help-28

[fallow thunder]

sure have fun

[jaunty iris]

does !close work here ?

.close

[full forum]

Channel closed

Closed by @jaunty iris

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[torn jolt]

Word Problem: How many days and MG I have left of my medication.

There are 30 squares of medication in a pack.

Each 1/4 is 15mg. The entire square is 60 mg. Half the square is 30mg.

I am allotted 30mg per day.

20 days have passed.

How many MG should I have left? How many days of treatment should I have left?

Thank you, my brain sucks and I've had the pen and paper out trying to figure this out.

[onyx glen]

well you eat 30 mg a.k.a. half of a square per day

20 days means 20 * 1/2 or 10 squares eaten

if you started with 30 squares, it means you have 20 squares left -- so these last you 40 more days

@torn jolt

[torn jolt]

40 doses left thank you

How do I calculate the 40 days into how many 30mg left

Thank you for helping me

[onyx glen]

40 days * 30 mg/day

40 * 30

it's just multiplication, nothing more

[torn jolt]

So I should have 1200 mg

[onyx glen]

yes

[torn jolt]

Thank you so much

[full forum]

@torn jolt Has your question been resolved?

[full forum]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sudden field]

prove that cos(2a)(1+16cos(2a)cos
2
(4a))=2cos
2
(a)+16cos
4
(2a)cos
2
(4a)−cos
2
(8a)

[onyx glen]

.close (other channel)

[full forum]

Channel closed

Closed by @onyx glen

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tranquil lantern]

part c

doesnt make sense with the graph shown in the solution:

i dont get why theres an asymptote at y = 300

[gritty rose]

$\lim_{t\to \y} e^{-0.5 t} = ?$

[glossy valve]

riemann

[tranquil lantern]

im not sure i never learnt this

[gritty rose]

how about limits?

[tranquil lantern]

i know that

[gritty rose]

do you know $e^{-at} = \frac{1}{e^{at}}$

[glossy valve]

riemann

[tranquil lantern]

yes i know this

[gritty rose]

and when a > 0, then at > 0 for t > 0

[tranquil lantern]

yes

[gritty rose]

what's the limit of 1 / (e to something large) ?

[tranquil lantern]

well

[gritty rose]

?

[tranquil lantern]

itll be very small

0?

[gritty rose]

yes this is 0

[tranquil lantern]

ohh ok i understand that now

[tranquil lantern]

what if it was

[gritty rose]

what

[tranquil lantern]

wasnt negative exponent like 0.5t instead of -0.5t

[gritty rose]

you should be able to figure that out

[tranquil lantern]

i get it tysm

.solved

[full forum]

Channel closed

Closed by @tranquil lantern

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[obtuse valley]

i need help to explain to solve this please

first i need to plug in 3 to find L which is 5

. close

[full forum]

@obtuse valley Has your question been resolved?

[full forum]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[crude lake]

Hi! I am struggling with letter E. They seem to be both on the right side of the bell curve and I cannot figure out if I should add them or subtract them... Can someone enlighten me? Thanks! ❤️

[austere cove]

If you have the cumulative distribution, CDF(x) this is P(score < x).

So if you do CDF(x) - CDF(y) = P(score < x) - P(score < y). This will give you (if x > y) the value P(y < score < x)

[torn jolt]

Hi

[austere cove]

Hi, if you're looking for help, please check #❓how-to-get-help

[crude lake]

Oh okay! So to clarify, I need to subtract them? I am not fond with the CDF 😢

[austere cove]

yeah

[crude lake]

Thank you! Last question! Will it be the same case with letter d?

[austere cove]

yeah exactly

[crude lake]

Thank you so much!

.close

[full forum]

Channel closed

Closed by @crude lake

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[marsh cloak]

im back with more data management. I got lost trying to figure out how I should continue

[honest hill]

use the frequency column

[marsh cloak]

the frequency coloum to find the variance or the standard deviation or both?

[honest hill]

both

[marsh cloak]

okie one sec

this is the answer key

[honest hill]

my brain is fried

sorry

idk how to help

[marsh cloak]

😭

<@&286206848099549185>

@gritty rose

[full forum]

@marsh cloak Has your question been resolved?

[marsh cloak]

.close

[full forum]

Channel closed

Closed by @marsh cloak

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[steel solar]

$\cos(\frac{x}{2})-\sin(2x)-\cos(\frac{7x}{2})=0$, solve for $x$.

[glossy valve]

;(

[steel solar]

!status

[full forum]

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

[steel solar]

2

Unfortunately, I can't show my work. My phone is dead and I don't have a charger.

So, just suppose I was at 1.

[keen vector]

ru sure you there's a closed form solution

[steel solar]

Yes, I checked desmos.

[steel solar]

Read above messages?

I need a new start anyways.

[keen vector]

i mean i think u can turn the whole thing into a polynomial in cos(x/2)

[honest hill]

i know you cant show your work, but you can at least explain what you did (and possibly where you got stuck)

[keen vector]

like a 7th degree polynomial?

[steel solar]

It's a quintic, in terms of sine. Not a very nice quintic at that.

[keen vector]

quintics are famously not soluble by radicals

(most quintics)

[steel solar]

cos(a)-cos(b) formula, some half-angle shennanigans, (I did prove a solution set was $x=\frac{\pi}{2}k$ for $k\in\mathbb{Z}$) but besides that I have nothing.

[glossy valve]

;(

[keen vector]

well u can factor that out of the quintic

[steel solar]

I already did.

Let me see if I can find my worksheet, it's a bit messy so it might take a while to transcribe it.

