help-13

[short vapor]

I'm struggling on how to describe it but if you were to hypothetically add 1 to every natural number on this list why wouldn't you eventually generate a number that is not on the list?

[crimson delta]

no

the list is infinite

try telling me a number which isnt on the list and I would tell you how it is generated

[short vapor]

I don't understand

Wym tell me how it's generated

😭

Sorry bruv the bulb in my head dim

But I couldn't give you a real number that isn't on the list

[crimson delta]

you are saying that you would somehow generate a new natural number which wasnt on the list before

or wait

is that what you are saying?

the initial list includes all natural numbers

[short vapor]

I guess I am

My answer would be last natural number +1 I know the list is infinite and doesn't really have a last number I guess I'm confused why you can add 1 to every position in the list of reals but you can't do the same with the list of naturals and create a new number yknow

Like I understand the reasoning

Or the process by which you can create new reals forever

You're right homie, you didn't have to delete I think I was misunderstanding, I was thinking like 1, 2, 3, ... , infinity but the size is infinite and there are infinitely many natural numbers

I'm a lil confused still but thank you I understand more

🙏

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[lyric phoenix]

Hello ! Dumb commutative algebra question : if $m$ is a maximal ideal in a ring $A$, and $mA_m$ the unique maximal ideal in the corresponding localization. I think $m^2$ should be the preimage of $(mA_m)^2$, but I can't prove it.
As far as I understand proving this is equivalent to proving that if $a.b = c.d$ with $a,b,c \in m$ and $d \not \in m$ then $c \in m^2$. Is there any way to finish this proof ?

[wraith dagger]

plougue

[cedar kiln]

@lyric phoenix Has your question been resolved?

[lyric phoenix]

I think I manage to be conviced by a proof with additional hypothesis I also had (A is a finite reduced k-algebra for a field k). Still curious whether or not this holds in general but I don't need it to advance in my work so I'll close the channel

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[small cloak]

anyone knows what is this theorm called

[small cloak]

that is used here

[tropic oxide]

"SAS congruence test" or some naming-related variation thereof, maybe?

[small cloak]

didnt find it

[tropic oxide]

can you perhaps show the entire question, just so we are clear about the instructions & what you are looking for

[cedar kiln]

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[twin mulch]

Our teacher calls it AB1 (Angle bisector theorem -1)… but I don’t see why you need it’s name as it is mainly used to prove angle bisector concurrency

Or you could just use congruence as @tropic oxide stated above

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[small tulip]

hello, i was given this one task of finding a value of a circle.. i dont understand this question at all... i may need some help

[weak ore]

circle theorems

Angles subtended from same arc of a centre is half of centre angle

[small tulip]

but what am i even searching for? an angle of a corner?

[crimson sedge]

find x

it would help to find angle DOF first

then use circle theorem mentioned above

[small tulip]

no, actually this is the thing i dont understand... what does "angle DOF" mean? because D,O and F are all corners

[crimson sedge]

find the angle on the corner of the middle term

[weak ore]

the verticle that connects d and f

[crimson sedge]

connected to the outer two corners

[weak ore]

it is basically o

[small tulip]

so basically i have to find corner O?

[weak ore]

angle O yes

[small tulip]

so beciaslly the first of the 3 letters is the corner that is specified?

and the others just point to the right triangle?

[crimson sedge]

the second of the 3 letters

e.g.
angle DOF - find angle O

angle DEF - find angle E

etc...

[weak ore]

The first and last letters are important in some cases however

[crimson sedge]

yes

the first and last letters show you which triangle you should look at

find the angle of the second letter inside that triangle

[small tulip]

ohhhhh

not the first

and what does this mean?

[weak ore]

angle

sometimes its a ^ shape on top of the middle letter too

[small tulip]

aha

but what is this 2y part? it isnt any corner name

[crimson sedge]

y is an unknown here

find the value of y

with the information provided

[small tulip]

no?

i thought i need to find this

[crimson sedge]

to find that, you need to do some steps first

use information given to find y, use y to find DOF, then use DOF to find DEF

[cedar kiln]

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[lavish turtle]

doesn

[lavish turtle]

doesn't the gradient take in a vector field and output a vector field?

[tropic oxide]

no, it takes as input a scalar field

grad: {scalar fields} -> {vector fields}

[lavish turtle]

ah ok

i see

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[rugged raven]

Help

[tropic oxide]

progress so far?

[cedar kiln]

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[azure vault]

I am learning conditional statements and in this question it asks me about bi conditional

Geometry

I dont completely understand what its asking?

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[lunar egret]

can someone walk me through this

[grand forge]

first one

integrate both sides of dy/dx = 0

what do u have

[lunar egret]

y=C

[grand forge]

is it?

[lunar egret]

im braindead so

[grand forge]

no your correct

but try y=-1 instead of -1

[lunar egret]

dude im gonna commit

the reason it kept failing

i forgot y=

i thought i was stupid

holy shit

[lunar egret]

ok anyway this one i genuienly dont know what im doing wrong

it says this when i put that solution in

[grand forge]

in this case

you want all the terms with a 'y' varaible on the side of the equation with dy/dt and all the terms of t on the other side

do that frist

its done it in the example

or the explanation i should say

[lunar egret]

ln|y|/g +C?

