help-19

[gentle flare]

how do i do that

[safe dock]

cross multiply it...

and factorize it

then

[onyx cloak]

the quadratic (numerator) doesnt have any real solutions

find the roots of denominator

then check for what intervals it holds true

[gentle flare]

so basically x = 3 and 7/9?

[onyx cloak]

those are the roots yes

now check for waht intervals the inequality holds

x<7/9
7/9<x<3
x>3

[odd edge]

@gentle flare Has your question been resolved?

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[forest sky]

on the left, we have (any real number, any real number) and on the right we have (any real number, any real number)

[weary pelican]

first of all, you can't do "(a,b) + (0,b) = (a,0) + (0,b)" just like that. every letter represents a potential different real number

so when you take an element of V+W

you can write it as $(a,b_1) + (0,b_2)$ for example

[clever fjord]

rafilou2003

[weary pelican]

this would correspond to the element $(a,0) + (0,b_1+b_2)$

[clever fjord]

rafilou2003

[weary pelican]

which is indeed an element of U+W

and vice versa, every element of U+W is an element of V+W

[iron bear]

,, V + W = \set {, (a, b) + (0, c) \where a, b, c \in \R ,} \
U + W = \set {, (d, 0) + (0, e) \where d, e \in \R ,}

[visual arch]

All of my maths life, I've taken as established convention that when we say b, then b = b. Why is it different in vector spaces?

Is it just so?

[iron bear]

,, \set {a \where a \in \R} = \set {b \where b \in \R} = \set {c \where c \in \R} = \set {d \where d \in \R} = \set {e \where e \in \R}

[iron bear]

i can keep going

[visual arch]

So that just means that a = b, but why then is it not valid to make my previous assertion? We distinguish by letters, normally, to indicate that we are referring to potentially different numbers

[iron bear]

it does not mean a = b

[visual arch]

in other words, it's confusing to ascertain that establishing that $ x \in U, x \in V$ does not mean that $x = x$

[clever fjord]

Alvaro Neto

[iron bear]

,, \set {(a, b) \where a, b \in \R} = \set {(b, a) \where a, b \in \R} = \set {(420a + 69b, 1337b) \where a, b \in \R}

[visual arch]

Am I missing something, or are we not changing the convention that is taught maths students up to university calculus?

[iron bear]

these are elements of a set

a and b have no meaning outside of the set

[visual arch]

okay

[iron bear]

they are purely used within the definition of the set, in the set builder notation

these sets all have the same elements

so the sets themselves are equal

[visual arch]

right. I think that clarifies things. Have to put it to the test

thanks!

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[jolly halo]

i cant seem to locate my mistake here
need help

[jolly halo]

this is the value of the integral

step 1-->2
i just did int(f(x)) from a to b = int(f(a+b-x) from a to b
step 2--> 3
i tried compliting the square in the denominator to form 1/1-x^2 like form

[mental lotus]

the constant outside the completed square is 11/4, so I think you have missed the sqrt(11)/2 and just put sqrt(11) instead

[odd edge]

@jolly halo Has your question been resolved?

[jolly halo]

oh yeahh ic it now

thanks a bunch!

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[weary pelican]

?

what is property P

[tawdry wagon]

Is P(S_i) the power set?

[mystic saffron]

no

is just a logic expression

for making clear what i mean lol

[weary pelican]

which one

[tawdry wagon]

Ah, okay. So P is a function that returns true or false?

[mystic saffron]

I am trying to find a property P that is worth studying

[mystic saffron]

No, P is just any property

[weary pelican]

well a property that is either verified or not depending on the subset

[tawdry wagon]

Ohhhhh, I think I see.

Thats so confusing

[mystic saffron]

for example
take a finite subset of the set of prime numbers, determine if for a k-partition of S, exists partition subsets such that the sum/product/difference etc of the elements are prime numbers
is an example of a problem like this

[weary pelican]

such that the sum/product/difference etc of the elements are prime numbers
one of these is not like the other

[mystic saffron]

why not?

[weary pelican]

ever seen a product of prime numbers being prime?

[mystic saffron]

oooh, yeah my bad lol

it was a example I come out

[weary pelican]

well even if you take the sum

the sum of two odd prime numbers is not prime

because it's even

[mystic saffron]

but I am trying to find a problem with this structure that is worth it

so I need to find this "insight" for a problem

[weary pelican]

if you don't even know what to do

I suggest for example

partitions of {1,...,p} such that the sum of each partition subset is divisible by p

[mystic saffron]

oh, nice ,interesting

[mystic saffron]

and how can be useful to know this? (in math of course)

[mystic saffron]

oohh, you giveme an idea. Looking for things that are fundamentals in math theories and asking about those with this structure, for example divisibility.
that may be helpful to the theory , right?

[mystic saffron]

so since divisibility is important to elementary number theory, asking questions about divisibility may be important too

what I am finding hardest Is to think in a problem that helps to understand better math, instead of just being a isolated result that is just interesting

mmmh, so in this case with number theory , also asking things about primes is more interesting that about n, since prime numbers have more impact in Number theory

[odd edge]

@mystic saffron Has your question been resolved?

[odd edge]

@mystic saffron Has your question been resolved?

