help-49

[sharp coral]

are you familiar with o notation?

[tidal turret]

what is O notation?

I pressume not

[sharp coral]

never mind them

so in the 1 variable case, if we have the remainder for the degree two polynomial is $R_2(x)$, then $R_2$ should only have $x^3$ and higher order terms, so that
[ \lim_{x \to a} \frac{R_2(x)}{(x-a)^2} = \frac{f'''(Q)/3! \cdot (x-a)^3}{(x-a)^2} = 0 ]
this is basically saying that $R_2$ approaches $0$ ``faster'' than any quadratic

[grand pond]

κλαουντ ☁ (cloud)

[sharp coral]

we're using essentially the same limit here except we replace [ (x-a)^2 \mapsto \norm{\vec x - \vec a}^2 = \norm{(x,y) - (a,b)}^2 ]

[grand pond]

κλαουντ ☁ (cloud)

[tidal turret]

this is basically saying that $R_2$ approaches $0$ ``faster'' than any quadratic

care to elaborate on that?

and why does everyone say that R2 should only have x^3 and higher order terms?

wtf with that

[sharp coral]

well you can imagine that if the limit approaches 0, then at some point $R_2$ must be massively bigger than $(x-a)^2$ (hence their ratio approaches 0)

[grand pond]

κλαουντ ☁ (cloud)

[sharp coral]

if the taylor series converges, then R_2 is just all the terms we chopped off of the taylor series (something x^3 + something x^4 + something x^5 + ...)

[tidal turret]

if the limit approaches 0, then should it be that the numerator is smaller and the denominator is getting larger . . . I dont think I follow AT ALL

[sharp coral]

oh sorry i got it backward

(x-a)^2 gets much bigger than R_2, because R_2 is getting smaller faster

[tidal turret]

yess, why is that

??????????????????????????????????????????????????????????????????????????????????????????????????????????????///////

[sharp coral]

well if we imagine a very simple 1D case, x^3 goes to 0 much faster than x^2 (thinking of x approaching 0) because it's multiplied by another factor of x

in this remainder, we basically have a bunch of cubic terms so they're going to act like x^3, and we're comparing it to something that acts like x^2

[tidal turret]

yes, cubic wins over quadratic all the time, so It doesnt make any sense to me how is this limit is approaching 0

you see what I mean? am I tripping hard? or WTF

[sharp coral]

in the limit as $x \to \infty$, $x^3$ gets much bigger than $x^2$

in the limit as $x \to 0$, $x^3$ gets much smaller than $x^2$

[grand pond]

κλαουντ ☁ (cloud)

[tidal turret]

ohh right

that detailed I missed Lmfao

however, why does everyone say that R2 contains only sum of polynomials of deg 3

@sharp coral you mentioned something about "chopping off" care to elaborate?

[sharp coral]

well in the 1D taylor series you have
[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \underbrace{\frac{f^{(3)}(a)}{2!}(x-a)^3 + \frac{f^{(4)}(a)}{4!}(x-a)^4 + \cdots}_{R_2(x)} ]
$R_2$ is literally all the other terms past the degree 2 terms

[grand pond]

κλαουντ ☁ (cloud)

[tidal turret]

I see

what about the multivariable case

we dont have f^{(3)} anymore, is that like the hessian multiplied by a hessian multiplied by a hessian?

just saying, lagrange reminder in 1 variable is already hard, now in multivariable is even harder, at least for me

is good that we had a brief summary of how it is in the single variable scenario, but how does it generalize for multiple variables?

like, we now have gradients, hessians, and jacobians or what? instead of normal derivatives

since we have multiple directions now we have to work with partial derivatives and mixed partial derivatives

correct?

also this @sharp coral equality, is Theoretical

like we cant find the error with the taylor reminder precisely

is just theoretical, in practice if we would have an equality with the remainder, it would be because we used a calculator of some sort

like, this is why we usually find an upper bound for the lagrange remainder and not the actual reminder Himself

you know what I mean?

or am I tripping hard again? WTFFF

[sharp coral]

well if you remember in the 1D case you could get a conservative error estimate by finding the maximum value of the derivative on the interval you're interested in (since Q must be somewhere between x and a)

here you can do the same except it would be more annoying because you have to maximize four different derivatives

[tidal turret]

dont tell me that, the following exercises ask me to estime the error for some taylor polis (multivar)

hopefully is not too hard

[sharp coral]

you can't really find the exact error because if you knew the exact error that would be equivalent to knowing the exact value

[tidal turret]

you mean an upper bound for the error?

[sharp coral]

yes "conservative estimate" = upper bound

[tidal turret]

yet, iirc, the taylor theorem says that f(x) = P2(x) + R2(x) how does he do it?

[sharp coral]

well yes

[tidal turret]

he is not using an approximation

[sharp coral]

because R_2 is the exact error

[tidal turret]

in theory, yeah

[sharp coral]

but in practice you can't actually find the exact value of R_2, you just find an upper bound for R_2 and call it a day

[tidal turret]

yeah, I think I get it now, but he says its the exact error because his series is infinite?

or why is the exact error R2(x)?

[sharp coral]

that's just the definition of R_2

[tidal turret]

care to elaborate?

[sharp coral]

like that's what R_2 is defined to be, is the exact error

[tidal turret]

this taylor reminder, how is it defined?

[sharp coral]

the formula you posted above is a consequence of this definition which lagrange derived

[tidal turret]

there is another reminder with an integral

iirc, no?

[sharp coral]

the definition is just $R_2(x) = f(x) - P_2(x)$

[grand pond]

κλαουντ ☁ (cloud)

[tidal turret]

whats the definition of P2(x) ?

[sharp coral]

there are various formulas for R_2 which include the one you posted above, an integral remainder, etc

[sharp coral]

second order taylor polynomial

[tidal turret]

yes but around what point? X and is an infinite taylor series or what?

a taylor series up to 2?

