help-43

[trail crater]

apologies

[kind viper]

ok right

so it looks like you didnt make any algebraic mistakes here as such

but the IH you've written out is not going to be of much help

[trail crater]

is the IH my assumption or prove step

[kind viper]

inductive hypothesis

the assumption

[trail crater]

oh

so...

[kind viper]

i think you may need to assume it works for two adjacent values of n

cause the sequence $a_n = (3n+1) \cdot 7^n - 1$ obeys a 2nd-order recurrence relation

[boreal girder]

Ann

[kind viper]

(rather than 1st as you would see with a*b^n + c or whatever)

[trail crater]

so ur saying to make multiple inductive hypothesis?

[kind viper]

im trying to avoid saying the words "strong induction" cause that would seem to go against the instruction to use "mathematical (weak) induction"

but yes basically

[trail crater]

ah ok nah our teacher be putting these ones idky

anwyas ty (:

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[kind viper]

"onesidky"?

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[raw portal]

i don't quite understand the direct transition from step one to 2

[raw portal]

how did he do it directly

the further factorisation i mean

[rigid perch]

on the U matrix, he "pulls 2 out" by multiplying the top row by 1/2

to counteract this, you would multiply on the left by the diagonal matrix [ \mat{2 & 0 \ 0 & 1} ] which corresponds to the elementary row operation of multiplying the first row by 2

[boreal girder]

cloud

[rigid perch]

he similarly "pulls 3 out" of the second row and combines the two effects into a single diagonal matrix

[raw portal]

how does this pulling and all work actually i am new to the idea , ive dealth with matrix multiplication in school but never this way

just not intuitive to me

can you elaborate

[rigid perch]

so if you were to multiply any matrix by the matrix
[ \mat{2 & 0 \ 0 & 1} ]
that would have the effect of multiplying the first row by 2, and otherwise leaving it unchanged. so for example,
[ \mat{2 & 0 \ 0 & 1} \mat{1 & 1/2 \ 0 & 3} = \mat{2 & 1 \ 0 & 3} ]

[boreal girder]

cloud

[raw portal]

interesting

so we basically just manipulate the identity

[rigid perch]

and in general if you do any elementary row operation to the identity matrix, then that generates what we call an elementary matrix, which has the effect of performing that row operation when multiplied

[raw portal]

yeah he got into elementary matrices

[rigid perch]

so to factorize a matrix, we can "multiply the first row by 2 and also divide it by 2" in order to really leave it unchanged, the same way as we can both multiply and divide a number by 2 to leave it unchanged

we just applied the division by 2 directly to the matrix and the multiplication by 2 in the form of an elementary matrix

[raw portal]

how did gauss think of it intuitively though

like manipulating the identity then multiplying it by a matrix

to perform the row transformation

[raw portal]

is this a thing to be learned in a lecture or to be thought ourself 💀

[rigid perch]

well that may or may not be how it's presented in lecture, just the way i like to think of it

it's sort of similar to the way that the LU decomposition itself is found

[raw portal]

does matrix multiplication get more intuitive as we move forward in linear aljebra

[rigid perch]

yes, you get more perspectives on it

[raw portal]

nice , thanks

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[analog sleet]

Hi Im confused the answer for b) says y=-5t^2 only

[analog sleet]

wait nvm

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[carmine garden]

I don't get why the underlined part matters here.
More specifically why is the minimal polynomial sq and not q

[hushed magnet]

cause q doesnt necessarily kill all of T?

range(qT) survives

s needs to kill that

[carmine garden]

I see

Thanks

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[warped oar]

Isn't this statement that "pi is the twice the smallest positive solution to cos(x) = 0" directly equivalently equal to saying that pi is the ratio of any circle's circumference and the diameter of it?

[acoustic yarrow]

how do u show the implication

[lime juniper]

depends how you define cosine. this form is preferred because cosine can be defined without circles

[acoustic yarrow]

they r tautologically the same statement because both define π but id like to see how u show if a then b and if b then a

[lime juniper]

for example, the unique solution to c''=-c, c(0)=1, c'(0)=0

[kind viper]

directly equivalent?

[warped oar]

direclty equivalently equal

[kind viper]

wait what the fuck does "equivalently equal" mean

[warped oar]

Because the input of cos(x) is taken in radians which are defined using circles

[acoustic yarrow]

what do u think directly equivalent means by the way

[warped oar]

The original statement implies pi is a straight angle.
This implies that pi = (half circumference)/radius = (2 * half circumference)/(2 * radius) = circumference / diameter

trivially the same statement = directly equivalently equal like the fractions five upon six and 10 upon 12

cos(x) = base/hypotenuse

<@&286206848099549185>

[tribal nacelle]

identical, equal with 3 lines instead of 2

(i think...?)

[soft bobcat]

? That would just mean equivalent

[warped oar]

very equal

[inner spoke]

It means you can go both ways

around the equal sign

if and only if

i guess

[warped oar]

Certainly, we can show that cosine(x) = a summation, but his is not directly equivalently equal

[strange ermine]

??

[soft bobcat]

What

[warped oar]

very equivalent*

[inner spoke]

cause it's infinite summation

[soft bobcat]

Still makes no sense

[strange ermine]

Still not making sense

[inner spoke]

@warped oar did u just make it up?

[warped oar]

no

[strange ermine]

!original

[compact pewter]

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.

[warped oar]

make up what

[inner spoke]

@soft bobcat and @strange ermine saying the same things

[warped oar]

npcs

[strange ermine]

Your question

[inner spoke]

the definition for very equal

[warped oar]

what is it

[inner spoke]

about the inifinte polynomial sum of cosine

i don't rlly agree

the infinite sum is by definition strictly equal to cosine

[warped oar]

oh

cos(x) = base/hypotenuse

[inner spoke]

so both way around the equal sign, they are very equivalent 😂

[warped oar]

This is my definition.