[wide sundial]

I was thinking you do the u = x/2 sub then use some trig tricks

Cos(u) - sin(4u) - cos(7u)

-2sin(4u)sin(-3u) - sin(4u)

[analog shale]

this works

[wide sundial]

Then just factorise and solve

[analog shale]

you just get 2sin(2x)sin(1.5x) so sin(2x)(1-sin(1.5x))=0

i think its something like that

[wide sundial]

Yeah that’s where im going

[analog shale]

oops mb

[full forum]

@steel solar Has your question been resolved?

[steel solar]

I see where I went wrong.

Yeah, made a dumb mistake. I just did $\cos(a)-\cos(b)=-2\sin(a+b)\sin(a-b)$ instead of $-2\sin(\frac{a+b}{2})\sin(\frac{a-b}{2})$. Silly me.

[glossy valve]

;(

[full forum]

Channel closed

Closed by @steel solar

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[raw salmon]

im confused about this question, how would i solve for any other exclusions?

[leaden ermine]

huh weird

[deep pulsar]

what im doing wrong

[summer fable]

it does work for 7, both is undefined

[full forum]

@raw salmon Has your question been resolved?

[raw salmon]

can you explain this please, im still a bit confused

[summer fable]

when you plug in x=7 in the original expression, it's undefined
on the simplified expression, it's still undefined
so it's still equivalent on x=7

if it's not undefined on the simplified expression, then we'd say that it's not equivalent at that x value

[raw salmon]

so would the answer be none?

[full forum]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[somber timber]

Check it

[somber timber]

No one are answering

[royal charm]

You waited less than one minute! Have some patience.

[somber timber]

I asked others for ur kind information

They didn't answered

[lime ether]

nope

asking in someone else’s channel doesn’t count

[somber timber]

Oh!

K

[alpine pecan]

u missed a term

[somber timber]

Which

[alpine pecan]

u missed two actually

5C3, 5C4

[somber timber]

Ok
After writing that ncn also we should include

[alpine pecan]

u already inculded that

thats the last term

which is -1

(a+b)^n always has n+1 terms if u expand it

[somber timber]

Notes

[alpine pecan]

,rcw

,rccw

[lime ether]

💀💀🙏🏻

[alpine pecan]

fucking hell

[lime ether]

bro

nevermind i won’t help

try again

[lime ether]

well done

👏🏻

[somber timber]

This are my notes

[alpine pecan]

what do u think wud come in the ..... part

[somber timber]

3 and 4

[alpine pecan]

write it

bruh

[somber timber]

3,4,5 but we should not write nth term

As nth term is not taken in my notes

[alpine pecan]

eh

[somber timber]

That's fine

[full forum]

@somber timber Has your question been resolved?

Channel closed

Closed by @somber timber

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[narrow seal]

I don't have a specific problem, but I'm revisiting integrals on a curve and I'm having a bit of trouble visualizing the concept

I'm used to seeing integrals as "area under a curve", and also integrating wrt time (i.e. integrating velocity wrt time to get displacement), but what's an example of a function that could be integrated on a curve like this, and what would the integral mean?

[narrow seal]

oh and why can't you just integrate wrt the parameter t, why does it have to be wrt arclength

[gritty rose]

t is meaningless

Arc length is descriptive

[full forum]

@narrow seal Has your question been resolved?

[green iris]

for this dont think of an integeral as the area under a curve

an integral is literally just an infinite sum of infinitesimals

literally infinitly adding infinitly small things

here the length of the arc is just the infinite sum of every little infinitly small piece of arc

[narrow seal]

that helps a bit more, thanks! it's my first time doing calculus in like a year so

.close

[full forum]

Channel closed

Closed by @narrow seal

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[summer cypress]

I'm currently a calc student and trying to comprehend continuity better, and #102 on my current homework just has me feeling incapable. If anyone would be willing to give me a nudge in the right direction, it would be greatly appreciated

[glossy valve]

Thoth

[summer cypress]

<@&286206848099549185> all i can think is that I can show that g(a) is continuous for some function like just x^2 adn f(x) as like x+1, just at something like a = 0, and complete the continuity checks and call it a day, but I'm worried that's not "proof"

alright yall well I gotta get to bed, any help in the next 6 hours is appreciated

[frosty monolith]

Try with a contradiction proof

You must demonstrate that f(g(a)) is continuous?

[summer cypress]

I'm not familiar with what that is 😅 I'm only in calc, and from assuming what that may mean, I wouldnt know how to go about trying to contradict an allegedly guaranteed statement

[summer cypress]

yes

[frosty monolith]

Assume that f is not continuous in g(a)

[summer cypress]

I mean, sure, but how would that go to prove that if they both are continuous then their composition will be?

not saying you're wrong, just trying to understand

[frosty monolith]

Wait I wrote it down and then I send it to you

[summer cypress]

sounds good, thanks

I currently have $f(x)=x^2-4,\quad g(a)=\dfrac{a}{2}$

[glossy valve]

Thoth

[frosty monolith]

If you have any questions don’t hesitate to

[summer cypress]

could you make it a bit bigger? 😅

[frosty monolith]

I’ll send you the file wait a minute

[summer cypress]

as long as it's something generic and ubiquitous

[frosty monolith]

Get Microsoft OneNote: https://aka.ms/GetOneNoteMobile

It’s all clear?

If you need any help don’t hesitate to write me directly in chat if you need

[summer cypress]

$\lim\limits_{x\to ?}f(g(x))$

[frosty monolith]

a^-

[glossy valve]

Thoth

[summer cypress]

ah, a

alright

I feel a bit uneasy about my professors reaction to that, but I trust your judgement

[frosty monolith]

To do what?

