help-36

[glossy zephyr]

Different velocity field

Look up to the example. 1.1

[manic helm]

uhh

[glossy zephyr]

And the question is slightly altered

Not respect to time

[manic helm]

ye

that's my qu like

how come here we know it's not wrt time

[glossy zephyr]

It asks for you to find the streamlines of the extensional flow

The original question asks for you to find an equation of the particle that has a IVP on R(0) for (x_0, y_0, z_0)

Based on time

[manic helm]

ohh

so in the first qu, the flow wasnt steady-state

?

[glossy zephyr]

Aka they are asking for general solution but just adapted to admit a starting value

[manic helm]

Because

i thought the flow was steady-state and thus used the method for finding streamlines instead

as streamlines = particle path for steady-state flows

[glossy zephyr]

They are, again, the problem is just adapted differently, the process remains essentially the same

it wouldnt be steady state if some of the equations of the vector form included t

[manic helm]

if it asked me to find the streamlines instead, would i have still used t?

[glossy zephyr]

Not in this case, no

Again, for this case, the difference between the two is just symbols

[final saddle]

@manic helm Has your question been resolved?

[manic helm]

i got some complicated functions 💀

yea in hindsight it rly wasnt that hard

i think i just wasnt expecting a coupled de

lol

one thing i will ask tho

which im trying to get my head around

if you have a steady-state flow

then u know its streamlines dont change wrt time

and for steady-state flows, particle paths = streamline

yet particle path can be time dependent ?

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[fickle holly]

Hi, I hope all is well. I'm a bit confused about which formula to use and how to set up the statistical problem # 2. I understand how to solve it when it's a single probability like Pr(ȳ > 17), but this one is throwing me off. I’ve already solved for the standard error but unsure of where to go from there.
N=3, mean=16.2,standard deviation=1.2

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@fickle holly Has your question been resolved?

[fickle holly]

<@&286206848099549185>

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@fickle holly Has your question been resolved?

[vernal vale]

likely sleeping

[fickle holly]

True haha thanks, and is your mom or dad jamaican

[vernal vale]

dad korean mom jamaican

i was born in korea though

[fickle holly]

LOL omg!! would u believe me if i said the same

[vernal vale]

no way, thats awesome though

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[patent pulsar]

why is det (A) = 1 wrong

I did the RREF to see if it could be simplified and I got an I3 Matrix so wouldn't det (A) = 1?

[odd seal]

rref doesnt preserve determinant

every invertible matrix can be rref'ed into an identity

doesnt mean all invertible matrices have det = 1

simple counterexample to your claim: 2I

[patent pulsar]

ok good to know I guess

I did it by cofactor expansion and got 16 which is right

[patent pulsar]

Is there a connection between RREF and determinants otherwise somewhere else?

[odd seal]

not really, you achieve RREF thru ERO, and EROs dont preserve determinant. But, if RREF has a form of identity matrix, then you can prove that the determinant is not 0

you can only use RREF to check if the det is 0, otherwise its not useful to calculate the determinant

[patent pulsar]

yeah if RREF (A) = I, det (A) is different of 0 because AA^-1 = I Ithink

[tiny gorge]

if you keep track of the row manipulations you can deduce the effect on the determinant

scaling a row by k multiplies the det by k
swapping two rows changes the sign of the det
and the "add scalar multiple of one row to another row" does not change the det

[odd seal]

yes, rref proves the existence of the inverse, so its same as claiming det is not 0

[patent pulsar]

I know EROs can be expressed as elementary matrices but does that have something to do with determinants?

[tiny gorge]

sure

RREF = (ERO_1)(ERO_2)...(ERO_n)A
so
det(RREF) = det(ERO_1)det(ERO_2)...det(ERO_n)det(A)

[patent pulsar]

In class I think we also saw that if Ek...E3E2E1A = I, then A = E1^-1E2^-2E3^-3....Ek^-1

[tiny gorge]

correct

[patent pulsar]

Also I notice determinants can get tedious really quickly, is there a fast method for 4 by 4 or more matrices

like whatever this is

there must be a short way for this

[azure wren]

row reduction is your best bet i believe, cofactor is a pain for 4x4

i doubt they would give you a 4x4 question in an exam unless it has a bunch of zeros in it (therefore making cofactor easy), otherwise just use matlab or numpy for determinants

[patent pulsar]

0s on a column or a row right?

[azure wren]

yeah

[patent pulsar]

cofactor doesn't care about column or row

[azure wren]

then you can do cofactor expansion down that row or column

yup

just be careful about the sign

[tiny gorge]

try subtracting row 1 from each of the other rows, most of the coefficients become zero

and that operation does not affect the determinant

[azure wren]

this thing

[patent pulsar]

yes the (-1)^i+j alternates the sign

I think I saw that kind of when I did telescoping series in Calculus 2

[azure wren]

yeah it's the same thing basically

checkerboard matrix makes it pretty easy to work out if (-1)^i+j is confusing

[patent pulsar]

at least the teacher made some problems really easy like this

[digital carbon]

lol

[patent pulsar]

0s = det(A) = 0

[azure wren]

hahaha that's great

[digital carbon]

It would have been really funny if he added a random different number in each row

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[fair sigil]

hi

[fair sigil]

2nd and 3rd

i did 1st by MPT

<@&286206848099549185>

@muted glacier

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[fair sigil]

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[tiny kraken]

Hey mate, do you want a moderator to be here?

[fair sigil]

sorrry

can u solve

[tiny kraken]

just XB' is parallel to CF, can you prove CF is perpendicular to XY?

[gray yacht]

Chill bro, my whole family might join this ticket if you keep tagging

[fair sigil]

yes

og

oh*

i got it

thanks man

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[fair sigil]

.close

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[thin thicket]

Is the multiplicative identity usually required in the definition of a ring?

[thin thicket]

I can't tell cuz I legit started today

is $(A,\cdot)$ a semi-group more common?

[soft zealot]

Julian

[desert mantle]

really depends on the book

[thin thicket]

my goal is to

construct the real numbers and i heard learning some basic algebra is nice

i mean R has the mult identity so it doesnt matter?

for my purposes?

