help-49

[chilly cobalt]

yeah i kinda see

then this implies bijection right?

[wheat nymph]

Yes, this is the bijection

[chilly cobalt]

woah

what topic of math is this im kind of curious now

[wheat nymph]

I would say ur problem is abt combinatorics

[chilly cobalt]

all of this??

thats crazy

anyways thank you!!

ill think about it more later

youre crazy smart

.close

[midnight plank]

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[midnight plank]

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[twilit jetty]

need help simplifying this integral to a series, given p > 0

[lethal path]

STOP

go home. you don't have a question.

[twilit jetty]

the screenshot was already done, it doesnt matter if it lasts a second or a day

[gaunt nimbus]

Math is done my g

[twilit jetty]

we've already won

[gaunt nimbus]

Wait for the new season

[lethal path]

no try to make it as long as possible

[twilit jetty]

that was also already done, 50/50 channels used

there was also already a record for a channel closed with no words exchanged

[twilit field]

I'm tempted to close this now

[lethal path]

really?

[twilit field]

.coose

[gaunt nimbus]

Those scammer channels

[twilit jetty]

not really

I think it was me

I asked an AMC level question I got stuck on then someone reacted to it with an emote pointing to the correct way

[lethal path]

oh so you had to type something to open the channel right

and then no one typed anything

[twilit jetty]

that someone had effectively closed the channel with 0 words

so no words were exchanged, but words were said

[lethal path]

yeah I remember, it was so frustrating

people would ask in each other's channels

[twilit jetty]

theres also a similar record for the quickest channel to be closed

I dont know what that record is, I know it at least has to be a few minutes

I was there to see an example of that

but shorter than that Idk

[lethal path]

yeah they just self-closed

[rigid depot]

👁 👁

[lethal path]

it happens all the time

[twilit jetty]

that works

[twilit jetty]

anyways back to this sum, this is a legit question

[flat spire]

.solved

[gaunt nimbus]

[twilit field]

.lcose

[twilit jetty]

the main concern here is twofold

first, Im having a -> infinity to have it integrate over a domain where the function is finite and not broken

[twilit field]

common blue role L, no close perms

[lethal path]

pk means 'hope you drop dead' in Cantonese

[twilit jetty]

I didnt know cantonese let alone that thered be slang for htat

[flat spire]

is u blind

where is my blue

[twilit jetty]

I gotta watch out for 4s too by this point

[twilit jetty]

now that means I begin with a limit like this

I now need to move the integral past the limit and the sum, so I need to justify that that works

after moving, it does converge, but note that swapping lim a -> infinity and lim N -> infinity diverges since the integral would be integrating a finite sum across (0, 1]

so its not always solid

Im personally thinking that if e^(something) converges in general, then maybe there should be some consistent method to show that it converges

swapping the integral and sum can be justified since N is finite and so its integrating a finite number of terms

the best I could figure out here is that its no worse then finding an integrable function g(x) that can surpass f(x)^k/k! instead of having to contend with the sum

not very plausible

so the first problem here is justifying swapping limit and integral

the second problem is this:

the middle sum always diverges if 1/p is an integer

however the first sum always converges as a finite sum,
and the last sum doesnt seem to indicate diverging for integer 1/p

also, we know the integral (integrand graphed in black) always converges and has area < 1/2 since 0 < x^x^p < x

so the sum has to converge

[twilit jetty]

the divergence is being canceled out somewhere for integer 1/p and I cant see where

with those two problems explained, I can show the work I have so far

not entirely sure what I can show for certain, its entirely possible theres a error hiding somewhere

first, we can try writing x^x^p as e^(x^p ln(x)) then as a sum, just as usual

by (1), somehow we swap the integral and limit

since the sum is finite, the sum and integral can be safely swapped

move the 1/k! out

then inverse u-sub x = e^t

using another u-sub to divide by p, simplify to this

split the sums into terms based on if
k - 1/p is negative
k - 1/p is 0
k - 1/p is positive

this can be attained by using these bounds for the sums

if 1/p is an integer, itll simplify to [0, 1/p - 1], then 1/p, then [1/p + 1, infinity
if 1/p is not an integer, ceil(1/p) > floor(1/p), so conventions state the sum is 0
the other bounds simplify to [0, floor(1/p)] and [ceil(1/p), infinity)

the first sum is finite,

and simplifies to this
this is I think the best we can get with a sum like this

the middle sum would be this, if 1/p is an integer, and 0 otherwise

since this sum has k - 1/p = 0, k is given to be 1/p in this expression

as a result, it simplifies to this

which you can observe diverges since a -> infinity

if k is not 1/p, this term does not appear and so no divergence happens

note this in advance,

then the right sum's integrals can be calculated

the result is not pleasant

if you combine the -1 inward, the sum also isnt very easy to compute, since just placing finite numbers in for infinity would I think have the sum diverge (itll swap the N and a again)

[twilit jetty]

and see here theres nothing about this sum that suggests itll behave differently for integer 1/p, despite the fact that the middle sum diverges as seen here

the best Ive done so far is to just try to see whether particular portions of these sums converge/diverge but not much progress there

for some reason its not even guaranteed that this has absolute value less than 1

nasty

so hopefully finding a solution to the two problems of swapping the integral and limit, then finding where the divergence is cancelling out given 1/p is an integer would help this along

then after that of course is finding the answer

Id also like to point out that, for the record, theres a mistake in this work since the final sum does not add up to what it should be, so theres a mistake hiding in this work and I cant find it

-248.7 for an area between 0 and x

oh whoops

shouldve had all of these images in dark mode

maybe I reopen this channel with all the images corrected

.close or not

[midnight plank]

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[twilit jetty]

(Im still stuck on this though)

[midnight plank]

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[chilly cobalt]

Let $ x_1, \ldots, x_n$ be $ n>1$ real numbers satisfying $ A=\left |\sum^n_{i=1}x_i\right |\not =0$ and $ B=\max_{1\leq i<j\leq n}|x_j-x_i|\not =0$. Prove that for any $ n$ vectors $ \vec{\alpha_i}$ in the plane, there exists a permutation $ (k_1, \ldots, k_n)$ of the numbers $ (1, \ldots, n)$ such that[ \left |\sum_{i=1}^nx_{k_i}\vec{\alpha_i}\right | \geq \dfrac{AB}{2A+B}\max_{1\leq i\leq n}|\alpha_i|.]

