarry Tell me what did u understand in this question

[sand quarry]

idk how to answer that as I haven't even developed a strategy to solve it

[mortal oak]

Ok

[light hazel]

bro is lex fridman

[mortal oak]

I think this might not be the complete question

We can't figure out what clubs are or what it's relation with citizens

[light hazel]

the question seems trivial? Every club must have at least 1 person so clearly number of people is less or equal to number of clubs

[mortal oak]

Read the 2 conditions

[light hazel]

2nd condition doesnt matter

[mortal oak]

First one states odd members required then what was the need for 2nd condition?

[light hazel]

2nd condition is trivially true without even needing to be said since odd number + odd number is even

[mortal oak]

Adding any pair of odd no. gives an even no

[mortal oak]

I dont think so

[vast zenith]

We don’t get any new info from the 2nd condition

[mortal oak]

I think clubs and distinct clubs are 2 different things

[light hazel]

2k+1 + 2n + 1 = 2(n+k+1)

[vast zenith]

We can’t have a pair of non distinct clubs: that doesn’t really make sense

[mortal oak]

I know this bruh

[sand quarry]

but the size of an intersection is not the same as the sum of the size, though

[untold ravine]

you have to prove that every club has at least one distinct member for that

[vast zenith]

Wait

Clubs can have the same members

So the 2nd statement says the intersection of any two (distinct) clubs is even

[mortal oak]

Nc

[vast zenith]

Not that the sum of all their members is even

[sand quarry]

yes

you got it

[mortal oak]

This was what we were missing

So intersection must be even

[sand quarry]

yes

[mortal oak]

But wait

To prove this, we need to take the extreme conditions only

So, we need as many clubs as possible

So we avoid intersection of clubs

Now to maximize it even further, take 1 citizen per club

[glossy dew]

ok so the scariest part of this question is that a member can belong to more than one club

i see

[mortal oak]

Hence proved ig?

[vast zenith]

this only shows an edge case

We want to show it generally

[glossy dew]

surely the question setter wouldn't have made it a trivial question?

[mortal oak]

We need to take the edge case only

[vast zenith]

And why is that

[mortal oak]

Because we need to maximize the no. of clubs?

[vast zenith]

I think I understand what you’re getting at: we would rather show the maximum number of clubs is n

[mortal oak]

If 2 clubs intersect, they share a minimum 2 citizens thus taking down no. of clubs by 2

[vast zenith]

But then I wonder if 0 is counted as an even number here: that is if each club has an individual number can we really say the number of shared members is even?

[mortal oak]

Yes we can

It is not specified that clubs have shared members

So, this question is straightforward logic or u could say common sense

I can't see any use of mathematical principles here

[glossy dew]

not really getting why you only considered that there should be one member per club if the number of clubs is greater than the number of members

[sand quarry]

ok let me write this in proper notation so we can get over this ambugity

[glossy dew]

surely there could be three members per club

[mortal oak]

So are empty clubs possible?

[celest frigate]

No because they have an odd number of members

[glossy dew]

a member can belong to two clubs

[mortal oak]

They need to have odd no. of members

[mortal oak]

Take 2 clubs

[glossy dew]

it doesn't seem like the question is designed to be trivial

[mortal oak]

Suppose 3 members in each club

2 members are shared

Here, no. Of citizens = 4 and no. Of clubs = 2

Now take 4 clubs and put 1 members in each

Here, no of citizens = 4 and no. Of clubs = 4

[celest frigate]

Well but as blurple galaxy pointed, we need reasoning as to why we cannot have more clubs

It is not necessary that a member belongs to one club only

A member can belong to any number of clubs

[mortal oak]

So to maximize no. Of clubs, we take 1 meber per club

[mortal oak]

Here is the reasoning

[celest frigate]

Why

[sand quarry]

Let $n\in\Z_{\ge1}$ and let $[n] = {1,2,\dots, n}$ be the set of citizens. also let $F = {C_1, \dots, C_m}$ be a family of $m$ distinct subsets of $[n]$.

\medskip
If the family $F$ satisfies the following two conditions:
\e{enumerate}{
\ii $\abs{C_i}\equiv 1 \q (\op{mod} 2) \q \forall i \in {1,2,\dots, m}$
\ii $\abs{C_i\cap C_j}\equiv 0 \q (\op{mod} 2)\q \forall i,j \in {1,2,\dots, m}, i\ne j$
}
then $m\le n$

[mortal oak]

What's this? The solution?

[sand quarry]

no

[celest frigate]

The question statement

[sand quarry]

i just restated the question to the best of my ability

[mortal oak]

Ok didn't see mod 2

[mortal oak]

Lemme write smth and send

I dont know how to use texit

[dry oxide]

#help-8 message i accept his proposal. step into pure's presence and become a LaTeX apprentice.

did you check out the cheatsheet pinned in #latex-help? like he asked you to

[mortal oak]

Yea

I will check it out

There you go @sand quarry

[sand quarry]

i dont think this is correct

[celest frigate]

I found another way. give me a moment

[glossy dew]

i is not j+1..

i is jC2

[sand quarry]

^

[mortal oak]

Why tough?

[sand quarry]

also your inclusion exclusion is not correct

[mortal oak]

Well whatever

It doesn't matter

[sand quarry]

You cannot just stop after subtracting the pairs. Without the triples, quadruples, etc., the equation does not yield the total number of citizens

[mortal oak]

Hmmm

Do you know one thing about inclusion exclusion??