[desert mantle]

while the cauchy sequence construction of R is technically a factor ring, that perspective does not help that much imo

so not sure who gave you that advice

[plucky rover]

Ig the perspective helps you with the stuff building up to ℝ

[desert mantle]

or what they intended with it

[plucky rover]

But still, I agree with Dena. This isn't very helpful

[thin thicket]

hm ok

[desert mantle]

you construct R from Q and thats already a field anyway

[thin thicket]

i wanted to construct Q too i forgot to mention\

and since fields are rings i figured learn what rings are

[desert mantle]

well ok sure then you need to know what a ring is

well, you need to know what Z is

[thin thicket]

and then i clicked all the links on wikipedia and started on magma

[desert mantle]

and all the basic axioms about it

[thin thicket]

got to*

[desert mantle]

and then you arrived at a ring

[plucky rover]

Are you trying to construct starting from basic set theory

[thin thicket]

yea ring monoid semiring magma

well i know how N is constructed

i havent looked at Z but im assuming its not too complicated

[plucky rover]

Are you familiar with equivalence relations

[thin thicket]

i learned about those a month ago and gave up

but yea

[desert mantle]

go back to those

[thin thicket]

i have the properties of homogeneous relations in myu notes

[desert mantle]

all constructions N->Z->Q->R are about equivalence relations

[thin thicket]

ok

[desert mantle]

all those constructions can be generalized but you dont need to know the general framework here

[thin thicket]

im doing an analysis class rn and we have a project where we construct a set that is isomorphic to R from an assumed Q

but i kinda just wanna fill in the details

thanks yall

[desert mantle]

i.e. N->Z can be generalised to constructing a group from a monoid, Z->Q can be generalised to constructing a field from an integral domain, Q->R can be generalised to constructing a completing of a metric space/completion of a linear order

[thin thicket]

good thing i learned about metric spaces

ok

.close

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[plucky rover]

I'm looking to prove something using the following axioms and modus ponens: /

\begin{enumerate}
\item $A \lor A \to A$
\item $A \to A \lor B$
\item $A \lor B \to B \lor A$
\item $(A \to B) \to (C \lor A \to C \lor B)$
\end{enumerate}

Using the above I want to prove that $A \lor (B \lor C) \to (A \lor B) \lor C$

[soft zealot]

Xavier 🌺

[plucky rover]

Not looking for a solution, just looking for a hint

The first thing I think I need is proving transitivity of implication

Also $A \to B$ is defined as $\neg A \lor B$

[soft zealot]

Xavier 🌺

[plucky rover]

(ie. We are working in classical logic)

Please tag me if answering

[craggy plank]

[lapis wedge]

could you use truth trees?

[plucky rover]

The problem is that I'm pretty sure I'm not allowed to use ∧ for this problem cuz it's Russell-Whitehead

[plucky rover]

I wish lmao

I want to start with $A \to B$ and $B \to C$ and get $A \to C$

[soft zealot]

Xavier 🌺

[plucky rover]

I'm just unsure what exactly I'm allowed to use

[gray yacht]

Start with axiom 2 and 4

Couple of modus ponens and then chain implications might work

[plucky rover]

I'm pretty sure I don't even need axiom 2 lol

Hold on Lemme write out what I have

[gray yacht]

You do ig

[plucky rover]

We don't have A to begin with though

So I don't see what axiom 2 yields

Actually no the only way axiom 2 would help here would use transitivity

And that's what I'm trying to prove in the first place

Ah nvm I figured out how to write ∧ in this system, the problem is trivial now

$(A \to B) \to ((B \to C) \to (A \to C))$

[soft zealot]

Xavier 🌺

[plucky rover]

.close

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[warm python]

<@&268886789983436800>

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[warm python]

Spam messages please check logs

oh, were the messages delted by mods

nvm

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[sharp wraith]

how do i rigorously show this

[pliant shore]

what have you tried?

also, what tools are you allowed to use?

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[static oak]

How do I do this? I did this over a year ago so it’s review but I forget how

[worldly mesa]

,rotate

[static oak]

Or actually I did this like two years ago

[harsh cobalt]

ax² + bx + c = 0

[worldly mesa]

well you have the formula

just plug in the values

[static oak]

But I don’t know what variables to plug in

I don’t know which ones are a which ones r b which ones r c

[worldly mesa]

.

[errant basin]

looking at your tag makes me want to ragebait swifties brb

[harsh cobalt]

Thanks.

[static oak]

So 3x^2 js a 4 is band 7 is c

Do I include X when I put them? If I don’t what do I do with the square for 3

[polar spruce]

no

[harsh cobalt]

The equation format is
ax² + bx + c
Just take a, b and c.
And plug then in the equation

[polar spruce]

only the coefficients

[desert mantle]

the thing in front of the x^2 is a. the thing in front of x is b. the thing on its own is c

[worldly mesa]

you say c is 7 but that is not the value of the coefficient

[polar spruce]

c is -7

[harsh cobalt]

In this case,
3x² + 4x - 7
Thus being,
a = 3, b = 4, c = -7

[bleak granite]

[\underbrace{3}{a}x^2+\underbrace{4}{b}x+\underbrace{-7}_{c}=0]

[soft zealot]

ΠαϳαμαΜαμαΛλαμα

[static oak]

On the top why is it a + and - sign how do I know which one to use

[bleak granite]

the form if a^2+bx+c, in the first equation we have a -7

[harsh cobalt]

+- = -
It's just you take the minus with it

[bleak granite]

note that since -7=+(-7), c=-7

[static oak]

Huh

So it’s just a - not a plus?

Why include the + then

[harsh cobalt]

Yes, but since the format is + (number)
If that number is negative it will be +(–n)
Thus c = –n

In this case, c = –7

[harsh cobalt]

^

[bleak granite]

wait - do you mean in the formula?

