help-26

[pseudo jetty]

yeah that's how you'd show (2,1,1) is in the kernel

[shut obsidian]

So A * s(2, 1, 1) = 0

[odd pagoda]

yes

which is supposed to be valid for all s

by linearity you know that s doesnt actually matter

so just A*(2,1,1)=0

[shut obsidian]

Yeah

[shut obsidian]

So we are done now..? How do we arrive at the same as what we got earlier though

[pseudo jetty]

make an A that kills (2,1,1)

[shut obsidian]

Ah, well we can just set A = [[a, b, c], [d, e, f], [g, h, i]] and obtain possible entries like that. Afterwards we should be done, right? We'll have a free variable again

[pseudo jetty]

you also have to make sure that A only kills the span of (2,1,1)

[pseudo jetty]

maybe too big of a matrix

you want only s(2,1,1) to lie in the kernel of A

so rank nullity tells you that choosing a 2x3 A is a good idea

[shut obsidian]

Alright

[shut obsidian]

This method feels a bit overkill though

[pseudo jetty]

its a general method

[odd pagoda]

understanding > simplicity

[pseudo jetty]

solutions to Ax = b always have this structure

[shut obsidian]

Thank you Snow and Denascite

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[odd pagoda]

.reopen

[topaz sinew]

✅

[odd pagoda]

also I am currently not sure how you want to use your method for something with two parameters?

maybe I'm stupid tho

its late

but my method easily generalizes to two or any number of parameters

[shut obsidian]

Two free variables?

[odd pagoda]

wait nvm I'm stupid

[pseudo jetty]

i think it works

[odd pagoda]

yeah it does

[shut obsidian]

Yeah.. And having to wake up at 6:30AM tomorrow sucks

[pseudo jetty]

low tech method

[pseudo jetty]

me today

[shut obsidian]

Thank you both

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[tepid glacier]

yo what

can someone help me I am so confused

am I wrong or is my teacher wrong

B is the vertex

[reef fjord]

you are correct if you said only CBA and ABC

[tepid glacier]

I did

so my teacher is wrong

?

[reef fjord]

the question was labeled wrong yes

[tepid glacier]

because they are saying the correct answer is BAC and ACB

neither A or C are the vertex

[reef fjord]

yes

[tepid glacier]

ok

had to make sure I am not an idiot

I'm going to email my teacher

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[distant linden]

why do they do matrices?

and why do they let y = 0

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[stray mason]

1. what are some books that regard the formal axiomatization of arithmetic? (I tried to ask in #book-recommendations , I hadn't received a response and doubt I will until I ask again; some amount of times.)
2. please recommend an ideal order formal mathematics and informal mathematics can be studied, specifically; for instance, set theory after arithmetic and logic.

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[stray mason]

someone please help

[stray mason]

<@&286206848099549185>

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[spiral phoenix]

Hello! So, I have to find the largest n that verifies the condition. I don't know where to start with it

[lapis sorrel]

,rcct

[coarse tusk]

,rccw

[coarse tusk]

$f(x)=x-\frac{n}{x}$ and $$\int_1^{\sqrt{2}} x^{n+1} f(x)^n dx \ge \frac{n}{2021}$$

[thorny flame]

kheerii

[coarse tusk]

is this the question?

[opal vault]

wouldn't make sense for f to depend on n

i think they're both 1

[coarse tusk]

yeah, that's why I was confused

[spiral phoenix]

That is the question

Yeah

[autumn plaza]

omg is that even math

[opal vault]

rewrite x^n f(x)^n together

[spiral phoenix]

No no

[opal vault]

then find an easy way to integrate x^(n+1)f(x)^n

[spiral phoenix]

f(x) = x- 1/x

Not x - n/x

[coarse tusk]

ah that makes more sense

[coarse tusk]

process is still the same though

$x^{n+1}f(x)^n = x\cdot (xf(x))^n$

[thorny flame]

kheerii

[spiral phoenix]

Okay, let me give it a try

See what I get

Oh, yeah, it's easy from here

Thanks! Have a good day!

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[night wharf]

Hi i am confused on why the negative version does not converge

[night wharf]

I would think it would just because of the alternating series test but idk

[loud oasis]

at x = - 1/9 it is not alternating

[night wharf]

if it is (-1)^n that is alternating right?

so wouldn't that twice be the same as like (-1)^n+1?

[loud oasis]

,, (-1)^n (-1)^n = (-1)^{n + n} = (-1)^{2n} = ((-1)^2)^n = 1^n

[thorny flame]

cloud

[night wharf]

yea but isn't that the same as (-1)^n *-1?

why is that one different?

[opal vault]

(-1)^n (-1)^n vs (-1)^n * -1

you can see the difference

[night wharf]

it does not giive the same result?

[opal vault]

(-1)^n is different from -1

[opal vault]

so of course those are different

[opal vault]

and one yields (-1)^2n = 1

so no alternating sign

[night wharf]

ok got it

thanks

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[pearl fog]

a kite has a shape 12 12 16 16 resepectively, if one of the diagonals is 20 long, what if the other length of the diagonal if it is a rational number?

[hot anchor]

use a diagram

And then pythagorean theorem

[pearl fog]

i checked and if i assume TV is is the 20, i get an irrational number so im pretty sure it has to be WU

but how do i find it then?