[grand pond]

Copter

[chilly cobalt]

i was thinking maybe a weighting manipulation could help?

one sec i will tex

Let $\pi(1),...\pi(n)$ be some permutation. Then let $k_{\pi}$ be some real number for all permutaions $\pi$. Then [ \left(\sum_\pi |k_\pi| \right) \max \left| \sum_{i = 1}^n x_{\pi(i)} \vec{\alpha_i} \right| \ge \sum_{\pi} |k_{\pi}| \left| \sum_{i = 1}^n x_{\pi(i)} \vec{\alpha_i} \right| \ge \left| \sum_\pi k_\pi \sum_{i = 1}^n x_{\pi(i)} \vec{\alpha_i} \right| = \left| \sum_{i = 1}^n \vec{\alpha_i} \sum_\pi k_{\pi} x_{\pi(i)} \right| ]

[grand pond]

Copter

[chilly cobalt]

sigh nvm

so like

which k should i pick so that this turns out pretty?

[pine wave]

Okay so nonzero sum, not all the same

What if we randomly pick

[chilly cobalt]

$\left| \sum_{i = 1}^n \vec{\alpha_i} \sum_\pi k_\pi x_{\pi(i)} \right|$

[grand pond]

Copter

[chilly cobalt]

is the one of the right

[chilly cobalt]

?

[pine wave]

Like the first thing I would try is randomly pick a permutation

[chilly cobalt]

hm

should we try out some value of n too?

[pine wave]

The second thing i would try is to pick a local maximum, that there are no local swaps which improve the left hand side

[chilly cobalt]

oh yea

that makes the ineq look nicer i guess

it still looks pretty weird

[pine wave]

The idea for local swaps looks easy, if you pick a particular direction you want the result to be, then sort the vectors by projection to that direction. Finally use rearrangement inequality trick to maximise

[chilly cobalt]

wait, what do you mean by "local swaps"?

[pine wave]

Like swapping two elements of the permutation, but i thought about how to do better than that

[chilly cobalt]

ohh

had an idea, dunno if its relavant but

so let |a_(pi(q))|be maximum vector size some permutation, then we can write |ai - aj| = t|a(pi(q))|?

crashes out

[pine wave]

hmm not sure... none of the obvious bounds seem to give something yet

[chilly cobalt]

gahh

wait just had an idea

nvm

😭

[pine wave]

I think suffices to show for alpha_i real numbers

Otherwise project into the direction of largest magnitude vector

[chilly cobalt]

how would we project? sorry i forgot all my vectors sob

[pine wave]

The largest magnitude vector (if ties pick any) defines a subspace, for each vector, take the closest vector in this subspace

Basically reduces 2d into 1d problem

[chilly cobalt]

oh right

[pine wave]

You can check we dont lose anything from doing this

The 1D problem is trivial to maximise

So we are on the right track

[chilly cobalt]

yippee!!

how would we right that formally though?

[pine wave]

Write it by saying "wlog alpha1 longest, define beta_i = alpha_i dot alpha1"

[chilly cobalt]

write* 😭

[chilly cobalt]

i forgot the dot product existed omg

[pine wave]

Something involving dot products

[chilly cobalt]

how would we rewrite the sum? i understand the idea though

[pine wave]

idk yet

[chilly cobalt]

hm

i dont think im well prepped in vector stuff for your approach yet 😭

ill try to think of my weight thingy

thought of an aproach

pick $k$ such that [ \sum_\pi k_\pi x_{\pi(i)} = \begin{cases} 1 \textnormal{ if } i = 1 \ 0 \textnormal{ otherwise} \ \end{cases}. ]

[grand pond]

Copter

[chilly cobalt]

then the sum earlier simplifies

[\left(\sum_\pi |k_\pi| \right) \max \left| \sum_{i = 1}^n x_{\pi(i)} \vec{\alpha_i} \right| \ge \vec{\alpha_1} , ]

[grand pond]

Copter

[chilly cobalt]

can we assume some constraint for the vector on the right?

wlog

<@&286206848099549185>

[midnight plank]

@chilly cobalt Has your question been resolved?

[chilly cobalt]

sob

[midnight plank]

@chilly cobalt Has your question been resolved?

[chilly cobalt]

.close

[midnight plank]

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[midnight plank]

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[snow dawn]

Two ships approach a lighthouse from opposite directions. They are 100m apart from each other and the angle of depression of the lighthouse to the two ships is 30° and 45° respectively. Find the height of the lighthouse

[pine wave]

!status

[midnight plank]

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

[snow dawn]

I will draw the diagram, hold up

Distributing 100m to ships A and B we have x + (100 - x)
Then what do we use from SohCahToa and why?

Oh I see

.close

[midnight plank]

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[chilly cobalt]

Let $n$ be a fixed integer with $n \ge 2$. We say that two polynomials $P$ and $Q$ with real coefficients are $block-similar$ if for each $i \in {1,2,...,n}$ the sequences $$\P(2015i),P(2015i-1),...,P(2015i-2014)$$ and $$\Q(2015i),Q(2015i-1),...Q(2015i-2014)\$$ are permutations of eachother. $\$ (a) Prove that there exist distinct block-similar polynomials of degree $n+1 \$ (b) Prove that there do not exist distinct block-similar polynomials of degree $n$

[grand pond]

Copter

[chilly cobalt]

i was given a hint to prove that P+Q is a constant polynomial

so let R(x) = P(x) + Q(x), S(x) = P(x) - Q(x)

ive gotten that S must have 1 root in the interval [2015i, 2015i-2014] for every i in {1,2,..,n}

ive tried R(x) + S(x) = 2P(x) but it doesnt seem to give anything

why would R be constant in the first place ;-;

<@&286206848099549185>

not a single human being in here 🥀

[icy herald]

i can help u

[chilly cobalt]

please

[icy herald]

ok so what is here...

first READ IT ALL 3 TIMES!

ju to make sure

just to make sure

ok?

bru

fine fine bye!

.close

.close

[chilly cobalt]

wat,

sigh

im being horsed around

[midnight plank]

@chilly cobalt Has your question been resolved?

[midnight plank]

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

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[chilly cobalt]

.reopen

[midnight plank]

✅

[lusty python]

wh do you need to reopen?

[chilly cobalt]

nobody helped me yet

[lusty python]

wait

how do you define 2 functions are permutations of each other

[chilly cobalt]

they defined the sequences as permutations, not the functions itself

[midnight plank]

@chilly cobalt Has your question been resolved?

[midnight plank]

@chilly cobalt Has your question been resolved?

[chilly cobalt]

sigh

.close

[midnight plank]

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[midnight plank]

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[keen harbor]

In the following description, shouldn't the underlined e_1 instead be e'_1?