[sharp coral]

any point really, just the R_2 and the P_2 are based on the same point

you can call the point a in 1D or (a,b) in 2D if you want

and P_2 is really just the taylor series formula written out to the first three terms (constant, linear, quadratic)

[tidal turret]

I am getting lost in between the actual taylor theorem and the approximation

another thing that is driving me crazy is this D^alpha. any idea what It means? the jacobian?

[tidal turret]

the thing is, for this R2(x) to continue like that, we should know f is of class C^infty, so like a polynomial function for example

usually we just know f is of class C^k or C^(k+1)

also, from the original image, we just know F is of class C^2, so how is that the remainder is using third order partial derivatives?

isnt this a typo and should have said F is of class C^3?

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

@tidal turret Has your question been resolved?

[sharp coral]

well yes this equation is only true if f is equal to its taylor series. it doesn't have to have a full taylor series to have a degree 2 taylor polynomial

[sharp coral]

f must be twice differentiable to have a second-order taylor polynomial approximation and three times differentiable for the lagrange formula for the remainder to be true

[sharp coral]

it explains what it means immediately above

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

seems kinda abstract

[tidal turret]

yeah so I pressume in the image I sent there must be a mistake or something, because otherwise we dont know if R2 is defined no?

is claiming f is C2

but using third order partials for R2

[tidal turret]

wdym?

[sharp coral]

the definition of R_2(x) is just R_2(x) = P_2(x) - f(x). so that only requires f be C2. Any formulas for R_2 other than that may require higher differentiability

so the formula of R_2 being the sum of all the other terms of the taylor series requires infinite differentiability, and the lagrange remainder formula requires the function be 3 times differentiable

[tidal turret]

.solved

[midnight plank]

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[rare maple]

#help-45 message

[lusty python]

Are you still stuck on that question?

[lusty python]

And secondly, is the latter fixed or not?

[lusty python]

?

Now we assume as normal, a <= b <= c where a, b, c are amount of money in each accounts

Now since c >= b then Peter can always transfer $b from account with $c to account with $b, as in doubled

And the remaining money can just be transfered to a

[midnight plank]

@rare maple Has your question been resolved?

[quiet hinge]

you need c-b to be equal to a to be able to do this

you cant always follow that strategy

[rare maple]

We can only consider odd numbers only

I conjectured earlier that GCD a,b can be achieved for 2 accounts which turned out to be false

Might be achievable

[midnight plank]

@rare maple Has your question been resolved?

[midnight plank]

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[night gyro]

hello can someone go through this with me step by step im really confused

[midnight plank]

@night gyro Has your question been resolved?

[civic gazelle]

$\overrightarrow{OD} = \lambda \cdot \overrightarrow{OB}$

[late rover]

$\overrightarrow{OD}$ looks better

[night gyro]

lamba times ob?

[grand pond]

(De)Carbonized

[night gyro]

what is going on

ok

i was thinking this

[civic gazelle]

you express $\overrightarrow{OB}$ in terms of $\overrightarrow{OA}$ and $\overrightarrow{AB}$ and you got your first equation

[grand pond]

(De)Carbonized

[night gyro]

ok how

[night gyro]

also is this right

[civic gazelle]

that's wrong

[night gyro]

why

dang

[civic gazelle]

because the problem states the ratio $\frac{OD}{OB}$ = $\lambda$ and not $\frac{OD}{DB}$

[night gyro]

ohhh

i see

what should we do instead

[grand pond]

(De)Carbonized

[night gyro]

we find a multiplier instead?

OD times lamda

?

[civic gazelle]

the problem wants us to express OD in 2 different ways with a, b, mu and lambda

[night gyro]

yh

[civic gazelle]

here's the first thing you get

[night gyro]

but it lready states OD is lmbda

[civic gazelle]

$\overrightarrow{OD} = \lambda \overrightarrow {OB} = \lambda (\overrightarrow {OA} + \overrightarrow {AB})$

[grand pond]

(De)Carbonized

[night gyro]

thats to find OB

[civic gazelle]

then you substitute OA and AB in

[night gyro]

why did it chnage it looked right before

[civic gazelle]

got your first equation

[night gyro]

wait

[night gyro]

i dont get this part could u explain it

[night gyro]

which is a+2b+2a-0.5b= 3a+1.5b

[civic gazelle]

$\frac {OD}{OB} = \frac{\lambda}{1}$
$\rightarrow OD = \lambda \cdot OB$
$\rightarrow \overrightarrow{OD} = \lambda \overrightarrow{OB}$ (since those two vectors points to the same direction)

[grand pond]

(De)Carbonized

[night gyro]

yes i see u just colved for OD

ok

[night gyro]

i got the first equation here

[civic gazelle]

you forgot to multiply lambda

[night gyro]

lambda (3a+1.5b)

like this?

[civic gazelle]

yes

[night gyro]

wait

why do we do this

[civic gazelle]

there's a lambda at the front here

[night gyro]

ok

our next eqaution

?

[civic gazelle]

it's a bit tricky

[night gyro]

why we doing OB when we need OD what

wht is this question

[civic gazelle]

we just got our first expression for OD here

lemme type out the second one

$$\overrightarrow{OD}$$
$$= \overrightarrow{OA} + \overrightarrow{AD}$$
$$= \overrightarrow{OA} + \mu \overrightarrow{AC}$$
$$= \overrightarrow{OA} + \mu (\overrightarrow{OC} - \overrightarrow{OA})$$

[grand pond]

(De)Carbonized

[night gyro]

our second 1?

[civic gazelle]

yea

[night gyro]

but mew AC does not go to D it foes all the way upto AC

[civic gazelle]

same explanation as before

AD/AC = mu -> AD = mu*AC

[night gyro]

how do u know ACis not given

only the ratios r givwen

[civic gazelle]

well vectors OA + AC = OC right?

