help-13

[vital jay]

Oh wait

No yeah you're totally right lmao

[shut nimbus]

that's what I'm confused

[vital jay]

In that case, (x + c)^2 should just touch the line since it's an even power, while (x + d)^3 should cross the line since it's an odd power

Does that make sense?

[shut nimbus]

for the (x+c)^2 and ( x+d)^3, they both touch the line. Only a= -2 crosses the line

[vital jay]

(x+d)^3 crosses the line, it just gets flat with the line at the same time

While (x+c)^2 touches touches the line, but leaves on the same side it started from

[shut nimbus]

oh wait, you r right

[vital jay]

Yeah, that's the main visual difference between even and odd powers

[shut nimbus]

but how come b = 2, cuz it didn't touch or cross the line 🥹🥹🥹

[vital jay]

Checking the difference between powers like 2 and 4 is a lot harder to do visually though

It does cross the line over at -2!

[shut nimbus]

Isn’t that a = -2

[vital jay]

No, since we have (x + b) in the equation but not (x + a)

The a is out front and it's just scaling the whole equation

[shut nimbus]

Oh yeah, wait so how come a = -2

Sr for asking too much😭😭😭

[vital jay]

Lol it's all good

Well this one's harder to analyze visually, I think it might be easiest to try plugging in values and seeing which match the graph

Lets say we know b=2, c=-4, and d=-1, so we have y=a(x+2)(x-4)^2(x-1)^3

What would we get if we plugged x=0 into that?

[shut nimbus]

is it -32....

-32a

[vital jay]

Yep!

Now based on this graph, do you think a should be a positive or negative number?

[shut nimbus]

negative

[vital jay]

Yup!

[shut nimbus]

cuz it goes down

[vital jay]

...You know what, that's not what I was getting at but it's probably a better explanation

What I was thinking is that on the graph we have that when x=0 that y is a positive number

So y = -32a when x = 0 would imply a is negative

[shut nimbus]

wait, I don't really get this part

[vital jay]

What I mean is that on this graph, at x=0 the line is above the x-axis

[shut nimbus]

ohhh

okok

so a has to be negative so that y can be positive 🤔

at x = 0

[vital jay]

Yep, since (x+2)(x-4)^2(x-1)^3 is negative when x=0

So the a out front has to be negative too, in order to make it positive

[shut nimbus]

oh it makes sense

[vital jay]

Great!

[shut nimbus]

but what if I want to know that specific number at a

like when I can define a is negative

but why it is number 2

[vital jay]

Well you'd probably need the graph to be a bit more precise to get that's it's exactly 2

But using x = 0 again, you can tell that y should be a bit over 50 there right?

[vital jay]

So in order to make -32a match that, a should be about -2, though you might also guess something like -1.75

[shut nimbus]

yeah yeah I got it

[shut nimbus]

ohhhhhh I understand now

omg thank you so much ✨✨✨✨😭😭😭😭😭

[vital jay]

No problem! I'm glad I could help :)

[shut nimbus]

.close

[cedar kiln]

Channel closed

Closed by @shut nimbus

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[torn marsh]

What have I done wrong

[torn marsh]

,rotate

[vital jay]

You're trying to take the derivative of 2x(x+1)^5?

[torn marsh]

correct

Using the product rule

[normal cipher]

Why did you try the quotient rule first

The rule you have used is the quotient rule

[vital jay]

So there's a couple of things, but I think you're mainly just using the wrong rule

The rule you're using is for u/v

[torn marsh]

Oh I used the wrong rule

[vital jay]

The one you want is u'v+uv'

[torn marsh]

Oh okay makes sense, I’ll re attempt the question

My other problem is

[vital jay]

👍

[torn marsh]

,rotate

[tropic oxide]

,rccw

[tropic oxide]

missing brackets

[torn marsh]

Using the chain rule

[tropic oxide]

you applied the chain rule correctly but you forgor the brackets

5(3x^2-x+1)^4 * (6x-1)

and then you correctly multiplied the 5 into that bracket even though it was still invisible

[vital jay]

Yeah pretty much

[torn marsh]

Ohh

We need to bracket itV

[vital jay]

Your first answer was right, just missing the brackets

[torn marsh]

?

[vital jay]

Yup!

So (30x - 5)(3x^2 - x + 1)^4

You could also just leave the 5 outside, 5(6x - 1)(3x^2 - x + 1)^4

It's the same answer either way, so which one you wanna use is up to you

[torn marsh]

So is this the answer

Or we need to expand it out

[vital jay]

That answer's right!

Expanding it out would also be right, but you probably wouldn't get anything new out of it

[torn marsh]

That’s true

@vital jay I redid the question

And still got it wrong

[tropic oxide]

how are you determining whether you got things wrong btw

[torn marsh]

I checked my ans with the solutions provided

[tropic oxide]

but also ok two things you did wrong

one is handwriting the other is actual math

[vital jay]

Oh you dropped the 2x on the right

[vital jay]

v'u is [5(x+1)^4]*(2x) right?

You dropped the 2x

[tropic oxide]

and the other hadnwriting thing is that when you wrote u'v you made your u and v indistinguishable

[torn marsh]

Oh I left out 2x

[tropic oxide]

yeah you forgor 2x 💀

[torn marsh]

Yea that’s what I’m trying to figure

[tropic oxide]

have you tried doing these questions with deliberate slowdown

[torn marsh]

I’ve been going slowly since I’m new to these qs

[tropic oxide]

anyway, as far as handwriting goes:

• v should have a sharp corner. perhaps stop your pen for a moment at the bottom point.
• u should have a tail on the right.

,, u

[wraith dagger]

Ann

[torn marsh]

How is it meant to be done

[tropic oxide]

like this bit on the letter u

it's non-negotiable

[vital jay]

Ultimately though, it just matters that you or a teacher can tell them apart

[tropic oxide]

https://www.youtube.com/watch?v=OMB6lVXtP4Q

[torn marsh]

That’s true but some teachers may not able to read it properly so less marks for me

[tropic oxide]

you should put extra effort into making sure all of your letters are distinguishable.

[vital jay]

I'd say your handwriting is better than mine so w/e ¯_(ツ)_/¯

I just make a final draft for all my assignments where I either type it up or rewrite it as clean as possible

My scratch notes though, are a sin against all proper handwriting techniques

[torn marsh]

I think I stilll got it wrong

I did what you guys told me to do

[vital jay]

That should be right? What answer are you expecting?

[tropic oxide]

same shit

they just didn't bother factoring

your answer is correct now

[vital jay]

Yeah, they just didn't pull out the (x+1)^4

[cedar kiln]

@torn marsh Has your question been resolved?

Channel closed

Closed by @torn marsh

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

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

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

[formal depot]

can anyone teach me equations and inequalities with trigonometric functions?

[oblique flare]

you got a specific question or example ?

[formal depot]

yea one sec

.latex

!latex

how do i use the latex command

[scenic narwhal]

$x = 3$

[oblique flare]

$x=y$

[wraith dagger]

@scenic narwhal

yoboiqimmah

[formal depot]

$|\sqrt3\sin(x)+\cos(x)|=\sqrt2$

[oblique flare]

\sin \sqrt \cos

[formal depot]

i barely know anything about sin and cos

i know that that are identities

but i know none

[tropic oxide]

$\sqrt{3}$ btw

[wraith dagger]

Ann

ori299

[oblique flare]

Start by squaring both sides

[formal depot]

every time theres a root in both sides?