[weak creek]

find area of triangle then ?

[hot anchor]

a^2 + b^2 = c^2

pythagorean theorem

Two sides of a right triangle = a,b
hypotenuse = c

[weak creek]

@hot anchor i think he is assuming WU=20

[pearl fog]

oh eya

lemme check

[viscid portal]

You notice a 3-4-5 triangle

Which is a Pythagorean triple

Like Astra said

And then area, height etc

[pearl fog]

oh

[weak creek]

but the hypo=16 not 20

[pearl fog]

so WTU ans WVU is a right angle

[viscid portal]

Yup

[weak creek]

not nessasarly

wait never mind

@viscid portal could we also find area of triangle and then compare with area of kite?

[pearl fog]

i got 19.2

[pearl fog]

it is

[pearl fog]

is this correct?

[viscid portal]

Yeah something like that

[weak creek]

alr

[hot anchor]

ur right sorry

I think you do need to find the area

[viscid portal]

Uhm so 12*16/20? 48/5

[pearl fog]

whar

yea tho

but like thats only TC we want to find TV

[viscid portal]

No that's TV

You divide by two for the area of the triangle

And then multiply by two to go from TC to TV

If you think about it if one diagonal is 19 and the other is 20 your kite would be nearly a square

[pearl fog]

find the sum of the squares of all integers n where $\sqrt{\frac{4n+25}{n-20}}$ is an integer

[thorny flame]

Skill_Issue

[pearl fog]

i concluded that n<=19

wait

i got n=-15, n=41, n=25

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[pearl fog]

how many numbers from 1 to 10000 that can be expressed as the sum if 5 and 7 consecutive integers but not 3 consecutive integers

[pearl fog]

ah wait nvm

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[sweet roost]

it says that rank is 3 if k is not equal to 1 and not equal to -1

[sweet roost]

k is not equal to 1 because otherwise the last row would be all zeros

but why k cannot be equal to -1?

[raven sparrow]

If k is -1 I'm pretty sure the rank is still 3

[sweet roost]

ok there must be a mistake then

thank you

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[topaz sinew]

Please don't occupy multiple help channels.

[balmy island]

.close

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[jade kindle]

<)ABC=130
R=10
AB=BC

[jade kindle]

I need to find AC

although cant find the way for it

<@&286206848099549185>

[wooden moon]

can you draw a perpendicular to chord AC

and also join OA and OC

[ancient belfry]

The triangle is isosceles

[jade kindle]

yeah

[jade kindle]

then what?

this is trigonometry

[wooden moon]

what would be <AOC?

[ancient belfry]

R mean radius?

[jade kindle]

yes

[ancient belfry]

Ok just checking

[jade kindle]

130/2?

or not

i dont remember the sentences so well

[wooden moon]

no

[ancient belfry]

Then this should be able to be solved like a parrellagram

Or that kite shaped I forget the term

[wooden moon]

what would be <AGC?

[jade kindle]

130

because its on the same line

i think

[wooden moon]

umm no

[jade kindle]

bru

i just need to find AC i dont want the idea to get it though

[wooden moon]

ABCG is cyclic quadrilateral

[jade kindle]

how?

oh i undertsand

oh

so

180-130

AGC is 50

[ancient belfry]

Since AB=BC and A to the center point is equal to B to the center point you can connect them up

[wooden moon]

yes

so what would be <AOC???

[jade kindle]

hmm

100

[wooden moon]

yes

[jade kindle]

because on the same line and i forgot how to say the sentence

[wooden moon]

angle subtended on the center is twice the angle subtended on the circle

[jade kindle]

ohh

[ancient belfry]

Btw what subject is this exactly I get that you use trig but there’s a lot more geometry in this so is this like real analysis?

[jade kindle]

and AO is the radius

which means 10

[jade kindle]

trigonometry

[ancient belfry]

Never had to do this in my trig class

[jade kindle]

OC also radius

so also 10

[wooden moon]

now draw perpendicular D to AC

[jade kindle]

which means AOC is isosceles

which also means OAC and OCA is 40

i got it from here

[wooden moon]

yeah you can do trig from here

[jade kindle]

thank you very much @ancient belfry and @wooden moon

.solved

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[dark wharf]

someone please help

[rigid ivy]

Numerically, what is the x coordinate?

[dark wharf]

x

[rigid ivy]

Numerically. What is the actual number? The value

[dark wharf]

1

[rigid ivy]

yes

And which of those trig values is equal to 1?

[dark wharf]

A

[torpid dagger]

nope

[dark wharf]

What

[willow cipher]

sin(0) is 0

[torpid dagger]

sin(0) = 0

[dark wharf]

The answer is sin(0)

[willow cipher]

its not

[dark wharf]

Oh

[willow cipher]

think of cosine being x and sine being y

so here you said the x coord is 1

so the answer also needs to equal 1

[dark wharf]

Is it C

[willow cipher]

no

[dark wharf]

1

[willow cipher]

what is cos(0) ?

[dark wharf]

0

[willow cipher]

double check that

[dark wharf]

Cos(0)=1

[willow cipher]

yes

[dark wharf]

Thanks

[willow cipher]

np

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[fiery crystal]

Need help with this congruence

[chilly walrus]

$32x \equiv 28\mod36$

[thorny flame]

artemetra

[chilly walrus]

is this the problem?