[inner spoke]

i was talking abt the Taylor Series but nvm

[inner spoke]

solid for geometry

[inner spoke]

oh wait u r 15 my bad

[tribal nacelle]

solid...?

[soft bobcat]

Are you trolling again philosopher

[inner spoke]

he's 15 bro

[soft bobcat]

Ive already reported u once

I wont hesitate to do it again

[tribal nacelle]

https://tenor.com/view/solidworks-epic-win-okina-matara-touhou-tony-hawk-gif-27690846

[strange ermine]

I believe he is the orange propic guy called Mathematician

[warped oar]

Yes.

please dont spam if youre not helping

[inner spoke]

anyways, was ur question answered @warped oar

[warped oar]

The definition of cosine(x) is base upon hypotenuse

Is the answer to my original question "yes"

[inner spoke]

A definition of cosine is based upon hypothenuse in a right triangle, yes.

[strange ermine]

Your original question makes no sense
At least redefine the term "equivalently equal"

[inner spoke]

consider it means identical

i think

[warped oar]

cosine(x) can be showed to be equal to a certain summation, but 'tis not that easy

[inner spoke]

and what

[warped oar]

So I think my question has been answered

[soft bobcat]

You should send a screenshot or picture of your original question as it is written so we understand it better

[inner spoke]

it is actually equal to that summation

[warped oar]

This was a question of mine

[inner spoke]

for now, just in keep in mind that cos x is base or adjacent over hypothenuse

u'll discover other definitions later on

[warped oar]

okay, thank you very much 😄
and y'all

[soft bobcat]

Base is wrong btw

How would you define the "base" of a triangle?

[inner spoke]

he means adjacent, but it seems clear in his head i think

[soft bobcat]

Yeah adjacent is a better word

[warped oar]

yes

[soft bobcat]

But anyway tomato tomahto

[warped oar]

thank you, again.

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[carmine garden]

can I have a hint, I really doubt I need to find the Linear transfromation here

[carmine garden]

I mean If I have to it would be $\begin{bmatrix} \cos(π/180) && -\sin (π/180)\ \sin (π/180) && \cos (π/180) \end{bmatrix}$

[short ferry]

geometrically, what does minimal polynomial even mean?

[boreal girder]

wai

[carmine garden]

It maps to the 0 vector

you can a linear combination of the operator raised to various powers, with the highest power. being monic

[short ferry]

oh, now that I think of it, what i wanted to do might not be correct. Sorry

[hushed magnet]

v=e1

[carmine garden]

Okie, will try to work with that

so I get T(e_1)=e_1+e_2

T^2(e_1)=2e_2

T(e_2)=-e_1+e_2

T^2(e_2)=0

e_1,T(e_1) are LI

so T^2(e_2) is not

so the minmal poly is x^2+2=0

ok, looks like I made a mistake

what am I doing wrong

ooh\

T(e_1)= ( \cos(1),\sin(1))

T^2(e_1)= (cos(1)-sin(1), 2 cos(1) sin(1))

or to express it better

T(e_1)= cos(1)e_1+ sin(1) e_2

Yea, I thik I got it

thanks

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[unique hill]

IS GETTING THE AREA OF THIS POSSIBLE??

[inland nacelle]

Try breaking it down into simpler geometric shapes first

[unique hill]

i did but...

[inland nacelle]

Like a rectangles and triangles if that helps

I do not know if the lines are 90 degrees

[unique hill]

this is so stupid

wtf how do i do this 😭

[inland nacelle]

i also dont know if they are parallel

[unique hill]

no angles r given NOTHING

just this

is it possible?

[inland nacelle]

what

thats a bit stupid

Unless the diagram is drawn to scale

Then we can approximate it

[unique hill]

it is

yes drawn to scale

[inland nacelle]

Ok

One way to do it is to plot it on a square grid and count the grid squares

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[charred sundial]

Completely lost, don't know how to start

[short ferry]

try similar triangles

[inland nacelle]

do you know intercept theorem

[charred sundial]

how tho

[charred sundial]

no

[short ferry]

try sliding the line with x+6 closer to the other one

[inland nacelle]

do you know ratios between parallel lines?

[short ferry]

that wont change any distances

[runic creek]

just use intercept theoremm

[inland nacelle]

He said he dosent know

[runic creek]

wait tthe pic i am sending is loading

[charred sundial]

nope

[runic creek]

wait just search it up on google

its taking so long for me to load and send

[hearty saffron]

5/6 = 12/x is intercept theorm I tink

[charred sundial]

ok hold on lemme check this theorem out

[runic creek]

yes

[inland nacelle]

Indeed, it is useful

However an alternative is @short ferry 's way

[short ferry]

alternatively, you can slide the lengths like this without rotating. Since you didnt rotate them, the angles stay the same and you got 2 similar triangles

x / 12 = 6 / 5

[hearty saffron]

nice one

[charred sundial]

ohhhhh i see

[charred sundial]

very nice method, didn't even consider that

[inland nacelle]

this is an alternative

using the fact that green guy similar to purple guy

[short ferry]

yeah, interesting

if a / b = c / d, then they are equal to (a + c) / (b + d) as well

funny little theorem which is a consequence of this

[charred sundial]

thanks guys

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[dawn palm]

guys

[dawn palm]

why geogebra show this for this

and wolframalpha show this

how i resolve

[azure vault]

you can see that WA's graph has a y axis with very very small values

so seeing it from Desmos's perspective

you won't be able to see a thing

but if you zoom in on the y axis

[keen granite]

https://www.desmos.com/calculator/8r4mazddwq

[azure vault]

surely enough, you'll see the very small but negative y values between 0 and 5

[keen granite]

fun fact you can scale the axes on desmos and get the same thing W|A is showing :)

[dawn palm]

oh ok i see its bcz its very small value

thanks

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[velvet moon]

I don't understand a)ii)

[kind crane]

the theorem/series that allows you to make the expansion makes assumptions on x.