[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}]

[harsh cobalt]

I believe it's minus b

[soft zealot]

ΠαϳαμαΜαμαΛλαμα

[static oak]

I don’t think I did it rigjt for the firsg one I got -1.23

[harsh cobalt]

@static oak
This is the formula,
The equation is 3x² + 4 -7
Let's simplify it.
3x² + 4 + (-7)
Because + - = –

[harsh cobalt]

,rotate

[bleak granite]

,rccw

[bleak granite]

don't write the sqrt(130) as a decimal

then it's no logner equal

because sqrt(130) has infinitely many decimals

[final tangle]

how are you getting 130

[static oak]

Ohh

Now I got -2.333

[harsh cobalt]

@static oak If you want to verify your answer, grab a calculator, press
Mode 5 3
And plug in the values

[static oak]

I’m not sure where to put the numbers 53 after I click mode

It just sends me back to the main page

[harsh cobalt]

Then u must have a different one. There should be a root equation in it

Just not in mode 5 3

https://www.calculatorsoup.com/calculators/algebra/quadratic-formula-calculator.php

[final tangle]

ideally leave the answers in exact form, with fractions / radicals depending on what you have
that would be the decimal approximation of one of the values

(-7/3 as shown in the calculator above, but the values are nice enough that you shouldn't be using a calc)

[final tangle]

you use both, the QF indicates two values.

$$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
indicates
$$x = \frac{-b-\sqrt{b^2-4ac}}{2a} \gray{\underbrace{\black{\vee}}_{\text{or}}} x = \frac{-b+\sqrt{b^2-4ac}}{2a}$$

[soft zealot]

ραμOmeganato5

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[late magnet]

(Mark scheme in red box) I don’t understand why you can just sub x=-1? Is it something to do with the factor theorem? My working is in black but it got me nowhere as seen lol

[late rose]

do you know the remainder theorem?

[late magnet]

No

Is it similar to factor theorem?

[late rose]

If $f(x)$ is a polynomial, then the remainder when $f(x)$ is divided by $(x - a)$ is $f(a)$

[late magnet]

Oo ok

OOO OKK processed it

That makes sense - thanks!

[late rose]

the proof of this is as follows

Let $Q(x)$ be the quotient and $R$ the remainder. Then $f(x) = (x - a)Q(x) + R$. To make the divisor 0, substitute $x = a$. Thus, when $x = a$, $f(a) = (a - a)Q(a) + R = 0 + R = R$

[late rose]

Similarly, the remainder when $f(x)$ is divided by $(ax + b)$ is $f\left(-\frac{b}{a}\right)$

[leaden moon]

It's bezout's if im not mistaken

[late magnet]

That’s so clean

[late rose]

it's different

[late magnet]

Thanks! For the proof and help 🙂

[leaden moon]

Or youcan use the traditional way

If $f(x)$ divided by $g(x)$ gives $Q(x)$ and a remainder of $r(x)$ then $f(x) = Q(x)g(x) + r(x)$

[soft zealot]

1 divided by 0 equals Infinity

[late magnet]

Yea

[late magnet]

But how would we know Q(x)

[leaden moon]

Oh yeah

[late magnet]

😭

But that looks helpful, I’ll probs use it in other q.s similar to this one so thanks still!

.close

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[sturdy cloak]

how do i calc the partial derivative of this wrt q dot?

[scarlet sequoia]

Well since it's wrt q dot, all "q" terms are like constants

so you just need to differentiate a dot product

[sturdy cloak]

so

is it the norm of q squared dotted with q?

ie the q dot just vanishes

[sturdy cloak]

yh

?

[formal trail]

how are you dotting a vector with a scalar

[sturdy cloak]

ah mb so here q is a constant vector right?

wrt to q dot

[formal trail]

yes

[sturdy cloak]

so a constant vector dotted with itself

is just a constant scalar?

[formal trail]

yes, i agree with |q|^2 being a constant scalar

[sturdy cloak]

so the derivative is just |q|^2 times q

ie (|q|^2 )q

[formal trail]

if you've established that $\pdv{}{\dot q} \dot q \cdot q = q$ then yes

[soft zealot]

κλαουντ ☁ (cloud)

[formal trail]

just make sure not to use dot to denote scalar multiplication

[sturdy cloak]

yes ok

[sturdy cloak]

this might sound like a dumb question but how would u calc pdv of q dot q wrt to q?

ie this

would u use product rule?

[formal trail]

yes, the product rule does apply to dot products (you can prove this if you want)

[sturdy cloak]

thx for the help

appreicate it

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[tidal birch]

whats the answer for unstable?

[tidal birch]

.close

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[silk wind]

What am I doing wrong here?

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[rocky tusk]

this doesn't make any sense

you aren't choosing epsilon

you choose delta as a function of epsilon

the epsilon is some arbitrary value

you know that you have control over |x| since we will have 0 < |x| < delta

we want to also bound 1/|x + 2|

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[simple gate]

help

[simple gate]

i need help with the end of year, year 9 exams in sydney australia

[vital crag]

[vital crag]

a math question specifically

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[vital kelp]

Darboux integral > riemann integral

[vital kelp]

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[static oak]

I’m really confused here how does 2•the square root of six equal the square root of 24 it would be the square root of 12

[true escarp]

what is sqrt(2) * sqrt(2)

or better yet whats sqrt(4) * sqrt(6)

[static oak]

The square root of 4?

[true escarp]

yes

[static oak]

Square root of 24?

[true escarp]

yes

[static oak]

Oh wait

[true escarp]

now can you simplify sqrt(4)

[static oak]

Is it because we are adding the twos together which equal four and then it’s 4• the square foot of six?

Sorry this concept is new to me

[true escarp]

we've established that sqrt(24) = sqrt(4) * sqrt(6) right

[static oak]

Yea

[true escarp]

but can you simplify sqrt(4)

do we know the square root of 4

is there a number that when squared equals 4

[static oak]

2

[true escarp]

okay

so

sqrt(24) = sqrt(4) * sqrt(6) = 2 * sqrt(6)

does this make sense

[static oak]

Ohhh

[true escarp]

we dont get 12 because we aren't factoring out 2 from 24

we are factoring out 4

and then we simplify sqrt(4) = 2

[static oak]

How did we get one after tho?

[true escarp]

the 1 + sqrt(6)?

[static oak]

Yea

[true escarp]

just divied both terms by 2

[static oak]

Then why does the square root stay the same?