[sand quarry]

also 2(C) + i > i is a tautology

it proves nothing

[celest frigate]

consider a club $C_1$, choose a member $a_1$ and call it the characteristic element of $C_1$. For any other set if $a_1 \in C_i, i \neq 1$, then it shares an even number of elements with $A_i$, so the set must have some other element and it must also have soem characteristic element (the remaining element ensures that it exists, but it may be different). Following this, we must have that each set has a unique characteristic element

[mortal oak]

It is that the first term is always greater than the final value

[celest frigate]

I could've written it better but I think it is rigourous enough. You get the idea.

[woven radish]

Adhi

[celest frigate]

If not then it contradicts the problem statement, each club has an odd number of elements and the union of two sets is even @sand quarry

[mortal oak]

So this basically comes down to inclusion exclusion
We need to prove that the first term is greater than or equal to the final value we get after applying inclusion exclusion

Am I right?

[celest frigate]

How would you apply Inclusion Exclusion though, you have to invoke the intersection of three clubs as well. A member can be in more than two clubs

[runic prawn]

say you swear

https://tenor.com/view/larry-larry-baby-ai-baby-laughing-ai-baby-baby-gif-17394836155970552795

[glossy dew]

ooooh this is interesting

[mortal oak]

That's why inclusion exclusion

[celest frigate]

But how do you represent the values of the intersection of three sets/four sets/...

[mortal oak]

U know the formula for intersection and union of 3 sets?

It came from applying inclusion exclusion

[sand quarry]

ok but question

[celest frigate]

I know that, but how do you put it rigourously in the context of this problem. It would become a mess. You would go to 4, 5, 6, .., m. It produces a long solution I suppose

[mortal oak]

Yea right

[runic prawn]

!nogpt

[devout snow]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[sand quarry]

from what i understand your argument is like, "if $C_i$ shares $a_1$ with $C_1$ the interesection is even so $C_i$ must have some other element which we can call $x$. So we can pick $x$ as the characteristic element for $C_i$"

\medskip
but what if $x$ is already the characteristic element for a third club $C_k$?

[celest frigate]

$x$ cannot be if it shares an odd number of elements with that set.

[woven radish]

Adhi

[glossy dew]

yeah that part is quite ambiguous

[celest frigate]

Go algorithmically, first do $a_1$, then do $a_2$...

[woven radish]

Adhi

[glossy dew]

but maybe the characteristic element part is possible

[celest frigate]

I proved it disjointly but using the algorithm, it implies that it holds for all sets. If two sets have the same characteristic element, they would share an odd number of elements. They cannot be equal sets because they share an even number of elements.

In the case there are less sets, we cannot define any element to be the characteristic one but there must be some element that holds

[sand quarry]

wait but like im still confused

if C_i is composed entirely of people who are already characteristic elements of other clubs your thing runs out of options even if m <= n. Am i misunderstanding smething?

[celest frigate]

@sand quarry I think I see why are confused. Suppose 1 is the characteristic element of set A. Consider set B,C,...D contain A. They share an even number of elements with A does not imply that the element leftover is the characteristic element. However, if there is no such element, then A=B because they would have all members equal, and if C and B had an entirely different characteristic element, and did not have any other elements, they would become equal. So two unequal sets must have different characteristic elements.

B and C must share a different pair of elements with A, if the remaining elements are same, the characteristic element would be chosen from the shared pair

Think of it as choosing $a_1$ as the characteristic element and then wiping out $a_1$ from all sets and also wiping out $A_1$, then the other sets must still be distinct because they shared an even number of elements with A. It cannot be that a set has elements 2,3,4 and another set has 1,2,3,4, this is by definition A which we wiped out

[woven radish]

Adhi

[celest frigate]

My statement above and here is just the general idea, not exactly a well-written proof. I am too lazy to do that right now.

[mortal oak]

So did you guys find a way to prove this mathematically?

[celest frigate]

@mortal oak what do you mean mathematically. The arguement I presented is mathematical

[mortal oak]

We already concluded this right? We just gotta prove this concept mathematically directly or indirectly

[celest frigate]

I still do not see a solution with PIE

[mortal oak]

U kinda explained it out

It works fine ig

[celest frigate]

We do not know how many elements intersect

We only know the congruence wrt 2

[mortal oak]

make it general

like m

[celest frigate]

How general

For intersections of 3,4,5...,m sets

[mortal oak]

yes

[celest frigate]

But how

[mortal oak]

that was the problem we were facing right?

[celest frigate]

Why did you think of PIE though?

What motivated it? I would be interested in knowing

[devout snow]

@sand quarry Has your question been resolved?

[mortal oak]

I dont know; maybe cause lex mentioned it??

[mortal oak]

here along with the question

[devout snow]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
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• 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.

[jaunty kettle]

Example 16 -

I have no idea what to do

[willow helm]

,rccw

[jaunty kettle]

Waiting...

[sturdy fiber]

RD SHARMA

oh its 10th

forgor bpt or thales theorem

[rare kernel]

well see example 13 on page 12.40

[sturdy fiber]

check page 12.40 and send photo

or dont

his choice

[devout snow]

@jaunty kettle Has your question been resolved?

[jaunty kettle]

9th

[jaunty kettle]

It's not useful

Further complex

I think it could be understand simply

[glossy dew]

use midpoint theorem

[jaunty kettle]

How?

[celest frigate]

What have you done so far @jaunty kettle

[jaunty kettle]

There is no use of midpoint theorem in it

[glossy dew]

okay do you have a diagram or something?

[jaunty kettle]

Wdym ;-;

[glossy dew]

just a quadrilateral with the midpoints connected

[jaunty kettle]

No digram

Ok lemme send photo

,rccw

[celest frigate]

@jaunty kettle this is a different question? Just clarifying

[inner sorrel]

pg no?