[grand forge]

Erm

[lunar egret]

no im braindead

[grand forge]

g(y)

[lunar egret]

im tryna do the thing

[grand forge]

Just mean the functiob of y

Not an actual ‘g’

g(y) could be y^2 + 4 for all we care

It’s like how f(x) is a functiob of f

(1-7t)dy/dt - y =0

(1-7t)dy/dt = y

[lunar egret]

so would it be the integral of 1-7t/dt = integral of y/dy

[grand forge]

not quite

so now im dividing by 1-7t

dy/dt=y/(1-7t)

divide by y

y dy/dt = 1/(1-7t)

times by dt

dy/y = dt/(1-7t)

now u can integrate

this is a method called seperaation of variables i beleive

[lunar egret]

y^2/2+C

i dont think thats right

[grand forge]

equals what

intergrate right side too

[lunar egret]

-1/7ln|1-7t|+C

[grand forge]

no

[lunar egret]

https://tenor.com/view/done-dead-deceased-crisis-existencial-crisis-gif-5649382

[grand forge]

you didnt integrate left properly

dy/y = y^2/2???

[lunar egret]

oh wait

holy shit im blind

[grand forge]

Yea

sorry

i made a lot of mistakes

[lunar egret]

lny+C

[grand forge]

hard to type on phone

Yea

so now

ln(y) = -1/7ln(1-7t) + C

solve foy y

and apply initial conditinos

[lunar egret]

y=-1/|1-7t|^1/7

wait what

im dumb

theres no other variable to solve

cause 3 subs in for t and -6 for y

[grand forge]

your + c is forgotten

let e^c = c

coz constant is a constanrt

e^c is just another constant

c will be a horrible number

just a heads up

[lunar egret]

But when you solve for y, C-C right

[grand forge]

wym?

[lunar egret]

Here isn’t it lny+c

[grand forge]

Bascially

what we do

is we combin both them

so we have +a on y side, +b on t side

and than we make c = b-a

Coz its just constants

so its just lny(y) = -1/7ln(1-7t) + c

e^ both sdies

y = e^c/(7x-1)^1/7

e^c = C

C being just another consant

[lunar egret]

Oh ok

[grand forge]

C is horrible btw

[grand forge]

remember

e^c = C

[lunar egret]

that didnt work either

[grand forge]

Yea it should

[lunar egret]

i mean for the answer when i put it in it said it was incorrect

[grand forge]

C = -6 * 20^1/7

[lunar egret]

ye thats what i got

[grand forge]

Yea

thats the answer

[lunar egret]

hold on maybe the softwares being stupid

so just to confirm

this is y= right

@grand forge

[grand forge]

mhm

that is correc

[lunar egret]

its not accepting it as the answer

[grand forge]

Its literally correct

[lunar egret]

bruh im gonna commit ive been working on this one problem for over an hour now and even when i get it right

it doesnt work

[grand forge]

https://gyazo.com/1574eef7f98fba98800e84edff0f9fec

lmao

[grand forge]

LITerally the correct answer LOL

[lunar egret]

anyone got chegg 😭

[grand forge]

you dont even need chegg

https://mathdf.com/dif/#expr=(1-7x)d(y)%2Fd(x)-y%3D0&func=y&arg=x&vals=3%3B-6

Thats a link to your eqaution

and the solution

[lunar egret]

im gonna cry

[grand forge]

with your initakl conditions

[cedar kiln]

@lunar egret Has your question been resolved?

[lunar egret]

<@&286206848099549185>

[cedar kiln]

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[silk patio]

?

[royal loom]

Do you have a question? If so, post it here

if not (.close) the channel

and read #❓how-to-get-help #rules

[silk patio]

I dont

idk why this made

.close

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[waxen crater]

quick question

i got (-x²-1)/(x²-1)²

but the answer is (-x²+1)/(x²-1)²

[bright bridge]

are you solving for the derivative

[dire geode]

,w diff x/(x^2-1)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by riemann#2529

[dire geode]

Show this

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[crimson sedge]

I first multiplied (hx+k)(x+j) --> hx^2+hjx+kx+kj

I found out that h=4. I don't know how to move on from there though.

[long swan]

equate coefficients

h = 4

b = (hj+k)

-45 = kj

[maiden tendon]

check out this for work

Let me know if it makes sense

[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.

[bright bridge]

¯_(ツ)_/¯

[maiden tendon]

Key idea is to know what h,k, and j are already integers (like -1,0,1,2,3...)

[cedar kiln]

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[crimson sedge]

.close

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[marsh rain]

Hi. This is a related rates question and the answer provided was 3 whereas I had gotten 4.5. Sorry if my handwriting is messy but I'd appreciate if someone pointed out where I went wrong and what to do instead.

[cedar kiln]

@marsh rain Has your question been resolved?

[muted cedar]

pls zoom in lmao

[cedar kiln]

@marsh rain Has your question been resolved?

[marsh rain]

Whoops mb

[cedar kiln]

@marsh rain Has your question been resolved?

[cedar kiln]

@marsh rain Has your question been resolved?

[dreamy sleet]

,rotate

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[crimson sedge]

is this linear equation
(x-5)(x+6) =0
because if I multiple these factors, then I will get quadratic equation 👀

[dire geode]

Correct it is not a linear equation

[crimson sedge]

@dire geode thanks

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[fading viper]

He simply multiplied the fraction by 1 right? Since 3x/3x=1

[fading viper]

For context, I'm watching https://youtu.be/YNstP0ESndU at 7m 57s

[jade sinew]

Yes

[fading viper]

Okay thanks

If it were an equation, would you also have to multiply the other side by 3x/3x or no?

