help-33

[marsh citrus]

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[fallow pewter]

,tex dom A = { x \colon \exists y \ni (x , y) \in A }

[elfin berry]

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

[fallow pewter]

what does the reverse epsilon mean here ?

[brave marsh]

No nvm sorry I was wrong

Think it's just "such that"in that case

[fallow pewter]

why not use : ?

[brave marsh]

Makes sense with the definition of Dom, it's the set of x values that map to a y

Must be just to avoid confusion with the previous :

[fallow pewter]

oh i see

is there another way to re-write the same ?

[marsh citrus]

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[brave marsh]

That would be the simplest way to write it I guess

[brave marsh]

.close

[marsh citrus]

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[tulip folio]

quick linear algebra question

[distant peak]

didnt we show you how to do it ?

[tulip folio]

The method doesn't work

That was shown

[distant peak]

???

[tulip folio]

yes

[distant peak]

its like the only method to find eigenvectors

:/

[tulip folio]

I did T - 3I_3

let's called that Z

then did ZX= zero vector

and solved for X

got no solution

so something is wrong here...

[distant peak]

its not wrong you didnt do it well

[tulip folio]

I double checked with wolfram alpha

my calculations were not wrong

[distant peak]

$\begin{pmatrix} 6 & 6 & 6 \ -4 & -4 & -4 \ -4 & -4 & -4 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix}$

[elfin berry]

Ｈｅｒｅｌｓ

[tulip folio]

yes

[distant peak]

we have two rows that are linear dependents, welp

$x_1 + x_2 + x_3 = 0$

[elfin berry]

Ｈｅｒｅｌｓ

[distant peak]

$x_3 = -x_1 - x_2$

[elfin berry]

Ｈｅｒｅｌｓ

[distant peak]

$\begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = \begin{pmatrix} x_1 \ x_2 \ -x_1 - x_2 \end{pmatrix}$

[elfin berry]

Ｈｅｒｅｌｓ

[distant peak]

$$=x_1 \begin{pmatrix} 1 \ 0 \ -1 \end{pmatrix} + x_2 \begin{pmatrix} 0 \ 1 \ -1 \end{pmatrix}$$

[tulip folio]

yeah I see where I messed up

yep

ty

[elfin berry]

Ｈｅｒｅｌｓ

[tulip folio]

I had 1 more question if u don't mind

[distant peak]

here are your eigenvectors

[tulip folio]

it's super quick

sorry

is it not diagonizable

Do u agree?

[distant peak]

can you show me the characteristic polynomial ?

[tulip folio]

sure

[distant peak]

it is diagonalizable

[tulip folio]

What makes u say that?

[distant peak]

for each eigenvalue, the algebraic multiplicity is equal to the dimension of their eigenspace

Thats why I recommand you to read your course before doing exercises

[tulip folio]

ok yeah

ye I thought I could do this without reading the notes lol

[distant peak]

If you dont learn your lessons you are kinda dead in linear algebra

[tulip folio]

it seems easy tho once I read the notes

[distant peak]

its very straightforwaded

[tulip folio]

yea

sounds good

ty brother

appreciate it

[distant peak]

np

[tulip folio]

.close

[marsh citrus]

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[marsh citrus]

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[copper drum]

Compute the Dirichlet density of the set of primes p such that p - 1, p + 2 and p + 3 are squares modulo p

[copper drum]

I don't even know where to start here

[half token]

Help me i am ponos

[copper drum]

What?

<@&286206848099549185>

[marsh citrus]

@copper drum Has your question been resolved?

[copper drum]

<@&286206848099549185>

[elder ridge]

Rewrite these conditions using legendre symbols

Recall the distribution of quadratic residues too

[copper drum]

So, correct me if I'm wrong. but I've always got confused about the significance of Legendre symbols; I know how to work with them, manipulate them, and the Quadratic reciprocity theorem; but I don't actually really get what it says, i.e. I know how to compute its value; but I don't really know what a legendre symbol tells us exactly

or really what a quadratic residue is

If I would have to guess, I think it says that, using it here; it tells us that ((p-1)/p) = 1, ((p+2)/p) = 1, and ((p+3)/p) = 1

Although, I guess it could be the case that the last two could be 0

Say if p = 2, then this would satisfy the conditions as 1, 4 and 5 are all squares mod 2 (as 5 = 3 = 1 (mod 2))

So ((p+2)/p) I guess could equal 0

Oh, so I guess here the whole point is we're trying to compute these values or whatever

((p-1)/p) = (-1/p)

((p+2)/p) = (2/p)

((p+3)/p) = (3/p)

I guess, we wanna know first for what values of p is (p-1)/p an integer

So, suppose it equals an integer, k, then (p-1)/p = k, this would imply p - 1 = kp iff p - kp = 1 = p(1-k) iff p = 1/(1-k), we need p at least 2; so k must be negative, the only value of which would give an integer p would be k = -2, which gives p = 1 which is NOT prime,all others give non-rational 1/(1-k), so therefore p doesn't divide p-1 here, so we have (-1/p) = 1 which means that p must be congruent to 1 modulo 4

<@&286206848099549185>

[copper drum]

I guess an easier way of noting this is just the fact that (p-1)/p = 1 - 1/p, which is an integer if an only if p = 1 or p = -1, which are neither prime

Similarly, p for which (p+2)/p is an integer, (p+2)/p = 1 + 2/p which is an integer if p = -2,-1,1 or 2; but 2 is prime so this works

So it could be 0

But wait actually, if p = 1 (mod 4) then p = 1 (mod 8) (I think)?? So ((p+2)/p) must be 1, but I guess since we know that p = 1 (mod 8) anyway this doesn't really tell us anything new

We now want p for which (p+3)/p is an integer

(p+3)/p = 1 + 3/p which is an integer if an only if p = -3,-1,1,3; so since p = 3 is prime we need to check this I guess

p + 3 is obviously a quadratic residue mod p

As we have 3 = 6 = 9 (mod 3)

Wait, the whole condition for ((p+3)/p) to be 1 is for p = 1 (mod 4), which we already know

