help-19

[loud palm]

hmmhmm

gotcha

[loud palm]

.close

[odd edge]

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[void niche]

Please Someone help

[bitter folio]

Welcome to mathcord btw

[void niche]

Help me with this

Prove that the shortest line segment that can be drawn from a point to given line is Perpendicular to the given line.

It's confusing for me

[bitter folio]

Ok

Draw it out

[void niche]

I tried many ways

[bitter folio]

Draw an arbitary line and a point

[void niche]

What did you thought I will not try to do this?

[bitter folio]

What?

@void niche what did you try

[outer wadi]

try using the pythagorean theorem to prove this

[warm dock]

take a line and any point, say A, lying outside the line

then draw rays extending outwards to all directions from that point A

consider only the rays that intersect the line

you'll find that only one ray exists that joins the line and the point and is also the shortest one. this ray is perpendicular to the line and the point

all other rays don't fit this criterion

note: this is not a rigorous proof

[outer wadi]

what do you mean by only one ray exists that joins the line and the point?

[warm dock]

oh sorry, i meant
only one such ray extending out from point A will exist that will intersect the line and also be the shortest ray

[outer wadi]

there are infinite many rays that will intersect the line

as long as the ray is not paralell to the line

unless Im misunderstanding

[warm dock]

yeah and the shortest one will be perpendicular to the line

[outer wadi]

yeah but they are trying to prove that

[warm dock]

this is a way to prove

at least i was taught this way

[odd edge]

@void niche Has your question been resolved?

[outer wadi]

lets look at it like this

for any point on the line and the given point A create a position vector v by subtraction of the two coordinates

When the magnitude of this vector is minimized we obtain the shortest distance to this line

what is this minimized vector and prove that this vector is perpendicular to the line

[exotic bobcat]

Euclidean axioms tell us that "The shortest path between two points is the straight line".
So this shortest path is also a straight line, we just have figure out to which point on the segment we're drawing a straight line

now of all the infinitely many points on the line segment, which one minimizes distance to the arbitrary point?

And what angle will the connecting line have w.r.t. the original line?

[odd edge]

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[tired wadi]

i need help with this

[tired wadi]

so what i did is i wrote sin3x with the formula 3sinx - 4sin^3x

so what i really need help with is how to solve integral pi/4 to 3pi/4 of 1/sinx dx

thats all

nvm i got it

.close

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[odd edge]

@mystic saffron Has your question been resolved?

[dense turret]

okay

[ripe chasm]

okay

jinx haha!

[dense turret]

Oh :o

[ripe chasm]

u can do it

[dense turret]

thanks

now you don’t need to overcomplicate things when we have difficult powers

Do you know the binomial expansion?

[ripe chasm]

im getting 843

i might have gotten it wrong idkkk :((

wait lemme re solve

[dense turret]

Well i’ll explain till that time 💀

Soo

Don’t overcomplicate things

First of all, we need to find the roots of the first equation.

Can you try yourself?

Or should I give you a hint?

Oh wow, well, i would give up on that question too

But ill give you a hint.

man we just have experience 😭

[ripe chasm]

frrr

[dense turret]

So we have:

shushhhh

class is in session

[ripe chasm]

ate

[dense turret]

So like u just want the answer or smth?

[ripe chasm]

i think 843 is correct

[dense turret]

guys come on 😭

[ripe chasm]

well im about to dig a hole and bury myaself

myself

soorry

[dense turret]

ok ok

So look at the above eq

Now solving a cubic is going to kill us

But what if we made it simpler?

Let’s call the x3 = u

So now we have u + 1/u = 18

Multiply both sides by u.

(GIVEN U ≠ 0)

quadrstic baby!,1!

yessir

Solve the quadratic now

🗣️‼️🗣️GIT OUTTT

duh

quadrstic formula

q u a d r a t i c. F o r m u l a

[ripe chasm]

real

[dense turret]

let anvesha do it then

[ripe chasm]

help no

[dense turret]

you’re indian aren’t you? im an indian too >:)

[ripe chasm]

yea

good!

I AM INDIAN TOOO

[dense turret]

CORRECTO

[dense turret]

Uk hindi?

[ripe chasm]

YES

sorry are we?

[dense turret]

We dont take any accountability

[ripe chasm]

anyways back to da topic

ima try

[dense turret]

ok ok

[dense turret]

You have 3 minutes.

yeah but that is besides the point

[ripe chasm]

9 plus 4root 5

and 9-4root5

[dense turret]

Clap !1!1

Same thing

Anyway its correct

[ripe chasm]

is it correct idk

YAY

are u fr

HELEP

HELP

[dense turret]

Hes not joking

[ripe chasm]

I WAS ABT TO SAY THAT WORD OMG

[dense turret]

Ok ok

[ripe chasm]

I WAS ABT TO SAY R U BEING DEADAHH WITH ME

YEAH

no9w

now

what

[dense turret]

Now did we solve for u or x?

S H U S H

[ripe chasm]

u

we solved for u

2 roots

[dense turret]

bro, we’re solving for a function of x 💀

[ripe chasm]

ate

[dense turret]

BUT, WE FOUND U.

we want x though.