[fathom onyx]

No? You're demonstrating how the old basis vectors can be written in terms of the new basis vectors

[keen harbor]

thanks

.close

[midnight plank]

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[midnight plank]

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[molten bay]

Z2×z4 is cyclic?

[molten bay]

Any heck for solving or should I make all elements?

z2={0,1}

z4={0,1,2,3}

Z2×Z4={(0,0)(01)(02)(03)(10)(11)(12)(13)}

[lyric charm]

Z_2 × Z_4 ?

[eternal pawn]

A cyclic group is a finite monogenous group right

[lyric charm]

if it's cyclic, an element of order 8 has to exist. does it?

[queen spade]

i really need help guys, im having a test tmr
supposedly speaking if i had two planes that were perpendiculared on the same line, would the distance between the two opposite apexes to the two opposite planes equal each other
english is not my first language so this could be a little hard to understand
hopefully you guys can help me

[oak nymph]

!occupied

[midnight plank]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[eternal pawn]

Type here #help-14

[molten bay]

Of course

[molten bay]

Is this to me?

[lyric charm]

yes it's to you

[molten bay]

Why it has to 8 order element?

[oak nymph]

Not necessarily. Z is not finite but cyclic

[molten bay]

z_2 is cyclic

Z4 too

[oak nymph]

Cyclic just means generated by 1 element

[lyric charm]

$|\bZ_2 \times \bZ_4|$ is obivously 8

[grand pond]

Ann

[lyric charm]

you're asking if this group is a cyclic group of order 8 or not

in order to be a cyclic group of order 8 it has to have a generator, which is an element of order 8

[eternal pawn]

It depends on where you live I guess

[molten bay]

Phi(8)=4

[lyric charm]

irrelevant

[molten bay]

I needed to check all the elements order

(0,0) order 1

(01)-->order 8

[lyric charm]

(0,1) does not have order 8

it has order 4

[molten bay]

(01)+(01)+....

[lyric charm]

(0,1)
(0,2)
(0,3)
(0,0)

it has order 4 not 8

[eternal pawn]

French convention is for cyclic groups to be finite

[molten bay]

Why did you divide when it is (0,4)

because it is z_4

I thought we do z_8😭

[lyric charm]

Z_2 × Z_4 ≠ Z_8

[molten bay]

Where is your hammer ma'am

[lyric charm]

what hammer

[molten bay]

Which can fix my brain

[lyric charm]

what did i tell you about implying that i beat my students?

[molten bay]

Hang on!! So tell me should I check all elements order?

[lyric charm]

no, you don't need to.

[molten bay]

any heck?

Hacking*

[lyric charm]

in a product group there is a trick for element order

[molten bay]

Yes. Please tell me the spicy recipe

[lyric charm]

given $g \in G$ and $h \in H$,

the order of $(g,h)$ in $G \times H$ will be $\text{lcm}(ord(g), ord(h))$

[molten bay]

gcd (m,n) 1?

[grand pond]

Ann

[lyric charm]

in Z_2 × Z_4,
the left coordinates only ever have order 1 or 2,
and the right coordinates only ever have order 1, 2 or 4

[molten bay]

LCM😔

[lyric charm]

there is no way to take lcm of these and get 8

so element of order 8 does not exist

[molten bay]

LHS-->(1,2)

RHS-->(1,2,4)

[lyric charm]

general: $\bZ_m \times \bZ_n \cong \bZ_{mn}$ \textbf{if and only if} $\gcd(m,n)=1$.

[grand pond]

Ann

[molten bay]

Great!!

[oak nymph]

Aka the chinese remainder theorem

[lyric charm]

weaker form thereof

[molten bay]

Huh??

[lyric charm]

we are talking only about the additive groups here

[oak nymph]

Ofcourse, there is a generalisation to rings

[lyric charm]

not the full CRT

[oak nymph]

Whats the matter?

[molten bay]

I solve CRT with wrong method

I forget the formula after 5 days

I mean that I first to solve two equations then I put it into third one

[oak nymph]

You solve CRT? sorry, i dont understand. Are you talking about solving CRT problenms?

Yes, there is a system of equationas statement of the theorem as well, which im assuming you are referring to

[molten bay]

Thanks

.close

[midnight plank]

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[oak nymph]

They are equivalent, you can try proving

[midnight plank]

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[wanton kraken]

Hello ,I'm stuck on this task, I don't know how to draw the picture correctly and how to continue the task, I don't know what to include in the integral, which substitute to take, if anyone can help, I'd be very grateful

[runic berry]

Hm... I'm going to make some observations and suggestions. And I might just leave it there.

C1 is the circle of the cylinder at z=3

c2 is the diagonal from a point on that ring to the center of the cylinder at z=5.
Its like if this is a cone with radius 4, base at z=3, and height of 2, c2 would be a segment at the outside of that cone.
That might make things easier to draw. I don't know.

I would split out the integral into 3 separate parts and think about each one changes as you transition from c1 to c2

[wanton kraken]

Thank you

[steel crest]

i would also convert to cylindrical

[runic berry]

Sorry, i was confused. Please ignore the word transition

I'd do 6 separate integrals and add them up
z^dx along C1
z^dx along C2
ydy along C1
ydy along C2
zydz along C1
zydz along C2

@steel crest convert to cylindrical coordinates and keep the dx and dy terms together?

feel free to correct/revise

[steel crest]

hmm actually it may not be necessary

like, it will work, but not sure how much easier it will make it

[midnight plank]

@wanton kraken Has your question been resolved?

[wanton kraken]

Is image alright,and please if you have time can you explain how can i make integral from this

[runic berry]

@wanton kraken Do you know how to parameterize around a circle?

[wanton kraken]

With rcos and rsin?

[runic berry]

yeah

[wanton kraken]

Okay so i maybe make projection of c1 on xoy plane ?

[runic berry]

yeah definitely treat it like a 2d circle

if you're used to integrating over circles in 2d

[wanton kraken]

Okay man thank you

Do you just have some hint for second area

C2

[runic berry]

One more thing on C1: I didn't notice that C1 is only from (0,.4,3 to -4, 0,3), meaning its only 1/4th or 3/4ths the way around the circle.
Maybe draw both curves in their own images? And just treat them as completely separate problems?

[midnight plank]

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[wanton kraken]

.reopen

[midnight plank]

✅

[runic berry]

C2 is just a linear curve so you can parameterize with
f(t) = t(-4, 0, 3) + (1-t)(0,1,5)

[wanton kraken]

Okay, thank you for your time

.close

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[dawn pine]

i dont get how the ratio becomes 1:4 in the solution

[marble epoch]

The side length of the smaller equilateral triangles is half of the side length of the bigger one. Area scales with the square of length, and 2²=4, so the area of the smaller triangles is a quarter of the area of the bigger one

Does the idea that area scales with the square of length make sense, or would you like me to explain that further?