[elder ridge]

this isn't right

[copper drum]

Oh damn that's not true wait

[elder ridge]

You shouldn't think of the legendre symbol as a fraction

the fraction is purely notational

Basically you're looking for primes such that the following is true:

[copper drum]

No, because (a/p) = 0 if p divides a

so you need to show p doesn't divide a

[elder ridge]

But this is obvious

[copper drum]

For it to be 1, it can't be -1 since we have it's a quadratic residue

[elder ridge]

Assume WLOG p>3 then you're alright

[copper drum]

I don't think it matters though

[elder ridge]

$$\legendre{a}{p}$$

[elfin berry]

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

[elder ridge]

does this work

nvm

[copper drum]

I've used the fact that p can't divide p -1 to show definitively that p is congruent to 1 (mod 4)

[elder ridge]

Note that $$\left(\frac{a+kp}{p}\right) = \left(\frac{a}{p}\right)$$ for k an integer

[elfin berry]

Jeeves

[elder ridge]

what do you mean here

p can't divide p-1 by definition

or do you mean $$\left(\frac{-1}{p}\right) = 1 \iff p \equiv 1 (\textrm{mod} 4)$$

[elfin berry]

Jeeves

[elder ridge]

which is true

The idea is that you can transform these Legendre conditions into congruence conditions

and then use the Chinese remainder theorem to get a single congruence condition

from then you're done

[copper drum]

Yeah that's what I've used, ((p-1)/p) = (-1/p) which is 1 when p = 1 (mod 4), and this is true since we know that ((p-1)/p) = 1 since p doesn't divide p -1 and we know p - 1 is a quadratic residue or whatever

[elder ridge]

You're just solving the simultaneous equations $$\left(\frac{-1}{p}\right) = \left(\frac{2}{p}\right) = \left(\frac{3}{p}\right)=1$$

[elfin berry]

Jeeves

[copper drum]

Yes

[elder ridge]

which will give you a condition mod 4, 8, and 12

then combine using the CRT more or less

[copper drum]

We don't really gain any more information from the = (2/p) though I think, because this equals 1 when p = 1 (mod 8), which is obviously true because p = 1 (mod 4)

[elder ridge]

exactly but you need to check this

[copper drum]

Right?

[elder ridge]

and when p = 7 (mod 8) but ofc this can be discounted

[elder ridge]

we do gain information my bad

we deduce that $$p \equiv 1 (\textrm{mod } 8)$$ of course

[elfin berry]

Jeeves

[copper drum]

But this is the same as saying p is congruent to 1 (mod 4)?

OH WAIT

It's not

it's stricter

5 = 1 (mod 4) but 5 doesn't equal 1 mod 8

So we eliminate some primes

[elder ridge]

Do the same for the last case to deduce that but I think there's a much faster way

[copper drum]

Quick question, when we know that (2/p) = 1 iff p = +- 1 (mod 8), I'm confused what that's saying, is that an p = 1 (mod 8) or p = -1 (mod 8) because surely it can't equal both right?

[elder ridge]

yes

[copper drum]

Okay, so we know that p = 1 (mod 8) so far

Wow, this is actually really fun maths lol

I'm tired of proofs this is actually like problem solvy lol

Um, so we have (3/p) = 1

[elder ridge]

This is a little computational

If you want to do a programming project on stuff like this you can try this:

https://www.maths.cam.ac.uk/undergrad/catam/II/15pt6.pdf

[elder ridge]

this is true when $$p \equiv \pm 1 (\textrm{mod } 12)$$ by Gauss' lemma

[elfin berry]

Jeeves

[copper drum]

idrk (3/p) = 1 tells us, obviously since p = 1 (mod 4), we also have (p/3) = 1 mod 4

[elder ridge]

Clearly discard -1, +1 gives us that $$p \equiv 1 (\textrm{mod } 24)$$

[elfin berry]

Jeeves

[elder ridge]

So you're done

[copper drum]

Can you explain this, my version of Gauss' lemma is Suppose $p \nmid a.$ Let $\mu$ be the number of least residues of the elements in ${a,2a,3a,...,\frac{p-1}{2}a}$ that are negative, then $(\frac{a}{p}) = (-1)^{\mu}$

[elfin berry]

LeftySam

[elder ridge]

yeah

[copper drum]

How does this tell us that p = 1 (mod 12)?

[elder ridge]

Try it yourself, it really isn't that hard

and it tells us that $p \equiv \pm 1 (\textrm{mod } 12)$

[elfin berry]

Jeeves

[copper drum]

Oh, I think I have the result in my notes "As an example, we can take a = 3. The above shows that 3 is a quadratic residue modulo p if and only if either
p = +1 (mod 3) and p = 1 (mod 4) or
p = -1 (mod 3) and p = -1 (mod 4). That
is equivalent to either p = 1 (mod 12) or p = - 1 (mod 12) by the
Chinese remainder theorem"

That's the result right?

[elder ridge]

yeah

this isn't really a proof but it's correct

[copper drum]

Yeah it's just an example so worth remembering
So we have p = 1 (mod 12) too, and p = 1 (mod 8)
Surely we can't use CRT here because (12,8) = 4?

[elder ridge]

you can do this by inspection

note the solution will be modlo lcm(12,8) = 24

[copper drum]

Ah okay, so p = 1 (mod 24)

So it's all primes of form p = 1 + 24k for integer k

[elder ridge]

yes

[copper drum]

and the dirichlet density of this is just gonna be 1/phi(24) which is 1/8?
Why did you say it was 0 earlier I'm confused

[elder ridge]

yeah I messed up earlier

I'm actually not familiar with dirichlet density

but I could guess what it was

[copper drum]

So 1/8th is right do you think?

[copper drum]

Yeah as far as I'm understand it's just a method by which you determine how big the set of a subset of the prime numbers is

[elder ridge]

yeah I'm sure

looks like it relies on a lot of stuff you probably haven't learnt though so I'm wondering why you're working with it

[copper drum]

How so?

