help-28

[grave elm]

maybe

or maybe he just wants you take a very rough estimate

which is actually a valuable skill to have

[winter patio]

Ugh

It just sucks

Well if it is by formula could u help me do part b please

[grave elm]

use the same formula as part a

just let the time be a variable (to later solve for)

and make the total population = 8400

that'll give you an equation to solve

[winter patio]

We don’t decide?

Divide?*

[grave elm]

divide by what? 2800?

[winter patio]

So time will be A right

[grave elm]

you could do that, as long as you know what you're doing

[grave elm]

sure, you could name it A if you're gonna do the equation approach

[winter patio]

Cus a is usually the population

[grave elm]

then name the time t

or n

the population after n years is
2800 * (1 + 0.15)^n as you already figured out in part a)

set that equal to 8400 and find n

[winter patio]

I’m kinda confused

[grave elm]

Population = 2800 * (1 + 0.15)^n

(after n years)

You want the population to be 8400

so you want 2800 * (1 + 0.15)^n = 8400

if you now solve for n, you'll get the time required for the population to reach 8400

[winter patio]

8400 = 2800(1+0.15)^t

Ok so it’s def division then

2800 on both sides

[grave elm]

yep

[winter patio]

Wait

What would that be im sorry

Cus 2800 doesn’t go into 8400 easily

Highest is 7400

Wait it’s 3

Byeee

I miss calculated

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[weak flame]

im stuck

i wanna factor

but uh

B is missing

[glacial stream]

$6x - 26 = x^2 - 6x + 9 \implies x^2 -12x +35 = 0$

[glossy valve]

@glacial stream

[weak flame]

im so dumb

omg

[glacial stream]

We all make mistakes lol

dw

[weak flame]

but like

in theory

if b was 0

what would i do

[glacial stream]

Oh then you just have an equation of the form $x^2 - a = 0 \implies x^2 = a \implies x = \pm \sqrt{a}$

[glossy valve]

@glacial stream

[weak flame]

mkay thx

[glacial stream]

!done

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[opaque cliff]

We are given a real $n \times n$ matrix A and $F_A(x)=x^TAx$ is the associated quadratic form. I have already proved that $B=\frac{1}{2}(A+A^T)$ is symmetric and has an equivalent quadratic form.

[glossy valve]

Techcable (Nicholas)

[opaque cliff]

However I now want to show that B is the unique symmetric matrix whose quadratic form matches A.

[vestal moat]

What do you mean by "equivalent quadratic form" ?

[opaque cliff]

To do that my teacher says I consider an arbitrary symmetric matric C, so that F_A=F_B=F_C all have the equivalent quadratic form

A quadratic form that behaves the same for any vector

This is my proof that $B=\frac{1}{2}(A+A^T)$ has the same quadratic form as A

[glossy valve]

Techcable (Nicholas)

[opaque cliff]

Where $F_M(x)=x^TMx$ is defined for any square matrix $M$

[glossy valve]

Techcable (Nicholas)

[opaque cliff]

So I assume an arbitrary symmetric matrix C so that $\forall y \in \mathbb{R}^n, F_A(x)=F_B(x)=F_C(x)$. I want to show that $C=B$.

[glossy valve]

Techcable (Nicholas)

[opaque cliff]

In office hours, my teacher said that I should prove that for any vector $y\in \mathbb{R}^n$ that $|By-Cy|^2=(By-Cy)^T(By-Cy)$ is zero

[glossy valve]

Techcable (Nicholas)

[opaque cliff]

And if two matrices behave the same on every vector, they are the same

So I manage to expand $|By-Cy|=(By)^T(By)-(Cy)^T(By)-(By)^T(Cy)+(Cy)^T(Cy)=y^TB^2y-y^TCBy-y^TBC+y^TC^2y$ using symmetry of $C,B$

[glossy valve]

Techcable (Nicholas)

[opaque cliff]

But than I get stuck, because we know by hypothesis $y^TBy=y^TCy$ but we have $B^2$ and $C^2$ here

[glossy valve]

Techcable (Nicholas)

[opaque cliff]

He said I should rearrange that to get $y^T(B-C)y=0$, then manipulate this further to get what I want

[glossy valve]

Techcable (Nicholas)

[opaque cliff]

I've tried right multiplying and left multiplying by B and C, but this doesn't seem to get me anywhere

[vestal moat]

Ok pause

Assume that you have a matrix M, such as for any x :
x^t.M.x = 0
What can you tell me about M

[opaque cliff]

very little?

[vestal moat]

Think

[opaque cliff]

it preserves orthogonality?

no. it does the opposite

because <x,x>!=0 usually, but in this case <x,Mx>=0

so M is a rotation matrix

so B-C is a rotation matrix?

hmm

I'm not sure where this gets us

[vestal moat]

You're supposed to prove that M = 0

[opaque cliff]

huh wait

[vestal moat]

Let me make this easier

assume M is symmetric

I'll go for 10 min, I let you think about it ...

Spoiler alert in next message

[opaque cliff]

okay I think you got me on the right track 🙂

[vestal moat]

use spectral theorem

[opaque cliff]

err I don't think he wants us to use that - he didn't introduce that and this is a multivariable calc class not a linear algebra class

but I will think about it

I think I'm close

[vestal moat]

I assume you've had your linear algebra class already with the spectral theorem in it, right ?

[opaque cliff]

yes

[vestal moat]

Cause dealing with quadratic forms, the spectral theorem is extremely useful. So I would be surprised if it was off limits

Did you find it ?

[opaque cliff]

It is diagonalizable?

I'm not sure why that would be good

also i guess it's diagonalizble with an orthonormal matrix O

[vestal moat]

Ok ... that's good ... keep going

[opaque cliff]

so C=OSO^T fo some S

where S is diagonal

[vestal moat]

No need to get into that. assume lambda is an eighenvalue now associated to x

Can you tell me what happens ?