[dawn pine]

i think i get it, so is it like that for all 2d objects?

[marble epoch]

Yes

Area always scales with the square of length, and volume always scales with the cube of length

[dawn pine]

ok i see, that makes sense

[marble epoch]

Great, glad to have helped

[dawn pine]

thanks for ur explanation!

.close

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[magic hinge]

can anyone help how do i find the constraint

[obsidian merlin]

b?

[pearl hull]

,rccw

[magic hinge]

thought so but it isn't b

[midnight plank]

@magic hinge Has your question been resolved?

[modest oriole]

why dont you ask in physics server

[midnight plank]

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[molten bay]

So here my doubt is when we find order two element in k4

We put 3 but can we apply phi thing in all group of z_+

[twilit jetty]

the 3 I think is from "members of K4 which are order 2"

since K4 is not like Zn, you dont need to use phi to calculate that 3 of its members are order 2 (and the rest arent)

for Z3, you also could see that two of its members are order 3 (1 and 2) but that could be done with phi(3)

[midnight plank]

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[molten bay]

.close

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[obsidian cliff]

context is judsons abstract algebra theory and applications

abt group theory

[obsidian cliff]

is the question asking like "prove an additive identity exists"

interesting indeed

what does this mean

[carmine sigil]

Well, mod n means that we are dealing with integers of the form an + b, where we only care about the b, and b is smaller than n but larger than 0

So we can rewrite this in that form using the definition, and then demonstrate that the identities hold under our normal notions of addition under the integers

[obsidian cliff]

thats it??

[carmine sigil]

So 0 + a => (bn + 0) + (cn + a) for some b and c ∈ ℤ

And you just need to show that 0 + a when expanded and rearranged gives an expression that you can use the definition of modulo the other way that gives a + 0

And then show either of these is a

[obsidian cliff]

huh

ok

i mean i guess that makes sense

[obsidian cliff]

sorry its just like it looks and seems obvious so idk what im allowed to assume

[carmine sigil]

You're allowed to assume integers work the way we know integers work. But integers mod n are a slightly different beast, so we need to show they work as we expect them to.

And we can do that easily using our regular integers.

[obsidian cliff]

right

thats fair

ok

[obsidian cliff]

then its very doable

thanks

.close

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[lethal steeple]

How to prove that the series 1+1/2+1/3+....+1/n is divergent? I don't want to use integration

[lavish venture]

cauchy condensation

[twilit field]

look at a smaller divergent series whose. divergence you can ascertain

[lavish venture]

1 + (1/2 + 1/3) + (1/4 + 1/5 + 1/6 + 1/7) + …

[runic hamlet]

consider the sum 1+1/2+1/4+1/4+1/8+1/8+1/8+1/8+1/16+1/16+...

[fresh sparrow]

also the standard argument of comparing it to 1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 +

[lethal steeple]

I don't get this

[lavish venture]

^

[runic hamlet]

compare it term by term to your sum

[lethal steeple]

I still don't understand?

[twilit field]

this is another way

you can prove this via induction

[autumn canopy]

Is Denascite's sum smaller or larger?

[lethal steeple]

Smaller?

[runic hamlet]

yes

why

[lethal steeple]

Cuz 1/4 is less than 1/3

[runic hamlet]

so your sum is bigger than 1+1/2+2/4+4/8+8/16+16/32+....

what is that value

[lethal steeple]

This will be infinity so my sums value would be infite too

Am I understanding it right?

[runic hamlet]

yes

[lethal steeple]

Thanks!!

.close

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[keen python]

frick chatgpt

[keen python]

uploadign the question wait

[sudden yacht]

!nogpt

[midnight plank]

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).

[keen python]

18bii

okay so compound founction of gf(x) is 1/(2x**2-x)

i found the domain to be

x cant be 0

and x cant be 1/2

[sudden yacht]

Awesome, correct

[keen python]

but for the range i need the maximum point

for that

i need to turn gf(x) into a perfect square

i can turn the bottom into a perfect square

do i just extract the maximum point from the bottom perfect square?

[sudden yacht]

This or you can use the vertex formula, being x_V = -b/(2a)

[keen python]

I FORGOT THAT EXISTED

wait so i take a and b from the bottom quadtratic and just ignore the 1 numerator?

[sudden yacht]

No, you don't ignore it

[keen python]

as in

like what i meant is

i will inmclude 1 obviously ewhen calculating the value for y

after i got x from x=-b/2a

[sudden yacht]

The fact is that since the function gf is the reciprocal of the function f, the vertex of f (which is the minimum) becomes a point of maximum for the function gf (local/relative max, not absolute though)

[keen python]

oh so the minimum of f becomes the maximum of gf

but incase

i come across a situation

where say gf is not a reciprocal

i use x=-b/2a

like zat

right

[sudden yacht]

Then you can't do these considerations

[keen python]

wot

[sudden yacht]

This is a special case, in general it's very difficult to find the range of composite functions, and you need calculus for doing it

[keen python]

im at the start of my book

calculus is at the very end

lol

okay well

thank u boss

the

[sudden yacht]

You're welcome 🤗

[keen python]

can i aks another question

[sudden yacht]

If it is a bit long, it's better to open another channel

[keen python]

i am planning to quit my math coaching
and just use this server if i have any questions
you think this is a good idea? im very average at maths

[sudden yacht]

Otherwise go ahead here

[sudden yacht]

Sure, that's the aim of this server

[keen python]

kinda scared

see i give my igcse's in 2 years

and the meta is

everyone does coaching for almost every subject

chem,physics,maths,bio,english ect.

im not doing any coaching because i feel like

if oyu study by yourself

[sudden yacht]

I think this might better be moved to #discussion

[keen python]

you actually learn the content

okay

.close

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[twin crow]

let the set of all unicorns be an empty set.

now, how can the following statements both be true at the same time and not contradict each other?

1."all unicorns are red"
2."all unicorns are green"

[runic hamlet]

welcome to vacuous truths

frankly just accept them and move on

they are weird

its not like there are counterexamples to those statements

thats how I like to think about them

[twin crow]

yeah, i understand that each statement is vacuously true, but what bothers me is how they can both be true at the same time

?

hmm, that sounds convincing

[twin crow]

alright

lol, i guess so

thanks guys

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[bright summit]

Can I prove that the smallest positive number in the subgroup of (Z,+), Za+Zb (a,b>0) is gcd(a,b) like this?
Proof by contradiction, suppose there's some number n<gcd(a,b) that is the smallest positive integer in Za+Zb
Then n must be gcd(a,b)-pa-qb for some positive integers p and q
but gcd(a,b)<=a and gcd(a,b)<=b
so n=gcd(a,b)-pa-qb <=0 for all positive integers p and q
contradiction since n>0

[visual tiger]

Then n must be gcd(a,b)-pa-qb for some positive integers p and q
No

[bright summit]

why not

[visual tiger]

because it could be +pa - qb, or -pa +qb for all you know

[bright summit]

but n needs to be smaller than gcd? and both a, b, p and q are positive?

