help-27

[magic thicket]

If you go to a subfield (that still contains the coefficients) or a field extension, it's conserved. But if you send it into a field of nonzero characteristic (if it was in Mn(Z)) then it may now

Z_n is a field iff n is prime

[frail igloo]

hey i knew that!

why is that the case, though?

[magic thicket]

In which case it's also common to write it Fp

[hazy totem]

because the two nontrivial factors of a composite are nonzero but multiply to 0 in Z_n

so that means not all nonzero elements are invertible

[magic thicket]

there's a b such that ab = 1 [n] iff there's b, k such that ab + kn = 1, i.e. iff a and n are coprime

which is true for all a in {1, ..., n-1} if and only if n is prime

also all finite integral domains are fields

[frail igloo]

i shouldnt have asked. i'm so far behind in linalg i'm barely hanging on. you guys r inspirations tho.

thanks for the help ;-;

[hazy totem]

it's okay, we've been doing math for a while

[frail igloo]

while we're here though, if you don't mind..

i've asked this before and i still don't get it

why is 2^-1=3 in Z_5

:C

i don't know how to find inverses

THEY DIDNT EVEN TELL US

[hazy totem]

because 3*2 = 6 ≡ 1 [5]

since all addition and multiplication is done modulo 5 by definition, you have 3*2=1 in ℤ₅

[magic thicket]

btw this tells you how to find inverses for high values of p by euclid's algorithm

but for small p, it can be faster to just try them all

[hazy totem]

As for how to find them, since it's a very small finite field, you can just try elements until one of them gives 1 modulo 5

likewise, 4*4 = 16 = 15 + 1 ≡ 1 [5]

[magic thicket]

look up something like
euclid's algorithm for solving Bezout's equation
if you're interested

[hazy totem]

(that's is unrelated btw, you don't need to know this here)

[magic thicket]

as we said

[frail igloo]

okay something about this i'm just not getting lol.

why does this tell me 2^-1=3 in Z_5?

[hazy totem]

the multiplicative identity is 1

[magic thicket]

what's the definition of an inverse ?

[hazy totem]

ok well you seem to be at it already i'll leave you to it

[frail igloo]

i know 2^-1=1/2

right?

[magic thicket]

no

that's in R

[long fog]

It depends on field

[magic thicket]

which is totally distinct from Z5

1/2 not in Z5

[frail igloo]

right so this is clearly where i'm so lost

how do i like... translate this

lmao

[magic thicket]

you don't

2^-1 doesn't even exist in Z4

[frail igloo]

Z_5?*

i remember you saying this last time

[magic thicket]

it doesn't exist in every Zn

[magic thicket]

because of this

[long fog]

It needs to be Zp

[magic thicket]

doesn't have to

just needs to be an odd n (> 1) for 2 to be invertible

[long fog]

For 2 but does it have inverse for all elements then?

[magic thicket]

we already went over this

[long fog]

Okay

That's good

[frail igloo]

welcome to the party 🙂

[hazy totem]

let's keep it to one helper

[long fog]

Yes

There won't be no parties no more

[magic thicket]

[frail igloo]

idk 1 helper can contain all my stupidness

im a bit of a handful

[magic thicket]

I contain quite a bit of stupidness myself

worry not

[frail igloo]

THATS THE SPIRIT

[magic thicket]

especially when trying to be a helper at 1 am
it didn't work out so well

but it's not 1 am rn so we're good

[frail igloo]

YEEHAW

so here we go

they say here 2^-1 = 3

i'm assuming that means in Z_5, right?

[magic thicket]

yes

this is a computation in Z5

[frail igloo]

i feel like you've already explained, but i don't understand

how do you perform that computation?

[magic thicket]

you just notice that 3 works

because 2*3 = 6 = 1 [5]

after having tried 1 and 2, which didn't work

[frail igloo]

https://www.youtube.com/watch?v=shaQZg8bqUM i think this is what you've been trying to explain

i just needed it to be explained like i'm a baby

[magic thicket]

that's the euclid algorithm version

for when trying every possible number is too long

but it's a lot more computational and harder than just trying every number from 1 to 4

so a very bad idea for n = 5

[frail igloo]

hmm..

[magic thicket]

because 2 is invertible, you know its inverse exists
is it 1 ? that makes 2 so no
2 ? 2*2 = 4 so no
3 ? 2*3 = 6 = 1 [5] so yes

[frail igloo]

so you're asking what gives me a remainder of 1?

[magic thicket]

yes

[frail igloo]

a definitely of an inverse is something with the remainder of 1???????

[magic thicket]

you know Zp is a field but didn't study basic structures ?

like groups and rings ?

[frail igloo]

i swear they taught us nothing about doing computations in fields in this course.

i'm a first year student and it's for linear algebra

[magic thicket]

yay US curriculum

[frail igloo]

canada* 😛

[magic thicket]

meh

[frail igloo]

same thing

[magic thicket]

the french one is very good at precisely that

[frail igloo]

i've been drawing my numbers in a clock-like circle for some computations like for Z_5 2+5 i'd start at 2 and add 5 counting around the clock 🙂

[magic thicket]

a is said to be invertible in a ring A if there is an element b in A such that ab = 1. In that case we call b a's inverse and write b = a^-1

[frail igloo]

see i don't even know what a ring is

:C

[magic thicket]

it has +, -, * (and distributivity, so a(b+c) = ab + bc, but that's a detail) and sometimes division

precisely division by invertible numbers

if you can divide by any nonzero element, i.e. every nonzero element is invertible, you call it a field