[jade sinew]

No, because multiplying by 1 doesn't change anything

[fading viper]

Right, I just find it cool that you can do that to manipulate algebraic terms

[jade sinew]

Yep, you'll find a lot of problems are solved by either multiplying by or adding 1

[fading viper]

Like while 3x/3x=1, multiplying by it still "changes" the expression

[jade sinew]

Exactly, as you keep going there'll be a lot of that, multiplying by 1 to be able to manipulate the expression

[cedar kiln]

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[wraith dagger]

marc

marc

marc

marc

marc

marc

[cedar kiln]

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[quasi jungle]

Hi, I am solving the b) of the following problem. Here is the translation:

A thin plate has the shape shown below. This plate can be modeled by the region S of the plane located between the polar curves r = 2 sin(θ) and r = 3 + sin(3θ).
a) find the surface area S
b)If the density of the plate is proportional to the square of the distance from the origin, what is the mass of the
plaque ? for part b), you can use software to evaluate trigonomic integrals.

[quasi jungle]

<@&286206848099549185>

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[crimson sedge]

Could someone please show me how they would work that out

And what’s the formula

[obsidian coral]

Well, it's a cylinder and you're looking for the volume

That should be enough info for you to look up the formula yourself

[crimson sedge]

I’ve tried I’m just not the best at maths

Just need someone to explain it to me

[obsidian coral]

Did you look up the formula?

[silent glade]

the formula is hr^2*pi

if i'm interpreting it right that the cylinder drawn is a pier hole, then u just plug in the numbers it gives u and multiply it by 15 for each pier hole

[silent glade]

where r is the radius of the base and h is the height

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[coral surge]

how many numbers between 99 and 1000 both excluding can be formed such that no digit is repeated

[coral surge]

my answer is 640

[surreal cave]

can you show ur work?

[coral surge]

but it is showing incorrect

sure

@surreal cave

[surreal cave]

$\begin{tabular}{|c|c|c|}\hline\8 & 8 & 10\ \hline\end{tabular}=640$ I have no clue what this means

[wraith dagger]

XxMrFancyu2xX

[coral surge]

i multiplied

[surreal cave]

but where did 8, 8, and 10 come from? (not tryna be rude, just curious)

[fair mortar]

he did like a trial error

probably he chose that first digit can have 10 possibilities, then the second can have only 8, and the third can have only 8 ( from ones to hundreds)

although i think it must me 8 9 10 idk

[surreal cave]

well think about is from all the numbers 100-999 the first digit can be anything 1-9 (9 possibilities), the next can be (0-9) excluding one, then the next is (0-9) excluding two

[clever sun]

@coral surge what's the answer

[coral surge]

9x9x8 = 648

[coral surge]

thank you for the explanation

[surreal cave]

yw!

[coral surge]

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[twilit crane]

Help

[cedar kiln]

Please don't occupy multiple help channels.

[obsidian coral]

Don't open multiple channels

.close

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[patent creek]

how to convert 5+i into trig form

[patent creek]

i know r=squareroot of a^2+b^2

r=sqr25+1

r=sqr26

and tan theta=5

thats not on the uniot cicrle

idk how to do

cus there infinte numbers of posiblites

5/1,10/2,15/3

...

[boreal epoch]

huh

@patent creek you find the angle (arctan of your answer) find the module value (Pythagorean theorem) and then you express it as |Z|*cos(å)+isin(å)

[patent creek]

so tan^-1 of 5

[boreal epoch]

or you can find the module and find sin cos first and find their arcs

[patent creek]

arctan is faster right

[boreal epoch]

it's a bit tricky when u use arctan since you need to take not at what side of the graph you are

meaning when a and ib are both positive you expect the angle to be 0-90

[patent creek]

yes

[boreal epoch]

and from there gl calculating

[patent creek]

ty

i got

root26(cos0.0239+isin0.0239)

ty bro

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[fringe mortar]

bit of a stupid question, but I like self studying but cant do it in an organized way and I'm fairly new to it. how should i approach a textbook (I'm doing Axler's Linear Algebra Done right, but I'm having trouble starting the first chapter as I do not know if I should take notes on the lesson pages or if that is wasting time)
and how should i organize the chapters in terms of time management

[fading narwhal]

I like to take notes on the key points

yeah

[fringe mortar]

thanks! how should i manage the amount of time i work on each chapter (i want to set goals for this)

[fading narwhal]

I guess the amount of time you should spend is based on how much total time you have and how familiar you are with the content

[silent glade]

ye just whatever time u need to understand the content

otherwise if u move on without understanding it there was no point in spending time on it at all

[fading narwhal]

hmm

[fringe mortar]

thank you all

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[limber ferry]

what would the name of the series be?

[shadow brook]

Telescoping series

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

[limber ferry]

ah okay thank you

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[pine dome]

I dont understand this

[pine dome]

I got a completely different answer

than the one in the solution sheet

This is what they said

[crimson sedge]

How did you get your answer?

[pine dome]

I did the length = theater times radius

i converted the 71 to radians

and times it by 1.6

[cedar kiln]

@pine dome Has your question been resolved?

[pine dome]

yolo

[cedar kiln]

@pine dome Has your question been resolved?

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[verbal hazel]

Open.