[opaque cliff]

its scaled

[vestal moat]

Yeah but use the particularity of M

[opaque cliff]

its zero

[vestal moat]

Ok

So ... You're done ...

M is diagonalizable and all its eighenvalues are 0. So M = 0 ...

[opaque cliff]

huh

that would certinally work

[vestal moat]

Now in your exercice. B-C is symmetric

let's call M = B-C

you have x^t.M.x = 0

[opaque cliff]

yes I see how it applies here

[vestal moat]

Ok that's it

[opaque cliff]

Are there matrices where <Mx,x>=0 but M\ne 0?

just curious

[vestal moat]

If you're in R2x2

A 90° rotation might work

[opaque cliff]

i see

yes

thank you 🙂

[vestal moat]

you're welcome

[opaque cliff]

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[sleek quest]

can someone help me do this? i dont know where to begin

[gritty rose]

what's the definition of convergent

[sleek quest]

i was hoping you could help me with that

i was quite confused

[gritty rose]

is that not in your notes or book

show how it was defined for you

[gritty rose]

yea not enough context for me to help you

[sleek quest]

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[brittle cliff]

I'm lowkey confused what this problem is doing isnt it finding tangent line equation at (-3,8)? how is that x=-3

[tight mortar]

why couldn’t that be a tangent line?

[brittle cliff]

ummm yea? but the slope i found by plugging x=-3 is 9 and if i use that its gon be y= 9(x+3) + 8? no?

[tight mortar]

how did u find 9?

I can’t help but notice you have 2t^3 instead of 3t^2 when differentiating t^3

[brittle cliff]

3t^2 / 2t-4 |t=-3 ---> 27/2?

[slender bane]

what is a parametric equation? just like x and y are identified seperately?

[tight mortar]

the question gives x=-3, not t=-3

(it’s easier to find the t value from the y coordinate)

[brittle cliff]

y=8 so t=2?

[tight mortar]

yes

[brittle cliff]

i plug this?

[tight mortar]

yes

whats the denominator when we plug in?

[brittle cliff]

dx/dt = 2t-4

dx/dt |t=2 ---> 0

[tight mortar]

can we divide a non zero number by 0 and get a finite value?

[brittle cliff]

no

[tight mortar]

soooooooooo whats the slope (and what does that mean for the tangent line graph)

[brittle cliff]

x=-3?

[tight mortar]

thats the equation but do u know how that happened lol

[brittle cliff]

wait is it like we have dy/dx = dy/dt / dx/dt but since dx/dt = 0 dy/dx blows up so we js take x value

[tight mortar]

uhhhh

yeah

it’s a vertical line

[brittle cliff]

is it like when we got vertical tangent situation?

[tight mortar]

yea

[brittle cliff]

wait what about horizontal tangent?

[tight mortar]

(heres the graph for better intuition ig?)

[tight mortar]

then you have dy/dt = 0

[brittle cliff]

vertical tangent --> slope is x value? and horizontal tangent --> slope is y value?

for parametrics

[tight mortar]

well it’s not really that the “slope” is the x/y value, the tangent line is

but yeah basically

[brittle cliff]

ok

thanks 🙂

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[river talon]

hiyaaa, found another proving problem; i did it differently from the hint though so im not sure if what i did was right;;;; please also see if the writing's good. thanks!!

[river talon]

whoops the top was cut-off a bit-- thats "If $a^3 > a$ then $a^5 > a$"

[glossy valve]

Plvzfq_rit

[lime ether]

why are you saying WLOG

[deep gale]

they didn't prove it for x < 0?

[lime ether]

exactly

proving it for x < 0 is a separate case that you must do

saying WLOG here makes no sense

[river talon]

whoops, ive used WLOG incorrectly, my bad;;;

i thought that if the proof structure was similar, then one could use WLOG; i guess not though;;;

alright lemme revise this

[modest obsidian]

real

[deep gale]

try prove that (-1)^2 = 1 hint: subtract 1 from both sides

[lime ether]

well what assumptions are you already working with

[modest obsidian]

its kinda trivial though dw x<0 so multplying by x flips the sign as - × - = +...

something like that

[lime ether]

for example do you have that multiplying by a number < 0 changes the sign

[modest obsidian]

t?

[river talon]

yeah i was already assumin that if one multiplied a negative on both sides of the equation, then the inequality flips

-- wait flip, do i need to prove that too T-T

[modest obsidian]

sorry, real - {0}

[lime ether]

you should have a list of ordered field axioms you’re working with here

did your teacher not provide any?

or the book

[deep gale]

|| (-1)^2 = 1 ==> (-1)^2 - 1 = 0 ==> -1(-1 + 1) = -1(0) = 0 ||

[river talon]

im just self-studying from the book;;;

[lime ether]

which book

[river talon]

Velleman's "How to Prove It"

[lime ether]

surely they listed some axioms for the real numbers

earlier in the chapter or whatever

[modest obsidian]

shouldve looked into axioms before doing any proving

strange choice... maybe

[river talon]

yeah-- that shouldve been my first destination on hindsight

[lime ether]

when doing the exercises you can obviously omit proofs for things the author has already proven or has listed as an assumption/axiom

[deep gale]

omit for axioms yes

Omit authors trivial proofs yes

Omit authors relevant proofs? No 😈

omw to prove measure-theoretic riesz representation theorem in daddy rudin once i complete baby fr

[river talon]

dang, massive respect

i hope to one day reach ur level orz

[modest obsidian]

everyone starts at zero

[lime ether]

everyone has done truth tables before (probably)

[river talon]

on the Velleman-axiom issue-- it did not seem like he listed any field axioms;;; he only has the basic sentiential and quantifier logic, a bit of set theory, but no field axioms explicitly listed

[lime ether]

what chapter is this

[river talon]

chapter 3, proofs

im in the first part of that chapter

i mightve possibly missed it

[lime ether]

alright let me get the pdf hold up

[river talon]

much thanks orz

[lime ether]

alright i’m looking now

ok so he actually does prove these things

[deep gale]

Wait what

What things

[lime ether]

like it’s completely botched

🤣

[deep gale]

as in he proves field axioms?