Z is a ring, but not a field. The smallest field of R is Q, the rational numbers

[magic thicket]

just perform the operation in Z, then take mod n at the end

3*4 = 12 = 2*5 + 2 = 2 [5]

though note you can take mod n at any and every step if it's more convenient

4^7 = 16 4^5 = 1 4^5 [5] for example

so you notice that 4 = 4^-1

4² = 1 [5]

so 4^7 = (4²)^3 * 4 = 1^3 * 4 = 4 [5]

[frail igloo]

ohhhhhhhh

[magic thicket]

which is also what they noticed there

[magic thicket]

more generally, by Fermat's theorem, a^p = a mod p

more generally, the smallest k such that a^k = 1 [n] is phi(n), the euler totient function
Then a^-1 = a^(phi(n)-1) so a is invertible
so if a isn't coprime with n, there's no power of a (besides 0) that equal 1

but you probably don't need to know that yet

[frail igloo]

ima be honest my mathematical maturity is so underdveloped

[magic thicket]

how long have you been in undergrad ? 6 months ?

[frail igloo]

yea

[magic thicket]

you're only just starting out

[magic thicket]

and it's not like these curriculums make you grow more mature much

[magic thicket]

proof based stuff much more so

[frail igloo]

i guess it's not obvious to you, because you assumed i had a bunch of background knowledge because maybe it's uncommon to work in fields in linear algebnra without this other background

so it took us a while to get to "when you see 2^-1 in Z_5 you should be thinking 2 multipled by what gives me a remainder of 1 in Z_5?"

[magic thicket]

I know americans take linalg early

[magic thicket]

you just made me realize that's a way of finding inverses

though quite a bad one in Fp

[frail igloo]

and good god hopefully i recapitulated it properly

lmfaooo

[magic thicket]

it's no faster than this

[magic thicket]

which is why tacking on some Fp stuff for the sake of it feels a bit off

[frail igloo]

what even is Fp stuff? xD

see i don't even know ;-;

[frail igloo]

like i have no idea what this means

[magic thicket]

Z5 stuff

[frail igloo]

ah ok.

may i ask where you are in your math career?

just curious

[magic thicket]

finished undergrad a month ago

taking exams to enter grad school this month

[frail igloo]

pure math?

[magic thicket]

math major, physics minor, likely headed into a more programming/algorithmic career though

[frail igloo]

very cool! i kinda wanna do physics.

i mean that's what i've taken courses for already so that's already been set in motion

we'll see how that pans out for me heh

and by kinda i mean i really (think) i wanna do physics

but not like you asked.

thank yyou for your help

and determination xD

[magic thicket]

don't sleep on calculus then

and remember that for a nontrivial solution to exist, the determinant must be zero

[frail igloo]

calc 2 has been much more enjoyable than calc 1

[magic thicket]

integrals, differential equations
vector calculus for electromagnetism too

[frail igloo]

ik ill need it all but calc 1 felt so abstract by comparison

calc 2 feels creative and fun

[magic thicket]

vector calculus which we got introduced to in physics class lol

[frail igloo]

linear algebra for quantum stuff ;-;

[magic thicket]

calc 1 is the simple stuff

[frail igloo]

and other physics stuff to probably

[magic thicket]

still high school really

just apply the algorithm

[frail igloo]

i never took calc in highschool

[frail igloo]

thats prob why it felt so boring xD

[magic thicket]

calc 2 requires a bit of thinking to be fast, but stays easy

just need to have an intuition/experience for what method to use

it's after calculus that things get real

like real analysis

actual linear algebra

[frail igloo]

i dont think ill need a real analysis course for physics but ive heard many people talk about how difficult it is

isnt it just nitty gritty calculus?

like with proofs n stuff?

[magic thicket]

oh poor little soul

[frail igloo]

i have no idea xD

[magic thicket]

it proves calculus, but damn the last time I took a derivative was a long time ago

it's a lot more interesting

[magic thicket]

the french undergrad curriculum is real analysis instead of calculus, but it also includes calculus

for first year students

that's why it's so good

and also why the prépas are meant for students who were pretty good at math

it's what I don't like about the american system, it's basically just a spoonfeeding of properties that have to be learned

when in fact they're obvious

the proof is elementary

what you understand you need not learn

[frail igloo]

that sounds intense wow

[magic thicket]

12h a week of math lessons, add to that some 12 or 15h of math at home

it's a lot of fun

but it's different

[frail igloo]

maybe i should just man up and move to france

[magic thicket]

the first assigment defined the theory of integrals on infinite domains and computed the integral of sin x / x
our 3rd assigment studied a parameterized integral, proved the irrationality of pi and computed zeta of 2 and 4

with 1 assigment a week

for math

• actual exercises (i.e. worksheets)

[frosty cradle]

what book(s) did you use?

[frail igloo]

was your first year physics course calculus based?

[magic thicket]

a teacher sir

no books

[magic thicket]

not really no, just defined a bunch of stuff in all fields of physics, sort of foundational stuff, and did basic stuff that only involved high school math

so only derivatives and integrals

[frosty cradle]

that in and of itself is better than the US. every class requiring a $150+ textbook

[magic thicket]

2nd year is where it got actually mathy, with partial differential equations, vector calculus in electromag, fourier stuff somewhat, etc

[magic thicket]

one is victim to capitalist consumerism and monopoly, the other one is state funded

[frail igloo]

they've gotten a bit better. most of our textbooks are online now, thankfully (so far, at least) but what i have bought really just felt like an opportunity for proffs to sell their stuff to students for a few extra bucks

i didnt even use my chem textbook and all of the solutions they had online outside of the textbook for free

🙂

feelsbad

[magic thicket]

chemistry ? pfff

feelsbad

[frail igloo]

i didnt take chem in highschool either so that was lots of fun

[magic thicket]

have some fun with this btw if you ever feel like it

[devout snow]

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[snow willow]

Short question. Why is it in a big brace now?