[ripe chasm]

YES WE FOUND U

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[dense turret]

I might swear

[odd edge]

✅

[dense turret]

listen up

We have u = 9 + or - 4 √5

And what is u?

x ^3

[ripe chasm]

x3

yeah that

im lazy to put all that stuff

and im on pc

[dense turret]

So we have x^3 = 9 + - 4 root 5

[ripe chasm]

(╯°□°)╯︵ ┻━┻

[dense turret]

But isnt this sus?

[ripe chasm]

why

would it be sus

wait yeah

[dense turret]

9 plus 4rt5

and 9-4root5

When i multiply these two

[ripe chasm]

WHY ARE WE GETTING 2 ANSWERS

[dense turret]

Its an identity…..

[ripe chasm]

81-180

[dense turret]

(9+4 √5)(9-4 √5)

[ripe chasm]

99

[dense turret]

Sure?

[ripe chasm]

yep

positive

[dense turret]

But its wrong?

[ripe chasm]

how

THE FRICK

IS IT WROBNG

[dense turret]

Now our identity is a ²-b ²

[ripe chasm]

YEP

[dense turret]

That’s (9) ² - (4 √5) ²

[ripe chasm]

81 - 80

s0

1

[dense turret]

Yes :)

[ripe chasm]

HOW TF DID I GET 99

[dense turret]

99 problems 1 solution

[ripe chasm]

IM DEAD BRO

[ripe chasm]

ATE AGAIN

[dense turret]

Gah lock in

Sooo

[ripe chasm]

IVE ADDED U BOTH IN MY FRIENDS THO

[dense turret]

We basically multiplied something times itself

[ripe chasm]

[dense turret]

DELULU SQUAD

[ripe chasm]

YEAH NOW

[dense turret]

Listen up

[ripe chasm]

YEP

i think we are going in the wrong direction

[dense turret]

There’s only 1 direction though

Oh wait they disbanded

Annnyyway:

Ok ill continue and u can read

Now we know that (9+4 √5)(9-4 √5) = 1

But wait! Both factors are x ^3!

So we have x3 * x3 = 1

x6 = 1

Now i’ll explain smth weird

Ok what happened?

[ripe chasm]

..

[dense turret]

Yeah because we found that (9+4 √5)(9-4 √5) is 1

[ripe chasm]

this is how we came to 1

[dense turret]

Now here’s smth rly weird for y’all to process.

[ripe chasm]

PLEASE

BUT I THINK WE ARE GOING IN A WRONG DIRECTION\

[south plume]

what's the problem guys 😭

[dense turret]

in reality x is a cubic root and thus has many solutions considering how many plus or minuses are there if we actually did the cubic

[ripe chasm]

THE ANSWER SHOULD BE 843 UGHHH

[dense turret]

The more we exponentiate the more complex our problem gets

[south plume]

no but like the original problem

[dense turret]

This explains why there might be many values for x

[south plume]

just curious

[dense turret]

You can read along from where we started ig since we’re too deep ;(

[dense moth]

Did you find x ?

[ripe chasm]

IF X3 + 1/X3 IS 18 HOW TF IS X6 + 1/X6 = 1????

[dense turret]

It is impractical

You need to solve a cubic for that

[ripe chasm]

THATSS 2

THATS 2

HELP

[dense turret]

Yes. You might find many solutions.

[ripe chasm]

LEMME TRY PLS

[dense turret]

But follow along

[south plume]

it isn't

[dense turret]

Exactly, its 2

[dense moth]

(x+1/x)³ = x³+1/x³ +3(x+1/x)
Or z³-3z = 18

[dense turret]

and it may also be 18 ²- 2

[dense moth]

So z = 3

[south plume]

it's not 2 either...

[dense moth]

,w solve x+1/x = 3

[clever fjord]

Results provided by WolframAlpha

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

Requested by xor_11

[dense moth]

These are a set of possible values of x

[dense turret]

Exactly

There are many solutions and the complexity increases w exponents.

[south plume]

guys don't solve for x lol

[dense turret]

That’s what im not doing

Look, i found x ^6 may be 1.

[errant gust]

how are you going to find x^7

if u find x

[dense moth]

x and 1/x are interchangeable notice carefully

[errant gust]

there has to be an easier way

[dense turret]

Now the question was x^7 + 1/x ^7

We know that x ^ 6 = 1

[ripe chasm]

GUYS IT CANT BE 1 or 2

I SHOULD BE WAY BIGGER THAN 18?!

[dense turret]

So x ^7 + x^6/x^7 .

[south plume]

it literally isn't ☠️

[ripe chasm]

SEVERAL PEOPLE WTH

[ripe chasm]

EXACTLY

[dense turret]

Ok ill explain how i got there ig if there’s smth wrong pls help

[ripe chasm]

i dont think we can multiply 2 dofferent values and call them x^2

[dense moth]

x³+1/x³ = 18
Or x⁶+1/x⁶ = 18²-2
z+1/z = 18²-2

[ripe chasm]

843 for me

[dense moth]

,w solve x+1/x = 18²-2

[clever fjord]

Results provided by WolframAlpha

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

Requested by xor_11

[south plume]

it's correct

[dense moth]

So this is x⁶ and 1/x⁶

[dense turret]

ooo

[clever fjord]

Results provided by WolframAlpha

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

Requested by f4nt0m.

[dense turret]

I’ll use wolfram to see the solution to the problem directly ig

Oh wait!