[mystic saffron]

.close

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[royal rose]

How can we find the generic formula for the following numeric series:
2/1; 4/1; 8/6; 16/24; 32/120

It seems to be 2^n for the numerators, and n! for denominators, but it somehow skips n=2

[odd edge]

@royal rose Has your question been resolved?

[royal rose]

<@&286206848099549185>

[lofty crater]

brro is stephen curry

[royal rose]

what💀

[mystic saffron]

That is so weird

[lofty crater]

what is generic formula

[mystic saffron]

Pattern

[royal rose]

as in x of n, where n is a natural number excluding 0

[lofty crater]

when x =0

[royal rose]

what

[lofty crater]

what is excluding mean

[royal rose]

without, so n would be given the values 1, 2, 3 and so on

[lofty crater]

i can help you but english is bad

write it in paper

summary it

[royal rose]

it seems like it needs a pattern that goes 0 1 3 4 5 so that the factorial gives the correct denominators

[lofty crater]

oh i difficullt

it

let me ask my teacher

[royal rose]

and yes, I copied it exactly from the book, so only other option is that its the author's fault

[lofty crater]

what boo

book

[royal rose]

its a moldovan book for math exercises

[lofty crater]

for fundamental ?

[royal rose]

general exercises for 11th grade

[lofty crater]

am grade 10

lol

[royal rose]

darn

[lofty crater]

how can you understand

that difficult as f

[royal rose]

where you from?

[lofty crater]

am from vietnam

[lofty crater]

you from us ?

[royal rose]

you might also get to it, the math programs are harder for us

[royal rose]

moldova, romanian at heart

[lofty crater]

romanian best robotic country

romanian dominate one of the biggest competitive

[royal rose]

never heard about that but aight

[lofty crater]

it just basic math problems but if it a advance problem it actually the same

[lofty crater]

give me advice

i want to study advance math

[royal rose]

i dont think its that advanced, what are you learning right now?

[lofty crater]

vectow

vecto

[royal rose]

r

numeric series are basically functions, you know that?

[lofty crater]

yes

[lofty crater]

dms me

[royal rose]

i'd just studied this a couple days ago so i'm also new at it

[royal rose]

ight

[royal rose]

imma close this soon cause it seems impossible

[lofty crater]

it is

[royal rose]

.close

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[ebon briar]

whats the behavior around the 0 for lim x-> 0 (4x^2 - 3x + 1) / 5x^2

[timid cairn]

what do you get if you plug it in

[odd edge]

@ebon briar Has your question been resolved?

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[tranquil void]

dy/dx= 1/2 e^3x-2

is this right?

[wanton bison]

No

[tranquil void]

sad

[wanton bison]

Everything remains the same

[mental storm]

apply the chain rule

[tranquil void]

so 1/2 e^3x-1

[wanton bison]

Now chain rule too

times (3x-1)'

[tranquil void]

could you drawn the brakets on this to help me see it properly please?

[wanton bison]

I dont understand your q

[tranquil void]

like this

or is it not the same thing we are doing

[wanton bison]

e^f(x) is a exponential function

So no

You cant do that

But exp. functions base e have the property to be the same when differentiated

But you need chain rule

[ y = e^f ] then [ y' = e^f \cdot f' ]

[clever fjord]

bacc

[tranquil void]

it says to just use chain rule

not the outher ones

[wanton bison]

wut

[mental storm]

thats the main method youll use

[tranquil void]

it only says to use chain rule

[mental storm]

a question is not gonna tell u every possible derivative rule u need to apply...

[tranquil void]

well the one under it tells me to use the outhers

,w diff 0.5e^(3x-1)

[clever fjord]

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[mental storm]

🤦

[tranquil void]

i put diff lol

[mental storm]

we alr told u the chain rule, whats stopping u from actually using it

[tranquil void]

are these things equal?

[mental storm]

yes

[mental storm]

now identify what the "f" is

[tranquil void]

how do you do it with the -1

is it this?

the -1 is putting me off a bit

[fervent hawk]

Do you know what is the derivative of e^x with respect to x?

[tranquil void]

xe^x

[fervent hawk]

$\dv{e^x}{x}=e^x$

[clever fjord]

Biscuity

[fervent hawk]

nope

[fervent hawk]

maybe this is why you got it wrong

[tranquil void]

e^k
ke^k

[fervent hawk]

?

,w diff e^x

[clever fjord]

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Requested by biscuityxd

[fervent hawk]

that k is a constant

[tranquil void]

like the 3 in ours?

[fervent hawk]

but we have -1

so it's better to use chain rule

[tranquil void]

yea thats what is putting me off

[fervent hawk]

have you learnt chain rule?

[tranquil void]

yes

but im not sure with this one

[clever fjord]

Biscuity

[fervent hawk]

agreed?

