help-13

[mental trail]

that's not what we're saying

what even is that supposed to mean

anyways

maybe you could use the fact

[fringe moth]

ln = log e

[mental trail]

that e^x is its own derivative

[fringe moth]

instead of log 10

[mental trail]

to come up with an antiderivative of e^(-x)

[mental trail]

ln = log_e, not log e

[fringe moth]

Im trying to think

Im re reading what you said

[mental trail]

idk, compute the derivative of e^(-x) if you're out of ideas

[fringe moth]

are you saying this is the right process?

[mental trail]

no xdddd

[fringe moth]

okay sorry

[mental trail]

e^(-x) is NOT of the form x^n

[fringe moth]

I was just doing a generic example

I was following the rule of anti derivatives

[mental trail]

you were following the rule of antiderivative of x^n

[fringe moth]

yes

[mental trail]

which doesn't apply to e^(-x)

[fringe moth]

so is it a different rule?

[mental trail]

I hope we're clear on that

[fringe moth]

okay thats clear

Yes

[mental trail]

ok

try computing the derivative of e^(-x)

[fringe moth]

so what's the rule in this case

[mental trail]

and that will give you the idea

for the antiderivative

[fringe moth]

you're asking me to find the derivative of e^(-x)?

[mental trail]

yes

[fringe moth]

isnt that -1 * e^(-x-1)?

[mental trail]

I'm gonna cry

[fringe moth]

what???

[mental trail]

what's the derivative of e^x

[fringe moth]

e^x?

[mental trail]

yes

[fringe moth]

okay

[mental trail]

so how do you compute the derivative of e^(-x)

remember

it's e^x

[fringe moth]

-e^x

[mental trail]

but applied to -x

[fringe moth]

?

[mental trail]

uuuuuuu close

[fringe moth]

wait

I know

[mental trail]

what is the rule of derivation for e^(f(x))

[fringe moth]

-e^-x?

[mental trail]

YES

ok

[fringe moth]

okay ty

so now Im supposed to do the F(a)-F(b) part?

[mental trail]

F(b)-F(a)

[fringe moth]

yes sorry

[mental trail]

if you found your antiderivative

[fringe moth]

so [-e^-(1)] - [e^-(0)]

correct?

1/e - 1 ?

[mental trail]

you messed up signs

[fringe moth]

wait if it's F(b) - F(a) then it's the opposite order

[mental trail]

integral from a=0 to b=1

F(1) - F(0)

[fringe moth]

isn't it 1 - 1/e?

[fringe moth]

okay yes

[mental trail]

well yes

[fringe moth]

hmm okay ty

[mental trail]

[-e^-(1)] - [-e^-(0)]

[fringe moth]

isn't it opposite order since it's b then a

[mental trail]

int from a to b = F(b) - F(a)

F(x) = -e^(-x)

[fringe moth]

okay so, I think it's more clear now, ty. But I still struggle with the "e" number, specially in natural logarithms, any advice?

[mental trail]

[-e^-(1)] - [-e^-(0)]

[mental trail]

not sure what to suggest to you

[fringe moth]

then it's 1/e - 1

but why

[mental trail]

[-e^-(1)] - [-e^-(0)] = -1/e - (-1) = -1/e + 1

[fringe moth]

oh wait

[mental trail]

.

[fringe moth]

I forgot the -e

so it's -e^-1 + e^0

Ill close now ty

.close

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[desert mica]

The following figure shows a block (m = 200g) fixed to extermity of a spring (k).
The spring is compressed with a maximum compression where the block (m) is placed at position A. The block performs
45 complete oscillations within 20s. Determine the spring constant k.

[desert mica]

chatgpt says it's 40N/m

so i'm very confused if chatgpt is wrong or i'm wrong

[dire geode]

!nogpt

[cedar kiln]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[desert mica]

well yes, thats wshy im asking. ive been relying on it for proofchecking and its been okay but stuff like this happens and most of the times i am wrong and i really dont see what's wrong this time D:

[summer silo]

What is this chatgpt babble.
The question is pretty simple. I don't see why your chatgpt replied with this

[summer silo]

Can you explain your equations here

[desert mica]

okay

T = t/N (time/cycles) to find the period

after we find the period - we use T = 2pi * sqrt(m/k)

since we have everything except k, we can isolate k

[summer silo]

Where does that come from

[desert mica]

that's a formula

hodl on

[summer silo]

Do you know how the formula applies?

because I can drop in T^600=m^330

[desert mica]

i don't quite understand

[desert mica]

it's more of a general formula(?)

i just derived uhh

[summer silo]

To what

[desert mica]

k = m/(T * 2pi)^2 from the origianl formula

[desert mica]

to find the period of something which is in simple harmonic motion

[summer silo]

How did you get
[T=2pi sqrt(m / k)]
to
[T * 2pi= sqrt(m / k)]

[wraith dagger]

fishwhale

[desert mica]

sorry

i just realized

i'm stupid

hold on with me for just a sec pls i'll resolve it

[summer silo]

My suggestion is

if you want to use automated tools

Use them all the way

You can use sympy to solve for an algebraic equation

Tell chatgpt to write you the sympy solution

You can run sympy locally

It should solve this equation without any problems

[summer silo]

Fact that chatgpt is not showing your error shows you cannot be using chatgpt for this purpose

[desert mica]

it should .. i think i was just failing to spot the problem or i was convinced i was right so i didn't really consider it

yeah i know about sympy ^^

[desert mica]

i got no one to rely on D:

but you're totally right

T = 2pi sqrt(m/k)
T/2pi = sqrt(m/k)
(T/2pi)^2 = m/k
(T/2pi)^2 * k = m
k = m/(T/2pi)^2

let me check..

[summer silo]

Anyway we all make mistakes from time to time. It's not proof of anything more significant.
And even if it were, it's not that glorious to be 'not stupid'

[desert mica]

okay i'm getting 40.05N/m now. thank you so much!!