It is 843

There are 2 other complex solutions

[ripe chasm]

THANK GOD

we gotta focus on fantom dude

[dense turret]

What grade are you?

[dense moth]

This is wrong

x+1/x = 3

[dense turret]

Bro what?!

[dense turret]

Yes it’s correct

[errant gust]

this is a question from school?

dang

[dense turret]

Usually junior high schoolers get withing x^3 nothing else

Wait there are no roots of unity

Like you would only get maximum x^3 ar that grade

Dont feel bad this question is smth few can solve in 8th grade

Hmm

But i have no idea where i went wrong

[dense moth]

(x³+1/x³)(x²+1/x²) = (x⁵+(x+1/x) + 1/x⁵)
= (x⁵+1/x⁵ +3)
x⁵+1/x⁵ = 18 × 7-3
(x⁵+1/x⁵)(x²+1/x²) = (18×7-3)×7
Or x⁷ + 1/x⁷ + (x³+1/x³) = (18×7-3)×7

,calc 1877-18

[clever fjord]

Result:

864

[dense turret]

I understood where i went wrong

[dense moth]

Is this wrong ?

There was an error fixed it !

[clever fjord]

Result:

861

[dense turret]

This taught me smth :o

Wait i think i’ll be able to explain it to you in dms

[dense moth]

861-18 = 843

[dense turret]

But not now

[dense moth]

X+1/x = 3

[dense turret]

Thanks guys :D

[dense moth]

Did it a long ago, just looking for an easier approach

[dense turret]

But in the expansion it doesnt completely cancel out right?

[dense moth]

This yeilds correct results

[dense turret]

For one to know the higher exponent, one has to also do the lower exponent

And then sub that in

Right?

Ohhh

Can i try generalizing given x + 1/x?

To find x^n + 1/x^n

[errant gust]

im sure its possible

[dense turret]

ookay

[ripe chasm]

IM BACK

SORRY WHERE ARR ME

ty for tagging

[dense moth]

(x³+1/x³)(x²+1/x²) = (x⁵+(x+1/x) + 1/x⁵)
= (x⁵+1/x⁵ +3)
x⁵+1/x⁵ = 18 × 7-3
(x⁵+1/x⁵)(x²+1/x²) = (18×7-3)×7
Or x⁷ + 1/x⁷ + (x³+1/x³) = (18×7-3)×7

It's solved

[ripe chasm]

so whats the answer

is it 834

[dense moth]

,calc (18*7-3)*7-18

[clever fjord]

Result:

843

[ripe chasm]

YAYY

YEAH THAT

843 my bad

YAY

[dense moth]

NP !

[ripe chasm]

ALRIGHT GN

[dense moth]

no

[dense moth]

...

[dense moth]

...

[dense turret]

Im continuing…

[dense moth]

(x+1/x) this has no power, how do you expand it using binomial

[dense turret]

Oh i forgot an n at the top for exponent

[dense moth]

If I consider (x+1/x)^n then we will have this pattern, which is also quite hectic to solve

[dense turret]

So im making a recursive function ig

[dense moth]

Yeah surely

[dense turret]

I’ll call maybe K(n,k,x)

Where K(n,0,x) has the property of being the binomial series expansion for (x + 1/x)^n

[dense moth]

Generating a recursive function is rather difficult

[dense turret]

But ill try :o

[dense moth]

Yeahh, generalised is better !

[dense turret]

K(n,0,x)=x^n + 1/x^n + K(n-1,1,x)

Gimme a sec there’s a redundancy

Middle parameter is useless

[dense moth]

It's difficult to figure out, as we will have the coefficients mess up with the simplicity of the eqn

[dense turret]

Mhm :(

[dense moth]

I better see

If we know, x³+1/x³ and x²+1/x²

We can reach to any desired result

Through repeated products

You may try to unfold that pattern

[dense turret]

Yeah, as shown by your solution for the orginal problem

[dense turret]

Hmm ill try

[dense moth]

Yeahh please let me know!

[dense turret]

Why do i think this will be related to beta/gamma functions?

[dense moth]

You need to level down

[dense moth]

This is linear algebra

[dense turret]

Yeah but like the recursiveness and the powers

[dense moth]

Yeahh I see ...

[dense turret]

Well i havent entered high school yet and i’ve only heard abt linear algebra;(

[dense moth]

Were beta, gamma functions taught earlier?

[dense turret]

No i just learned it for integration

[dense moth]

Okk !

[dense turret]

I’ll try what you recommended…

[odd edge]

@mystic saffron Has your question been resolved?

[odd edge]

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[odd edge]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[woven pier]

cos(x) = x
solve for x

[teal glen]

we'll cos(x) is an odd function or even ??

cos(-x) = ?

[woven pier]

cos(x)

[frigid canopy]

!status

[odd edge]

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

[teal glen]

okay so there's two solutions

[frigid canopy]

uh, no

[teal glen]

my bad

[woven pier]

,w cos(x) = x

[clever fjord]

Results provided by WolframAlpha

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

Requested by anshul6368

[woven pier]

This is only an approximation

[frigid canopy]

there is no analytical solution

at least as far as I know

You'd have to expand cos(x) using it's series expansion and then solve for x

[odd edge]

@woven pier Has your question been resolved?

[woven pier]

.close

[odd edge]

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• 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!

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[humble burrow]

can all natural numbers be represented by $2^n+3^k$, where n and k are whole numbers?