[snow willow]

why not: 4x^2 + 3x -28x -21

Sometimes without and sometimes with

i dont understand

[mystic oak]

The size of the braces have no significance as far as i know, the coloring is just to signifiy the correlation, that the expression on top is the same as the expression on the bottom

And the braces exist because the - in front of the braces at the top hasn't been distributed through the expression yet so you need some way to show it applies to the whole thing

[snow willow]

Oh okey. Imagine, if there would be a * or /. Does there still have to be a big brace?

At + and - a bracket must be set

But at * and /?

Are you german actually?

[mystic oak]

Norwegian

You can imagine the -(4x^2+3x-28x-21) being the same as -1*(4x^2+3x-28x-21)

[snow willow]

I know I know

[mystic oak]

The point being yes, if there is - in front of brackets, you will have to take it into account regardless of operator as the - applies to everything inside the brackets

[snow willow]

Thank you.

.close

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[alpine reef]

lets say a pie of length 1 in half precisely in a hypothetical scenario where flaws and miscalcualtions cannot happen you cant have two pieces of 0.5 each because the middle shared boundary cant be given to both of them and if thats the case then how can 1 = 1/2 + 1/2

[mystic oak]

What is the question here? The boundary between theoretical math and real world interactions where perfect fractions are hard to do?

[cosmic gust]

"you cant have two pieces of 0.5"?

[alpine reef]

not precisely

[cosmic gust]

Not sure what you mean

[alpine reef]

can you theoretically cut something of unit 1 into two precise pieces of 0.5 each?

[cosmic gust]

Yes?

[alpine reef]

but if thats the case then the middle part (0.500000) should't it be attached to one of the 2 pieces?

you see where i am going with this?

[cosmic gust]

Aren't you talking about a case like this?

[alpine reef]

thats a good example

[cosmic gust]

So what's the problem here?

[alpine reef]

for instance if we where to cut it in half

where would the black line belong?

which of the 2 pieces

[cosmic gust]

It's a hypothetical line

It doesn't exist

Or 1D(doesn't have a height, just length)

[alpine reef]

even in this case

[cosmic gust]

And if that line has a height you can always divide it by two

[alpine reef]

even then

lets say the diameter is exactly 1

0.50000000.... is the exact middle

[cosmic gust]

Let's just use square since I guess there is no need for a circle

Let's say you splitted this into two pieces

And let's say it has a side length of 1 and so area of 1

What is the problem here

[alpine reef]

if you want to cut it in half each piece should theoretically be 0.500000....

[cosmic gust]

Yes

Or just 0.5

[alpine reef]

but if the first piece is 0.5000000....

the second one cannot be exactly 0.50000000... because the former has taken the middle part which is the infinitely precicse 0.500000

[cosmic gust]

There is no middle part

[alpine reef]

can you see what i mean?

[cosmic gust]

It's an imaginary line

Even if there is a line

You can just divide that line to two pieces

[alpine reef]

i can both see and not see what you mean

[cosmic gust]

Okay let me visualize it

[alpine reef]

the fact that i got this question on my own by studying algorithms and time complexity

[cosmic gust]

It's not really a question

It's a logical process

[alpine reef]

elaborate if you may

[cosmic gust]

Like

Saying 2 + 2 is 4

We accept that as a truth

And build on it

[alpine reef]

ok

you mean the same goes with fractions

including the one i stated above

[cosmic gust]

You can always divide things into half

Even the line itself

[alpine reef]

we consider the line to be infinitely thin?

[cosmic gust]

And in most cases those lines are hypothetical

[cosmic gust]

Normally we do

[alpine reef]

but even if we cut the line in half

the same problem occurs

[cosmic gust]

And in mathematics we call them 1 dimensional, heightless

[alpine reef]

where does the exact half of the line go

[cosmic gust]

If it doesn't have a height 0/2 = 0

[alpine reef]

so its basically a concept

[cosmic gust]

Yes it's just imaginary

You can draw these imaginary lines everywhere

So does that mean nothing can be splitted? No.

It's one of those questions when you randomly get a weird idea and think that it's true

I've been there

[alpine reef]

yeah thats exactly the case

lives here rent free

[cosmic gust]

Like randomly thinking, how do we even know if 2 + 2 is 4

[alpine reef]

yeah its a bit of a (math) existential crisis

[cosmic gust]

Yeah we can call it that :p

[alpine reef]

anyway thanks for your interest lets hope it sinks in as an idea in my head

[cosmic gust]

My pleasure

You can close with .close

[alpine reef]

.close

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[torn wave]

this is more of a question for your teacher because there is no general convention for how to make results like these look "nice"

personal preference plays a role here too

[restive river]

I dont wanna so it

Because then

It will become

• (3-2root3)

[torn wave]

i would advise to distribute the -1 because it would be the same as flipping the order of subtraction

[earnest haven]

the only thing that you should do is just put the negative sign in numerator
thats a convention but people follow it only in earlier classes

[torn wave]

-(3-2root5)=2root5-3

thats about it

[earnest haven]

explicitly writing the 1*(3-2root5 somehow makes it worse)

[earnest haven]

yea that should be it

[restive river]

So

[earnest haven]

so do whatever you want

its the same thing either way

if you teacher wants you to do it this specific way so do it like this

[restive river]

Wha

[pearl moon]

If you teacher asks you to do that change the teacher

Nahh

[earnest haven]

ok dont do that definitely

[pearl moon]

2√5-3/11

[restive river]

Iforgot to put the -

[earnest haven]

you just removed the negative sign altogether

[pearl moon]

Multiply numerator and denominator by -1

[earnest haven]

do this in the numerator

[restive river]

Dis is wat I meant

Wait

Yes

[pearl moon]

Nope you need parentheses

[restive river]

Mk

But like

How is

[pearl moon]

Like -(3-2√5)/11

[restive river]

Yea

[earnest haven]

instead of just a floating negative remember it is actually -1 so when you multiply that by (3-2root5) it becomes -1*3 -1 times -2root5

so it becomes -3 + 2root5

[restive river]

Wait

Yes

[earnest haven]

exactly

just write 2root 5 before -3 if you want

just makes it look neater

[restive river]

So dats the final answer

[pearl moon]

Yup

[earnest haven]

yes

and in third step you need not put the parenthesis
the parenthesis is used to convey that that negative sign is being multiplied to the whole numerator and not only to the first number to the left in the numerator

[restive river]

Coz of the fact -(a/b) = -a/b = a/-b ?