[true escarp]

because you have 2 * sqrt(6)

if you divide that by two then its just sqrt(6)

you are just undoing the multiplication by two

like if I did 2 * 5

[static oak]

I thought we would cancel the terms here so we get rid of the 2 next to the + and - so we would get 2 +- square roo of six

[true escarp]

but then i divided it by 2

i would just get 5

because i first doubled it and then I halved it

[true escarp]

no you need to divide both terms by two

just how you expand multiplication across each term

[static oak]

I know in class we talk about cancelling so how would I know when to divide and when to cancel?

[true escarp]

its the same thing

cancelling is for when you divide by something you already multiplied by

you can 'cancel' them out

because they are inverse of each other

for example

say I have 5 times some number x, so 5 * x

if I divide this by 5

I know I will just get x

because im multiplying by 5

but then dividing by it as well

so I can cancel the 5s

(5x)/5 = x

[static oak]

Ohh I think I got confused because in the answer key it never kept the thing we are multiply the square root of six by since it is 1

[true escarp]

yes

[true escarp]

exactly

but writing 1 times something is redundant

because 1 * sqrt(6) is still sqrt(6)

so you dont need to write the 1

just like how you would write 2x but not 1x

you just write x

[static oak]

Honestly I write 1 with X when I’m doing problems because it messes me up with just having X in the middle of solving a problem. Sorry if that doesn’t make sense

[true escarp]

it makes sense

and that's fine

but generally its considered than 1 * something can just be written as something

because its not providing any additional information

so just make sure you do that for the final answer

[static oak]

Kk

Ty

[leaden moon]

Ngl the table method is the method I HAVE to use later on. Haven't heard of it before lol

[final saddle]

@static oak Has your question been resolved?

[static oak]

Ok I’m doing smth wrong for the second long deviation should I use (X+2) instead

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[smoky fog]

I would like to confirm if my graphs are correct

[smoky fog]

I am unsure about the last one

[tiny gorge]

yes this is right

the last one is not hamiltonian because any cycle that touches every vertex has to pass through v5 twice

[smoky fog]

Awesome tysm

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[jagged flare]

ok i havent solved this yet, i just want to know if this is something you can actually do

ok nvm im silly

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.reopen

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✅ Original question: #help-36 message

[jagged flare]

wait im double silly

ok so

if we do P(x0,y0) such that y0=1/4 and x0 such that x0^2+1/4=x0, then we are left with f(f(x0)-y0)=0, let f(x0)-y0=a, so there exists a such that f(a)=0

plug in x=a, we get f(a^2+y)=f(-y)

by guessing we get that f(x)=x^2 is a solution

ok heres the iffy part which i dont know is possible or not, but can i do like
f(a^2+y)=f(-y)
(a^2+y)^2=(-y)^2
a^4+2a^2y+y^2=y^2
a^2(a^2+2y)=0, since a is fixed and y can be anything, a^2+2y cant be 0, so a^2 must be 0 which implies a=0

is this something i can do?

i think from this i can only get that f is even, but eh anything helps atp cause im hella stuck

[worldly mesa]

Looks fine

[jagged flare]

so, if we guess a solution, we can use it for the functional equation bit?

[worldly mesa]

Knowing a solution doesn't tell you much about other solutions just by itself

But it can be a guideline on what function are you dealing with

[jagged flare]

so by knowing one of the solution, the fact that i can deduce a=0 can be used yes?

[worldly mesa]

Yeah

[jagged flare]

okayy thank you!

.solved

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[supple heath]

Im struggling a lot with linear algebra, i dont understand what does the notation M(...) mean and how to start this problem

[worldly mesa]

M(f,C,B) is the matrix representation of f with respect to the basis C and B

Where the i'th column is made of the coefficients when you write f(c_i) as a linear combination of b_i

[supple heath]

i'th column?

[worldly mesa]

Yeah, like the first column, second column etc

[supple heath]

hm im not gettint it :(

[final saddle]

@supple heath Has your question been resolved?

[final saddle]

@supple heath Has your question been resolved?

[blissful stream]

Hi Erika :) If you haven't yet I would recommend you watch the three blue one brown 'Essence of linear algebra' I feel like it helped me a lot to develop some intuition. https://www.youtube.com/watch?v=fNk_zzaMoSs&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab

https://www.youtube.com/watch?v=P2LTAUO1TdA this video is on change of basis so kind of the problem you are working on right now

Let me try to explain (but take everything with a huge grain of salt since honestly I've forgotten like all about it). But M(f, C_3, B_2) is as has been said the matrix representation of the mapping f with respect to the basis C_3 and B_2. That means to get this matrix what they did is take the unit vectors (1, 0, 0)^T, (0,1,0)^T, (0, 0, 1)^T then write it in the basis of C_3 here that's the canonical basis of R^3 so nothing changes. Then you apply f to these vectors and get the two dimensional ones. Which are then written as the linear combinations of the other basis. Here that's B_2 or C_2 (in C_2 you have the canonical basis again which means you don't really do anything)

So to start try finding out what f does

[supple heath]

yes i already wtched those videos and helped me a lot, but even that im still struggling

[blissful stream]

the second matrix should tell you

[blissful stream]

yeah I totally feel that! I also struggled with linear algebra a lot to be honest

[supple heath]

the second matrix is telling me that to change of C_3 to C_2 its using (2 -3 2, 4 -1 24)?

[blissful stream]

yes

so as far as I remember that is the f

[supple heath]

but how is that represented? i mean, is a multiplication right?

[blissful stream]

I don't know if you have the same notation as we did but this is basically how I would always write it to understand

[supple heath]

is that phi?

[blissful stream]

yes

do you understand what phi does?

[supple heath]

why the inverse of C3?

[blissful stream]

cuz that took very long for me to understand

[supple heath]

no, ive never seen that phi

[blissful stream]

oh okay yeah that makes sense

basically phi is writing a vector you have to another basis

so for phi_c2 you take whatever vector you have and write it in the basis of c2

[supple heath]

oh okay, for me its like this i think

[blissful stream]

the inverse is basically taking that vector from c3 and writing it in the canonical basis

[supple heath]

could be u, v, w whatever

[blissful stream]

interesting yeah that seems to be the same then

[supple heath]

ok but why are you doing the inverse?

i dont understand this part, its like ure returning to canonical basis applying the inverse but what does that (1 0 0) has to do

[blissful stream]

that's the unit vector

since we're already in the canonical that doesn't do anything (I just added it because it's a step that you do need to do in case you start out in another basis like B3 for example)

[supple heath]

that (1 0), (-1 1) and (3 4) from what basis are?