[fiery crystal]

Yeah

[shrewd horizon]

that's not quite valid, you can't divide by 2 (or any even number) mod 36

$0 \cdot 2 \equiv 18 \cdot 2 \mod 36$, but $0 \not\equiv 18 \mod 36$

[thorny flame]

bee [it/its]

[shrewd horizon]

but you can divide both sides by 4 and also divide the mod by 4

so you actually get 8x = 7 mod 9

[fiery crystal]

Isnt there a property that says
a≡b(mod c) = ka≡kb(mod c)?

[shrewd horizon]

the forward direction of that is true

the backwards direction isn't necessarily true unless k is coprime to c

[fiery crystal]

Hold on, im lost

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[fiery crystal]

.close

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[hoary hedge]

i dont understand how to do this?

[topaz sinew]

@hoary hedge Has your question been resolved?

[merry phoenix]

well, the two intercepts give you the values of x and y at two points, the turning point tells you something about the value of the derivative at the point, and it is also another point where you know the value of x and y, so you should be able to form enough equations by substituting in the values

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[teal patio]

<@&286206848099549185>

[rigid ivy]

!15m

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[teal patio]

I need all of the hits but I am a bit confused

[rigid ivy]

We can't help with exams

[teal patio]

It’s not an exam

It’s just a practice question

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[rough vapor]

a = 10
d= log7

idk bc e

[reef fjord]

,tex .log rules

[thorny flame]

🫎 Moosey 🫎

[rough vapor]

so B is pie log e

100/5

[reef fjord]

$\log(x)=\log_{10}(x)$

[thorny flame]

🫎 Moosey 🫎

[reef fjord]

$\ln(x)=\log_{e}(x)$

[thorny flame]

🫎 Moosey 🫎

[rough vapor]

what do I do with the 2 on e

[reef fjord]

what is the question? does it just want you to simplify or?

[rough vapor]

so e

has a 2 in front of one of the logs

what do I do with it?

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[oak remnant]

<@&286206848099549185>

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@oak remnant Has your question been resolved?

[oak remnant]

<@&286206848099549185>

[raven sparrow]

Have you tried anything?

[oak remnant]

Yes I have

[raven sparrow]

The second component of the vector field is arctan(y) + 3x^2.
Differentiating with respect to x yields 6x, not 1/(1+x^2) + 6x

[oak remnant]

Oh okay thanks!

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[wooden tendon]

what is MINIMUM

[wooden tendon]

in relation to #49 which is what I’m doing

I’ve plugged in 1 and 3

and have gotten points

but how do I find the minimum

do I need to use MINIMUM

what is MINIMUM

[fallow igloo]

First of all, do you agree that the function looks like a quadratic function when you graph it?

[wooden tendon]

an odd looking one yeah

isn’t that because it’s a rational function

[fallow igloo]

yeah, it’s not a quadratic function

But looks pretty alike

[wooden tendon]

what is MINIMUM

[fallow igloo]

And you’ll find the minimum

[fallow igloo]

The smallest value you can get

[wooden tendon]

do you know why it’s in all caps

as opposed to minimum value

[fallow igloo]

?

Wat

[cursive patrol]

the instructions make it seem as if you're supposed to use a calculator

[wooden tendon]

I’ll check what my teacher said about it

oh yeah I’m supposed to use a graphing calculator

is MINIMUM a button on most calculators?

[cursive patrol]

no

[wooden tendon]

where would it be on a graphic calculator

[cursive patrol]

i don't know what calculator you have

[wooden tendon]

It’s a ti-84

should I just plug it into Desmos?

[cursive patrol]

you can just look up a manual for your calculator

i'm assuming this works https://www.cocc.edu/departments/math/calculator-videos/files/finding-maximums-minimums.pdf

[wooden tendon]

alr

[wary tulip]

new mqnic pfp

[cursive patrol]

yos whenever i happen to have nitro this happens

[wooden tendon]

I’ve got it figured out

Thanks

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[cursive patrol]

you can

(though for in-class work i assume your teacher will want you to know how to use your calc)

[cursive patrol]

yw

[wary tulip]

mine is about to run out. unlucky

[cursive patrol]

misfortune

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[ionic oar]

Problem is I think that the set is countably infinite

Because you can pair each element to a binary natural number right

[verbal path]

you can use cantor's diagonal argument if youve seen that before

(he used it to show that R is uncountable)

[ionic oar]

Wait are the arrays finite length here

[strange whale]

what would ...1111111 correspond to though

[verbal path]

theyre infinite

[ionic oar]

Yeah well it is under that section

[strange whale]

they never said you couldn't have infinitely many 1's

they even state (1, 1, 1, 1, ...) is an element of that huge set

[ionic oar]

hmm

fair

Wait can this just be exactly the diagonalization argument

But base 2 instead of 10

[strange whale]

yea

[ionic oar]

Well that was simple enough I guess

🫠

I'll try to mimic the proof style of the argument in (0, 1)

Thanks for your help

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[modern mural]

1+logx^yz + 1+logy^xz+1+logz^xy

[modern mural]

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[wintry coyote]

Don't know how to set up the integral

Why isn't it the same as the first one it's just that the bounds are different

<@&286206848099549185>

[honest lion]

Hmm if you're carrying only half the rope then shouldn't the weight be halved too?