[desert mica]

aw thank u man

for being so patient with me and saying that. thanks ;D

.close

[cedar kiln]

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[summer silo]

no worries

[cedar kiln]

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[kindred ice]

Hi, what did they do to get rid of the '-1'

[kindred ice]

oh its just cross multiplied

one sec

[paper bluff]

yes

[kindred ice]

ok wow great i did it

fuhh yeah

thanks

.close

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[paper bluff]

good job

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[marsh warren]

for b)

[marsh warren]

i got 4/145

but the answer says 2/75

sorry in the pic i meant to write 4/145

[cedar kiln]

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[cursive crater]

I’m self studying Rudin’s Principles of Mathematical Analysis and I have a couple questions about the proof of Theorem 2.35:

\begin{thm}Closed subsets of compact sets are compact.\end{thm}\begin{proof} Suppose $F\subset K\subset X$, $F$ is closed (relative to $X$), and $K$ is compact. Let ${V_\al}$ be an open cover of $F$. If $F^c$ is a joined to ${V_\al}$, we obtain an open cover $\Om$ of $K$. Since $K$ is compact, there is a finite subcollection $\Phi$ of $\Om$ which covers $K$, and hence $F$. If $F^c$ is a member of $\Phi$, we may remove it from $\Phi$ and still retain an open cover of $F$. We have thus shown that a finite subcollection of ${V_\al}$ covers $F$.\end{proof}

Is $F^c$ the complement of $F$ in $K$ or $X$? By my understanding, the proof works if $F^c$ is the complement of $F$ in $X$, but not if it is the complement in $K$.

Also, does ${V_\al}$ refer to a cover of $F$ in $K$ or in $X$? Again, I think it has to be $X$ but am not sure.

[wraith dagger]

Nacho Boi

[dusky panther]

F^c is the complement of F in X.

[cursive crater]

thank you

[cedar kiln]

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[void sand]

{V_alpha} is a cover of F in X

[cursive crater]

thanks

[cedar kiln]

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[random oriole]

i have a question about this equation. now i know im supposed to get rid of the 1 first and then divide it by 2 (leaving 4 on the right side) before squaring the whole thing and then subtracting the final 9, resulting in x = 7. problem is when i first saw that my thought process was "ok lets start by squaring the whole thing to get rid of the radical" and if i do the result would be 4x + 4(x + 9)^(1/2) + 37 = 81, except thats not what im getting. im getting the 4x + 4(x + 9)^(1/2) on the left but instead of the 37 im getting 19. yes i know this doesnt actually help at all with solving this but thats not whats bothering me. i need to find out what im doing wrong before im making the same mistake in a different context

[hollow trail]

can you show your work?

[random oriole]

lol as i was typing it out i figured out my mistake. i remembered to square the 2 for the 2x to make it 4x but i forgot to do it again with the 9 and got 18 instead of 36. problem solved i guess

.close

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[split cedar]

Hi uhm im struggling with like all of these questions and i dont know where to start so i dont have any work shown im just confused on everything...

[cedar kiln]

@split cedar Has your question been resolved?

[split cedar]

<@&286206848099549185>

[polar bronze]

hey! have you gone over factoring out the common factor in an expression?

[split cedar]

uhmm i believe so but it was months ago and i dont remember

[polar bronze]

oki! so typically these problems look something like

simplify 4 + 12x

and you're trying to pull the common factor out (the factor all of the terms share).

in this case, what factors do 4 and 12x share?

[split cedar]

4 ,2, 1 im ps

[polar bronze]

yes! so you take the largest one (greatest common factor, in this case 4), divide each term by that, and rewrite as

4*(1+3x)

make sense so far?

[split cedar]

yes

[polar bronze]

awesome

so in 4a, that's all you're doing really

you can expand further and you can have variables in your greatest common factor (gcf). in this example:

$4x^3 + 2x^2 - 6x$

[wraith dagger]

plutonat

[polar bronze]

what's the gcf here?

[split cedar]

2?

[polar bronze]

2 is a common factor for sure

what else does each term (4x^3, 2x^2, and -6x) have in common?

[split cedar]

x or 1?

[polar bronze]

x!

so you can pull out 2x from the expression. what would it look like after you factor it out?

[split cedar]

2x (2x^2 + x - 3x) ?

[zealous timber]

Ding ding ding

[split cedar]

omg that felt like so ez

[polar bronze]

nice job!!

so in 4a, you have

$\frac{6-10x}{15x-9}$

[wraith dagger]

plutonat

[polar bronze]

look at the numerator and denominators separately, what can you factor out?

[split cedar]

denominator = 3?, numerator = 2?

[zealous timber]

So what does the numerator and denominator become once u factor out 2 and 3 respectively?

[split cedar]

3 - 5x / 5x - 3?

[polar bronze]

close, don't forget that you need to be multiplying by the 2 and 3

[split cedar]

12 - 20x / 45x - 27?

[polar bronze]

when you pull out the factors, they don't go away, they're just pulled up to the front of the expression

[split cedar]

ohh

[polar bronze]

here you divided each term by 2x and then had 2x * (new terms)

[split cedar]

2 (3-5x) / 3 (5x-3)?

[polar bronze]

in the words of @zealous timber, ding ding ding

[split cedar]

yyaayayay

[polar bronze]

there is one more thing tho. what do you notice about (3-5x) and (5x-3)?

[split cedar]

theyre the same just arranged different?

[polar bronze]

yes! they're different by a factor of -1: (3-5x) * -1 = 5-3x

so pulling out a negative one, we get

$\frac{-2 \cdot (5x-3)}{3\cdot(5x-3)}$

is this clear?

[wraith dagger]

plutonat

[split cedar]

so if u flip the numerators to be the same as the denominator the factor u took out turns negative?

[polar bronze]

pretty much. we pulled out a negative one, so it became
2 * -1 * (5x - 3)

and 2 * -1 = -2

[split cedar]

oooo that makes sense

[polar bronze]

so, now last step lol

do the numerator and denominator share any factors?

[split cedar]

5x?

[polar bronze]

a little more than that - they share the entire expression (5x - 3)

[split cedar]

oooo so could u condence the whole fraction? -2+3 (5x-3)^2?