[blissful stream]

we know that if we write the vector (2,4)^T in the basis B2 we get (1, 0)

what does that tell you?

[supple heath]

is this correct?

hm i dont think so

[blissful stream]

I don't quite understand what you are trying to do tbh

[supple heath]

we get (1, 0) in the canonical base right?

[blissful stream]

we have (2, 4) in the canonical base

and that is (1, 0) in the b2 base

so you're only using 1*( first basis vector of b2)

[supple heath]

uhm

is this right?

[blissful stream]

that looks good

(a, c) and (b, d) are the basis vectors of B2 in this case

[supple heath]

yes

[blissful stream]

so you have found the first one :)))

[supple heath]

but if i do the same with (-3 -1) and (2 24) i get different a and c

[blissful stream]

a and c should be fixed now. then you use that to find b and d

[supple heath]

i dont know how to get b and d

[blissful stream]

try solving both equations

for b and d

you should get the same values

[supple heath]

yes, theyre the same value

-1 3

[blissful stream]

I got that too 🥳

So do you know what B2 is now?

[supple heath]

but how do i get to that conclusion, when i see the exercices i dont know where to begin, and i dont know how did u get to that equations

[blissful stream]

yeah I feel that

have you ever seen something like this?

[supple heath]

yes

[blissful stream]

then that's where you need to look for answers

once you actually understand what's happening there you can apply it to basically anything involving this

at least it worked like that for me. took forever to actually get a feel for

[supple heath]

the writing is so aesthetic

[blissful stream]

but is very useful

[blissful stream]

aw thank you ☺️

[blissful stream]

usually all the blue things are given. you know all the bases and f and want to figure out how to use f to map from B3 to B2

you're trying to build M(f, B3, B2)

in this exercise you were given M(f, C3, B2) and f (although it was written confusingly as M(f, C3, C2) which means the phis are just the identity function and you can see how that just gives you f)

[supple heath]

i see

[blissful stream]

instead you were looking for the basis B2. and for that you basically just use the vectors you get out of f when you're in the canonical basis and compare them with the ones the matrix spits out in B2

[blissful stream]

that's how you get this

[supple heath]

this is insane

[blissful stream]

we were super lucky that the first equation was basically super lucky we got the a and c directly

[blissful stream]

it is!!!

for my linear algebra class we actually got that problem in the exam when all we ever did before was find M(f, Bn, Bm)

[supple heath]

oh

[blissful stream]

it was horrible 😭

[supple heath]

🥹

i hope the exam will be easy

[blissful stream]

i'm keeping my fingers crossed for you!

wish you all the best luck

[supple heath]

I’ve had a really bad streak. It’s exam week, and I hadn’t been able to study either algebra or calculus because I was busy with other exams. I was planning to start studying from Friday to Monday for the exams, but I got sick and have been in bed all these days. Today I finally feel a bit better, so I started working on algebra, but I didn’t understand anything :/

[blissful stream]

oh noooo that sounds really bad

I hope you get better soon

[supple heath]

thanks a lot, now I understand a little bit more about this.

[blissful stream]

there's tons of videos online with professors explaining it on other problems

going through them helped me tons personally

so I can only recommend that

[supple heath]

yes, thank you so much :)

[blissful stream]

no problem

happy to help :)

[supple heath]

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[potent shuttle]

hey can someone give me a simple proof for f(x) = e^x not being a surjective (onto function)

[jagged flare]

e^x=-1 does not have a solution

[zealous storm]

what's the codomain you're working with?

[jagged flare]

real

[pliant shore]

I bet it's the reals too

[potent shuttle]

(0, ∞). f: ℝ → (0, ∞)

[amber holly]

In that case it is surjective

[zealous storm]

it is tho 😭

did you mean [0, inf)?

[potent shuttle]

yes

[zealous storm]

yeah well there's no value of x such that e^x = 0

[potent shuttle]

🥲 i have made my assignment for proof of a function being oneone but not onto with this e^x function thinking id be cool. i was a fool

made graph and everything realised it is onto for this

imma do f(x) = 2x ig

[pliant shore]

the point we're all making is that 'onto' depends on your codomain

your codomain can be anything you want

[pliant shore]

change your codomain to this, and you have e^x being injective but not surjective

[delicate sorrel]

But whats the proof to this

[pliant shore]

it's easier to think about 2^x = 0

[delicate sorrel]

Huh?

[pliant shore]

or rather, 1/2^x = 0

[pliant shore]

cause 2^x = e^((ln 2) * x)

[delicate sorrel]

Oh

Yes but whats the mathematical proof

[pliant shore]

what's your definition of 'proof'

[delicate sorrel]

That anything i the power of something is never 0

[delicate sorrel]

Using axiomes to conclude somthg

?

Nvm we dont have to go that deep

[pliant shore]

we can use the fact that the product of two non-zero numbers is never zero

so start with 1/2 * 1/2 = 1/4: that's not zero
then we consider that 1/2 * 1/4 = 1/8 can't be zero either

so proceeding inductively, we've shown this is impossible for an integer x

then hopefully if you agree that 2^x is an increasing function, then 1/2^x is a decreasing function

so for 1 <= x <= 2, 1/2^x is in [1/4, 1/2]
for 2 <= x <= 3, 1/2^x is in [1/8, 1/4], and so on

[delicate sorrel]

Or we can suppose that there is an x wich satisfies e^x = 0 that means for any y in R e^(x + y) = 0 e^y anything times 0 is 0 so for any y + x = k’ in R e^k = 0 but since e^1 =e by definition so there is a contradiction to e^k=0 hence e^x is never 0

I used the fact that any real number k could be represented as the sum of two real numbers

[pliant shore]

I don't understand your proof

[delicate sorrel]

What part

[pliant shore]

how does that imply that for any y in R, e^(x + y) = 0?