[velvet moon]

What is the answer??

[kind crane]

!noans

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[velvet moon]

I got x=2

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[foggy ruin]

Hello! I’m new here 🙂
i’m a teen but I’m not very good at math and am falling behind drastically in this subject! I need help!

[kind crane]

you can just follow the bot message above and post your question now

[kind viper]

we can't really help you in any more general way

but we can recommend places for self-studying

such as khanacademy

if you have a specific question you'd like us to help with, you're welcome to post it here

[astral trellis]

can yall help me with a question ?

[kind viper]

!occupied

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[astral trellis]

all of them are

[kind viper]

read #❓how-to-get-help for instructions. there are four channels which are free rn.

[foggy ruin]

Got it, thank you guys! I don’t have any specific questions but thank you 🙂

[kind viper]

ok in this case you can .close this channel

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[runic verge]

hi all, again me
How did he take a square root of 2 in this method, from where did he get it at all? Thanks!!

[dusky nymph]

he wants to get it in the form 1/(t^2 + 1)

so looking at the denominator of the original, t would have to be the square root of this:

[runic verge]

this is the same as 1 / (1 + x^2) dx? right

thats what he is trying to get

[dusky nymph]

yea, that general form (he ends up with a scale factor but that's fine)

[runic verge]

could you maybe explain please why does square root of 2 help?

[dusky nymph]

it's not that it helps or hurts, it's just necessary if you're gonna get the denominator in the form t^2 + 1

you basically pattern match and say, ok i have (something) + 1

so t^2 = (something)

and therefore t = sqrt(something)

and since the something has a 2 coefficient out in front, sqrt(something) will have a sqrt(2)

[runic verge]

i am really thankful for your fast help, i wish you all the best Bungo!

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[untold pelican]

Hello you great math fellows.
I need some help regarding a calculus/linear-algebra problem please.
So let say we have a cylinder with the equation x^2 + y^2 = 2 in R^3 (where z is free), and a plane like x+z=4.
So their intersection result an ellipse, and I like to find the equation of the ellipse, but I don't know how?
I just found the x in terms of z from the plane, and plugged in cylinder equation which gives us a perpendicular cylinder to the original cylinder. I'm not sure how to find the equation of the ellipse (the intersection of plane and the cylinder)
Any hint is appreciated.
Thank you for your time and consideration.

[kind crane]

What does "x in terms of z" mean? There's no z in either of your shapes

[untold pelican]

Oh, sorry, there was a type, I fixed. the plane is x+z=4

[winged lion]

You mean that?

[civic quartz]

maybe we can start by finding the angle between the plane z = 0 and x+z=4

then trig for the major axis lenght?

[untold pelican]

Yes sir, the equation of the green ellipse.

[winged lion]

If you wanna find a parameterization, on the xy-plane it's a circle with radius 2, so you can set x = 2cost and y = 2sint. Now the z-component you would take the equation x+z=4 and solve for z.

So overall z = 4-x = 4-2cost

Oh wait, you asked for an equation, actually

[untold pelican]

Thanks, and how to find the equation of the ellipse in cartesian? in the std form of (x-h)/a^2 + (y-k)/b^2 = 1 ?

[winged lion]

Well the thing is the ellipse is on the plane and not cartesian plane

soo

Unless you are asking for the projection, which I dont think you are asking for

[untold pelican]

The projection of ellipse should be the circle I guess?!

[winged lion]

Yes

[untold pelican]

So, no that doesn't help actually. I'm thinking to rotate the plane and the circle for 45 deg, as the angle between the plane and xy plane is 45, then finding the ellipse will be easier I think(no sure), and finally rotate back the ellipse once it's found. But I don't know if it's even logical.

[winged lion]

Might as well rotate the parameterization with a rotation matrix I guess

If we see (2cos(t), 2sin(t), 4-2cos(t)) as a vector function

[civic quartz]

or we can just eliminate t directly no?

cos t = x/2

sint =y/2

cos t = (4-z)/2

[civic quartz]

converting this to cos t , and equating all

like if you want that equation form

i found my new love btw , parametric equations , thanks anti algebraist

[untold pelican]

The correct equation should be (sqrt(2)cos(t),sqrt(2)sin(t),4-sqrt(2)cos(t)) as I calculated(I hope I did it right).

[civic quartz]

oh yeah just read your question again

the radius is sqrt 2

[winged lion]

https://www.desmos.com/3d/7wqzybzkgj

I rotated it with R_y matrix around y-axis

Then I think you can insert the components into the ellipse equation.

[civic quartz]

did you want this type of equation?

[untold pelican]

Yes, first I like to solve it without rotating if it's not very devil in calculation. let's see. And yes, rotating along by y axis could help I suppose.

[winged lion]

Oh, how did you do it?

[civic quartz]

used your parametric equations and eliminated t

[winged lion]

Ah

smart.

[civic quartz]

finding the locus of your parametric thing

[untold pelican]

Not, really, but it looks very cool.

[civic quartz]

what are you looking for exactly ?

maybe just an ellipse in the xy plane with the dimentions of the ellipse you cut?

[winged lion]

Well that's like the green one, but scaled down a bit?

[untold pelican]

No, that's the exact ellipse.

[untold pelican]

The ellipse in standard form.

[winged lion]

Well it doesnt look like

[civic quartz]

uhh like the xy ellipse?