[jaunty kettle]

Yes, but in the example 16 - it's written that the solution is in 13th example

[jaunty kettle]

12.4

[inner sorrel]

12.4 what

[jaunty kettle]

Page no.

[celest frigate]

@jaunty kettle is your question this one or the first one?

[jaunty kettle]

My question -

[glossy dew]

this doesnt seem to be the right diagram

[celest frigate]

@jaunty kettle the midpoints form a rhombus, what does this imply. Hint: ||Definition of Rhombus||

[jaunty kettle]

Ah wait

Lemme think a little

[jaunty kettle]

You didn't get it brother/sister

See the solution area of example 16

[celest frigate]

Also, construct a diagram while youre at it (you dont have to share it) and ignore the solution you have provided. We will go by reasoning.

[jaunty kettle]

I have created one. It don't have something useful. We can just draw a common parallelogram shapes and label it

Nothing else to do

[celest frigate]

Good

[celest frigate]

Now think about this

[jaunty kettle]

I don't understand that

[celest frigate]

Huh?

What does a rhombus mean?

[jaunty kettle]

A four sided parellogram with equal sides, equal angles

[inner sorrel]

Equal angles?

Which angles are equal

[sturdy fiber]

that is how the numbering works in rd sharma

[jaunty kettle]

Aren't all angles equal

[celest frigate]

No

[jaunty kettle]

Fr I don't either understand it well

[inner sorrel]

ik but 12.4 for me doesn’t have this

[sturdy fiber]

you will get used to it lol

[inner sorrel]

It has only properties

[jaunty kettle]

You may have second book. Rd have two hardcopies

[sturdy fiber]

well the 9th and 10th ones have 2 hardcopes

[inner sorrel]

I have the bigger one, maybe you have the smaller omega

[celest frigate]

@jaunty kettle you may check the definition of a rhombus from your text or internet and then answer

[jaunty kettle]

Yes

[inner sorrel]

aha

[sturdy fiber]

a shape with all 4 sides equal

one sec

[jaunty kettle]

Done

[celest frigate]

Yes, so what is a rhombus

[sturdy fiber]

I think i remember now

[jaunty kettle]

A quadrilateral with equal sides, opposite angles equal, diagonals bisect each other

[celest frigate]

@sturdy fiber please avoid giving solutions. Let OP work through it

Yes

[sturdy fiber]

Sry

[celest frigate]

Now what condition would be necessary to imply that the sides have equal lengths @jaunty kettle

[jaunty kettle]

Equal mid points

Of the outside

Quadrilateral

[celest frigate]

What do you mean

[jaunty kettle]

Midpoints of PQRS quadrilateral must be equal

[celest frigate]

???

[jaunty kettle]

So sides of rhombus will be equal also

[celest frigate]

But what is necessary so that is true

[jaunty kettle]

Ok

Wait

[celest frigate]

In other words, how do you show it

[jaunty kettle]

I don't know

Therefore asking for help

[celest frigate]

Draw the diagonals

Try and deduce a relation between the sides and diagonals

[jaunty kettle]

Ahm diagonals perpendicular?

Or bisect each other

[glossy dew]

surely you would have done some questions with quadrilateral midpoints connected?

[celest frigate]

I need a relation between the sides and diagonals

[jaunty kettle]

Yes

[glossy dew]

what did you do there

[jaunty kettle]

I don't know

[celest frigate]

Ok try and consider triangle PQR

[glossy dew]

like proving the quadrilateral formed by joining midpoints is a parallelogram

[celest frigate]

Do you see anything

[jaunty kettle]

Ah maybe

[celest frigate]

Have you studied similar triangles or the basic proportionality theorem

Either will do

[jaunty kettle]

The diagonal is perpendicular to the base of triangle "PR" side

[celest frigate]

I need a relation with the sides

Not angles

[celest frigate]

@jaunty kettle this is a hint

[jaunty kettle]

Diagonal parellel to the side

[celest frigate]

Yes

And what else. I need one more thing

Hint: What is the ratio of line joining midpoints and base diagonal

[jaunty kettle]

Ahm

[jaunty kettle]

Perpendicular?

[celest frigate]

Use similar triangles or basic proportionality

No

The ratio

[jaunty kettle]

Don't know these

[celest frigate]

Also they are parallel

They cannot be perpendicular

[celest frigate]

Trigonometry?

[jaunty kettle]

No

[glossy dew]

okay the lack of diagram is really weighing down

[celest frigate]

We would ideally need similar triangles or basic proportionality theorem

[glossy dew]

the conversation

[celest frigate]

OP has a diagram he says

[glossy dew]

where

[celest frigate]

On his paper

[glossy dew]

yeah but how do i explain it on his diagram

[celest frigate]

Fair

@jaunty kettle share your diagram

A picture will do

[jaunty kettle]

Ok

[celest frigate]

@glossy dew we would still have to invoke similar triangles. OP has not studied them.

[jaunty kettle]

Try to tell me something. Maybe I know it

I know some higher concepts also

[celest frigate]

Do you see that the line joining the midpoints is half of the diagonal parallel to it

[jaunty kettle]

Yes

[celest frigate]

Why is it so

[glossy dew]

awesomeee

now apply midpoint theorem like you normally do for these kinds of questions

starting with triangle PQS

[jaunty kettle]

Ok

[celest frigate]

Now we have shown that it is a parallelogram

[jaunty kettle]

Don't know the proof 😭. I agreed because they seems that way

[jaunty kettle]

How?