[oblique flare]

It’s the reasonable approach bcz sin and cos can’t be related directly without squaring
And note that squaring would still keep the left side positive so no missing solutions

[formal depot]

also idk radians and all that

[oblique flare]

And solving trig equations hell nah

[oblique flare]

At least is x in degrees and restricted to some domain ?

[formal depot]

wdym some domain

[oblique flare]

Some questions restrict some values of the angle you are solving for

Anything like
x in (0,360] next to the question ?

[formal depot]

it says find all real solutions

i mean x could be more then 360

[oblique flare]

The question say x In degrees or nah ?

[formal depot]

nope

just says find all real solutions and then that

[oblique flare]

Then it’s radians

[formal depot]

why does it matter

x is a number

[oblique flare]

x radians is also a number

But let’s not get that side tracked

[oblique flare]

Have you squared both sides ?

[formal depot]

yea

$|\sqrt3\sin(x)+\cos(x)|^2=2$

[wraith dagger]

ori299

[formal depot]

$|\sqrt3\sin(x)^2+2*\sqrt3\sin(x)*\cos(x)+\cos(x)^2|^2=2$

[wraith dagger]

ori299

[oblique flare]

Wait maybe not I found and easier method

[formal depot]

i think thats right

[oblique flare]

You know angle sum and difference identities ?

[formal depot]

nope

i dont know anything

[oblique flare]

At this point just revise your trig man

[upper ruin]

sin(x + y) = sinxcosy + sinycosx

You haven't been taught this?

Because if you divide by 2 and multiply by 2 your LHS that formula shows up nicely and can be pretty useful

[formal depot]

lhs?

[upper ruin]

Left-hand side

[formal depot]

i think if i want to learn all the stuff i have to learn in the time i need to il have to do smth else

ty for the help tho

im just gona read all notes and get past exams and try questions

[cedar kiln]

@formal depot Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[rustic gust]

I need to find the ratio between the areas of ABD and BCD triangles. This came up in aptitude test, can anyone help?

[rustic gust]

Here are the possible answers

[dim lotus]

Uh i could try and help but i just woke up lol

Okay so what do you know about the question and what do u need help with?

[sacred tusk]

are there any more info given?

[south tundra]

Start by expressing the areas of ABD and BCD in terms of the length of BD

[cedar kiln]

@rustic gust Has your question been resolved?

[sacred tusk]

nvm

[sacred tusk]

Hint: BD is a common side

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[steady citrus]

I've gotten up till the second photo,
CE = CD = 6
BF = BD = 8
And then the radius is 4
I've taken AE and AC as x (I don't know if that's correct) but now I'm completely stuck
I need to find the length of AC and AB

[steady citrus]

Pls help

[gritty galleon]

is this ncert

[steady citrus]

Yes

[vagrant gorge]

you should probably join the center to the three vertices and find the area of the triangles and add them and then equate it to the area of the whole triangle found by herons formula

[steady citrus]

I saw a solution online but i couldn't understand anything

[gritty galleon]

what do you think we should do

[cedar kiln]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[steady citrus]

So the area of triangle AOC, COB and AOB?

And then equate that to the area of ABC

[vagrant gorge]

thats for you to think

[vagrant gorge]

ohhh

[steady citrus]

Alr alr lemme try, one sec

[rustic gust]

Sorry I was doing the test, I couldn't respond

[steady citrus]

I got x = 7

It took so long

Let me see if its correct

[rustic gust]

I think I'm interrupting, Ill just post on another channel

[steady citrus]

It's correct 🙏

[steady citrus]

Thank you

.close

[cedar kiln]

Channel closed

Closed by @steady citrus

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[pseudo merlin]

excuse me how do I do dis

i was trying to use this methjod

I get this

[twilit escarp]

ln(u)' = u'/u

Not 1/u

[pseudo merlin]

i dont understand

[twilit escarp]

You integrate 1/(1-2x) as ln|1-2x| which isn't the case

[pseudo merlin]

ohh

[twilit escarp]

Same for 1+2x

[pseudo merlin]

how would i do it?

[twilit escarp]

You need to have the derivative of 1-2x (or 1+2x) on top

[pseudo merlin]

oh

[twilit escarp]

In order to integrate it

[pseudo merlin]

so i have to make it 2?

[twilit escarp]

Or -2

[pseudo merlin]

oih

w2ait let me try

[twilit escarp]

But yeah

Remember to compensate, if you multiply by 2, you need to divide by 2 aswell

[pseudo merlin]

gulp

i was just gonna multiply by 4

4 and -4

[twilit escarp]

Ok but so you need to divide by 4

Same result

[pseudo merlin]

oh ok

[pseudo merlin]

thx sao much

.close

[cedar kiln]

Channel closed

Closed by @pseudo merlin

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[gentle frost]

Hellur - I'm doing some fluid mechanics questions and I was wondering if I could go through my solutions to verify them 😄

[normal cipher]

If you got the solutions

You can send here and we can check your

[gentle frost]

Oh yes sure. 1 moment

This is the first question. I'll write down my steps and answes.

[cedar kiln]

@gentle frost Has your question been resolved?

[gentle frost]

Based on Bernoulli's Principle and Continuity, the energy loss between point A and B is caused only by friction assuming steady state :

\begin{align*}
\frac{P_a}{\gamma} + \frac{v_a^2}{2g} + z_a &= \frac{P_b}{\Gamma} + \frac{v_b^2}{2g} + z_b + H_f\
\frac{P_a - P_b}{\gamma} &= H_f
\end{align*}

In a parallel system, the pressure drop in each pathway is equal to the energy drop across the entire system. Hence, the equation can be derived.
\begin{align*}
H_f &= H_{f1} = H_{f2} = H_{f3}\
f(\frac{L_1}{D_1})(\frac{v_1^2}{2g})& = ... = f(\frac{L_3}{D_3})(\frac{v_3^2}{2g}) \text{- Using Darcy's equation for friction loss}\
&\frac{64}{Re}(\frac{L_1}{D_1})(\frac{v_1^2}{2g}) \text{- Can be derived as the system is Laminar. f = 64/Re}\
\frac{64 \mu}{\rho vD_1}(\frac{L_1}{D_1})(\frac{v_1^2}{2g}) &= \frac{64 \mu}{\rho vD_2}(\frac{L_2}{D_2})(\frac{v_2^2}{2g})\
\frac{L_1v_1}{L_2v_2} &= \frac{D_1^2}{D_2^2}\
\frac{v_1}{v_2} &= \frac{D_1^2L_2}{D_2^2L_1}\
\therefore v_1 &= \frac{11}{36}v_2 \text{- I have substituted the diameters here}
\end{align*}

Now, since the ratios are found - $v_1 = \frac{11}{36}v_2 ; and ; v_3 = \frac{33}{64}v_3$ then, based on material balance -

\begin{align*}
Q &= A_1V_1 + A_2V_2 + A_3V_3\
(33 GPM \times 0.002228 \frac{ft^3 s^{-1}}{GPM}) &= \frac{11}{36}A_1V_2 + A_2V_2 + \frac{33}{64}V_2\
\therefore V_2 = 9.87 ft/s
\end{align*}

Now, we can substitute velocity back in.
\begin{align*}
V1 &= \frac{11}{36} \times 9.87 = 3.01 ft/s\
V3 &= \frac{33}{64} \times 9.87 = 5.09 ft/s\
Q_1 &= \frac{\pi}{4}(\frac{0.5}{12})^2 \times 3.01 = 4.104 \times 10^{-3} ft^3/s\
Q_2 &= \frac{\pi}{4}(\frac{1}{12})^2 \times 9.87 = 0.054 ft^3/s\
Q_3 &= \frac{\pi}{4}(\frac{.75}{12})^2 \times 5.09 = 0.0156 ft^3/s\
\end{align*}

😭 thats the best latex i can do

[wraith dagger]

Maddie

[gentle frost]

For the pressure drop, I think I can derive it myself as I have the fluid velocity. So i have omitted it in the above calculation:) (Also I'm limited to only Bernoulli's, Darcy's and Continuitys.. So yeah..)