[untold pelican]

https://www.desmos.com/3d/n52amddhcd
Maybe we are using vary coefficients?

[civic quartz]

he wanted the equation for the circle having radius sqrt2 , we assumed he wanted for the radius two

[winged lion]

Oh wait

nvm

I thought radius was 2 instead it's sqrt(2)

well then we are done

[civic quartz]

oh . but this is 3d, uhhh

[winged lion]

you do that but with radius sqrt(2)

[untold pelican]

The radius is sqrt(2) by x^2+y^2=2

[winged lion]

yes, I originally read x^2+y^2=4 somehow

I apologize for any inconvenience.

[untold pelican]

Yes, so z is also involved I believe. pretty interesting.

[untold pelican]

No worries, some days ago I did 6+2=7, this is fine.

[civic quartz]

i mean yeah its on R3, its along the x+z plane

is this any specific problem you need it for

or a thought exercise

[winged lion]

Well 6+2=7 mod 1

[civic quartz]

amen

[untold pelican]

This is just very interesting. I have to do this. Than you great fiends and math fellows helping me regarding this, really appreciate it. It seems I have to study some more calculus regarding this, and I'll do(to forget it again).
Math bless us all.

🍻🤘🤓

[civic quartz]

nice to meet you too , have a good one

[untold pelican]

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[ruby lava]

im tryna solve this by graphing it into demos, the assignment was talking about using the change of base formula. im confused on what y2 is going to be, i think the notes i took are confusing me, what would be y2 or am i wrong on both of them?

[rigid perch]

neither of them is quite correct

the left one is close

[ruby lava]

i knew i did smth wrong

so this is the formula right

[rigid perch]

yes

[ruby lava]

so b=5 and a=3+1, would be 4 then or??

im confused

[rigid perch]

the log only applies to the 3

[ruby lava]

so it would be (log3)+1?

[rigid perch]

yes

[ruby lava]

so y2=(log3)+1/log5 ?

[rigid perch]

no you should only divide the log part

[ruby lava]

wait is this is it

[rigid perch]

yes

[ruby lava]

ive been reading it as log5^(3+1) not (log5^3)+1

thats on me

thank u very much

what did i do wrong in this

nvm i think its cus no parentheses on x-1

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[glacial pike]

I hate geo

[glacial pike]

I don’t even know where to begin here

[glacial pike]

That’s it. I’m losing

Thank u

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[rocky shadow]

Hi guys, just wanted help for this particular question.
What I've tried:
Calculating the area of the hexagon by dividing it into two trapeziums and getting a total area.
Dividing the hexagon into equal triangular slices based on the edges (40 deg).

Where I'm stuck: I have no idea how to get to the area of the square or the circle. It feels like there is not enough information given to get it.

Would appreciate any help or hints towards the right direction!

[inland nacelle]

@rocky shadow the circle is tangent to the hexagon right

So draw the radius at the tangency point and what can we say about the triangles formed?

[rocky shadow]

but can we really say that if we dont know if its drawn to scale?

[inland nacelle]

Its inscribed

So internally tangent

[rocky shadow]

ah

okay so since since its the tangent, we can say that thats the radius of the circle?

[inland nacelle]

Wait lemme draw

[inland nacelle]

What can we say about the angles in the red triangle?

[rocky shadow]

it equals 180

orange and green = 120

[inland nacelle]

what can we say about the relationship between orange and green?

[rocky shadow]

supplementary?

wait

adjacent?

sicne they share a common side

[inland nacelle]

There the same angle

[rocky shadow]

OH

[inland nacelle]

Do you know why?

[rocky shadow]

since the length opposite the hypotenuse is the tangent on the hexagon side

sorry that didnt make sense

the longer length is the tangent to the hexagon side

[inland nacelle]

its because the diagram is symmetric

do you see it?

[rocky shadow]

yeaaa

[inland nacelle]

yay nice

[rocky shadow]

thats a better way to describe it

[inland nacelle]

Now can you solve?

You might want to use the ratio of sides in a 90-60-30 triangle

[rocky shadow]

okay

i think i can do it

i will do

[inland nacelle]

nice

[rocky shadow]

thank you man

[inland nacelle]

Np

[rocky shadow]

youre a genius

do i have to close this btw

[soft bobcat]

Yes

[rocky shadow]

.close

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[tribal tinsel]

Hi

[tribal tinsel]

I need help urgently

[tribal tinsel]

This is the questions

[tribal tinsel]

Hello

How do I solve this to get the answer

It's it's probably really easy and I tried it multiple times

But I get stuck

At the last part when I have to put everything together

[soft bobcat]

!status

[compact pewter]

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

[soft bobcat]

show your work

[tribal tinsel]

👍

Is this clear enough

[atomic panther]

what are you finding difficult the multiplication?

[tribal tinsel]

Like what do I do at the end

Yeah

[soft bobcat]

where did x+2 go from the denom

look at your solution, it needs x+2 in denom

[tribal tinsel]

I know

[soft bobcat]

maybe recheck your work a bit, your method is correct

[ember crystal]

just try rewriting the the whole calculations again neatly

[soft bobcat]

find out area of bigger rectangle and smaller one

and resultant is bigger- smaller

[tribal tinsel]

Do I times the three binomials

Or do I cancel them out or are you even allowed to cancel them out if there is a subtraction sign

[atomic panther]

no you can not cancle

cancel*

only when there is multiplication

for example x+3/x the x can not cancel

but (x+3)(x)/x the x can cancel

[tribal tinsel]

Oh ok

[compact pewter]

@tribal tinsel Has your question been resolved?

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[solar folio]

how can I find the length L in terms of R

[solar folio]

without using trigonometry

[bleak dock]

you need trigonometry

[solar folio]

without sin, cos, etc.