[wanton bison]

||technically 1/e too||

[fervent hawk]

Then we can let u=3x-1, so we have:

[clever fjord]

Biscuity

[tranquil void]

are you able to show me the workings out for a similare question to mine

and then ill see if i can do mine

[fervent hawk]

oh sure

[tranquil void]

thank you

[fervent hawk]

let's make an example

y=e^(2x+1)

Let u=2x+1, so we have y=e^u

a moment please

,align
&\dv{y}{x}\=&\dv{e^{2x+1}}{x}\=&\dv{e^{u}}{x}\=&\dv{e^u}{u}\cdot\dv{u}{x}\=&e^u\cdot\dv{u}{x}\=&e^{2x+1}\dv{(2x+1)}{x}\=&e^{2x+1}(2)\=&2e^{2x+1}

there we go

[clever fjord]

Biscuity

[fervent hawk]

do tell if you're stuck on particular step @tranquil void 😺

[tranquil void]

oh

[tranquil void]

so its just 3/2 e^3x-1

[fervent hawk]

yep

3/2 e^(3x-1)

[tranquil void]

ok

thank you

🫡

.close

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[frigid canopy]

Trying to find $\lim_{x\to \infty} arctan(x^2-x^4) = \frac{-\pi}{2}$

[clever fjord]

Veni, vidi, perii

[frigid canopy]

rather prove it

So the definition for a limit at infity is

[quasi sparrow]

do some simple bounding on the arctan argument

[frigid canopy]

$\forall \varepsilon> 0 \exists N | x>N \implies |f(x)-L|<\varepsilon$

[clever fjord]

Veni, vidi, perii

[frigid canopy]

hmm?

[quasi sparrow]

x^2 - x^4 can be bounded more simply

[frigid canopy]

yes

[quasi sparrow]

and arctan is bounded

[frigid canopy]

Let me try doing this first

so $|x-1|<\delta$

[clever fjord]

Veni, vidi, perii

[frigid canopy]

$|x+1|<\delta+2$

[clever fjord]

Veni, vidi, perii

[quasi sparrow]

what

x is going to infinity

[frigid canopy]

yes

the maxima is 1/4

is that what you meant

[quasi sparrow]

no

,w arctan(-inf)

[clever fjord]

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[quasi sparrow]

arctan is bounded between -pi/2 and pi/2.

[frigid canopy]

yes

and x^2-x^4 is bounded between (- \infty , 1/4)

I think I should do some easier proofs on limit at infty first

.close

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[frigid canopy]

thanks

[quasi sparrow]

it's not

[frigid canopy]

,w range of x^2-x^4

[clever fjord]

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[quasi sparrow]

Ah my bad I read this wrong

[quasi sparrow]

You're right, carry on

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[weary vapor]

can i ask for help here?

[wanton bison]

ye

[tight crescent]

Of course.

[weary vapor]

a,b,c are elements of N numbers. a, b and c arent the same number.
Whats 2a - b + 3c IF a x c = 12 and b/c = a?

i got that b is 12
cause we know what a is, if we put it in the a x c = 12
its b/c x c = 12. c cancels out and b = 12.

but i dont know where to go from this

[summer cradle]

[weary vapor]

sorry if its a stupid question, im not that good at mathematics.

[fossil hawk]

Form a diophantine in a and c

[summer cradle]

there is no equation

[fossil hawk]

Hmm well then one way would be to factorize 12 and consider all possible situations

[summer cradle]

knowing a*c = 12 doesn’t narrow down the possible values of 2a + 3c to 1

there aren’t many

[fossil hawk]

Find all

[summer cradle]

but still more than 1

[fossil hawk]

Give the examinar all and let him choose

From the one he wants

[weary vapor]

i got the numbers, but i thought maybe i need some sort of equation cause my teacher wants it like that.
if we say
a = 6 b = 12 c = 2 it works

cause a x c = 12 . 6 x 2 = 12

[fossil hawk]

3×4 is also 12

[weary vapor]

and a = b / c so 6 = 12 / 2

[fossil hawk]

1×12 is also 12

[summer cradle]

poorly posed question if that’s what’s intended.

[fossil hawk]

my brother 4 = 12/3

[low locust]

well 1x12 is the only one it cant be

[fossil hawk]

Unless there is a constraint on a and c like they one of them is a even prime or there is some other relation given there can't be a unique solution

[weary vapor]

the question is.
What's 2a - b + 3c IF a x c is 12 and a = b/c

[low locust]

can you give an image of the question

[weary vapor]

yeah if you know Albanian lol.

[low locust]

post it anyway

[weary vapor]

If a,b,c are elements of N and a,b,c aren't the same number. What's 2a - b + 3c if a x c = 12 and b/c = a

theres no numbers given only that sentence

[fossil hawk]

That's not Albanian

[weary vapor]

Ok ill send it in albanian

Nese a,b,c jane elemente te N dhe a,b,c nuk jane numra te njejte. Sa eshte 2a-b+3c nese a*c=12 dhe b/c = a

[fossil hawk]

..

We need the image

[weary vapor]

brother, the teacher wrote it on the black board, we put it down, theres nothing else but that sentence

[low locust]

uhh. a-c=12 or a*c=12

[weary vapor]

sorry, a*c=12. Misstype

[low locust]

hmm well then it just doesnt have a unique answer

[weary vapor]

i guess its not a equation but you just have to give the numbers you want and see if they apply

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[grim copper]

any idea how to integrate this?