💀

[lime ether]

😭😭😭

[river talon]

dang... did i pick up the wrong book lmaO

[lime ether]

bro this dude just uses the theorem to prove the theorem

[modest obsidian]

proving axioms?

[lime ether]

P implies P

qed

[deep gale]

the bruh

[river talon]

💀

[modest obsidian]

cinema

"so if we assume that P is true, P is true" Q.E.D.

[deep gale]

I wish i knew of good math logic intro book to recommend rn

[modest obsidian]

mathematical logic is overrated, just look at it!

[lime ether]

like he doesn’t explicitly state the axioms he’s using but all he does is prove the contrapositives of the standard axioms

this book is high key buns bro

[river talon]

dang;;;

[deep gale]

Will michael spivak's calculus be an overkill or nah?

[lime ether]

i’ve heard good things about book of proof by hammack

but honestly yea spivak has a good intro

any analysis text really

first chapter of munkres is good

[deep gale]

rudin is...

Different...

[lime ether]

well yea rudin is rudin

i mean like

abbott

[deep gale]

hehehehehehe

[lime ether]

ross

tao

[deep gale]

I heard tao is bad

[river talon]

yea--- i tried reading a rudin and it took me a while to read one page 😔

[lime ether]

hella words

[deep gale]

Dont

[lime ether]

yea don’t

[deep gale]

Do michael spivak

You won't understand rudin unless your autisn level is greater than or equal to mine

[lime ether]

🤣

[river talon]

😆

Calculus, by Michael Spivak yeah?

[lime ether]

honestly though this basic logic stuff doesn’t require a super in depth book

like you don’t need to waste your time with simple number theory proofs for a whole month

[deep gale]

Yea

[lime ether]

what’s your background though?

like what other math have you done

and what’s your goal

[river talon]

ah uhh i did some single-variable calc, some linear algebra

[lime ether]

did you use books for those or no

[river talon]

havent done any multi-variable stuff, and forgot the sequence stuff in smth like calc 2

[river talon]

uhh i think i did

but they were like

the calculus 7

[lime ether]

yea stewart is shit ngl

if you want to do math math

there are way better choices

[deep gale]

It still won't be an easy read, especially when you new to math, but it is something that can be worked with, also if you want a very easy supplementary textbook to spivak, like say you get stuck on a very hard problem in spivak and wanna practice some easier stuff, then i heavily recommend bartle's introduction to real analysis. It has lots of easy exercises and problems to practice on

[lime ether]

btw hammacks book is free online if you still need like the basic set theory stuff

covers all of the foundations youll need really

[river talon]

dya think it like rly needed tho, or does it like depend on the person

[deep gale]

Stewart is for computational calculus mainly, so it doesnt have much theoretical stuff

[river talon]

yeah;;;

[deep gale]

I like stewart tho

[lime ether]

i mean idk your ability tbh

so i can’t say

[deep gale]

The applied math problems are cool, i think stewart introduces for example the fitzhugh nagumo model iirc

[lime ether]

like if you don’t know what a function is or a cartesian product you might want to go with hammack or something

[river talon]

hmm a function is.. a relation such that... each element in the domain... is mapped to one and only one element in the range... innit?

it has been a while 😔

[deep gale]

Not really

A function is a rule assigning elements from one set to the other, more abstractly it's gonna be a map f:X --> Y between two sets X,Y

Injective function f:X --> Y is one such that for y in f(X) , the function maps one and only one element x from X to it, that is y = f(x)

[deep gale]

This can be written equivalently as demanding that whenever x1 ≠ x2 , then f(x1) ≠ f(x2)

[lime ether]

you mean ≠

[deep gale]

Ah yea mb

[lime ether]

x_1 = x_2 —> f(x_1) = f(x_2) is by definition of a function

[river talon]

ah dang... thanks for correctin' me+

but yeah i should prolly give hammack a quick read through on this before takin on spivak

[lime ether]

this would be a good idea

take a few weeks

are you in high school?

[river talon]

nope im in college

[lime ether]

math major?

[river talon]

comp sci unfortunately 😔

[lime ether]

tough

wait if you’re in cs then isn’t this like the last math class you take

proofs/logic

if you’ve already taken discrete

[river talon]

in my university, we do like discrete math and logic yea

but after that

its full of computation, aside from that side show on vector spaces in linear algebra

which is uh... quite @ss to say the least 😔

[lime ether]

linear algebra is lame

i agree

going through LADR is a chore

are you interested in learning further math for fun or are you just doing this for school

[river talon]

im doing this for fun yes

quite interested in pursuing this avenue

in like my free time

[lime ether]

if you’re on break soon you can bang out hammack relatively quickly

and spivak is a good intro to analysis tbh

you’ll be surprised how little you feel like you know from calculus

[deep gale]

To me rudin is better 😈

[lime ether]

🤮

[deep gale]

I can see where this is coming from, but tbh

[lime ether]

dawg like i just don’t give a fuck about upper triangular matrices man

who cares about nilpotent operators

snooooooze

[deep gale]

Even the elementary theories such as every mxn matrix with entries in K being row equivalent to a row-reduced matrix are kinda cool, but nothing beats the abstracter linear algebra problems, they lookin hella hot

[lime ether]

i liked ladr before matrices lmao

i just think matrices are so lame

[deep gale]

By abstracter linear algebra i mean like

R-modules

Lots of cute r-modules

[deep gale]

Matrices have definition in abstracter linear algebra..