[bleak dock]

yeah you need those

[solar folio]

not necessarily

[winged lion]

use their taylor series

[solar folio]

lol

[winged lion]

wait sin = 1/csc

[bleak dock]

you could try approximating the polygon as a circle

but that won't be an exact answer

[junior crater]

what is "a"? the area of the small triangle?

[solar folio]

a is apothem

I don't know how to get to L=r * (sqrt5 -1)/2

[bleak dock]

oh wait are you given the length of a already?

[solar folio]

No I am only given the radius

[bleak dock]

oh okay

[solar folio]

and the fact that the inscribed polygon is a decagon

the answer is this

but I cant use trig

only geomtric constructions

and triangles' similarities perhaps

[bleak dock]

ah yes the decagon is constructible with a ruler and compass

[bleak dock]

okay you should have said that from the very beginning

but thanks for clarifying

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

did you not read the Wikipedia page then?

[solar folio]

I think all polygons are constructible with ruler and compass

[bleak dock]

false

[junior crater]

7-gon is not

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

[solar folio]

?

[bleak dock]

yeah and there's another version here

[bleak dock]

read the page: it has the steps

[solar folio]

ok I will, but one thing

if the polygon was not a decagon how would I find L(r) for the polygon also

[slate sand]

Holy moly gauss is goated he figured out how to draw the 17-gon

[solar folio]

is there a general "method"

[slate sand]

You should try to figure it out

[solar folio]

like a hexagon is very easy because the angles are 60 30, 90

[slate sand]

It’s a good exercise

[solar folio]

but a decagon is harder

[slate sand]

Consider triangles

From the centre of each polygon, draw lines to each vertex

How would you describe the angles and the side lengths?

[bleak dock]

the method for the decagon works for polygons with 2 * F(n) sides

where F(n) is the nth Fermat prime

3, 5, 17, 257, 65537....

[bleak dock]

I doubt so

you can find general methods for specific cases

but not every constructible polygon at once

[solar folio]

I understand

btw is a heptagon constructible or no?

[bleak dock]

that's the one with 7 sides

so no

[junior crater]

"the regular heptagon is not constructible with compass and straightedge but is constructible with a marked ruler and compass"

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

[solar folio]

so this is an approximation

ok ok, thanks then

[bleak dock]

yeah, there are lots of really clever constructions that approximate well

or they marked the ruler, which is not allowed

that would be called a neusis construction

[solar folio]

such an interesting topic tho

[bleak dock]

indeed

that's why the ancient Greeks and a lot of other mathematicians studied this

[junior crater]

and thats why there is galois theory in algebra

[solar folio]

alr thanks

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[next epoch]

this does not give you a vector space since multiplying by zero gives you a matrix not in V

[kind viper]

really?

[dusky nymph]

it doesn't say a and b have to be nonzero

[next epoch]

if it doesnt say that then we have a vector space

right

[kind viper]

sure we do.

in general you shouldn't assume letters have to be nonzero ex nihilo

[next epoch]

perfect thank you

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[forest token]

im getting 2 solutions , a = 2 and a = 0

how do i know which one to take

[brazen quiver]

which is the largest non-negative integer value from that list of options?

[forest token]

2

but thats wrong 😔

im not that dumb bro

[brazen quiver]

then you've made a mistake

[soft bobcat]

Yes this one is cursed

[brazen quiver]

!show

[soft bobcat]

I remember

[compact pewter]

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

[soft bobcat]

Wait let me show u

the solution

this is adv problem right?

[forest token]

yea

[soft bobcat]

Yeah its hella cursed

[forest token]

oh 😭

[soft bobcat]

i have the link to a chat from before

wait

[kind viper]

ok important question: does {} mean fractional part or are those just brackets

[forest token]

just brackets

if its fractional part they will mention

[soft bobcat]

#help-44｜stanley-🌲-v2-dans message

This is just one message in the chain, i think the original question is higher up

[forest token]

LMAO thats the exact page im looking at in my book rn

hes allen too

[soft bobcat]

Yeah

Just read the convo

It's all been done

I couldn't do it at first too 😭

[forest token]

i lost track

of what u guys were talking about

like u were saying negative numbers cannot be raised to fractional powers??

[brazen quiver]

negative numbers can be raised to fractional powers, but you wind up getting complex numbers out of it.

and you have to deal with branches.

it's a mess.

[forest token]

not always

[soft bobcat]

Yeah there are some specific cases though like (-1)^1/3

like the guy pointed out

[forest token]

yeah

[soft bobcat]

But they generally can't be

If you want a real result

[brazen quiver]

even then (-1)^1/3 does have a real solution, but it's not the principal solution

[forest token]

(-2)^1/3

[brazen quiver]

the principal solution is exp(pi i / 3)

[soft bobcat]

O

I didn't know that

[forest token]

broo why tf will u think of putting a value back in the limit

when it says largest value 😭

[soft bobcat]

Exactly right

It's such a random question

[forest token]

yeah

i got it now tho

ty @soft bobcat @brazen quiver

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[brazen quiver]

https://en.wiktionary.org/wiki/principal_root

(mathematics) A complex number which, when raised to the power of n, yields the radicand of its nth degree radical, and which has the greatest real part among all such numbers, and positive imaginary part in case of equality of the real parts.

[soft bobcat]

.reopen

[compact pewter]

✅

[brazen quiver]

just the first thing google pulled up

[soft bobcat]

I see

Alr I'll read

Ty

[brazen quiver]

👍

[soft bobcat]

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[hexed hemlock]

Can someone explain to me how this equals each other, this is putting it in geometric series form a/1-r

[solemn remnant]

They don't equate one another

[kind viper]

those... don't look equal. context?