[summer cypress]

to submit that as my answer

not verbatim obviously

but the general proof by contradiction

[frosty monolith]

There’s only one truth in mathematics at this level

And there are infinitely many solutions for one problem

[summer cypress]

fair enough

[frosty monolith]

So if he doesn’t want another approach to same problem he is not a good professor

Anyways I’m also a Calc 1 student

[summer cypress]

to be fair, she might want something from the textbook, and I've never been great at absorbing textbook material

[frosty monolith]

What do you study that you’re doing Calc without knowing how a contradiction proof works

[summer cypress]

this is my first year of college

and ee/cs

where would I have learned how a contradiction proof works prior to this, is my question in return, hah

[frosty monolith]

The logic induction proving and contradiction proving are the foundamentals to understand all the calculus theorem

You can’t understand a theorem if you don’t have the tools that allow you to understand it

[summer cypress]

this is week 3 of class

we've not covered anything related to proofs or logic

[frosty monolith]

But you covered the continuity

Lol

Where do you study?

[summer cypress]

"covered" is a strong word

[summer cypress]

a local college as a transfer student to the University of Texas

[frosty monolith]

Ah okay usa that’s normal

[summer cypress]

:P

I don't much care for the education system here

but c'est la vie

too broke to escape

.close

[full forum]

Channel closed

Closed by @summer cypress

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[raven mauve]

Hi

[raven mauve]

Can i get some help please.

[raven mauve]

I know. Im just tryna find the question

lol

[honest hill]

!da2a

[full forum]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[raven mauve]

Ok anyway

So basically in class we are starting quadratics/parabolas. i know how to use the quadratic equation and complete the square and factorise etc

But then im getting confused on using features of the graph

Like turning point, line of symmetry etc

And the 3 forms of quadratics

So i was wondering if i could get help on stuff like that

[honest hill]

do you have an example?

[raven mauve]

The quadratic relationships is what i need help on the most

[raven mauve]

Here is an exampl

I done question a wrong

So can we start there please

<@&286206848099549185> its been 15 minutes

Should i delete this conversation?? I think its been forgotten

Oh well. :((

<@&286206848099549185> one last ping, and if not then ill probably ask tomorrow or something

Oh well thanks anyway.

[full forum]

@raven mauve Has your question been resolved?

[raven mauve]

.close

[full forum]

Channel closed

Closed by @raven mauve

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[marsh wedge]

How do I simplify this? I know n! = 1* 2 * .. * n * (n+1)*n

[spiral vigil]

look at a_6 for example

$\f{2\cdot2\cdot2\cdot2\cdot2\cdot2}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

[glossy valve]

hayley is NOT BRITISH

[marsh wedge]

okay the multiplication downwards I understand

but 2 * 2 * 2 *... 2

[spiral vigil]

what about it

[marsh wedge]

I dont know

everything

I though 6! is only 6 x 5 x 4 x 3 x 2 x 1

[spiral vigil]

it is

which is what I wrote

[marsh wedge]

the term upwards equals 64

[spiral vigil]

and the top is 2^n

[spiral vigil]

yeah but like

[marsh wedge]

ahh

but its infinte because the squared n ?

[spiral vigil]

squared?

there's no squaring happening here

[marsh wedge]

2^n

[spiral vigil]

that is not squared

[marsh wedge]

okay

mb

[spiral vigil]

n^2 would be n squared

anyway

[leaden ermine]

[marsh wedge]

https://tenor.com/view/sal-lurking-sal-lurk-sal-vulcano-sal-lurk-gif-21668959

[nocturne shard]

ratio test goes brr brr

[marsh wedge]

https://tenor.com/view/skeleton-reaction-information-my-reaction-to-that-information-my-honest-reaction-to-that-information-gif-613275461876515673

me:

[pseudo cape]

Are you trying to argue that the limit is 0?

[marsh wedge]

I have to show that the limit is 0

https://tenor.com/view/tmn-tmn-podcast-tumeniaises-podcast-tmn-gif-tumeniaises-gif-gif-17180912

math is not mathing nowadays

[pseudo cape]

Let $n \in \mathbb N$ be fixed first. Then we have $$0 \leq \frac{2^n}{n!} = \frac{2}{n} \cdot \underbrace{\frac{2}{n - 1} \cdots}{\leq 1} \underbrace{\frac{2}{4} \cdot \frac{2}{3} \cdot \frac{2}{2} \cdot \frac{2}{1}}{\leq 1} \leq \frac 2n$$ for $n$ large enough (say $n \geq 5$). Now take limits, then the RHS goes to $0$ and the LHS is $0$, so the limit must be $0$ too.

@marsh wedge

[marsh wedge]

ah okay yes I understand

but the 2/n at the end I dont

[pseudo cape]

The first factor is 2/n

[marsh wedge]

yes

[pseudo cape]

Everything else we said is <= 1

So that makes the first factor <= than what it is

TeX bot is down..

[marsh wedge]

It is probably mad because I dont understand easy stuff

[pseudo cape]

No it's fine, there surely is an idea behind this and after you get familiar with that you will be able to use it in the future

The argument is that 2/(n - 1) is <= 1 for every n >= 3. Also 2/4 * 2/3 * 2/2 * 2/1 = 2/3 <= 1

So for n >= 5 (or any other big enough n), we have that the whole thing is <= 2/n

[marsh wedge]

okay yes it gets everytime smaller until it reaches 1

but how is 2/n = 1 ?

[pseudo cape]

It's not and I never claimed that

I'm just claiming that the other factors make the first factor which is 2/n smaller than itself (so are less than 1)

If you multiply any number by something less than 1, it becomes smaller

[pseudo cape]

I'm basically just saying $a \cdot (\text{something something something}) \leq a$ because something something something is $\leq 1$

We can surely say that. As an example, if a = 5 and something something something = 0.8, then surely 5 * 0.8 <= 5 (even strictly less)

[marsh wedge]

What I understood now is that, 2"n is smaller than n! which means the number is getting smaller which means we reach 0

is this the same?

[full forum]

@marsh wedge Has your question been resolved?