[visual tiger]

so?

you don't need BOTH coefficients a and b to be positive

you just need one of pa or qb to be bigger than the other, and that one is negative

[bright summit]

hm yeah true

[visual tiger]

that's not how you're gonna do it

[bright summit]

okay i'll think about it

[visual tiger]

did you prove gcd(a,b) is in Za+Zb?

[bright summit]

i can do that with bezout's identity

?

[visual tiger]

yes

so why not just prove that gcd divides all numbers in that set

that's my hint to you

[bright summit]

ohh okay

gcd(a, b) is in Za+Zb by Bezout’s identity which states there are integers p and q st pa+qb=gcd(a, b), and pa+qb is in Za+Zb by defn
since gcd(a, b)|a and gcd(a, b)|b, then gcd(a, b)|ma+na for all integers n and m
And hence gcd(a, b) is the smallest positive integer

[midnight plank]

@bright summit Has your question been resolved?

[midnight plank]

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[viral dagger]

i labeled the children as like a_1, a_2, ..., a_6, assuming a_1 starts with the candy

for a_odd to change, then either they give out candies, or they each recieve candies

second thing means a_1-a_3 is constant, then mod 4 (since they get rid/change 4 at a time) gets that a_1-a_3 mod 4 is invariant

im p sure

if n=u mod 4, then a_1-a_3=u mod 4

if all of them have the same candy then a_1-a_3=0 mod 4

since its invariant then u=0

if m is a solution then mk is also a solution cause you can just repeat the moves k times

trying numbers out i got that n=28 works which implies 28k works

any hints to continue this?

im like 99% sure that 28 is the smallest solution

i was hoping you can somehow reduce n mod 28

[merry folio]

1+1

[midnight plank]

@viral dagger Has your question been resolved?

[oblique ridge]

It seems to me like it's not too hard to think of a strategy that works for 28k, so what's left is showing that no other value is possible

One way of doing that might involve showing that the number is both divisible by 7 and 4. You can show 7 by counting the number of even vs odd moves, and how many sweets are in those spaces

n-4e+3o=-4o+3e
n=7e-7o=7(e-o)

[viral dagger]

how did you geet this?

[oblique ridge]

o is the number of moves on odd seats

e is the number of moves on even seats

moves take 4 away from their side and add 3 to the other side

so even seats start with n and then lose 4e and gain 3o

and odd seats start with nothing and then lose 4o and gain 3e

[viral dagger]

ohh

what about -4o+3e?

[oblique ridge]

that's the number of sweats remaining on odd seats

and I'm setting that equal to the even seats and then solving for n

[viral dagger]

wait i think im getting this

alright, thank you!

.solved

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[molten bay]

Number of subgroups in z_p^n

[molten bay]

I know that tau function which tells about subgroups

z_p has 2 subgroups

p^n will have n+1?

[lyric charm]

do you mean $\bZ_{p^n}$ or $\bZ_p^n$?

[grand pond]

Ann

[midnight plank]

@molten bay Has your question been resolved?

[molten bay]

I don't know the right one

I was asking about left one

I guess right one is like z_p×z_p×.....n times

[lyric charm]

yes the right one is the product of n copies of Z_p

but ok

yes you'll have n+1 subgroups then

[molten bay]

Hmm

Thanks ma'am

Btw which book did you read for group theory?

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[umbral viper]

To me it looks like they treat the modulus as brackets, why are they allowed to do that?

[umbral viper]

i dont quite understand how they get there

[sudden yacht]

Because of this

[umbral viper]

Okay so that part applies, but why? How does it work?

[lethal path]

cause a - x^2 will be nonnegative there

[sudden yacht]

By definition of absolute value

[lethal path]

for |a - x^2| = a - x^2, you need a - x^2 >= 0

[umbral viper]

So they just fall away and i can then use it like a normal brackets?

Because the value HAS to be positive?

[lethal path]

for example also, |-5| = -(-5)

so that's the definition of |x| when x < 0
-x

but then when x > 0, |x| = x

[lethal path]

if you set that condition, sure

[umbral viper]

Just trying to make sure i can apply it myself

So in this case because a-sqrt(a)^2 always has to be a positive number the abs falls away and i can use them as brackets instead?

thus multiplying all by x

[lethal path]

you should check yourself that's what happens when you set a - x^2 >= 0

hint: ||difference of two squares||

a-sqrt(a)^2 is only when x = sqrt(a)

oh and the question states x >= 0 so that's an extra condition

[umbral viper]

Yeah but anything below sqrt(a) is smaller thus still has to be positive right?

[lethal path]

or yeah, the graph or whatever reasoning also works

cool just wanted to make sure you knew the reason why

[umbral viper]

Cool thanks. That makes sense

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[eternal pawn]

given a strictly increasing integer sequence $(a_i)_{1 \leq i \leq n}$ show that

$\prod_{1 \leq i < j \leq n} (X^j - X^i) $ divides $\prod_{1 \leq i < j \leq n} (X^{a_j} - X^{a_i})$ in $Z[X]$

[grand pond]

bloubbloub

[eternal pawn]

if somehow the set of divisors of every $a_j - a_i$ contained the set of divisors of every $j - i$ it would be ok, but I'm not sure this is the right path

[grand pond]

bloubbloub

[midnight plank]

@eternal pawn Has your question been resolved?

[blissful compass]

interesting problem

[midnight plank]

@eternal pawn Has your question been resolved?

[little anvil]

Maybe a Path could be factorising the polynôms in C[X]

Into prime factors

[midnight plank]

@eternal pawn Has your question been resolved?