[zealous timber]

Note: when we ask abt what factors they share, we are referring to the specific number or expression that's being multiplied both on the top and bottom, never anything dealing with addition or subtraction

So if u have a 5x-3 both in the numerator and denominator, then they share a factor of 5x-3 itself and not just 5x or -3 since those are the terms being added together

[polar bronze]

yes good point

[polar bronze]

not quite, since in this case we're looking at a fraction

what is

$\frac{5x-3}{5x-3}$

[wraith dagger]

plutonat

[split cedar]

oooo so if somethings like 3x - 4 and 3x + 6 would 3x be the only factors they share? what would u do with the -4 and +6?

[zealous timber]

Nope so that means they don't share anything because (3x-4) and (3x+6) are completely different things

But u can factor 3x+6, because the 3 and a 6 have 3 in common so u can factor out a 3 getting u 3(x+2)

[split cedar]

oooo

[polar bronze]

@split cedar try problem 4 when you get a chance and feel free to ping me or @zealous timber for help!

[split cedar]

@polar bronze im sorry to ping again so soon but i tried 4b and now im a little stuck on what to do adter i take out the CF...

[polar bronze]

how can you factor x^2 - 4? it doesn't have to be a common factor, but do you know how to factor it into something like (x - _)(x + _)

[split cedar]

uhmmm i remember something like that but i cant remember how to actually put the factor into that form..

[polar bronze]

that's fine! so in the quadratic
x^2 + bx + c
we want to find two numbers that when added, equal b, and when multiplied, equal c

so for example
x^2 + 5x + 6
we want to find two numbers that add up to 5 and multiply to 6. in this case it's 3 and 2:
(x + 3) (x + 2)

the idea here is that when you multiply (x + 3) (x + 2) out, you will get x^2 + 5x + 6 back. these expressions are equivalent, it's just easier to see the zeroes in one of them.

does this make sense?

[split cedar]

uhhh sort of?

[pastel vault]

you mean (x + 3)(x + 2)

[polar bronze]

yes thank you

[polar bronze]

I'm gonna link two videos on factoring from my favorite help channel, organic chemistry tutor, it might be helpful for you to review!

https://www.youtube.com/watch?v=qeByhTF8WEw&ab_channel=TheOrganicChemistryTutor
https://www.youtube.com/watch?v=9MUDAr1PnlE&ab_channel=TheOrganicChemistryTutor

[split cedar]

thank you!!

im really struggling lol my exams tmr and i only have 4/8 units done...

[polar bronze]

oh no

good luck 🫡

[split cedar]

thank u sm🩷

[polar bronze]

of course!! happy to help:)

[cedar kiln]

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[brave copper]

lol

mods help help hleplhe pl

im just gonna ping uh @rustic coyote hi

[twilit escarp]

Rather ping @ moderators if you need them and not a single one

[cedar kiln]

@urban coyote Has your question been resolved?

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[queen plaza]

This is the question

[queen plaza]

This what I came up with

But when I try to put answer, it says incorrect

Creator did mention there's decimal point

Am I doing something wrong

[honest field]

how many decimal points

[queen plaza]

They never mention...

[honest field]

also you should not approximate along the way

only put everything into the calculator at the very end and then approximate that result

[queen plaza]

Oh ok so put all calc in calculator in one go
And then approx the final answer?

[honest field]

yes

[queen plaza]

Ok will try

[honest field]

other than that all your steps to find x are correct

[queen plaza]

Alright, I guess I'll need to convince or persuade the creator to do something about the answer input

Thank you for the help

[honest field]

np

[queen plaza]

.close

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[earnest garden]

how is this ever 1/4

ok i am so stupid omg i just notice

forget this please

.close

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[lavish vine]

How many ways are there to write 12 letter words using only A B and C such that they begin and end with A and no pairs of the same letter exist in the word

[lavish vine]

!status

[cedar kiln]

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

[lavish vine]

1

[tropic oxide]

hmm wait this needs clarification

you mean that ADJACENT letters can't be the same right

[lavish vine]

Yes

[tropic oxide]

ok show your progress thus far?

[lavish vine]

I have tried several things but none of them seem right

So 1 is more appropriate

[tropic oxide]

mkay

so our words look like A----------A

omg i just had the craziest idea

but it involves matrices and a bit of graph theory

[lavish vine]

I am not aware of those

[tropic oxide]

bleh ok

well ok i think for a start we may wanna count the words with no adjacent duplicates and only the initial A condition

but then... hm. how to incorporate the final A

[lavish vine]

We get 2¹⁰ with the initial A

[tropic oxide]

2^10? not 2^11?

if we just ignore the restriction on the final letter it should be 2^11

[crimson delta]

I feel like you should be able to just count how many of all total strings end in each letter

let a(n) be the number of strings that start in A, have no pairs, end in A and have length n. then similar b(n) and c(n)

easy enough to set up a recurrence

[blazing zephyr]

im p sure you can use genfunc on this

[tropic oxide]

if op does not know matrices it's a high chance he doesn't know genfunc either

[dusk depot]

whats genfunc

[blazing zephyr]

generating functions

[tropic oxide]

generating functions

this isn't your channel though

[blazing zephyr]

neither do i 🤷

[lavish vine]

I mean i do know matrices but not graph theory

[dusk depot]

sorry idk cuz i just joined

[tropic oxide]

read #rules and #info then

[dusk depot]

ok thx

[lavish vine]

But yes I don't know generating functions

[blazing zephyr]

let ^0 mod 3 be A, ^1 mod 3 be B, ^2 mod 3 be C, let the next term be multiplying by x+x^2, thus f(x)=(x+x^2)^11, and do some trickery, lthough recurrence is prolly better yea

[tropic oxide]

idt this works

[lavish vine]

I don't think I can continue this further

[crimson delta]

the recurrence is simple enough

[tropic oxide]

for reference: i was going to recast strings meeting op's condition as walks of length 11 in a triangle graph with nodes A, B, C

and compute their number via the adjacency matrix

[crimson delta]

and then compute adj matrix ^11 ?

oh well, not much easier, is it

[honest field]

ummm I got an idea

[crimson delta]

funnily enough you could just guess that all of "ending in A,B,C" are nearly equally likely and take a third of 2^11

and then round

[honest field]

if between 2 As there are no A then there are only 2 options to fill in B and C

[crimson delta]

fifty-fifty that you get the correct one

[lavish vine]