[leaden ermine]

Wenn du die Konvergenz nachweisen musst, könntest du es auch mit der Taylorreihe von e^x argumentieren, die für alle R konvergiert

[full forum]

@marsh wedge Has your question been resolved?

[sacred rivet]

scam, do not click

<@&268886789983436800>

[marsh wedge]

https://tenor.com/view/multiversx-x-xportal-egld-crypto-gif-11578576393793628563

https://tenor.com/view/sml-brooklyn-guy-i-never-have-to-work-again-now-that-im-a-millionaire-millionaire-gif-27597475

[full forum]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[marsh wedge]

.reopen

[full forum]

✅

[marsh wedge]

Okay this I understand

But now how n! can be bigger than 2^n

In my mind ^n makes the number way bigger than n-1

[onyx glen]

But now how n! can be bigger than 2^n
n! outgrows any exponential

think about it

when you look at 2^n for high values of n, you're throwing more 2's into the product

but for n! you throw bigger and bigger factors into the product

[modest obsidian]

progressively multiplying n numbers is greater than multiplying by only one number n times

[onyx glen]

so in fact $\frac{u^n}{n!} \to 0$ for \textbf{any} constant $u$ (and $2$ is not special at all here)

[glossy valve]

ann.in.a.teacup

[modest obsidian]

think of it this way, 2x2x2x2<2x3x4x5 since 2<3,4,5

now extend that to n terms and you have 2^n<n!

[onyx glen]

note that for big u's it may take n! quite a bit of time to "catch up" to the pure exponential. but once it does, it's already joever

[marsh wedge]

Is there a way to show me maybe an example?

[modest obsidian]

let's say for 3^n and n!

[minor crater]

here's 10^x (red) and x! (blue) plotted on a logarithmic scale. as you can see, x! overtakes after around 25

[modest obsidian]

that works too

[onyx glen]

here it is for 10^n vs n!

yellow is when the factorial takes over

[modest obsidian]

there's always a point where rate of growth of exponential is less than rate of growth of n!

or something like that

[copper zenith]

true

[modest obsidian]

which means n! will outgrow any a^n

[marsh wedge]

Okay this I get, but if I do n-1, n-2… how can it get bigger than u^n

[marsh wedge]

I get this too

[onyx glen]

ok like how about this

in terms of n we are always speaking in the LONG RUN

and if you compare n! vs. u^n

[copper zenith]

fr

[onyx glen]

where n is much larger than u

then n! and u^n will both be products of n things

but the things that go into n! will be much, much larger than u

[marsh wedge]

For me it only works with n**+** 1

[onyx glen]

i think you are kind of overthinking it

or maybe you are overthinking the concept of the factorial

[copper zenith]

ahh

so we are discussing factorial

[marsh wedge]

That’s what my psychologist always says

[onyx glen]

i am saying that you are overthinking this one particular mathematical concept

[copper zenith]

🇼

[marsh wedge]

Yes yes

[modest obsidian]

just think of it as multiplying with only one number n times vs multiplying with +1 number n times

[onyx glen]

i think there are too many cooks in the kitchen

[marsh wedge]

Okay so I just save in my mind, that n! Is bigger than u^n I think this will work

[onyx glen]

n! outgrows u^n

[modest obsidian]

the more the merrier

or something

maybe

[copper zenith]

well

[marsh wedge]

But I feel like there is something missing, I didn’t mentioned a_n = 0

[copper zenith]

this argument last a while

[marsh wedge]

How do I say „yes this true“ in math language

[leaden ermine]

T

1

[marsh wedge]

https://tenor.com/view/t-pose-świat-według-kiepskich-gif-21563317

[modest obsidian]

a=1

trivial stuff

[full forum]

@marsh wedge Has your question been resolved?

[onyx glen]

a_n itself is never 0

its LIMIT as n -> +∞ equals 0

[marsh wedge]

Okay we have done a)
but now I did b) have I could do this for a and b?

oh imagine the k´s are n´s. Did this with a youtube video and she wrote k my bad

[onyx glen]

yes this is correct

or you could also like

[grim chasm]

Yeah you just did the ratio test

[onyx glen]

realize that $\sum_{n=0}^{\infty} \frac{2^n}{n!}$ actually has a known value

[glossy valve]

ann.in.a.teacup

[marsh wedge]

okay thank you guys

https://tenor.com/view/kirby-cute-kirby-right-back-at-ya-nintendo-kirby-nintendo-gif-2104911732742207285

[full forum]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fresh dawn]

Suppose S is a simply connected open set and 0 does not belong to S.
Is there a unique continuous logarithmic branch on S? (z0 in S, log(z0)=0)

[full forum]

@fresh dawn Has your question been resolved?

[full forum]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[drifting summit]

am i leaving out any steps in this proof?

[full forum]

@drifting summit Has your question been resolved?

[full forum]

@drifting summit Has your question been resolved?

[full forum]

@drifting summit Has your question been resolved?

[rotund ibex]

I think it's right. I would say the first paragraph is a bit confusingly worded but maybe that's just me

[drifting summit]

yeah

this is techicalyl my first upper level math class we're im kinda writing proofs freely

so im always worried about how i wor dstuff

[rotund ibex]

You should probably state what you are assuming for the sake of contradiction first. 'Suppose a < \sup(S), then...'

[drifting summit]

but what i was trying to say if a < sup S and a is an upper bound it directly contradicts the definition of the supremum

ah

that makes alot more sense

[rotund ibex]

Yeah. I would write

Let a < sup(S). Then a is not an upper bound for S. But by definition of S for all x in S, x < a, so a is an upper bound for S.

You don't really need to do contradiction for that bit to be honest.

[drifting summit]

ohh okay

[rotund ibex]

You can just say 'for all x in S, x is less than a so a has to be greater or equal to the supremum'.