[wintry coyote]

but isn't that the rate

it's lb/ft not lbs

or am i confused

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[jaunty dawn]

has anyone got an idea on how to start this task?

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@jaunty dawn Has your question been resolved?

[jaunty dawn]

<@&286206848099549185>

[strange whale]

you better try asking in #real-complex-analysis @jaunty dawn

[jaunty dawn]

u sure?

its like second semester uni

[strange whale]

I wouldn't find it out of place there

or go to #multivariable-calculus if you want

not sure you'll get much help in multi-calc either

@jaunty dawn

[jaunty dawn]

ill try it

[jaunty dawn]

you got no idea

?

[strange whale]

i have no idea indeed

i wouldn't redirect you otherwise

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Still there?

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[rotund hawk]

DId I do this right? and if so what am i supposed to do now?

[rotund hawk]

do i multiply the first side by the denominator and then add them together?

wait what, how am i supposed to get this

this is the question

how did they simplfy this

[lone raven]

Your b' got a little messed up

[rotund hawk]

also I was wrong here, the 10x should be 20x

[rotund hawk]

ye

[lone raven]

ye, then they got like denominator so they could add the numerators. With algebra manipulation it simplifies to the above. But if the problem doesn't ask for simplification then besides b' what you got would've been a good answer.

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[rotund hawk]

I think I need to know how to simplify

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[ivory sorrel]

which problem

[tawny elm]

Both

[ivory sorrel]

is this from a past jee adv paper

[crimson kayak]

Then answer is in terms of variables right?

[tawny elm]

Idts

Or theyd have written it there

[tawny elm]

No

Its 3

[ivory sorrel]

can you determine the sum of roots

do you know vitea's formula

[tawny elm]

Magnitude of beta is double of that of alpha

[tawny elm]

Whats that

[ivory sorrel]

problem 7 first, right

https://en.wikipedia.org/wiki/Vieta's_formulas

[tawny elm]

yes

[tawny elm]

yea i do know this

if the roots are alpha and beta

then alpha plus beta=-b

but beta=2alpha

so 3alpha=-b

but answer is 3??

[ivory sorrel]

that sounds wrong to me

[tawny elm]

how

median divides base into equal parts iirc

okay so alpha is 1

[ivory sorrel]

wait, I didn't notice that part of the question

sorry

yes

[tawny elm]

damn lmao i need to try questions before getting scared from diagrams

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[icy rapids]

How do I see if ac and at are perpendicular?

[icy rapids]

AC= centripetal acceleration, AT= tangential acceleration

[ivory sorrel]

The centripetal accleration is perpendicular to the tangent by defn

[icy rapids]

Ah

Thank you

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[tepid anchor]

My approach thusfar: get the eigenvalues for the matrix A which are -30, 10i, and -10i
then we see that the rescaling to be with in the stability region of Runge-Kutta 4 should be a rescale of about 0.09

however the format the question wants you to answer in only allows for 1 decimal leading me to 0.1 which wouldn't work for RK4

Anyone know what I'm doing wrong here?

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[vocal rose]

hypixel

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[tepid anchor]

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[weak creek]

What is ans of 13
I'm getting x=-2
Y=5
m=-1

[ivory sorrel]

,rotate

[ivory sorrel]

,rotate

[unkempt dragon]

Ello

[ivory sorrel]

,w solve 2x+3y=11 and 2x-4y=-24

[thorny flame]

Results provided by WolframAlpha

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

Requested by physicsrocks

[ivory sorrel]

so you have $5=-2m+3$

yeah

[thorny flame]

ƒ(Why am. I here)=I don't know

[ivory sorrel]

your answer checks out

[unkempt dragon]

What's the question here

[ivory sorrel]

Q13

[unkempt dragon]

Got it

[weak creek]

Thx

[unkempt dragon]

I second this.

Seems to check out

[weak creek]

Thx

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[rigid laurel]

is this correct?

[near storm]

Are you proving?

[rigid laurel]

yes

[near storm]

The difference of two squares factorisation I think you forgot to square root the power

[dense crescent]

Yes correct

[near storm]

Oh wait

It's multiplong by 1

[versed cairn]

I would write step number 3rd first and then step number 2

[near storm]

Yeah the order was a bit wack

[rigid laurel]

okay!!

thank you guys 😭🙏🏼

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[sonic temple]

hey! quick wording question

[sonic temple]

Define a function T : R3 → R2 by
T (x, y, z) = (x + y + z, x + 2y − 3z).

is the overarching problem

one of the sub problems is:

Show that T is onto

what is that supposed to mean? onto what?

[normal tapir]

do you know what surjective means

[clear crescent]

Here the codomain of T is ℝ². We say that T is onto (or surjective) if the image of T equals the codomain

[sonic temple]

yes i know what surjective means, basically no missing outputs

[normal tapir]

onto = surjective

[sonic temple]

ohhh i see

ive never seen that word used liked that before in this class

[clear crescent]

I hate when it's used

[prisma mesa]

why?

its shorter

[sonic temple]

what is the image of T? is it the transformation?