We've done like L-functions, dirichlet densities, dirichlet characters etc

[elder ridge]

It looks like it relies on Dirichlet's theorem for primes in arithmetic progressions

oh have you?

nevermind then

[copper drum]

Yeah, this is what I used to get 1/phi(24)

[elder ridge]

Yeah I get that

You know way more about algebraic number theory than elementary number theory then lol

[copper drum]

This could be because I'm failing to recall some results we did last year because last year I stopped going to lectures and did really bad lol

[elder ridge]

it happens

fair enough

[copper drum]

Hurt me pretty bad, got 22% on a stats module lol

[elder ridge]

ow

this is why I don't do stats

[copper drum]

a 3 hour exam the morning after a 3 hour afternoon exam lol I walked out after an hour, I averaged 40% for the year and had to retake modules. Not the best, also suffered in my jan 3rd year exams because although I did better (50%) I struggled because I had to focus on my group project

Only need 75% average to get a 2:1 now but I'm doing really with these modules so hopefully doable

[elder ridge]

Oh I thought you were in the US with all those percentages haha

I mean that seems fair - I did similarly bad in my first year

Yeah it looks like the specialisation is working out

[copper drum]

The 40% was my second year lol, I actually did great first year, I remembered getting 93% on a module lol

But exams were online so it's different

[elder ridge]

Last time I got over 90% was in school 😦

or anywhere near tbh

[copper drum]

and last year I had a bit of a mental breakdown (weirdly, unrelated to uni) so I basically was left with a week to learn a year's worth of content, most modules I would have like an afternoon to learn from scratch lmao; so the fact I passed any I'm taking as a W lol

[copper drum]

what uni do you go to, if u don't mind me asking; saw you link a Cambridge thing so Cambridge?

[elder ridge]

yeah

[copper drum]

Yeah I mean no wonder you don't get 90% it's literally Cambridge lol

[elder ridge]

Would be nice though

[copper drum]

You getting a 2:1 from Cambridge is gonna be more impressive than a first from where I'm at tbh

Except if ur at like idk UCL, LSE etc.

I'm at Southampton so it's russell group but like decent

[elder ridge]

Russell group doesn't matter

at least for maths

[copper drum]

I think it does to an extent

Like certain non-Russell groups are good

[elder ridge]

It's an arbitrary cutoff point - if you're looking at course quality/performance standards there are many better points to draw a line

but none of it matters anyway

[copper drum]

Yeah, I mean I picked the wrong degree I think; I might end up doing another ug in econ because that's what I wanna end up doing, so 2:1 would be huge because it'll mean I can convert from maths ug to econ masters

Anyway

Thanks for the help.

.close

[marsh citrus]

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[elder ridge]

lol fair enough

no worries

[marsh citrus]

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[marsh citrus]

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[fierce pewter]

I'm confused on how to start

[marsh citrus]

@fierce pewter Has your question been resolved?

[fierce pewter]

.close

[marsh citrus]

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[hollow cliff]

given an equilateral triangle with side length 40, find the area of the triangle to the nearest tenth

[hollow cliff]

Would I just do 1/2(40)(40)

[marsh citrus]

@hollow cliff Has your question been resolved?

[hollow cliff]

what would i do from there

[tropic kindle]

have you learnt pythagoras theorem

[hollow cliff]

if its an equilateral triangle i suppose all sides are 40

[hollow cliff]

i havent

[tropic kindle]

oh

[hollow cliff]

any suggestions?

[tropic kindle]

what about the formula to calculate equilateral triangle areas

have you learnt anything like that

[hollow cliff]

tell me the formulka

and ill lyk if i know it

[hollow cliff]

no

[tropic kindle]

what about trigonometry? sin cos tan?

[hollow cliff]

i do know thaty

[tropic kindle]

do you know how all angles in equilateral triangles are 60 degrees

[hollow cliff]

yes

[tropic kindle]

how do you find h

its 40, not 90

[hollow cliff]

i actually do not know

[tropic kindle]

what is sin60

[hollow cliff]

1/2

sin 60

i think its

(radical 3)/2

[tropic kindle]

I mean in the triangle I showed you

[hollow cliff]

sin = opp/hyp

i believe its h/40

[tropic kindle]

yes

$\sin60^{\circ} = \frac{h}{40}$

[elfin berry]

Tangerine

[tropic kindle]

you can then find h

[hollow cliff]

34.64

[tropic kindle]

exact answer

$sin60 = \frac{\sqrt{3}}{2}$

[hollow cliff]

would i do 40 times that divided by 2

[elfin berry]

Tangerine

[tropic kindle]

yes, but its better to express h in a more accurate way to reduce error

h should be $(40)(\frac{\sqrt{3}}{2}) = 20\sqrt{3}$

[hollow cliff]

i see

[elfin berry]

Tangerine

[tropic kindle]

then use that to calculate the area

[hollow cliff]

20 radical 3) (40)

[tropic kindle]

it should be 692.8, corrected to the nearest tenth

[hollow cliff]

/2

yesd

it makes a lot more sense now

thank you

[tropic kindle]

npnp

[hollow cliff]

.close

[marsh citrus]

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[still temple]

Hello

[marsh citrus]

Please don't occupy multiple help channels.

[late hare]

@still temple go the the previous channel you said hello, ask your question there and here do .close

[still temple]

My previous channel for some reason doesn’t show

[tall umbra]

click this shi bro

it'll show

[still temple]

.close

[marsh citrus]

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[wooden grotto]

not sure what I'm doing wrong here

[wooden grotto]

I wrote down in my notes for the process to get this answer as follows:

we separate the variables and integrate both sides as follows:

∫(1/e^(2u)) du = ∫e^(4t) dt

To integrate the left-hand side, we use the substitution v = 2u, dv = 2 du, which gives:

(1/2) ∫(1/e^v) dv = -(1/2) e^(-2u)

Integrating the right-hand side, we get:

∫e^(4t) dt = (1/4) e^(4t)

Substituting these results back into the original equation and simplifying, we get:

-(1/2) e^(-2u) = (1/4) e^(4t) + C

To find the value of C, we can use the initial condition u(0) = 9. Substituting t = 0 and u = 9 into the above equation, we get:

-(1/2) e^(-18) = (1/4) + C

Solving for C, we get:

C = -(1/2) e^(-18) - (1/4)

Substituting this value of C back into the above equation, we get:

-(1/2) e^(-2u) = (1/4) e^(4t) - (1/2) e^(-18) - (1/4)

Simplifying and solving for u, we get:u(t) = -(1/2) ln[2(e^(4t) - e^(-18) - 1)]

yet somehow, it's wrong..