[cedar kiln]

@gentle frost Has your question been resolved?

[wintry monolith]

Do u know flowrate formula?

[gentle frost]

Q = av?

[wintry monolith]

Densityareavolume

Oh

Its not mass flowrate

Volume flow rate

[gentle frost]

HAHA

[wintry monolith]

Is the fluid non ideal 💀?

[gentle frost]

Wdym?

Its not stated.

The only properties that are stated is that the fluid flow is laminar.

[wintry monolith]

[cedar kiln]

@gentle frost Has your question been resolved?

[cedar kiln]

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).

[cedar kiln]

@gentle frost Has your question been resolved?

[modern estuary]

Is z real, or any complex number?

[dusty wren]

Has been solved already

Multipost

[empty epoch]

"Less than" doesn't make sense if z is complex.

[modern estuary]

Thats why i asked

Ofcourse, there is no natural total order on C

Usually the variable z is reserved for those of complex, so to be sure, i asked

[empty epoch]

Or something in three dimensions, of course.

[cedar kiln]

@gentle frost Has your question been resolved?

[flat mica]

my first inclination was to say none are correct, for this reason

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[main terrace]

How do I solve ∫ 1/( ( (√t) + 1 )(t²+1) ) dt

[main terrace]

Or

[ocean fiber]

i would try u-substitution with u equaling the term with the sqrt

[main terrace]

Yes but I'd get u⁴+1 in the denominator

Wouldn't partial fraction be my only choice here

That's somewhat tedious

[buoyant latch]

1/(u⁴ + 1) is a known integrand im pretty sure

And it’s very tedious to do

[main terrace]

Oof

.close

[cedar kiln]

Channel closed

Closed by @main terrace

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[earnest girder]

Claim

[earnest girder]

Can't get 4cos4a cos5a cos 6a in answer i am getting 4cos4a cos8a cos6a

Anyone here to helpe or hint me??

[modern estuary]

Hwo did you get 2cos(5A)cos(4A) in the third line

[earnest girder]

Spilit sin11a into 5a and 6a

Cos(11a)=cos(6a+5a)

And then applied cos(a+b)

[earnest girder]

Did u got it?

[modern estuary]

I think it should be cos(2A) insread of 4A

[earnest girder]

Ye thx 👍 🙏

I think now i can't take cos 4a common

[modern estuary]

Thats what i was asking about

[earnest girder]

Ok then i will do with another method thx

.close

[cedar kiln]

Channel closed

Closed by @earnest girder

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

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

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

[fallen musk]

In question is 4 (a) answer {-2n,.....-6,-4,-2,0,2,4,6.......,2n}

[crimson delta]

no

[fallen musk]

Oh

[crimson delta]

what is n in your answer

[fallen musk]

Natural no.s

[crimson delta]

do you mean that n is a natural number?

[fallen musk]

I mean n is a element of N

[crimson delta]

the problem is that the expression $\bigcup_{i\in\bN} A_i$ doesnt depend on the variable/letter $n$ at all

[wraith dagger]

Denascite

[crimson delta]

so your answer also cant depend on it

[fallen musk]

then whats the union of indexed sets A_i

[crimson delta]

well thats for you to figure out

but if n was for example 4, then your answer would be {-8,-6,-4,-2,0,2,4,6,8}

which is false

[fallen musk]

Isn't the same what I did answer above

Wait

[azure swift]

Hint: ||the set will be infinitely big||

[fallen musk]

Oh thanks

I don't know of that

I'm doing wrong something to decide what will be no to represent at end

[crimson delta]

do you mean that you dont know what infinite sets are?

[fallen musk]

to both side

[crimson delta]

do you know set builder notation

[azure swift]

Any even number can be in that union

[fallen musk]

I was joking

[crimson delta]

well that didnt land

[fallen musk]

Honestly I don't know alot of stuff

In maths

[azure swift]

Definitions is the key in the early stages imo!

Don’t take them for granted

[fallen musk]

Yeah that

Any even no.

[modern estuary]

And then note that any number in that union is even

So, now you can say what that union must be equal to

[fallen musk]

Like in question 3 I was confused whether 0 comes in natural no.s or not

Lol

Sometimes I doubt myself I'm doing/thinking correct or not

[fallen musk]

Kk

[fallen musk]

I don't think is there any representation for any even no.s ?

[modern estuary]

There is, using this notation
${2k | k \in \mathbb{Z}}$

[wraith dagger]

Herzog

[fallen musk]

Oh right

2k where k is Integer

[modern estuary]

You may also write {...,-2n,-2(n-1),...,-4,-2,0,2,4,...,2(n-1),2n,...} but i dont like this

[fallen musk]

Cool

Thanks

.close

[cedar kiln]

Channel closed

Closed by @fallen musk

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

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

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

[dusky orchid]

so theres 12C8 combinations

which is 12! / (8! * 4!)

which simplifies to 45*11 which is 495 combinations

but uh

idk what to do past this point

[dusky orchid]

thats the amount of ways the passengers before the couple could have boarded

idk how to find all the ways the couple would find two adjacent seats though

wait so theres 4 seats free

do i just need to find 4C2 or something

[heady hatch]

uh its ez

8 passengers boarded random

2of4 seats goes to them right/

Number of ways to choose 2 seats out of 4 is 6

[dusky orchid]

oh ok

so it was 4C2?

[heady hatch]

we can say each row having 1 st and 2nd and 2nd and 3rd

no

[dusky orchid]

okay yeah

[heady hatch]

Even if there are 4 empty seats only some arrangements of those 4 will have adjacent seat pairs

[dusky orchid]

so there can either be 2 rows with 2 adjacent seats free

[heady hatch]

You can’t just pick any 2 of 4 and expect them to be adjacent

[dusky orchid]

yeah thats the thing

[heady hatch]

could be

total combis would be 495

[dusk finch]

If you want a hint, instead count the number of arrangements whee they can't find two adjacent seats

that seems easier to me

draw a few examples of that btw

[heady hatch]

and 306 could be favourable

306/495 there u have it

uh so is there a answer triy/

its b or d

idk just simplify it

[dusky orchid]

okay

[dusky orchid]

where did 306 come from

[heady hatch]

uhm all cases try to find for one seat then mulply

[cedar kiln]

@dusky orchid Has your question been resolved?

[dusky orchid]

yeah im uh still confused as to how i find the empty seats

and the unfavourable and favourable outcomes

wait so for there to be no adjacent seats

there'd have to be 2 adjacent seats used by each and every passenger

actually no

not 2 adjacent

just two seats for every row

[cedar kiln]

@dusky orchid Has your question been resolved?

[cedar kiln]

Channel closed

Closed by @dusky orchid

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

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

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

[lilac minnow]

For which value of k does the polynomial function given by p(x) = 8x³ − 12x² − 2x + k have two opposite zeros?

[lilac minnow]

so you can ofc plug the values in but Im not allowed to use a calcualtor on the exam so that isn't really productive

[oblique flare]

Opposite ?

[lilac minnow]

if (a,0)

then (-a,0)

[crimson delta]

that means you have two factors (x-a)(x+a)

[oblique flare]

I see

[lilac minnow]

oh true

[oblique flare]

Then there should be some kind of
(x^2-a^2) somewhere in here when factoring by grouping

Sub any value of the choices for k
And try factoring by grouping

[crimson delta]

grouping is always risky cause you are assuming the values are nice

[lilac minnow]

hmm you can rewrite it as
-k = 2x * (4x^2 -6x -1)?