[night gyro]

we can do AO then OC

to find AC

[civic gazelle]

yep

[night gyro]

which is AC=

OA+OC so -a-2b+3a= -2b+2a

[civic gazelle]

great

now what's OD

[night gyro]

we did OD already no?

lambda (3a+1.5b)

[civic gazelle]

we are trying to get the second equation for OD

[civic gazelle]

This is our first equation we've established before

[night gyro]

yh

wht we do now

why we findin 4 equation my teacher just did OB and AB find 2 equations and solve

[civic gazelle]

you replace OA = a + 2b and AC = -2b + 2a into this equation

[night gyro]

what is the questiion even asking we already found OD

[civic gazelle]

reread the problem again

we have to express OD in two different ways

[night gyro]

u wnt me to reread the whole thing?

ok 1 sec

ok done

we found 1 way

[night gyro]

this

ok got it

[night gyro]

why does OC-OA give us AC

dosent make sense

[civic gazelle]

because OA + AC = OC

you just switch OA into the other side

[night gyro]

AC= OC-OA?

[civic gazelle]

yes

[night gyro]

3a-a+b?

[civic gazelle]

3a - (a+2b) you mean?

[night gyro]

yh ok

3a-a-2b

2a-2b?

[civic gazelle]

yes

[night gyro]

next

a+2b+ mew (2a-2b)?

[civic gazelle]

correct

you can simplify this further

[night gyro]

expand the mew?

a+2b+mew2a-mew2b?

[civic gazelle]

yes

now factor a and b

[night gyro]

wouldnt it jst give this

[civic gazelle]

to get (1 + 2mu)a + (2 - 2mu)b

[night gyro]

what

oh yes

got it

[civic gazelle]

you'll see why we group it like this later on

[civic gazelle]

and that's our second expression for OD

[night gyro]

yes

[civic gazelle]

now we have:
(3 lambda)a + (1.5 lambda)b = (1 + 2mu)a + (2 - 2mu)b

we're asked to find mu and lambda to satisfy this equation

what do you get from this

[night gyro]

idk

[civic gazelle]

remember that a and b are not algebraic numbers, they're vectors

[night gyro]

jst a eqaution with bunch of variables tht dont match

3 lambda= 1+2mu

[civic gazelle]

exactly

[night gyro]

1.5 lambda= 2-2mu

simulatnos equation?

[civic gazelle]

yes

[night gyro]

can we do it like

[civic gazelle]

why is it 1.5k at the top and 3h at the bottom

[night gyro]

oh mb

ok let me fixt

[civic gazelle]

but I get the idea

yeah go solve for it

[night gyro]

yh so

idk wht to do here

sry

[night gyro]

anyway ty i got it

.close

[midnight plank]

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[ionic thicket]

when doing dot product (for example: (ai+bj).(ci+dj)), is it = (ai)(ci) + (ai)(dj) + (bj)(ci) + (bj)(dj) = ac + 0 + 0 + bd = ac + bd due to i.i and j.j = 1 and i.j and j.i = 0 from the cos theta = a.b/|a||b|

[ionic thicket]

like is this the logic behind dot product

because i and j are orthonormal

[lyric charm]

yes

[ionic thicket]

thanks

.close

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[twilit field]

Determine if $(M+N)^0= M^0 \cap N^0$ is true when $M,N$ are not both finite dim

[grand pond]

wai

[nova yoke]

what does ^0 mean, annihilator?

[twilit field]

yea

I suspect it's true

[nova yoke]

yea i think so, what's your reasoning?

[twilit field]

One min

Let $f \in (M+N)^0$ then $f(v) \forall v \in M+ N$. $v \in M \implies v \in M+ N$. so $f(M)=0$ similaarly for F(N)

[grand pond]

wai

[twilit field]

excuse my abuse of notation

f(M)+=0 means f(v)=0 for any v in M

[nova yoke]

yes, so that implies $(M + N)^0 \subseteq M^0 \cap N^0$, how about the converse

[grand pond]

Bungo

[twilit field]

follows from linarity of the functional

[nova yoke]

yep

both directions pretty trivial really

[twilit field]

ywa

*yea

simialry for (M \cap N)^0 = M^0+ N^0

[nova yoke]

yea that one also seems easy if i'm doing it right in my head

if f kills M and g kills N, then f+g kills anything that's in both M and N

and conversely, if f kills M + N then in particular it kills M, so f = f + 0 is in M^0 + N^0

[twilit field]

yup

[twilit field]

I think follows simialrly for (M+N) perp

[nova yoke]

you mean if we assume we're in an inner product space?

[twilit field]

yup

[nova yoke]

yea i would think so offhand, orthogonal complements act a lot like annihilators

[twilit field]

I think by reisz they are the same ( practically)

[nova yoke]

yea in finite dimensions, i haven't had enough coffee to think about infinite dimensions haha

of course annihilators are more general since they exist even if there's no inner product

[twilit field]

I just hope I'm not asked to prove |V^*>|V| in my exam 🙏 [ in inf dim]

[nova yoke]

oh yea the dual and double dual stuff gets fun in infinite dimensions

[twilit field]

yea , it does

[nova yoke]

do you have an exam soon?

[twilit field]

this sat

[nova yoke]

ah best of luck then!

[twilit field]

It's quite extensive

[twilit field]

tq

[nova yoke]

final exam?

[twilit field]

midsem, but of a high level course

so we;ve covered a decent bit

[nova yoke]

ah yes, makes sense

[twilit field]

like we'll start studying operators after midsem [ on fin dim spaces extensively and inf dim wherever possible]

[nova yoke]

i have probably asked before but i don't recall, is this a second course in LA or an honors course or smth?

[twilit field]

Operator theory

[nova yoke]

oh nice

[twilit field]

but with no analysis so really a ton of LA

Another question

If $y,z \in V^*$ , and $[x,y]=0 $ whenever $[x,z]=0$ . Then $\exists a: y=az$

is the whenevr here a iff

nvm

[nova yoke]

wait, x is in V?