[eternal pawn]

the "chapter" is cyclotomic polynomials

my idea was that once you factor all the X^something you would be left with a product of (X^(j-i) - 1) on the left side and (X^(a_j - a_i) - 1) on the right

but for all integer $n, m$, $X^n - 1$ divides $X^{nm}-1$ so maybe with some kind of pigeonhole principle I could get away with this

[grand pond]

bloubbloub

[eternal pawn]

Else I would need to use $X^n - 1 = \prod_{d | n} \Phi_d$ where $\Phi_d$ is the d-th cyclotomic polynomial but I'm not sure how to

[grand pond]

bloubbloub

[eternal pawn]

<@&286206848099549185>

[gleaming cloud]

Maybe

you say
$X^{j-i} - 1 = \prod_{d \mid j-i}\Phi_d(X)$
and
$X^{a_j - a_i} - 1 = \prod_{d \mid a_j - a_i}\Phi_d(X)$

[grand pond]

Amiso_

[eternal pawn]

ye ?

but how do we know it will divide?

[gleaming cloud]

First you have
$P(X) = \prod_{1 \leq i < j \leq n} (X^j - X^i)$

[grand pond]

Amiso_

[gleaming cloud]

$Q(X) = \prof_{1 \leq i < j \leq n} (X^{a_j} - X^{a_i})$

[grand pond]

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

[gleaming cloud]

Then you have that
$X^j - X^i = X^i(X^{j-i} - 1)$
$X^{a_j} - X^{a_i} = X^{a_i}(X^{a_j-a_i} - 1)$

[grand pond]

Amiso_

[gleaming cloud]

and we know that $X^i$ divide $X^{a_i}$ therefore we do not need to take those anymore, we then create new polynomial
$P'(X)$ and $Q'(X)$

[grand pond]

Amiso_

[gleaming cloud]

$P'^(X) = \prod_{1 \leq i < j \leq n} (X^{j-i} - 1)$
(same for $Q'(X)$)

[grand pond]

Amiso_

[gleaming cloud]

then you can use the identity here (you will have a double product)

[eternal pawn]

I'm not sure what you mean by that

[gleaming cloud]

you have that $X^{j-i} - 1 = \prod_{d \mid j-i} \Phi_d(X)$

[grand pond]

Amiso_

[gleaming cloud]

then you have $P'(X) = \prod_{1 \leq i < j \leq n} \prod_{d \mid j - i} \Phi_d(X)$

[grand pond]

Amiso_

[eternal pawn]

yes

but the divisors might not be the same as in d | a_i - a_j

no?

[gleaming cloud]

hm

is the question

a divisor d of j -i might nod divide a_j - a_i?

[eternal pawn]

yes

[gleaming cloud]

we don't really need the implication $d \mid (j-i) \implies d \mid (a_j - a_i)$

[grand pond]

Amiso_

[gleaming cloud]

what we need is more that every $\Phi_d(X)$ in $P'(X)$ appears also in $Q'(X)$ (like the pigeon hole)

[grand pond]

Amiso_

[gleaming cloud]

the question is what is enough to show this

[eternal pawn]

yes

[gleaming cloud]

and from here I am not really sure

I can ask chatgpt

So

what is said is:

it is enought to prove

$X^{j-i} - 1 \mid X^{a_j - a_i} - 1 ; ; \in \mathbb{Z}[X]$

[grand pond]

Amiso_

[gleaming cloud]

because then
Every $\Phi_d$ dividing $X^{j-i} - 1$ also divies $X^{a_j - a_i} - 1$

[grand pond]

Amiso_

[gleaming cloud]

and that is true if $j-i \mid a_j - a_i$

[grand pond]

Amiso_

[eternal pawn]

thank you Mr GPT

[gleaming cloud]

and we don't need to prove that every pair is dividing

I'll screenshot

[eternal pawn]

that's what we've been saying though

[gleaming cloud]

we only need to ensure:
for each $d$, the numbner of $(i, j)$ with $d \mid j - i$ is less then or equal to the number of $(i, j) $ with $d \mid a_j - a_i$
And that can be show to hold, because $a_j - a_i \geq j - i$, and the sequence is stricly increasing, the set ${a_j - a_i}$ "spreads out more" than ${j-i}$, So each divisor $d$ appers at least as often among the $a_j - a_i$ as it does among the $j - i$

[grand pond]

Amiso_

[eternal pawn]

welp still nothing though

[gleaming cloud]

yeah

[eternal pawn]

I'm trying on small examples hold on

[gleaming cloud]

yes

[eternal pawn]

something weird

ok so with a = 1, 5, 6, 8, 12 I showed that you can't have only the j-i dividing a_j-a_i

you need to look at the divisors

perhaps it's possible by induction

waaaiiiit

[gleaming cloud]

maybe

[eternal pawn]

actually its not

[gleaming cloud]

hmm

let me try

[eternal pawn]

if $k$ is an integer, the number of $\Phi_d$ contained in $Q$ (or $Q'$) has a lower bound of

$\binom{\lfloor \frac{n}{k} \rfloor}{2} \left( \lfloor \frac{n}{k} \rfloor - 1 \right) + \binom{\lceil \frac{n}{k} \rceil}{2}$

[grand pond]

bloubbloub

[eternal pawn]

and the number of $\Phi_d$ in $P$ is $\sum_{i=1}^n \lfloor \frac{i}{k} \rfloor$

[grand pond]

bloubbloub

[eternal pawn]

so maybe it is enough to bound that sum

by something like $\frac{(\lfloor \frac{n}{k} \rfloor)^2 (\lfloor \frac{n}{k} \rfloor - 1) }{2}$

[gleaming cloud]

what we need to prove is that

[grand pond]

bloubbloub

[gleaming cloud]

$\forall d, \text{multiplicity of } \Phi_d(X) $ in $P(X) \leq $ multiplicity in $Q(X)$?

[grand pond]

Amiso_

[gleaming cloud]

because if this is the case

[eternal pawn]

yes

[gleaming cloud]

we can use the lemma

[eternal pawn]

which one

[gleaming cloud]

Let $a_1 < a_2 < \cdots < a_n$ be a stricly increasing sequence of integers.
Then for any integer $d \geq 1$, the number of pairs $(i, j)$ with $i < j$ and $d \mid (j-i)$ is less than or equal to the number of pairs $(i, j)$ with $d \mid (a_j - a_i$)

[grand pond]

Amiso_

[gleaming cloud]

this leads that for every $d$ the number of $(i, j)$ such that $d \mid j -i $ is $\leq$ the number of $(i, j)$ such that $d \mid a_j - a_i$ which directly implies that

[grand pond]

Amiso_

[gleaming cloud]

$\Phi_d(X)$ appears in $Q(X)$ at least as many times as in $P(X)$

[grand pond]

Amiso_

[eternal pawn]

I don't get why

[gleaming cloud]

from where?