[celest frigate]

We know opposite sides are parallel and equal because of similar triangles

[celest frigate]

Thats interesting

Its not ideal but whatever. Do you see the parallelogram

If yes, try figuring out the next step, it is ||Now it suffices to show that any two adjacent sides are equal. Why?||

[jaunty kettle]

h wait guyz, I am understanding from someone

[jaunty kettle]

.close

[devout snow]

Channel closed

Closed by @jaunty kettle

Use .reopen if this was a mistake.

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[rough nova]

What tools can i use to find the minimum of complex expressions?

[uncut crow]

wolframalpha?

[rough nova]

Well for some reason it didn't work for my expression

5((a²/(b²+c²)+b²/(c²+d²)+c²/(d²+e²)+d²/(e²+f²)+e²/(f²+a²)+f²/(a²+b²))+6(abcdef/(a+b)(b+c)(c+d)(d+e)(e+f)(f+a))^(1/3) for positive a,b,c,d,e,f

It's 6-variables, maybe that's a problem?

[celest frigate]

wow

[rough nova]

I'm trying to make an inequality and i'm guessing that this has a minimum of 9 but i'm not sure yet

[lyric hornet]

,w FindMinimum[{5((a²/(b²+c²)+b²/(c²+d²)+c²/(d²+e²)+d²/(e²+f²)+e²/(f²+a²)+f²/(a²+b²))+6(abcdef/(a+b)(b+c)(c+d)(d+e)(e+f)(f+a))^(1/3), a>0, b>0, c>0, d>0, e>0, f>0},{a,b,c,d,e,f}]

[woven radish]

Sending query to Wolfram Alpha, please wait.

Something went wrong!

An unexpected error occurred while processing your command!
The error has been reported and should be fixed soon.
If the error persists, please contact our friendly support team at our support guild!

Exception

KeyError: 'success'

[lyric hornet]

KeyError: 'success'

[willow helm]

Too good to be true I guess

[rough nova]

Lol

[lyric hornet]

maybe put that into mathematica?

do you work/attend an educational institution?

then they might offer mathematica to you

[uncut crow]

they are hiding the answer from us

[lyric hornet]

Might've foudn it but ra out of computation time?

but that's doesn't make any sense

[rough nova]

Any other tools?

[devout snow]

@rough nova Has your question been resolved?

[rough nova]

.close

[devout snow]

Channel closed

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[crude spruce]

Are you sure?

[rough nova]

Do you have a counter example?

[crude spruce]

I forgot but there was an inequality that generalized nessbit's

for n = 6

[rough nova]

Shapiro

.reopen

[devout snow]

✅ Original question: #help-27 message

[rough nova]

Oh nvm

I see it now

[crude spruce]

yeah that does seem like it

so your min is at least 15

[rough nova]

I'm dum dum

.close

[devout snow]

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[devout snow]

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[simple grotto]

Okay, so I have $P+(O-P)t+(X-O)x+(Y-O)y=Q$ where capitals are known points and lower case are scalar unknowns and $Q=B+sum((Q_i)q_i)$ with same deal, upper case known, $q_i$ unknown. I will later do bounded $q_i$ like >=1 and <=2 per $q_i$. How to solve stuff like this in 4d+? What math like generalized Fourier Motzkin elimination or any other can you recommend? First step whole space, second step, >= and <=. This is for computing software so best algorithms? X E.

[woven radish]

maybeJosiah

[simple grotto]

Note, never more $q_i$ than space dimensions. X E.

[woven radish]

maybeJosiah

[simple grotto]

Note $Q_i$ are vectors. X E.

[woven radish]

maybeJosiah

[simple grotto]

I can take $m>=q_i>=n$ and set the other variables to $q_i$ in the original and get two inequalities per each equation. X E.

[woven radish]

maybeJosiah

[simple grotto]

I can do Fourier Motzkin elimination with just inequalities from this. X E.

Always with an equal. X E.

What would be most efficient for this?

Fourier Motzkin elimination is overkill, better things?

[devout snow]

@simple grotto Has your question been resolved?

[devout snow]

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[fickle magnet]

excuse moi

[fickle magnet]

this is absolutley wrong ig

this is derived from bernoulli trials, original question is

2nd part

healp plaeij

[devout snow]

@fickle magnet Has your question been resolved?

[devout snow]

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[fossil moth]

I need help with finding the volume of a base semicircle thing

[hardy trail]

Like a cylinder cut in half? (With a semi-circle base)?

[fossil moth]

so its like a polygon base with the cross section is perpendicular to the x axis

i can send a picture wait one second

[hardy trail]

Please do :)

[fossil moth]

Yeah and like

idk where to start

[lunar harbor]

orz

[fossil moth]

I know that i eventually have to intergrate 1/2(pi)(r)^2

Of like 2 sections

[0,3] and [3,6]

[young spade]

Tbh theres two possible solutions, i take it we consider the semi-circles run parallel along the y axis.

[fossil moth]

idk guys saying theres 2 answer kills me

[young spade]

We will assume the semi circles are oriented like this

[hardy trail]

They say the semi-circles are oriented perpendicular to the x axis, so wouldn't the above diagram be the correct choice?

[fossil moth]

yesssss

[hardy trail]

Oh wait

Nvm it doesnt say that, my bad.

[fossil moth]

its like that

thats what my teacher did

last time

[young spade]

ill also assume this cause the other way is absolutely fucked up, lmao

Well, lets try to clear this up, would you be expected to actually do an integral, or you just want to find the volume and thats it?