[clear crescent]

Surjective is ✨fancy✨

[clear crescent]

All the values that T maps to

[normal tapir]

the image of a function f:A->B is the set im(f)= {f(a) | a in A}

[sonic temple]

ok thank you !

ill probably come back later for help with this problem lol i just needed to know what it meant lmao

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[sonic temple]

heya! im back to ask about the surjectivity of a matrix (like i said i would lol)

[sonic temple]

so the matrix itself is :

waut

i lied hold on

not a matrix

this

ok back to what i was saying

should I first find the matrix representation of its basis from R3 and R2?

[loud oasis]

that would be a good start

note that we expect this matrix transformation to be surjective if it has rank 2 (the rank is the dimension of its range, and we need that to match the dimension of its codomain, which is R^2)

[sonic temple]

yup

so im thinking first i find the matrix representation

put it in reduced row-echelon to get it to rank 2, yes?

idk why they put finding the matrix representation as part 4 when its needed for part 3 kekw

[loud oasis]

well we also expect the rank to be the number of linearly independent rows (or columns). we can see by inspection that the 2 rows of this matrix would not be scalar multiples

[sonic temple]

the matrix should be

( 1 1 1 )
( 1 2 -3 )

right?

[sonic temple]

yes but if we put it in row echelon form we shuld still get rank 2 i believe

[loud oasis]

yes, row echelon form is still the usual way of finding rank, but for small matrices there are shortcuts

[sonic temple]

by preforming R2-R1 on that we get (1 1 1)
(0 1 -4)

thats rank 2, is it not

so its the same as the codomain

[loud oasis]

yes

[sonic temple]

so that means its surjective

[loud oasis]

yes

[sonic temple]

yes!

ok so im working on another problem now if you'll hold just a moment

does the matrix

(1 1 1)
(0 1 1)
(0 0 1)

have rank 3?

the question is:

Show that B = {(1, 1, 1), (1, 1, 0), (0, 1, 1)} is a basis for R3. Find
the coordinate vector of (1, 2, 3) relative to the basis B.

so i set the vectors as a matrix and preformed row operations to reduce the rows

if the rank is 3 from my transformation, the same dimension as R3, its a valid basis correct?

[loud oasis]

yes

[sonic temple]

and the coordinate vector is 2(1,1,1) -1(1,1,0) + 1(0,1,1)

right?

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[vestal plume]

Need help with the 6 marker

What i got so far

I set up and tried to solve some equations with resolving forces and moments but i got nowhere

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[fallow igloo]

Have you solved it?

[vestal plume]

Yes

[fallow igloo]

Cheers

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[winged lichen]

I need to prove that L = $(0^n1^m | n\equiv m+1 \pmod{3} \land a \neq b)$

[thorny flame]

Megu

[civic schooner]

whats your grammar

and what is a and b

[winged lichen]

$n \neq m$

my abd

[thorny flame]

Megu

[coarse tusk]

!original

[topaz sinew]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[winged lichen]

{0,1}

[molten raven]

you defined a set L, what do you want to do with it?

[civic schooner]

n equiv m+1 mod 3 implies n neq m

[winged lichen]

wait Ill type the full question

[civic schooner]

no i mean do you have productions

[coarse tusk]

you can also just send a screenshot/picture of the original question

that'll be better actually, if you can provide one

[thorny flame]

Megu

[winged lichen]

the brackets left but whatever

[coarse tusk]

0^n 1^m...?

[winged lichen]

I need to know if its regular, cfl or netiher

[civic schooner]

what does that have to do with proving it

[winged lichen]

exactly what I sent, just need to classify it

because I need to classify with proof

[civic schooner]

the second part is unnecessary innit

[coarse tusk]

am I just not smart enough to understand this

[winged lichen]

well yeah, just to clarify

[coarse tusk]

what branch of math is this?

[civic schooner]

$n\equiv m + 1 \mod 3 \implies n \neq m$

[thorny flame]

esca (@ with reply)

[civic schooner]

languages grammars stuff

[molten raven]

binary stuff i think

[coarse tusk]

umm..

[civic schooner]

basically 1 and 0 are strings, not numbers

[coarse tusk]

i shall leave

[civic schooner]

0^2 = 00

[winged lichen]

theoretical computer science ig

[coarse tusk]

ahh okay

[winged lichen]

and formal language theory

[civic schooner]

yeah that

[civic schooner]

so for it to be regular there needs to be only one production rule for every nonterminal right

first come up with a set of productions

the other way is to i guess make an fsm and see if its first order or second order

[winged lichen]

what is an fsm?

oh I googled it

its just a normal automata ok

the only way I can think of doing this is creating a CFG that represents the language

[civic schooner]

so can you express this language with just a set of left linear or right linear productions?

[civic schooner]

yep you write a set of productions

and if they all satisfy the above linearity conditions its regular

[winged lichen]

but how do I prove the set of productions generates that language?

[civic schooner]

well the production kinda defines the language lol. but if you wanna say the grammar spans all words in the language and nothing more you can use double inclusion i think

ie,
given arbitrary m and n satisfying the mod 3 condition, 0^m1^n can be generated from the grammar, and any word generated by the grammar can be expressed as 0^m1^n

[winged lichen]

$S \to 0T111 | 000T1$
$T \to 0T1 | 00S11 | 000S111 | 000S | S111$

[thorny flame]

Megu

[winged lichen]

shit how do I newline

[civic schooner]

lol press enter twice

[winged lichen]

whatever, how does that look to you?