[orchid torrent]

Oh god that’s gonna be tedious

Hint: complete the square

[grim copper]

uh sorry for being dumb but how do I do that

got it

[odd edge]

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[scenic lodge]

.close

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[scenic lodge]

.reopen

[odd edge]

✅

[scenic lodge]

.reopen

Why did he use Pythagorean theorem instead of power reducing formula to simplify

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[steep aspen]

5^a = 25^1/2

the 1/2 is a fraction

whats ab

[steep aspen]

5^a = 25^1/b

sorry

wrote it wrong

1/b is a fraction

whats ab

[quasi sparrow]

use 25 = 5^2

,tex .exp rules

[clever fjord]

[lethal spoke]

Oops

[lethal spoke]

That was fast

[steep aspen]

huh.

im confused

what u send that for

[quasi sparrow]

rewrite this using the table and what i said above

[steep aspen]

5^a = (5^2)^1/b

what do u do after that

is that waht u meant

[quasi sparrow]

keep using the table until you get 5^a = 5^(something)

then a = something

[steep aspen]

my friend solved it

but im still abit confused

how did he get it

to

2/b

and a = 2

[tacit wasp]

$$25 = 5^2$$

[clever fjord]

Alberto Z.

[tacit wasp]

And also $${\left(a^b\right)}^c = a^{b\cdot c}$$

[clever fjord]

Alberto Z.

[tacit wasp]

So, putting those two observations together, you get: $$25^\frac{1}{b} = {\left(5^2\right)}^\frac{1}{b} = 5^{2 \cdot \frac{1}{b}} = 5^\frac{2}{b}$$

[clever fjord]

Alberto Z.

[steep aspen]

alright yeah i understand

thank you

.close

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[summer sage]

PLEASE

[summer sage]

UH

I DON'T KNOW

[errant jewel]

Me neither

lol

[summer sage]

NOOOOOOOOOO

[errant jewel]

WE NEED HELP

[summer sage]

HELP US

[errant jewel]

FR

But like look though how’s ur day?

[summer sage]

I think a) is going to be: B will be (fixed) along the x axis in the direction of 8i

[summer sage]

it's unproductive

I just

I just

man

[errant jewel]

Procrastination?

[summer sage]

I was doing differential equations earlier and then my mom made me go to the post office

I thought we were driving back home but

she drove me to costco and it got me all messed up

cause Iw as gonna get back to diff eqns RIGHT after I got back from the post office

[errant jewel]

Oh

[summer sage]

so when I got home I just got on roblox

😭

and it's been like that all day

[errant jewel]

That’s literally me

But I learned that no matter what you are going through whether it’s relationship, life just remember that suffering comes before success

A sword must be in the fire b4 being made

👍

[summer sage]

words of wisdom indeed

I just hopped off of roblox

I think ima do some mroe diff eqns

😄

[errant jewel]

Aight imma get back to bench press LOL

[summer sage]

lol

[errant jewel]

AND ROBLOX

[summer sage]

YE

XD

[summer sage]

I think a) is going to be: B will be (fixed) along the x axis in the direction of 8i

<@&286206848099549185>

[errant jewel]

I feel bad for u man😭

No one helping u

[odd edge]

@summer sage Has your question been resolved?

[errant jewel]

Just play Roblox while u wait man

[summer sage]

it's a sphere

that's why

because B has three components but it equals 8 no matter what you do

and A lies on the x-y plane

being the foundation or horizontal axis of the sphere

so I'm trying to maximize the z component

while keeping rho at 8

and r=<5cos(53), 5sin(53), 0> = 5

[tulip marlin]

oh nah

may chatgpt save your soul 🙏

[summer sage]

what did u have to change ot not get banned

[tulip marlin]

the world may never know

[brittle cargo]

@summer sage what level math is this?

My friend and I just started vector calc and our best guess is 143 but I have no idea how to actually work it out

Sorry for false hope

For a^

[summer sage]

physics

basic physics

so this is a 2D system

where the z component of A is 0

and the z component of B is also 0

it would reference a sphere, however there is no z component to make of it

so this would instead make a system of 2 vectors along the x-y plane

and the cross product needs to be in the direction of z

working this out rn

😭

[errant jewel]

Still no help…

[errant jewel]

Recommend me a Roblox game

[summer sage]

I've been on mm2 trading all day every day

lol

but a good one is tower of hel

I was literally at the top

and I

just

fell

eh

not the ebst

best*

I haven't played the good ones in a while

I might tell u some whne I'm not so busy

[errant jewel]

Mm2 is a game?

[summer sage]

😭

murder mystery 2

[errant jewel]

Oh

[summer sage]

lol

[errant jewel]

U should play rogue demon

It’s a fighting game

Even though I don’t. Watch anime

It’s a very good fighting game

Though

U should try it

[summer sage]

bro

I missed work today

I totally forgor

bro

I don't wanna get fired

what do I tell them

😭

[odd edge]

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[hexed sluice]

How do I find the lim x-> 0 f(g(x))

I first evaluated the lim x-> 0 g(x) and I got 2

but f(2) is undefined

[brittle beacon]

While f(2) is undefined [defined differently], you know that as x gets closer to zero, g(x) gets closer to 2...