[pearl moon]

Yup

[restive river]

Thanks guys

[pearl moon]

Welcome

Sooo

Close it

@restive river

[restive river]

.close

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[devout snow]

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[exotic stump]

what is your question

i think there is a sign error in this

[restive river]

a/b/c

How do solve that

I

[exotic stump]

not sure what that means

[restive river]

@exotic stump

[exotic stump]

solve it for x?

[restive river]

There is a fraction (1/ 7-2root10 )that is getting divided by 3

How do I divide a fraction by a number

Thats what im asking for

[exotic stump]

dividing by a fraction is equivalent to multiplying by the reciprocal

[restive river]

Can u draw it

[exotic stump]

$\frac{a}{\frac{b}{c}} = a\cdot\frac{c}{b}$

[woven radish]

a disappointing son

[restive river]

Wtf

[restive river]

Just look at this

Thats my prob

<@&286206848099549185>

[gilded axle]

ok, im not good with this but i'll explain some of it

they switch x into 1/x, which causes the signs to flip (second part)

so you can see '-' on the bottom, '+' on the top on the second preview

[restive river]

Im just asking why

They didnt simplify it

Yo @gilded axle

[gilded axle]

hm

[restive river]

Hm

weird grapes

Cool doe

lmao

::RogerThat:Roge

:Perfect:

girls

[jaunty musk]

They didn't simplify just because there isn't a need for it

[restive river]

But

whats wtong with simplifying it

[jaunty musk]

And what the exercise wants you to find is x+1/x, knowing what x is

[jaunty musk]

Nothing ?

[restive river]

But simplifying x

Can help us find 1/x quickly

I think

[jaunty musk]

There's nothing wrong with simplifying x. It's just that x+1/x will be an addition of fractions

In order to add fractions into a single one, what do we do ?

[restive river]

Add fracs into a single one?

Wat

[jaunty musk]

Yeah, like

Imagine you have

[restive river]

Im imagining

[jaunty musk]

Imagine you have to calculate $$\frac{3}{5} + \frac{4}{7}$$

[woven radish]

Epsilia aka Mellow

[restive river]

Ik make denos same

[jaunty musk]

Exactly

So in order to calculate x+1/x

Since they both are fractions

The smort thing would be to try to see how you would make denos same

Once you add them

[gilded axle]

ok why do you need to simplify it?

[jaunty musk]

And when you have these square roots, you already know it will be about multiplying by the conjugate of the denominator

Luckily enough, the denominator of 1/x is exactly the conjugate of the denominator of x

So this makes things a lot simpler

[restive river]

Right

U right

But

[jaunty musk]

You just have to multiply each fraction by the conjugate of the denominator of the other fraction

Which is exactly what your prof did

[restive river]

U right

But

[jaunty musk]

Yup ?

[restive river]

U ARE SO RIGHT DAM

but

Like ive been simplidying in the whole exercise right so in mah head im like I gotta simplify it too ya know and so I think ill simplify it but wat I learned from dis is tat u dont always have to simplify it and also the reason why i didnt go with the prof its coz, to add fracs u shud make denos same and here dey just combines both fracs and their denos arent even same so my brain cant accept that

[jaunty musk]

Yeah you don't always have to ! Exactly that's a good thing to memorize. There is a reason why we simplify, and it's not always that same reason we have as an objective in a problem.
I guess what feels a bit problematic to you, which I totally understand, is about how you are not having this agility with fractions yet. It will come soon enough after observing the mechanism behind it every time 👍

If you don't get why something is being calculated a certain way, try your best at understanding why things are the way they are

It's imo the best way to get your mind more flexible with these

If we've finished, you can close the channel with ".close"

@restive river

[restive river]

No like

[devout snow]

@restive river Has your question been resolved?

[devout snow]

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[vague stag]

What is the primitive of 1/(1+e^x)

[haughty bolt]

Let u = e^x, then du/dx = e^x and dx = du / u

substitute these values

into the integral

∫ 1 / (1 + e^x) dx
= ∫ 1 / (1 + u) * du / u (substitute u = e^x, and dx = du / u)
= ∫ (1/u) / (1 + u) du

(1/u) / (1 + u) = A/(1+u) + B/u

Multiplying both sides by the denominator (1+u)u

1 = A u + B (1+u)

we have

Setting u = 0, we get B = 1. Setting u = -1, we get A = -1/2.

(1/u) / (1 + u) = -1/2 * 1/(1+u) + 1/u

no subastitue back to intgral

∫ 1 / (1 + e^x) dx
= ∫ (1/u) / (1 + u) du
= ∫ (-1/2) * 1/(1+u) du + ∫ 1/u du
= -1/2 * ln|1+u| + ln|u| + C
= -1/2 * ln|1+e^x| + ln|e^x| + C

the primitive of the function f(x) = 1 / (1 + e^x) is:
F(x) = -1/2 * ln|1+e^x| + x + C, where C is the constant of integration.

