help-10

[warm shale]

Derivative

[timid silo]

x still is 2 times y so this is a subspace

is this correct?

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[shut lagoon]

It's correct intuition. Maybe just make it clear that you pick (x,y,z) in S and all but that's it.

[obtuse pebble]

@timid silo Has your question been resolved?

[wary vigil]

iirc any plane through the origin is a subspace of $\mathbb{R}^3$

[warm shale]

Katharine

[wary vigil]

similarly any line and the origin itself are subspaces too

although i don't think that will help you if you need to prove it by showing it fulfils the properties of a vector space

[obtuse pebble]

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[magic herald]

do we take the value of PQ = 1 as ([1,0],[0,1]) because i do not know how to approach this question

[daring sorrel]

It is an I (for identity matrix) and yeah it is that matrix

[magic herald]

so i'm etting the answer as a = 4 and b = 1

is my answer right?

i first mulitplied the P and Q matrices and got ([a-3, -a+4],[3-3b,-4+4b])

which is equal to I

so then a - 3 = 1

therefore a = 4

and 3b - 3 = 0

oh wait 3-3b = 0

3b = 3

b = 1

[daring sorrel]

Yeah that is correct

[magic herald]

thanks alot

appreciate it

[daring sorrel]

np

[light raft]

The two matrices have to be inverses so you can check quickly by taking the inverse of one of the matrices and see if you get the other matrix

[magic herald]

good point

alr thank you somuch

so much*

.close

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[burnt otter]

i dont know how to do part b

[obtuse pebble]

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[burnt otter]

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[static abyss]

How do I sketch the set of points (x, y) in the plane corresponding to the inequality?

[static abyss]

yeah so

what I got is a hyperbola with center at (1,1)

like this

and then a restriction y <= 1 and y cannot be y = -x + 3..

and idk where to go from there

[stiff umbra]

thats all

[static abyss]

wdym?

[stiff umbra]

its (probably, didnt check your calculations) the hyperbola you found with the restrictions y>x, y>-x+2

then the area above the hyperbola in that region (honestly or below idk try plugging in 1 point to see if the hyperbola is positive or negative

[static abyss]

so like how would u sketch the solution?

[obtuse pebble]

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[ember musk]

How to write this as piecewise function?

[timid silo]

piecewise function i assume you mean?

[pseudo raft]

start with if x > 0

[timid silo]

okay so piecewise

do you know what |x| is as a piecewise function?

[ember musk]

The root inside the absolute value is zero, but it is not defined when it is in the denominator

What I learned in school is that we get the root of the absolute value and put X greater than the root and smaller than the root and write the formula in this way. But the question is that the root is not defined here, so what should we write next?

[high lily]

you don't include 0 in your pieces

and only have stuff for
x>0
x< 0

[ember musk]

Aha , thank you

.close

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[keen rune]

Please explain.
What is that??

[timid silo]

what is what

[keen rune]

what is going on here

This 2 lines like || what is that

[robust raven]

k || l means that lines k and l are parallel

[keen rune]

and the one under that

[timid silo]

perpendicular

[keen rune]

How it will look like on the graph

[robust raven]

you need to read in your school book about the geometric interpretation of solutions to a system of linear equations

[keen rune]

nah

it is too complicated

[robust raven]

so leave it

[timid silo]

like that r smthn

[keen rune]

No, I need to know how it works, I have an exam tomorrow

[robust raven]

ah ok 🙂

[keen rune]

And these kinds of equations will be in the exam mostly

[keen rune]

Strange

[timid silo]

Have u never seen perpendicular lines before??

tf

schools are failing 😭

[keen rune]

🥲

It is something called parabola

is not it?

[timid silo]

its a st. Line

Parabola is a curve

perpendicular lines are 2 lines that form a 90° angle at their point of intersection

[keen rune]

Alright then, just tell me how they make these formulas

[timid silo]

For parallel lines, the gradients of both lines must be equal

[keen rune]

Why it is like this

What is a gradient?

[timid silo]

For perpendicular lines, the gradients of both lines should multiply together to get -1

gradient = slope

[timid silo]

Rise over run, change in y divided by change in x

?? 😭😭

👆

[keen rune]

What?

[timid silo]

for example a straight line forms on it, it has a gradient of 1

now we substitute y=mx+c, assume c =0

y=1x

[keen rune]

m is a gradient?

[timid silo]

if x is 1, y is 1, it forms as a point on the gradient/slope, obeys the same with anypoint in the curve

[timid silo]

yea

and c is like if you move the slope/gradient up or not

[keen rune]

yo bro now I get it

[keen rune]

In that case at the top, it is “a”gradient

or “a” and “b”?

[timid silo]

the line across a and b is gradient.

gradient is a line

[keen rune]

but you said it is “m”

coefficient

[timid silo]

m means gradient

or slope

they are the same

[keen rune]

But “a” and “b” they are not gradient

They are coefficients ?