No there can be As

[honest field]

I said if there's no A between 2 As

[blazing zephyr]

i mean i got a simmilar question but with 11 lettwrs and i got it right

[tropic oxide]

it's not that bad actually

the minpoly of the adj matrix is quadratic

so you can raise it ^11 easily

[honest field]

there are 8 positions left to fill As in

so we can split into 5 cases

[crimson delta]

well I mean ^11 is not that hard anyway. just ^8 * ^2 * ^1

[blazing zephyr]

also, if you really want to be cheezy then the possible 12 letters with A as the start is 2^11, divide by 3 and use the fact that the ammount for ending with A and B, and A and C shouldnt be far, and the possivility of wnding in B and C is equal

[crimson delta]

yes but you dont know if you have to round up or down

which actually flips for every n

[crimson delta]

but if only someone had mentioned that before

[blazing zephyr]

using the fact that B=C you can know

[lavish vine]

I am trying to find the recurrence buy no obvious pattern is emerging

[blazing zephyr]

oop

[crimson delta]

a string of length n+1 which ends in A either ended in B or C before

[honest field]

I think I solved it

@lavish vine can I share my solution here

[lavish vine]

Sure but spoiler it

[honest field]

actually do you have the answer

[lavish vine]

No

[honest field]

the idea: ||if between 2 As there are no As then there are only 2 ways to fill B and C in the gap, so if we figure out how many As are there and how many ways to place the As then we solved the problem||

[lavish vine]

I think I have found the recurrence

[honest field]

||general formula: if there are n As (the two As at the ends don't count) then the number of ways will be (9-n)Cn*2^(n+1)||

[lavish vine]

Nevermind my recurrence was wrong

[lavish vine]

Since there can be a maximum of 4 As

We apply this 4 times ??

[honest field]

||I should add furthur explanation the reason why there are (9-n)Cn ways to place As:
Let's say I have 3 As to put in 8 places marked from 1 to 8 and As can't be adjacent. I'll call the positions a_1, a_2 and a_3 (a_1<a_2<a_3). So we have:
3<=a_1+2<a_2+1<a_3<=8
So there are (8-3+1)C3 ways to put 3As||

[honest field]

5 times

there may be 0 As to 4 As

[lavish vine]

Oh

[honest field]

you prob got the gist of my sol now

[lavish vine]

Yeah

[honest field]

(I never properly learnt deep combinatorics btw)

[lavish vine]

Yep it matches the recurrence thing

For the cases I have tried

[honest field]

what is recurrence btw

[lavish vine]

I don't know that

I got it wrong twice

But I have tried several manually

I think this is the recurrence

a_r = 2^r - a_r-1

[honest field]

hmm what is r and what is a_r

[lavish vine]

r is the number of letters between the As and a_r are the combinations for r letters

[honest field]

ye seems like you got it correctly

same answer as the other one

actually guessing A,B, C appearing the same chance at the end is really close to the exact answer

[crimson delta]

the recurrence I meant is:
a(n+1)=b(n)+c(n)
b(n+1)=a(n)+c(n)
c(n+1)=a(n)+b(n)

and then you just compute all three numbers the whole way through

takes 30 seconds max

[lavish vine]

O

So the one I got is wrong ???

[crimson delta]

no

[lavish vine]

How do we get this ?

[crimson delta]

.

[crimson delta]

btw this is also just (a(n+1),b(n+1),c(n+1)) = A * (a(n), b(n), c(n)) where A is the adjacency matrix of the graph Ann was talking about

so in that sense I computed A^11 * (start values)

[lavish vine]

I don't understand adjacency matrices yet

[crimson delta]

well ignore it then. just wanted to add it for completeness

[lavish vine]

Thanks for the help

.solved

[cedar kiln]

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[opaque sedge]

why (1+2+3...n) = n(n+1)/2 but int_0^n(x)dx = n(n)/2 ?

[pastel vault]

the integral is the sum when the rectangles have infinitesimal width

[quiet plover]

an error of n/2

[pastel vault]

consequently there are infinitely many rectangles in the integral being summed

[crimson delta]

just because they are roughly similar there is no reason that they should behave completely the same

[quiet plover]

a bit of rectangle slices will be sticking up the curve y = x

and the area of all those remnants is n/2

[crimson delta]

geometrically, the integral is half of the area of a square nxn. but the sum is half the area of a rectangle nx(n+1)

[opaque sedge]

if u join the points (1,1), (2,2)... (n,n) and compare to y = x
we get the same graph

so area under the curve should be same

uh wait

[dusk finch]

this is how the actual plot should look like

[opaque sedge]

hmmmmmmmmm

i seeeeeeee

[dusk finch]

now 1 + 2 + ... + n is area under blue (till n) and red is the integral

you can even see that you get half square overflow per number

so blue is > red by n * 1/2

[opaque sedge]

interesting

[dusk finch]

if you wanted to express the discrete sum with a single integral, it would have to be x + 0.5

then the discrete pieces would cancel out perfectly

[opaque sedge]

why did u add 0.5?

ohh

[dusk finch]

because now there is as much extra blue as missing blue

[opaque sedge]

ohh

[opaque sedge]

but bue is still larger than red right?

[dusk finch]

nope, not really

if we cut it at e.g. n = 4, then blue is more than red by 4 little triangles and red is more than blue also by 4 little triangles

which cancels out perfectly

oh maybe you didnt notice the overlapping area

[opaque sedge]

there was one triangle per unit now there's half triangle per unit

[dusk finch]

this is with the +0.5 shift

[dusk finch]

in the original one, yes, blue is larger than red

by exactly n/2

[opaque sedge]

ahh

u r right

so

that means.. (0.5 + 1.5 .... n+0.5) = int_0^n (x+0.5)dx ?