[hexed sluice]

wait, but how does this work. once i find the inner limit, the expression yeilds lim x->0 f(2)

how do i find the limit as x approaches 0 of f(2)

[nimble blaze]

f(2) isn't actually undefined, there's a point at (2,3) but you don't care about that for the limit here

[hexed sluice]

wait, but look:

it isnt clicking for me. the thing is, what in the expression tells me to find the limit of f(x) after finding the inner limit?

[nimble blaze]

there should be some caveat saying stuff about f being continuous

[hexed sluice]

pls explain

[nimble blaze]

that would apply if f is continuous at:
lim as x→a g(x)

[summer sage]

the different conditions of the derivative test also apply to limits I believe

[nimble blaze]

here the value of the inner function is approaching 2, so you'd consider
$$\lim_{\this \to 2} f(\this)$$
(or $\lim_{x \to 2} f(x)$)

[clever fjord]

ℝαμOmeganato5

[hexed sluice]

so if f isnt continuos at that point, i should just do lim x-> 2 f(x)

[nimble blaze]

for stuff removeable continuity here, yes

and it gets slightly more complicated when there's stuff like jump discontinuities
where you'd need to consider one sided limits

[hexed sluice]

So, to summarize, for a composite function (f(g(x)), if the limit of (gx) as x approaches a exists, and if f is continuos at that limit, I can find the limit of the composiute function by evaluating f at the limit of g(x)? And if f is not continuos at the limit of g(x), then I need to evaluate the limit of the composite function by considering how the outer function f bevhaves as the inner function g(x) approaches its limit?

[nimble blaze]

yeh

[hexed sluice]

okay thanks!

appreciate it

.close

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[cyan echo]

What does the 2n change here from a normal n in this induction proof? It is stated that n belongs to all positive whole numbers. So wouldn't n and 2n be the same?

[forest sky]

,rcw

[forest sky]

if n is a positive integer, then that implies that 2n can only be a positive even integer

[cyan echo]

But if I try plugging in k/n=2 on both sides I get
k(k+1) = 2(2+1) = 6
And 1/3 * 2(2+1)(7*2+5)
= 3/3 * 2(14+5)
= 38
And 6 ≠ 38 ?
Shouldn't both sides be equal?

Since 2 is an even number?

[forest sky]

so we are saying that n = 1. so we get
(sum from k = 1 to k = 2 of) k(k+1) = 1(1+1) + 2(1+2) = 1(2) + 2(3) = 2 + 6 = 8
on the right side, n = 1:
1/3 n (n+1) (7n+5) = 1/3 (1) (1+1) (7+5) = 1/3 (2) (12) = 24/3 = 8
so both sides match

[cyan echo]

I see, that does make more sense. But what if we say that n = 3, would you then have to add all the sums of n = 3, n = 4, n = 5 and n = 6 to get the same result on the right side?

Since 2n = 6 if n = 3

[odd edge]

@cyan echo Has your question been resolved?

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[dark shore]

help

[dark shore]

need help asap

ao they want the force on the front or back of the tank

i first knew we had to integrate force since its not horizontal

then i knew force is (pressure)(area)

and further more pressure is (density of liquid)(acceleration due to gravity)(y) y is for every point y i guess

then for area which is the second part of the force integral (f=PA), i found area to be length * width, which can be expressed in this situation as 15/12 (since the very top of the tank is 15 meters wide and it decreases as we go down since its shaped like a triangle) so 15 width divided into the 12 total meters multiplied by y because thats for every y point going up and down the triangular shape, so 15/12y for width then multiplied by dy for length (which i assuemd the small change in y was the length of each small slices we take from water to all areas together) then after i found the area and the pressure i multiplied them together, so p = 1000 for water density * 9.8 for gravity acceleration * y * 15/12 * y * dy

@icy vault

did i do something wronhg

lol no worries

[odd edge]

@dark shore Has your question been resolved?

[half fulcrum]

$F_{net}=PA$

[clever fjord]

hrdxpqurcxktdbanql

[half fulcrum]

$P(y)=\rho g y$

[clever fjord]

hrdxpqurcxktdbanql

[half fulcrum]

$A(y)=\pi r(y)^2$

[clever fjord]

hrdxpqurcxktdbanql

[half fulcrum]

$r'(y)=\frac{15/2}{12}$

[odd edge]

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[clever fjord]

hrdxpqurcxktdbanql

[half fulcrum]

we know $r(0)=0$

[clever fjord]

hrdxpqurcxktdbanql

[half fulcrum]

Thus,

$r(y)=7.5/2y$

[clever fjord]

hrdxpqurcxktdbanql

[odd edge]

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[half fulcrum]

Therefore

[half fulcrum]

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[odd edge]

@novel venture Has your question been resolved?

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[glacial hearth]

How do you begin to evaluate the limit of e^x - 1 / x as x approaches 0? Can't seem to find any examples of this problem in our library's calculus books

[cursive field]

l'hopital's rule

if f(x)/g(x) is indeterminate, evaluate f'(x)/g'(x) instead

[signal yacht]

I wonder how you could do it without tht

[glacial hearth]

Are there any other way other than l'hopital? Or is it the only possible solution?

[signal yacht]

You can see the expansion.