[civic schooner]

for the second production you only need one

T to 0T1 | S

[winged lichen]

oh wait

$T \to 0T1 | 000T | T111$

[thorny flame]

Megu

[winged lichen]

looks better

[civic schooner]

also check 0001 against your condition

[civic schooner]

yep

[winged lichen]

and epsilon is needed

[civic schooner]

and make sure T includes just an S

or S can convert to T

[winged lichen]

yeah I did it on paper but usually use S' and S

had a brain fart transcribing it

[winged lichen]

yeah just fix the congruence and everthing else checks out

thanks I think you fixed it!

[civic schooner]

no i mean 0^31^1 doesnt satisfy n equiv m + 1 mod 3

other than that it seems fine

[winged lichen]

oh but back to my initial problem

is there any need to prove that this production generates L

[civic schooner]

well theres no general method

for an arbitrary grammar and language its an undecidable problem, meaning theres no algorithm to show $G\cong L$

[thorny flame]

esca (@ with reply)

[civic schooner]

for this one i think it should be sufficient to show that any production satisfies the given rule

[winged lichen]

well, theres no algorithm but there IS a way

we can do a proof by induction

[civic schooner]

yeah

[winged lichen]

but I haven't been able to do it yet

[civic schooner]

your inductive hypothesis would be that if T or S is in the form 0^m1^n ... then you show each possible production involving those is also in that form

and then have base case epsilon

and then another base case for T

dojt quote me on this though because i have not done enough formal proofs with language theory lol

[winged lichen]

assume it goes for any k < n and see if $n \equiv m+1 (\mod 3)$ still stands for any n?
its easy enough to show that T does not change those values

[thorny flame]

Megu

[winged lichen]

yeah the hard part here is the formal proofs

[civic schooner]

yeah thats the thing

you do a proof by cases with weak induction for each case

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[winged lichen]

just a quick sketch I did

$S \to T11 | 0T \newline T \to 0T1 | 000T | T111$

[thorny flame]

Megu

[winged lichen]

easy to prove n != m too

since congruence is done on m+1

[civic schooner]

you have

$L\coloneq {0^n1^m:n\equiv m+1\mod 3$

$G\coloneq$

$S\to 0\mid 00011 \mid 001 \mid T$

$T\to 0T1\mid 000T\mid T111 $

Proposition. For any word $w$, $w\in L \iff w$ can be generated by G.

Proof. To prove $w$ can be generated by G $\implies w\in L$, proceed by induction.

Base cases: All terminal combos that can produce S.
Inductive Hypothesis: $T\in L\implies 0T1, 000T, T111\in L$

this is how id do it for the backward direction. the forward direction is a little trickier i believe

the other direction i believe you can prove by just reversing the productions. but idk if thats a logical step

[winged lichen]

0T0 seems wrong

since 00 doesnt belong to L

[civic schooner]

sorry that should be 0T1

[winged lichen]

oh okok

[thorny flame]

esca (@ with reply)

[winged lichen]

I changed S to this

and with epsilon on T I think everything is covered

[civic schooner]

idt that works

[civic schooner]

epsilon is 0^01^0

so you cant include it as a terminal

best you can do is Tepsilon or epsilonT

which is meaningless

[topaz sinew]

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[winged lichen]

.reopen

[topaz sinew]

✅

[winged lichen]

why so?

[civic schooner]

for one you need to cover the case 001

i guess it works otherwise

actually nvm it doesnt work

[winged lichen]

ffs

[civic schooner]

latex

[winged lichen]

$S \to T11 | 0T \newline T \to 0T1 | 000T | T111 | \epsilon \newline$ but since T's purpose is not changing the $n \equiv m + 1 (\mod 3)$, wouldn't it be fine to keep $\epsilon$ there? otherwise we can't end a word

[thorny flame]

Megu

[civic schooner]

anyway $0^n1^m : n\equiv m\mod 3$ are valid by this grammar

[thorny flame]

esca (@ with reply)

[civic schooner]

for example you can say 01 and be done with it

[winged lichen]

well, you can do it by $ S \to 0T \to 00T1 \to 00\epsilon1 -> 001$ can't you

fuck me

[civic schooner]

ah yeah it does i guess

but this includes words that are invalid

[winged lichen]

for example?

[civic schooner]

^

01 is prefectly valid, unless you say S is the only valid final word

the better way to do it is enforce $n \equiv m + 1\mod 3$ in the production that deals only with terminals i think

[thorny flame]

esca (@ with reply)

[winged lichen]

I don't know if there's any difference but in my university we define S as the starting point of an automata
so, if you convert that CFG to a PDA, you would start with

[thorny flame]

Megu

[winged lichen]

since you can't get to S from T, and T never changes the modulo operation's value, $(\forall n)(\forall m)(n \equiv m+1 (\mod 3))$

[thorny flame]

Megu

[civic schooner]

well youre thinking of it in terms of the starting point of an automata, but it needs to be valid in the other direction too

ie, the grammar cant generate words not in the languahe

[winged lichen]

oh does it? where I study there's never been been that problem

I guess because we're just assuming the CFG is describing a PDA, basically

here, from the notes

on how to construct a PDA from a CFG

[civic schooner]

ah okay

well i mean

i guess for your purposes it doesnt matter then? but in this case its not too hard to get a grammar that generates only the words in your language

but yeah you can prove in the forward direction that 0^m1^n can only consist of 0^c(000)^a(111)^b1^c for arbitrary a, b and csurrounded by what you have in your production for S

[winged lichen]

yeah I guess it makes the proofs less tedious

thanks for the help man

got it figured out now!