[eternal pawn]

why is it less/equal

like I see that it's true but what's a convincing argument

[gleaming cloud]

for the lemma

we can consider two sets

(here is more chatgpt than me)

we consider two sets

$S = {(i, j) \in [1, n]^2 \mid i < j, d \mid j - i}$

[grand pond]

Amiso_

[gleaming cloud]

$T = {(i, j) \in [1, n]^2 \mid i < j, d \mid a_j -a_ i}$

[grand pond]

Amiso_

[gleaming cloud]

the claim is that $\mid S\mid \leq \mid T \mid$

[grand pond]

Amiso_

[gleaming cloud]

(this is just what we said before but in a proper way)

I got it

[eternal pawn]

yes

[gleaming cloud]

for exemple if the sequence $a_i$ is arithmetic for instance $a_i = ki + c$ then
we have $a_j - a_i = k(j-i)$ which is like a one to one mapping

[grand pond]

Amiso_

[gleaming cloud]

which is a bijection which leads that
$ d \mid j- i \iff d \mid a_j - a_i \implies S = T$

[eternal pawn]

sure but what about all other sequences

[gleaming cloud]

then if it is not arithmetic it is still strictly increasing sequence

we have the the mapping $(i, j) \mapsto a_j - a_i$

[grand pond]

Amiso_

[eternal pawn]

and?

[gleaming cloud]

which is increasing faster than $j-i$

[grand pond]

Amiso_

[gleaming cloud]

I thought I got it but

it just kept in the same place

[eternal pawn]

Mr GPT is really good at that

[gleaming cloud]

lmao

[eternal pawn]

<@&286206848099549185> second try?

[gleaming cloud]

Sorry I am kind of lost here

[eternal pawn]

I mean T_k doesn't even use the k variable

and the argument is wrong

or maybe it's in the following paragraph

[eternal pawn]

may I ask what's the injection?

[eternal pawn]

it's either horribly written or wrong

either way I did a little python script to compare the values for the lower bound I found vs the sum, and incredibly, they match for k = 5, 6 and N< 100

thus I have a strong suspicion that

[fallow scarab]

!nogpt

[midnight plank]

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).

[eternal pawn]

(number of $\Phi_d $ in $P =$ ) $\sum_{i=0}^{n-1} \lfloor \frac{i}{k} \rfloor = \binom{\lfloor \frac{n}{k} \rfloor}{2} \left( k - (n mod k) \right) + \binom{\lceil \frac{n}{k} \rceil}{2} (n mod k)$ ($\leq $number of $\Phi_d$ in $Q$)

[grand pond]

bloubbloub

[eternal pawn]

pretty sure a combinatorial proof would show this

but howww

[eternal pawn]

I did it

suppose $k\geq 2$, there are two cases to consider: $ k | n$ and $k \nmid n$:

if $n = kq$ both sums equal $k\frac{q(q-1)}{2}$

if $n = kq + r$ both sums equal $\frac{kq^2 - kq + 2qr}{2}$

in each case the equality holds. Now the only thing I didn't prove was something like $\binom{a - b}{2} + \binom{b}{2} \leq \binom{a}{2}$ which seems obvious

[grand pond]

bloubbloub

[eternal pawn]

in case you wanted to see the answer (even though I did not explain that much sorry 😅 dm me if you have questions)

.close

[midnight plank]

Channel closed

Closed by @eternal pawn

Use .reopen if this was a mistake.

[pseudo totem]

Hello!! Back again with some more confusion and another question.

I did: 146^2+95^2-2(146)(95) COS(70)

I got 70 from subtracting 20 from 90.

[pseudo totem]

This comes out to 25047.958 which I took the square root to get 158

[eternal pawn]

I got 113

wait no

maybe I made another mistake?

[pseudo totem]

What did u do pi for?

[eternal pawn]

because google is in radians

[pseudo totem]

I'm using desmos which happens to be in degrees?

[visual panther]

You entered 79 degrees, when calculating it ...

[eternal pawn]

I got the same answer on desmos though

[pseudo totem]

Omg ur right

[eternal pawn]

how did you find out 🤣

[visual panther]

30 years of experience as a maths teacher.

[blissful compass]

bunch of nonsense

[pseudo totem]

Well thanks teach, I got 144 which is right and I thank you 🙏🏼

.close

[gleaming cloud]

hm

I think it was already clear that this was wrong before but thanks for making it clearer

[midnight plank]

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[ivory prism]

Ok im very confused on this ive been using ai atp because i seriously dont have time to figure this out rn i need this problem done or im actually gonna fail my class today. Its probability and statistics. I havent been paying attention and now im biting the bullet

[ivory prism]

I would js graph it but i dont know how to because i havent yk been paying attention. Scold me but i seriously js need this done

[midnight plank]

@ivory prism Has your question been resolved?

[hoary pulsar]

What is the residual calculated from?

[blissful sentinel]

He left the server

[hoary pulsar]

Lmao????????????

[midnight plank]

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Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• 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.

[blissful trench]

yo

[mortal mirage]

you have a question?

[pearl hull]

yo

[blissful trench]

ya wait i think my laptop is lagging

[mortal mirage]

mk

[blissful trench]

is this website sucks ass or its just me?

[scenic wyvern]

the range of cos is [-1, 1]

this was never a possible answer

[pearl hull]

How did you get that? 😭

[fathom onyx]

it's you AND the website sucks ass

[scenic wyvern]

how did you get this in the first place

[fathom onyx]

I think he just divided by 3?

[scenic wyvern]

wow fam don't have to do him like that HAHAH

[fathom onyx]

/300?

[mortal mirage]

i challenge you to weave apples into this /jkjk

[blissful trench]

first of all can one person stays here?

[blissful trench]

i did 2954.43/3000 and got 0.98481

[mortal mirage]

yes this is correct

so uh

[fathom onyx]

-# If I place an apple on the slope of this triangle, then the rotation can be mapped onto the cosine by "how upright" it is; we can treat the cosine as a function of this "uprightedness" and...

[astral canyon]

I was about to say "he missed one dp and yall are destroying him for that" but dividing by 300 would give 9.84.., not whatever he got lmao

[mortal mirage]

why didnt you put that?

[blissful trench]

omg bruh

5 helpers at one help channel is crazy

i just need one person

[fathom onyx]

oh, did you then do inv cos? @blissful trench

In degrees?

[mortal mirage]

alr ill leave

[blissful trench]

let me continue

[pearl hull]

@mortal mirage you stay, everyone else leave

[fathom onyx]

https://tenor.com/view/oogway-my-time-has-come-gif-8019684

[pearl hull]

Since you’re the first one responded

[mortal mirage]

tyty

[mortal mirage]

ok, continue

[blissful trench]

I know how it exactly works but i just wanna make sure if this is the correct number

i did cos^-1(0.98481) and got 0.1745

[mortal mirage]

but it was just asking for the cosine

your answer was the inverse cosine of that (aka you found the angle)

[blissful trench]

are u saying i need to write cos 0.98481?