But i aint understanding 1% of it rn

[lime ether]

abstracter?

more abstract than axlers definition?

[deep gale]

Determinant is a homothety of the wedge type shit

[lime ether]

wtf is homothety

[deep gale]

Idk about that

[river talon]

true that is one funny lookin word

[lime ether]

analysis flavored shit is so much better than algebra flavored shit

[deep gale]

Both shall work together

• topologi

[lime ether]

i’ve only been working through point set but i like it

i’m just not looking forward to going through like dummit and foote next year

maybe i should suck it up and do it over the summer

to get ahead

[river talon]

Can confirm that studyin ahead makes life easier, did that for machine learnin and it helped me not die from the mess that is neural networks

[deep gale]

ive only been looking at munkres

Lots of stuff are boring to me so i kinda just take a look rather than read

Tho i will admit i lagged at the product and box topology subject

It looked cumbersome so i just kinda said alr and skipped box topology entirely xD

[lime ether]

yea i tend to skip things or just stop reading a book entirely if i come across a subject that bores me to death

bad habit

[deep gale]

And the R^w topology rly aint my shit

Like i get its interesting cuz its cool new set , but like

I aint got no reason to study ONLY its basic topological properties from the beginning

i would rather just do rudin

[narrow path]

??????????

[river talon]

Alrighy then, time to Fahrenheit 451 that Velleman book and crack open Hammack

Thanks for yalls help uwu

[deep gale]

Np

[river talon]

I also send a friend request if yall dont mind, yallre good people uwu

Ill close this thread then

See yall round+!!

.close

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[atomic vortex]

why is that there is no arrow from x2 to z1?

but for x3 theres 3 arrows (z1, z2, z3)

is that done on purpose or a typo

x2 should have 3 arrows too right

[full forum]

@atomic vortex Has your question been resolved?

[kindred grove]

if you look carefully at the W they give you, you see that z1 doesn't depend on x2, it ain't a typo

[atomic vortex]

😭 how do you tell

Im confused

nvm. im wrong

im not sure how to tell

wait idk

oh

wait

im slow

i get it

[kindred grove]

well compute Wx + b

[atomic vortex]

z1 = ax1 + 0x2

[kindred grove]

yea

[atomic vortex]

smh

ty

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[spice pilot]

An $n\times n$ matrix has the decomposition $A=A_1+iA_2$ where $A_1=\frac12(A+A^\ast)$ and $A_2=\frac1{2i}(A-A^\ast)$. I'm facing a question in my book whether or not it is reasonable to define $A_1$ to be the real part of the matrix $A$ and $A_2$ to be the imaginary part. I don't understand I think. Is it reasonable?

[glossy valve]

psie

[vestal moat]

Hello, how is the question asked in the book?

[vestal moat]

Hmmm

[onyx glen]

well

do we know that A1 is a real matrix

[spice pilot]

Yes, probably the problem is a bit ambiguous. Hmm.

[vestal moat]

Try to determine if A1 and A2 are real.
And try to check if : A = A1 + iA2

[spice pilot]

well, A* is the conjugate transpose, so actually I think A1 is not real, because in the ij-entry of the matrix, we are adding Aij+conjugate(Aji). If the transpose business were omitted and we only took conjugates of the entries of A, then obviously we'd have a real matrix.

[onyx glen]

yeah so

why would we call A1 the real part if it is not itself real

[nova basin]

yes it's the typical real-imaginary decomposition if we use conjugate instead of adjoint

[spice pilot]

Ok, thanks everyone. Have a great day! 🙂

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[spice grail]

can someone please check my answer and where ive used brackets(if they're used correctly)

[spice grail]

Check your use of brackets
Verify your answers
Always include units
Check that you have used the correct variables in your answers

have i done these correctly

because

should it be written as this

[robust slate]

Eh ig that could’ve been clearer, but it would be interpreted like that by default imo

If you want the square to apply to the log as a whole, you write ln^2 (v) or (ln(v))^2

[spice grail]

right

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[bold whale]

Using Gauss-Jordan, what do I do next?

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[sweet sky]

like usual just choose a pivot in the next column that isnt in a row with an existing pivot, and use that to eliminate the rest of the column

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[torpid perch]

AX - B = CX
AX - CX = B
(A-C)X = B

,w {{0,1,0},{1,1,3},{-1,1,-2}} - {{-2,0,1},{-1,1,2},{0,0,1}}

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[torpid perch]

3x3 . (nxm) = (3x2)

n = 3

m = 2?

,, \begin{pmatrix} 2 & 1 & -1 \2 & 0 & 1 \ - 1 & 1 & -3 \end{pmatrix} \cdot \begin{pmatrix} x_1 & x_2 \ x_3 & x_4 \ x_5 & x_6 \end{pmatrix} = \begin{pmatrix} 2 & 1 \-1 & 1 \ 6 & -2 \end{pmatrix}

[glossy valve]

938c2cc0dcc05f2b68c4287040cfcf71

[torpid perch]

,, \begin{cases} 2x_1 + x_3 + x_5 &= 2 \ 2x_1 + x_5 &= -1 \ -x_1 + x_3 -3x_5 &= 6 \ 2x_2 + x_4 - x_6 &= 1 \ 2x_2 + x_6 &= 1 \ -x_2 + x_4 - 3x_6 &= -2 \end{cases}

[hallow walrus]

are you sure A-C is not invertible?

[glossy valve]

938c2cc0dcc05f2b68c4287040cfcf71

[torpid perch]

ohh you are genius

lets check that!!