[cobalt sentinel]

$- \frac 12 e^{-2u} = \frac {e^{4t}}{4} - \frac {1}{2e^{18}} - \frac 14$

[elfin berry]

Stephen

[cobalt sentinel]

@wooden grotto wanna go thru it step by step after the line above?

Wait a second, Lemme try it out myself ig

Ok I’m not seeing how u got ur answer

[cobalt sentinel]

Could u show the steps after this @wooden grotto

[marsh citrus]

@wooden grotto Has your question been resolved?

[marsh citrus]

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[still temple]

so im doing this problem im getting 1/sqrt3x for my answer

but the answer key says this is the answer

[stoic saddle]

y = (1/sqrt(3))x, you mean?

[still temple]

yea

[stoic saddle]

yeah there's a caveat

show the book's answer

[still temple]

k

[stoic saddle]

sqrt(3)/3 and 1/sqrt(3) are one and the same.

your teacher's three is a bit wonky.

[still temple]

oh so i did it right

[stoic saddle]

yes

[still temple]

where does he even get a three from though

[stoic saddle]

wdym

are you asking how 1/sqrt(3) becomes sqrt(3)/3?

[still temple]

so to get my answer i did this

and multiplied by reciprocal

to get 1/sqrt(3)x

[stoic saddle]

yes and ive been trying to tell you that your answer is correct

[still temple]

so where did the teacher get the three from ?

nvm

they equal one another so its the same thing just an extra step thank you

he rationalized the denominator

anyways thanks

.close

[marsh citrus]

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[woven harness]

can someone explain this please

[marsh citrus]

@woven harness Has your question been resolved?

[woven harness]

<@&286206848099549185>

[woven harness]

.close

[marsh citrus]

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[tribal jungle]

I seem to be off by exactly 1. I can’t really spot where I went wrong though. Maybe while reindexing the sum, or using the evenness?

[marsh citrus]

@tribal jungle Has your question been resolved?

[marsh citrus]

@tribal jungle Has your question been resolved?

[marsh citrus]

@tribal jungle Has your question been resolved?

[marsh citrus]

@tribal jungle Has your question been resolved?

[marsh citrus]

@tribal jungle Has your question been resolved?

[marsh citrus]

@tribal jungle Has your question been resolved?

[wild ginkgo]

i can't read your image tho

[tribal jungle]

Mh is my writing really that bad?

It's a discrete time signal

$x[n] = a^{|n|} cos(2 \pi f_0 n T)$

[elfin berry]

Learath2

[tribal jungle]

$a = - 0.8$ and $f_0 T = 1/12$ find the energy of the signal $E{x[n]} = \sum_{n = - \infty}^{+ \infty} |x[n]|^2$

[elfin berry]

Learath2

[odd valley]

The error is with this step since the sum starts from n=1, the usual result for a geometric series starts from n=0:
$$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$
so you should have:
$$\sum_{n=1}^\infty \qty(\frac{16}{25})^n = \frac{1}{1-\frac{16}{25}}-1$$
where the -1 will cancel the first term, hopefully solving your discrepancy

[elfin berry]

Tymelord14

[tribal jungle]

God, that is trivial. I was just so focused on how I broke up the second sum, I didn't even think of that. Thank you so much

[odd valley]

No problem, these kinds of errors are the worst to parse after a long calculation 🙂

[tribal jungle]

.close

[marsh citrus]

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[hazy swan]

why does this equal 4 instead of -1

[knotty trellis]

why should it equal -1?

Did you do 4-(2.5+2.5)?

[hazy swan]

because 2.5 + 2.5 is 5

yea

[knotty trellis]

yes, but you have to do it from left side to right side

so 4-2.5 is 1.5. Then 1.5+2.5 is 4 so it's 4

[sleek lake]

pemdas is a troll acronym

it's not the convention, no one actually does that and it leads to this problem if you do

[hazy swan]

so should I trust left to right or pemdas

[knotty trellis]

Pemdas says that addition and subtraction are on same level

[sleek lake]

if you wanted you could do pedmsa

at least it's the right answer, although it's not what people do either

[hazy swan]

if you do pemdas its -1

[knotty trellis]

1. parenthesis
2. exponents
3. multiplication and division
4. addition and subtraction

[dire spire]

im curious to know how you got -1

[knotty trellis]

in pemdas, addition comes first. So they first added 2.5+2.5 and then did 1-5

[knotty trellis]

this is how it works btw

[dire spire]

but isnt it -2.5+2.5

[mighty solstice]

Even if you use pemdas incorrectly like that the negative sign stays with the first 2.5 so -2.5+2.5

[stoic saddle]

addition and subtraction are on the same tier

[hazy swan]

but then wouldnt it be 0 times 4

[stoic saddle]

so you read and perform them left to right

[sleek lake]

no?

[stoic saddle]

where do you see a multiplication sign

[stoic saddle]

he's saying that you can rewrite the expression as 4 + (-2.5) + 2.5

[hazy swan]

becaause if you are saying the negative sign stays with the 2.5 that means you have to multiply the 4 and 0

[dire spire]

multiply??

[stoic saddle]

???

[dire spire]

lmao

[stoic saddle]

no, there's no multiplication sign in here and no multiplication going on and there never was any. what are you talking about?

[mighty solstice]

0 is neither positive nor negative so either 4-0 or 4+0 will work. There’s no multiplication here

[knotty trellis]

btw are you 13 or older NubMan?

[dire spire]

but -2.5 + 2.5 still produces 0, right?