[timid silo]

a and b is a point forms on the line

or we call it coordinates

it obeys slope formula y=mx+c

[keen rune]

Alright, gradient it is the distance between two points, for instance, A(-3;6) and B(5;9) if we do y2-y1/x2-x1 we find the distance, or a gap between these two points?

and it will be like a line

[timid silo]

yea, 9-6/5-(-3) = 3/8

slope is 3/8 in this case

[keen rune]

slope means hypotenuse?

[timid silo]

yea you can call it hypotenuse

to classify slope is a straight line with a certain formula , like 3/8 as mentioned above, and points forms according to the rule y = mx+c, but hypotenuse is the distance between each point, where we can use pyth theorm to find it

[keen rune]

Ok, but what will be parallels on the graph, I mean

AB and CD are they parallels?

[timid silo]

in this case find their slope

which is infinity cause they just extends

if slope of ab = cd then yes

[keen rune]

And how will look like perpendicular one?

[keen rune]

Oh this is perpendicular

[timid silo]

no thats just the gradient of a line

[keen rune]

Why m1m2=-1?

[timid silo]

its the case if it’s perpendicular

[keen rune]

What if parallels

[timid silo]

Parallels = same slope

or gradient

m1=m2

[keen rune]

But why in that equation it is said that a1a2+b1b2=0 and they did not use m1m2=-1?

[timid silo]

rearranged it is

by=c-ax

And yp get it in y=mx+c

u divide through by b

so ur gradient is -a/b

[timid silo]

then u do some trickery w that and u get what u need

[keen rune]

ahahha trickery

[timid silo]

Idk u shld start studying before its ur exam

instead of cramming it in the day before

[keen rune]

To solve that shit

I just need to use that formula

a1a2+b1b2=0?

[keen rune]

I was busy

I completed gta vice city for 100%

🤩🤩

@timid silo

[toxic glacier]

yes

[keen rune]

I think that is enough for this equation, but hehehehe there a lot more questions about other equations, especially parabolas, I hate parabolas

.close

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[low dune]

On a segment AB = 2a, taken as a base, three isosceles triangles ACB, AC'B and AC"B are constructed, with respective heights a, 2a and 3a. Prove that C + C' + C" = 180º.

[robust bloom]

Draw the figure

[low dune]

it's here.

[robust bloom]

!status

[obtuse pebble]

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

[spice citrus]

what did you do?

[low dune]

I calculated with Pythagoras the length of BC, BC' and BC''. And after I use the sin and cosine functions to calculate the angles using Arcsin and Arcos, but I obtain something like 340º.

[spice citrus]

actually trig isn't necessary, do you know about similar triangles?

(also, you should have probably used arctan)

[low dune]

Thales.

[low dune]

Yes. I try with all trigonomety functions.

[spice citrus]

Can you prove that CC'B and CBC'' are similar

[low dune]

I think no.

CB is fixed, so you need to manipulate the triangles.

[spice citrus]

look carefully at the order of the points

(because they are similar)

[low dune]

How?

[spice citrus]

C stays in place, C' maps to B and B maps to C''

[low dune]

Ah, it is like rotating the triangles.

[spice citrus]

yeah, mirroring them

You can prove with sas that they are similar

[low dune]

What is sas?

[spice citrus]

side-angle-side

[low dune]

Ok, thank's for help. I will try again.

[obtuse pebble]

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[keen rune]

Please someone explain.
I don’t really understand what does E and D mean and how to solve such kinds of equation. I also don’t know what is “y.max and (x^2+2).min. This is an example.

[keen rune]

It would be great if you explain what they have done step by step.

<@&286206848099549185>

[obtuse pebble]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[keen rune]

what

What is wrong?

[obtuse pebble]

@keen rune Has your question been resolved?

[bright pumice]

In this case it looks like e and d are just the function symbols

like the f in f of x

y max is the maximum value of your curve on the y axis

[keen rune]

E and D ?

[bright pumice]

yeah sorry

and the min = 2 is essentially telling that the function x^2+2 has a minimum value of 2

on the y axis

[keen rune]

Why y>0?

How they found out

If I face such kinds of equation I wanna know how to spot that y>0 or y<0

[boreal compass]

basically you can see it's all above y axis

it never touches y =0 or goes beneath it

so it always >

[keen rune]

Yes I can see from that graph, but how to draw that?

[boreal compass]

wdym?

[keen rune]

How he drew that graph

[bright pumice]

yea when you put (0,2] you’re saying 0<y ≤2

oh

I drew that by inputting the formula into desmos’

desmos.com

[keen rune]

Right

But I can do that?

In my exam for example

[bright pumice]

oh absolutely not

unless it’s online

then go ahead i guess

[keen rune]

then how to identify y>0 or y<0

When I do not know how to draw the graph

[keen rune]

Hahahaha

[bright pumice]

so what they did

is that they extracted potential restrictions on x

so they solved for x^2+2 which is the denominator

with a y minimum value of 2

when x=0

and the maximum value for 4/(x+2) is 2 when x=0

[keen rune]

is not it 2

[bright pumice]

yes sorry 2

[keen rune]

ok, we should put x=0 and look for the max and the min?