[solemn remnant]

GPT is awful

GPT is your context

[kind viper]

you sure this is from GPT? :P

[hexed hemlock]

this is original question

[solemn remnant]

I'd recognize that output among thousands

[kind viper]

ok and show your work?

[hexed hemlock]

i had it righ t

i dropped a negative

[kind viper]

,rccw

[hexed hemlock]

my bad lol im being gaslighted by gpt

[solemn remnant]

Hint: $$\frac2{4x+3}=\frac23\times\frac1{1-(-4x/3)}$$

[kind viper]

you want it in powers of x+1 not x

[solemn remnant]

Oh you're right I misread

[kind viper]

it's much easier actually: $\frac{2}{4(x+1) - 1} = \frac{-2}{1 - 4(x+1)}$

[boreal girder]

Ann

[kind viper]

a/b = -a/-b

[hexed hemlock]

.close

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[atomic aspen]

The inequality should flip to x <= - 4. But I can't understand what I'm doing wrong.

[kind viper]

in general going from (x-a)(x-b) ≥ 0 to some combination of x ≥ a and x ≥ b and then trying to figure out which one should "flip" based on arcane rules is a bad approach

[atomic aspen]

Okey, what is the best approach. Been struggling with these inequality where there is something like |x^2 + ax + b|

[dire fiber]

You did the wavy curve wrong

x>=3 and x<=-4

But you wrote x>=-4

[kind viper]

you should draw explicit wavy curves/number lines for every single case and do the sign analysis thing

[atomic aspen]

Hmm.. Okey will try some more. Thank you guys!

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[dark pasture]

How to factor this

[glass turret]

well

have you tried anything

you can start off by factoring each term as far as you can

[dark pasture]

ok, but then i get to the exponents, and i can't understand why the exponents are simplified like this : 3/4 + 1/4 and not the other way around

i mean like 3/4 - 1/4

[jagged blade]

What is your first step?

[dark pasture]

sorry wait i got it confused

hm i simplify the -2/4

to -1/4

[jagged blade]

?

[unkempt sail]

?

[short leaf]

?-?

[jagged blade]

2 =/= 1

[short leaf]

It isn’t simplified to -1/4

[dark pasture]

omg i am so tired sorry, -1/2

[jagged blade]

Go to sleep then >.<

[dark pasture]

i just want to understand this and then i will do ahaha

[jagged blade]

$5x (3x + 2)^{-\frac{2}{4}} + (12x + 8)^{\frac{3}{2}}$

[boreal girder]

💪 Greenie The Power Queenie 💪

[jagged blade]

Okay so you've simplified -2/4 to -1/2

$5x (3x + 2)^{-\frac{1}{2}} + (12x + 8)^{\frac{3}{2}}$

[boreal girder]

💪 Greenie The Power Queenie 💪

[jagged blade]

What next?

[forest token]

y^-1 = 1/y

use that on the -1/2 power

[dark pasture]

ok then (12x + 8)^3/2 becomes (4(3x+2))^3/2

[forest token]

yes

[jagged blade]

Yep

What's 4^(3/2)?

$5x (3x + 2)^{-\frac{2}{4}} + (4(3x + 2))^{\frac{3}{2}}$

[boreal girder]

💪 Greenie The Power Queenie 💪

[dark pasture]

then (4^3/2(3x+2)^3/2) becomes 8(3x+2)^3/2

then i can factor taking out the lesser exponent

[jagged blade]

$5x (3x + 2)^{-\frac{1}{2}} + 8(3x + 2)^{\frac{3}{2}}$

[boreal girder]

💪 Greenie The Power Queenie 💪

[dark pasture]

and i cant understand why when factoring, the exponents become 3/2+1/2 instead of 3/2 - 1/2

[jagged blade]

In order to combine these, we need the two exponents to be the same

[jagged blade]

I'm not sure what you mean by this

Can you write down what you've got?

[dark pasture]

okok

(3x+2)^-1/2 * (5x+8(3x+2)^3/2+1/2) instead of (3x+2)^-1/2 * (5x+8(3x+2)^3/2-1/2)

[jagged blade]

Right

[glass turret]

when you divide

you subtract exponents

for example (2^4)/(2^2) = 2^(4-2)

[jagged blade]

So let's take it in two steps

[glass turret]

but if you're subtracting a negative exponent then you'd end up adding it

for example (2^4)/(2^-2) = 2^(4-(-2)) = 2^6

[jagged blade]

We want to make it so that the second term has a factor of $(3x + 2)^{-\frac{1}{2}}$

[boreal girder]

💪 Greenie The Power Queenie 💪

[jagged blade]

Right?

[dark pasture]

yes

[dark pasture]

ok yes

[glass turret]

and when you factor out a term from an expression, you divide the rest of the expression by that term

right?

[jagged blade]

$(3x + 2)(3x + 2)^{\frac{1}{2}} = (3x + 2)^{\frac{3}{2}}$

[boreal girder]

💪 Greenie The Power Queenie 💪

[jagged blade]

Are you happy with that?

[dark pasture]

yes

[dark pasture]

oooooh ok

[jagged blade]

So what do we need to do next?

[dark pasture]

(3x+2)^-1/2 * (5x+8(3x+2)^2) the it becomes like this

ok so i-ve got the rest figured out, but i couldn't figure thatù

[jagged blade]

You're alright now then?

[dark pasture]

thank you very much

.close

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[jagged blade]

Just to clarify, when you take out a factor from something, you have to divide the rest by that factor. If you're dividing by (3x + 2)^(-1/2), you have to subtract the powers, and when you subtract -1/2, you get +1/2

[dark pasture]

.reopen

[compact pewter]

✅

[dark pasture]

ok because it would be like this right? -(-1/2)

[jagged blade]

$\frac{(3x + 2)^{\frac{3}{2}}}{(3x + 2)^{-\frac{1}{2}}} = (3x + 2)^{\frac{3}{2} - (-\frac{1}{2})}$

[boreal girder]

💪 Greenie The Power Queenie 💪

[jagged blade]

Yep

[dark pasture]

ok now i understand it better, thanks a lot

[jagged blade]

Np

[dark pasture]

so now i close this with . close right?