[verbal hazel]

Given f(x) = x at x<1,
x^2 at 1<=x<=4
8(x)^1/2 at 4 < x
Then find the inverse of f.

How should I proceed?

[crimson sedge]

To find the inverse of a function, set y=f(x) and try to solve for x in terms of y. You can do this by considering each case

Although we also need to verify that this is a bijection, but I'm going to assume that's given

[verbal hazel]

Ok so taking cases I get inverse in each case.

Ok, I get it. Tnx

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[primal spruce]

i dont get the opposite side and adjacent

why is there two columns seperating it

i kinda get what an opposite and adjacent side is

[crimson sedge]

I think its asking which pair is opposite to angle C and which one is opposite to angle T

[primal spruce]

this the guy from yt said

[primal spruce]

but why angles though?

i dont get it

[crimson sedge]

wdym

[primal spruce]

wait wait idk how to explain

[livid hound]

the opposite and adjacent are relative to the angle you're using

[primal spruce]

can u explain?

im kinda confused

[crimson sedge]

Like if you drew an angle on that little C corner, which one would be opposite to it and which one would be adjacent to it

[hasty aurora]

An angle can only be adjacent to one side that isn't the Hypotenuse if its not the angle directly opposite to the hypotenuse

[primal spruce]

can u try to answer the first triangle so that i can understand it better

as an example

[hasty aurora]

For the C angle AC would be the adjacent and AT the opposite

[primal spruce]

ahhhhhhhhhhhhhh]

i get it

thanks

[hasty aurora]

np

[primal spruce]

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[solemn basin]

why are the brackets not (2-lambda) and (-2-lambda)

[short blade]

they pulled out a negative from (-2-lambda)

and multiplied it to (2-lambda)

if that’s not clear let me know

[solemn basin]

what would it be if that wasn't done? i got lambda = +/- sqrt

-1

[short blade]

whether you do that or not you should get the same eigenvalues

[solemn basin]

i didn't but i got lambda^2 + 1 = 0

[tropic oxide]

show work

[solemn basin]

I can't cos this isn't on my phone

i did (2-lambda)(-2-lambda) + 5 = 0

lambda^2 + 2lambda -2lambda - 4 + 5 = 0

lambda^2 + 1 = 0

[short blade]

why did you do + 5

[tropic oxide]

^

[solemn basin]

because 5 x1

oh wait its the determinant right?

so -5

[frosty ocean]

Yes

[solemn basin]

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[lunar crystal]

Can someone help me work this out?

[cedar kiln]

@lunar crystal Has your question been resolved?

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[peak prism]

Hey all, I need help proving the following -
let $G$ a set and $\varphi: G \to G'$ homomorphism. Show that if $H\leq G$ then $\varphi^{-1}\varphi(H) = \langle H,ker \varphi \rangle$
I'm really confused of how I'm suppose to use the inner product here, any help will be appreicated

[wraith dagger]

meitar5674

[cedar kiln]

@peak prism Has your question been resolved?

[cedar kiln]

@peak prism Has your question been resolved?

[crimson delta]

thats not an inner product

if I were to guess its the subgroup generated by H and ker phi

btw G should be a group, not just a set

[dreamy sleet]

that notation doesn’t mean “inner product”, it means “subgroup generated by”. Then you can consider the elements of the inverse image of H. What can you say about them?

[peak prism]

Yeah she is a group sorry

[peak prism]

So thats the subgroup generated by H, ker phi? meaning $h^n \cdot \phi^m$ for some $h \in H, \phi \in ker \varphi$?

[wraith dagger]

meitar5674

[dreamy sleet]

Yes

[peak prism]

Ok cool I'll give this a try then

I wasn't sure at all for the meaning of this

[cedar kiln]

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[peak prism]

I'm writing it down now, please let me know what you think 🙂

[crimson delta]

only if the group is abelian

otherwise it also includes terms like h_1^n phi^m h_2^k etc

[peak prism]

so that's my proof but guess it's not valid cuz we dont know whether it's abelian?

top is the other direction

[crimson delta]

(of course you could use a certain property about kernels to simplify that tho)

[peak prism]

and I'm not sure it even works if it's not abelian

[crimson delta]

your first row nearly works. but you would have to show that h^m phi^n is in the set. not what you did

ignoring the whole "the element might not have that form" issue

[peak prism]

Why? I mean, ain't that the definition?

oh uhm

Yeah

[peak prism]

Yeah I'm assuming this set is an abelian set, no?

[crimson delta]

what properties do you know about kernels

[peak prism]

thats $\varphi(\phi) =1$?

[wraith dagger]

meitar5674

[crimson delta]

those are the elements in the kernel, yes

I mean what do you know about the kernel as a subset of G

[peak prism]

I'm honestly not sure what you mean by that, maybe we haven't learned yet

[crimson delta]

subgroup? normal?

but ok you dont need it. you can use the same approach as before but just write the element more generally

it is some finite product of elements in H and ker phi

in some order

[peak prism]

but then how do I get rid of those $\phis$ and prove it belong to H?

[wraith dagger]

meitar5674
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[crimson delta]

the same way you did in your current way

[peak prism]

Oh it will be the same cuz $\varphi(h \cdot \phi \cdot h \cdot \phi) = \varphi(h)\varphi(\phi)\varphi(h)\varphi(\phi)$ right?