[mortal mirage]

no, your answer is 0.9848

(4 decimal places rounding)

[blissful trench]

that worked thanks and i think I am the suck one

[mortal mirage]

nah nah your ok

that website does suck ass tho

wtf is adjacent(hypotenuse)

[blissful trench]

fair

[mortal mirage]

💀

[blissful trench]

omg LMFAOOO

[mortal mirage]

are they saying to multiply the adjacent by the hypotenuse

[astral canyon]

it identifies as a hypotenuse don't judge

[blissful trench]

.close

[midnight plank]

Channel closed

Closed by @blissful trench

Use .reopen if this was a mistake.

[blissful trench]

.reopen

[midnight plank]

✅

[blissful trench]

I have other one

[pearl hull]

yay

[scenic wyvern]

rounding error

[blissful trench]

2209.87/350?

[scenic wyvern]

why are we doing it that way though

it's opp/adj, not adj/opp

[blissful trench]

ok

,calc 350/2209.87

[grand pond]

Result:

0.15838035721558

[blissful trench]

do i just put 0.1583?

[scenic wyvern]

you need to round, not truncate

the digit after the 3 is an 8

you can't round down here

[blissful trench]

that would give 3 digits

[scenic wyvern]

huh

it's still four

[fathom onyx]

the 8 AFTEr the 3

Means you round the 3 UP

[blissful trench]

I literally put this 4 digits and still gives an error

0.1583 right?

[fathom onyx]

What number DID you put there?

[mortal mirage]

the three becomes a four

[scenic wyvern]

you put 0.1583 or 0.1584?

[scenic wyvern]

no

the number after the 3 is an 8

8 is more than 5

you round up

not down

[blissful trench]

ah rounding

[mortal mirage]

0-4 round down, 5-9 round up

[blissful trench]

i see i see

[fathom onyx]

(talk about getting there in a roundabout way)

[scenic wyvern]

(generally true in math)

[fathom onyx]

(...I'll see the door)

[scenic wyvern]

don't just see it, open it sir, make it swing round

...guess i'm walking out with you

[fathom onyx]

join me in the merry land of shitty maths puns

[mortal mirage]

[scenic wyvern]

so OP gets it now?

[blissful trench]

ya but i might ask another one lol

[mortal mirage]

kk

[scenic wyvern]

i'm on standby

[mortal mirage]

were literally here 24/7 lol so

[blissful trench]

u know what I will close this and open a new one

[pearl hull]

Good choice

[blissful trench]

im just refreshing my memory from hs stuff

[scenic wyvern]

aight

[blissful trench]

.close

[midnight plank]

Channel closed

Closed by @blissful trench

Use .reopen if this was a mistake.

[midnight plank]

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.

[kindred wraith]

Question 14

I skipped some steps in between

[vivid yoke]

How did you get -16(y^2-1)^2 in the third line?

[kindred wraith]

I assumed the function to be equal to some constvalue say y

Oh wait

[vivid yoke]

Yes b^2-4ac

[kindred wraith]

Nvm i read wrong

[vivid yoke]

c=4(y+1) no?

[kindred wraith]

Took 4 common from "c"

[kindred wraith]

a is y-1

(Y+1)(y-1)=ysq -1

[vivid yoke]

Oh yeah i misread ito

[kindred wraith]

Then into 16 for the 4x4

[vivid yoke]

I thought you wrote (y^2-1)^2

[kindred wraith]

It all good

[kindred wraith]

I think it might be from the x-1 i did on 5th step

But that shouldn't affect the roots tho

So idk how they getting -ve sum

Thier final answer is b
-6

[vivid yoke]

,w 7x^2 +18x -25 =< 0

let see

[kindred wraith]

Oh shit

The 2nd factor would be

[vivid yoke]

7x^2 +18x -25 = (y-1)(7y+25)

[kindred wraith]

7y+25

[vivid yoke]

yup

[kindred wraith]

Man

Ight thx

[vivid yoke]

lol I'm not gonna do the problem like this

[shadow schooner]

it works though

[kindred wraith]

?

[kindred wraith]

Yea that's the graph of it

I made mistake in 2nd factor

[vivid yoke]

$\frac{x^2-3x-4}{x^2-3x+4}=1+\frac{-8}{x^2-3x+4}$

[kindred wraith]

Wait -3-2-1+0+1=-6?

[vivid yoke]

no lol, the answer isn't right ig

[kindred wraith]

That pretty rare for this book

[shadow schooner]

is 1 in the range of f?

[kindred wraith]

Yes

Because it's less than equals 0

[vivid yoke]

no it's not lol

Yeah I see

[kindred wraith]

Huh why?

Oh shit

I see

[grand pond]

Alexis_Fx

[kindred wraith]

Denominator becomes 0

Dang alr

Thx ima go do sum more questions

[ancient ridge]

If you have access to calculus, I would just differentiate instead of doing all this fiddling

[vivid yoke]

$\frac{-8}{x^2-3x+4} \in [\frac{32}{7},0)$

[grand pond]

Alexis_Fx

[kindred wraith]

We haven't learnt calc yet

We r on algebra;/ so I cant do that

[vivid yoke]

$\Rightarrow \frac{x^2-3x-4}{x^2-3x+4} \in [\frac{32}{7}+1,0+1)$

[grand pond]

Alexis_Fx

[vivid yoke]

you should be able to do this, no?

[kindred wraith]

Yea

But i don't really like taking the 1 out and making linear

So I just solve quadratic by quadratic

[vivid yoke]

it's okay ig, using limit here it's more efficiency than quadratic

[kindred wraith]

Ight well thx im go now

[kindred wraith]

Tbh i don't get the though of taking out 1 during exams

So I try to avoid it and just solve nornally

Thx again

.close

[midnight plank]

Channel closed

Closed by @kindred wraith

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[midnight plank]

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.

[frail coral]

can someone please solve this

[mystic condor]

Divide it into triangles

[frail coral]

i tried to do it

it looks easy but i cant solve it🙁

[full leaf]

is it related to trigonometry

[mystic condor]

Theres no angles so definitely no

[frail coral]

it might be cos of corresponding angles

[mystic condor]

With things like this i usually turn it into line functions so coordinate geometry would suffice

[frail coral]

maybe a third parallel line in between the triangles

[pearl hull]

Simple and straightforward 👍

[mystic condor]

Why is ur name the scholars mate lol

[pearl hull]

cuz I’m simple and straightforward