[civic schooner]

npnp

[winged lichen]

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[rose bough]

I have a small question

[rose bough]

when finding the equation of a straight line

what is the reason to continue to 3x - y = 2?

why not stop at y = 3x - 2? As that fits y = mx+b

[loud oasis]

the last one is in "standard form" which is ax + by = c (less useful for graphing but can be useful for solving simultaneous equations)

[rose bough]

got it, thank you

.close

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[still dirge]

Hello there.

[still dirge]

I would like to ask a few questions relating to these problems.

[neon iron]

What are they?

[still dirge]

A: If I am correct, they wish to factor the whole polynomial for the sake of simplifying the operations for easier solving, yes?

B: The process of checking answers most likely involves checking whether or not
C:```

Well, I am honestly quite confused as to the exact nature of B due to lack of understanding of the actual flow of the problem.

Been stuck on this for, what might be, 20-50+ hours, so.

[neon iron]

For A, it can vary, though typically its easier to find the zeroes. Say we had x^2 + 4x + 4 and I asked when this is 0, you'd have to do math and work this out, but if I gave you (x+2)(x+2) you can just look and see oh, it's -2, besides doing this is much easier than working through the entire pythagorean theorem

And can you elaborate on B?

And C didn't paste :p

[still dirge]

Ok, yeah, I'm not going into CS.

I yield.

[neon iron]

lol whats ur issue?

[still dirge]

It's very hard to put into words but....

A: The purpose seems complex, and multi-faceted. There are not only a million reasons and differences in interpretation with this problem, but also an extreme level of complexity where layers stack upon layers.

[brave coral]

Huh

[still dirge]

The only smart move here is to just take what I can.

[still dirge]

So you're saying there are other methods to solve this? Is the factoring actually useless?

[neon iron]

There are other methods, but factoring is easier

[still dirge]

I don't understand when solving it why they broke it into binomials, and .....

.....

[neon iron]

like 1 sec

[brave coral]

No it's not. It's probably the fastest method to do by hand

[rose bough]

you can solve it with this formula

[neon iron]

Like this isn't the only method but like would you rather do that or this

Hey I already got that >:(

[rose bough]

haha

[still dirge]

[neon iron]

It's a process, it's hard to say if it has a purpose

It's important for math though

[rose bough]

the purpose depends on what you wanna do with it

[rose bough]

yes

what here got you stuck for 20-50+ hours I don't really understand

[neon iron]

Are you having any troubles with the actual factoring, or just its significance?

[brave coral]

(x - a)(x - b) = 0

for x to satisfy that equation, x = a or x = b
because (a - a) = 0 and (b - b) = 0

not sure if that's what you were asking

[still dirge]

A: thought X values were unknown values rather than input values that had to be solved for

B: thought they were solving for the problem which they arent (they are only finding values of X at a point and they are actually solving for the functions result)

C: I dont understand why they would factor and there were multiple interpretations of how they were factoring and why they would be done

D: I was having extreme difficulty thinking in straight lines or weaving any past experience to the present causing me to be unable to solve or properly process things  which caused me to be unable to properly process the binomial multiplication for some time.```

There were multiple nightmare class problems within this that caused me severe difficulties for long hours.

E: It took an absolutely rediculous ammount of time to find even a single possibility or outcome that would explain how they got this.

Until I realised there probably isnt an order they're probably doing literally whatever theyw ant

and in reality i got into a logical trap assuming they were doing something

when in reality they werent doing anything at all and were just reforming the problem however they wanted so long as it was equivalent

i.e ghost problem logical trap

and even then i still cant understand why exactly they would factor it in that way

[neon iron]

A. Variables are kind of all about input/output, like if we had 2x = 4, you need to input the right value to make the equation true, so it is simpler as x just happens to be a mystery value, but like with your problems x can be multiple things and still generate a true solution

Does that make any sense for starters?

[still dirge]

yes : ( (why would I not know that oh the years...)

[neon iron]

No :(

We :)

[still dirge]

https://cdn.discordapp.com/attachments/1181451878849060936/1251321008225521675/image.png?ex=666e2700&is=666cd580&hm=960202517ba9888cbaff4b9b15f971ca488c49d58429881b6589a600032acde8&

Ok, fine.

So, what next?

[neon iron]

B. When given a quadratic like this, where f(x) = (quadratic xd), we consider the solution to be a value of x where f(x) = 0
So like for just x^2, you would say the solution/zeroes/x-intercepts/whatever-term-teacher-uses to be 0, since 0^2 = 0

Of course usually there will be a question more than just terms which you seem to only have, so you just do what the question asks

That is kind of on whatever platform you're using since, unless you cut out some of the info, pretty inspecific of them

But does that make sense thus far too?

Tryna go 1 by 1 so I don't just type you an essay and none of it make logical sense lmao

[still dirge]

https://tenor.com/view/pov-you-giga-chad-chad-meme-gif-25615024

[neon iron]

We build up like Fortnite (that goes hard)

[still dirge]

problem with 99% of teachers

[neon iron]

Well it's just me and you on Discord so at least we have the priveledge to do that lol

But everything good on B?