[drifting summit]

makes alot of sense!

since its just the definition anyways

[rotund ibex]

Yep.

I think your second part is fine though.

[drifting summit]

thanks then!

.close

[full forum]

Channel closed

Closed by @drifting summit

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dusk zealot]

Hey yall, I am not 100% fluent in english so my problem is translated originally from my language, but I am not sure if correctly

[full forum]

@dusk zealot Has your question been resolved?

Channel closed

Closed by @dusk zealot

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[flint lodge]

What intervals does the function increase and which ones does it decrease in

[jagged robin]

Just look at the graph and write down what parts have the line pointing down and what parts have the line pointing up

[full forum]

@modest spoke Has your question been resolved?

[flint lodge]

it means x is between -1 and 6

so its the region of the graph bounded by x = - 1 and 6

not including

yes

1 to infinity

because the graphs going to keep going infiniately

also negative infinity to -3

yeah

you can write it as (1, infinity)

(-infinity, 3)

you can have negative infinity and positive infinity

negative is alll the way at the left side of the graph

positive is the right side

sorry its -3

thats where the function is decreasing

at -3 then it starts to go back up

yeah that works as well

[full forum]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[distant scroll]

why couldnt the model be V=Ae^kt + B? where B is a constant

[buoyant plaza]

because it is not an exponential model then

[distant scroll]

does the B not just translate it up

or down

[buoyant plaza]

it does

[distant scroll]

oh wait i understand

sort of]

thanks

.close

[full forum]

Channel closed

Closed by @distant scroll

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wooden aurora]

what would the domain and range be for this?

[wooden aurora]

is it just D = { X E R }

how would u do for range

[hot lily]

The right arrow shows that the function approaches positive infinity

But we have a horizontal asymptote below

So what does that tell you

[wooden aurora]

yes

hmm

that

wait idk tbh

that it goes up inf

but it doesnt pass -2

[wheat garden]

ye

[hot lily]

So our range would be something like -2, infinity

[wheat garden]

the only values of y are from -2 and above

not including -2

[hot lily]

Do you know how to use brackets/parentheses in ranges

[wooden aurora]

wait but it doesnt pass -2 so could it be -2?

[wheat garden]

it never touches -2

[wooden aurora]

ye ( is for open dot or inf and [ is for closed dots

[wooden aurora]

ye it never touches so why is the range from -2 and not like -1.999999

cuz isnt it where the range starts

[wheat garden]

there will always be a value closer to the asymptote than the -1.9… no matter how many 9’s you use

that’s why it’s conventional to use the actual number -2

[wooden aurora]

oh oka

i get it now

ty

.close

[full forum]

Channel closed

Closed by @wooden aurora

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[zenith heath]

Algebra: Polynomials

[long sky]

Yes that is a topic that does exist

What about it

[full forum]

@zenith heath Has your question been resolved?

[full forum]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[languid gazelle]

might sound dumb but why is the J unit vector for the cross product -J and not +J

[umbral dome]

that's how determinants work

[languid gazelle]

i thought it was always i + j + k

multiplied by some scalar

[umbral dome]

when you expand a determinant in the top row, it always goes + - + (for cross products and any other type of determinant)

[languid gazelle]

damn how did i miss that in class

so component 2 when calculating determinants is always (-)

not +

i - j + k

[umbral dome]

yes

[languid gazelle]

appreciate that

[honest hill]

$\begin{pmatrix}1&-1&1&-1&\cdots\-1&1&-1&1&\cdots\1&-1&1&-1&\cdots\-1&1&-1&1&\cdots\\vdots&\vdots&\vdots&\vdots&\ddots\end{pmatrix}$

[glossy valve]

Bonk

[honest hill]

wow thats ugly

[languid gazelle]

sure is

[honest hill]

but thats what you have to multiply them by

as you can see, it alternates

[languid gazelle]

ohhhh wait

[honest hill]

so if you were gonna do expansion from the 4th row for example

[languid gazelle]

damn that makes a lot of sense

[honest hill]

youd start with -1, and alternate

[languid gazelle]

here i thought the + - + thing only applied to 3x3 matrices

i se

e

so its just + - + - + - ...

[umbral dome]

it also applies to 2x2s, that's why the 2x2 determinant formula is like that

[languid gazelle]

also for 3x3 lets say

is J the opposite sign because cross product finds a vector thats orthogonal to the other 2 vectors

so its a different Y coordinate

and it cant be perpendicular if all 3 directions face the same way as the other 2 vectors

[full forum]

@languid gazelle Has your question been resolved?

Channel closed

Closed by @languid gazelle

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[granite shadow]

Vertex formula proof pls

[granite shadow]

Why h=-b/2a and why k=(b²-4ac)/(-4a)???

And why those numbers will make h and k the coordinates of the vertex?

PLS ANSWER MEE

[timid iris]

lets say the quad is y = ax^2 + bx + c

[granite shadow]

Yes

[timid iris]

can you "complete the square"?

[granite shadow]

Yes

I ended up with a(x+b/2a)² - (b²-4ac)/4a

[honest hill]

stay calm

[granite shadow]

h=-b/2a and k=(b²-4ac)/(-4a)

<@&286206848099549185>

[honest hill]

!15m

[full forum]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[timid iris]

!15m

[full forum]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[honest hill]

yes

thats correct

[timid iris]

lmao emoji change

[granite shadow]

Yeah but why h and k are also the coordinates of the vertex

[honest hill]

wdym?