[wraith dagger]

meitar5674

[peak prism]

and again those \varphi(\phi) are just 1

[crimson delta]

yes thats the idea

[peak prism]

Cool, thank you 🙂

Maybe I need some induction there but I think I'll leave it short

thank you!

[cedar kiln]

@peak prism Has your question been resolved?

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[jagged pine]

Can you convert the equation x^2 + 2xy + y^2 - 12x - 4y + 20 = 0 into something like the y = a(x-h)^2 + k form or standard form

[vagrant elbow]

Maybe you can factorize it

,w plot x^2 + 2xy + y^2 - 12x + 4y + 20 = 0

[wraith dagger]

Results provided by WolframAlpha

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Requested by NEONPerseus#2128

[vagrant elbow]

Nvm it's a parabola

[jagged pine]

So its not factorable?

[umbral hull]

in general it won't have the form that you stated, in particular because it wouldn't make sense to have something of degree 1 equal to something of degree 2 when the y^2 is in your equation

my algebraic geometry is a little rusty but there is a "standard form" or something akin to it for conics in two variables

let me look it up

I'll list a few theorems from my algebraic geometry book on conics that may help to answer your question:

"For any given conic P(x,y)=ax^2+2hxy+by^2+2gx+2fy+c, it is affinely equivalent to a prenormal form y^2=q(x) where q(x)=Ax^2+2Bx+C and A, B, and C are in your field K"

in this case that just means A B and C are real numbers

In particular another theorem
"In R^2 (reals in 2 dimensions) any conic is affinely equivalent to one of the nine normal forms."

The nine normal forms are as follows:

y^2=x - parabola
y^2=-x^2+1 - real ellipse
y^2=x^2-1 - Imaginary ellipse
y^2=x^2+1 - hyperbola
y^2=x^2 - Pair of real lines
y^2=-x^2 - imaginary line pair
y^1=1 - pair of real parallel lines
y^1=-1 - imaginary pair of parallel lines
y^2=0 two repeated lines (lines coinside)

conics is one of the few areas of polynomials where we pretty much know everything of interest

so if you want to know more I would just read up on conics

you can actually classify this as one of the following with basically no work, using delta invariants

[crimson sedge]

hi

are you helping rn blue guilmon

[umbral hull]

yes I'm answering the question of a standard form

for this sort of problem

[crimson sedge]

ok

[umbral hull]

oh these rooms are occupied, you'd need to go up to the available room

[crimson sedge]

oops

[jagged pine]

thanks--mostly my confusion is just from what is required (like whether I need to simplify it into a certain form)

[umbral hull]

well it seems to be a hyperbola

so probably yes

hyperbolas, like ellipses and circles have foci and can be written in terms of their foci

a hyperbola is what happens when you take the foci of an ellipse and cross them opposite of each other, the ellipse then sort of "circles" around infinity giving it that look where it shoots off in two directions about the focal points

circles are the speacial case when the foci are the same point

[jagged pine]

its actually just a parabola

sideways

[umbral hull]

ah ok, well parabolas also have focal points!

[jagged pine]

this is what it looks like

[umbral hull]

not entirely convinced that's a parabola by that graph

alone

lemme check the delta invariant rlly quick

yup it is a parabola

[dreamy sleet]

In this case you can always try rotating it by substituting rotated coordinates for x and y

[cedar kiln]

@jagged pine Has your question been resolved?

[umbral hull]

I do not believe this is factorable

https://www.wolframalpha.com/input?i=is+x^2+%2B+2+x+y+%2B+y^2+-+12+x+-+4+y+%2B+20+irreducible%3F

[dreamy sleet]

from here, if you want to watch the whole video: https://www.youtube.com/watch?v=h9OWnuarYuc

[umbral hull]

yup but we'd need to solve for theta

On the bright side, if you were to find theta you could write this out in a simpler way, it happens to be invariant under affine transformations (that's the one useful thing that my delta invariant calculation gave me) so in particular under rotations

sorry if none of this really answered your question, but no it isn't factorable over the reals it seems

[dreamy sleet]

the original question was how to convert it into standard form

to do that, you can rotate it

[umbral hull]

by what

[cedar kiln]

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[dry rampart]

I'm doing this function:
f(x) = x⁴ - 2x², within the interval of [-1, 1]
The assignment is to apply either Lagrange's Theorem or Rolle's Theorem, depending on the conditions.
The function has no discontinuity, so first condition is satisfied. The derivative's dominium also has no discontinuity, so that's the second condition.
The interval is [-1, 1], therefore

f(-1) = -1
f(1) = -1

Same result, therefore Rolle's Theorem.

[cosmic steppe]

What's the issue

[dry rampart]

These are the solutions I found.
My question is, knowing the interval is [-1; 1], are all three solutions acceptable, or just 0, considering it's the only one that's in between them?

I'm not exactly sure what [-1; 1] means in comparison to, let's say, (-1; 1)

My math teacher's horrible.

[cosmic steppe]

[-1,1] is the set of all real numbers between -1 and 1 inclusive; (-1, 1) is the set of all real numbers between -1 and 1 exclusive

So -1 and 1 are not elements of (-1,1) while they are elements of [-1, 1]

[dry rampart]

Which means all three solutions are acceptable?

X=0; X=1, X=-1

[cosmic steppe]

Well you're trying to use rolle's theorem or Lagrange's MVT?