[twilit field]

it's <-

[twilit field]

yes

[x,y]=y(x)

halmos' notation

[nova yoke]

do you mean y = az at the end then?

[twilit field]

oops

yes

[nova yoke]

ah ok

[grand pond]

wai

[nova yoke]

oops bad wording

[twilit field]

This tells us ker(z) \subseteq ker(y)

Well, it has to be iff.y=az would mean they both have the same kernel

[nova yoke]

yea why would that imply y = az for some a

[twilit field]

which isn't necessarily true

[twilit field]

.

[nova yoke]

right (unless a = 0)

[twilit field]

so that's it?

[nova yoke]

what's the problem statement, prove or give a counterexample?

[twilit field]

prove

[nova yoke]

but we just established it's not true, no?

unless you assume they really meant "iff" instead of "whenever"

[twilit field]

yea

That's what I thought

halmos is wacky, maybe this is old lang?

*language

[nova yoke]

hmm lemme take a peek at halmos, what section/page is this?

[twilit field]

sec 14

page 22 in my book

[nova yoke]

ah i think it works without assuming iff

these are linear functionals, so their kernels must have dimension n-1 or n

[lethal path]

rank-nullity theorem strikes again

[lethal path]

should be just n - 1 then?

[nova yoke]

from the context i'm assuming that V is finite dimensional here (since that is after all the title of halmos's book)

[twilit field]

I'm assuming infinite here

Should have mentioned

[nova yoke]

well we have ker(z) contained in ker(y)
so nullity(z) <= nullity(y)
if nullity(y) is n, then y is the zero functional, so we can take a = 0
otherwise we have the nullities equal, so the kernels are equal

[nova yoke]

hmm, might actually need "iff" for the infinite case, lemme see if i can think of an example

[midnight plank]

@twilit field Has your question been resolved?

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[ionic thicket]

how do i expand this?

[ionic thicket]

i get something like this

but idk how to proceed

not very familiar with distributing derivative operators

[midnight plank]

@ionic thicket Has your question been resolved?

[dawn dagger]

is this from physics

[ionic thicket]

nope

[dawn dagger]

is this backwards notation

[ionic thicket]

what is that

[dawn dagger]

nvm i think i get the notation

I think this is off what you got

How are r and θ defined?

You should first treat this as a dot product

Derivative operators act from the right so dr dr -> dr^2

[nova yoke]

you will probably want to use these facts:

[dawn dagger]

yea that too, you can derive that also easily

[ionic thicket]

ohh thanks its just some really really annoying algebra but in the end i got it

also really useful to know rhat.rhat and thetahat.thetahat = 1

and rhat.thetahat = 0

[nova yoke]

yep definitely need those

(i.e. rhat, thetahat are orthonormal)

the thing that makes them tricky is that they both depend on theta

[ionic thicket]

what theorem/identity states that dxdy = |det J| ds dt, where J is the jacobian matrix?

[ionic thicket]

yeah

[nova yoke]

afaik it's just called the (multivariable) change of variables theorem

[ionic thicket]

hxx

thxx

[midnight plank]

@ionic thicket Has your question been resolved?

[midnight plank]

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[rancid vigil]

,, \sin(x) \leq \frac{1}{2}

[rancid vigil]

how to solve

[mystic condor]

Pretend it equals 1/2 first

Then visualise it on the unit circle

[rancid vigil]

,, x = \frac{\pi}{6} + 2k \pi

so the solution should be an interval

i draw some lines in the circle , but idk how to group the solution into one line

[clever sedge]

This notation is a bit unusual

[rancid vigil]

dont mind lol

[clever sedge]

, k \in \mathbb{Z}

[mystic condor]

Yes, but theres another value of x, then you need to figure out, then think about where sinx will be less

[rancid vigil]

it should be all the way down from pi/6 to 5pi/6

[silent dock]

<@&268886789983436800>

[midnight plank]

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[leaden seal]

can the first derivative test be used to find a change in sign and therefore whether a point is a POI or POU

[gleaming latch]

what is POU?

also no

[sharp coral]

the criterion for a point to be an inflection point is whether the second derivative changes signs

you can't see that just by looking at the first derivative

[leaden seal]

point of undulation

so for this question would i find the domain

and stationary points

[midnight plank]

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[native wharf]

how do I do this 💀

[native wharf]

Please no really confusing terms

[chrome swift]

What does “coordinates are switched” mean to you

Or what’s that synonymous with

[native wharf]

sorry i didnt know u replied

I forgot to say to pign me

but

i think it means inverse

[chrome swift]

Ye so just find the inverses and see what satisfies both conditions

[native wharf]

also for this wuestion answer is a) right

[chrome swift]

Yeah I don’t see what’s wrong with A

[native wharf]

ure saying its a right

[vague seal]

yes

[native wharf]

.close

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[viral dagger]

ive only spoiled myself this far, how did they get -1 is in the range of f?

ok... i may have misread the subtitution as (1000,1000)

.solved

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[rich leaf]

the context in the question: we have defined the function as a relation of the Cartesian product of two sets, and two sets are equal when they are the subset of each other

though its pretty clear to me that if two sets are equal, then the first element of the ordered pairs are all equal, and as all elements of a set is captured by the first element of the the odered pairs then the domain(f) = domain(g), but can someone help me write this clearly and formally?

[midnight plank]

@rich leaf Has your question been resolved?

[torpid apex]

two sets A and B are equal if there is a bijection f between A and B; if you define a functions as sets (i would have to see exactly how your text does this), then two functions f and g would be equal if there is an invertible map between them; you would then be able to show that dom f = dom g , and f(x) = g(x) forall x implies f=g

[rich leaf]

they define it like this

[bold peak]

Well start writing out what you have

We can help you make stuff more precise, but the overall answer should still be yours

[midnight plank]

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[spiral forum]

is this channel closed?

