help-23

[stoic dune]

Oh that's a necklace group

Uh, 1 sec let me look into that. I don't deal with them often

[cobalt thunder]

My first idea would be to bruteforce it with the inclusion-exclusion principle. The shape has a total of 3 symmetries: flipping accross horizontal axis, flipping accross vertical axis, and 180 rotation. The number of combinations would then be

[total number of combinations] - [number of combinations preserved by at least 1 symmetry] + [number of combinations preserved by at least 2 symmetries] - [number of combinations preserved by all 3 symmetries]

[stoic dune]

Looking at https://en.wikipedia.org/wiki/Necklace_(combinatorics)

[pastel vault]

so theres a problem with that in that its a shape without rotational symmetry if thats the right term, it can only be rotated 180 degrees not 90

[cobalt thunder]

Oops that wrong, this gives the number of combinations not preserved by any symmetries

[pastel vault]

ima be honest i am not able to process this sentence at all at my current energy level T_T

[stoic dune]

Ah I see. Tricky. Let me think

It's still a group, so those methods still work. Could look into Burnside. But that's complicated

[cobalt thunder]

I'm thinking about some simple application of the orbit-stabilizer theorem

[pastel vault]

this has very quickly left my league 😭

[cobalt thunder]

Oh I think I was on the right track

So I'm just calling a combination which doesn't change when a symmetry (flipping horizontally/vertically / rotating 180) is applied to it "preserved" by that symmetry

[cobalt thunder]

What I calculated here is the amount of combinations not preserved by any symmetry; this means that when you apply a symmetry to a combination counted in here, you get a different combination. By the exercise's requirements, such a combination would be counted 4 times: 3 times for each symmetry, and 1 time for when no symmetry is applied

[solar hazel]

burnside's lemma is really helpful for counting stuff like this

you don't really need to know any group theory to use it

[cobalt thunder]

Right, I forgot the statement of Burnside's lemma, it seems what I was doing is exactly Burnside's lemma

[solar hazel]

just consider necklaces without rotations and reflections and allat being equivalent to start. so there are 2^6 'necklaces' here. call the set of these S. then you have something called an action that takes the objects in S and performs literal physical processes on them, like rotations or reflections. some objects will be equivalent under these actions. meaning that you can obtain one from the other by applying a process to them. each class of equivalent objects is called an orbit. and then what you want you want to count is the number of orbits, since these account for whatever equivalence you're trying to capture

burnside's lemma gives you a formula for the number of orbits

in terms of the number of fixed points for each physical process. a fixed point is a necklace such that when you do the process, the necklace doesn't change

[pastel vault]

im trying to read the wikipedia page for burnside's lemma but i dont understand what half of the symbols mean so im very confused

[solar hazel]

like if you have the necklace
r
b b
r
if you rotate twice, you get the same necklace. so this is a fixed point under the 'rotate by 2' process

it is also a fixed point under the rotate by 4 process

and the do nothing process

necklace
r
b b
b
is not fixed by 'rotate by 2', because rotating twice gives
b
b b
r

[pastel vault]

i understand it conceptually i just dont understand how that translates into a formula

[solar hazel]

ok can you post the formula you're looking at and i can translate it to plain english

[stoic dune]

I'm cool with actually tackling the original problem. Burnside seems to be a good way to go about it. First, we want to identify a group that acts on your points.

I think you are only allowing horizontal flips, vertical flips, and 180 degree rotations. If so, that's the dihedral group with 4 elements.

[pastel vault]

i mean im looking at the whole wikipedia page but i dont understand what's relevant and whatisnt

because i dont know what anything means

[solar hazel]

just this is fine then

[pastel vault]

that is correct i would like to understand how to apply this more broadly though

[stoic dune]

Burnside can be applied very generally haha. You'd want to learn some group theory to fully use it

This question is a bit different from a lot of combinatorics, which is maybe unlucky if you are just learning combinatorics

On the other hand, if you are interested in some really cool math, go grab a group theory book! A seriously cool study.