[stoic saddle]

sure does.

[dire spire]

thats what i was saying

there is no way to get -1

or multiply

nubman you need to grind harder

[late geode]

whoever is getting -1 is misinterpreting pedmas

[knotty trellis]

this is the order of operations, some operations are on the same tier

[stoic saddle]

@hazy swan are you still here?

[hazy swan]

ye

[stoic saddle]

ok, first off,

as was asked earlier, are you 13 or older?

[hazy swan]

Im older

the highest math I took was pre calc lol

[stoic saddle]

k

so you're revisiting basic arithmetic now

okay, now tell us: do you have any questions left to ask?

[hazy swan]

Not really

[stoic saddle]

ok

then you can .close this channel.

[hazy swan]

.close

[marsh citrus]

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[marsh citrus]

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[celest finch]

can someone help me solve this one please idk if im supposed to like divide or something

[still temple]

u have to use sin2x=2sinxcosx

i guess

[stoic saddle]

you don't have to, but it's a good first move.

[still temple]

after u move cos to the right side

left**

the take it as a comon factor

then solve as it equals to 0

[stoic saddle]

first off

!nosols

[marsh citrus]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[stoic saddle]

second, "move" is bad

[celest finch]

my teacher wants me to do it this way

[still temple]

why is that

[stoic saddle]

"move this to the other side" is sometimes ambiguous, and even if it isn't, leaves room for error.

[still temple]

i didnt devide

if u devide u lose answers

[celest finch]

ok so how would i rewrite it

using double angle formula

[still temple]

its sin(2x)=2sin(x)cos(x)

so

[celest finch]

right

[still temple]

6sin(x)cos(x)=cos(x)

6sinxcosx-cosx=0

cosx(6sinx-1)=0

[stoic saddle]

and i didn't say you did.

[celest finch]

why are u guys arguing

lol

[celest finch]

ok thx!

[still temple]

oh

i thought u meant that

sorry

[marsh citrus]

@celest finch Has your question been resolved?

[marsh citrus]

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[burnt heart]

Can someone help me understand why the Answer on the left is incorrect but the answer on the right IS correct?

[cobalt sentinel]

Try plugging in a point like x = 2

Is it positive or negative

[burnt heart]

Negative?

[cobalt sentinel]

There u go

The one on the left is positive at x = 2

[burnt heart]

Ohh so above the x axis is positive?

[cobalt sentinel]

y values above the x-axis are positive yes

[burnt heart]

Hmm

So I can plug in any numbers into the equation? Like 3 or -4?

And whatever it gives me means the graph will be negative or positive?

[cobalt sentinel]

Yea u can plug anything in

As long as it’s in the domain yea

[cobalt sentinel]

Yep, or 0

[burnt heart]

Ooooh okay

Tysm!

[cobalt sentinel]

Np

Nice pfp

[burnt heart]

Thanks lol

.close

[marsh citrus]

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[marsh citrus]

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[slow dock]

What am I doing wrong, the denomenator should be -x^3/3 +O(x^4)

[main idol]

Doesn't look like you multiplied and instead added

[slow dock]

Where exactly?

?

<@&286206848099549185>

[cursive leaf]

How can I help?

[slow dock]

idk where my mistake is

[slow dock]

.

[cursive leaf]

Okay, let me take a look.

It looks like instead of multiplying you added instead.

[slow dock]

where+

?

[cursive leaf]

The denominator should be -x^3/3 +O(x^4).

I can understand that.

[slow dock]

I too get that XD

[slow dock]

just where did this happen

werll

plz help

<@&286206848099549185>

<@&286206848099549185>

[quasi garden]

?

[slow dock]

.

if you can walk me throug the entire process would be better too

[quasi garden]

oh sorry im not very experienced in calculus yet
just wait for a while someone should come along

[slow dock]

sure and thanks

<@&286206848099549185>

<@&286206848099549185>

[dire spire]

don’t spam tag

[slow dock]

sorry...

[keen leaf]

i cant get ur handwriting

[slow dock]

Xd

well it is...

cosx - (1-x)e^-x = (1- x^2/2 + O(x^4)) - (1-x)(1 -x + (-x)^2/2 + (-x)^3/6 + O(x^4)

=1 - x^2/2 + O(x^4) - 1 + x - x^2/2 + x^3/6 + O(x^4) + x - x^2 + x^3/2 - x^4/6 + O(x^5)

=2x - 2x^2 + 2x^3/3 - x^4/6 + O(x^5)

$\cos x - (1-x)e^{-x} = \left(1 - \frac{x^2}{2} + O(x^4)\right) - \left(1-x\right)\left(1 - x + \frac{(-x)^2}{2} + \frac{(-x)^3}{6} + O(x^4)\right)$

[elfin berry]

Shift

[slow dock]

$1 - \frac{x^2}{2} + O(x^4) - 1 + x - \frac{x^2}{2} + \frac{x^3}{6} + O(x^4) + x - x^2 + \frac{x^3}{2} - \frac{x^4}{6} + O(x^5)$

[elfin berry]

Shift

[slow dock]

$2x - 2x^2 + \frac{2x^3}{3} - \frac{x^4}{6} + O(x^5)$

[elfin berry]

Shift

[slow dock]

so where is my problem ??? <@&286206848099549185>

[marsh citrus]

@slow dock Has your question been resolved?

[slow dock]

Well

I guess i will never find out

[summer stream]

$x^{3}$

[elfin berry]

HAMADY SAKANOKO

[summer stream]

$-\infty$

[elfin berry]

HAMADY SAKANOKO

[summer stream]

$-infty$

[elfin berry]

HAMADY SAKANOKO

[twilit grove]

stop spamming here, use #latex-testing

[slow dock]

.close

[marsh citrus]

Channel closed

Closed by @slow dock

Use .reopen if this was a mistake.

[slow dock]

.reopen

[marsh citrus]

✅

[slow dock]

.close

[marsh citrus]

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• Show your work, and if possible, explain where you are stuck.
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[still temple]

Curious about this, but is surjectivity arbitrary?