[dusk finch]

nope

[dusk finch]

the blue sum is still 1 + 2 + 3 + ... + n

[opaque sedge]

ahhh

yes

blue is same

[dusk finch]

so that means that
1 + 2 + 3 + ... + n = int_0^n (x+0.5)dx

[opaque sedge]

i see

[dusk finch]

x integrates to x^2 / 2 and .5 integrates to .5x

so youre gonna get n^2 / 2 + n/2

which is (n^2 + n) / 2

and n(n+1) / 2

[opaque sedge]

thanks @dusk finch

[blazing zephyr]

!done sorry to be that guy

[cedar kiln]

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[opaque sedge]

!done

[cedar kiln]

If you are done with this channel, please mark your problem as solved by typing .close

[opaque sedge]

.close

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[crimson delta]

well clearly important enough that its used in the proofs

what kind of answer do you expect?

those are basic and foundational things, yes

the groups Z/nZ are very fundamental

[blazing zephyr]

whats d&f

[crimson delta]

given that they are the only cyclic groups

so they appear often

dummit and foote presumably

[ripe ether]

Dummit Foote Abstract Algebra

Can you screenshot a few examples?

Of what thay use that you don’t know

[crimson delta]

I dont know if d&f includes a section at the beginning with what prerequisites they assume

if so then they also often recommend some other books

but I mean, this basic modulo and divisibility stuff can be found easily on the internet as well

[ripe ether]

It’s "u divides <right side of |>"

[crimson delta]

yes

yes

[ripe ether]

Or in other notation? (Question for you to figure not find, it may be obvious)

Yes
Or also using mod notation? (3 ways)

Noooo

Noooo

Let’s do an experiment if you’re down

[crimson delta]

fwiw, the notation a|b is explained on page 4 of d&f

if you hadnt skipped that

[ripe ether]

I have a few further questions for you to future out on your own to know the divisibility stuff to some extent as nature to you

[ripe ether]

Lemme cook for a min

Is there a collab writing option here?

Would be easier

Like whiteboard app or smth

[dusty wren]

Mfw they get the right ans and say gpt cooking

[ripe ether]

It’s not beyond anyone’s observation capabilities. No need to dwell

It should be the hated

You read this

And still can’t

Ok

Ok I’ll start

Focus and dont use anything external it’s not about getting it right

Suppose you have a function ψ(n) that returns the number of integers up to n that are relatively prime to n

It’s that they don’t share a divisor or gcd = 1

[dusty wren]

You may also know it as coprime

[ripe ether]

Yes

But we’ll go back for a sec

Write down the natural numbers up to 25 in a column

On paper haha

Like 20 = 4 x 5

Just in next column 4x5
Save the ink lol

Now go from the bottom and for each number write it as a product of two natural numbers. Ommit 1 times the number itself

Ok

So the primes are those you didn’t put anything next to

Now

Yes

But that’s just convention

Now for each multiplication you wrote

Rewrite it writing the numbers that are multiplied themselves as a product of natural numbers

Yes

But also it can show you something you I suspected don’t know because of what you said you don’t know ☺️

That’s all?

It’s not

[ripe ether]

Write it maybe I think

You’ll see and you’ll remember forever

The other thing about them you forgot

I honestly cannot get the sentence for some reason but that’s my block

Now see wherever the first time you wrote more than one product

Like for 24

3 expressions

Imma write it down and try to figure

[ripe ether]

In those cases

Then you factored again

Compare all the results for a given number looking at them as the lower factorization terms matter

Then they share that divisor and are not relatively prime right?

[ripe ether]

You get what I mean to do?

Yes

Like

I found

You have to let that sink in as if there was no decimal representation and every number is expressed only such

It gives you what the number really is if I had to put it

So now

Euclids lemma

@austere garnet do you understand the statement?

p divides

p is prime

When you see |

You just read from the left " a | b" "a divides b"

Do you know the law of contraposition?

[cedar kiln]

@austere garnet Has your question been resolved?

[ripe ether]

Apply it to the implication

So
Given arbitrary a,b and a prime p
p ∤a and p ∤ b then p ∤ab

So if neither a or b have p in their factorization

a new p will not appear with a times b

If it didn’t in either a or b

Ok

Relatively prime

For every prime number p, every natural number is relatively prime to it as p is prime

[ripe ether]

This

Euclids lemma and the equivalent contraposition are trivial if you think or write a and b as products of primes in the first place

[ripe ether]

What I meant partially

Know what I mean?

Like p is prime so is expressed as such already and then on the right side of "∤" (;= does not divide) or "|" (;=divides) you have a list of primes - a product !but because of the fundamental theorem of arithmetic (that every number is a unique product of prime numbers)
it just means p appears on the right if "p | right" or doesn’t appear on the right - "p ∤ right"

Expressing numbers as their prime factorization translates the question of divides to just does it appear in the product

It’s trivial one could say but seeing how that can only work because of the fundamental theorem of arithmetic gives a lot of intuition

[quiet plover]

i cant even see where the question is

can you please repost it

[crimson sedge]

Pins?

[quiet plover]

so we're still at the same qn?

[crimson sedge]

Oh nevermind sorry, I assumed that

[ripe ether]

Show me what you’ve written

[quiet plover]

i don't even know if a question is being discussed

[ripe ether]

The original is far gone

[quiet plover]

yeah fair

[ripe ether]

We’re learning about primes, divisibility etc

[quiet plover]

yeah that much I can follow

not sure what exactly is the argument

[ripe ether]

It was meant to provoke just an observation @austere garnet

Gimme 4 minutes to write what I meant

Write down what you’re trying to show

[quiet plover]

can you just repeat what you were struggling with

it's hard to make out with a roughwork image

alr

[ripe ether]

One can prove Euclids lemma but it’s not what I meant

[quiet plover]

i think you know what 'divides' means?

as an example 7 divides 35

you can write 35 as 5 * 7

and wow

[ripe ether]

Once in class I got out of that step by showing the "it’s visible as such" "proof" via the contraposition

[quiet plover]

there's a 7 in there

you can basically factorise any number like that

into a product of prime numbers

78 = 2 * 3 * 13
44 = 2² * 11
510 = 2 * 3 * 5 * 17

and here you can factorise a and b as a product of primes

this is what it would look like if you tried writing it in a proof

[ripe ether]

@austere garnet

[quiet plover]

so if these two are a and b

how would you write ab from this?