[hallow walrus]

either that or just use row transformations

[torpid perch]

,w det {{2,1,-1},{2,0,1},{-1,1,-3}}

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[hallow walrus]

there we go

[torpid perch]

its invertible

[hallow walrus]

I'd still use row transformations instead of finding the inverse

[torpid perch]

,w inverse {{2,1,-1},{2,0,1},{-1,1,-3}}

?

I dont get it

[hallow walrus]

do you know what elementary row transformations are?

[torpid perch]

yes, but I need you to elaborate, whats the plan of action?

how does that relate?

[hallow walrus]

Row transform the matrix A-C here to the identity matrix (this is possible because A-C is invertible) and do the same row transformations on the right hand side

Since row transformations amount to premultiplying by some matrix, you can apply the same row transformations on the left-most matrices on two sides of an equation to get an equivalent one

[torpid perch]

I think I will just find the inverse, I barely have any braincells

thats already good enough for me, I was planning on solving the 6 variables system

I cant keep track of the elementary row operations in my head and putting that in a matrix, I dont think so

,w {{2,1,-1},{2,0,1},{-1,1,-3}}^(-1) * {{2,1},{-1,1},{6,-2}}

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[torpid perch]

I guess is simple

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[strange basalt]

what do you think of the question

any thoughts?

[torpid perch]

well

if A is inversible the det is non zero

so solve for det(A) = 0 and the k we found we exclude it from R

,w det {{1,0,1},{-1,k,1},{0,3,2}} = 0

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[torpid perch]

A is invertible when k is not 3

[strange basalt]

yep

[torpid perch]

the second condition is what troubles me

[strange basalt]

|kA| = k^n|A|

[torpid perch]

we need to know properties of determinant

[strange basalt]

think about it

[torpid perch]

det(AB) = det(A).det(B)

[strange basalt]

thats true too

[torpid perch]

ok

[strange basalt]

imagine a determinant

[torpid perch]

A is 3x3, so 1/8

[strange basalt]

I of order 3x3

[torpid perch]

det(I_{3x3}) = 1

[strange basalt]

now imagine 3I as a matrix

[torpid perch]

det(3I) = 3^3

[strange basalt]

instead of 1 3 would be at the aij , where i = j

yeahh

almost there now ig

[torpid perch]

in this case the determinant is not a diagonal matrix tho

[strange basalt]

you can apply it to any

[torpid perch]

det(1/2 . AB) = 2det(A^-1 . B)

[strange basalt]

lemme give an example ig

[torpid perch]

1/8 . det(A) . det(B) = 2 . det(A^-1) . det(B)

@strange basalt

[strange basalt]

yeah

[torpid perch]

is there any property for det(A^-1)

[strange basalt]

how is inverse defined

at root level

A A-1 = I?

[torpid perch]

who the fuck knows

[vestal moat]

There are 2 cases. If B is not invertible, the equation is always true

Find the cases where B is invertible

[torpid perch]

good observation lmfao

,w det {{k,3,0},{4,k,1},{1,0,1}} = 0

[strange basalt]

genuinely forgor

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[torpid perch]

when k := {-3, 3} B is not invertible

[strange basalt]

but

[vestal moat]

Yeah so we also need A invertible so k is not 3

[torpid perch]

A and B share the same k ?

[vestal moat]

Naturally

Now assuming B is invertible

Can you simplify the equation

[torpid perch]

lets recap

when k:={3, -3}
B IS NOT INVERTIBLE

when k = 3:
A IS NOT INVERTIBLE

so we need k to be R - {3, -3}

[vestal moat]

No

[torpid perch]

?

elaborate

[vestal moat]

You need k to not be 3. -3 works

[torpid perch]

right, when k = -3 the equation trivially is satisfied.

[vestal moat]

Now

For k not -3 and not 3

B and A are invertible

Can you simplify the equation

[torpid perch]

1/8 . det(A) . det(B) = 2 . det(A^-1) . det(B)

det(A) . det(B) = 16 . det(A^-1) . det(B)

det(A) = 16 . det(A^-1)

now what?

is there a determinant property I am missing?

[vestal moat]

det A^2 = 16

[torpid perch]

can you elaborate

[vestal moat]

det A = +- 4

[vestal moat]

I used this : det(A^-1) = det(A)^-1

[torpid perch]

det(A)^-1 = det(A^-1)?

[vestal moat]

Of course

[torpid perch]

in general: det(A^n) = det(A)^n?

[vestal moat]

Yes

If n positive

[torpid perch]

?

in this case n = -1

[vestal moat]

Yeah but A is invertible so it works

[torpid perch]

oh ok, lmfao

very interesting

[vestal moat]

det(A.A^-1) = det I = 1

So det A = 1/ det(A^-1)

[vestal moat]

Now Let’s go back to this

[torpid perch]

now?

,w det {{1,0,1},{-1,k,1},{0,3,2}} = 4

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[torpid perch]

,w det {{1,0,1},{-1,k,1},{0,3,2}} = -4

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[torpid perch]

k = {-3, 5, 1}

@vestal moat

final answer?

[vestal moat]

Yup it seems to work

You can check

[torpid perch]

thats all the Ks?

[vestal moat]

Yup

[torpid perch]

ok thanks

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[autumn raptor]

Anyone know how to do this?

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@autumn raptor Has your question been resolved?

[velvet sedge]

try drawing a picture :)

[autumn raptor]

And?

[velvet sedge]

label important parts of the picture, set up equations that are important (namely the distance between the man and the woman)

[autumn raptor]

Could u like show me the calculations

[narrow path]

dude just draw it💔💔

[autumn raptor]

Wdym?