[crimson delta]

better to assume the third factor is (x-p) and multiply out

and then compare coefficients

[oblique flare]

Yea I get that but the values are nice

[lilac minnow]

so do (x^2-a^2) * (x-p)?

or wdym

[crimson delta]

yes

times 8

[lilac minnow]

I see because x^3 is 8 right'

[crimson delta]

the coefficient of x^3 is 8, yes

[oblique flare]

The coefficient of x^3 is 8
Not that x^3 is 8 but yea

[lilac minnow]

yeahyeah sorry english isnt my first language

8 * (x^2-k^2) * (x-p)

gives 8x^3 -p * k^2 * x + k^2 * p
and then I equal that to -k right? so
-k = 8x^3 -p * k^2 * x + k^2 * p

[crimson delta]

you messed up k^2 and a^2

and the x^2 term is missing

[lilac minnow]

oops

[crimson delta]

and the 8 should appear in every term

[lilac minnow]

yeah I got it here on paper correctly oops

my fault

but you gotta equal it to -k?

[crimson delta]

no

you compare it with 8x^3-12x^2-2x+k

and then compare coefficients

write down again what you got for 8(x^2-a^2)(x-p)

[lilac minnow]

8x^3 -8px^2 -8k^2x + 8k^2p

[crimson delta]

a^2 instead of k^2

but yes

from this you can first solve for p and a^2

and then you can compute k

[lilac minnow]

oh yeah true

so -8p = -12 <-> p = 4/3

[crimson delta]

close

check again

[lilac minnow]

oops... 3/2

[crimson delta]

happens :)

[lilac minnow]

so then 8 * a^2 * 3/2 = k
12 a^2 = k
a = sqrt(k/12)?

[lilac minnow]

assuming I did that right

[crimson delta]

use the other term first

the -2x

[lilac minnow]

hm how so?

[crimson delta]

-8a^2x = -2x

[lilac minnow]

so a = 1/2

[crimson delta]

(or a=-1/2)

[lilac minnow]

yep true

so +- 1/2 = sqrt(k/12)

which then makes k = 3

because sqrt(3/12) = sqrt(1/4) = +-1/2

[crimson delta]

or the first line

[lilac minnow]

8 * 1/4 * 3/2 = k

so 3 = k

[crimson delta]

yes

[lilac minnow]

thx a lot

[crimson delta]

yw

[lilac minnow]

.close

[cedar kiln]

Channel closed

Closed by @lilac minnow

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

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

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

[modest swift]

Find the area of the shaded area. Given: All angles that appear to be right are right. Round to the nearest tenth.

[slender ginkgo]

,rccw

[slender ginkgo]

u've separated them into shapes

so total area is the sum of the area of these shapes

[dusty wren]

yards

[cedar kiln]

@modest swift Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[hasty shuttle]

hi can anyone tell me if this proof is correct

[worldly chasm]

I don't see an error

[hasty shuttle]

thank you

[dusk finch]

do you call (a or -a) == T negation law btw?

never seen that name before

[swift locust]

school sent this sheet

[dusk finch]

oh ic

[swift locust]

is this correct, ive never seen the step used from 3 to 4, not seeing it on the sheet of laws we were given

it doesnt look like the distributive law ive used before

[dusk finch]

it seems kinda weird

it is correct, but i dont see how it directly follows from the distributive law

[hasty shuttle]

i hate these

[dusk finch]

youd have to do it in multiple steps

[swift locust]

ok, what other steps would work? i was trying to distribute first, but i dont see where i can go after that to get
(~p ∧ ~s) ∨ (p ∧ s) as a result

[dusk finch]

distribute twice

what did you get after the first distribution?

[swift locust]

(~p ∨ s ^ ~s) v (~p ∨ s ^ p)

thats what i got

[dusk finch]

use parenthesis

(-p v s) ^ (-s v p)
((-p v s) ^ -s) v ((-p v s) ^ p)

it should be this, with parenthesis like this

now you can use distributive law on the 2 sub-expressions

[hasty shuttle]

can you speak to us like we are apes

we have no clue what were doing

[dusk finch]

use distribution on these 2

[hasty shuttle]

God bless you

[swift locust]

like distribute the -s into the (-p v s) ?

[dusk finch]

the motivation behind that is that you'll get s ^ -s, which will cancel

and in the second expression, youll get -p ^ p

so lots of stuff will cancel

and now you have v in the middle instead of ^, so if youre lucky, you should just end up with what you need

[swift locust]

what else do i do with -p ^ p besides = to F tho?

i dont see what else i can do with that

[dusk finch]

it will evaluate to F and then if youre lucky, youll get to use some identity law

[swift locust]

ahh

[dusk finch]

whenever you have F or T in ur expression, it's very likely you can somehow make it disappear through identity (or domination)

[hasty shuttle]

i see

[swift locust]

so this is correct now?

[dusk finch]

seems ight

idk how strict ur teacher is about e.g. stuff like commutativity

can you apply commutativity and distributivity in one step?

or do you have to do it in separate steps?

[dusk finch]

because the distributivity here wants it to be of form p ^ (q v r)

[dusk finch]

while here it's (q v r) ^ p

so its basically equivalent up to commutativity

[swift locust]

im not sure how strict the marking is honestly but ill switch it around to be safe

[dusk finch]

i just dk whether youre supposed to do it in extra step, or whether you can just do it at once (its not a big deal anyway)

[hasty shuttle]

thank you bro

[swift locust]

alr thx a lot for the help

[dusk finch]

np

[cedar kiln]

@hasty shuttle Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

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

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

[sonic fossil]

If \theta \in \left(0, \frac{\pi}{4}\right) and

[
t_1 = (\tan \theta)^{\tan \theta}, \quad
t_2 = (\tan \theta)^{\cot \theta}, \quad
t_3 = (\cot \theta)^{\tan \theta}, \quad
t_4 = (\cot \theta)^{\cot \theta}
]

Then which of the following is correct?

1. t_1 > t_2 > t_3 > t_4

2. t_2 > t_1 > t_3 > t_4

3. t_4 > t_3 > t_1 > t_2

4. t_1 < t_3 < t_2 < t_4

[wraith dagger]

BlackidoZ
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[sonic fossil]

what can be the first idea that would help to work out this problem

[tropic oxide]

write z for tan(θ)

you are comparing:

• t_1 = z^z
• t_2 = z^(1/z)
• t_3 = (1/z)^z
• t_4 = (1/z)^(1/z)

knowing 0 < z < 1

[unique copper]

You can also use the fact that the inequality will be true for all theta in (0, π/4) → tan theta in (0, 1).

So you can substitute tan theta = 0.5 and the inequality order should hold.
Now, just find t1, t2, t3, t4 for tan theta = 0.5 and you will get the order.

But if it's not about fast solving in exams, you should try to observe its behaviour as Ann said.

[tropic oxide]

i would put everything in terms of one base, either z or 1/z depending on preference

[sonic fossil]

Whydo you exactly put 0.5

[unique copper]

you can put any number but i felt square root is easier to calculate nothing else

[sonic fossil]

Oh yes I got it

[tropic oxide]

1. t_1 > t_2 > t_3 > t_4
   i wonder what these weird boxes are

[sonic fossil]

Lol

like I use chatgpt to write latex because I don't know the format

Option 3

.close

[cedar kiln]

Channel closed

Closed by @sonic fossil

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

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

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

[crimson sedge]

How does sin^(-1)(sin(θ)) = (θ); (-π/2) ≤ θ ≤ (π/2)

[glossy inlet]

By definition?