And after we found them, what then?

[bright pumice]

no it depends what kind of equation you have but in this equation it works since you have no horizontal displacement

[keen rune]

what is that?

Horizontal displacement

[bright pumice]

the equation isn’t to the left

or to the right

like this

[keen rune]

How it will look like when it is to the left?

[bright pumice]

that is to the left

and the minimum value for y in this case is at x=-2 y=2

(-2,2)

have you learnt polynomials and ax^2+bx+c

you can use the -b/2a formula to find the vertex

[keen rune]

yes

[obtuse pebble]

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[restive fog]

yeah... not 'clearly' to me lol can someone explain how they got that last sum

[restive fog]

there's more context to the problem obvi but im just specifically asking about that last step

the coefficient thing

[obtuse pebble]

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[obtuse pebble]

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[restive fog]

.reopen

[obtuse pebble]

✅

[restive fog]

help

[obtuse pebble]

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[versed stratus]

what topics would I need to know to solve this, I don't want a solution , just want to kow what I need to know to solve this. Thanks ! source:_ jee advance 2023

[rich plume]

S=70+700+7000+70000.....+5+55+555+5555+55555......+2+2+2+2......

[versed stratus]

so it's elementary factorisation of sorts ?

[regal tundra]

Geometric series

[rich plume]

not exactly a gp

[regal tundra]

Well you would write is as sum of geometric series

[versed stratus]

thanks ! Also, profile picture checks out, given how fast you solved it.

sorry if that's insulting

[rich plume]

thats actually a compliment

[rich plume]

also note that you must very carefully determine number of terms of each of the three series

[versed stratus]

aren't they all 98?

[rich plume]

no

[versed stratus]

ok, that's enough of a hint, I'll figure out where I'm going wrong from here. Thanks so much!

[rich plume]

welcome

[obtuse pebble]

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[hoary idol]

Need help with 10(10)

[versed stratus]

what have you tried?

[hoary idol]

I tried dividing by cos^2x

@versed stratus

<@&286206848099549185>

.close

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[fervent dirge]

Can anyone help in this prob

[serene fossil]

!occupied

[obtuse pebble]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[nimble gulch]

i did this

i think i proved that if a_n is absolutely convergent then a_n+ and a_n- both converges

but

if a_n is conditionally convergent which means that a_n is convergent and |a_n| is not then i am getting that a_n- is convergent by comparison test

where is my mistake?

[timid silo]

yes this is right

if |an| converges it implies an+ converges and an- converges

[nimble gulch]

what about (b)

[timid silo]

what is b?

[nimble gulch]

b question

[timid silo]

i cant really understand the question you are asking

can you clarify a bit more?

[nimble gulch]

here (b) question

[timid silo]

do you know what means conditionally convergent?

also an is supposed to be |an| ?

[timid silo]

?

[nimble gulch]

no they are different

[timid silo]

ok if an is conditionally convergent, it means the series |an| converges but the series an itself diverges

[nimble gulch]

where is mistake here where i underlined lines with red?

[nimble gulch]

ni

[nimble gulch]

it is completely opposite

[obtuse pebble]

@nimble gulch Has your question been resolved?

[nimble gulch]

<@&286206848099549185>

.close

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[timid silo]

is there a way to graph trig transformations using the 5-point method IF there is a restriction?

[timid silo]

I find it's taking longer + more resources to figure out where the graph ends with the 5-point method

versus just using a table of transformations

https://www.youtube.com/watch?v=Vw-RwPBWS8g << Is what I'm referring to with 5 pt method

[obtuse pebble]

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[sleek folio]

Hi can somebody please explain what this expansion is?

[sleek folio]

Taylor?

[final thunder]

Just binomial theorem

[robust raven]

it looks, it is being played with Poisson distribution here

[final thunder]

[(x+y)^n = \slm{k=0}n\binom nk x^ky^{n-k}]

[warm shale]

Pure

[sleek folio]

ok thank you

🫡

[final thunder]

[sleek folio]

.close

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[languid fern]

Hi, how would I solve this step by step

[robust sleet]

Just like how u would solve for z

x*

Isolate z

[languid fern]

So I move everything to the other side

[robust sleet]

Yes

Happy Christmas!

[languid fern]

Ah okay

Remind me what would happen to the (3+7i) if moved to the other side

[robust sleet]

U need to realizify the denominator

By multiplying the denominator and numerator by the conjugate of the denominator

Then split Complex(z) into Re(z) +i*Im(z)

[languid fern]

Stuck here

[robust sleet]

.

When u divide both sides by 3+7i

[languid fern]

You mean 3-7i?

Since it’s the conjugate

[robust sleet]

First this then multiply nom and denom by the conjugatw

[languid fern]

Conjugate of what

3+7i?

[robust sleet]

Yes

[obtuse pebble]

@languid fern Has your question been resolved?