[vague stag]

Can’t be that complicated to find a primitive

[haughty bolt]

i dont know much 😅

[vague stag]

Well thanks for trying to help anyway

.close

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[inner void]

Exercice 1

[inner void]

Exercice 1

[rugged sparrow]

multiply both sides by 7

[inner void]

The thing is I don't even understand the question compare what?

How am I supposed to compare two unknown numbers ???

[rugged sparrow]

probably a linear equation in 2 variables

[inner void]

The value of a and b is unknown

[rugged sparrow]

or draw graph!?

[rugged sparrow]

if you take a=something you can find b with that respect

[wooden veldt]

Probably wants you to say which one is bigger

[inner void]

Hold on

This is the original exercise

As you can see in the first case we have a = 8/15 and b =9/20

Pretty clear but in the following cases It just doesn't make sense to me

[wooden veldt]

You need to say whether a is bigger or b is bigger just using the equation a-b = -4/7

[inner void]

Mhm and do you have any idea on how I can proceed to prove which one is bigger

[devout snow]

@inner void Has your question been resolved?

[devout snow]

@inner void Has your question been resolved?

[inner void]

Ummm does someone know what to do?

[frosty cradle]

you have a - b = -4/7? and it is asking which of a or b is bigger?

[devout snow]

@inner void Has your question been resolved?

[inner void]

Yes

That what I'm asking for

[frosty cradle]

if you subtract one number from another, and the answer is negative, do you have an intuition for which is bigger?

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[little helm]

wait can someone tell me,why this has no extreme point?

[devout snow]

Please don't occupy multiple help channels.

[chrome girder]

it's only defined on an open interval (-3,3)
it has no point where derivative is 0
it's differentiable everywhere

so, no extreme value

[little helm]

but i got -6

would it be right when the intervall would be longer

[chrome girder]

the interval can't be longer

ln not defined

[little helm]

ik but i mean if i would try to find an extreme point would i get an answer?

[frosty cradle]

how did you get -6

[chrome girder]

setting derivative = 0

[chrome girder]

this is the whole function it goes off to infinity

[frosty cradle]

,w calc 1 / (3 + -6) + 1 / (3 - -6)

[woven radish]

Results provided by WolframAlpha

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

Requested by cwatson#5438

[little helm]

i multiplite with 3-x and then i multiplite with 3+x

[frosty cradle]

I think you forgot to change signs then

[chrome girder]

anyway this is the real problem

[little helm]

but i dont know if it is allowed to do that

[little helm]

what do you mean?

[chrome girder]

$\frac{1}{3-x} + \frac{1}{3+x}=0,$
$3 + x + 3 - x = 0,$
$6 = 0$

bad texit

[woven radish]

zfnQRZJT

[frosty cradle]

right so where does -6 come from?

[chrome girder]

probably arithmetic error

[chrome girder]

.

[little helm]

true

[little helm]

but if i have this

and i would multiply it with 3-x

what would i get

[chrome girder]

$1 + \frac{3-x}{3 + x} = 0, x \neq 3$

[woven radish]

zfnQRZJT

[chrome girder]

for example

if the function was $\ln(3+x) - \ln(3-x) + x$

[woven radish]

zfnQRZJT

[chrome girder]

and the derivative was $\frac{1}{3+x} + \frac{1}{3-x} + 1 = 0$

[woven radish]

zfnQRZJT

[chrome girder]

then you would think that has a solution

at $x = \pm 3.873$

[woven radish]

zfnQRZJT

[chrome girder]

but again, the function is not defined there

[little helm]

ok

[chrome girder]

my advice is always look at desmos

to see if it actually has a solution

[little helm]

kk

but one question

[little helm]

if i have this do i have to multiply with 3+x

[chrome girder]

if you want

$3 + x + 3 - x = 0$

[woven radish]

zfnQRZJT

[chrome girder]

no solution

[little helm]

but it would be right

and the solution for that would be -6

oh

[frosty cradle]

...no

[little helm]

i see what you mean

but i have to calculate x-x first

before i do something else?

[frosty cradle]

you have to compute the entire left side

it doesn't matter which you combine first

[little helm]

so first i should right 6=0 which cant go

and then i can do other steps if necessary

[frosty cradle]

what other steps?

you are done, that shows there are no extrema

[little helm]

i know but if there were some avaible i mean

But i did understand it i thank both of you+

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[rugged sparrow]

Do you what tv size means

[devout snow]

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[halcyon parrot]

Anyone know how to get number 4

[urban hornet]

Differentiate it

[halcyon parrot]

y = 5x^4

[urban hornet]

Teah

Now put x=-2

[halcyon parrot]

but it wants whats perpindicular

[urban hornet]

Slope of tangent = 5×16 = 80
m1×m2=-1
m2=-1/80

Slope of perpendicular = -1/80

Point passes through (-2,-32)
Y+32=-1/80 ( x+2)

[halcyon parrot]

right cause its just the opposite

@urban hornet ??

[devout snow]

@halcyon parrot Has your question been resolved?

[halcyon parrot]

<@&286206848099549185>

[urban hornet]

m1×m2=-1

[halcyon parrot]

how did you get to that sorry

[urban hornet]

Product of slope of two perpendicular lines =-1

[urban hornet]

tan90= |m2-m1/1+m1m2|
m1m2+1=0
m1m2=-1

[halcyon parrot]

ok and where did our -2,-32 come from

[urban hornet]

Tangent touches at x=-2 i.e. when y=-32

[halcyon parrot]

i understand that but i dont get how you put that together if that makes ense

[devout snow]

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[gloomy valve]

Can I walk through this problem with someone, genuinely have no idea where to start

At least for part a

[restive river]

Where in question a did you get stuck?