,w expand e^x

[clever fjord]

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[cursive field]

good idea

but honestly l'hopital's is just generally simpler

[signal yacht]

some instructors don't like it

which makes sense

[cursive field]

but yes, i agree

[signal yacht]

it's more or less a crutch

[cursive field]

this is a more logical way of looking at the limit

[glacial hearth]

True but my instructir doesnt want us to use it

[cursive field]

ah okay

[signal yacht]

sometimes evaluating it otherwise is easier than taking derivatives

[cursive field]

you can use percy's method of looking at the taylor series

[signal yacht]

use the expansion then

seems easiest

[glacial hearth]

I'll do that. Thanks!

.close

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[odd edge]

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[sterile blaze]

whats it called, the set of a+bsqrt2 where a and b are rational numbers. field extension?

[cursive field]

Q[sqrt 2]

[pastel dew]

$\mathbb Q[\sqrt 2]=\mathbb Q(\sqrt 2)$

[clever fjord]

everg

[pastel dew]

$\mathbb Q\subset \mathbb Q(\sqrt 2)\subset \mathbb R$

[clever fjord]

everg

[pastel dew]

the middle field is the smaller one which contains the rationals number and $\sqrt 2$

[clever fjord]

everg

[sterile blaze]

is there a way to say it colloquially

like in conversation

[cursive field]

Q adjoin root 2

adjoin not adjoined

[sterile blaze]

also is Q[sqrt2] the same as Q[x] with the relation x^2=2

[sterile blaze]

thanks

[cursive field]

mhm

you can also say

Q[x]/x^2 - 2

[sterile blaze]

adjoint?

[cursive field]

no

adjoin

not adjoint

[sterile blaze]

oh. identifying x^2-2 with 0 which is the same as x^2=2 yea?

[cursive field]

mhm

[sterile blaze]

thanks ren

[cursive field]

np

[sterile blaze]

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[pastel dew]

yes as you have the following isomorphism $\mathbb Q[x]/ (x^2-2)\to\mathbb Q[\sqrt 2], x\to \sqrt 2$

[clever fjord]

everg

[cursive field]

btw this is akin to how we can write C = R[i] = R[x]/x^2 + 1

[pastel dew]

the map without the first quotient is trivially surjective

and you quotient the first ring so that you will get also an injective mao

.close

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[sterile blaze]

im back

[sterile blaze]

sanity check but 46 is false? taking R2 under addition. the x axis and y axis are subgroups. but their union isnt closed

[cursive field]

how isn't it closed?

[sterile blaze]

when is the first map not an isomorphism?

[cursive field]

basic linear algebra: the two basis vectors in R^2 span the entire plane

[sterile blaze]

(1,0)+(0,1)=(1,1) which isnt in the x or y axis

[cursive field]

oh wait i misread ur statement

[sterile blaze]

but it says union. not product

[cursive field]

lmfao mb

yeah ur correct

mb mb

[sterile blaze]

yay

what about what everg said

[cursive field]

?

what about it

[sterile blaze]

^

is the first map ever not an isomorphism?

[pastel dew]

I did say that it was an isomorphism

[cursive field]

it's always an isomorphism lol

[sterile blaze]

oh

oki ty

[cursive field]

those are quite literally the same ring

[sterile blaze]

everg n ren

[cursive field]

np

[sterile blaze]

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[sterile blaze]

same guy with a different hat

[pastel dew]

$\mathbb Q[x]\to\mathbb Q[\sqrt 2]$ this actually is only surjective..so that you need to consider the quotient ring

[clever fjord]

everg

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[sterile blaze]

i need some help with 51

[sterile blaze]

im not too sure how to go about it

i got x^2=x^-2 as well

not sure what else i can do

ive to show xy=yx

[lyric marlin]

Have you proven that inverses are unique

[sterile blaze]

isnt that always true in groups

[lyric marlin]

yes it is

so I assume you've proven it

[sterile blaze]

yea

is there any hints

that isnt outright the solution haha

[lyric marlin]

I suspect that there is some element that both xy and yx are inverses for

(I've not found it yet, so this might be dead end)

[sterile blaze]

haha

might be tricky

[cursive field]

i'm getting x^2 = y^-2

[sterile blaze]

ill leave it here for now and continue trying

[cursive field]

and stuck

[sterile blaze]

i got similar

[cursive field]

wait i must've messed up

[sterile blaze]

i dont think this is really helpful (

hahaha

[pastel dew]

$y=1\implies x=x^{-1}$

[clever fjord]

everg

[sterile blaze]

hi everg

ohhh

wait i might be able to do it now

[pastel dew]

hence $xy=(xy)^{-1}=y^{-1}x^{-1}=yx$

[clever fjord]

everg

[sterile blaze]

yea

i was writing exactly that

thankss

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[mystic saffron]

does this be true?

[mystic saffron]

,, \frac{a}{b} = \frac {b^{-1}}{a^{-1}}

[clever fjord]

anjali

[tulip dew]

y3w

[mystic saffron]

oki

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[lament sonnet]

https://docs.google.com/forms/d/e/1FAIpQLSdBbZAn6dQtmkScWb490UzlInkuAr6QVmwnA_ir_YN4NWW0_w/viewform?usp=sf_lin

can you fil this in

this is for school

[mystic saffron]

nuhu h

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[cold briar]

what is a zeta function

[low locust]

it would be easier for you to google it and then if you dont understand something come back with a more precise question

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@cold briar Has your question been resolved?