[royal herald]

are you including 0 in natural numbers

[humble burrow]

wait a minute i said that all weird

[clever fjord]

christmasisoversus

[timid sky]

even if it was, how do you represent the number 1?

[royal herald]

you cant

[humble burrow]

why?

[royal herald]

look at what Aero said

[humble burrow]

0 is a whole number

[royal herald]

how do u represent 1

[humble burrow]

that was my mistake

[timid sky]

you still can't

[humble burrow]

why?

[royal herald]

find n and k such that 2^n + 3^k = 1

[humble burrow]

n, k=0

[merry kestrel]

nope

try plugging

[humble burrow]

oh true

[royal herald]

2^0 = 1
3^0 = 1
2^0 + 3^0 = 1 + 1 = 2

[humble burrow]

then n and k would have to be rational numbers

that's actually a good question hold on

[merry kestrel]

why do you think this will work?

[frigid canopy]

well, one is always odd, and one is always even

[humble burrow]

find all rational numbers $n$ and $k$, such that $2^n+3^k=1$

[frigid canopy]

so...

[fickle silo]

even then it cant

pretty sure at least

[merry kestrel]

3^{no real number} = 0

[clever fjord]

christmasisoversus

[paper onyx]

No real solution(s) of this

[humble burrow]

the humble limit as k approaches negative infinity:

[merry kestrel]

thats not a specific choice of real number

[paper onyx]

You mean -infinity

[humble burrow]

it just isn't my day today, huh

damn

[hallow pelican]

unreasonable crashout

[paper onyx]

Lol

[elfin zodiac]

<@&268886789983436800>

[paper onyx]

<@&268886789983436800> take care of this pls

[humble burrow]

he is very amusing

[steel viper]

Ain't no way bros crashing out

[paper onyx]

Tru

[humble burrow]

i find this very funny i am laughing so hard

hahahahahahaha

[steel viper]

I just joined too

[merry kestrel]

notice your goal is to find a choice of k

[viscid flint]

yep

[humble burrow]

oh man that was a good one

woah why are you orange

hold on

[humble burrow]

yeah

[elfin zodiac]

That is not orange

[humble burrow]

their latex text is orange

[merry kestrel]

||you can change the color in preamble||

[humble burrow]

anyway, this question is pretty cool

[paper onyx]

What's the constraint over k and n

[humble burrow]

rational

[paper onyx]

Hmm

[clever fjord]

christmasisoversus

[elfin zodiac]

Does not exist, with real numbers is possible

Not with rational

Unless i am missing something

[humble burrow]

why

[elfin zodiac]

Because 2 and 3 are coprimes

[humble burrow]

you can reach the set of all natural numbers with the exception of 1, when n and k are whole numbers

[elfin zodiac]

You can have a real solution like 2^(-1)+3^(-ln2/ln3)

[crisp wadi]

||6||

[humble burrow]

oh wait true

hold on how did i overlook this

[elfin zodiac]

When the exercise is wrong from the begining is normal to overlook stuff

[humble burrow]

$\frac{1}{2}+\frac{1}{3^{\frac{ln2}{ln3}}}$

unfortunate

[crisp wadi]

One more } at the end

[humble burrow]

where?

[paper onyx]

$\frac{1}{2}+\frac{1}{3^{\frac{ln2}{ln3}}}$

[clever fjord]

lex.in.a.teacup

[humble burrow]

i see

[clever fjord]

christmasisoversus

[humble burrow]

alrighty then

.close

[odd edge]

Channel closed

Closed by @humble burrow

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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>.
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• 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.

[patent loom]

We have a triangle, height 2 meters, every angle is 60°, what's the length of every side?

[odd edge]

Please don't occupy multiple help channels.

[patent loom]

Ok

[dense moth]

Equilateral triangle ...

[covert zinc]

root 3

i really need to learn latex

[patent loom]

Root?

[covert zinc]

eh wait im confused

[patent loom]

U mean √3

[covert zinc]

ah yea i think so

[patent loom]

1,732= √3 ?

[covert zinc]

oh wait it is √(16/3)

mbmb i got confused

here, evaluate x

[patent loom]

Wait

[covert zinc]

umm... what are you doing

sin 90????

[patent loom]

Where is my fault

[covert zinc]

lil dude

[patent loom]

We are studying trigonometry

Idk

[covert zinc]

this is an easier method imo, it skips the trig

[covert zinc]

if you really wanna do that, sin60 = 2/x

[patent loom]

Omg it's basic algebra

Ur right

[covert zinc]

yes

ofc im always right

[patent loom]

Do u really think?

[covert zinc]

hmm?

[patent loom]

Lemme see

Are u 100% sure?

Then answer to me
What is the general solution for a system of second-order nonlinear partial differential equations with variable coefficients in a non-trivial three-dimensional domain, subject to complicated and time-dependent boundary conditions, and what is its asymptotic behavior for large times?