[narrow path]

basically draw a coordinate plane and place man and woman on it

[autumn raptor]

That is a calculus question

We don’t need coordination

[narrow path]

are you trying to do this without calculus or something

[autumn raptor]

Ofc using caculus

Plus derivative

Ok

[narrow path]

a bit flawed cause it only works for s>300 but thats all the problems asking for

so it doesnt really matter

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[frank gulch]

can someone give me some hint, kind of lost even after reading my text

[rapid rain]

Is it easy to show it's at most countably infinite?

(you've proved Q is countably infinite right?)

[frank gulch]

this was given as an example in my text

but its my first time doing a problem like this

i think i missed something though i need to find a map $f:\mathbb{N} \rightarrow S$ that is a bijection right

[glossy valve]

nilo

[frank gulch]

that shows that set S is countably infinite?

[languid junco]

did you ever learn a theorem about subsets of countable sets?

[frank gulch]

is it this?

wait whats the difference between countable and countably infinite

i should probably take some time to read

nvm

ill be back

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[onyx glen]

<@&268886789983436800> ayo this is getting out of hand a bit innit

[austere cove]

@red scaffold can you show your work and share your thoughts on this problem?

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[stray pelican]

help in graphing functions over two periods

[stray pelican]

these questions

using chunks and period

<@&286206848099549185> please

[gritty rose]

!15m

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@stray pelican Has your question been resolved?

[placid oak]

1. determine the period of the function
2. using the period, define the values for x - should start at the start of your period, have some number of points, and end at the end of your period
3. use the function to map each x value to it's corresponding y value
4. plot all your (x,y) pairs on a graph
5. draw the curve which goes through all your (x,y) points

[stray pelican]

how to do steps 1 and 2

[stray pelican]

aren’t the steps different for sin cos tan cot?

[manic oasis]

period of the funciton for tangent is Pi, and if you have a shift on the inside (3x) you'd do pi/3

do u get that?

[stray pelican]

somehoe

w

[manic oasis]

do u get wwhat a period is

it can be a bit tricky w tangent

[stray pelican]

if possible can anyone solve it for me and send me the answer with steps

[stray pelican]

so that i practice it

if possible with steps

finding the period and chunks

[placid oak]

---

For example, suppose I want to graph y = sin(x).

1. the period is 360 degrees
2. let's use x values
   {0, 30, 60, 90, 120, 150, 180, 210, 240, 270, 300, 330)
3. using the function, we have corresponding y values
   {0, 0.5, sqrt(3)/2, 1, sqrt(3)/2, 0.5, 0, -0.5, -sqrt(3)/2, -1, -sqrt(3)/2, -0.5)
4. now we plot these (x,y) pairs
5. now we draw a smooth curve going through all these points

[stray pelican]

which question is this?

lemme show you how my teacher solves it

[placid oak]

It's an example I made up, so you can follow the method for your questions

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[stray pelican]

.reopen

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✅

[stray pelican]

smth like that

@placid oak

<@&286206848099549185>

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[viscid ledge]

what is the difference between a satisfiable set of wffs and a consistent set of wffs

[viscid ledge]

@calm nacelle @grave elm help pls

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[placid oak]

What's a "wff"?

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[cobalt pelican]

How do I solve

[cobalt pelican]

Is it the same as solving a right angle triangle

[sharp osprey]

that is a right angle triangle so yes!

[cobalt pelican]

Oh my goodness you’re so right

Hypothetically if it wasn’t could I still use tan cos and sin to solve it

[mortal sigil]

(you labeled the hypotenuse wrong, tho)

[cobalt pelican]

Yeah I see that now

[mortal sigil]

you can use the law of sines'

[cobalt pelican]

Oh okay

[slender bane]

I mean it is a right triangle, just use like basic trig values

[mortal sigil]

given you have one angle and its correspondent side, plus another angle, you can figure out the side lenght, and the other angle too

[mortal sigil]

but yeah, this too ig... (boring)

[slender bane]

boring simple, that is my principle of doing math

[cobalt pelican]

Better than complicated

[mortal sigil]

sometimes the simple solution doesnt tell you everything...

[cobalt pelican]

You know what you’re right

[mortal sigil]

but yeah, just figure out the last angle and use trig funtions

[cobalt pelican]

You should join a poet server too

[mortal sigil]

this isn't a very interesting problem either way

[cobalt pelican]

I’m gonna try and solve it hold on watch

[mortal sigil]

I do write epic poems, but it's not in english tho

[cobalt pelican]

Alright bilingual

[calm nacelle]

(if you open a new help channel I can help you)

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[tall rivet]

When did I join this group?

[gritty rose]

#discussion if you want attention

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[gentle brook]

what is this asking for? i’m not sure how to go about solving the problem

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@gentle brook Has your question been resolved?

[languid junco]

if my initial deposit is 2000 and I double it, how much will my future balance be?

[gentle brook]

oh wait 4000

[languid junco]

yep!

[gentle brook]

so i did this and changed it to ln and made it 2.75 years bevause ir said round to the nearest quarter year, but it says it is wrong

[languid junco]

okay, well your first step should be to divide both sides by 2000

[rough tundra]

where did ln(2000) come from?

[gentle brook]

this is what i did

[rough tundra]

not quite, we seek to find the time our money has doubled, that is analogous to solving the equation (2*2000=2000 e^{2.8t}) not solving (2000=e^{2.8t})

[glossy valve]

PajamaMamaLlama

[gentle brook]

i can’t tell what the difference is and how the answers would be different , can you elaborate

[rough tundra]

the first equation 4000=2000e^... is this

the second equation 2000=e^... means if I start with $1, and I solve for t, that's asking how long until that $1 turns into $2000

(also sorry for ping)

[languid junco]

side note: your exponent needs to be 0.028t since we need to convert to a decimal

[rough tundra]

Nota bene: 2.8% = 0.028

[gentle brook]

but i can’t simplify this for my calculator i’m so confused i’m sorry 😭

i don’t understand because if you divide 2000 from each side it turns into 2000=e^2.8(t)

[rough tundra]

not quite

if 4000=2000x, if we solve for x what do we have?