[ripe ether]

Is that readable? @austere garnet

[quiet plover]

can you write it a bit briefly

on a paper if you like

it looks like you missed the powers

[cedar kiln]

@austere garnet Has your question been resolved?

[cedar kiln]

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[frosty pilot]

is it not -1/6*(-4/...) so - was supposed to cancel out and leave me with +?

[dire geode]

,w expand (-1/6) * (-4/(x-5/6))

[wraith dagger]

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[dire geode]

multiply wolfram's result by -2 over -2

[frosty pilot]

ahh, i guess this works

but idk why this doesn't

or are they equivalent

ok i think i just chose a wrong answer

.close

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[young ledge]

,tex \mathrm{P}(t)= -3t{^4}-2t{^3}-2t{^2}\
\mathrm{R}(t)= t-1\
\
\mathrm{P}(t)-2\mathrm{R}(t)\
\
\mathrm{P}(t)=-3t{^4}-2t{^3}-2t{^2}-0t+0 \
2*\mathrm{R}(t)= \hspace{4cm}2*t-2 \

[young ledge]

okay quick question i cant find the solution, i feel its pretty obvious but i dont get it

so i have to substract those Polynomials

it originally is R(t)=t-1

but on the substraction it is

so i think it should become 2t + 2

but my teacher told me when doing substractions of polynomials

[spring forum]

@young ledge Do you have a picture of the original problem? I don't understand it because at first you wrote $P(t)=-2 \cdot t^4 \dots,$ and now it's $P(t)=-3 \cdot t^4 \dots$

[wraith dagger]

Marianne

[spring forum]

wait

,tc color white

[wraith dagger]

You have switched to the white colourscheme.

[spring forum]

@young ledge Do you have a picture of the original problem? I don't understand it because at first you wrote $P(t)=-2 \cdot t^4 \dots,$ and now it's $P(t)=-3 \cdot t^4 \dots$

[wraith dagger]

Marianne

[spring forum]

there that's better

[young ledge]

mb

[spring forum]

Okay, and the question is asking for…?

[wraith dagger]

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

[young ledge]

my question is about if this should be positive or negative

[spring forum]

OK. Basically whether or not $2R(t)$ is negative?

[wraith dagger]

Marianne

[young ledge]

yes

[spring forum]

OK. So that means we can just see whether or not $R(t)$ is negative, correct?

[wraith dagger]

Marianne

[spring forum]

because 2 is positive

[young ledge]

yeah but i am not sure cuz my teacher told me that when its an substraction the second term should be inverse

[spring forum]

wdym>

?

[young ledge]

in this case R(t) should have their negatives and positives inversed

[spring forum]

Also, is there any other information in the problem? Because it's hard to determine whether or not 2t-2 is negative without knowing the value of t or something like that

[young ledge]

nope not anything else this is the whole table

[spring forum]

Ah, I get the problem now.

OK, so we start with $P(t)-2R(t).$ Since we know the values of $P(t)$ and $R(t),$ can you plug in the values of $P(t)$ and $R(t)$ into this expression?

[wraith dagger]

Marianne

[young ledge]

yes, did so in the original image

just re arranged and completed the polynomials

[spring forum]

OK. Now, when you subtract it, you first multiply the thing you're subtracting by $-1.$

[wraith dagger]

Marianne

[spring forum]

Because it's basically $P(t) + 2R(t) \cdot (-1).$

[wraith dagger]

Marianne

[young ledge]

so R(t) becomes -2*t-2?

[spring forum]

Careful! R(t)=t-1, so 2R(t)=2t-2. What do we get when we multiply that by -1?

Remember, 2t-2 is just 2t+(-1)*2 😉

[young ledge]

but 2R(t) wouldnt be positive cuz -1 * -1 = +2

so then it applies the -1

which would change it to negative

[spring forum]

Maybe I'm missing something here. Is it given in the problem that R(t) has to be positive?

[young ledge]

like i mean this is the original operation

multiplied by 2 it would be -1 * -1 = +2

[spring forum]

wait, how did you get that?

you have 2(t-1)

and then you distribute, I think

[young ledge]

i was trying to apply the negative times negative equal positive on the -1

like

(2.t) (2 . -1)

[spring forum]

OK, 2.t+2.(-1), and after you multiply by -1, you get (-1)*(2.t+2.(-1))

putting this in math mode to make things look better:

$(-1) \cdot (2 \cdot t + 2 \cdot (-1))$

[wraith dagger]

Marianne

[young ledge]

WAIT i think i got why i had this issue

i was thinking 2*1 was -1 multiplied by -1 😭 i mistook it as a power

[spring forum]

No problem! 🙂

Can you finish it from here?

[young ledge]

yeah i think so im sorry for wasting your time

.close

[cedar kiln]

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[spring forum]

No, it's fine!

[young ledge]

.reopen

[cedar kiln]

✅

[young ledge]

found another issue

i dont understand what does the -S(t) minus would change on the S(t) and the operation overall

this

[cedar kiln]

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[void sand]

okay, I'm confused by this example

[void sand]

it comes directly after this Proposition, so I believe the author is trying to show how this formula is used to do concrete calculations

but my confusion lies in the issue that it seems like they didn't use the formula at all?

I don't understand how their computation is the same as the indicated coordinate formula given by
[\sum_{j = 1}^{3} \left(\sum_{i = 1}^{3} X^i \pdv{Y^j}{x^i} - \sum_{i = 1}^{3}Y^i \pdv{X^j}{x^i}\right)]

[wraith dagger]

higher!

[void sand]

I even checked the formula by performing the calculation using that formula myself

and its the same answer

but idk what the author did

it sure as heck doesn't look the same

my current suspicion is that they applied the definition somehow

but I have no idea how they did, if that was indeed what happened

[void sand]

oh, oops. there's supposed to be a \pdv{}{x^j} at the end there

[mental trail]

huh

lemme try something

$X = X^i\partial_{x^i}$, $Y = Y^j\partial_{x^j}$

[wraith dagger]

Raphaelisius Maximus MMIII

[mental trail]

and so

$$[X,Y] = X^i\partial_{x^i}(Y^j\partial_{x^j}) - Y^j\partial_{x^j}(X^i\partial_{x^i})$$

eeeeeh _

[wraith dagger]

Raphaelisius Maximus MMIII

[mental trail]

so

[void sand]

me notto andastand

[mental trail]

I feel like I'm forgetting something

how do you compute something like $\partial_{x^i}(Y^j\partial_{x^j})$

[wraith dagger]

Raphaelisius Maximus MMIII

[void sand]

isn't that equal to $\pdv{Y^j}{x^j} \pdv{}{x^j}$?

ah

[wraith dagger]

higher!