[jagged blade]

Yep

[dark pasture]

thanks again

.close

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[glass turret]

yes

[compact pewter]

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[elfin abyss]

If, A + B + C = π
Prove, sin2A + sin2B + sin2c = 4sinA sinB sinC

[elfin abyss]

Someone help

If, A + B + C = π
Prove, sin2A + sin2B + sin2c = 4sinA sinB sinC

[subtle helm]

Let’s try simplifying

[elfin abyss]

Alr

[subtle helm]

Not sure. But applying double angle first looks good

Lemme try wait

[elfin abyss]

Okayy

Sure thanks

[jagged blade]

We're starting with a sum, and ending up with a product

So we probably want to use the sum-to-product identities

[subtle helm]

i went from rhs to lhs using this too

my initial approach wasnt helpful

notice c = pi - (a+b)

[elfin abyss]

Cool

U guys are so helpful only if i could repay u giys

[subtle helm]

[ \sin(\pi - (A+B)) = \sin(A+B)]
and
[ \cos(\pi -(A+B)) = -\cos(A+B)]

[boreal girder]

k

[elfin abyss]

Wait isnt the first one supposed to be -sin (a+b)?

Oh noo wait

Sorry its in 2nd quadrant

[compact pewter]

@elfin abyss Has your question been resolved?

[elfin abyss]

No

[compact pewter]

Channel closed

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[jagged blade]

.reopen

[elfin abyss]

Miss is that a command?

[compact pewter]

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[jagged blade]

Yeah you'll need to use it keep your channel open

[elfin abyss]

Greenie can u dm me pls?

[elfin abyss]

Dm pls

[soft bobcat]

[compact pewter]

@jagged blade Has your question been resolved?

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[compact pewter]

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[crisp ingot]

wth

[compact pewter]

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[ebon crane]

1+1=

[ebon crane]

it

[quartz yoke]

2

[short lantern]

<@&268886789983436800> im not gonna bother trolling back with this one

[quartz yoke]

Pls help me

[short lantern]

no

[sturdy hill]

do not invade random help channels to ask for help

you have your own, wait patiently

.close

[compact pewter]

Channel closed

Closed by @sturdy hill

Use .reopen if this was a mistake.

[quartz yoke]

Ok 😭 mb sorry

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[sleek wind]

can anyone help me with this?

[ember marten]

what have you tried?

[sleek wind]

using variables

[ember marten]

you might need to elaborate what you're trying to achieve with that?

cause like the equation already has variables lol

[sleek wind]

im trying to find x

im using a,b

[ember marten]

right, so how are you trying to isolate it?

[sleek wind]

but doesnt work

[ember marten]

okay, what are a and b

[sleek wind]

how do i write swuare root

[ember marten]

you could do sqrt(...)

[sleek wind]

but 1-x^2 = (1-x)(x+1)

[ember marten]

right, so that makes sense

so you let a = 1 - x and b = 1 + x, or something like that?

[sleek wind]

this is a

idk how to write root

[ember marten]

okay, and the other root is b?

you could do (...)^(1/3) if you want to do cube root

or like uhh

[sleek wind]

yeah

[jagged blade]

I'd suggest multiplying it out as a first step

cbrt() would be a reasonable notation

[ember marten]

i actually think their substitution is much easier, so they don't have to worry about distributing inside the radicals

cause they'd end up with (ab) * (a + b) = 2

[jagged blade]

Or alternatively, $\sqrt[3]{1 - x^2}$ is actually what you're looking for

[boreal girder]

💪 Greenie The Power Queenie 💪

[vale star]

Maybe try substituting (1+x) as a and (1-x) as b?

[ember marten]

^^ this is what i thought was slightly easier

[jagged blade]

I think that's what he's done already

[ember marten]

he's tried this

[jagged blade]

After this step though, you're probably gonna have to undo it

[ember marten]

right, so i guess a = 1 - x and b = 1 + x is probably better

[jagged blade]

Mmm?

[ember marten]

@sleek wind still here?

[sleek wind]

ye

[ember marten]

if you tried a = 1 + x and b = 1 - x, what would your equation become?

[ebon halo]

I'm reading all of this, I'm in form 1 nd.. sorry, I'm not here to help but is this what I'll have to suffer through in the future?? Cuz I need a headsup so desperately right now.

[sleek wind]

still wont work

or we can switch ab to the side

we will have a+b = 2/ab

[vale star]

Finding a relation between a and b might?

[jagged blade]

In what sense is it suffering?

[sleek wind]

1-x = a => 1+x = a+2 maybe?

oh wait wrong

[jagged blade]

-a + 2

[sleek wind]

1 - x= a then 1+x will have to be a+ 2x

feels wrong

[jagged blade]

It could also be -a + 2

Do we know how many roots there are supposed to be?

[ember marten]

bro i'm ngl, just cube both sides it'll make things easier

you know the expansion formula for (a + b)^3, right?

[sleek wind]

ye

[ember marten]

try cubing both sides of the original

what do you get?

[sleek wind]

1-x^2 (___)^3 = 8

[vale star]

Is this is ryt then u will get a quadratic eqn ,so 2 roots?

[ember marten]

right

so expand the stuff in the cube, simplify

distribute the (1 - x^2) into that expansion

etc

[jagged blade]

I think something useful to note would be that 2 can be written as a^3 + b^3

So we have ab(a + b) = a^3 + b^3

[ember marten]

oh that is handy

though going from there...