Rolle's theorem requires you to find two points such that f(a) = f(b) and f(x) is continuous [a, b]

[dry rampart]

Yeah. And considering f(-1) and f(1) are equal, and not different, I need Rolle's.

... at least that's what I understood.

[cosmic steppe]

Okay yeah that's fine then ye just making sure you're on the right track

[dry rampart]

Kinda needed to know, myself lol

[cosmic steppe]

Well 0 is an element of [-1, 1]

[dry rampart]

That much, I could tell 😅

[cosmic steppe]

So it's kinda included when you state $\exists c \in [-1, 1] : f'(c) = 0$

[wraith dagger]

Umbraleviathan

[dry rampart]

My question wasn't the 0

It was 1 and -1

Since I didn't know whether [] meant inclusive or exclusive.

[cosmic steppe]

[] is inclusive, () is exclusive

[dry rampart]

Yeah, so x=1 and x=-1 are both acceptable solutions in [-1; 1]. Right?

[cosmic steppe]

Mmhm

[dry rampart]

Perfect, that's all I needed to know.

Thank you for your help.

Really needed it for tomorrow's test.

.close

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[solemn mesa]

Anyone know how to do the 3rd question?

[clear berry]

how did you do the first one

[solemn mesa]

First one you just plug them in the equation and you get 1, second is according to bezout's thingy that 2 numbers or coprime if we can find number a and b that solve the equation ax+by=1

[clear berry]

But do you actually get 1?

5(6n + 1) - 3(10n + 3) = 30n + 5 - 30n - 9 = -4

[solemn mesa]

Oh wait I typed it wrong the numbers are 6n+2 and 10n+3 sorry

[clear berry]

ah okay

[solemn mesa]

Yeah so any idea

[clear berry]

a moment

[solemn mesa]

Alright

[clear berry]

wait, 2nd question asks you to show that they're coprime

which means their gcd is 1 right?

[solemn mesa]

Yeah

[clear berry]

then how does the 3rd question even make sense

[solemn mesa]

You're supposed to prove that d is also equal to 41

[clear berry]

huh? but gcd will be 1 right

gcd is always unique

[solemn mesa]

Well Idk but that's what my teacher is saying

wait so

what my teacher did is

[clear berry]

I just wrote a small code to check their gcd uptil a million, it always comes to be 1

[solemn mesa]

since a and b are divisible by d then 3a and 10b are divisible by d aswell and 10b-3a is divisible by d after calculating 10b-3a he got 41 so 41 is divisible by d and since 41 is a prime number then d=1 or d=41

I have no idea whats going on

[clear berry]

seems like you wrote the question wrong then

[solemn mesa]

Well no that's literally the question I double checked

[clear berry]

cause in the question there's no mention of 3a or 10b or 10b-3a

[solemn mesa]

Well b = 6n+2 and a=10n+3

[clear berry]

a=10n+3?

and just in the previous question you showed that 6n+2 and 10n+3 are coprime

[solemn mesa]

yes

[clear berry]

which means so are a and b

which means their gcd is 1

[solemn mesa]

mhm

I just dont know anymore

[clear berry]

probably ask your proff for clarification

[solemn mesa]

Yeah ill do that next week

thx for the help tho

[clear berry]

[solemn mesa]

.close

[cedar kiln]

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[solemn mesa]

.reopen

[cedar kiln]

✅

[solemn mesa]

3rd question

[stuck walrus]

Euclidean algorithm

[solemn mesa]

How

[clear berry]

got it

[solemn mesa]

Ello

[stuck walrus]

hold on what

d = 1 and d = 41 what

d = 1 should be the only answer

[clear berry]

take n=1 then you have a=8 and b=13 so gcd = 1

[solemn mesa]

Mhm

[clear berry]

let n=12 then a=41 and b=123 so gcd = 41

[solemn mesa]

Oh there we go

[clear berry]

anyway

[stuck walrus]

oh brain fartes

[clear berry]

so, do you know that gcd(a, b) = gcd(b, a) = gcd(a + mb, b)

[solemn mesa]

Yeh

[clear berry]

alright then gcd(10n + 3, 3n + 5) = gcd(10n+3 - 3(3n+5), 3n+5) = gcd(n-12, 3n+5)

[solemn mesa]

Ok yeah

[clear berry]

then you can repeat the process till you get 41 on one side

(it's just one more step)

[solemn mesa]

Ooooh

I see

[solemn mesa]

Ty

.close

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[teal charm]

how is this wrong?

[stuck walrus]

you flipped x and y

remember x = (constant) is a vertical line

and y = (constant) is horizontal

[cedar kiln]

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[blissful glen]

Can someone help me with this equation? I don't understand what happened and what method was used?

[blissful glen]

Here it is:

Here is the solution:

What happened in the second step?

[crimson delta]

they subtracted twice the first equation from the second one

and added three times the first to the third one

[cedar kiln]

@blissful glen Has your question been resolved?