[delicate sorrel]

I just multiplied both sides by e^y

[pliant shore]

you can multiply both sides by e^y I guess

sure

ah, then that violates the fact that e^x is injective

nice one

[delicate sorrel]

Yepp

[pliant shore]

that's what you wanted to say but you didn't say it

anyways, I get it now

[delicate sorrel]

The most rigorous proofs are those who countains the simplest notions

[pliant shore]

indeed

[final saddle]

@potent shuttle Has your question been resolved?

[drowsy epoch]

When you "multiply" by e^y you gotta make sure y!=x or else you multiply by 0

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[solar crest]

can someone check if this is correct or not

[tiny gorge]

oh i'm probably misinterpreting your diagram

the arrow is misleading

should just be the line segment itself

[solar crest]

oh what

wait

but it's a vector tho?

also like

[tiny gorge]

no, it's a set of points

just like U is

[solar crest]

what..

omg this is my friend's answer

she got full marks

💀

[tiny gorge]

well it's right except for the arrow

[solar crest]

okay look

[tiny gorge]

in fact every point on the line segment is part of T(U)

[solar crest]

i did this

[solar crest]

oh

is my thing wrong??????????????

[tiny gorge]

if i were grading it and that's all you wrote, i would take some points off probably

because it looks like you're saying that only the vector (2,2) is in T(U)

[solar crest]

no just diagram wise

[solar crest]

wait is it not

[tiny gorge]

when in reality everything of the form (x,x) with 0 <= x <= 2 is in T(U)

[solar crest]

is [2,2] not then?

cuz i thought it's <=

[tiny gorge]

it is, it's included in (x,x) with 0 <= x <= 2

[solar crest]

ookay

and it's a unit square right

so shouldnt it be shaded in

[tiny gorge]

U is the unit square with the interior (should be shaded)

[solar crest]

i dont get how myu friend got full marks for doing just an arrow

[tiny gorge]

T(U) is not a square, it's just the line segment from (0,0) to (2,2), including both endpoints

[tiny gorge]

generous grading probably 😁

[solar crest]

how is that possible? if U is already a square itself... after applying transformation it no longer becomes a square??

[solar crest]

lucky her

[tiny gorge]

right, because T is collapsing everything down to a single line (one dimensional)

you can see that because both coordinates of (x1+x2, x1+x2) are the same

[solar crest]

oh what

[tiny gorge]

so T maps everything onto the line y = x

[solar crest]

omg

[tiny gorge]

then it's just a matter of working out which part of that line is actually in T(U)

[solar crest]

wait how did you even know that the dimension collapsed

[tiny gorge]

well if you don't notice it immediately you can plug in some points to get the idea

[solar crest]

this is a bit confusing

[tiny gorge]

if you know about matrices you could also try to find the matrix for T, and notice that its columns are linearly dependent

[solar crest]

😭

how do i find the matrix

i feel like the range is throwing me off

like the 0<=x1<=1

[tiny gorge]

well the easiest way to find the matrix is to find where T sends (1,0) and (0,1)

the first column of the matrix will be $T\begin{pmatrix}1 \ 0 \end{pmatrix}$ and the second column will be $T\begin{pmatrix}0 \ 1 \end{pmatrix}$

[soft zealot]

Bungo

[solar crest]

ohhh

thanks

how do i get better at lin alg in general bruh

it's so confusing

[tiny gorge]

you get more practice

all math is confusing at first 😁

[final saddle]

@solar crest Has your question been resolved?

[final saddle]

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[gentle zephyr]

how do I calculate this partial derivative?

[gentle zephyr]

,, \pdv{\text{}}{x} \left[-2x\right] \cdot \left[x^2 + y^2 + 1\right]^{-2} = \text{??????}

[soft zealot]

Renato

[ornate knot]

consider y as a constant

and use product rule

its the same as
d/dx(g(x)*h(x))

[gentle zephyr]

yep

,w second partial derivative with respect to x of 1/(x^2+y^2 + 1)

[soft zealot]

Results provided by WolframAlpha

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

Requested by renatoxbz

[gentle zephyr]

looks good to me

[gentle zephyr]

I appreciate the help, thank you

yeah, product rule in disguise and then chain rule

.solved

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[dusk parrot]

since theyre all integer coefficients, doesnt it force the polynomial to either have all real roots or if complex, have the conjugate too as a root?

[tired walrus]

sure

but why did you select that (a+ib)^2 is never also a root?

[dusk parrot]

same reason

[tired walrus]

i do not follow your logic

[dusk parrot]

wait

[tired walrus]

the roots of a poly with real coeffs are either real or come in conjugate pairs

[dusk parrot]

you'd have two complex roots to balance

instead of one, and with n = 3

that's not possible

[tired walrus]

how exactly would it follow from this that raising a root of your polynomial to the second power always gives you a non-root

[rocky tusk]

n >= 3

[dusk parrot]

for n =3 specifically

[tired walrus]

for polynomials of degree 3?

[rocky tusk]

it doesn't say n = 3

[dusk parrot]

which means it should be valud for n =3 too

[rocky tusk]

no

[dusk parrot]

?

[tired walrus]

i can name you a polynomial of degree exactly 3 with a root of the form a+ib and with (a+ib)^2 also a root.

[worldly mesa]

if 1 is a root to your polynomial, then 1^2 = 1 is also a root contradicting B

[rocky tusk]

a, b \neq 0

[dusk parrot]

which is?

[tired walrus]

p(x) = (x^2+1)(x+1)

[dusk parrot]

not true because b =/=0

[worldly mesa]

its not written, even in that case just take a cyclotomic polynomial

[tired walrus]

it has i as a root,

and it also has i^2 a.k.a. -1 as a root

[dusk parrot]

that.... is a conjugate

+-i

[tired walrus]

ok and

so what

[rocky tusk]

what

[dusk parrot]

its talking about only one complex solution

[rocky tusk]

it says a and b are nonzero real numbers

[dusk parrot]

it is given that a, b =/= 0

[tired walrus]

it's talking about only one complex solution but B does not say it is the only one

[tired walrus]

it is not even given

you are imagining it

[worldly mesa]

oh right

[dusk parrot]

first sentence says theyre non zero....?