[languid fern]

Done thanks

[obtuse pebble]

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[obtuse pebble]

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[muted delta]

There are two fixed points on a triangle - A and B above the x-axis. Where do you position the third point, C on the x-axis such that the angle ACB is the largest possible.

[obtuse pebble]

Please don't occupy multiple help channels.

[timid silo]

Consider a circle passing through points A and B. To maximize the angle ACB, we want the line segment AB to be the diameter of the circle

[muted delta]

What

But

That doesn't always work

Because C has to be on the x-axis

<@&286206848099549185>

Can I use trig in any way?

[obtuse pebble]

@muted delta Has your question been resolved?

[muted delta]

<@&286206848099549185>

[obtuse pebble]

@muted delta Has your question been resolved?

[obtuse pebble]

@muted delta Has your question been resolved?

[lethal breach]

well, logically, C would be right in the middle of A and B with its y-coord = 0

not sure how i'd do it with algebra or trig tho

actually, forget that too. it doesnt seem to work when A and B both have the same x, but different y

[muted delta]

Nope

That's not true

[muted delta]

They can't have same x

[polar fossil]

i think if you reflect one of the points across the axis and then connect them with a straight line you get your answer

that one minimizes distance at least

[muted delta]

One second

What

That doesn't make sense

[polar fossil]

take one of the points and reflect it across the x axis

[muted delta]

Ok

And then connect it to the other point

[polar fossil]

then connect that new point to the other one yeah

[muted delta]

Lemme try this

[polar fossil]

where that line intersects the x axis

is your third point

[muted delta]

I don't think thats it

[polar fossil]

hmm

[muted delta]

I'm using a calculator

To check

But I want a mathematical way to do it

[polar fossil]

uhhh

set up an expression using law of cosines maybe

[muted delta]

How

You have one side one angle

Well really

You have one side

[polar fossil]

yeah hmm

i guess you could use calculus

[muted delta]

to optimize

The angle

?

[polar fossil]

calculate it for C = (x,0)

and find the maximum

[muted delta]

Sure

[polar fossil]

i feel like you'll get a very simple answer out

[muted delta]

I should

I hope

What could I do with this

I don't know very advanced calc

I tried using inverse tan but it becomes too complex

[lethal breach]

idk if this will work, but you can try it this way:
make a circle passing through A and B, and has x-axis as its tangent.
wherever the circle touches the x-axis, is C

[muted delta]

But where is that

It's funny that was my original problem

And I made it into this

[muted delta]

@me if you got ideas, cant figure out

[obtuse pebble]

@muted delta Has your question been resolved?

[obtuse pebble]

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[timber island]

!status

[obtuse pebble]

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

[timber island]

Hint: notice that the numbers $\frac{-1+\sqrt{-3}}{2}$ and $\frac{-1-\sqrt{-3}}{2}$ are the roots of a quadratic equation

[warm shale]

ItzKraken

[royal basin]

that, or notice they are cube roots of unity.

[timber island]

Ah yes that's perhaps easier

Well use that property

U can write it as $(\left(\frac{-1+\sqrt{-3}}{2}\right)^{48})^2$

[warm shale]

ItzKraken

[timber island]

Oof i wrote it wrong

[pine sail]

That'd be the both of you.

[timber island]

But well w^50 = w^48 * w^2

[pine sail]

that's not omega^1/3 but OMEGA.

[royal basin]

did you actually mean this

[pine sail]

or not

You need the sum mentioned in the question.

Let it be omega^2

And start simplifying the other term.

The other one isn't omega^2 though?

Is that your sum?

Yeah that's wrong.

Well not quite

If $\f{1+\sqrt{-3}}{2} = \omega$ then $\f{1-\sqrt{-3}}{2} = \omega^2$

[warm shale]

! What the hell am I doing here?

[pine sail]

yeah

Does that someone know De moivre's theorem?

[obtuse pebble]

@glad dawn Has your question been resolved?

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[obtuse pebble]

✅

[royal basin]

no, that's de morgan's.

a few typos in that, but yes.

[obtuse pebble]

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[obtuse pebble]

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[graceful iris]

I was having trouble calculating how much stretch foil do I need to wrap a single pallet, here are the basic pallet and stretch foil info:

Pallet dimensions:

H: 2.4m
W: 0.94m
L: 1.34m

Foil:

L: 1300m
W: 0.5m

Pallet would be wrapped 3 times

[royal basin]

is the pallet just a box?

and you want to cover it with 3 layers of foil?

[graceful iris]

Yes

[graceful iris]

Yes, 3 layers, but every layer goes over the half of the previous one, to make it more firm

But 3 full stretches in total

[obtuse pebble]

@graceful iris Has your question been resolved?

[royal basin]

not sure i understand this

i'd need to see some kind of diagram

but like the way i see it rn you're covering a box with 3 layers

so just find its surface area and multiply that by 3?

[graceful iris]

I explained it poorly, since the foil width is 50cm whenever a layer is laid the next one is put over the bottom half of the previous one, until the full pallet is stretched

[royal basin]

im gonna need some kind of visual here.