[blissful glen]

subtracted how? how do you add an equation to the other one? What is that method called

[crimson delta]

you know how you can subtract on both sides of the equation the same value and the equation still holds, yes?

well the first equation says that x1+x2+3x3-x4 = -1 so on both sides is the same value

which we can subtract from equation 2

first we multiply by 2 tho cause that will work out nicely in a second

so 2x1+2x2+6x^3-2x4 = -2

and now we can subtract that from equation 2

on both sides we subtract the same value

on the left hand side we get 2x1+x2+x3+2x4 - (2x1+2x2+6x3-2x4) = -x2-5x3+4x4

and on the right hand side we get 8-(-2)=10

so that gives the new equation -x2-5x3+4x4 = 10

the nice part here is that suddenly the x1 is gone

one variable less to worry about

[blissful glen]

thanks you so much...now I see

but how would I know to multiply by how much or do I just improvise

also in the 3rd eqaution like you said it was 1st equation + 1st +1st + 3rd = 1st + 1st + 1st + 3rd

just checking

[crimson delta]

the key idea is to multiply the first equation by something so that the x1 will cancel

after you add/subtract

[blissful glen]

I see

[cedar kiln]

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[crimson delta]

.reopen

[cedar kiln]

✅

[blissful glen]

so I can just make anything up as I see fit

like 3rd - 1st

2nd + 3rd and so on

for example

in 2nd + 3rd

only the 3rd changes? or the 2nd?

[crimson delta]

well whichever one you choose to replace by the result

you could phrase it as adding the second to the third -> third changes

or adding third to second -> second changes

you get the same equation either way

[blissful glen]

Thanks now I understand...so I just do anything I want and add anything I want how many times I want

with the goal of eliminating x1

and so on

[crimson delta]

first x1

then x2

and so on

until you are at the end

[blissful glen]

Thank you

.close

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[crimson sedge]

hey, in the theory of linear momentum we've never used the coordinates with î etc, i dont understand why it's necessary

[crimson sedge]

also why are we only looking at v in the x direction and not in y?

[proven thistle]

î simply represents the direction of positive x axis

[proven thistle]

It's your wish

Choose your axis

But make that you draw the other planes with respect to your original plane

[crimson sedge]

okay i find it confusing still but thank you for your time!

[proven thistle]

It's better you stick with the convention

The time will preach on its own

[crimson sedge]

I'll remember ty

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[slow wedge]

How this is not 1/5

[slow wedge]

The answer is 2/5 but why?

[bright bridge]

2 and 4 are even

[slow wedge]

Ok and

[bright bridge]

how did you get 1/5

[slow wedge]

1 devided by 5

[dapper raven]

how many numbers are there in total?

[slow wedge]

5

1.2.3.4.5

[dapper raven]

right

and how many even numbers are possible?

[slow wedge]

2,4

[dapper raven]

right

[slow wedge]

So 2

[dapper raven]

yes

when you spin it and it lands on a number

it could land on any

so what is the chance that it lands on an even number out of all of them?

[slow wedge]

Ok thanks

2 over 5

I didn't read the question probably

[cedar kiln]

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[idle cedar]

Hi everyone,

Can someone explain to me what is supposed to happen in the convolution formula when m > k ?
X and Y are random variables on (Omega , P ) and they are independent.
$P(X+Y = k) = \sum_{m \in X(\Omega)} (P(X=m) \cdot P(Y=k-m))$
My concern is that when m>k then that shoudlnt be allowed , should it ?

[wraith dagger]

barış

[idle cedar]

$P(X+Y = k) = \sum_{m \in X(\Omega)} (P(X=m) \cdot P(Y=k-m))$

[wraith dagger]

barış

[idle cedar]

is it just supposed to be 0 then ?

[cedar kiln]

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[tulip crow]

A man has 1,000,000 dollars and buys baseball cards for 325 dollars, every 10 same cards he gets will max out his card and the rest he will receive are extra, how many extra cards will he receive?
Help?

[flint plinth]

i'm not sure what this means: "max out his card"

can you give a simple example that illustrates?

[tulip crow]

like once he recieves 10 cards, the rest would be extra

[flint plinth]

10 copies of the same card, or 10 of any card?

[tulip crow]

like if he got 10 cards of x baseball player, after 10, they are all extra

[flint plinth]

is it really possible to answer this without knowing how many of each card exist, and how many total cards?

i mean what if every card in existence is unique?

[tulip crow]

yea, thats what im assuming

if the pack has specific cards then this problem would be difficult tho

no?

because then we have to deal with percentages of those cards

[flint plinth]

yes, i think you have to know something about the distribution of possible cards

at one extreme, maybe the company making the cards went crazy and all the cards it created are identical

at another extreme, they're all distinct from each other

or any possibility in between those extremes

[tulip crow]

should i ask my teacher?

[flint plinth]

yes, i would ask what assumptions you should make

[tulip crow]

should i get back to you? or just repost the question?

[flint plinth]

can just repost it, no guarantee that any particular helper will be online at a particular time

[tulip crow]

alr

.close

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[tame granite]

The diffraction angle produced by a 780 Hz sound is 35 degrees. What is the diameter of the circular opening it passes through?

[tame granite]

like ik the equation imma use is sin(theta) = 1.22(wavelength/diameter)

but where do you get wavelength from the frequency given

ik this is physics and not reeally a math class but was thinking someone would prolly know anyway

[foggy merlin]

wavelength = cT

if I remember

c is the speed of light

[dapper raven]

no

[foggy merlin]

T period

[tame granite]

what

[dapper raven]

sound

so speed of sound in the material

idk what medium it is

[tame granite]

which means that is not the way to go about it

[foggy merlin]

hmm

yea its sound my bad

so $\lambda = cT = \frac{c}{f}$

[wraith dagger]

Herels

[foggy merlin]

f the frequency

c the speed of the sound

[tame granite]

i think i got it

i think u assume that V is 343 m/s

[dapper raven]

probably

[tame granite]

and do V = lambda * f

[dapper raven]

speed of sound in air

[tame granite]

ok yeah thats it

i forgor how to close a channel

[foggy merlin]

I didnt assume anything

.close

[dapper raven]

.close

[cedar kiln]

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[tame granite]

.close

lmao

[dapper raven]

interesting

i can close a channel?

that wasn't mine?