[gentle brook]

OMG ITS 2 💔

i’m sorry it’s very late where i am

[rough tundra]

no worries, mistakes happen

then it becomes rather easy to solve (2=e^{0.028t})

[glossy valve]

PajamaMamaLlama

[gentle brook]

YES i did it thanks guys

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[neon quartz]

Having issues with solving this, I can do every part besides the last part and how they get the 1/1/4 at the end, Im confused on how they get a 1

I tried dividing,multiplying, and all sorts of things but cannot figure out how they get this answer.

[onyx glen]

you should not be writing 1 1/4 as 1/1/4. people won't understand you.

[neon quartz]

best way to communicate then, Im new to fractions still.

[onyx glen]

in plain text there is no better way to write it than to put a space like i did

anyway

[neon quartz]

Ok

[onyx glen]

do you know how to think about fractions in terms of pizza

[neon quartz]

in what ways, like if I cut a pizza in 8 slices and take away 1 7/8

[onyx glen]

imagine cutting a pizza into 4 slices

each of those slices represents 1/4 or one quarter of a pizza

are you with me so far

[neon quartz]

1/4 in what way

[onyx glen]

"way"?

[neon quartz]

nonono my bad I misread it

go on

[onyx glen]

now the fraction 5/4 means you have five slices like this,

which can be put together into one whole pizza (1) and there will be a single slice left over (1/4)

[neon quartz]

so the 5/4 would be broken down to the mixed number

[onyx glen]

yes

i am explaining this particular instance of conversion from improper fraction to mixed number

[neon quartz]

yes of course

Now im confused how they get 2/4 at the end before getting the final 5/4

[shadow flame]

1 1/4 doesn’t mean one one fourth, it means one + one fourth

[shadow flame]

1/2 = 2/4

[onyx glen]

they rewrote 1/2 as 2/4 to get it over the same denominator as the other fraction

[neon quartz]

thats confusing

[onyx glen]

ok

back up

[neon quartz]

I can get everything else now besides that part

[onyx glen]

you know how when you add two fractions you want them to have the same denominator, yes?

[neon quartz]

yes so it makes it easier to solve

[onyx glen]

yes, so

3/4 + 1/2

4 is not the same as 2

so the second fraction, 1/2, needs to be put in a form where its denom is 4 instead

[shadow flame]

it’s the only way to solve (unless if you already know what the answer means) actually

[neon quartz]

I didnt notice the dennom was 4

my bad

if the denom is the same it doesnt change

the upper number above it though is coming from where exactly

[onyx glen]

the tool we have for transforming fractions is to multiply the numerator and denominator by the same thing

in this case, we make $\frac{1}{2}$ into $\frac{1\cdot{\color{red}2}}{2\cdot{\color{red}2}}$

[glossy valve]

Ann

[neon quartz]

ookay so you always just times the second fraction

[onyx glen]

no

not always the second

sometimes you have to do this with only the first, or with both -- it depends on what you get for your lowest common denominator

https://youtu.be/5fbzZEK77DQ

[neon quartz]

ah okay

so if the denominator is higher than the numerator you would times it but if the numerator ishigher than denomator you times the numerator

[onyx glen]

no

this has nothing to do at all with which numerator is higher than what

or with numerators at all

you should watch the video and maybe practice a hundred different fraction addition problems instead of trying to see patterns that aren't there

i can tell you for a fact that trying to look for patterns as you are now will only make you more and more confused

[neon quartz]

I understand

I will atch the video

before I depart im assuming there are a variety of different forumulas for solving a fraction

[onyx glen]

there are no "formulas"

[neon quartz]

I was pointing towards just different equations or ways of solving a fraction

[full forum]

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[rain merlin]

for inequalities, our teacher taught us not to square root on both sides, i assume from the fear of someone writing for example x>+-2, and instead told us to sketch a concave up graph with 2 roots and see what range to take wrt x axis

however i just came across a qs where the inequality was cos^2(x) < 1/4 and since knowing what the cos^2 graph looks like was not required in the syllabus, i took |cosx| < 1/2, drew a cosx graph and a line of y=1/2 and got the ranges.

my question now is have i stumbled upon the way to solve for example x^2 > 4 so |x| > 2?

[rain merlin]

(this is just a yes or no answer i believe, ping when replying)

[gritty rose]

What do you mean "got the ranges"

Yes x^2 > 4 implies |x|>2

[rain merlin]

for what angles x is cosx< 1/2

[rain merlin]

alright brilliant

oh wait no mb

wrote the qs wrong

it was cos(x-pi/2) so basically sin

that's why there were rangeS

is that what you meant?

[gritty rose]

I still have no idea what you mean

[rain merlin]

lemme just show u the graph

i had an overall range that theta lies bw =-pi/6 and 7pi/6 inclusive and there was this gp question that boiled down to |cos(theta-pi/2)| < 1/2

so from -pi/6 to pi/6

and 5pi/6 to 7pi/6

2 disjoint ranges

for where x can lie to be less than half

@gritty rose

here's the exact original qs btw

[gritty rose]

What does what you're doing have to do with question b

[rain merlin]

r= 4cos^2 (theta-pi/2)

0<=r<1

for there to be a sum to inf

we dont take the -1 cos it's a square

then simplify

4cos^2 (theta-pi/2) < 1

|cos(theta-pi/2)| < 1/2

i couldnt find a diff way to do it so i had to self teach myself this by dry running values

[gritty rose]

This looks correct yes

[rain merlin]

yep

i just ran u thru it so you could see

pretty fun qs ngl

alr thanks

.close

[full forum]

Channel closed

Closed by @rain merlin

Use .reopen if this was a mistake.