[bold peak]

!status

[midnight plank]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[near geyser]

I got 750gm?

Actually I don't know how do you start

[bold peak]

Well we know he bought it for 30/kg

And sold it at a profit of 66.67%

So what's his actual selling price

[midnight plank]

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[tidal turret]

need some help with exercise 9

Describe the values of gcd(7a+8, 7a + 3) as a function of the values of a in Z

[midnight plank]

@tidal turret Has your question been resolved?

[near geyser]

What?

[near geyser]

Which language?

[near geyser]

5a not 7a

[tidal turret]

Describe the values of gcd(5a+8, 7a + 3) as a function of the values of a in Z

[near geyser]

Language name?

Europe?

a belongs to z

Put a=0
Gcd(8,3)=1

a=1--(13,10)=1

a=2---(

18,17=1

[supple pilot]

use gcd(n, m) = gcd(n, m-n)

[near geyser]

a=3

(23,24)=1

(5a+8,2a-5)

I guess n-m?

[tidal turret]

gcd(5a+8,7a+3) = gcd(5a+8,2a-5)

gcd(5a+8,2a-5) = gcd(a -2, 2a -5)

[near geyser]

-2a+5 if we do n-m

[tidal turret]

gcd(a-2,2a-5)=gcd(a-2,-1)

@supple pilot

you here?

[tidal turret]

spanish

[supple pilot]

yea

[tidal turret]

check my work

pls

[supple pilot]

gcd(5a+8,2a-5) = gcd(a +18, 2a -5)

[tidal turret]

= gcd(a+18,-41)

[supple pilot]

yea and 41 is prime

[tidal turret]

so?

whats your point

[nova yoke]

that considerably narrows down the possibilities for the gcd, yes?

[tidal turret]

either 41 | a + 18

[supple pilot]

so the gcd will be 1 unless (a+18) is multiple of 41 yea

[tidal turret]

right

[tidal turret]

took me a bit to think it through

[tidal turret]

so either a + 18 is coprime with 41

or 41 | a + 18

you here?

[supple pilot]

yup, for any given prime p and other integer x they are either coprime or x = kp

[tidal turret]

now what? what should i do now

HELP

[supple pilot]

nothing, that's the function

it's 1 for all integers except multiplies of 41

where it's 41

[tidal turret]

what now?

[nova yoke]

just write out the condition

a+18 is a multiple of 41

a+18 = 41k

a = -18 + 41k

[supple pilot]

so for a+18 = k*41 you have 41
a = k*41 - 18 => f(a) = 41
and f(a) = 1 for other integers

[tidal turret]

what?

[supple pilot]

okay, the question asks you to provide the value of gcd as a function of a, right?

[supple pilot]

yes

so the answer you need is some function f: Z -> Z
that takes an integer, plugs it into a and returns the value of gcd(5a+8, 7a+3)

[tidal turret]

is just that if the input of this function is divisible by 41

[supple pilot]

not the input, the input + 18

[tidal turret]

then A is of the form a = -18 + 41k

[supple pilot]

yes

when a is of that form, the function returns 41

otherwise it returns 1

[tidal turret]

what

[tidal turret]

ahh the function is tje GCD

i see

[supple pilot]

yes

[tidal turret]

yup

[tidal turret]

so if 41 | a + 18 then GCD = 41

[supple pilot]

yes

[tidal turret]

otherwise GCD = 1

[supple pilot]

yup

[tidal turret]

simple as dat

ok, so a quick recap of the procedure done @supple pilot @nova yoke

how would it be?

1. do X
2. do Y
3. do Z
4. cleanup and celebration

[nova yoke]

math in a nutshell

[supple pilot]

1. take the gcd(5a+8, 7a+3)
2. subtract one of the expressions from the other until a is only in one of the expressions
3. look at the expression that doesn't contain a anymore (so it's a number, in this case -41)
4. determine the divisors of that number (let's call it x), if it's prime then gcd = x when the other expression is a multiple of x, 1 otherwise

if it's not prime then it's a bit more complicated

the 2. point is basically what is called the euclid's algorithm

[tidal turret]

i just like doing it so it sticks the algorithms easier is a learning hack

[tidal turret]

wdym?

in this case 41 is prime

we were lucky, you are saying?

[supple pilot]

if we had something like 4 instead of 41 then we would have to consider multiples of 2 too

so we would have gcd = 4 for multiples of 4, gcd = 2 for multiples of 2 that are not multiples of 4, 1 otherwise

[tidal turret]

what if we have a gazillion divisors, like 400

we just use prime factorization and consider those

[nova yoke]

then it starts to become faster to write a computer program that writes the answer for you

[supple pilot]

then we would have to consider them all
but also considering that divisors can divide other divisors so it becomes a mess

[tidal turret]

what about finding lots of solutions for x^n + y^n = z^n for n > 2 can computer find those?

[tidal turret]

ah damn it, i see

[nova yoke]

yea try writing that and get back to me when it finishes haha

[supple pilot]

yes it's quite easy, but the program won't fit on the page so it's left as an exercise for the reader

[nova yoke]

i'll get you started, x = y = z = 0

[tidal turret]

this is derailing to a sitcom

i appreciate the help Lads and Gals

[tidal turret]

.solved

[midnight plank]

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[indigo fulcrum]

have a calc 1 exam tmrw and related rates is on it. this is from prac exam but also how would i solve this but also go abt solving other related rates problems

[idle geyser]

sketch and label variables, write one equation linking them (geometry/volumes), take d(...)/dt on both sides plug the snapshot numbers, solve for the unknown rate, and report sign + units

[indigo fulcrum]

for this case im thinking pyth theorm. so a^2 + b sqr = c srq. so then im thinking 34 sqr plus 3 sqr = unkown. but lets say we find that answer we differentiate that or ewhat