Because this one can be easily solved appealing to some 3d geometry

[fossil moth]

um i have to find the intergral

like i have to do the

find a slice

then yeah

(the area of the slice)(change in x or y)

[young spade]

Well, pretty obvious thing to say, but you can obviously see that this shape is symetric with respect to the x axis, right?

[hardy trail]

Fair enough lol

[fossil moth]

yea its symmetric

[young spade]

And you also agree that semi circles are also symetric, placed as they are in this problem

[fossil moth]

and like from [0, 3] the slopes are the same one just negative the other not so 2/3 and -2/3

huh

im confused

i know the circle from [3, 6] is like al the same

and the one that goes from 0 to 3 is progressively getting bigger

[young spade]

right now you can kinda ignore the figure, just keep with me here, do you agree that circles are symetric with respect to x axis?

[fossil moth]

yes the semicircles are going to be symmetric

so can we jut find the whole thing then divide by 2 later?

[young spade]

Well, for the integral, we will then define a function for just half the side of this base

[fossil moth]

okay

[fossil moth]

oooo yes its symmetric

[young spade]

In reality, this will give us half of the volume we want

At the end we will have to multiply by 2

[fossil moth]

yes that makes senseeee

[young spade]

Now, we in reality dont have to deal with the integral of half (or now quarter) circle.

[fossil moth]

yes

[young spade]

You can see that if, earlier, the diagram showed as the diagram, our new function actually tells us the radius

[fossil moth]

oh yeahhhhhhh

which is 2

oooooo

[young spade]

for the flat part, yes, for the sloped line, we will have to do integrals technically

[fossil moth]

yeahhhhh

[young spade]

Well, remembering these are quarter circles, what will be the area at each little slice? in the flat part.

[fossil moth]

1/4 pi r^2?

no thats the area

oh

[young spade]

yeah, thats we need

[fossil moth]

oh yes the sliceeeee

[young spade]

When we want volume, we do area * length or height

[fossil moth]

yess

[young spade]

Now, from 3 to 6, the function is a constant, so we actually dont need to do integrals

[fossil moth]

oh yeahhh we just multiply and then add the cone part later right?

[young spade]

Yep.

Can you tell me the volume from 3 to 6

[fossil moth]

12pi

?

[young spade]

make sure you use this formula

[fossil moth]

3pi

[young spade]

yea

now, for the sloped part, we do need to do an integral

Using the formula pi/4 r^2

r is now a function

[fossil moth]

OOOOOO yessssss

and like

[young spade]

and we integrate from 0 to 3 then

[fossil moth]

2/3x-(-2/3x)?

for the integrand?

beecause like top-bottom

[young spade]

remember we are just using half the figure

ç

[fossil moth]

Ooooohhhhh omg

so we dont care about the -2/3x

holy make senseness

[young spade]

in this case, r = 2x/3

[fossil moth]

i put it in desmos

it gave me pi

lol

[young spade]

plug in this formula, and integrate from 0 to 3

well, now,you can just add all the values we got pi+3pi = 4pi

and multiply by two to get our original volume

[fossil moth]

waitttttt

woah making it into quarters makes it so easy

okay wait and then after that we multiply by 2 why? i thought by 4 because its in 4 parts

oh wait no because thisis a semi circle

so final answer is 4pi?

wait no

8pi

[young spade]

4pi is what we got, but you have to remember we were calculating for exactly half what they were asking

[fossil moth]

Woahhhh people in this discord is crackeddddd

[young spade]

When you do this kind of manouvers id usually advice you actually draw your process so you can recall what you did and dont miss correction factors

[fossil moth]

yep yepppp and then i have 1 more question

Wwhats side a

im confused

[young spade]

yk what an equilateral triangle is?

[fossil moth]

yeah a triangle with all even sides

right?

[young spade]

side "a" is any of its sides since they are all even

to be more clear, the numerical value of its length

[fossil moth]

wait

so the equation they gave us

whatre we supposed to do wit it...?

[young spade]

remember how i mentioned that volume = area * length

well, knowing the base of a figure and its length/height, you can find the volume of it

[fossil moth]

wait so like the base is that equation whats the hieght?

do i do a^2+b^2=c^2?

[young spade]

Your "length" is the fact youre integrating through x

But remember that the area is also a f(x)

[fossil moth]

can you showw me?

hellppppppp

ahhhhhhhhhhh

.close

[devout snow]

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[inner sorrel]

eh

[devout snow]

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[pine kernel]

I need to integrate this equation for my math project, but I've only learnt the simple integrals (sin, cos), integration by parts, and integrating factor. how do i do this one or is there no way other than using a calculator?
$L=\int_{0}^{20\pi} \sqrt{1+(\frac{2\pi (1.15)}{1})^2\sin^2(\frac{2\pi}{1}x)}dx$

[woven radish]

Amby

[fossil locust]

are you sure it's $+(2 \pi (1.15))^2$ or is it $-(2 \pi (1.15))^2$?

[woven radish]

south

[dire iris]

this is an elliptic integral, unsolvable in elementary functions

https://en.wikipedia.org/wiki/Elliptic_integral

[fossil locust]

no, it has the wrong sign as written

[pine kernel]

it is a +

[lucid lily]

Any shortcut for trig identities for sin cos tan cosec sec cot for values 30 45 60’

[fossil locust]

still seems to be an elliptic integral somehow

[pine kernel]

short snippet

to provide more ocntext

but idk how to proceed

[lucid lily]

M thanks

[dire iris]

what are you trying to find with L?

[pine kernel]

arc length

of the function

[dire iris]

an analytic value, a numerical value?