[timid iris]

this is incorrect i think

[granite shadow]

The coordinates of the vertex are (h,k)

[honest hill]

yes

[granite shadow]

Why is it like that

[honest hill]

idk what you mean

[timid iris]

nvm its correct

[honest hill]

#help-39

[granite shadow]

imagine if you discover a(x+b/2a)²-(b²-4ac)/4a

And then how do you know that b/2a can be transformed into -h and the other one into k

Whereas you know damn well that h and k are the coordinates of the vertex

[honest hill]

its not that you transform it, you define h to be -b/(2a)

[granite shadow]

Yeah that's what I mean

How do ppl know that the coordinates of the vertex are (-b/2a,(b²-4ac)/(-4a))

[honest hill]

do you know how to translate an equation

like, move it up down left and right

[granite shadow]

Umm no.

[honest hill]

youve never seen smth like this?

or this

[granite shadow]

Oh I know that

[granite shadow]

But I don't really understand how this one works

[honest hill]

sure

if you want to move the graph to the right by s units

then you replace x with x-s

and if you want to move the graph up by t units, you add t to the equation

[granite shadow]

Yeah but why?

Why is it x-s

[honest hill]

wdym?

[granite shadow]

Why is moving a graph by s units requires replacing x with x-s

[honest hill]

im not sure how to explain it

but ill try

imagine, instead of moving the graph, youre moving the grid

with the axes and origin and stuff

actually

idk

sry

[granite shadow]

it's alright

I'll figure it out on my own

If I can't

[honest hill]

assume for now that its true

if you have a function f(x)

[granite shadow]

Then I'll just do tests

[honest hill]

and want to move it to the right

you transform it to f(x-s)

with s the shift to the right

[granite shadow]

And to move it to the left it's f(x+s)

[honest hill]

yes

now

lets start with ax^2

the most simple quadratic

this clearly has its vertex in (0,0), right?

[granite shadow]

It's to determine the shape and the direction of the parabola

[honest hill]

f(x)=ax^2

[granite shadow]

Yes

[honest hill]

okay, now we shift it to the right by h

f(x-h)=a(x-h)^2

and our vertex follows along to (h,0)

still following?

[granite shadow]

Yes

[honest hill]

great

now we have a(x-h)^2

and we want to shift it up by k

which gives us

a(x-h)^2+k

and once again, our vertex follows to (h,k)

[granite shadow]

A(x-h)² + k

[granite shadow]

All right

I'll try it

[honest hill]

$a(x-h)^2+k$

[glossy valve]

Bonk

[honest hill]

so, now we have shown that f(x)=a(x-h)^2+k has its vertex in (h,k)

by simply shifting the original function ax^2 up and down and left and right

[granite shadow]

Ok thx from here I'll do it on my own

Appreciate the help

[honest hill]

you sure you got it?

we can conclude it in 1-2 sentences

[granite shadow]

Somewhat

I can do tests on my note book

To see if it's true

[honest hill]

we have shown that a(x-h)^2+k has its vertex in (h,k)

[granite shadow]

Yeah and I'm writing it down rn

Plus the explanation

[honest hill]

and also know that we can rewrite ax^2+bx+c into a(x+b/(2a))^2-(b^2-4ac)/(4a)

from here we can find h=-b/(2a) and k=-(b^2-4ac)/(4a)

[granite shadow]

No need to tell me

I understand perfectly how to get k and h

[honest hill]

thus we can conclude that ax^2+bx+c has its vertex in (-b/(2a), -(b^2-4ac)/(4a))

[granite shadow]

Yes

[honest hill]

ik, i just wanted to conclude it fully

(also partly for myself :))

[granite shadow]

Ok thx for helping me

Bye

.close

[full forum]

Channel closed

Closed by @granite shadow

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[safe fulcrum]

how am I supposed to do this when they won't give me f(x)?

[umbral dome]

by reading the graph

as stated in the problem

[safe fulcrum]

I've found delta x and gotten the areas of the rectangles but I can't calculate them without the equation for the graph

[onyx glen]

you are supposed to eyeball them from the graph

[safe fulcrum]

wait do they literally just want me to multiply the width of the rectangle with the height for each one

I'm an idiot

.close

[full forum]

Channel closed

Closed by @safe fulcrum

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cobalt flicker]

Hi, had a question about a simple calculation.

I can't for the life of me (relearning math) understand why I cannot do
8 / (1/2 * 1/2)

(as in, treat it like a fraction of 8 OVER 1/2 * 1/2 = 8 / 1/4)

as it results in
8 / (1/4) = 32

instead of the "correct" answer of
8 / 1/2 * 1/2 = 8

When we multiply or divide, I assume they carry the same importance, so I can perform 4/2 * 8 = 16 --> 4 * 8/2 = 16

Is this just the importance of working left to right?

[uneven canyon]

When dividing by a fraction you’re essentially multiplying by the reciprocal.

[torn jolt]

When you're working with expressions like in the picture above it is indeed important to work left to right, this is just because the ÷ and × operators doesn't work so smoothly together, like as you said, (8÷1/2)×1/2 ≠ 8÷(1/2×1/2). What can help is seeing the ÷ always as a fraction, and then calculate always multiplication of fractions that are quite intuitive

[uneven canyon]

8 divided by a half is like 8*2

[torn jolt]

With multiplication it's legit to do a×(b×c) = (a×b)×c

[uneven canyon]

Yeah

[slate violet]

it's called the associative property

compare a + (b + c) = (a + b) + c

but if you replace addition with subtraction or division, it doesn't work

you don't even have a - b = b - a
or a/b = b/a

[torn jolt]

@cobalt flicker

[slate violet]

yes, for subtraction and division it's really really important to work from left to right

it's cause you have a division sign

[uneven canyon]

This.

[full forum]

@cobalt flicker Has your question been resolved?

Channel closed

Closed by @cobalt flicker

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[pseudo flare]

for part c, i found the value of x

i dont have access to that document so i dont know the answers, and im not sure what dimensions it’s looking for

but is my working out correct?