[idle geyser]

use variables first then differentiate

[indigo fulcrum]

can u show how

[idle geyser]

sure

let x(t) = boat–dock distance

s(t) = rope length

geometry gives s^2 = x^2 + 3^2 (the 3 ft is constant)

differentiate both sides 2s * ds/dt = 2x * dx/dt so dx/dt = (s/x) * ds/dt

at x = 24 s = sqrt(24^2 + 3^2) = sqrt(585)

given ds/dt = -3 ft/s then dx/dt = (sqrt(585)/24) * (-3) = -sqrt(585)/8 which is abt -3.02 ft/s

so the negative means the boat is moving toward the dock at 3.02 ft/s

[indigo fulcrum]

o

oh

ok

Thank you

[idle geyser]

np

[indigo fulcrum]

.close

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[indigo fulcrum]

still dont get

[midnight plank]

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[dusty oracle]

how can i do aii?

i'm not sure where to start

[grand pond]

chilledsofa

chilledsofa

[dusty oracle]

<@&286206848099549185>

mb

i know how to do it

.close

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[shut canyon]

If determinant of augmented matrix is 2021, then the system of linear equations is regular.

[grim vector]

Determinant of augmented matrix ? How does it exist

[shut canyon]

I don't yet know

Oh!
"if the determinant of a system of linear equations is 2021, then it is regular"

Si le déterminant d'un système d'équations linéaires est égal à 2021, alors ce système est régulier. maybe in francaise

[grim vector]

Le déterminant fonctionne que sur des matrices carrées et la matrice augmentée n'est pas souvent carrée

Si tu as AX = B avec detA != 0 alors oui il y a des solutions, même une unique solution. Parcontre si A n'est pas carrée et donc que le det n'est pas définie, ça ne veut PAS DIRE qu'il n'y a pas de solution, il y en a soit une infinité, soit aucune

[shut canyon]

∀ sys règulier ∃! solution

[grim vector]

Si système régulier signifie det != 0 alors oui

[shut canyon]

système linéaire régulier = matrice carrée avec det ≠ 0

[grim vector]

Par contre, le screenshot que tu as posté ne porte pas sur ça

[shut canyon]

Je suis très reconnaissant·e pour votre aide.

[midnight plank]

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[spiral rock]

First isomorphism theorem!

[twilit field]

Without that :(

As far as I can recall we haven't done this in class

( first iso theorm)

[dense harness]

what is V‘

[spiral rock]

Dual

[twilit field]

dual of $V$

[grand pond]

wai

[dense harness]

Ok

usually it’s like *

And what’s M0

[spiral rock]

Anyway, try to get a linear map
V' --> M' with kernel M0

This same maps gives an isomorphism

Just instead of
f --> g
f + M0 --> g

[twilit field]

Let ${f_i + M^0 \mid i \in \Omega}$ be a basis of $V'/M^0$. let ${f_i \mid i \in \Lambda}$ be basis of $M'$

[twilit field]

wait what

[grand pond]

wai

[spiral rock]

Dont use basis

Its ugly

[dense harness]

You shouldn’t use basis

Because of infinite dimension

[twilit field]

F, no wonder I didnt get this in my assignment

[dense harness]

So what exactly is M0 btw

[twilit field]

annhilator

[twilit field]

I don't get this

[spiral rock]

Can you give such a map?

[twilit field]

I don't even get what you've written

T:V'->M' such that ker(T)=M^0?

[spiral rock]

Yeah

[dense harness]

If you get this map you can use the induced map

[twilit field]

$M \subseteq V \implies M' \subseteq V'$. So $|V'|≥|M'|$. Further $M^0 \subseteq M'$. So $|V'|≥|M'|≥|M^0|$

[runic hamlet]

forget cardinalities

[grand pond]

wai

[runic hamlet]

M' is also not a subset or V'

[dense harness]

M‘ is not a subset of V‘ too

[twilit field]

but M is a subset of V

[runic hamlet]

yes

what are the elements of M' ?

[dense harness]

Dual are functionals on whole space

[runic hamlet]

what are the elements of V' ?

[dense harness]

And the whole space is different

[twilit field]

functionals on M'

functionals on V'

[runic hamlet]

and what does that mean?

[twilit field]

these functionals aren't defined on V\M

[runic hamlet]

yes

[twilit field]

Define $T(f)= \begin{cases} 0 & f \in M^0 \ f \vert_{M} & \text{ otherwise } \end{cases}$\

[grand pond]

wai

[runic hamlet]

why piecewise

[dense harness]

Your first case is the same in The second

[runic hamlet]

what is f|M if f is in M0

[twilit field]

so just $T(f)= f \vert_{M}$

[grand pond]

wai

[spiral rock]

Yep

Very natural map

[twilit field]

0

ngl in hindsight this shouldn't have taken me so long to come up with

[dense harness]

So what can you say about the induced map

[twilit field]

it's surjective for sure

injective too, but that will require some work I suppose

[visual tiger]

Injective?

[dense harness]

U just showed its injective

Well

[twilit field]

oops

[dense harness]

Not showed

[twilit field]

one-one

[dense harness]

But u constructed it so its injective

[spiral rock]

Its not injective

[twilit field]

then how is it an isomorphism

[dense harness]

The induced map is

[visual tiger]

Right induced map

[twilit field]

what

I'm so confused

so I've constructed a map that takes any element in V' to M', fine. with kernel M^0

now what

This isn't injective is it

[dense harness]

No

do you know what the induced map is tho

[twilit field]

no

[dense harness]

It’s the map [x] -> [f(x)]

And it’s only well defined if [0] is in Kern f

[twilit field]

okay...

I don't see how this ties back to my OG question

[dense harness]

Well the induced map is now a map V‘/M0->M‘

[twilit field]

sure, that makes sense

and how is this injective

[dense harness]

Show that the kernel is 0

[twilit field]

it is M^0 yes

oh

🤦

[dense harness]

yeah M0 is 0 in the quotient

[twilit field]

quotient spaces are wacky creatures

got it

thanks

[twilit field]

Just wanted to know, is this supposed to be easy?