[twilit escarp]

By definition

[glossy inlet]

That's its whole purpose

[crimson sedge]

i mean i dont understand the definition

[twilit escarp]

Oh ok

To define a bijective inverse

We need the function to be bijective

So we say that we take the [-pi/2,pi/2] interval and see that sin is strict decreasing

So is continious and so is making a bijection from [-pi/2, pi/2] -> [-1,1]

And so the inverse is defined on [-1,1] and send back to [-pi/2, pi/2]

Noted arcsin

[crimson sedge]

ohh, now i get it, thanks

.close

[cedar kiln]

Channel closed

Closed by @north ether

Use .reopen if this was a mistake.

[twilit escarp]

The [-pi/2, pi/2] is important cuz else, its not bijective and we need some tricks to do so

[crimson sedge]

.reopen

[cedar kiln]

✅

[glossy inlet]

Although you could just as easily choose a different interval to establish a bijection, this one is conventional

[twilit escarp]

Even tho it could have been defined from the start on any intervals where sin is bijective, but since sin is odd, its quite cool to take it centered in the origin

Ah i got sniped

[quiet python]

yo, whats the difference between all the maths? like math 30 and 31 and etc

[crimson sedge]

im sorry but i dont really understand 'bijection/bijective'...just so you know, im just a 14 year old

[glossy inlet]

Are you familiar with "one-one" and "onto" functions?

[twilit escarp]

Yo no idea but its occupied, please open a new help channel, the ones avaible just above ! Ty

[glossy inlet]

Or functions at all?

[crimson sedge]

fuctions yes

one-one, no

[glossy inlet]

[twilit escarp]

Bijective == one to one which means that you have a unique output for a unique entry

[glossy inlet]

Injections and surjections?

Maybe it's a translation thing

[crimson sedge]

🫠 nope

[crimson sedge]

ohh

[glossy inlet]

Is this just a question you had from looking at stuff or was this taught to you in class?

[ember heath]

you can derive it by plotting the graph

[crimson sedge]

taught to me

[glossy inlet]

Huh that's odd then

[crimson sedge]

imma share what my teach told me, just 1 minute

[twilit escarp]

So you have proprety such as, a continious, strict monotonic function is bijective

[glossy inlet]

You've been taught about inverses without the conditions for what makes a well defined inverse?

[crimson sedge]

exactly, the teacher did something with graph

went straight over my head

[ember heath]

look at what i sent

[twilit escarp]

I admit that this is not a good thing

[glossy inlet]

I blame your teacher

[ember heath]

yeah we can assign blame

but thats not going to help

[crimson sedge]

wait, i am sharing what he told us

[glossy inlet]

It might not be good to proceed in this case, is what I was getting at

Without establishing the basics

[crimson sedge]

Something like this

[glossy inlet]

It seems that he was restricting the domain of f(x) = sin(x) to produce a bijection as was said by someone else earlier
nvm

[crimson sedge]

this is from the lecture

[upper ruin]

Yeah, that's this

[glossy inlet]

Only problem he hasn't done that lmao

[upper ruin]

Indeed

[calm sierra]

just chipping in a bit here, an intro to injectve/surjective/bijective functions https://www.mathsisfun.com/sets/injective-surjective-bijective.html

[upper ruin]

This is the restricted domain of sine, i.e. from -π/2 to π/2

For defining arcsin

[crimson sedge]

ohhh

[crimson sedge]

finally something makes sense to me 😭 , im sorry

[upper ruin]

Sorry for the horrible drawing of mine lol

[upper ruin]

Don't worry, we're all here to help

[crimson sedge]

its fine, i understand "restricted domain"

[upper ruin]

And we do that really gladly when people show real interest, such as in your case 🤗

[crimson sedge]

thanks 🙌

so the same applies with inverse functions of cos(θ)?

like not the exact same

but similar

[upper ruin]

Yep

[crimson sedge]

if you can, could you explain this too, by graphical method, it is called i suppose?

[upper ruin]

For cosx the restricted domain is from 0 to π

[upper ruin]

The fact is that cosx is bijective, if you consider x between 0 and π

[crimson sedge]

just to confirm, pi represents measurement in radian right?

[upper ruin]

π is just the number 3.1415...

And π (rad) = 180°

[crimson sedge]

im sorry for the poor terminology

[twilit escarp]

Don't be sorry, you're here to learn

[upper ruin]

All good 👍

[twilit escarp]

Rather do the mistake here than in an exam paper

Nobody is judging

[crimson sedge]

yeah, im aware of that

[crimson sedge]

this too

[crimson sedge]

i meant in this scenario

[crimson sedge]

throughout the conversation, not just this message

[upper ruin]

Also because me and many other here are not English natives

[crimson sedge]

thanks 🙏

[upper ruin]

You have been understood perfectly, which is the hood point

[crimson sedge]

me neither, although im bilingual, im not familiar with mathematical terminology

[upper ruin]

I'm not sure what you mean with representation

[crimson sedge]

where?

[upper ruin]

π is a number like all the other ones, so you can have π degrees, or also 56 rad

[upper ruin]

In your previous sentence about π

[crimson sedge]

not represents

i meant, pi here means pi (rad), right?

[upper ruin]

Ohh yeah yeah

In the plots you meant, right?

[crimson sedge]

yeah

[upper ruin]

Yep those values on the x-axis are in rad

[crimson sedge]

thanks

im grateful for all of your help

imma close this then, ig?

.close

[cedar kiln]

Channel closed

Closed by @north ether

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[vapid jolt]

For context this is intro level stats, the second image shows what the test specifically covers.

Why is this wrong? I thought that the ≠ means its a two tailed test, and I should use 2*norm.dist(z, 0, 1)

[cedar kiln]

@vapid jolt Has your question been resolved?

[cedar kiln]

@vapid jolt Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[heady lava]

This might be stupid, but my Calculator always gives slightly the wrong number, every time, I've done dozen of exercises and I don't know what's wrong, I have an exame in 2 days so pls help haha

[fathom gulch]

Rounding errors? Also, is it using the correct method? (least squares vs logistic regression)

[heady lava]

impossible since im just copying the table from the book

[fathom gulch]

Yeah but when you choose to make a regression of the data, do you have a choice of which method the calc should use?

[heady lava]

´my bf did the same and we have the same calculator and his was right

[fathom gulch]

If you don't have any algs stored, you might wanna factory reset the calc then?

[heady lava]

oh yeah but the other one is not what theyre asking

in the question

[heady lava]

how do i do that sorry

[fathom gulch]

Do you have the manual for the calculator? In the menus, do you see a "reset" option somewhere?

[heady lava]

i found it

nope

still doing it

[fathom gulch]

I suppose you triple checked you entered the values correctly? Did you clear the tables too?

[heady lava]

yep

i did it with three exercises

[fathom gulch]

If your bf doesn't take the exam at the same time as you, you could always borrow his?
Unfortunately, I won't be of any more help, maybe some other Helper will know better!

Best of luck to you

[heady lava]

hahahha he is leaving the country tomorrow, i will try asking my friends

thank you tho!!