[pseudo flare]

i used a calculator for this

.close

[full forum]

Channel closed

Closed by @pseudo flare

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[turbid breach]

https://cdn.discordapp.com/attachments/1339153793719205939/1339153794126315552/IMG_8215.jpg?ex=67adafa8&is=67ac5e28&hm=e960a7aaee70a6f0e9ccccef60a446b3f0817fac1dde0f6e5314ad1dc68a8099&

[vestal moat]

Tu es bloqué ou ?

[turbid breach]

Je voulais vérifier si ça ça marche attend:

prendre x = 0 pour annuler le premier scalaire puis x = pi pour annuler le scalaire de la fonction identique puis un x quelconque (non congru à 0 ou pi module 2pi) pour annuler le dernier

[vestal moat]

Ça marche

[turbid breach]

Ok merci beaucoup

Pour le reste on est d’accord qu’il suffit de prendre une fonction strictement monotone

(Pour montrer que ça n’engendre pas tout f(R,R)

[vestal moat]

Pour la question 2 ?

[turbid breach]

Oui

[vestal moat]

Y a plus simple

Si une famille de 3 vecteurs est libre, une famille génératrice a au moins un cardinal 3

Donc une famille de 2 vecteurs ne peut pas être génératrice

[turbid breach]

Ah oui

Merci

.close

[full forum]

Channel closed

Closed by @turbid breach

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[rotund ibex]

This isn

[rotund ibex]

oh

This isn't really a maths question, but how on EARTH do you draw a nu without making it look like a u or a v??

$\nu, v, u$ are basically the SAME

[honest hill]

$\nu$

[glossy valve]

Bete Puttigieg 🐢

[onyx glen]

give it a bit of a flourish

[glossy valve]

Bonk

[onyx glen]

like when you do its second half

extend it like a )

[rotund ibex]

Well it ends up just looking like I was trying to draw a v but then I forgot how to write in English

[honest hill]

i draw it more like this

v, u and nu

[rotund ibex]

That's a v that's running very quickly to the right. Away from a predator of some sort

I suppose I shall have to do this

[torn jolt]

I just leave my answers in latex even when I am writing

[austere cove]

I make my nus have a floppy bit on the left, us have one on the bottom, and v is just two straight lines

[onyx glen]

that's not a nu

[rotund ibex]

Righto

Guessing game time

The top-right is my attempt to do a nu

[honest hill]

cuz i drew it shit

[torn jolt]

looks like a face....this image

[rotund ibex]

Oh my

[torn jolt]

with two eyes and a mouth

now you can't unsee it can you?

[rotund ibex]

Certainly

Goodness. X and chi is the other one

(I've no idea why it's pixelated)

.close

[full forum]

Channel closed

Closed by @rotund ibex

Use .reopen if this was a mistake.

[onyx glen]

.reopen

[full forum]

✅

[rotund ibex]

That psi is truly something

[onyx glen]

what about it

[rotund ibex]

It's more like a trident

[onyx glen]

psi is inherently trident shaped

[rotund ibex]

Well yes but I just did it like this

One moment

You seem to have extended it very far

[onyx glen]

i gave it both an ascender and a descender yeah

$\psi$

[glossy valve]

ann.in.a.teacup

[honest hill]

$\Psi$

[glossy valve]

Bonk

[rotund ibex]

I suppose that's how it is

Here are my attempts at zeta and xi

They are also pretty sad

Anyway thanks for your help.

[honest hill]

when i first saw someone use this in a lecture i got soooo confused lol

cuz when writing theyre sloppy and if oyu dont know it, makes no sense

[rotund ibex]

The worst thing to see: Xi / Xi

.close

[full forum]

Channel closed

Closed by @rotund ibex

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[regal gorge]

2180

[regal gorge]

Solve algebraically

[shy cedar]

Vad har du testat?

[regal gorge]

Har inte kommit så långt fram

men började med att få bort nämnaren

i första ledet

yea

Stockholm

[shy cedar]

Nice, Stockholm also

[shy cedar]

Skicka gärna bild på vad du gjort

[regal gorge]

En sekund

[shy cedar]

I’m studying at KTH

[regal gorge]

@shy cedarkan jag stryka nämnaren med första täljaren

[shy cedar]

Om jag förstod dig rätt, nej det får man inte

[regal gorge]

varför inte

[shy cedar]

Har lite svårt att visa just nu, men basically för att det är två termer i nämnaren

[regal gorge]

aha okej

så ska jag multiplicera bort nämnaren?

@shy cedar

[shy cedar]

Multiplicera med nämnaren på båda sidor ja

@regal gorge

[regal gorge]

ska jag också få bort 25

från andra ledet

[shy cedar]

Aa

[full forum]

@regal gorge Has your question been resolved?

Channel closed

Closed by @regal gorge

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wooden marlin]

903486480985489478

.close

[full forum]

Channel closed

Closed by @wooden marlin

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[marsh pilot]

For what values of 'r' , (4-11r)/12 is an integer. Where r is also an integer.

[full forum]

Please don't occupy multiple help channels.

[onyx glen]

i'm gonna just close your previous help channel there...

[marsh pilot]

yeah

[marsh pilot]

many things... came up with nothing

[onyx glen]

nothing at all? not even a single value of r that works?

at the very least, if you are really completely dry for ideas, you could tabulate the values of 4 - 11r for r close to zero

[marsh pilot]

i mean yeah i got some... values of r but by substituation... i mean i tried every single values of whole number.. soo its "8"...

[onyx glen]

so you found r=8 works.

you could notice more generally that $4 - 11r \equiv 4 + r \pmod{12}$ --- which should make it easier to see what the situation is more fully

[glossy valve]

ann.in.a.teacup

[marsh pilot]

but question says r belongs to intervel of (1000 , 2000)

[onyx glen]

this is the first time you have shared this with us

[marsh pilot]

how may i find that.. i can't find by substitution everyvalues from whole numbers...