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[frozen phoenix]

"Given three linearly independent vectors (A,B,C), create a vector X, which has the same angle towards A,B,C". I've found a formula for the vector X with cycle properties but im stuck at finding the coefficients.

[frozen phoenix]

"(HInt) Write X as a linear combination of A, B,C and find coefficents"

k is just some variabel so I could think

[odd edge]

@frozen phoenix Has your question been resolved?

[cold briar]

NO

.close

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[mystic saffron]

Can someone help me solve it:

[odd edge]

@mystic saffron Has your question been resolved?

[meager juniper]

Looks gross but perhaps straightforward using L'Hopital's rule. Have you tried that?

[mystic saffron]

I tried that let me show you what I got, this is disgusting really

What now? lol

[meager juniper]

Well, limit as x -> 0 on this thing is still 0/0 it seems

[mystic saffron]

Yeah, I can use L'Hopital again but to what end?

[meager juniper]

Madness mostly

How about using the product rule of limits first?

lim f(x) g(x) = lim f(x) * lim g(x)

[mystic saffron]

I can try

[meager juniper]

Oh

@mystic saffron the derivative of the bottom is slightly incorrect. You forgot the rest of the chain rule, but it doesn't affect the fact that it's 0/0

[alpine hawk]

I used l'hopital rule and had got 0/0 form

[mystic saffron]

I got this now, can I continue from here somehow?

[quasi sparrow]

Each of those fractions in the product have a limit

,w lim x to 0 of log(1+cx)/x

[clever fjord]

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@mystic saffron Has your question been resolved?

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[cursive hedge]

i am stuck on 2c
what ive come up with is that there are 4! arrangements for each couple, then because each couple can swap seats, this makes it 4! * 2^4. is this correct?

[onyx cloak]

yes

[cursive hedge]

how do you know

[onyx cloak]

the same reason you know

[main yacht]

Hi

[onyx cloak]

4! cuz 4 couples, then each member of a couple can swap seat with each other so and there are 4 couples that can do that so 2^4

so 4!*2^4

[cursive hedge]

ic

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[distant umbra]

how can I solve this?

[distant umbra]

this is a non exact ode

so i tried making it exact

by multiplying with a function

but thats not working

[outer kiln]

ik

[outer kiln]

dy/dx?

right ?

[distant umbra]

yea-

[outer kiln]

wait ...

[sand horizon]

Mdx + Ndy = 0

[distant umbra]

its not in this case

[sand horizon]

oh

but if you distribute the y in the parentheses ?

and same for x ?

[distant umbra]

del M/del y = 1 + 2xy
del N/del x = 1 -2xy

[sand horizon]

so f_1(xy)ydx + f_2 (xy) x dy = 0

1/Mx - Ny = 1/2x²y²

there's probably (for sure) a yt video about

[distant umbra]

i dont get what you did here

[sand horizon]

https://www.youtube.com/watch?v=Oe8DXI1kwH8

[distant umbra]

thankyou

[sand horizon]

no problem, i've watched it and he's telling good

[distant umbra]

okay but how did he find the integrating factor

he just said that should be the integrating faactor

is there a proof or something for this

[hallow crystal]

ydx+xdy=xy(xdy-ydx)
d(xy) = x^3y d(y/x)
d(xy) = (xy)^2 x/y d(y/x)
(d(xy))/(xy)^2 = (d(y/x))/(y/x)
-1/xy = ln(y/x) +c

it can be converted into an exact

@distant umbra

[distant umbra]

fair

thankyouu

.close

[odd edge]

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[odd edge]

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[cunning tinsel]

What sre these called

Cus my professor wont help us and gives us exams w thjs shi

[mystic saffron]

exponents

[cunning tinsel]

is there a way to find those type of exponent equatinss thkugh

cause thers nothing we get ffom the professor

[limber glade]

you just distribute powers over

and simplify

[cunning tinsel]

easier said than n done when weve got no explanation from our professor

[odd edge]

@cunning tinsel Has your question been resolved?

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[odd edge]

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

[granite zinc]

1

[lethal spoke]

Do you know what greater than is

[odd edge]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[mystic saffron]

n<x<m

remember this visually

that x lies between n and m where n<m

[odd edge]

@granite zinc Has your question been resolved?

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[blazing quiver]

I need help

[nimble blaze]

~~expand and ~~ rearrange until you get
stuff > 0
or stuff < 0

[blazing quiver]

oh alrthanks

[nimble blaze]

factorise

[blazing quiver]

Uh

Wait

Do I expand em

And simplify? Or x can be -3 1 or -1

[nimble blaze]

you don't need to expand here, ignore what I said about that earlier

[blazing quiver]

Cus math is bad idk why

[nimble blaze]

both terms have a common factor,
factor that out

[blazing quiver]

X

So x(1+3)(1-1)-(1-1)

<p

0

[nimble blaze]

no

[blazing quiver]

Am I right?