[graceful iris]

Okay hang on

Every layer goes around the whole pallet, but also over the previous layer, at least 50% of it, so it holds better, until the whole thing is stretched with foil, 3 times

[royal basin]

so you keep going until the SIDES are covered in 3 layers of foil

and you don't do the top and bottom

right?

(also this looks like clingwrap and not foil)

[graceful iris]

Stretch film, not foil correct

The top is also covered

Bottom same as in the picture

[graceful iris]

Correct

[royal basin]

ok but wait

then you also put the thing on its side and do the same 3 layers stuff right

[graceful iris]

So you go around the pallet like the picture, covering it gradually from bottom to top

https://tenor.com/view/wrapping-machine-gif-21082319

[royal basin]

yeah but do you also wrap its top face and bottom face

and if so how

[graceful iris]

Bottom like this

Top is not covered fully, just the edges

[royal basin]

sorry, i am confused.

ok, so the face that touches the floor is not covered.

and neither is the top.

the area of the other four faces is 2HW + 2HL

so you need 3 times that

[graceful iris]

Correct

[graceful iris]

No it's fine, I explained the whole thing terribly

[obtuse pebble]

@graceful iris Has your question been resolved?

[royal basin]

ok so what is troubling you now?

[obtuse pebble]

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[tulip sky]

i have a conjecture

[tulip sky]

suppose you have two integers- (a, b)

the conjecture is that

for some (x, y) belonging to Z

ok nvm i got it

ty ig lmao

.close

[obtuse pebble]

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[graceful marten]

Solve the equation

You can just do

1x^2 + 3x + 7

you get two x values

And then how do i turn it back?

e^x = value 1 e^x = value 2

?

[high lily]

firstly not an equation
secondly don't use the same variable to represent different things

if you want to make a substitution to get a quadratic equation, use something that isn't x

[graceful marten]

ah ok

I understand what you meant now

A = e^x

A^2 -5A +4

A = 4 and 1

e^x = 4 e^x =1

and then from there.. right?

[versed stratus]

from that find x?

[graceful marten]

yeah

ln both sides

x = ln 4

x = ln 1

[high lily]

yes

[graceful marten]

.close

[obtuse pebble]

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[royal basin]

missing =0

[obtuse pebble]

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[fossil turtle]

Can anyone help me solve this

[fossil turtle]

By using similar triangles I got 3y = 2x

I need another equation to get x and y

[obtuse pebble]

@fossil turtle Has your question been resolved?

[tulip sky]

a good way to imagine it is by "unfolding" the pool tabke

instead of visualising the ball is bouncing back and reflecting

imagine it keeps continuing

so it first goes up by 3 units

then you imagine it doesnt bounce back/reflect

and it keeps going up 4 more units, outside the table

and then lastly it goes up 2 more units

so in total it went up by9 units

or delta y = 9

and horizontally it travelled 4 components

so delta x= 4

and slope = delta y/delta x

so the slope of the initial line is 9/4

and you should be able to work from there

@fossil turtle

[fossil turtle]

Thank you

[tulip sky]

<3

[obtuse pebble]

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[timid silo]

A device is designed in such a way that by chance it takes one of the two letters A or B from the input and after going through three stages
Prints 1 character in the output. In each step, the input letter goes to the next step with a probability of 1/4 without changing, or it changes into another letter and enters the next step. If the probability of choosing the letter A is 2 times that of the letter B, with what probability is the letter A printed by the machine the same as the input letter?

[timid silo]

Ok guys you didn't solve the previous one. Don't let me down on this one

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

Nope. It has not

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

Bruh how long must I wait?

[obtuse pebble]

Helpers are just people volunteering their time to help you. Be polite and patient.

[timid silo]

its christmas, u cant expect everyone to be on here

Oh that's a good point

when it “changes to another letter” can that letter be any letter in the alphabet or is it limited to just A and B?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

No just A and B

Im guessing the last part saying A is double as likely as B is only referring to the first part?

the initial selection

And are we only calculating the probability of the input and output being A?

[obtuse pebble]

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[celest musk]

hi guys, how do you prove rigorously that {(-1)^n, n∈N} doesn't have accumulation points?

[celest musk]

i'ts easy to see that E={1, -1, 1, -1, ...} but how do you prove it rigorously?

[robust raven]

consider two subsequences: n = 2k and n = 2k - 1, that is all

[royal basin]

did you mean this as a sequence?

or as a subset of R?

bc as a subset of R this is just {-1, 1} is it not?

[celest musk]

yes

[royal basin]

and you're meant to prove that this sequence doesn't have accumulation points at all?

[robust raven]

and your question had been set wronglty troo

[royal basin]

why do -1 and 1 not count as accumulation points for you? @celest musk

[robust raven]

your sequence has two different accumulation points and is therefore divergent

[royal basin]

the question was not (or at least did not seem to be) about convergence or lack thereof.