[onyx glen]

until now i did not know you specifically wanted to find all r strictly between 1000 and 2000

as opposed to just all possible integer values of r

[marsh pilot]

now you know...

Soo.... you got any idea?

[onyx glen]

this still stands

unless you know nothing about modular arithmetic and this made no sense to you

[marsh pilot]

yeah i know nothing about that

[onyx glen]

in which case i will instead say this: $\frac{4-11r}{12}$ is an integer if and only if $\frac{4-11r}{12} + r$ is an integer

[marsh pilot]

anyother method/?

[glossy valve]

ann.in.a.teacup

[onyx glen]

and $\frac{4-11r}{12} + r = \frac{r+4}{12}$

[glossy valve]

ann.in.a.teacup

[onyx glen]

so at least now you know r has to leave remainder 8 when divided by 12

[marsh pilot]

will ya please.. find some values of r??

pls?

[onyx glen]

no, you can do that yourself.

i have already given you a huge chunk of the solution

[marsh pilot]

ok bro

[onyx glen]

bruh for real???

[marsh pilot]

fr

[onyx glen]

<@&268886789983436800> this is the third time ive gotten called "bro" by this helpee

after asking him twice not to do that

[marsh pilot]

kiddin dude...

stay calm...

[nocturne shoal]

Do not call people things after they already tell you not to

[marsh pilot]

ohk ohk ohk

[nocturne shoal]

If you're kidding, stop kidding

[marsh pilot]

@onyx glen chess match rn?

[onyx glen]

no.

[marsh pilot]

ok.

[full forum]

@marsh pilot Has your question been resolved?

[full forum]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[stuck fiber]

I think I almost know how to solve this one but im a bit stuck

[honest hill]

whats your question

[onyx glen]

stuck where

[stuck fiber]

Oh wait

💀

I posted the wrong pic lmao

[honest hill]

open a new channel then

[stuck fiber]

Oke

.close

[full forum]

Channel closed

Closed by @stuck fiber

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[swift bridge]

I’m pretty lost for this question. Im trying to find a recurrence relation to get the inequality for limsup

[swift bridge]

But I can’t seem to relate the sup in anyway

[minor crater]

are you talking about this part?

[swift bridge]

Yes

I’m thinking of maybe doing something like s_n = (s_n / s_n-1) * … * (s2/s1)

Or smthing like that

But I don’t think that works out

[charred raft]

u forgot a s1

$s_n = s_1 \cdot \frac{s_2}{s_1} \cdot ... \cdot \frac{s_n}{s_{n+1}} = s_1 \cdot \prod_{k=1}^{n-1} \frac{s_{k+1}}{s_k}$

[glossy valve]

Goëtia

[torn jolt]

it will

[swift bridge]

Oh ok so am on the right track then?

I just don’t see how that recurrence/equation relates to sup

[charred raft]

what recurrence?

[swift bridge]

Sorry I mean this equation

[charred raft]

just apply ln

[torn jolt]

a relation between sn and sn+1

but wont

[torn jolt]

yep thats it

and use limsup\

[swift bridge]

This natural log?

[torn jolt]

yes

[swift bridge]

Wait wha… so somehow applying natural log makes things work

Why apply ln tho? Shouldn’t it be 1/n?

[charred raft]

do s_n ^(1/n)

[torn jolt]

u will get that 1/n ln sn equal something just try

[swift bridge]

Ah and then apply natural log?

[swift bridge]

yeah i think i get $\frac{1}{n} (\ln s_1 + \sum_{k=1}^{n-1} \ln \frac{s_{k+1}}{s_k})$

or smthing like that

[torn jolt]

what dat

nice

[glossy valve]

LXDL

[swift bridge]

maybe i take limit as n goes to infinty now?

[charred raft]

ye

[torn jolt]

yes take n>=k

where is k the apcr

[swift bridge]

but isnt right hand side 1/n ln s_n

[swift bridge]

whats apcr?

[swift bridge]

but we wantn limit of s_n ^1/n

[torn jolt]

yes just eleminate 1/n ln s1

cause it will converge to 0

[swift bridge]

ye sry i mean the left hand side,

where we have 1/n * ln (s_n)

[torn jolt]

yes

now

[swift bridge]

the right hand side diverges i think

since we are summing non-zero terms

[charred raft]

no

[swift bridge]

wait sry i missed the 1/n

mb

[charred raft]

we dont care bout convergernce now

[swift bridge]

okay

we care about lim sup right

[charred raft]

ye

[torn jolt]

the condition well just make us get rid of lns1

[swift bridge]

hmm okay

[charred raft]

what?

[torn jolt]

in other words

[swift bridge]

maybe i take sup o f$\frac{1}{n} \sum_{k=1}^{n-1} \ln \frac{s_{k+1}}{s_k})$

[glossy valve]

LXDL

[torn jolt]

well take n so big that lns1 approx to 0

[swift bridge]

ye i see that part, but i dont see why we take ln and how this relates to lim sup

[torn jolt]

wait ill write

[swift bridge]

i think i am getting $\sup { \ln \frac{s_{k+1}}{s_k} : n \geq 1 }$

[glossy valve]

LXDL

[swift bridge]

the problem is that it is only n >= 1 not n >= N

so i cant make a seuqnec of these sups

[charred raft]

what r u doin?

[swift bridge]

Smthing like this

This is just scratch work but like the last two lines

[charred raft]

oke try this

multiply by s_n

bound s_(n+1)

u get $(L-\varepsilon)^{n-N}s_{N} \leq s_n \leq (L+\varepsilon)^{n-N} s_N$

[swift bridge]

ye

[charred raft]

telescope through it till N