[gloomy valve]

Sorry I should’ve clarified

Part a

Oops I misread

Like I don’t know where to begin to find a basis

Given U

Can I begin by substituting the restraints of x1 and x3?

[restive river]

Sure

[gloomy valve]

Okay

[restive river]

Wait, don't you mean x1 and x2?

Oh no I see

[gloomy valve]

Also the question was updated, here is the most recent version

There is no + 3

I apologize for the confusion

[restive river]

Hmmm yeah that +3 had me do some serious mental gymnastics

[gloomy valve]

Yeah I didn’t realize the image I had was outdated I’m so sorry

Anyways, lemme right out the constrains, I’ll send an image right now

[restive river]

@gloomy valve Did you just die?

[main tree]

what math are u taking?

[gloomy valve]

Here

Linear algebra proof based

[restive river]

Yep that works

[gloomy valve]

Sick okay

And from here

Do I begin the proof to show that it is a basis ?

Given U is what I have above

[restive river]

Sure

[gloomy valve]

To show that this is a basis, I need to show that span U is equal to R^5 and it is linearly independent. To show that span U equals R^5, will I need to show all of R^5 is in U and vise versa ? (Subset both ways implies equality)

Do I have the right idea or should I rethink what I’m doing 😭

[restive river]

I am gonna be honest I am sorta new to this subject myself and your words blew right past me

I need to show all of R^5 is in U
But... Not all of R^5 is in U though

To show this forms a basis in U, you just need to show:

• that all the vectors generated by your basis are in U
• that all the vectors in U can be generated by your basis

I think I get what you were trying to do, it was almost right

[arctic field]

[restive river]

Why? Did I say something stupid?

[arctic field]

a basis is spanning and linearly independent

that's what you need to check

[restive river]

Snow if you're available I'll let you take over

[arctic field]

i am not

[restive river]

sad

[gloomy valve]

How would I prove that our basis of U spans r^5?

[restive river]

You need to show your basis spans U

Forget about R^5

Plus I'm pretty sure U does not span R^5

[gloomy valve]

Okay that makes more sense

Yeah not all of U is in r^5

[restive river]

No, all of U is in R^5, but not all of R^5 is in U

[gloomy valve]

Fuck yeah ur right

U is a subset of R^5

What I sent above, with the 3 vectors, would that a good start to begin the proof

Yes right because we need 3 vectors to form a basis of U

[restive river]

Mhm

Now just show that these 3 vectors do indeed form a basis and you're all set

[gloomy valve]

Okay I will do that right now, for the sake of the channel, should I close it and ping you when I am ready ?

So you can check my proof

If you don’t mind ofc

[restive river]

No need to close this channel

[gloomy valve]

Ok

i haven’t made much progress, here is where im stuck

Because like I don’t know what to start with my proof statement

Like are the 3 addition thingys individual vectors ?

[devout snow]

@gloomy valve Has your question been resolved?

[gloomy valve]

Here is the progress I have so far

[devout snow]

@gloomy valve Has your question been resolved?

[devout snow]

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[broken sand]

.close

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[tardy vapor]

Help

[sonic smelt]

I think you need to apply law of cosines

[tardy vapor]

I tried

[restive river]

apply sine rule on A and B and then square both the sides

Try to simplify from there onwards

[tardy vapor]

Three variable whole square?

[restive river]

no need of squaring

just simplify

a / sin A = b /sin B

we know A = 2B

[tardy vapor]

Ok

[restive river]

yep

apply sin 2B formula here

[tardy vapor]

2sinbcosb

[restive river]

you will end up getting value in cos B since sin B gets removed

then apply cosine rule on angle B

simpify and you will get your answer

[restive river]

yep

[tardy vapor]

Sinb should cancel

[restive river]

yep

[restive river]

good

try to simplify

[tardy vapor]

Any hint

[restive river]

bring all the terms on LHS and then try to factorize

this is the hint

[tardy vapor]

Equating to zero

[restive river]

yes

[tardy vapor]

Not possible answer

How to factorise them

What if b²-c²=(b-c)(b+c)

@restive river

<@&286206848099549185>

[earnest haven]

move the b(b2 - c2) to rhs so it becomes a2 (c-b) = b(c2-b2)

then apply difference of square identity so (c-b) cancels from both sides

then youre left with a2 = b(b+c)

so option A

[tardy vapor]

Thank you

.close

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[chrome girder]

induction?

[tardy vapor]

<@&286206848099549185>

[chrome girder]

^

[tardy vapor]

Induction+ combustion

[chrome girder]

you can even check easily which answer is correct by setting k to some low value

like k = 1

[tardy vapor]

For k=1 , 1+(n+1)!/n!

[chrome girder]

= n + 2

[tardy vapor]

☝️

It's not simplifying

[soft wren]

the answer is C

[tardy vapor]

Explain

[soft wren]

you can check it easily with $\binom{n}{m} = \binom{n - 1}{m} + \binom{n - 1}{m - 1}$.

[woven radish]

Z_char

[soft wren]

and the reason i guess C is the answer is based on combination.

$\binom{n + t}{t}$ is like using t marks to split n objects.

[woven radish]

Z_char

[soft wren]

you know what i mean? what you choose is the place you put boards on. other places are objects. and the objects between two boards are in the same group.

notice $\binom{n + k + 1}{n + 1} = \binom{n + k + 1}{k}$.

[woven radish]

Z_char

[tardy vapor]

Board? Object?

[soft wren]

oh sorry the answer should be C.