[foggy merlin]

helpers role

can

[cedar kiln]

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[ornate ridge]

Objective: To get the maximum reward from the project from the reward pool, these will be distributed based on the two factors outlined below. A transaction strategy needs to be defined: how many transactions need to be made and what the amount of each transaction is.

[cedar kiln]

Please don't occupy multiple help channels.

[ornate ridge]

How can I solve this exercise? Please I need help 😭

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[bold ginkgo]

I dont have a written question but id like to know if powers of complex numbers are possible, if so what happens and does this allow for logs to be negative

[vernal shell]

Yes they are, Euler exponential form of complex numbers is useful for that

[bold ginkgo]

What exactly does that do?

Is there any application for it as such

I just cant imagine what it might be needed for thats all

[vernal shell]

I don't know 😵‍💫

[bold ginkgo]

Haha thats ok it was just a query

[vernal shell]

At least for a complex number to the power of an integer or real number, it might be useful for some mappings in control theory and electronics

I don't know of any applications of complex exponents

[bold ginkgo]

Yeah i can inderstand this but my head cant even wrap around the answer of a complex power

Like how that even works

[vernal shell]

Oh then you should watch 3b1b videos about complex numbers

I think he explains a good visualization in the Riemann hypothesis video

[bold ginkgo]

Okay thanks ill have a look

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[neat cargo]

is this correct?

[crystal raptor]

plug your value for X back into the top and find out

[neat cargo]

didnt think about it

thanks!

.close

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[surreal cave]

How do I evaluate $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}=1+\frac{1}{9}+\frac{1}{25}+\ldots$

[wraith dagger]

XxMrFancyu2xX (anti glomed)

[short blade]

do you have more context

i think you can use parseval's theorem

[surreal cave]

yea I was trying $\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}$ cause math is fun then I broke it up into $\sum_{n=1}^{\infty}\frac{1}{(2n)^2}-\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$ the first sum is easy but the second boi idk

[wraith dagger]

XxMrFancyu2xX (anti glomed)

[surreal cave]

p-series?

[iron saffron]

you gotta relate with 1/n^2

[surreal cave]

my question is 1/(2n-1)^2

[iron saffron]

yes

[short blade]

they're not trying to show convergence

they want to compute it

[surreal cave]

"evaluate" 😅

[iron saffron]

but u have information about 1/n^2 which is pi^2/6

so u can relate it with yours

[surreal cave]

yes but idk how to relate the Basel Problem to the (2n-1)^2

[short blade]

how would you suggest to continue archimonde

how would you relate them

[iron saffron]

was just a guess, let me think

[surreal cave]

W|A says it's $\frac{\pi^2}{8}$

[wraith dagger]

XxMrFancyu2xX (anti glomed)

[surreal cave]

but I wanna compute it myself

[short blade]

you could maybe invert this as a fourier/taylor series

i think that should work. i recall doing a problem like this

[surreal cave]

idek anything about Fourier transforms, series, etc. 💀 but it's upcoming on my list of math to learn

but as for a Taylor if that works how would I convert to a Taylor Series? (idk what that would even mean)

[short blade]

so this can be seen as the fourier series $$f(x) = \sum_{n=1}^\infty \frac{1}{(2n-1)^2}\cos(nx)$$

evaluated at x = 0

[wraith dagger]

maximo

[surreal cave]

ohh this looks kinda like Feynman's Technique of Integration

except in series not integrals

[short blade]

it could be

[surreal cave]

but how does this help me evaluate?

[short blade]

then you have two conditions imposed on this. a_n = 1/(2n-1)^2

and b_n = 0

and each of those has a specific definition in terms of integrals

in particular

[surreal cave]

sounds like Parseval's Theorem I've heard about this somewhere

[iron saffron]

i will paint cause i don't know about latex

wait

[short blade]

$a_n = \frac{1}{\pi}\int_{-\pi}^\pi f(x)\cos(nx), dx$

[wraith dagger]

maximo

[short blade]

ok i think this is it

[short blade]

and b_n is the same but with sin(nx)

[iron saffron]

odds + even numbers

[short blade]

also i mentioned this earlier haha

but i don't think it helps directly

[surreal cave]

wait

I have an idea

[iron saffron]

and that's pi^2/6

[surreal cave]

the sum of 1/n^2=pi^2/6

the evens are pi^2/24

subtract the evens from the total

[short blade]

was this known?

[iron saffron]

yes

[surreal cave]

the evens can be generated by taking the sum divided by 4

so then pi^2/6-pi^2/24

= pi^2/8

[iron saffron]

correct

[short blade]

ah i see, nice

[iron saffron]

i think that's it

[surreal cave]

but Maximo your idea definitely had merit it was the solution on the stack exchange so thank you

ohh that's cool

$\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}=-\frac{1}{12}\cdot\pi^2$

[wraith dagger]

XxMrFancyu2xX (anti glomed)

[surreal cave]

@short blade think you might enjoy that -1/12 appearance 😂

anyways

thank you too @iron saffron ❤️