[still temple]

a function $f: A \longrightarrow B$ is said to be onto (surjective) if Range($f$) = $B$

[still temple]

Aka for every $b \in B$ there exists an $a\in A$ such that $\map f a = b$

[still temple]

but the codomain is not unique

you can define the codomain to be anything as long as you have Range($f$) $\subseteq B$

[still temple]

Well this also makes me think

are two functions with different codomains, but the same range, equal?

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

<@&286206848099549185>

[still temple]

everything you've said makes sense @still temple

[still temple]

I would say no

a function requires a domain, codomain and rule

oh hm thats really interesting

so like

even if it is like

imagine those two functions okay

<enumitem>
\

oops

[still temple]

\begin{align*} &f: \R \longrightarrow \R &&f: \R \longrightarrow \R\setminus \set{\R_{< 0}} \
&x \mapsto x^2 &&x\mapsto x^2
\end{align*}

[still temple]

sorry i had a case of "status: i found a solution but my latex wasnt rendering"

[still temple]

so anyways like

for those two functions

the domain and range are inevitably the same

but i guess the two functions themselves are not the same like you said right

[still temple]

:xd:

[still temple]

yes

[still temple]

yes, on a technicality

Im not sure when it actually matters though

mm yeah

also why not R_≥0

:p

because that would exclude 0 would it not haha

thats like

$\mathbb{R}_{\geq0}$

something u can get from the first function

since 0 is in the image of f

so

[glacial hedge]

That's just R+

[still temple]

wait

https://math.stackexchange.com/questions/871553/functions-with-different-codomain-the-same-according-to-my-book

oh

so they would be the same

so surjectivity is really just

"it depends" lmao

some bozo's thing

[still temple]

okay

so i will just ask a follow up question

because

i dont get cartesian products

[glacial hedge]

The set of throws of 2 dice is just the cartesian product of 2 sets of dice throws

[still temple]

[
f: ; \Z\by\Z \longrightarrow \Z \
(n,m) \mapsto n^2 - m^2
]

so this was an exercise for us

to figure out if it is bijective

[glacial hedge]

How is that actually defined ?

[still temple]

oh yeah i remember that

[glacial hedge]

How is n^2 = m^2 an element of Z

[still temple]

oh woops i messed up

[glacial hedge]

Ok

[still temple]

so

like one issue i have with this

first question: how is the m supposed to still exist with the mapping?

[glacial hedge]

Honestly pretty simple

[still temple]

since we are going from z^2 to z

[glacial hedge]

There exist bijections from N to N^2, why not Z^2 -> Z ?

[still temple]

yeah i am just trying to rationalise it i guess haha

[glacial hedge]

Actually since N ~ N^2 and N ~ Z, N ~ Z^2 ~ Z

[glacial hedge]

They're both countable so...

It exists, but is it this function ?

You don't have a clue ?

[still temple]

dont think i am following along particularly no

[glacial hedge]

Could it be injective ?

[fair fulcrum]

there's a bijection from the integers to the positive integers
and then you can just plug that in to the square function

because N+ <=> N+^2 is bijective

maybe I'm not understanding :(

[glacial hedge]

We weren't really asking about that

[fair fulcrum]

ah

[glacial hedge]

It's known don't worry

[still temple]

okay before we proceed

let me just ask one thing

why is it not
[\Z^2 \longrightarrow \Z^2]

[glacial hedge]

n^2 - m^2 is clearly an integer

[still temple]

oh right it is not some array

so like

if you wanted to show that it is injective

[glacial hedge]

f(2,4) = -12 for example

[still temple]

then you have to show that
[
\map f {a,b} = \map f {c,d} \implies (a,b) = (c,d)
]

somehow?

[still temple]

i am guessing?

[glacial hedge]

If you think it's injective yes

[still temple]

oh hmm alright so like

[
a^2 -b^2 = c^2 - d^2
]

[still temple]

[
(a+b)(a-b) = (c+d)(c-d)
]
But now, how does one proceed

[glacial hedge]

There are several ways

[still temple]

oh hmm

any good nudge in the right direction?

[glacial hedge]

What makes you think it's injective ?

Maybe it isn't after all

[still temple]

I don't particularly think it is, i am just trying to find it out using this

if i can achieve some contradiction

then well like, it is not injective

[glacial hedge]

Then consider both possibilities

For example, how could these two things be equal

In practice

Maybe an edge case where they're conveniently both equal to something special

[still temple]

a+b = c+d and a-b = c-d being one possibility i suppose?

[glacial hedge]

Or maybe it is injective

How could that happen?

Do you want a bigger nudge ?

[fair fulcrum]

are you trying to find that this implies (a,b) = (c,d)

[still temple]

the simplest case i can think of is for if both sides are equal to 0

[glacial hedge]

Yes

[still temple]

oh hmmm

[glacial hedge]

How do you make a product 0

[still temple]

if either one of the terms is 0

[glacial hedge]

Hence

[still temple]

So if a = -b or c = -d i suppose?

[glacial hedge]

For example

So can you give an example that shows it's not injective ?

[still temple]

an example that shows it is not injective hmm

[still temple]

i mean to find such an example it means i have to prove that this implication is indeed wrong right

oh wait

[glacial hedge]

A counterexample is enough

[still temple]

we have just done it?