[tired walrus]

oh no yeah ok sorry you're right

i couldnt read

okay so a, b ∈ R \ {0}

[rocky tusk]

lol

[tired walrus]

i still have one up my sleeve though

[worldly mesa]

1+x+x^2

[tired walrus]

$p(x) = (1+x+x^2)(x-7)$

[rocky tusk]

n >= 3

[soft zealot]

Ann

[dusk parrot]

isnt that also +-i

[worldly mesa]

no

[tired walrus]

this one has degree 3

it has a root equal to -1/2 + sqrt(3)i/2

[dusk parrot]

complex number with its conjugate

[worldly mesa]

its has roots of unity as roots, which are not +-1, +-i

[dusk parrot]

its roots are conjugates...

[tired walrus]

yes

and as it happens

the SPECIFIC conjugate pair of roots that i picked just HAPPENS to be each other's squares!

[dusk parrot]

which is what im saying

it's nessecarily a conjugate pair

[tired walrus]

and nobody is contradicting you on that

nobody said that the roots of our poly didnt come in conjugate pairs

[rocky tusk]

you can have (a + ib)^2 = conjugate of (a + ib)

[tired walrus]

they do

they do!!

[dusk parrot]

(a+ib)*(a-ib) =/= (a+ib)^2 for a, b =/= 0

is what im saying

[dusk parrot]

if b was 0 yes

but not allowed

[tired walrus]

ok i think you're either trying to over-generalize or like

overthink somehow

let's backtrack shall we

or maybe you would like to issue a refusal and tell me to disappear

know that that's always an option

if at ANY POINT WHATSOEVER you wish to tell me to fuck off then tell me to fuck off

[dusk parrot]

i cant think of a counter example

[tired walrus]

are we understood thus far

[dusk parrot]

sure

[tired walrus]

ok

so you have this theorem which states

if polynomial p has real coeffs and some complex number z is a root of p then conj(z) is a root of p as well.

are we on the same page thus far

[worldly mesa]

as I said earlier 1+x+x^2 is a counterexample

[rocky tusk]

$(a + ib)^2 = a - ib \iff a^2 - b^2 + 2abi = a - ib \implies a^2 - b^2 = a, \quad 2ab = -b$ and from here since $b \neq 0$ you get $a = -\frac{1}{2}$ and so $b = \pm\frac{\sqrt{3}}{2}$

[dusk parrot]

doesnt work because pair of conjugates

[tired walrus]

ok let me handle this y'all

[dusk parrot]

oh shit

[tired walrus]

@dusk parrot are we in agreement so far

[soft zealot]

knief

[dusk parrot]

im an idiot i was thinking of a^2 as a 💀

youre right theres nothing stopping (a+ib)^2 from being conjugate of a+ib 💀

[tired walrus]

Let a polynomial p have real coefficients, and let z be a root of p.

A) We can SAY FOR CERTAIN that z^2 ISN'T a root of p, because z^2 isn't equal to conj(z).
B) We CAN'T SAY what happens with z^2 as to whether it's a root of p. It could be another root of p by sheer coincidence, or it could also not be -- the theorem doesn't say.

what it sounded like to me is that you were asserting (A) when in fact (A) is false.

and the true statement among these two is (B).

[rocky tusk]

i just don't know why you didn't think D was correct

🤔

[dusk parrot]

tbh the questions are supposed to have singular solutions

so

[rocky tusk]

yea but like D is obviously true

[tired walrus]

@dusk parrot do you agree or disagree w this

[dusk parrot]

i didnt read D first

[dusk parrot]

hmm fair enough i guess

[tired walrus]

what part remains unclear or doubtful?

[dusk parrot]

nothing i made a stupid while solving the question

[tired walrus]

yeah but for some reason you didnt give me a clear "i agree with you" or "i disagree with you"

i'd like to know why

[dusk parrot]

i was doing THIS

but

my idiotic brain decided a^2 = a

😄

and hence the assertion that b must be 0

[worldly mesa]

just saying there are more counterexamples then those that (a+bi)^2 = a+bi

[dusk parrot]

this is the case where the statement is false, well that it has to be the conjugate specifically

[worldly mesa]

take 1 + x + x^2 + x^3 + x^4

[dusk parrot]

its okay, my problem's solved

[worldly mesa]

and in general $\sum_{n=0}^{p-1} x^n$ for an odd prime p

[dusk parrot]

never seen this tho

[soft zealot]

ExpertEsquieESQUIE

[worldly mesa]

these are cyclotomic polynomials

[dusk parrot]

im pretty sure this is above hs maths tho

[worldly mesa]

yeah, but you can work out some specific cases just by simple calculations

notice for $p = 5$ the roots are of the form $e^{\frac{2 \pi i k}{5}}$ for k=1,2,3,4

[soft zealot]

ExpertEsquieESQUIE

[dusk parrot]

in polar coords?

[worldly mesa]

yeah

[dusk parrot]

then what?

[worldly mesa]

take $e^{\frac{2 \pi i}{5}}$, squaring it we get $e^{\frac{4 \pi i}{5}}$ which is also a root

[soft zealot]

ExpertEsquieESQUIE

[dusk parrot]

ah getting back to the same thing but with polar coords

ill keep it in mind, maybe it'll be useful somewhere else then

thanks ❤️

.stop

[worldly mesa]

its .close

.close

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[fading pumice]

can someone explain epsilon delta

[gleaming jetty]

which epsilon delta?

[hybrid heath]

Limits, i assume

[gleaming jetty]

i know but there are like 4-5 definitions that use epsilon and delta

the continuity limits sequences idk

[fading pumice]

yes

[gritty drift]

https://math.libretexts.org/Bookshelves/Calculus/Calculus_3e_(Apex)/01%3A_Limits/1.02%3A_Epsilon-Delta_Definition_of_a_Limit

Read through that and describe any parts you don't understand. You can't expect anyone to tutor you on a whole topic from scratch.

[fading pumice]

but like why

why is this necessary

[tulip coyote]

(why is the epsilon-delta definition of a limit necessary?)

[final saddle]

@fading pumice Has your question been resolved?