[pine kernel]

numerical

[fossil locust]

yeah you can't do it by hand

[pine kernel]

gg

[dire iris]

then compute the elliptic integral with a CAS

[pine kernel]

okay but i have to kind of explain why i can't do it by hand

do i just find a source on elliptic integral and say it can't be solved in elementary functions

[fossil locust]

yes

[dire iris]

you can say "L is an elliptic integral of the second kind, and cannot be solved in elementary functions"

[pine kernel]

alr thx

[fossil locust]

it's weird cause the parameter here is $k^2 = -1$, and we normally expect $k$ to be real (it's the eccentricity of the ellipse this integral comes from)

[woven radish]

south

[fossil locust]

I imagine there's a generalisation

[pine kernel]

hm

i found this on stack exchange but i don't get any of it lol

maybe it makes sense to u

[fossil locust]

https://math.stackexchange.com/questions/910597/int-sqrt1-sin-2-x-dx-an-elliptic-integral

[pine kernel]

oop

[pine kernel]

so does this mean that it can be solved?

or i still need CAS

[fossil locust]

no, this just shows you it is an elliptic integral after all

the sign, + or -, doesn't change that

[pine kernel]

ohhhh

[fossil locust]

you still need CAS

[pine kernel]

okay thanks

ib bro

gg

.close

[devout snow]

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[fossil locust]

I missed this

this also explains why (sin and cos are just horizontal shifts of each other)

[devout snow]

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[dense solar]

im not sure how to move on from here, i was thinking tuat for (C⊃C) but i need a line so i guess i cant do it, not sure what to do next

[dense solar]

i could maybe start another sub proof to for (C⊃C) but im also getting stuck here

[polar chasm]

can u somehow prove C⊃C without assuming it?

if u could prove that, then you could deduce G and apply MP on 2.

[polar chasm]

what rules are available

[dense solar]

A​C​P
A​d​d
A​I​P
A​s​s​o​c
C​D
C​o​m
C​o​n​j
C​P
D​i​s​t
D​M
D​N
D​S
E​q​u​i​v
E​x​p
H​S
I​m​p​l
I​P
M​P
M​T
S​i​m​p
T​a​u​t
T​r​a​n​s

[polar chasm]

Taut is for tautologies?

does it let u derive C⊃C?

or impl could

[dense solar]

no, i tried it and it says it needs a line,

ooh i didnt try impl

[polar chasm]

is that proof writer available online btw? If so, could u send a link?

[dense solar]

https://www.webassign.net/question_assets/hlogic14/LogicTool/prod/completeTool/completeTool.html

does this work?

[polar chasm]

I'd try this one

C⊃C can be proved by assuming C, and then repeating it in the subproof

[dense solar]

this one is graded for school

[polar chasm]

then by CP, you get C⊃C

[polar chasm]

wait so u cant use CP?

[dense solar]

i can

but y would i

would i use it as a subproof? this has to be a indirect i think

[polar chasm]

the whole proof will be indirect

but you can probably prove C⊃C via a very simple CP (Assume C, derive C and by CP you get C⊃C)

apparently there isn't a rule that allows u to simply repeat stuff, so u gotta do it like this

[dense solar]

ok, let my try to work through this with that

this is what i got so far

[polar chasm]

great, if u look at the 2 parts of line 10, what can u say about their truth / falsehood

These 2 parts H⊃(E⊃H), (K•∼K)

[dense solar]

(K•∼K) impossible contridiction

H⊃(E⊃H true

[polar chasm]

Indeed

so the strategy will be to prove H⊃(E⊃H) and then conclude (K•∼K), which will finish the indirect proof

To prove H⊃(E⊃H), u can again use a short direct proof

or 2 direct proofs

. Assume H
Assume E
Repeat H
Conclude (E⊃H)
Conclude H⊃(E⊃H)

[dense solar]

for repeat H are you saying to use reiteration?

[polar chasm]

ideally, if there is such a rule

[dense solar]

i dont think so unfortuantly

[polar chasm]

then u have to do the long way and derive
H * H by tautology
and then
H by simplification

[polar chasm]

I don't think this will work btw,

[dense solar]

this is the final problem and it is kicking me lol

[polar chasm]

As I said, try proving H⊃(E⊃H)

If u manage to prove that, then you are basically done, bcs MP gets u to K*~K which is a contradiction, so u can conclude the IP

[polar chasm]

And how do we usually prove implication? We use CP.

So we start by assuming H, what we need to prove now is E⊃H. And this is an implication. How do we prove implications? By CP. So we assume E. Now we have to prove H. We already have H at our disposal and we just need to repeat it, so we use the H*H trick and first derive H*H by tautology and then H by simplification

[polar chasm]

the structure will look like this

[dense solar]

ahh

thank you so much

my exam is next week over this, same format and everything, all online

gonna be tuff

.close

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[lilac pelican]

Is 2π/3 counted in the interval ]-π/2 , π/2[?

[lilac pelican]

(-π/2 , π/2)

[heavy current]

well, what do you think? :p

[dire iris]

is 2/3 > 1/2?

[lilac pelican]

I think yes

[lilac pelican]

Ye

[dire iris]

so is 2/3pi > 1/2pi?

[heavy current]

is it the case that -pi/2 < 2pi/3 < pi/2?

[lilac pelican]

Wait no

Okay how about 2π

[dire iris]

yeah 2/3pi is bigger, so not in interval

same with 2pi, its bigger than pi/2

[heavy current]

ask yourself the same question!