[celest musk]

because if you choose a small neighborhood like 1/2, then 1-1/2 = 1/2 is not an element of E

same for -1

[royal basin]

hold on

hold on, hold on.

that doesn't seem to mesh with any kind of definition of "accumulation point" that i know...

do you have one on hand? y/n

[celest musk]

i haven't studied those things yet, just topology of the real line, surroundings and accumulation points

[celest musk]

uhmm yes wait

[royal basin]

ah.

ok, right.

so we're considering {(-1)^n : n ∈ N} as a set after all.

so it is just the finite set {-1, 1} of two points.

and it does indeed not have any accumulation points.

and neither does any finite set.

[celest musk]

yeah

[royal basin]

every point of a finite set is an isolated point.

you confused us earlier bc i thought you were talking about the sequence (-1)^n, which has 1 and -1 as limit points / points arising as the limit of a subsequence

[celest musk]

ok so how do you prove this using the definition of neighborhood?

[royal basin]

let $E \subseteq \bR$ be finite, let $x_0 \in E$, and let $E_1 = E \setminus { x_0 }$ (for convenience). we wish to show that $x_0$ is not an accumulation point of $E$, i.e. we wish to show that $$\exists \ep > 0 : B(x_0, \ep) \cap E_1 = \varnothing.$$ do you follow so far?

[warm shale]

Ann

[royal basin]

@celest musk

[celest musk]

yes

[royal basin]

ok right

$\ep := \frac{1}{2} \min_{x \in E_1} |x_0 - x|$ will do the trick.

[warm shale]

Ann

[royal basin]

as long as you have access to the following facts:

• the minimum of a finite set is a thing that exists
• the minimum of a function over a finite set is also a thing that exists
• the minimum of finitely many positive numbers is itself positive
• half of a positive number is still positive

[celest musk]

sorry i don't understand this

[royal basin]

as in you can't read it?

[celest musk]

yeah

[royal basin]

compute the distance from x_0 to all points in E_1

(i.e. to all other points of E except x_0 itself)

take the smallest of them

and divide it by 2

[celest musk]

can you give me an example please?

where x_0 is 1 or -1

[royal basin]

you want me to do it specifically for E = {-1, 1}?

warning, it will be trivial

[celest musk]

these things are hard for me not trivial lol

[royal basin]

i will need you to type the exact words "Yes, I want you to do it for E = {-1, 1}" before proceeding. otherwise i will not do it

[celest musk]

Yes, I want you to do it for E = {-1, 1}

please

[royal basin]

ok

let x_0 = 1, then E_1 = {-1}

then $\min_{x \in E_1} |x-x_0| = |(-1)-1| = 2$

[warm shale]

Ann

[royal basin]

the minimum distance from x_0 to a point in E_1 is the distance from x to the ONLY point in E_1.

and so this is 2

and so the epsilon given by my proof would be 1.

[robust raven]

Tommy, your proof is that if you take any real number x other than -1 and not 1, then if you surround "x" with a ball = surroundings of any sufficiently small radius epsilon, then you will not find either -1 or 1 in that ball

and Ann has shown it for you

[obtuse pebble]

@celest musk Has your question been resolved?

[obtuse pebble]

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[celest musk]

.reopen

[obtuse pebble]

✅

[celest musk]

took me a while but i think i got it, thank you very much

[celest musk]

okay got it thank you very much

.close

[obtuse pebble]

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[fervent nexus]

!help

[obtuse pebble]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[final thunder]

lol

What have your tried?

@fervent nexus

[fervent nexus]

I thought of using the properties of cubes

like a^3 + b^3

[final thunder]

This problem is pretty much proof by construction unfortunately.

Note tho that

[\wrb{(n+1)^3+(n-1)^3 = 2n^3+6n}]

Maybe you can go from here

[warm shale]

snow

[final thunder]

Every integer divisible by 3 can be expressed as 6n or 6n+3

[fervent nexus]

What do you mean by proof by construction

[robust raven]

read here, no need to copy it and paste in discord 🙂 https://en.wikipedia.org/wiki/Sum_of_four_cubes_problem

[final thunder]

If you want to try the problem yourself, try adding two more cubes here so that you get 6n.

[fervent nexus]

thanks

.close

[obtuse pebble]

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[muted delta]

Animals at a zoo are assigned into many different groups. Each group must have at least two animals, and every group shares at least one animal with every other group. Prove that you can split partition animals into exactly three "splits" such that each group is represented in at least two splits, and each animal is in exactly one split.

[obtuse pebble]

@muted delta Has your question been resolved?

[muted delta]

<@&286206848099549185>

[muted delta]

<@&286206848099549185>

[muted delta]

Yes

btw @ me if you have any hints/ solution

[spice citrus]

@muted delta Maybe you should post this in MODS, because this seems like an olympiad problem

[muted delta]

mods?

[spice citrus]

Math olympiad discord server

[muted delta]

"Mathematical Olympids"?