U are lil coocked

[covert zinc]

Idk lil bro that kinda just looks like a random mashup of words to me

[patent loom]

Xd

Then u don't know it all

[covert zinc]

Well I recognise the words by themselves

I've never seen them put together this way

[patent loom]

Haha

Xd

[covert zinc]

And I'm pretty sure they're not meant to be put together this way either

[patent loom]

I'll need to be able to answer that next year of school

It's a real question

[covert zinc]

Well good luck lil dude cuz ur having trouble with Pythagoras

[patent loom]

I just translated

[covert zinc]

Wow

[patent loom]

I know

[covert zinc]

Wait what year are you

[patent loom]

Not easy to answer

I'm from a baccalaureate

[covert zinc]

Oh dang ur in 4th year uni

Wait how are you having trouble with Pythagoras

[patent loom]

I'm not but our teacher wants us to answer

[covert zinc]

4th year of aerospace engineering bro

[patent loom]

DON'T GODAMM ASK :_(

Yeah I know

[covert zinc]

Okok I'm sorry

[patent loom]

Our teacher is crazy

Help

[covert zinc]

Wow so ur 3rd year aerospace engineering

Hmm I suppose I could try to solve a second order partial differential equation

Sounds like fun

The rest of it... Not fun lmao

[patent loom]

I know

I'm coocked

[covert zinc]

Lmao

Ok I gtg now

[covert zinc]

I'm not allowed to give you an answer, but I can give you the working and let you do it yourself

[patent loom]

Wait

The solve is not ok

[covert zinc]

Wdym

[patent loom]

Wait

I think I've done something wrong

Where's my fault?

[covert zinc]

Minus x^2 not +

You should get 3x^2 on the other side

[patent loom]

Oh shit

I fucked up on something that easy

Where's my mind ong

[odd edge]

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[whole hazel]

How did we judge to choose that interval based on that period?

[hardy panther]

the tangent function has asymptotes at n*pi +- pi/2

this one is compressed horizontally by a factor of 1/2 (or, another way, it's progressing twice as quickly)

so instead of the "default" interval of -pi/2 to pi/2, it's -pi/4 to pi/4

[whole hazel]

Ahhh right right

Ty

.solved

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[odd edge]

@obtuse herald Has your question been resolved?

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[wanton bison]

it is manifest lol

[raw palm]

did the solution work based on $s_{n+1}=\sqrt{2+s_n}$ or am i imagining things

[wanton bison]

[ s_n < 2 ]
[ \Rightarrow \sqrt{s_n} < \sqrt{2} ]
[ \Rightarrow 2+\sqrt{s_n} < 2+\sqrt{2} ]
[ \Rightarrow \sqrt{2+\sqrt{s_n}} = s_{n+1} < \sqrt{2+\sqrt{2}} < 2]

[raw palm]

so what

[wanton bison]

It seems like so

[raw palm]

$s_{n+1}=\sqrt{2+\sqrt{s_n}}\implies s_n=(s_{n+1}^2-2)^2=s_{n+1}^4-4s_{n+1}^2+4$ so $s_n-s_{n+1}=s_{n+1}^4-4s_{n+1}^2+4-s_{n+1}\leq s_{n+1}^4-5s_{n+1}^2+4=(s_{n+1}^2-1)(s_{n+1}+2)(s_{n+1}-2)<0\implies s_n<s_{n+1}$ so that ${s_n}$ is increasing and thus convergent since $s_n<2\ \forall n\in\mathbb{N}^*$

[cerulean vortex]

on page 29?

[past rover]

tf is this

this cant be real bro

😭

[past rover]

??

this isnt even maths that people do

[raw palm]

thats a solution manual

[past rover]

are u supposed to solve that or smth??

[raw palm]

it is a solution manual for rudin's book

[past rover]

oh bru

[clever fjord]

𝔸dωn𝓲²s

[past rover]

yo question what are the arrows for

[wanton bison]

I don't understand how they derived it but it seems still right

[cerulean vortex]

derived what?

[wanton bison]

s_(n+1) < 2

[cerulean vortex]

oh

i think i know

[raw palm]

they used a wrong closed form of s_n

[raw palm]

this did this

but you can still reach the same result even with the correct closed form

by induction

[cerulean vortex]

huh?

i think it's right

[raw palm]

you have $s_1=\sqrt{2}<2$ now suppose that $s_n<2$ then $s_n<4$ which means that $s_{n+1}=\sqrt{2+\sqrt{s_n}}<\sqrt{2+\sqrt{4}}=2$

[raw palm]

what is right

[wanton bison]

oh yea

[cerulean vortex]

their reasoning for s_{n+1}<2

[raw palm]

it is right if they do this

but from what comes after it is clear that they were using the wrong closed form of s_n

[cerulean vortex]

really?

wait i will type what i think they meant

[raw palm]

$s_n^2-2=s_{n-1}\implies s_n=\sqrt{2+s_{n-1}}$

[raw palm]

so that $s_{n+1}=\sqrt{2+s_n}\neq\sqrt{2+\sqrt{s_n}}$

[cerulean vortex]

$s_{n+1}=\sqrt{2+\sqrt{s_n}}$\
since $s_n < 2$ we can substitute to get $s_{n+1}<\sqrt{2+\sqrt{2}}$
and since $\sqrt{2} < 2$ we can substitute to get $s_{n+1}<\sqrt{2+2}$

[clever fjord]

Axe

[wanton bison]

I mean yea haha 3rd time now

[past rover]

can u whole lot stop being ignorant and tell me what the arrows are for sh

smh

[raw palm]

but this line shows that they werent using the same closed form that you are using

[raw palm]

the arrows mean implies

so like you have a statement

and then from this statement you deduce another statement

[past rover]

ok

[raw palm]

where the truth of the first means that the second is true

but the truth of the second doesnt mean that the first is true

[past rover]

can anyone rate this

[raw palm]

doesnt this really show that the sequence used in the solution is different than that given in the question

[cerulean vortex]

oh yeah i can't explain that one

[raw palm]

thats why i said that they used this same sequence to reach that 2 is an upper bound

[raw palm]

otherwise its illogical to write this after that

tbh its not entirely illogical , they could've remembered the sequence in a wrong way only here

i cant dive to the author's mind to check what he was thinking at that moment

all i can do is to speculate

but i am curious , is this solution manual written by rudin himself ?