Oh

Sh

Mb

I'm too dumb

Wait

Lemme figure it out

Do I factor with (x+1)

Wait nah

[nimble blaze]

you'd factor out (x**-**1)

[blazing quiver]

How come

[nimble blaze]

that's what's present, not (x+1)

[blazing quiver]

Oh ok

[nimble blaze]

note that x-1 = -(1-x)
you can factor out either (x-1) or -(1-x) depending on personal preference

[blazing quiver]

I'll use with -tive sign

Whats next

Wait isn't - (x-1) (x+1)

[nimble blaze]

no

[blazing quiver]

Uh

I'm lost

[nimble blaze]

first apply

note that x-1 = -(1-x)

[blazing quiver]

Ok

[odd edge]

@blazing quiver Has your question been resolved?

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[little cedar]

i need help with some algebra please help

[little cedar]

<@&286206848099549185>

[robust wyvern]

well any start on it?

or like, are u stuck at the beginning?

[little cedar]

kinda i know its not no solution and not all real numbers and it is an or not an and

[robust wyvern]

sure

ur whole aim in this situation is to abandon that unknown "u" by itself on one side

right now, u just have a mess: -3|2-4u| + 5.5

So we need to simplify this inequality step by step

any idea how to start off?

[little cedar]

distribute the -3

to the numbers in the bar

[robust wyvern]

thats not allowed

[little cedar]

really?

[robust wyvern]

also its called the absolute value lol

yeah

[little cedar]

ik its what my teacher says

[robust wyvern]

the rule of multiplying or distributing with absolute values is |A| × |B| = |A×B|

but in this case, u have -3

which is negative

so u cant

[little cedar]

your right

[robust wyvern]

any other ways?

to start?

[little cedar]

do 2-4w

and take the absolute values of the sumbers

(numbers)

[robust wyvern]

no

u dont know the sign of 2-4u

it could be negative or positive, depending on the u

[little cedar]

i mean does it matter its absolute value

[robust wyvern]

yes it does matter

[little cedar]

?

[robust wyvern]

what if 2 - 4u is negative?

then its absolute value would be its opposite, 4u - 2

oops mb

[little cedar]

your good

[robust wyvern]

thank u ser

Anyway, u rlly tried everything but the one thing that u have ignored, bud

[little cedar]

the 5.5

[robust wyvern]

yes

how do u remove it from that side?

[little cedar]

-5.5 and then do it the other side?

[robust wyvern]

perfection

i assume u know how to count so

[little cedar]

ye

-1.1

[robust wyvern]

-3|2-4u| < -9.6

[little cedar]

no im slow

your right

[robust wyvern]

welp i did it for u anyway

Now, whats ur next move?

[little cedar]

do whats in the absolute value bars

???

[robust wyvern]

u definetly can do that

but its a bit advanced

just always leave the absolute values at last when it comes to inequalities

[little cedar]

im advanced man im doing alg in 8th grade

[robust wyvern]

js try my way

[little cedar]

yes sir

[robust wyvern]

as i said, absolute value is always at last to be done

so what do u do next?

-3|2-4u| < -9.6

[little cedar]

get rid of the -3

[robust wyvern]

yes, how to?

[little cedar]

+3

on both sides

[robust wyvern]

nope

-3|2-4u| < -9.6

[little cedar]

?

[robust wyvern]

in this case, the -3 is being multiplied

not added/subtracted

[little cedar]

oh yeah

so divide

ahhhhh

[robust wyvern]

good, but anything to keep in mind before dividing a negative number in an inequality?

[little cedar]

you always switch the inequality

[robust wyvern]

u mean sign of inequality

yes

[little cedar]

yea

i listen a little you know

[robust wyvern]

as i said, ill assume u can count, so it becomes |2 - 4u| > 3.2

[robust wyvern]

eh, some pointless details, its alright

[little cedar]

okay lets do the absolute vale nect

[robust wyvern]

yes

the thing youve been nagging about non stop since the start

we can finally do it, since its all absolute value on one side

now do u have any rules or formulas in mind to advance ?

|2-4u| > 3.2

[little cedar]

subtract then take the absolute vale of the number

[robust wyvern]

no

[little cedar]

damn

[robust wyvern]

sir

do you know anything about the u?

[little cedar]

nope

i said i listen sometimes

not all

[robust wyvern]

well just a reminder yk

never remove absolute values if you know nothing about the unknown

[little cedar]

hmmmmmm

[robust wyvern]

theres a formula u have to apply rn tho

[little cedar]

hold up one sec

[robust wyvern]

|2-4u| > 3.2

[little cedar]

there

[robust wyvern]

a formula to apply

ill give u them four just in case yk

[little cedar]

yea your irght

hold up i just noticed you pfp wtf

[robust wyvern]

|x| > a means x > a or x<-a id a is strictly greater than 0

|x| > a means all real numbers if a is less oe equal than 0

|x| < a means -a < x < a if a is strictly greater than 0

|x| < a means no solutions if a is less or equal to 0

[robust wyvern]

dont mind it

[little cedar]

okay

[robust wyvern]

which of these are to be applied on |2-4u| > 3.2?

[little cedar]

the first one

[robust wyvern]

bravo