[mental trail]

I'm thinking product rule though how would it apply

nah it's weird

[void sand]

hmm

[mental trail]

I'm staying away from ts 🥀

[void sand]

do you have any idea what the author did, raf?

[mental trail]

eh

[void sand]

like it seems like the coordinate formula works, but the author didn't even use it?

[mental trail]

I would start from the coordinate formula

and use X^j = X^k = 0

and Y^i = Y^k = 0

[void sand]

so did John Lee use some mystical method he's hiding from us all or what

[mental trail]

idk

lemme try coordinate formula

[void sand]

sure

[void sand]

I did the calculation here though

it yields the same result that he got

it's just that idk how his thing is the same as my thing

[mental trail]

$$\sum_{l = 1}^{3} \left(\sum_{m = 1}^{3} X^m \pdv{Y^l}{x^m} - \sum_{m = 1}^{3}Y^m \pdv{X^l}{x^m}\right)$$

[wraith dagger]

Raphaelisius Maximus MMIII

[mental trail]

ok

"l = 1" means l = i

first sum is 0

[void sand]

wait wait, wdym l = 1

[mental trail]

i'm summing from l = 1 to 3

I wasn't gonna use i and j

so

l = i -> first inner sum is 0, second inner sum is:

Y^j dX^i/dx^j

will count as negative

l = j -> first inner sum is X^i dY^j/dx^i, second inner sum is 0

l = k, everything 0

$[X,Y] = X^i \pdv{Y^j}{x^i} - Y^j \pdv{X^i}{x^j}$

[wraith dagger]

Raphaelisius Maximus MMIII

[mental trail]

did I get smthg wrong?

[void sand]

I uh, kinda don't follow

[mental trail]

also why the frick is there an extra d/dx^j at the end

[void sand]

when I did the calculation, the interior sum yielded a -1 for l = 1, a 0 for l = 2, and a -y for l = 3

[mental trail]

oh wait

bruh

X^i actually refers to the i component of the X diff operator

so it's actually X^i d/dx^i

[void sand]

cause it's a vector field

[void sand]

yes

I thought you and I were on the same page there

my bad

[mental trail]

no I'm bad xdd

[void sand]

I still have no idea what Lee did in his calculation

[mental trail]

no but this is exactly what we get

[void sand]

like, what the heck went on here

[mental trail]

I'll rewrite it correctly

[void sand]

okie dokie

[mental trail]

$$[X,Y] = X^i\pdv{(Y^j\partial_{x^j})}{x^i} - Y^j\pdv{(X^i\partial_{x^i})}{x^j}$$

[wraith dagger]

Raphaelisius Maximus MMIII

[mental trail]

this should be easier to read

[void sand]

what does \partial_x^j mean?

I suck at notation

[mental trail]

$\partial_{x^j}$ is a short for $\pdv{}{x^j}$

[wraith dagger]

Raphaelisius Maximus MMIII

[void sand]

why is this what we get?

I don't understand how this is the same as the coord formula

[mental trail]

I just replaced XY - YX by what they are

[void sand]

I'm still kinda confused, particularly on what this gets us

[mental trail]

I'll be honest with you

[void sand]

is this what the author used, or is this like, something unrelated?

[mental trail]

I don't know how to continue

[void sand]

fair enough

[mental trail]

idk what the author used

maybe it's this

[void sand]

god I hate this book

[mental trail]

but I don't know how to develop the partial derivative of some product or smthg

[void sand]

perhaps

I think I might just run with this coord formula and pretend the author did nothing

or I'll ask another person if they know

[void sand]

cause it worked

[mental trail]

xddd

go for it

[void sand]

okie dokie

thank you for your time raf

I appreciate it

[mental trail]

I am impotent in this situation

xddddd

[void sand]

issok

most of my time spent reading this book is being confused

so I'm used to it by now

[mental trail]

xddd

[void sand]

I'll close this now

tanq

.close

[cedar kiln]

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[cedar kiln]

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[cursive cedar]

Am I understanding this right?
Inputting the basis column vectors will return the column vectors of [1 -3 8; 1 -2 5; 4 -11 30]

We want to find what happens when we transform e2

However e2 is already part of our basis
So is it just the second column vector? [-3; -2;-11]?

Wait no

[void sand]

I think you've mistakenly used the rows instead of the columns

[void sand]

yes, <0, 1, 0> gets mapped to the coordinate representation <-3, -2, -11> under T

you can verify this with matrix multiplication if you so desire, but this is pretty much definition of the matrix representation of T

but that coordinate representation really means that <0, 1, 0> is getting thrown to -3B_1 - 2B_2 - 11B_3

[cursive cedar]

0,1,0 gets mapped to -3,-2,-11 in basis B

What we want is the transformation of 0,1,0 in the standard basis?

[void sand]

so you can use that to compute T(0, 1, 0)

[cursive cedar]

so we do -3b1 -2b2 -11b3?