[jagged blade]

So we have a^3 - a^2b - ab^2 + b^3 = 0

[ember marten]

oh i see an (a-b) in there

[jagged blade]

Not sure if this is going anywhere though

[ember marten]

a^2(a - b) - b^2 (a - b) = 0?
so (a+b)(a-b)(a-b) = 0

[jagged blade]

I can also eyeball 1 solution straight off

[ember marten]

right, the only solutions could be if a = b or if a = -b

which should be enough to justify that the eyeballed solution is the only one

[jagged blade]

Ayyy we did it

[ember marten]

idk if papaya is still here :(

[jagged blade]

So we have
1 + x = 1 - x -> x = 0
or
1 + x = -1 + x -> no solutions

[molten badger]

Prove that x=1 is not a solution, divide both sides by 3th root(1-x^2). Rewrite (3throot(1+x)+.....) as a fraction so that we only have 1+x+1-x=2 on the numberator so now we have 2 same numberator on both side

[jagged blade]

?

The proof that x = 1 is not a solution is very short lol

If x = 1, then the LHS = 0

And the RHS = 2

[molten badger]

This is my way to solve it

[jagged blade]

0 =/= 2

[molten badger]

Yess I know

[jagged blade]

How are you going to rewrite it as a fraction?

[molten badger]

But we need to prove since we want divided both side with number that different from 0

[ember marten]

we never needed to divide both sides by that quantity

[molten badger]

Try to make the numerator have form a^3+ b^3 by multiple both by a^2-ab+b^2

[ember marten]

cause you'll end up having to cube both sides and then multiply it back over anyway, and then expand

might as well just cube both sides

[jagged blade]

$\sqrt[3]{1 + x} + \sqrt[3]{1 - x} = \frac{2}{\sqrt[3]{1 - x^2}}$

[boreal girder]

💪 Greenie The Power Queenie 💪

[molten badger]

$\frac{1 + x + 1- x}{sqrt[3]{(+x)^2} - \sqrt[3]{1-x^2} + \sqrt[3]{(1 - x)^2} = \frac{2}{\sqrt[3]{1 - x^2}}$

[boreal girder]

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

[molten badger]

Damn

[sleek wind]

sorry guys

[boreal girder]

Rajvir

[sleek wind]

im doing another problem so i wasnt here

sorry

[jagged blade]

We have succeeded

[sleek wind]

is this the answer?

[pseudo shale]

Chat

[sleek wind]

tysm guys

[pseudo shale]

No

[jagged blade]

No. He just turned up now lol

[sleek wind]

oh ok

[pseudo shale]

I was helping a person solve smtg

[jagged blade]

It starts here

[pseudo shale]

Who did not even know (a+b)^3

Took 2 hours

Ts too tuff

[molten badger]

$\frac{1 + x + 1- x}{\sqrt[3]{(1+x)^2} - \sqrt[3]{1-x^2} + \sqrt[3]{(1 - x)^2}}= \frac{2}{\sqrt[3]{1 - x^2}}$

[boreal girder]

Alexis_Fx

[molten badger]

This is what i mean

[sleek wind]

how

[molten badger]

Sorry im on phone rn so i typing slowly

[sleek wind]

ah yes

[jagged blade]

$a = \sqrt[3]{1 + x}$

[boreal girder]

💪 Greenie The Power Queenie 💪

[jagged blade]

$a^3 = 1 + x$

[boreal girder]

💪 Greenie The Power Queenie 💪

[sleek wind]

i understand

it now

ty

[jagged blade]

Oki

[sleek wind]

goated

[jagged blade]

And then we did some rearranging and factorising

[sleek wind]

from there on ik

do i close it now?

[jagged blade]

If you're happy, yes

[sleek wind]

.close

[compact pewter]

Channel closed

Closed by @sleek wind

Use .reopen if this was a mistake.

[compact pewter]

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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.

[jovial widget]

very quick functions question.. for the transformations if the parent function's values are -2 would the -5 only mean it transforms downward by 3 units or by 5?

[jovial widget]

same with the 1

would that mean it moves right by 3 or 4

[timid pawn]

think of the transformations relative to the original function sqrt(x)

[jovial widget]

so -5 would represent the function translating down by 3?

[old mica]

yeah, it's relative to the original function you were given

[jovial widget]

thank youu

even with the -1?

[old mica]

mhm.

well... I guess the question you've been provided is a little ambiguous

[jovial widget]

the question is just stating the tranformations done

[old mica]

it's not clear if f(x) and g(x) are both transformed versions of the original function sqrt(x), or if g(x) is a transformed version of f(x)

[jovial widget]

i didnt include the original question 😅 maybe i shouldve wait

" Describe the transformations to the graph of the reciprocal parent function,
to produce the function g(x)=-\frac{1}{2}(x+1)-3"

oh wait

;emme just take a ss

[old mica]

huh. but that's not the same as the question you initially provided

[jovial widget]

noo the question i originally provided was me just trying to figure it out

[old mica]

ah

well, function transformations are all relative to a parent function

[jovial widget]

more of a personal question

oh wait

THATS WHY]

[old mica]

if sqrt(x) is your parent function, then you ought to say that g(x) is translated to the right by 4 units, and down by 5

[jovial widget]

oops wrong question'

mb

😅

[old mica]

hm, they haven't said what the parent function is

[jovial widget]

f(x) would be

and g(x) represents the child function i think

[old mica]

the wording makes me think they want you to describe the transformations that you need to apply to f(x) to turn it into g(x)

it sounds like you have the same idea

[jovial widget]

yes i think so

[old mica]

you should be careful with your transformations when doing this though :p