.close

[cedar kiln]

Channel closed

Closed by @heady lava

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[torn marsh]

Correct or incorrect

[torn marsh]

,rotate

[torn marsh]

because its different to my answer solution, but not sure if its the same thing

[pure mulch]

can u guys help me with a math problem its kinda fast not that complex

[torn marsh]

!help

[cedar kiln]

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

[pure mulch]

maaan

i dont feel like doing that brah

[dusty wren]

You problem

[torn marsh]

@dusty wren

[dusty wren]

Checking

Well its definitely not the same thing lemme see what went wrong

[torn marsh]

alright

[dusty wren]

Can u tell whats wrong here

@torn marsh

[torn marsh]

hmm

oh

chain rule

[dusty wren]

Yeah

[torn marsh]

dammit

how do i stop forgetting

[dusty wren]

Start remembering

Just practice lol

[vagrant moat]

"the derivative of the outside (inside) multiplied by the derivative of the inside"

[torn marsh]

ikk

[vagrant moat]

I'm saying if you get that line into your head, you won't forget

[torn marsh]

yeah

i get the line

[west path]

i mean are they wrong

[torn marsh]

ok thanks guys

[vagrant moat]

And for the product rule:
"the derivative of the first times the second + the first times the derivative of the second"

[dusty wren]

I mean theres nothing to gain from that statement tho it was a joke lol

[torn marsh]

.close

[cedar kiln]

Channel closed

Closed by @torn marsh

Use .reopen if this was a mistake.

[vagrant moat]

They have a nice rhythm to them so that should make them easier to remember

[torn marsh]

.reopen

[cedar kiln]

✅

[torn marsh]

I re attemped the qs, is this correct now?

,rotate

[dusty wren]

No

You forgot to write the second function in multiplication with -4(1-x)^3

[torn marsh]

so like this?

,rotate

[dusty wren]

Yes

[torn marsh]

it still doesnt

match the answer

The first line

[dusty wren]

Answer wrong

[torn marsh]

which one?

[dusty wren]

Textbook

Lemme check just in case am hallucinating

,w differentiate 1/3 x^2(1-x)^4

[wraith dagger]

Results provided by WolframAlpha

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

Requested by executor.1011

[dusty wren]

Yeah yours is right

[split ice]

as a sanity check, this looks correct to me

[torn marsh]

what about my answer

[split ice]

also the same, your -4 and x^2/3 are just far apart

[torn marsh]

which one is preffered

[split ice]

in this case? id say both work

and are preferred

[dusty wren]

!done

[cedar kiln]

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

[torn marsh]

.close

[cedar kiln]

Channel closed

Closed by @torn marsh

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

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

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

[mortal yarrow]

what does it mean for a set to be open in R^n

[mortal yarrow]

this is extrememly confusing

extremely*

[glossy inlet]

Well, do you know what an open ball is?

What part is confusing you?

[mortal yarrow]

nvm

i understand this completely

.close

[cedar kiln]

Channel closed

Closed by @mortal yarrow

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

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

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

[viscid dust]

We consider $A_0 \in M_n(\mathbb{R})$ such that $A_0 + A_0^T$ has eigenvalues $\leq 0$, and we study the matrix equation:

$$
A'(t) = A(t)^2 + (A(t)^T)^2
$$

Show that $A$ tends to a finite limit as $t \to +\infty$ to be determined. A suggested approach is to solve the problem for the symmetric and skew-symmetric parts of $M_n(\mathbb{R})$ by setting:

$$
S = \frac{A + A^T}{2}, \quad D = \frac{A - A^T}{2}.
$$ i find D cste now how I work with S ?

[wraith dagger]

Clint

[hidden mural]

what did you find for D

[cedar kiln]

@viscid dust Has your question been resolved?

[viscid dust]

A0-tA0/2 I Forget to write it but A(0)=A0

We define ( D(t) = \frac{A(t) - A(t)^T}{2} ).
Then its derivative is:

[
D'(t) = \frac{A'(t) - A'(t)^T}{2}
]

From the equation:
[
A'(t) = A^2 + (A^T)^2
\Rightarrow A'(t)^T = (A^2)^T + ((A^T)^2)^T = (A^T)^2 + A^2
]

So:
[
D'(t) = \frac{A^2 + (A^T)^2 - (A^T)^2 - A^2}{2} = 0
]

(\boxed{D(t) \text{ is constant over t}})

[wraith dagger]

Clint

[viscid dust]

We define the symmetric part as:
[
S(t) = \frac{A(t) + A(t)^T}{2}
]
Differentiating:
[
S'(t) = \frac{d}{dt} \left( \frac{A(t) + A(t)^T}{2} \right)
= \frac{A'(t) + (A'(t))^T}{2}
]
Using the equation ( A'(t) = A(t)^2 + (A(t)^T)^2 ), we get:
[
(A'(t))^T = (A(t)^T)^2 + A(t)^2
\Rightarrow S'(t) = \frac{A(t)^2 + (A(t)^T)^2 + (A(t)^T)^2 + A(t)^2}{2}
= 2S(t)^2 + 2D^2
]
Thus:
[
\boxed{S'(t) = 2S(t)^2 + 2D^2}
\quad \text{where } D = \frac{A_0 - A_0^T}{2} \text{ is constant.}
]

[wraith dagger]

Clint

[cedar kiln]

@viscid dust Has your question been resolved?

[cedar kiln]

@viscid dust Has your question been resolved?

[cedar kiln]

@viscid dust Has your question been resolved?

[cedar kiln]

@viscid dust Has your question been resolved?

[split ice]

next step would be to see if S converges (goes to a finite limit)

also, can you write A in terms of S and D?

[split ice]

hint: ||the way you wrote S' isn't helpful imo, try what you did for D||

[cedar kiln]

@viscid dust Has your question been resolved?

[viscid dust]

I tried this but it doesn't help A=S+D

[viscid dust]

[
\boxed{S'(t) = A(t)^2 + \left(A(t)^T\right)^2}
] I didnt help to

[wraith dagger]

Clint

[cedar kiln]

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).

[split ice]

the idea is, if you can show that $\lim_{t \to \infty} S(t)$ is finite, then you can show that $\lim_{t \to infty} S + D = \lim_{t \to \infty} A$ is finite as well, right?

[wraith dagger]

haseeb

[split ice]

i will say i'm not sure how to take the limit S, i thought i did but it was wrong ;-;
what about the fact that the eigenvalues of $A_0 + A_0^T$ are nonpositive?

[wraith dagger]

haseeb

[viscid dust]

y that's the point we use that Mn(R)=Sn(R)+An(R) that why we use S and D to find the limite of them ton conclude

[viscid dust]

great question idk where we can use this hypothesis we should find somthing like S=exp((A0+TA0)*t) so lim of S = 0 but idk

[cedar kiln]

@viscid dust Has your question been resolved?

[hushed sonnet]

I do think the forum is better for these kinds of problems

People are used to easier stuff here 😉

[viscid dust]

so where i can ask on this server ?

[dusty wren]

Well if you want to ask on the forum, that would be #1021175428326633542 . But it's not the case that your doubt won't get resolved here- do keep in mind that multiposting isn't allowed, so try to keep your question limited to only either a channel or a forum post at a time.

[viscid dust]

I made a mistake, I deleted the post on the forum, I thought.

[cedar kiln]

@viscid dust Has your question been resolved?

[potent fractal]

You accomplished a nice intermediate fact, let's finish it

Once can show that for a matrix-valued function $F(t)$, we have
$$\frac{d}{dt}(F(t)^{-1}) = -F(t)^{-1}\dot{F}(t)F(t)^{-1}.$$

[wraith dagger]

spindle

[potent fractal]

Now let's solve the following equation: $\dot{S}(t) = 2S(t)^2$

[wraith dagger]

spindle

spindle

[potent fractal]

Nvm this doesn't help with S' = 2S^2 + 2D^2

[viscid dust]

^^

[potent fractal]

Yeah I'm now thinking about using either series or derivatives of functions of matrices

[proper prism]

whats the question?