[void sand]

the columns in a matrix representation of a linear map T with respect to two bases tells us how T maps each basis vector in the domain to a linear combination of the basis vectors in the codomain via its columns: the ith column is the coordinate representation of T(b_i) in the basis of the codomain

in your example, the second column is <-3, -2, -11>, which means that the coordinate representation of your second basis vector <0, 1, 0> in terms of the basis B after applying T is -3b_1 - 2b_2 -11b_3

i.e. T(<0, 1, 0>) = -3<-2, -3, 1> - 2<0, 1, 0> - 11<-3, -2. 1>

[cursive cedar]

tells us how T maps each basis vector in the domain to a linear combination of the basis vectors in the codomain

So it's not the point -3,-2,-11, it's the linear combination

That makes sense now

[void sand]

exactly

[cursive cedar]

okay thank you

.close

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[thick stream]

I wanted to ask is (a) solvable ? I solved (b) through youtube but i am stuck on (a)
I feel they haven’t given us enough stuff to solve the perimeter and area of trapezium

[lavish vine]

What have you tried

[thick stream]

Not much i tried gemini but it gave the wrong answer,

[honest field]

!nogpt

[cedar kiln]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[lavish vine]

So we need a construction

We know r=5

[thick stream]

The semicircle i can solve as r is given

Im only stuck on the trapezium part

[lavish vine]

So now we construct a line from the centre of circle to centre of line xy

What will that give us ???

This one

[thick stream]

5 i guess ?

[lavish vine]

The length of this line is 3

And OX is 5

[thick stream]

Okay ?

[lavish vine]

So now can we find XY

[thick stream]

The thing is this 8th grade problem and i cant use advanced stuff

[honest field]

this is not advanced

[lavish vine]

Are you aware of pythagoras theorem

[honest field]

it's just basic pythagoras

[thick stream]

Yes i am

[lavish vine]

We just use that

[thick stream]

How are we using pythagoras to find XY ??

[honest field]

just marked it's midpoint

[thick stream]

Oh ..

[honest field]

ABYX is an isoceles trapezoid anyway

[thick stream]

Oh that makes much more sense sorry for not understanding 😭

[honest field]

nah it's okay

any cyclic quadrilateral that has two parallel sides is an isoceles trapezoid

[lavish vine]

Lag

[thick stream]

Ok i think im grasping on something
let me go and try sumthing , ill come and tell if i could solve it 👌?.

[lavish vine]

@honest field did you prepare for any math competition

Just asking out of curiosity

[honest field]

no

iirc

I did participate AMC 8 in 8th grade

and BEBRAS a year later

but yeah that's about it

[lavish vine]

So is stuff like number theory covered in your regular school curriculum

[honest field]

not really

but in vietnam schools usually have an additional class in the afternoon for more advanced maths

[lavish vine]

Our math curriculum is so shit

Compared to this

[honest field]

they banned doing it just this year btw lol

[lavish vine]

O

[lavish vine]

All they teach us is coordinate bashing calculus and extremely computational math

[honest field]

but most of the stuff I know about modulos are from advanced math books I bought

I did remember some horrible problems

proving 2^100 has 31 digits

in an advanced 6th grade book btw

[lavish vine]

This isn't too bad

For 6th grade it is

[honest field]

it's in one of the first pages iirc

in the 7th grade one you are required to prove Wilson's theorem

[honest field]

funny how you saw this while I had a crazy idea

dropping perpendicular from X then calculate each section lmao

[lavish vine]

Better than mine I would have coordinate bashed it if it had come in an exam

[honest field]

iirc I have never used coordinates for a problem like this

[gritty galleon]

is that always true

[honest field]

it is

you can prove that easily

[gritty galleon]

i am struggling to prove that, do i need a construction for this?

[honest field]

no

[gritty galleon]

alr

[honest field]

you just need very basic knowledge about cyclic quads

[gritty galleon]

quick question

[honest field]

many people made the classic mistake of proving the two sides have equal side length to prove an isoceles trapezoid

[gritty galleon]

if i drew the diagonals, would YAB = XBA

[gritty galleon]

i am not supposed to do that?

[honest field]

no

[honest field]

useless

[gritty galleon]

cmon atleast hear me out

[honest field]

ok

[gritty galleon]

yeah so?

[honest field]

you have the sum of angles from the parallel lines

[gritty galleon]

okay wtever i proved its isosceles

yeah now its doable

[honest field]

you just need to prove the two side angles are the same

@thick stream have you figured it out yet?

[cedar kiln]

@thick stream Has your question been resolved?

[cedar kiln]

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[analog swift]

hi i would use some help on this question

[analog swift]

<@&286206848099549185>

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[oblique flare]

4x+y=2x+2x+y this would help

[blazing zephyr]

cross multiply, get it in the form of ax=by, and divide by a and y to both sides to get x/y=b/a

[analog swift]

thanks, for the help

.close

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[fast barn]

do I use cos = 5/13 as a key to find cot + cosec = 1.5? or their is a way to find 1.5 from cot+cosec

[tropic oxide]

it's the other way around.

"show that X" means X is your goal.

[wintry monolith]

Componendo dividendo

Oh nvm

[wintry monolith]

Do u know ur identities?

[fast barn]

i don't remember much....

[wintry monolith]

The golden identities

[fast barn]

only (a+b) stuff right now

[wintry monolith]

Trig identities

[tropic oxide]

i don't think you need anything much more complicated than pythagoras here.

like at minimum you should write things in terms of cos(θ) and sin(θ) to start with

[wintry monolith]

U can just use cosec^2 - cot ^2

[floral flax]

hello

[wintry monolith]

And find cosec from it

[floral flax]

i will help u

[wintry monolith]

[floral flax]

so ronaldo is sin theta

and messi is a dancer

so messi dances with sin theta

and sin theta doesnt enjoy

[wintry monolith]

[floral flax]

so he kicks the dancer

[tropic oxide]

what is this analogy...

[floral flax]

and he flies

and hits tan theta

who is neymar

he cries n rolls

n rolls into the hospital

[fast barn]

@tropic oxide do I use cos theta = 5/13 to find cot+cosec=1.5?

[wintry monolith]

No

Ur supposed to prove it

[tropic oxide]

as i said before, no.

[fast barn]

ok

[tropic oxide]

cos(θ)=5/13 is your goal.

[wintry monolith]

U know if cosec+cot is x

[fast barn]

ok

[floral flax]

no you have to hit tan theta

[wintry monolith]

Cosec - cot is 1/x

[tropic oxide]

was this all serious or a joke

[wintry monolith]

@fast barn

[fast barn]

huh

lemme check

[floral flax]

i mean no offense but a person who actually blvs this is most likely an idiot

[wintry monolith]

Do u know cosec^2 - cot ^ is 1