[odd edge]

@raw palm Has your question been resolved?

[odd edge]

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[proven viper]

can someone explain what happened here he some how used these two equaltions to get the a/20 =k thing

[proven viper]

this is the question

[odd edge]

@proven viper Has your question been resolved?

[upbeat drum]

is that all the working out you got
you lost me on the second line chief

[odd edge]

@proven viper Has your question been resolved?

[proven viper]

oh no theres more but i just edited it to be the only part i didnt understand

with only these two equations how did i get

thats what confuses me

cuz if you equate them you just get

2b=a

[odd edge]

@proven viper Has your question been resolved?

[proven viper]

<@&286206848099549185>

[errant gust]

where did k come from

i dont get whats happening

whats the question

[nimble garden]

Me too

[proven viper]

i snet the question above

sent*

[proven viper]

i think its just a temprary variable since it gets canceled out later on

[odd edge]

@proven viper Has your question been resolved?

[proven viper]

.close

[odd edge]

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[plucky hatch]

hi why is 21st century 2000-2099?

[plucky hatch]

Why not 20?

Because of 2000.

[dense moth]

What was 1st century?

0-99 right ?

[plucky hatch]

Yes.

[dense moth]

They wouldn't have called it zeroth century

[dense moth]

So it's like it ...

[plucky hatch]

Why isn’t it 0th century?

If it’s 0

[woven pier]

It's the convention

21st century means the 21th century.

If it were 21XX, then clearly 21 centuries have passed after 0 AD and 21XX should be considered to be the 22nd, so 21XX is the 22th century.

It's like time:
0s - 1s means 'in the first second'
Similarly, 20s - 21s means 'in the 21st second'.

[dense moth]

What's the probability we will see the 22nd century?

[woven pier]

1/75

[dense moth]

Ah so I and you, both see is

1/75²

[plucky hatch]

0-99 - 1st
100-199 - 2nd

[dense moth]

Yeahhh

[woven pier]

yes?

[plucky hatch]

Can you tell me all Roman numbers? @dense moth

[woven pier]

Roman numerals have a few rules, but here are all the symbols:
I, V, X, L, C, D, M.

[south plume]

https://www.calculatorsoup.com/calculators/conversions/roman-numeral-converter.php

[plucky hatch]

And in normal numbers?

[south plume]

normal lol

[south plume]

1, 5, 10, 50, 100, 500, 1000

[plucky hatch]

But what is for example 2?

[odd edge]

@plucky hatch Has your question been resolved?

[odd edge]

@plucky hatch Has your question been resolved?

[weary pelican]

for roman numerals, you usually place them from greatest to lowest

that way they add

so if you want to do for example 2023

you would start preoccupying with 2000

so MM (M+M = 1000+1000 = 2000)

then 20

XX

then 3

III

MMXXIII

now the rule to remember

is how to write numbers with "4" or "9"

we don't want to write 4 times the same numeral in a row, it makes it lenghty

so instead

we take the unit above

and subtract 1 from it

example:

to write 40

I could do XXXX but that's too lengthy

instead, I take 50, so L

and to subtract 10 from it

I put an X before the L

XL

[weary pelican]

it breaks the pattern of greatest to lowest

so we know it means "subtract the small one from the big one"

kinda same thing with 9s

so for example 90

it could be LXXXX (50+10+10+10+10)

but too lengthy

so go 10 above

we get 100, which is C

subtract 10 from it

XC

how this can apply multiple times in the same number:

to write 99

we again start from greatest to lowest

so write 90

which is 100 - 10

so XC

then 9, which is 10-1

IX

99 -> XCIX

[plucky hatch]

ok

so what is four?

@weary pelican

[odd edge]

@plucky hatch Has your question been resolved?

[odd edge]

@plucky hatch Has your question been resolved?

[odd edge]

@plucky hatch Has your question been resolved?

[languid jetty]

Four is IV as you have V (5) minus the I (1) before it

2024 would be MMXXIV for example

[odd edge]

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[dense stone]

What is radius of circle when area of ABC is 24 ,AB=6,BC=6 and AC=4

[bitter folio]

Circumcircle or incircle

[dense stone]

I don't know either

[chilly ravine]

did you write the question word for wrod

[bitter folio]

!original

[odd edge]

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.

[bitter folio]

Incircle then

[dense stone]

Isn't it just 1 = A÷s ?

[bitter folio]

No we want r

[odd edge]

@dense stone Has your question been resolved?

[odd edge]

@dense stone Has your question been resolved?