[glacial hedge]

Just show such a case actually exists

And is not just an abstract construct lol

[still temple]

okay i mean like, with this in mind, you do get the first equality that f(a,b) = f(c,d), but to make this injectivity hold we have to show that (a,b) = (c,d), and by considering a = -b and c = d case, then you get (-b,b) = (c,d) but -b is never equal to c under the integers

thats my logic i hope i didnt frick it up

[glacial hedge]

Well it could

[still temple]

really?

oh i see

[fair fulcrum]

does f(4,4) = f(3,3) not disprove injectivity immediately

[glacial hedge]

But it seems you have not understood the importance//meaning of the word example

[glacial hedge]

Yes it does

[fair fulcrum]

idk why I picked 4 and 3

was thinking about pythagorean triples

[glacial hedge]

Every (n, n) and (n, -n) get sent to 0

[still temple]

ah so i guess i should've done what zfnQRZJT did

[glacial hedge]

Because it directly disproves injectivity

[still temple]

yeah i was trying to think of a general approach to it

but this is way simpler i guess

[glacial hedge]

Rather than showing you can have some abstract counter-example, just show it

[still temple]

(anyways, in general, I dont think contradiction works well in noninjective proofs. much easier to find a concrete counterexample)

[glacial hedge]

When you can at least

But that's most often the case

[still temple]

thats a good thing to know, idk why but i always feel like counterexamples or specific cases are not complete enough

[glacial hedge]

They are

[still temple]

this thinking is completely wrong i know

[glacial hedge]

"look, here's the truth, right in front of you, you can't deny it, it's right there"

[fair fulcrum]

call it an inconsistency in zfc

[still temple]

yeah haha

anyways

surjectivity

[glacial hedge]

Now if want a bit of a more challenging one, you can try and find whether f is surjective

Yeah exactly

[still temple]

alright so being onto is defined as this

but for this case

it would mean

[glacial hedge]

Though do note I'd consider that to be quite a step up in difficulty

[still temple]

there exists f(a,b) = c?

for all a,b

[fair fulcrum]

finding whether all integers are the difference of 2 squares?

[glacial hedge]

For any c

[glacial hedge]

Basically yes
But pls

[fair fulcrum]

ok I will shh

[glacial hedge]

!nosols

[marsh citrus]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[still temple]

okay i did not read that

i will just

imagine its not there

[glacial hedge]

Which I may be taking a bit too at heart rn but anyways

[glacial hedge]

There's nothing you can't read

[still temple]

oh wdym?

[glacial hedge]

Autocorrect making typos

[still temple]

wait what i said was directed at what zfnQRZJT wrote for how to do it haha

but regardless back on topic

[glacial hedge]

He just rephrased it as something not in terms of f

[still temple]

oh hmm

okay so just i dont get to confused in my thoughts,

We are trying to confirm that for every $c \in \Z$ there exists $a,b \in \Z^2$ such that $\map f{a,b} = c$

Meaning we have to prove that
[
a^2 - b^2 = c
]
holds for all integers

[glacial hedge]

That's what he said yes

[still temple]

yeah

i just rationalised it

okay

to prove this...

lets see

oh okay so we have to try to hunt a counterexample for c

well

[fair fulcrum]

a,b in z^2 is an interesting notation
I would say a, b in z

but yk valid

[still temple]

i cannot seem to understand how i should prove that

[glacial hedge]

Agreed

[still temple]

i cant come up with a good counterexample

[glacial hedge]

Try to do some theory first

Get an understanding of the problem

Then see if it works or if you have an idea of how to build a counter-example

[still temple]

'do some theory' is pretty vague haha

maybe i should come back to this later on? after i get a better understanding?

[glacial hedge]

How about you write m = n+k

Try to relate the quantities somehow

[still temple]

wait a question

[glacial hedge]

Or n = m+k if that's simpler

[fair fulcrum]

isn't that the same

[still temple]

what are those 3 variables supposed to be? haha

[glacial hedge]

k = n-m

Trying to rewrite f to something that may be simpler to understand

[still temple]

oh hmm

simpler to understand

it is more or less just the difference of squares

meaning

(a-b)(a+b)

assume k = a-b

then

k(k+2b)?

this turns into k^2 + 2kb

[glacial hedge]

wlog k >= 0
Then f(n,m) = k^2 + 2mk = k(k+2m)

[still temple]

which is just a quadratic by then, and quadratics are defined over the reals (which encompasses the integers), but they are only defined over the reals for a non-imaginary discriminant

[glacial hedge]

Hint: ||it's about integers||
||Look at some arithmetic property||
||Like parity||

[glacial hedge]

I cut it into 3 parts purposefully btw

[still temple]

okay i clicked the first hint

its about integers hmm

so we have to see if k^2 + 2mk = c holds for all integers c

[glacial hedge]

Don't you think the product is a more useful form ?

[still temple]

my mind was thinking of quadratic properties so i sticked to the first haha

but lets think about this

k(k+2m) = c

[glacial hedge]

Although if you want to plug that into the quadratic formula and manage to get something out of it, I respect that. But I doubt it

[still temple]

seems like a really convoluted way to go about it anyways

😅

[still temple]

now hmm

what do i do with this guy hm

okay i will click the second hint

arithmetic property

[glacial hedge]

Hopefully you see what I mean by that

[still temple]

i can only see distributivity being applicable there

[fair fulcrum]

2m is even

k is odd or even

[still temple]

hmm how does that end up helping me now

oh!

wait

odd + even = odd

and odd(odd) = odd

so whatever we get is going to be odd

so we can rewrite c as

[fair fulcrum]

waitwait

[still temple]

2m+1 for some integer m?

[fair fulcrum]

holdon

[glacial hedge]

Arithmetic property such as prime decomposition or parity yes

[fair fulcrum]

not if k is even

[glacial hedge]

Do we ?

[fair fulcrum]

welp

[still temple]

well

if k is even then it will be even

so i mean

[glacial hedge]

Yes

[still temple]

do we split it into cases

of either

[glacial hedge]

Yes

[still temple]

oh my god this problem is actually amazing lmao

[glacial hedge]

Do we reach every even number ?
We can do that case first

[still temple]

alright

[fair fulcrum]

how so

[glacial hedge]

Gets them to think

[fair fulcrum]

fair

[still temple]

(fun fact: any bijective function f from Z^2 to Z can prove that Q is countable)

[glacial hedge]

Indeed

[fair fulcrum]

any true statement can prove that Q is countable

/joke

[glacial hedge]

Any false statement as well

[fair fulcrum]

good point

[glacial hedge]

Also don't mind us, we're just having our own little fun

[still temple]

so assuming $k =2u \textss{for} u \in \Z$ and knowing that $k+2m = 2w+1 \quad w \ in \Z$ as well
then [
2u(2w+1) = c
]
but can we assume that $w = u$