Is tht what it's called

[spice citrus]

https://discord.com/invite/mods

Yeah

[muted delta]

Ok

I'll try

[spice citrus]

https://discord.com/channels/533153217119387658/533157665430568981 this channel probably

[muted delta]

alr

Im going to still keep this open

In case anyone knows

[drifting wraith]

i probably don't understand the rules, so there's like a venn diagram and a bunch of animals

and we need to color all animals in 3 colors

you can just use 2 colors however you want, and then use 3rd color for one animal in the middle

then each group will automatically have 2+ colors in it

weird question

[obtuse pebble]

@muted delta Has your question been resolved?

[muted delta]

Yes

[muted delta]

well no

Because they can be arranged in any way

[drifting wraith]

we're guaranteed 1+ animal in the middle

[muted delta]

No

we’re not

[drifting wraith]

by reserving one color for one of middle animals, we solve the problem

[muted delta]

yeah but that’s only for the three groups case

this is not general

Given a specific set it’s really easy to come up with it

How do you prove it in general is kinda hard

[drifting wraith]

well it kinda says like there's an animal shared by every group, you think it means something else?

it could say "with some other group"

but it says with every other group

[muted delta]

It means that

Every group shares at least one animal with every other group

So let’s say the groups were A, B, and C

You know that A and B share at least one, B and C share at least one, and A and C share at least one

But that doesn’t imply all of them share one

[drifting wraith]

maybe

[muted delta]

I’m pretty sure we have to make an algorithm

That sorts the animals from the groups into splits

and show that when it ends

You have the right arrangement in the splits

[obtuse pebble]

@muted delta Has your question been resolved?

[obtuse pebble]

@muted delta Has your question been resolved?

[muted delta]

Well the groups are predecided along with the animals

If you only had 2 animals you could have max 2 groups

Yeah do you could have two groups with 0

and they share two

0 by themselves

[obtuse pebble]

@muted delta Has your question been resolved?

[timid silo]

Bro I waited for so long. I turned off the phone since no one was answering

[obtuse pebble]

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[twilit meteor]

can someone help me with this question

[stone radish]

Allen?

[twilit meteor]

lol yea

[stone radish]

alright

[twilit meteor]

orthogonally means 90 degrees at vertex or focus?

[twilit meteor]

u r also in allen ?

[stone radish]

no I'm in 10th ehh

[twilit meteor]

how do uk then?

[stone radish]

olympiads

I think m = -1?

@twilit meteor

that's what you need to do ig

[twilit meteor]

can u ans this

[stone radish]

wdym by vertex / focus?

[twilit meteor]

nvm

[stone radish]

oh

[twilit meteor]

how did u get that

[stone radish]

vertex is the point at which the graph is at min or max

ryt?

where y' = 0 ?

[twilit meteor]

yea

[stone radish]

alr

I think the tg is perpendicular

to the st line

[twilit meteor]

oh

so u mean a normal?

[stone radish]

yes

[twilit meteor]

oh k

[stone radish]

normal to that point

now I think m = -1

[twilit meteor]

tks i shall do it

[stone radish]

yes

(m1) x (m2) = -1

might as well just solve it

[twilit meteor]

i was only confused abt whether orthogonally meant subtending angle at vertex or smtg else

tq @stone radish

[stone radish]

alr

[twilit meteor]

.close

[obtuse pebble]

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[stone radish]

np

[obtuse pebble]

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[robust bloom]

In a singles tennis toumament, 2n players are paired off in n matches; an
denotes the number of different ways in which this pairing can be done.

[robust bloom]

!status

[obtuse pebble]

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

[robust bloom]

1

[stone radish]

ill try

[robust bloom]

okay

[stone radish]

so the total number of ways to choose a pair is

[robust bloom]

2n choose 2

[stone radish]

2n(2n-1)/2

so

n(2n-1)

[robust bloom]

how did u get that

[stone radish]

ok

lets say you have 4 elements in a set

or 4 letters

A B C D

so A will form 3 pairs (n-1)

(AB, AC, AD)

B will form 2 pairs (n-2)

(BC, BD)

and C will form 1 pair (n-3)

(CD)

so in total

[robust bloom]

6

so we used 4c2

[stone radish]

yes

uh

no?

[robust bloom]

we just found out the combination

[stone radish]

wait

yes

[robust bloom]

I was trying to do this with recursion

[stone radish]

so total number of ways to choose k items from n items is nCk

[robust bloom]

yes i get that

Thank you!

@stone radish

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[royal basin]

$\int f(ax+b) \dd{x} = \frac{1}{a} F(ax+b)$, where $F$ is a known antiderivative of $f$

[warm shale]

Ann

[royal basin]

you can also view it as a $u$-substitution: in $\int \cos(3x+1) \dd{x}$, take $u := 3x+1$ and it'll fall out

[warm shale]

Ann

[obtuse pebble]

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[grand marsh]

how do you do this

[timid silo]

Translate, please

[grand marsh]

a) Show that x_n >= 2 \forall \in \mathbb{N}.
b) Study the monotonicity of the sequence (x_n).

c) Study the convergence of the sequence (x_n) and determine its limit.