[mystic saffron]

@dense stone

뭐가 이해가 안되시는거임

[dense stone]

저 식은 그냥 1 = A÷s 라는 결론밖엔 안나와서

어떻게 풀어야하는지도 모름

[mystic saffron]

님이 이 문제를 어떻게 접근하셨는지 설명해보세요

근데 왜 여기에 한국 문제를 올림? 한국어로?ㅋㅋ

[dense stone]

피타고라스 정리를 사용하여 AC의 중점과 B 사이의 길이를 1:2로 비례배분해서 풀려고함

근데 그건 오답이나옮

[mystic saffron]

개념서는 읽어보심?

내접원 문제를 푸는데 피타고라스를 사용해서 풀려는건 좀 신기하네

@dense stone

내접원의 반지름 r로 두고 삼각형 넓이 공식 만드시는 법은 아시나여

[dense stone]

읽었는데도 이 문제 푸는데는 도움이 되는게 없음

[mystic saffron]

@dense stone

[dense stone]

몰라요

[mystic saffron]

개념서에 안나와 있어요?

[dense stone]

네

[mystic saffron]

사진 찍어 보내보세요

저게 무슨 책인데요

[dense stone]

책에서 출제한게 아니라서 모름

[mystic saffron]

그러면 저 문제 출저가 뭔데요

책이 아닌데 개념서는 어떻게 있음

@dense stone

[dense stone]

같은 책에서 본 개념서가 아님

[mystic saffron]

님이 본 개념서를 사진 찍어주셈

저 문제 출저도 알려주고

개념도 하나도 모르는데 어떻게 문제를 풀 생각을 하는거지..

@dense stone

@dense stone

[dense stone]

어디서 출제한건지 불명임

[mystic saffron]

님이 어디서 저 문제를 가지고 왔냐고요

그냥 저 사진 하나만 온라인에서 퍼온거임?

[wild stump]

Chat is decorated with these letters

[mystic saffron]

그러면 그 퍼온 링크를 달라고요

[dense stone]

마타수학에서요

[mystic saffron]

내접원의 반지름이 r이면

그 삼각형의 넓이는 1/2 곱하기 내접원의 반지름 곱하기 삼각형 세변의 길이의 합

이 공식을 아예 모르는것 같은데

이걸 모르는 상태에서 저런 문제를 푼다는것 자체가 님이 공부를 드럽게 안한거임

기본도 안잡혀있는데 왜 저런 문제 풀 시도를 함?

맨날 한국어 문제를 여기에다 올리고 민폐 끼치고 부끄럽지도 않나..

[dense stone]

풀긴 해야하는데 어떻게 풀지 모르겠어서 결국 올린거임

[mystic saffron]

왜 여기에다 올림

다른 곳도 아니고

그리고 보통 상대가 풀이를 알려주면 감사합니다라고 말을 하지 않나요?ㅋㅋ

[mystic saffron]

이거 쓰셈

@dense stone

@dense stone

@dense stone

[dense stone]

오답이 나왔는데 공식은 맞는건가요?

[mystic saffron]

님이 한 풀이를 보여주셈

모르겠으면 제발 본인이 좀 찾아보고

애초에 저걸 모르고 이 문제를 푼다는것 자체가 미친행동임

[dense stone]

24 = r * (1/2) *6 + 6 + 4 *
24 = 3 + 6 + 4 *r = 13 * r
r = 24/13

[mystic saffron]

삼각형 세변의 길이의 합을 설마 못구함..?

6 더하기 6 더하기 4를 모름..?

[dense stone]

6 + 6 + 4 인거 구한거 맞아요

[mystic saffron]

16이잖아..

16을 2로 나누면 뭐야 8 이잖아

8에 반지름 r 곱하면 뭐야 8r이잖아

8r이 뭔데 24잖아

그러면 반지름은 뭐야 3 이잖아

진짜 이런 컨셉충은 또 오랜만이네ㅋㅋㅋㅋㅋㅋ

[dense stone]

.close

[odd edge]

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[vague oxide]

part b) I dont get how you get the 6y(dy/dx)2 when you d/dx part a

[timid sky]

well, can you show how you'd have done it?

We could tell you what went wrong in your approach first. That'd be more constructive

[vague oxide]

its messy but its the line below part c

[timid sky]

shouldnt this be dy/dx (6y - 1)

[vague oxide]

eh how come

ithought if you make the external dy/dx into d2y/dx2 you dont have to do that

sorry i havent done math since my break started so i've forgotten a bunch of things

[clever fjord]

Aero

[vague oxide]

oh i didnt realise that needed product rule

[timid sky]

yeah its two functions being multiplied are they not

[vague oxide]

i see okay thank you!

.close

[odd edge]

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[timid sky]

oh but do you understand the answer to your original question now, though?

[vague oxide]

yep i do thanks!

[timid sky]

aighty

goodluck

[vague oxide]

.reopen

[odd edge]

✅

[vague oxide]

.close

[odd edge]

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[keen mist]

(\mu = 0,71)

Giải thích các bước giải:

(\begin{array}{l}
\sin \alpha = \frac{{0,2}}{1} = \frac{1}{5}\
\cos \alpha = \sqrt {1 – {{\frac{1}{5}}^2}} = \frac{{2\sqrt 6 }}{5}\
\vec N + \vec P + \vec F + {{\vec F}{ms}} = \vec 0\
N = P\cos \alpha \
F – {F{ms}} + P\sin \alpha = 0\
F + P\sin \alpha = \mu P\cos \alpha \
\mu = \frac{{F + P\sin \alpha }}{{P\cos \alpha }} = \frac{{1 + 0,2.10.0,2}}{{0,2.10.\frac{{2\sqrt 6 }}{3}}} = 0,71
\end{array})

[clever fjord]

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

[keen mist]

.close

[odd edge]

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[low locust]

rational root theorem

[frozen quarry]

huh....

oh using 2i?

x = 2i and -2i?