[spiral rock]

The solution we got to is pretty simple

[twilit field]

in infinite dim, I'd replace dim with cardinality, right

[twilit field]

yea true, I should have been able to come up with this myself 😭

[spiral rock]

What is Ran

[twilit field]

range

[spiral rock]

Oh

[dense harness]

Don’t think it is

[twilit field]

then what would dim become

[dense harness]

idk if there’s a replacement

[twilit field]

yea, same

Let $T: \mathcal{P}(\R) \to \mathcal{P}(\R)$ be the differentiation map . $Ran(T) = \mathcal{P}(\R); Ker(T) = k; k\in \R$

[grand pond]

wai

[twilit field]

The intersection is not {0}

Does this work?

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

<@&286206848099549185>

.close

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[spiral rock]

<@&268886789983436800>

[midnight plank]

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[violet pike]

.close

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[rich yew]

[civic gazelle]

thua

[midnight plank]

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[sharp dust]

i was doing percentile and i need to find P10 but it came out at 0.6th item and now idk what to do

[lethal path]

that doesn't look right

should be 5 - 1 = 4 instead

considering 100/4 = 25:

8 is the 0th percentile
30 is the 25th percentile
32 is the 50th percentile = median
67 is the 75th percentile
72 is the 100th percentile

[stuck olive]

Math

[midnight plank]

@sharp dust Has your question been resolved?

[midnight plank]

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[near geyser]

If I have two sets

A=m, B=2 then how many number of into functions?

[near geyser]

2^m would be total number of functions

[winged maple]

what is an into function?

[sacred folio]

subtract the number of onto functions from the total

[winged maple]

is that surjective or injective?

[sacred folio]

that should be easier i think

[near geyser]

Injective

[sacred folio]

btw here the answer would be just 2 as there are only two possible values the function could take

assuming B is the range set ofc

[midnight plank]

@near geyser Has your question been resolved?

[near geyser]

.close

[midnight plank]

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[ionic thicket]

can someone help me make sense of this? thxx

[sacred folio]

basically it has simplified u.dS to u.n(dS)

where n is the direcn of the area vector

and dS is the magnitude

[ionic thicket]

what is the dot product doing

like geometrically?

[midnight plank]

@ionic thicket Has your question been resolved?

[sacred folio]

hmmm

lemme thnk

you can maybe see it as calculating the electric/magnetic flux for the surface

since we only want the field lines escaping the surface we use a dot product to remove any parralel ones

so in summary you can say the dot product is used to find or isolate the vector lines escaping the surface

[vague seal]

The dot product is to only take the part of u that aligns to the normal vector of the surface

aka perpendicular to the surface

as only that part of u matters for the surface integral

[sacred folio]

ye that is better....

[midnight plank]

@ionic thicket Has your question been resolved?

[ionic thicket]

how is this done? i am not very very familiar with how dot product works

i only know the basics of it

[vague seal]

A feature of the dot product is that if one of the vectors is a unit vector it turns into a projection operator

that means

if i multiply a vector by the unit vector in the x direction

i am projecting that vector onto the x direction

taking the x component so to speak

in this case instead of the x direction you take the direction perpendicular to the surface

meaning you project u to the normal vector

taking only the parts of u that are along the normal vector

to do it in detail requires details of the normal vector n and also the vector field u

[midnight plank]

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[sour hull]

alright so.. say that I have 13 people on a room handshaking eacch other, without repeat the handshake, how many handshakes will occur, how do I do that?

[spiral rock]

if this is a homework question, can you send a picture?

[pine wave]

(of the homework problem)

[sour hull]

there is no picture

I just translated it

do you mean like

you want to see it

untranslated?

[spiral rock]

this is better than nothing

[sour hull]

ok

give me a sec

I don't have a phone, I'll try to see if the webcam does the job

[pine wave]

[hint 1: can you count the number of ordered pairs of people?]

[sour hull]

wdym with that

"ordered pair of people"

*pairs

[pine wave]

erm... okay guess i have to explain what "ordered pairs" are

okay simpler hint
[hint 0: try it for a smaller number of people, and count by hand]

[sour hull]

I did tho

[pine wave]

what did you get

[sour hull]

I though like

12+11+10+9+...

I did it on a whiteboard and counted the nodes

not right tho

[pine wave]

what did you get

[sour hull]

90

ops

sorry

[pine wave]

90?

[sour hull]

78

sorry

typo

not on the keyboard

[pine wave]

strange, 78 seems right

[sour hull]

oh

no wait

it's one of the options

still, I solved that with sigma notation

I was supposed to solve it

with combinatorics

so I though I did something wrong

[pine wave]

have you heard of binomial coefficients

[proud violet]

You still did solve it with combinatorics. Just in a different way than the “simplest” way

[sour hull]

maybe the name is diferent, is it something like, (x+3)(x+2)(x+1) and the coefficient is -3, -2, -1 (that also reminds me of roots, but is it something like that?)

No mean

Í know that's the roots

when you talk with coefficients

[pine wave]

that's closer to stirling numbers of the first kind iirc

but what you want to consider are the coefficients of $(1+x)^n$

[grand pond]

Element118

[sour hull]

umm, yeah I have no idea

hold it

I'll see it in the textbook

[sour hull]

yeah no it's not here

maybe I learned it last year and forgot about it

[sour hull]

plus, what does "iirc" mean

ok yea sorry, makes sense now, 13 people, subgroups of 2, how many subgroups can you make

let me see if that matches

yeah, that's it, I swear if it wasn't for the tip on that I wouldn't figure that out

.close

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[midnight plank]

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[leaden seal]

Hi, I am stuck on the assembly drawing

[leaden seal]

As you can see I've drawn the 3d projections for each part but I am unsure as to what it means by 2 view assembly drawing

[midnight plank]

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[crude zealot]

!help

[midnight plank]

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

[midnight plank]

Please don't occupy multiple help channels.