[obtuse pebble]

@grand marsh Has your question been resolved?

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[torn river]

anyone knows how to solve these?

[timid silo]

do you know about Gaussian reduction

or simply known as solving systems of equations by elimination

[torn river]

umm

can u show me how to do it pleaseee

wait

i got x = 4/15

is that correct?

[tame narwhal]

give both x and y solutuions, specify which problem

[torn river]

is this correct?

shoot

i wrote down the quesrtion incorrectly

[timid silo]

yes

[obtuse pebble]

@torn river Has your question been resolved?

[torn river]

hmm ok wait

[obtuse pebble]

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[lyric mural]

I am doing some operations and at a certain moment I get this: 594!
And I put it in the calculator and I get infinity, but it doesn't make sense.
What comes out then?

[lyric mural]

Help please

[royal basin]

594! is just a fuckoff big number

,w 594!

[warm shale]

Results provided by WolframAlpha

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

Requested by ann_dec

[royal basin]

are you doing combinatorics of some kind?

[lyric mural]

No, I'm doing Chudnovsky's formula for π

[royal basin]

enlighten me?

[lyric mural]

Why?

[royal basin]

i'd like to know what that formula is

[lyric mural]

Wait

[royal basin]

but if you think it below you to enlighten lowly plebs like me then there's no need

[tardy epoch]

https://en.m.wikipedia.org/wiki/Chudnovsky_algorithm

Quite boring

[lyric mural]

$[\frac{1}{\pi} = \frac{426880 \sqrt{10005}}{\sum_{n=0}^{\infty} \frac{(6n)!(3n)!(n!)!}{(3n)!^3 (n!)^3 640320^{3n+1/2}}}]$

[warm shale]

H^Ψ(r)=EΨ(r)
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[obtuse pebble]

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[timid silo]

is the symbol for implication commonly used?

First time i see the notation of a set meant to denote implication

.close

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[rocky pebble]

Need help with this functional equation: f(x^2 + f(y)) = f(f(x)) + f(y^2) + 2f(xy)

[rocky pebble]

I have gotten that f(0) = 0

f(x) = f(-x)

f(x^2) = f(f(x))

and f(x^2 + f(x)) = 4f(x^2)

[timid silo]

what do u have to work out

the function?

[obtuse pebble]

@rocky pebble Has your question been resolved?

[rocky pebble]

yes, the solution is f(x) = x^2 and f(x) = 0

but it has to be proven

[obtuse pebble]

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[unique shell]

Havin trouble with part c

[unique shell]

I got that 15a + 10b + 6c = 120 through manipulation

b should be greater than or equal to 12

and idrk where to proceed

[haughty coyote]

it gives n(a/2+b/3+c/5) = n/30 * (15a+10b+6c), so I agree
But why b >= 12 ?

[ebon yoke]

notice that c is divisible by 5 => c = 5k for int.k => 3a+2b+6k=24, here notice that b is divisible by 3 and etc.

[ebon yoke]

after this you will get equation that is much more easy

[unique shell]

cause wouldn't 10b = 120 - 15a + 6c

10b = 3(40 - 5a - 2c)

When a = 1 and c = 1 (minimum values i think)

10b = 3(33)

10b = 99

since b is divisible by

b is greater than or equal to 12

[haughty coyote]

no

you've proved here that a=1, c=1 leads to no solution because 10b=99 has no solution

if you consider, say, a=2, b=4, c=10, that's a solution

it has b =4, not >= 12

[unique shell]

oh

ic ic

OH

yeah

[unique shell]

but how can b = 4 if b is divisible by 3

[haughty coyote]

nvm I miscomputed

b=3

[unique shell]

b = 3z for some integer z > 0
3a + 2(3z) + 6k = 24
a + 2z + 2k = 8
a = 2m for some integer m > 0
3(2m) + 2z + 2k = 8

3m + z + k = 4

Have I made a mistake cause the last equation isnt working

😭

[ebon yoke]

in pre-last line

it should be 2m+2z+2k=8

[unique shell]

OH yeah

m + z + k = 4

ty

so for possible triples (m,z,k) are
(2, 1, 1)
(1, 2, 1)
(1, 1, 2)

Meaning possible triples (a,b,c) are
(4, 3, 5)
(2, 6, 5)
(2, 3, 10)

ty

.close

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[graceful stirrup]

So I have done some math and I've come to:

wind (vector) = [-13.01, 1.94]

|wind| = 13.16 km/h ---> this is the speed of the wind which is the correct answer

θ = arctan (1.936/-13.01)

θ = -8.46 degrees

if i do add 270 degrees + 8.46 degrees i get a bearing of 278.46 degrees which is the correct answer

i dont understand why my θ is negative.

to solve for the wind vector I did (ground velocity - heading) after converting these two to cartesian vectors

[marsh geyser]

arctan cannot differentiate between the second and fourth quadrants because it only sees the ratio of y to x, not their individual signs

[graceful stirrup]

you're right

thanks for your help