[low locust]

no

[frozen quarry]

oh

please explain then

[low locust]

if p/q is a rational root (fully reduced), then p is a divisor of the last term, i.e. 9, and q is a divisor of the first coefficient, aka 4

so p is one of 1,3,9,-1,-3,-9 and q is one of 1,2,4,-1,-2,-4

that gives you a few options to check

(preferably start with q=1)

after you find a root, use polynomial division to get rid of it

after you found two roots you can use the quadratic formula

[long tinsel]

would it not be easier to just construct a polynomial with roots $u_i = 4 + x_i^2$ and find the constant term?

[clever fjord]

neon

[low locust]

imo rrt is easier here but up to you

[long tinsel]

because then you just sub u = 0

(divide by 4 throughout first)

[frozen quarry]

i am sorry but i havent used this theorm.... i'll check it out and try

[frozen quarry]

you would get a power of 3/2 which i have i'm not sure how to work with....

[long tinsel]

take the odd terms to one side and square

[frozen quarry]

huh...

You mean like this neon...?

[long tinsel]

$u = 4 + x^2 \implies x = \pm \sqrt{u - 4}$
Substituting this into the polynomial we have:
[4(u - 4)^2 + 8(u - 4)^{3/2} - 17(u - 4) -12(u - 4)^{1/2} + 9 = 0 ]
Taking the non-natural indices terms to one side and squaring we have:
[ [4(u - 4)^2 - 17(u - 4) + 9]^2 = (u - 4)[-8(u - 4) + 12]^2 ]

[clever fjord]

neon

[frozen quarry]

ohhh

[long tinsel]

yeah like that

[long tinsel]

So now you need to find the constant term/leading coefficient

do you understand why we're doing that?

[frozen quarry]

no...

to find the sum of the roots...?

[long tinsel]

product

yeah

[frozen quarry]

sorry product

ahh thank you

this was the giiven solution

which i understand

but i dont think i would ever think of a solution like this in a exam

its way easier tho

thank you neon and Denascite

[long tinsel]

damn

[low locust]

(the reason why I said rrt is that rrt is at least a procedure to follow, with the disadvantage that it only works if the roots are nice enough)

[frozen quarry]

im gonna check it out now

[low locust]

(it may not be quicker if you can come up with this solution, but you dont have to come up with this solution in exchange)

[frozen quarry]

i've never learnt it before

[low locust]

you will see that its slower

[frozen quarry]

yep i prefer not having to come up with this solution..😆

thanks!

.close

[odd edge]

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[frozen quarry]

i actually checked out the theorm

didnt go too deep into it

but are one of these rational numbers always a root

like there is a 100% chance for one of the numbers to be a root?

[low locust]

no

if none of them are a root then all roots are irrational and probably suck

[frozen quarry]

ohh

so if it is rational we do get the root

[low locust]

yes

if there is a rational root we will find it

[frozen quarry]

thats so cool

thank you

its gonna be really useful

when does a person actually learn about this tho?

[frozen quarry]

lol im glad i asked this question today

[odd edge]

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[peak maple]

How do i do e?

thats the curve btw

[odd edge]

@peak maple Has your question been resolved?

[brittle notch]

Eliminate t

You have x = cost

And y = sin2t

So write something like sin2t is 2sintcost and then it is 2root(1-cos²t)cost

Like that

And simplify

[peak maple]

Oh

We alr have a Cartesian equation in the last question tho

Only thing that’s changed is the domain

[brittle notch]

Yep

[hollow quarry]

the domain of cartesian equation is the range of x = cos t (pi <= t < 2pi)

[peak maple]

i dont ge tit still

they just made the last equation negative

oh wait

cause they added pi onto the domain?

and cos becomes negative after 180 degrees or pi?

[hollow quarry]

Yeah pretty much

[peak maple]

cool thanks

[hollow quarry]

is this alevel maths btw?

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[peak maple]

ye lol

[hollow quarry]

Edexcel?

[odd edge]

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[peak maple]

ye

[peak maple]

.close

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[south meteor]

Guys I need help

[upper onyx]

you are searching for binomial formula.

[south meteor]

yhhh

Explain me 🙏

[fierce jackal]

give solution to this question please

[upper onyx]

!occupied

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[upper onyx]

where should i start? do you know binomial formula?

[south meteor]

I don’t know what it is

[upper onyx]

https://letmegooglethat.com/?q=binomial+formula

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[cloud hatch]

QUESTION: I'm working with Stiefel variety (We define $V_p(R^n)$ as the subspace of matrices with n lines and p columns such that it's a orthogonal family of vectors), if p<n-1, I have to show that $V_p(R^n)$ is 'simply connected', and here we mean that every covering of it is trivial. I tried multiple things but couldn't really show it, and I don't find indications online. Maybe It's homotopic to S^2?

[clever fjord]

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

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