[chrome girder]

nosols and if you had to give a solution make sure it's the right one

/lighthearted

[soft wren]

use this to check it.

$$\begin{align}
\binom{n+k+1}{k} &= \binom{n+k}{k} + \binom{n + k}{k - 1} \
&= \binom{n+k}{k} + \binom{n + k - 1}{k - 1} + \binom{n + k - 1}{k - 2}
\end{align}$$

[woven radish]

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

[soft wren]

notice you have (n+k)(k), (n+k-1)(k-1), if you continue to do that, you will have (n+t)(t)...

[soft wren]

im sorry, the answer is C.

[tardy vapor]

I didn't get the procedure

[soft wren]

ok, so do you understand how to prove the answer?

i mean, if you know the answer, use the procedure above to prove it?

[tardy vapor]

I'm just confused

[soft wren]

can you understand this?

[tardy vapor]

Did you choose option to prove question

[soft wren]

okay, you'd better to do this procedure on paper by yourself. that's the proof.

[tardy vapor]

Do i need to solve option 3

[soft wren]

yes.

and i can tell you how i think to get the answer.

$\binom{n+k+1}{k}$ is you have $n+k+1$ things, how many ways you have to choose $k$ things.

[woven radish]

Z_char

[tardy vapor]

Ok

[soft wren]

but you have another way to consider this, to assume you have $n + 1$ apples, and you need to put them on $k$ boxes, boxes are different and apples are the same. how many ways do you have?

[woven radish]

Z_char

[soft wren]

my bad, you have k+1 boxes.

[tardy vapor]

One box can contain how much apples

[soft wren]

k+1 different boxes and n+1 apples.

[soft wren]

infinite apples.

the answer is also $\binom{n + k + 1}{k}$.

[woven radish]

Z_char

[soft wren]

assume you have $n+k+1$ blanks at first, then you put on $k$ boards on $k$ blanks to divide the apples from different groups.

[woven radish]

Z_char

[soft wren]

you have 5 blanks.

[tardy vapor]

Ok

[soft wren]

you choose one.

put one on the first box, put three apples on the second one.

[tardy vapor]

Ok

[soft wren]

okay, then $\binom{n+k+1}{k}$ means n+1 apples in k+1 boxes.

[woven radish]

Z_char

[soft wren]

because 5 choose 1(above picture) is four apples(1 + 3) in two boxes.

[tardy vapor]

Why 3+1 why not directly 4

[soft wren]

because in the picture i showed is one in the first and three in the second.

im afraid that you do not understand, so i use 1+3 to relate to the picture.

[tardy vapor]

Let me tell
1 apple in first box

3 apple in second box

Two box contains 3+1=4 apple

[soft wren]

that's ture.

[soft wren]

so we have this.

[tardy vapor]

One thing that why second box put three apples instead 2

[soft wren]

emm, you can put two.

[soft wren]

if you choose the third blank in this picture!

so how many ways you choose blank = how many ways you put apples.

[tardy vapor]

For k Box (k+1) apple?

[soft wren]

im afraid i do not understand.

[tardy vapor]

Limit the number of box and apple so I can understand

For ex 20 box 20 apple

[soft wren]

okay, $\binom{n + b}{b}$ means put n apples in b+1 boxes.

[woven radish]

Z_char

[soft wren]

like the example in the picture, you have five blanks. you have to choose one. it's $\binom{5}{1}$.

[woven radish]

Z_char

[soft wren]

but anyway you choose one. you can consider this as a way to put four apples in 2 boxes.

im afraid i have to sleep. however, i can quickly go through the rest things and you can wait others for help.

you have the meaning about the answer now.

then consider the meaning about the question.

it's just another way to count the ways. you consider the first apple is on which box. on first box we have n apples on k boxes. on second box we have n apples on k-1 boxes. and we plus them together. that's the formula in the problem.

[tardy vapor]

Can you consider me any video solution from yt

[soft wren]

and they present the same meaning - the ways to put n+1 apples on k+1 boxes.

[chrome girder]

that is, distinct boxes but with nondistinct apples

[soft wren]

im sorry i can't... i don't use it very often and then i don't know too much about that.

[tardy vapor]

Ok

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[devout snow]

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[fierce matrix]

how do you reason deductively to explain how the number of cubes are produced?

[pseudo basin]

you have 5 layers of cubes each laid out in a 5 by 5 grid.

[woven radish]

Pro_Hecker

[fierce matrix]

5^3 is deductive reasoning also?

[tall stirrup]

nope

you have 5 layers of cubes each laid out in a 5 by 5 grid.

[fierce matrix]

so this is the reasoning that leads to 5^3? or this means the same as 5^3? 🙂

a cube has 8 corners
a cube has 12 edges
a cube has 6 faces

the cube is cut up into smaller cubes
the smaller cubes are the same size

in one corner there is 1 cube
in one edge there are 5 cubes
in one face there are 5 x 5 cubes

this cube has 1 cube for each corner
3 more cubes for each edge
9 more cubes for each face

there is a smaller cube inside it that
has 1 cube for every corner
1 cube for every edge
1 cube for every face

there is one cube inside this smaller cube

[devout snow]

@fierce matrix Has your question been resolved?

[fierce matrix]

what are different ways to reason this?

[tall stirrup]

Deductive Reasoning Example: All dogs have ears; golden retrievers are dogs, therefore they have ears.

[pseudo basin]

you could reason that there we 8 corner cubes, 3 cubes in each of 12 edges (excl. the corner cubes you've counted), 9 cubes in the middle of each of 6 faces, and 27 cubes in the interior.

you will get the same number.

[fierce matrix]

thank you @pseudo basin and @tall stirrup

this is the rest of this task if you are interested:

[devout snow]

@fierce matrix Has your question been resolved?

[devout snow]

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[torpid roost]

I need help with the 5th one please

<@&286206848099549185>

[devout snow]

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[woven radish]

Result:

35

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[devout snow]

Please don't occupy multiple help channels.