[viscid dust]

[
\text{Find } \lim_{t \to +\infty} A(t) \text{ where } A(t) \in \mathcal{M}_n(\mathbb{R}) \text{ and}
]
[
\begin{cases}
A'(t) = A(t)^2 + \left(A(t)^T\right)^2, \
A(0) = A_0, \
\sigma\left(A_0 + A_0^T\right) \subset \mathbb{R}^-.
\end{cases}
]

[wraith dagger]

Clint

[viscid dust]

My work for the moment :
[
\text{We use the decomposition } \mathcal{M}_n(\mathbb{R}) = \mathcal{S}_n \oplus \mathcal{A}_n \text{ with }
S(t) = \frac{A(t) + A(t)^T}{2},\quad D(t) = \frac{A(t) - A(t)^T}{2}.
]

[
\text{We have shown that } D(t) \text{ is constant. We now study the evolution of } S(t).
]

[wraith dagger]

Clint

[proper prism]

find A & A^t in terms of S & D_0 @viscid dust

[potent fractal]

He already did that and established the following system:
$$
\begin{cases}
\dot{D} = 0, \
\dot{S} = 2(S^2 + D^2), \
(S+D)(0) = A_0,
\end{cases}
$$
where $S(t)$ is symmetric, $D(t)$ is skew-symmetric, and $\sigma(A_0 + A_0^T) \subset \mbb{R}^-$.

[wraith dagger]

spindle

[snow nacelle]

yeah so youre gonna want to add 7

[proper prism]

good now they can diagonalize D into 2x2 blocks

[potent fractal]

Well, diagonalization doesn't really simplify the problem, does it? Because the transition matrix also depends on t and therefore the equation gets messy
Even if we diagonalize S or D^2 (they are symmetric and hence diagonalizable)

[proper prism]

whats the issue, why r we even dealing with the transition matrix explicit form, we know it exists thats it

[potent fractal]

I'm not saying that we should look at the explicit form

I'm just saying that I don't see how it simplifies the problem

[proper prism]

S(t) converges to a block-diagonal matrix S_infty

[potent fractal]

Why?

[proper prism]

as i said

each S_i converges to -b_i I_2 as t->infty

[potent fractal]

Why?

Also where did the transition matrices go

[proper prism]

u know riccati equation?

[proper prism]

consider S tilde instead of S

[potent fractal]

We have D^2 = C B^2 C^{-1} where B is the block-diagonal matrix, okay

Now we should get rid of C and C^{-1}, as you said

[proper prism]

S tilde = P^t S P

[potent fractal]

Yes sure, but now compute d/dt (P^T S P)

It's not equal to P^T d/dt S P

Oh nvm D is constant and hence P is constant

[proper prism]

we needn't concern ourselves with that spindle

[potent fractal]

Yes I see I see

In my head I was diagonalizing S and that idea failed because the transition matrices were not constant

So when I saw your idea with diagonalisation of D I assumed that it had the same problem

Because I forgot that D is constant and we're fine

Also we can just diagonalize D^2 (because it's symmetric) and simplify even further

[proper prism]

exactly

K = -D_0² to be exact

[potent fractal]

Anyways, I'm not the OP so let's wait for them to answer

[proper prism]

yes for sure , anyways i just wanted to hint out that the way to the proof is considering that new non linear equation as a riccati equation

we need a solution that satisfy the equilibrium basically S'(t) = 0 as t->infty

$S_i(t)' = 2(S_i (t)^2 - b_i ^2 I_2)$

S_i is a 2x2 matrix block of S tilde the diagonalized form of S

[wraith dagger]

Goëtia

[potent fractal]

But we're not guaranteed that P^T S P will have a block-diagonal form, or are we?

Because P makes D block-diagonal, not S

[proper prism]

the equation is invariant under orthogonal transformation

[potent fractal]

I agree

[proper prism]

i think we can assume that S is invertible x)

sec let me think of smth

[potent fractal]

I understand that your idea is to reduce the problem to 2x2 matrices

But applying a constant orthogonal transformation to the whole equation does not make both S and D block-diagonal at the same time

[proper prism]

S^2?

[potent fractal]

Also S(t) changes with t so this is another problem

[proper prism]

sec

D0 and D0² commutes

as in since D0 is antisymmetric

there exists P: P^T D0 P = (B1 0 ...
0 B2 ..
...
0 ... Bk ..
0 ... 0_{mxm})

where B_i = (0 b_i
-b_i 0) b_i > 0

0_{mxm} is the kernel of D0

K = -D0² is block-diagonal too in the same basis

[potent fractal]

Furthermore, it's diagonal

[proper prism]

yes

apply that transformation to the original non linear equ of S

u will end up with invariance

[potent fractal]

Yeah we get S' = 2S^2 + J, where J is a diagonal matrix with non-positive numbers on the diagonal

S(t) is symmetric but not necessarily diagonal or block-diagonal

Anyways, we simplified the problem slightly, which is nice

[proper prism]

however 🙂

it decouples asymptotically

[potent fractal]

damn nice

[proper prism]

the off-diagonal elements satisfy homoge linear equation with lyapunov neg exponents

thus they decay to 0

anyways i gotta go watch some anime, ping me whenever is needed, take care!

[potent fractal]

I think I got the idea, it's time to wait for OP's response
Take care

[viscid dust]

Since $D \in \mathcal{A}_n(\mathbb{R})$ is a real skew-symmetric matrix, it admits a block-diagonal canonical form under an orthogonal change of basis. That is, there exists an orthogonal matrix $P$ such that:
[
P^T D P =
\begin{pmatrix}
\begin{matrix}
0 & b_1 \
-b_1 & 0
\end{matrix} & & \
& \ddots & \
& & \begin{matrix}
0 & b_k \
-b_k & 0
\end{matrix} \
& & & 0_m
\end{pmatrix}.
]

Squaring this expression yields:
[
P^T D^2 P =
\mathrm{diag}(-b_1^2 I_2, -b_2^2 I_2, \dots, -b_k^2 I_2, 0_m),
]
where each $b_i > 0$, and $I_2$ is the $2 \times 2$ identity matrix.

Now let us define $\widetilde{S}(t) = P^T S(t) P$. Since $P$ is orthogonal and $D$ is constant, the transformed system becomes:
[
\widetilde{S}'(t) = 2\left( \widetilde{S}(t)^2 + P^T D^2 P \right).
]
In this basis, the system is decoupled into independent blocks. Each $2 \times 2$ block $\widetilde{S}_i(t)$ satisfies the matrix Riccati differential equation:
[
\widetilde{S}_i'(t) = 2\left( \widetilde{S}_i(t)^2 - b_i^2 I_2 \right).
]

This shows that the full matrix evolution reduces to analyzing a collection of identical Riccati-type equations in low dimension.

[wraith dagger]

Clint

[viscid dust]

Have I summarized your progress well?

[potent fractal]

You probably overlooked the part where we discussed that \tilde{S} need not be block-diagonal

Goëtia claims that \tilde{S} decouples asymptotically due to off-diagonal considerations, but it's still not obvious

We reduced the problem to the following:
$$\begin{cases}
\dot{\widetilde{S}} = 2\widetilde{S}^2 - J, \
\widetilde{S}(0) \le 0,
\end{cases}$$
where $J = 2\text{diag}(b_1^2, b_1^2, \dots, b_k^2, b_k^2, 0, \dots, 0)$ and $\widetilde{S}(t)$ is symmetric.