[lilac pelican]

Sure why not

It lands in it thr interval

,rotate

[heavy current]

an element is contained in that interval if and only if it satisfies this inequality

[heavy current]

this gives you a very easy way to check if a given number is in the interval

[lilac pelican]

Ah

[restive river]

hi hatim

[lilac pelican]

Yo

[dire iris]

the value 2pi is not in that interval. do you mean (-1/2pi*n, 1/2pi*n)?

[lilac pelican]

I thought it just had to land in the area between those 2

[dire iris]

so periodic interval

[lilac pelican]

When drawing it

[lilac pelican]

No no

[lilac pelican]

When it comes to angles I check like this

I draw it

[dire iris]

yes, angles are periodic, so thats what u mean right?

[lilac pelican]

No I my original question is what I meant

[dire iris]

the actual value 2pi is greater than 1/2pi, so not in the literal interval

[lilac pelican]

Okay noted

Thanks

[dire iris]

but 2pi - 2pi = 0 is in the interval, so it is on the periodic interval

[lilac pelican]

What

Whyd you subtract it from itself

[dire iris]

the angle is in the interval, right?

that is what it means

the actual value is not though

[lilac pelican]

So its not in the interval

I got i got

.close

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[fickle magnet]

.reopen

.reopen

[solid osprey]

where did i go wrong?

[solid osprey]

ugh i regret writing this on paper

[solid osprey]

oop wait the sqrt should be 324^2+432^2

ok... what the actual fuck

[solid osprey]

i triple checked i inputted this into the calculator and it did NOT show 540

what the fuckk

.solved

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[foggy iron]

im not sure where i went wrong for part b

[sand quarry]

I think your solution is pretty much correct

But you made a silly arithmetic mistake

It should be 3.25 not 3.75. Can you see why?

[foggy iron]

omg yeahh i see it now, i shouldve done 4-0.75

thankss!

[sand quarry]

Mhm

You're welcome!

[foggy iron]

.close

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[timid obsidian]

Holy shit this is harder than it looks
I've tried some methods
My answer says that the area is 89m² assuming that BCDE=AFGH

[hollow ice]

your problem says its a regular octagon but doesnt look like it is, coz if it were, then that length given as 4m wont be 4m for an octagon of side 5m

[hollow ice]

but, you can may be get by by assuming that the sides AB and FE are parallel to the diagonal HC (which they would be for a regular octagon)

and divide the octagon into more manageable pieces such as two isosceles trapezoids and a rectangle HCDG

but then again, that area comes out to be 32+32+55=119m^2

[devout snow]

@timid obsidian Has your question been resolved?

[timid obsidian]

Ok I'm back

[timid obsidian]

Correct

[timid obsidian]

That is the correct answer somehow

I got 89m²

Wait let me check where I went wrong

Here it was just assumed that AB=5m

[hollow ice]

can you describe how you got that? maybe that can be explained by some other combination of simple figures

but I would not take this problem too seriously, since it clearly is not geometrically correct

[timid obsidian]

Yeah I was wrong

Arithmetic mistake

[timid obsidian]

Yours and this method was correct

Only thing missing here is AB=5m

Thank you

.close

[devout snow]

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[idle pewter]

what i done so far

[rain summit]

Pull up the unit circle

[idle pewter]

i dont use that

is it not an exact value?

[lavish sigil]

Keep going

[woven radish]

jewels!

[idle pewter]

The 2x is confusing me, i know that has something to do with the amount of times the graph repeats but i dont know how itll affect my x values

[lavish sigil]

Ah right okay

If x lies between 0 and 180 degrees, then 2x lies between 0 and 360 degrees

Agreed?

[idle pewter]

Yes

[lavish sigil]

And I assume you know the value between 0 and 90 degrees for which the cosine is 1/sqrt(2)?

[idle pewter]

i dont understand what you mean

[lavish sigil]

If $0 \leq x^\circ \leq 90$, then can you solve $\cos x^\circ = \frac{1}{\sqrt{2}}$?

[woven radish]

jewels!

[idle pewter]

45 degrees

[lavish sigil]

Yep

And on the full circle, 0 to 360 degrees, do you know in which quadrant the cosine is positive again?

(Other than the first)

[idle pewter]

4th

[lavish sigil]

Yep

So 45 degrees into the fourth quadrant would be what angle?

[idle pewter]

i dont understand again

315?

[lavish sigil]

Yes

[idle pewter]

if thats what ur asking idk

[lavish sigil]

So those are your two values for 2x

[lavish sigil]

Because of this

We know that some angle's cosine is 1/sqrt(2), that angle is between 0 and 360

That angle is 2x, so 2x = 315 or 45

[final tulip]

you should really use radians for questions involving inverse trig

[fossil locust]

it's not necessary to

[idle pewter]

the question is a degrees question, id lose marks for doing it in radians

[fossil locust]

yes all of these questions are in degrees

[idle pewter]

im still having a hard time understanding this

are there any visuals so i can understand it better?

[lavish sigil]

Well you kind of rejected the unit circle

[idle pewter]

i dont know what it is at all

[fossil locust]

go back and learn that

[vast zenith]

khan academy is a good starting point

[final tulip]

I find it more intuitive to use radians for this

[fossil locust]

x-coord = cos(theta), y-coord = sin(theta)

[idle pewter]

my teacher didn’t teach it so i thought it wouldn’t be a necessary

[fossil locust]

well your teacher expects you to learn this by osmosis...

it is very necessary

[lavish sigil]

Did your teacher use some form of circular aid with rays to show the angles?