[sullen wave]

What the fuck

[rain merlin]

ugh r u just gonna go around being like this in every channel

[slender onyx]

@sullen wave i already warned you to not post randomly in others help channels. come back in 24h

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[worthy tapir]

need to check if (e^x+1)*e^x is a possible solution to y'=e^2x-y

[worthy tapir]

how

[onyx glen]

plug it in lol

[fast peak]

compute the derivative

[worthy tapir]

what did i do wrong

[void nova]

,rccw

[void nova]

,w y' = e^{2x} - y

[glossy valve]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by zuanon.alberto

[void nova]

Here, nothing
It simply means that the function you found is wrong

[worthy tapir]

yay great

so I didn't do anything wrong

it just did not fit 🙂

thx

[void nova]

Yep

[worthy tapir]

damn your good at LaTeX

Wish I could throw out commands like that too

[void nova]

I wrote nothing in LaTeX 😅

[worthy tapir]

what is it then

TeXit

,w y'=e^{2x}-y

[glossy valve]

Results provided by WolframAlpha

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Requested by allnamesaretaken2342329_72708

[worthy tapir]

uuuuuh

that is awesome

[worthy tapir]

,rccw

[void nova]

That's not latex

[worthy tapir]

,rccw

[void nova]

That's the command for running Wolframalpha

[worthy tapir]

how to I turn it anti clockwise?

[void nova]

,rccw

[worthy tapir]

,rccw

[worthy tapir]

,rccw

[worthy tapir]

ok

have a great day

,rccw

[worthy tapir]

,rccw

[worthy tapir]

,rccw

[void nova]

You're welcome 🤗

[worthy tapir]

.close

[full forum]

Channel closed

Closed by @worthy tapir

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[idle jolt]

i need help with a geometry task:
there is a rhombus ABCD with an acute angle a. A belongs to a line that divides the angle a by a ratio of 1:3 and |CD| by a ratio of 2:3.
prove that cosa = 1/8

[idle jolt]

so i ended up with the last equation and earlier i decided to do something like cos(3*1/4a) = cos(3b) but it didnt really work and it made things a lot messier

[full forum]

@idle jolt Has your question been resolved?

[full forum]

@idle jolt Has your question been resolved?

[boreal adder]

wow this question is hard

or im stupid

[hot gyro]

The ratio of CD 2:3 so shouldn’t it be the other way around like o is where the line lands :do=3x and oc =2x?

[idle jolt]

uh I don't know to be honest

I'll redraw it in a minute

the angle doesn't look like it's divided by a ratio of 1:3 now

oh wait

you were referring to what's said in the task lmao

well I suppose it makes slightly more sense that way

[hot gyro]

So is it like this or like the old one lol 😭

[idle jolt]

well if we're going with your interpretation of the task then it's this way

[hot gyro]

Yea

[hot gyro]

Watch out here

[idle jolt]

yeah the formula goes a^2+b^2-2ab

therefore after transforming it to -cosa its +

[hot gyro]

Oh okok

Now try solving it with 3x instead of 2x

I can’t help rn I’m not home

[idle jolt]

yeah it doesnt really give me anything

[hot gyro]

Hm

[full forum]

@idle jolt Has your question been resolved?

[boreal adder]

consider triangle ACD only. now the line AP is the angle bisector of angle CAD which is a/2.
by the angle bisector theorem, PC/PD = AC/AD = 3/2

@idle jolt

[idle jolt]

uhh hang on

i need to process this

[boreal adder]

try from there

[idle jolt]

why are you deleting those?

[boreal adder]

ill leave the full solution in spoilers

||now consider this diagram||
||DO bisects AC and so AO/AD = 1.5/2 = 3/4. since AOD is a right angle cuz ABCD is a rhombus, AO/AD is also cos(a/2)||
||now by the double angle formula, cos(a) = 2 * (3/4)^2 - 1 = 9/8 - 1 = 1/8||

[idle jolt]

alright i wont click just yet ill try to figure this out

[boreal adder]

the crucial step is the angle bisector theorem, it should be real easy from there
idk why i couldnt think of it before

[idle jolt]

where are you getting AO/AD=3/4?

[boreal adder]

did u understand why AC/AD = 3/2?

[idle jolt]

that's what I did with the theorem

[boreal adder]

u should not be getting sines and cosines at this stage lol

[idle jolt]

lol but I did

[boreal adder]

this is a very simple problem, just using the angle bisector theorem should work

do u know what the angle bisector theorem is?

[idle jolt]

well I just wrote it on the paper

[boreal adder]

oh u um

the first drawing was correct

[idle jolt]

are you sure

[boreal adder]

its not this

yes

[boreal adder]

its this

[idle jolt]

well I only wrote it this way bc this other person told me lmao

[hushed barn]

Also have to comment nice solution!

[idle jolt]

my first drawing was like yours

I'll redo it for the nth time

[boreal adder]

lmao alright

[boreal adder]

thanks

cant believe it took me that long to figure out tho

[idle jolt]

alright so AC = 15x/2

[boreal adder]

no just think about it in terms of ratios

stop trying to get absolute lengths

[idle jolt]

lmao

so 15/2?

[boreal adder]

no

ill try and explain

[boreal adder]

take triangle ADC. what does the angle bisector theorem say

[idle jolt]

I got cosa/2=3/4

[boreal adder]

oh

ok

then double angle formula

[boreal adder]

honest to god idk where u get these sines from

[idle jolt]

oh I'm just comparing the areas

but skipping 1/2