[fair fulcrum]

I'm trying to figure htis out actually lol

[fair fulcrum]

oh z^2 is not the set of perfect squares it's the set of 2 integers

forgor

[glacial hedge]

Yes

[fair fulcrum]

that makes sense

[still temple]

wait yes to which statement mateo haha

[glacial hedge]

zfn

[glacial hedge]

What nonsense is this

[still temple]

😭 sorry

[glacial hedge]

k just became 1 out of nowhere

[fair fulcrum]

how come 2u + 2m = 2w + 1

[glacial hedge]

Exactly

k is even

[still temple]

oh sorry

yes i jsust realised

[glacial hedge]

We said it's even -> even and odd -> odd
Hence looking at 2 separate surjectivity problems

[glacial hedge]

For the parity of k ofc

m is just a bystander

[still temple]

yes i did that wrongly, sorry, so with $k(k+2m) = c$, we can assume that $k = 2u$ and $k+2m = 2v$ assuming k is even

[still temple]

that would be correct in this case i suppose?

[fair fulcrum]

it's weird to me that texit takes all the words into consideration

[glacial hedge]

It's beautiful

[still temple]

okay so in this case

we have 2u(2v) = c

meaning 4uv = c

[fair fulcrum]

true

[still temple]

wait

[glacial hedge]

Hence...

[fair fulcrum]

hence we are 50% done

[still temple]

problem is that uv is never odd, so this never can be odd

[glacial hedge]

Because k is even

[fair fulcrum]

uv can be odd just not 4uv

[glacial hedge]

.

[glacial hedge]

We know it's even. We want to see if it reaches every even number

[still temple]

yeah! because even (4) by a potential odd is even

[still temple]

i mean

this is where the counterexample comes in i suppose?

[glacial hedge]

You didn't finish the theory yet

[fair fulcrum]

well you didn't do the k odd case but you could

[glacial hedge]

You were one sentence short

[still temple]

one sentence short..

[glacial hedge]

The one coming after that

[still temple]

hence the even case of k does not hold for all integers c, but only solely for the evens?

thats the only thing i can say after that

[glacial hedge]

We know that

Do we reach all even numbers ?

We didn't do all that for nothing

[glacial hedge]

What does that tell you about the possible values of c ?

[still temple]

oh i dont think u can get to 2

[glacial hedge]

Why

[still temple]

because well c can be 0, but then there is a jump to -4 and below and 4 and above

right?

like

assume u = 1 v = 1

you get c = 4

u = -1 v = 1

you get c = -4

[glacial hedge]

In short, we only reach multiples of 4 !

[still temple]

but there is no way you can get to c=2?

[still temple]

oh my god i am so dumb haha yes

[glacial hedge]

I'm actually impressed that Toby is still here

[still temple]

so now to fully figure out if it is surjective, we need to see if the odd case can eat up the rest of the integer possibilities other than multiples of 4

if it doesnt then this isnt surjective

right?

[still temple]

im impressed any of you are still here. Honestly much appreciated because this is a big learning experience

[glacial hedge]

We already showed it isn't surjective

It doesn't reach even numbers that aren't multiples of 4

[still temple]

oh the odd case can never reach those numbers

so we dont even have to do that

[glacial hedge]

Now ofc you can take this exercise a bit further

You can ask if every odd integer is reached, and if every multiple of 4 is reached

[still temple]

hmm isnt the second pretty clear? i mean it is simply 4uv after all

[glacial hedge]

I'd say they're both fairly easy yes

[still temple]

u can just hold one of them to be 1

and then

the other to be any integer

and boom

[glacial hedge]

But also technically I thought injectivity would be easy lol

[still temple]

sorry my foundations are horrible since i haven't really built on this all that much

[glacial hedge]

We've all been there at some point

[still temple]

yeah i am trying to build on this hopefully, since i will start doing counting problems which tbh are just sets extra haha

but anyways

this concludes this incredibly stretched out help channel haha

big, big thanks to you three lads @glacial hedge @still temple @fair fulcrum

have a great day!

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[marsh citrus]

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[glacial hedge]

Note : all odd numbers are reached

Proof: exercise (easy)

[still temple]

i will try doing it now

i am actually gonna write all of this cohesively on a separate document

[marsh citrus]

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[still temple]

helps me solidify the idea much better haha

[fair fulcrum]

I love how you can write all this working out while counterexample c = 2 existed

[glacial hedge]

Yeah but you gotta prove it

Prove 2 can't be reached without doing a bunch of theory

(n+m)(n-m) = 2
n+m = 1 and n - m = 2: impossible
n+m = -1 and n - m = -2: also impossible
n+m = 2 and n - m = 1: impossible again
Last case: impossible too
Feels much less fun to me

[fair fulcrum]

true I just mean there's a mundane and small solution

.close

[marsh citrus]

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[marsh citrus]

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[signal umbra]

Linear algebra

Kinda forgot

<@&286206848099549185>

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[marsh citrus]

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[marsh citrus]

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[oak void]

I can't see where i did wrong here

I've tried to make both x = 1 and y = 1

and got different answers

so it is definitely wrong

[glacial hedge]

first line of implicit differentiation should be -2y rather than +2y

[oak void]

how come

ah

yeah yeah i get it

-(2x dy/dx +2y)

yeah

thanks

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[marsh citrus]

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[smoky marsh]

.reopen

so for number 3, after finding a basis, is there a reason that they made z=-t? it would be the same as z=t right?

[glacial hedge]

of course not, it's not the same vector space

Span((1, 1,-1)) != Span(1, 1, 1)

[smoky marsh]

but why, since its just a free variable it doesnt matter what it equals to as long as everything else is right

[glacial hedge]

no

it's vectors of the form (t, t, -t)

they're not independent

[smoky marsh]

so if z=t then my answer would be [-1,-1,1] instead of [1,1,-1]

[glacial hedge]

no that would be the same vector space

[smoky marsh]

ok, so how do i choose the right t

what step am i doing wrong.