[wraith dagger]

spindle

[viscid dust]

that's what i said but idk how i deal with the off-diagonal elements

Even with diagonal éléments honestly

How do u solve S’=2S^2

We start from the equation:
[
S'(t) = 2S(t)^2, \quad \text{where } S(t) = (s_{ij}(t))_{1 \leq i,j \leq n}
]

This yields, for each ( i, j \in {1, \dots, n} ):

[
s'{ij}(t) = 2 \sum{k=1}^{n} s_{ik}(t) , s_{kj}(t)
]

\textbf{Diagonal entries} (( i = j )):
[
s'{ii}(t) = 2 \sum{k=1}^{n} s_{ik}(t)^2
\quad \text{(since } s_{ki}(t) = s_{ik}(t) \text{ by symmetry)}
]

\textbf{Off-diagonal entries} (( i \neq j )):
[
s'{ij}(t) = 2 \sum{k=1}^{n} s_{ik}(t) , s_{kj}(t)
]

[wraith dagger]

Clint

[viscid dust]

We have somthing like that [
S(t) = S_0 (I - 2t S_0)^{-1}
]
Mmh we can generalise for n=1 if S0 commutes with S but we dont have this hypothesis

[wraith dagger]

Clint

[viscid dust]

« it decouples asymptotically » wdym ?

[viscid dust]

?

U have done that ok ? Let ( D \in \mathcal{A}_n(\mathbb{R}) ), so ( D^2 \in \mathcal{S}_n(\mathbb{R}) ) is symmetric and negative semi-definite. Then there exists an orthogonal matrix ( P \in \mathrm{O}_n(\mathbb{R}) ) such that:

[
P^T D^2 P = \Lambda = \mathrm{diag}(\lambda_1, \dots, \lambda_n), \quad \lambda_i \leq 0
]

Let ( \widetilde{S}(t) = P^T S(t) P ). Then ( \widetilde{S}(t) \in \mathcal{S}_n(\mathbb{R}) ) satisfies:

[
\widetilde{S}'(t) = 2\left( \widetilde{S}(t)^2 + \Lambda \right)
]

In components:

[
\begin{cases}
\sigma_{ii}'(t) = 2 \left( \sum_{k=1}^n \sigma_{ik}^2(t) + \lambda_i \right), \
\sigma_{ij}'(t) = 2 \sum_{k=1}^n \sigma_{ik}(t)\sigma_{kj}(t) \quad \text{for } i \neq j.
\end{cases}
]

[wraith dagger]

Clint

[viscid dust]

I may not have any knowledge of this type of non-linear system, I don't know if you have any information, in any case it goes beyond the scope of my course, I think.

especially information near infinity

[coral pond]

how long does it take to write those god dawm

[worldly chasm]

Not that long, it's a skill you develop

[pastel vault]

https://texnique.xyz/
there's this LaTeX game you can try to practice

warning: it's hard

you can draw a symbol here and it will tell you the LaTeX also:
https://detexify.kirelabs.org/classify.html

[raven shard]

this game is not a good way to learn latex

[livid hound]

how are you supposed to enter differentials

[edgy spade]

??

[twilit escarp]

mathrm{d} x

Else it cries cuz its not vertical

[livid hound]

i ended up using \text
\dd didn't work

[twilit escarp]

I think its a cool way to discover some topics and some formulas used in it at least

[livid hound]

whoops didn't realise this wasn't a social channel

[inner thistle]

I just do \text d

[viscid dust]

$, \mathrm{d}x$

[wraith dagger]

Clint

[old hornet]

$,help$

[proper prism]

@viscid dust what exactly is the question status right now

[viscid dust]

Let ( D \in \mathcal{A}_n(\mathbb{R}) ), so ( D^2 \in \mathcal{S}_n(\mathbb{R}) ) is symmetric and negative semi-definite. Then there exists an orthogonal matrix ( P \in \mathrm{O}_n(\mathbb{R}) ) such that:

[
P^T D^2 P = \Lambda = \mathrm{diag}(\lambda_1, \dots, \lambda_n), \quad \lambda_i \leq 0
]

Let ( \widetilde{S}(t) = P^T S(t) P ). Then ( \widetilde{S}(t) \in \mathcal{S}_n(\mathbb{R}) ) satisfies:

[
\widetilde{S}'(t) = 2\left( \widetilde{S}(t)^2 + \Lambda \right)
]

In components:

[
\begin{cases}
\sigma_{ii}'(t) = 2 \left( \sum_{k=1}^n \sigma_{ik}^2(t) + \lambda_i \right), \
\sigma_{ij}'(t) = 2 \sum_{k=1}^n \sigma_{ik}(t)\sigma_{kj}(t) \quad \text{for } i \neq j.
\end{cases}
] now how we continue

[wraith dagger]

Clint

[dire geode]

Just post this to math stack exchange

[lofty sand]

I’m sorry I’m late to the party so I went back to the original problem. What’s the relationship between A(t) and A_0 ?

[upper laurel]

A(0) = A_0

[upper laurel]

tried solving this for a 2x2 matrix and couldnt get anywhere

the best I have so far, which is unchecked, is
[
\begin{cases}
a_{11}'(t)=2\left(a_{11}(t)^2+C_+^2\exp\left(4\int_0^t(a_{11}(x)+a_{22}(x)),dx\right)-C_-^2\right)
\a_{22}'(t)=2\left(a_{22}(t)^2+C_+^2\exp\left(4\int_0^t(a_{11}(x)+a_{22}(x)),dx\right)-C_-^2\right)
\a_{11}(t)-a_{22}(t)=C_1\exp\left(2\int_0^t(a_{11}(x)+a_{22}(x)),dx\right)
\end{cases}
]
for (t\ge0) where (\textstyle C_-=\frac{a_{12}(0)-a_{21}(0)}2,C_+=\frac{a_{12}(0)+a_{21}(0)}2,C_1=a_{11}(0)-a_{22}(0))

[wraith dagger]

mtt

[upper laurel]

this is assuming C_-, C_+, and C_1 can be zero, so replacing the exp integrals with a11 - a22 is not allowed

you can do so by assuming $C_1\ne0$, but the result doesnt look any easier
[
\begin{cases}
a_{11}'(t)=2\left(a_{11}(t)^2+\frac{C_+^2}{C_1^2}(a_{11}(t)-a_{22}(t))^2-C_-^2\right)
\a_{22}'(t)=2\left(a_{22}(t)^2+\frac{C_+^2}{C_1^2}(a_{11}(t)-a_{22}(t))^2-C_-^2\right)
\end{cases}
]

[wraith dagger]

mtt

[upper laurel]

at least we have that
[
a_{12}(t)=C_-+C_+\exp\left(2\int_0^t(a_{11}(t)+a_{22}(t)),dt\right)
]
and also $a_{21}(t)=a_{12}(t)-2C_-$

[wraith dagger]

mtt

[proper prism]

@viscid dust okey great, u've done enough there, you just need to have a look at each case separably

at the equilibrium S_tilde ' = 0 thus S_tilde ² = - diag(lambdas)

sigma_ii(t) >= 2lambda_i trivial

making sure sigma_ii converges to -sqrt(-lambda_i)

for the off diagonal elements, you might wanna consider an extra decomposition of that sum

$\sigma_{ij}'(t) = 2(\sigma_{ij}( \sigma_{ii}+\sigma_{jj}) + \sum_{k \neq i,j} \sigma_{ik}(t) \sigma_{kj} (t))$

[wraith dagger]

Goëtia

[proper prism]

the last term = 0

since we said that off-diagonal elements must be 0

the diff equ turns out to look like $\sigma_{ij}'(t) = -2c \sigma_{ij}(t)$