[still dirge]

https://cdn.discordapp.com/attachments/1181451878849060936/1250915709010182246/image.png?ex=666dff09&is=666cad89&hm=4aba72a0e87afb08d58d44e7a3179d5e3f5552da75a453d73048543aba949ed2&

To be thoughtless...

Ah, ok.

I just... needed to eliminate the rest of the distractions.

Ok, so, why is zero considered a truthy (coding term) solution?

oh, I remember

Because x were just input values, so you want to find where the function equates to zero.

I.e. the problem is inverted (you'd probably assume you're solving for something related to x)

Correct?

[neon iron]

Yep, sounds right!

[neon iron]

Though it's not 0 that's the solution, it's x when f(x) = 0

[still dirge]

Does f stand for function affecting x?

[neon iron]

It reads "f of x" but it's function f with parameter x, since you mentioned CS stuff you can probably recognize this notation a bit

[still dirge]

oH

yes, ty

[neon iron]

Np

[still dirge]

[neon iron]

For the rest I'd probably reccomend you find resources on factoring quadratics, I can try to explain it for you though at a point it might just be beneficial for you to watch youtube

It's kinda tricky at first but as you get used to it you'll be doing it mentally fast (in most cases, not saying you can do EVERY ONE instantly but those at least you can do with ease if you just keep pushing)

[still dirge]

Well, ok, thank you!

[neon iron]

Np

[still dirge]

ima go and try to

https://media.discordapp.net/stickers/1251309312090046494.png?size=160&name=Meta+Calculations

https://media.discordapp.net/stickers/1231850950302367795.png?size=160&name=Calculations

https://media.discordapp.net/attachments/1015759484708208792/1045835008012714017/a_3a6633f55a6525e045c618e50b97782a12.gif

[neon iron]

https://cdn.discordapp.com/attachments/944709788569309214/1245188975698903050/attachment.gif

[topaz sinew]

@still dirge Has your question been resolved?

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[ocean terrace]

whats the original question

[shadow shadow]

Quadratic are also polynomial, there are exponential equations, trigonometric, differential equations

[ocean terrace]

i dont think u can easily write out exhaustively a list of all algebraic expressions

one example is exponential to the x

and u can nest exponential to the x to create new operations and so on

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[sly shoal]

Hi! I'm kinda confused by the cos, how do I solve this? Thank you!!

[mellow arrow]

do you wish to evaluate f(x) at x=0.1?

[sly shoal]

Yes

[restive inlet]

horrid notation

$f(0.1) = \frac{\cos(2(0.1))-\cos(0.1)}{0.1^2}$

[thorny flame]

ℝαμΩℕωⅤ

[restive inlet]

if you want a decimal approximation, just eval with your calculator

and 0.1^2 isn't 0.1

[sly shoal]

I don't have a calculator

What does cos do though?

[restive inlet]

you'd be replacing ALL x with your desired value,
including the x in f(x)

cos is a trig function

have you done any trig before?

[sly shoal]

a little bit a couple years ago

😭

Is this just something I have to have a calculator for?

.close

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[restive inlet]

you can use stuff like taylor series to get a decent approximation

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[ashen tiger]

Hello.

[ashen tiger]

Need to know the formula to find the jump of percentage.

🥺

I have made the summation already.

270 and 375 respectively.

[topaz sinew]

@ashen tiger Has your question been resolved?

[ashen tiger]

No.

Can't believe I am not getting any help for such a simple question.

I am a Humanities student but for most this should be a breeze.

[ashen tiger]

[shadow shadow]

The first image you sent is that related to the question? It looks like a different question

[ashen tiger]

So sorry.

Yeah, I made a mistake.

Let me just reopen a new thread.

Solved.

Close.

.solved

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[ashen tiger]

.close

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[lapis ibex]

Hi, I need help with a combination question. I am still fairly new to the topic so I don;t really understand how to do this question, everytime I do it I seem to get the same wrong answer. The question is the image.

[lapis ibex]

I did 6C2 * 8C2 * 5C1 * 14C1

I dont get how its wrong

[neat stratus]

🤔

[ivory sorrel]

that isn't right?

[lapis ibex]

Nope

[ivory sorrel]

maybe you made a calculation mistake ?

[lapis ibex]

the answer is 11200

[ivory sorrel]

,w c(6,2)*c(8,2)*c(5,1)*c(14,1)

[lapis ibex]

I also got 29400

[ivory sorrel]

,w 29,400/3

[thorny flame]

Results provided by WolframAlpha

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

Requested by physicsrocks

[ivory sorrel]

I think your answer is right

I could be wrong though

[lapis ibex]

Ah ok thats good

Thank you very much for your help 🙂

[ivory sorrel]

I may be wrong

[lapis ibex]

its all good

[ivory sorrel]

don't take my word for it

[lapis ibex]

Alright but still thank you

[ivory sorrel]

found an answer on quora

[lapis ibex]

What did it say

[ivory sorrel]

essentially consider it case wise

you either choise a subject from A

, for from B, or from C

[ivory sorrel]

https://math.stackexchange.com/questions/1410213/in-how-many-ways-can-a-student-select-six-classes-from-three-groups-if-they-must