[fossil locust]

the unit circle is the easiest way to understand why cos(-x) = cos(x) and sin(-x) = -sin(x)

[idle pewter]

Exact value triangles?

we don’t get given questions that aren’t exact values on our non calculator paper

[lavish sigil]

When we deal with values beyond 180 degrees, a triangle without the unit circle doesn't really help

[fossil locust]

beyond 90, in fact

[deep abyss]

u used ASTC here, tht comes frm unit circle.

[idle pewter]

Well im not really sure what that is, just repeating what my teacher told me to draw for these

[deep abyss]

where exactly r u stuck tho

[idle pewter]

how cos2x affects my x values

[vast zenith]

it multiplies then by 2

[idle pewter]

instead of if it was cosx

[vast zenith]

say x=45, cos(x)=cos(45)=1/sqrt(2)

[deep abyss]

cos(2x)=1/rt2 what values does this give for 2x

[vast zenith]

cos(2x)=cos(2*45)=cos(90)=0

[idle pewter]

then so

i need to double my values?

[vast zenith]

What do you mean?

[tall comet]

it basically changes the period of cosine graph if you know what i mean

[idle pewter]

Yes i understand that

It would be 2 revolutions of the cosine graph in 0-360 right?

[vast zenith]

Yes

I’m kind of confused by this discussion though

[idle pewter]

you multiplied it by 2

[vast zenith]

Because the question specifies x is between 0-180

[idle pewter]

Yes

wait

the question is simply asking where on that yellow line

The value is 1/root2

?

[tall comet]

you see there are multiple points where the pink and yellow line intersect thus at all those points the value is 1/root2

[lavish sigil]

The graph there is in radians, that's all, but yes

[tall comet]

for that you need principle solutions

[idle pewter]

and the points where its 1/root2 on the regular cos x graph would be different

[tall comet]

yes

[idle pewter]

so then if i solve
cosx = 1/root2
and
cos2x = 1/root2
why do i get the same values or am i wrong?

[tall comet]

check the graph clearly the points where cosx and cos2x = 1/root(2) are different

[idle pewter]

Yes but im saying if i solved those them i would get the same values right?

I can see on the graph its different

[tall comet]

for cosx = 1/root2 and cos2x = 1/root2 the solutions will be different

[idle pewter]

what would the solutions be

[tall comet]

as i said for that you need principle solutions formula do you know it?

[idle pewter]

Nope

[tall comet]

ok how many solutions does your question ask

[idle pewter]

all within 0 and 180 degrees

[tall comet]

check the cosx = cos theta you need to put values for n (n is an integer) and make sure your answer doesnt exceed pi (180 degrees) as thats the limit

i recommend you search it up on youtube for a better explanation things will get much easy

[idle pewter]

Okay yeah, just one final question

could i not just divide 45 and 315 by 2

so my values for cos2x divided by 2 to get cosx

[tall comet]

yes you can do that but to initally find that the angles are 45 and 315 use principle solutions

[idle pewter]

Ok ill take a look at that

[devout snow]

@idle pewter Has your question been resolved?

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[restive river]

heyyyy i needed help

[restive river]

this is from the study guide to my final exam on monday

i needed to make sure i had everything figured out perfectly

[sand quarry]

So which part do you need help with and what have you done if anything

[restive river]

ok ok so i was just strarting actually

part a... is it as simple as it looks

[sand quarry]

Yes

Just plug and chug

[restive river]

bet

alr got it

soooo

b...

itd be the exact same right

@sand quarry

[sand quarry]

Yuh

[restive river]

b would be the exact same right

so

1 1 2 5 14 42 132 429

[sand quarry]

I havent calculated but you should check

[restive river]

yea it was

and part c is just an inductive proof

so i just gotta show that c0 = 1

[sand quarry]

For basis step yes

[restive river]

ok bu tin the problem its requiring cn

and cn isnt defined in the inductive proof for 0

[sand quarry]

I dont quite get what youre saying?

[restive river]

like how do i

so itd n+1 = 1

times Cn

but Cn isnt defined

or wait no

i gfotta divide right

[sand quarry]

Divide what

Also C_n is defined

[restive river]

wait let me do it rq and show u?

[sand quarry]

Sure

[sand quarry]

Yeah

That's fine

[restive river]

alr alr

[sand quarry]

So for your inductive hypothesis can you retrieve the expression for C_k?

[restive river]

hmmm let me think 😭

[sand quarry]

yes

[restive river]

ok i need serious help with the inductive step tho

[sand quarry]

ok so

using the recursive definition

[restive river]

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

[sand quarry]

no

[restive river]

no?

[devout snow]

@restive river Has your question been resolved?

[devout snow]

Channel closed

Closed due to timeout

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[devout snow]

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[copper spoke]

cana nybody help me here

[copper spoke]

how does this calculation work with 1 * 2 * 3 **** (2n+1)

[stone stump]

write out all the factors for eg n=4

[copper spoke]

ok ill try

i dont understand what to write out

should i use

should i write it out in the formula | a_n / a_n+1|

hm?

[stone stump]

yes

[copper spoke]

i wrote it out but it quite dont understand where the (2n+1)^2 in the numerator comes from

the rest i get

hmmm?

[stone stump]

well what did you write down

[copper spoke]

i think because the numerator has 1 * 3 **** (2n+3) , that somehow (2n+1) is somewhow included

[stone stump]

what did you write down when you did this

[copper spoke]

n=4 :(2n+1) = 1 * 3 * 5 * 9

[stone stump]

what about 7

and all the other rest

[copper spoke]

n=4 :(2n+1) = 1 * 3 * 5 * 7 * 9

its just a product of all uneven numbers

thats what i get