[final saddle]

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[radiant carbon]

hey

[radiant carbon]

Not sure where I went wrong for question 95

the actual answer has a -105/2 instead of -175/2

[west silo]

Not sure what I'm doing wrong, I need help

[dusk parrot]

have you tried U sub>

wait stupid question you already sent your work 💀

[celest crane]

!help

[final saddle]

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

[radiant carbon]

yh

i see it

my 4/3 turned into 4/5

lol

bad handwriting

[dusk parrot]

i was gonna say

xD

that the answer is in fact correct(the given one)

[radiant carbon]

yeah

.close

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[austere elbow]

I am wondering if anyone would be able to help me with "differential equations as models" which is section 1.3 in my textbook

[tiny gorge]

do you have a specific question or problem you're working on?

[austere elbow]

just looking to work through a problem with someone. I can find a specific problem. one second

[tiny gorge]

sure

[austere elbow]

i can try and work through this one on my own first and ill share what i come out to

[tiny gorge]

ok

[austere elbow]

i understand that A(0) = 50 and dA/dt = salt / time but i dont exactly understand how to write it into a diff eq as it gives me an initial volume.
Just not exactly sure how that translates into an equation

and it doesnt tell us the max volume, which also confuses me

nevermind, i think i got it. dA/dt = - A/100

.close

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[severe verge]

find A^n

[severe verge]

A^2= $\begin{bmatrix}
a^2 & 2a & 1\
0 & a^2 & 2a\
0 & 0 & a^2
\end{bmatrix}$

A^3= $\begin{bmatrix}
a^3 & 3a^2 & 3a\
0 & a^3 & 3a^2\
0 & 0 & a^3
\end{bmatrix}$

[soft zealot]

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

[severe verge]

some obvious patterns here: the main diagonal is a^n and A^n is always an upper triangular matrix

dont you tell me it's the first 3 terms in (a+1)^n expansion

[cursive bough]

are you expected to use induction? if so i'd keep taking powers, make a guess, and try to prove it

diagonalization is also an option if you've learned that

[severe verge]

what is a diagonalization

[cursive bough]

a method used to raise matrices to a power consistently. you need to find 3 independent eigenvectors though

since it's a 3x3 matrix

[severe verge]

we're only introduced to matrices so i dont know what eigenvectors and stuff are

[cursive bough]

then induction it is lol, which is probably easier anyway

[cursive bough]

but your guess here seems plausible

[severe verge]

one thing i see is that A = a*E + Y

where Y = $\begin{bmatrix}
0 & 1 & 0\
0 & 0 & 1\
0 & 0 & 0
\end{bmatrix}$

[soft zealot]

(De)Carbonized

[severe verge]

and then (aE)Y = Y(aE)

so A^n = (aE + Y)^n

fk my phone ran out of battery, I will hop on laptop later

[final saddle]

@severe verge Has your question been resolved?

[severe verge]

since this is commutative for matrix multiplication i think we can do binomial expansion

(aE)^n is just a^n *E

Y^2 = $\begin{bmatrix}
0 & 0 & 1\
0 & 0 & 0\
0 & 0 & 0
\end{bmatrix}$

[soft zealot]

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

[severe verge]

Y^3 and further is just a 0 matrix

so our expansion only has 3 terms

A^n= $\begin{bmatrix}
a^n & n \cdot a^{n-1} & nC2 \cdot a^{n-2} \
0 & a^n & n \cdot a^{n-1} \
0 & 0 & a^n
\end{bmatrix}$

[soft zealot]

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

[severe verge]

seems consistent with my results

induction is still prob faster lmao

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[compact socket]

The x-intercept is 2. The y-intercept is −6. The
function is increasing for all values of x. The
function is positive for x > 2. As x → +∞,
f(x) → +∞ and as x → −∞, f(x) → −∞.

what does x>2 mean?

[rocky tusk]

the function is positive when x is greater than 2

[compact socket]

ohh I got it now

I see okay thank you

[drifting plank]

Visual representation

Off scale

[compact socket]

the answer is linear right?

just making sure

[drifting plank]

Oh I thought it was a cubic function

Mb this would be linear:

[compact socket]

okay thats what I got

thanks so much sam

[drifting plank]

Yea sorry np

😅

[compact socket]

all good, have a good one

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[whole halo]

quickly want to make sure: [1, 1] is not necessarily the same as [i, i], right

[whole halo]

I havent needed [1, 1] = [i, i] when proving that <x + z, y> = <x, y> + <z, y>, that <x, y> = conjugate of <y, x>, and <x, x> ≠ 0 if x ≠ 0
also, the definition of [., .] seems to mirror the dot product in R^2, and that dot product can be defined as (a, b) * (c, d) = ac + 2bd and still fit the necessary axioms

only seem to need it when proving that <cx, y> = c <x, y> and there isnt a way around that

itd be a typo on the problem's part to leave out that [1, 1] = [i, i] for this purpose

[final saddle]

@whole halo Has your question been resolved?

[odd seal]

unless I am horribly confused and misremember,
[i,i] is real. And we can write [i,i] = i[1,i].
This means [1,i] is purely imaginary.
That makes [1,i] = - [i,1], or,
[i,i] = -i[i,1] = -i^2 [1,1] = [1,1]

[whole halo]

[1, i] is 0, so no

please read the question closely

[odd seal]

oof

[umbral hamlet]

what is 1?

are you fixing an element of V?

[whole halo]

the multiplicative identity of both C and R, so the identity of V

1v = v, that 1

[umbral hamlet]

but it wont belong to V

it belongs to the field but it isnt going to be in the vector space

[whole halo]

I think I gotta review how Im using the word identity here, that was very messy to say

[whole halo]

1 is the real number 1 I mean

[umbral hamlet]

right

[whole halo]

and i is the complex number i

[umbral hamlet]

but V is not R or C

V is some abstract vector space

[whole halo]

our assumption in the course allows us to say that V is R or C

this is for example there in conj(<x, y>) = <y, x>

thats given as an axiom for a vector space in this course

[umbral hamlet]

i really doubt that V is R or C, because then you are only dealing with rank 1 or 2 vector spaces