[runic pollen]

?????

I have no ideaaaa

Pls helpl

Where did a and z come from?

[vast lance]

A and z are arbitrary variables

Like x

[devout snow]

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[restive river]

https://media.discordapp.net/attachments/1096908705364189264/1100038364859736095/Rotated.gif

[restive river]

Is that meme wrong

ping on reply thanks

[frosty cradle]

#help channels are for homework type questions/problems

[restive river]

That is a question

[wheat pebble]

that meme is not wrong

[restive river]

and I got my answer

done

[wheat pebble]

you that smite player

[devout snow]

@reef fox Has your question been resolved?

[devout snow]

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[feral agate]

!status'

[devout snow]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[untold nova]

I tried using the cos inv x + cos inv y formula

And then it just got messy

Maybe i can try converting this into tan?

[feral agate]

pretty sure the first 2 terms resolve to pi/2

[untold nova]

How?

[feral agate]

try creating a right angled triangle

with the non hypotenuse sides as sqrt(p) and sqrt(1-p)

[untold nova]

Yeah i got that but

Actually no I'll try that

I think i got it

Thanks

[devout snow]

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[devout snow]

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[devout snow]

Available help channel!

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[slender wigeon]

Hello I'm trying to do these 3 integrals but i all of them are wrong, in the first one the solution from the book is 1/3sqrt(2x+1)(2x+1) i don't know how he got that
The 2nd one the solution from the book is Ln(sqrtx-1)^2 here i don't know why he didn't put the ^2 on the square root
For the 3rd one i don't really get the solution because it says 2/3sqrt(x+2)(x+5)

[devout snow]

@slender wigeon Has your question been resolved?

[slender wigeon]

<@&286206848099549185>

[restive river]

Yes

[slender wigeon]

Can you help me please?

.close

[devout snow]

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[devout snow]

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[gray phoenix]

Is the statement true that there are 12 odd three-digit numbers where the hundreds digit is twice as large as the tens digit?

[novel sedge]

No, the statement is not true that there are 12 odd three-digit numbers where the hundreds digit is twice as large as the tens digit. In fact, there are no odd three-digit numbers that meet this criteria.

To see why, let's suppose that there is such a number with hundreds digit h and tens digit t. Since the number is odd, its units digit must be 1, 3, 5, 7 or 9. However, since the hundreds digit is twice the tens digit, we have h = 2t.

Now, we can express the number as 100h + 10t + 1. Substituting h = 2t, we get 100(2t) + 10t + 1 = 201t + 1. Since we want the number to be odd, we need 201t to be even, which means that t must be even as well. But then the tens digit can only be 0, 2, 4, 6 or 8, none of which satisfy the condition that the hundreds digit is twice as large as the tens digit. Therefore, there are no odd three-digit numbers that satisfy the criteria.

[restive river]

please refrain from using ai to answer questions

[novel sedge]

k

[devout snow]

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[devout snow]

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[digital jacinth]

$e^{-x}=1$

[woven radish]

putridplanet

[digital jacinth]

solve the exponential, answer to nearest hundredth

idk where to start

[dark dawn]

nearest hundredth

[polar chasm]

take ln of both sides

[digital jacinth]

$lne^{-x}=ln1$

[woven radish]

putridplanet

[polar chasm]

ln(e^-x) is by definition -x

do you know what's ln(1)?

[digital jacinth]

$lne^{-x}=0$

[woven radish]

putridplanet

[polar chasm]

yep

and ln(e^(-x)) is by definition -x

or you can apply some log rules to get -x*ln(e)

which is just -x*1

[digital jacinth]

how

[polar chasm]

ln is inverse of e^x

[patent marsh]

$ln(e^{-x}) = -xln(e) = -x(1) = -x$

[polar chasm]

meaning ln(e^x)=x and e^ln(x)=x

[woven radish]

MellowDramaLlama

[digital jacinth]

how is ln(e) =1

[polar chasm]

ln(15) would mean: to what power do I have to raise e to get 15?

ln(e) means: to what power do I have to raise e to get e?

Which is obviously 1

[digital jacinth]

why ?

what does e equal

[polar chasm]

e is just a number

2.718281828 or sth

I dont remember it

and I cant. Since it has infinitely many nonrepeating digits

[digital jacinth]

why 15 ?

[polar chasm]

just arbitary example to demonstrate what ln means

[digital jacinth]

oh i get it

so raising the power of something to one is like saying

[polar chasm]

raising something to the power of 1 would be the correct terminology

[digital jacinth]

the same thing as one times the number

[patent marsh]

if you have time this video does a good job explaining what e is, its origins, and why it's important

https://www.youtube.com/watch?v=pDFcu_wLOzo&ab_channel=TheOrganicChemistryTutor

[polar chasm]

e is actually not so important in high school. The problem is that people rather derive definition of log from it's properties than properties from the definition

and have no idea what ln actually means

[patent marsh]

oh that's true

[devout snow]

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[devout snow]

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[devout snow]

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[slow magnet]

what is this formula?

[frosty cradle]

expected value? for a continuous random variable

[wooden zodiac]

we need consent

[restive river]

the definition of the expected value of X, a continuous v.r. whose density function is $f$, and usually $f$ is lowercased, but that's not a rule ofc

[woven radish]

shell.exe

[slow magnet]

ty

[slow magnet]

ty

[wooden zodiac]

youre welcome

[slow magnet]

if we changed it to E(x^2) it would still be expected value?

[restive river]

X^2 is also a random variable, so yes

[slow magnet]

ok ty

[devout snow]

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[devout snow]

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[frosty cradle]

draw them both as regions "bounded by" a line

maybe

[devout snow]

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[devout snow]

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[restive river]

So is this channel available?