[crude zealot]

is this right

what i did?

<@&286206848099549185>

hi??

[cerulean oyster]

No, it isnt

Try to explain your logic about these tho

[crude zealot]

I believe i fixed it?

@cerulean oyster

.close

[midnight plank]

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[crude zealot]

.reopen

[midnight plank]

✅ Original question: #help-49 message

[crude zealot]

hi

[silent dock]

<@&268886789983436800>

[crude zealot]

<@&286206848099549185>

hi is anyone here?

please

been waiting for around 40 mins

.close

[midnight plank]

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[rich yew]

More scams here as well

[midnight plank]

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[lime dagger]

Can someone help me w questions 1 and 2

[midnight plank]

Please don't occupy multiple help channels.

[pearl hull]

.solved #help-35 message

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[sonic mesa]

yall know about quadratic equations like finding roots and vertex??

[sonic mesa]

i need help to understand the steps

[oak kelp]

please show an example of a question you're stuck with.

[plucky lark]

Ok what do you need help with?

[near cliff]

Finding roots of quadratic:
$$x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$$
Sometimes, the roots of the quadratic will be complex.

or, (sometimes not possible) factor directly from quadratic.

Vertex:
$$x = -\frac{b}{2a}$$
Sub. this in to find the y-coordinate.

[sonic mesa]

lets say: f(x)= 10x^2-11x-6

find root and vertex

[oak kelp]

are you allowed to use the quadratic formula?

[sonic mesa]

yess

[oak kelp]

then toss it straight in there.

[sonic mesa]

yeah but wht abt the ones with factoring?

[plucky lark]

What method did your teacher teach you?

[sonic mesa]

factoring and plugging into the formula

[plucky lark]

I don’t think that’s a factoring method lol

[oak kelp]

what formula? after factoring you no longer need the quadratic formula, no?

[near cliff]

she means two methods probably

[plucky lark]

Ohhh

[oak kelp]

oh, factoring, and formula.

[sonic mesa]

no you dont. i meant the other method

[sonic mesa]

yeah

[plucky lark]

Wait I’m confused :/

[sonic mesa]

mb guys

[oak kelp]

in complicated cases like this, you can try to factor by splitting the middle term.

maybe you have heard of this being referred to as the "ac method", but I personally call it splitting the middle term.

[near cliff]

both methods of solving are acceptable.

if your calculator has an automatic quadratic solving function, use it: if the roots are fractions are integers, then you can solve by factoring

otherwise, just use quadratic formula

[sonic mesa]

alr lemme find a good example

[near cliff]

we call it the criss-cross method in my country lmao

[oak kelp]

this example actually has rational roots, so it's actually not too bad of an example to see the middle term split method.

[sonic mesa]

wdym by if roots are fractions are integers??

[near cliff]

if your calculator cant automatically solve quadratics, dont need to worry about that

[sonic mesa]

1. 10*6
2. factor :x^2-11x-60 right?

first two steps

[sonic mesa]

alr my teacher doesnt allow us to use graphing calculators so i dont think it will automatically solve

[oak kelp]

that's a new idea I have not seen before. my way involves finding two numbers that multiply to ac and add to b.

[oak kelp]

you don't need a graphing calculator to solve quadratics.

[near cliff]

you can use it to check if the roots are rational. if so, try the factorisation method. otherwise, it is a surefire sign for you to use the quadratic formula

[sonic mesa]

plugging in is a different case

[near cliff]

like your teacher disallows it but it's not like he or she is gonna somehow telepathically know you used it to determine which method is easier to use

all the teacher sees is the working step

[sonic mesa]

i mean im good with the ones where we just plug in

problem is where you gotta find by factoring method i cant simplify it with the calculator we use

and also where this "i" stuff comes in

when theres a negative square root

[oak kelp]

if the roots are complex, abandon factoring entirely.

[near cliff]

okay okay first of all does your calculator have a function that takes in a, b, and c of the quadratic and output its roots?

[oak kelp]

in fact, perhaps do a quick check with the discriminant before even trying to factor. if it's negative, just don't bother factoring at this level and throw it into the quadratic formula.

[near cliff]

this is also a good tip

to add, if it's irrational, just use quadratic formula

[sonic mesa]

no it cant solve for roots.. i can use it as a help with calculating stuff

[oak kelp]

and that. the way you'd know there would be trouble is to see if the discriminant is a perfect square.

if it is not, again just throw away factoring.

[near cliff]

well honestly

given everything

i think your best bet is to always stick to quadratic formula in exams

unless the factor is like really really obvious

calculating the discriminant is already like 3/4-th of the way to solving by quadratic formula

[sonic mesa]

so like i can use the formula for any type? cuz our teacher was like use factoring in these problems and use the formula for the others..

[near cliff]

it applies to any quadratic, yes

[sonic mesa]

alr!

thank you!

ah also

i need help simplifing and dividing square roots

[near cliff]

do you have an example problem?

[sonic mesa]

gimme a sec

nvm!

[near cliff]

also, for the other part to your problem on finding the vertex, you can either use x = average of the roots you got in part 1, or use x = -b/2a. Then substitute this value of x into the original equation to find the y-coordinate.

[sonic mesa]

finding AOS?

[near cliff]

finding the vertex of the quadratic equation

[near cliff]

^

[sonic mesa]

yeah like we use tht formula to find axis of symmetry and then use it for vertex

its the same way

[near cliff]

yeah

[sonic mesa]

alr thts it

thanks yall!!

[near cliff]

no problem

thank @oak kelp too

[oak kelp]

no need. I did nothing.

[sonic mesa]

you did help me understand! thanks

[midnight plank]

@sonic mesa Has your question been resolved?