[solar hazel]

X is your set of necklaces without considering any equivalent under actions (so if you have m beads and n colors then there are n^m necklaces in this set). G is the 'set' of physical processes that you are going to do the necklaces. like rotations and reflections. (so they would look like do nothing, rotate 180, reflect, reflect then rotate 180 here i think). |G| is the number of these processes. so 1/|G| = 1/4. for each process g in G, |X^g| is the number of necklaces that are fixed by g. so the ones that g does not change the appearance of, as in the examples i posted earlier. the sum over g in G part just means you do this for each g and add the number of fixed points up

[stoic dune]

Burnside can count how many objects there are, where two objects are the same when symmetric.
Where "symmetric" is given by a group, which is a "description of symmetry"

In your case, we want to know how many sets of points there are, where two sets of points are the same, when symmetric
Where "symmetric" is given by 180 degree rotations, horizontal/vertical flips.

[solar hazel]

wait is it any rotation or just 180 degree rotations? i can't tell from the old messages

[pastel vault]

for my shape only 180 degrees

[solar hazel]

the processes would look a little different here if 180 degree rotations only, but everything else is still fine

in that case there would just be 4 elements in G, no matter how many beads and colors

[cobalt thunder]

any transformation which preserves this shape, so 2 axial symmetries and 180 rotation

[solar hazel]

do nothing, rotate 180, reflect, reflect then rotate 180

actually i'll just edit that in

[pastel vault]

wait so is it (X^g) repeated G times?

[cobalt thunder]

It is the set of elements in X which are fixed by the symmetry g

[solar hazel]

you count the number of fixed points for each of these 4 processes, and add them up

that's what the sum in the formula is

[pastel vault]

let me try to find a visual example

[solar hazel]

in your example here there are 2^6 ways to color the points. let's look at fixed points under the 4 processes now. the do nothing process fixes every point. so there are 2^6 fixed points there. the rotate 180 process fixes some things, e.g.
b b
r r
b b
but it doesn't fix e.g.
b b
r b
b b
because rotating 180 gives us something else:
b b
b r
b b

you want to count how many fixed points there are for each process and add everything up. so it will look like 2^6 + ... + ... + ...

and dividing allat by 4 will give you the answer

[pastel vault]

ok a visual example made it make sense thank you

[safe radish]

@pastel vault Has your question been resolved?

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[safe radish]

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[ivory schooner]

how can i prove that FEC ~ BFA

[ivory schooner]

i know they are similar i just cant prove it

https://tenor.com/view/watching-you-observe-seeing-i-saw-you-gif-5740825

[lone void]

you know both of those triangles have right angles from the figure

[honest idol]

You mean DEC?

[ivory schooner]

yeah

[ivory schooner]

yea

and ab = cd

but i need 1 more angle for asa

[honest idol]

My take is, you need to examine the stipulations that yield 2 equal triangles

What are the conditions that need to be met in order for 2 triangles to be the same

[ivory schooner]

Wdym?

[honest idol]

.

[ivory schooner]

side angle side i think

or like a congruence theorum

sas, asa, hl

[honest idol]

Yes sure

[ivory schooner]

ohh i see

got it tyy

[honest idol]

You’re welcome man

[ivory schooner]

.close

[safe radish]

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[safe radish]

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[twilit cargo]

Integral of sec^5x dx

[twilit cargo]

my question is should i open the bracket up and distribute first or should i use integral = I

this is where i tend to get confused about using I

[visual linden]

sorry but how did you go from sec^5x to sec^3x?

[twilit cargo]

reduction formula

[visual linden]

I mean

[visual linden]

yeah checks out

[twilit cargo]

wait so its correct?

[visual linden]

I'll check it again

[twilit cargo]

wait im also double checking i might be

i accidently put in sin^5x in integral calc to double check

[visual linden]

yeah looks fine

Honestly a little bit different from how I do IBP

[twilit cargo]

i think using by parts in the second integral made it messy,

oh also i forgot to add teh final answer i got one sec

[visual linden]

yeah a bit cramped up

well, let's just do the honours then

,integrate sec^(5)x

damn

,w integrate sec^(5)x

[flat frigate]

Failed to get a response from Wolfram Alpha.
If the problem persists, please contact support.

Feb 22, 2026, 3:13 AM

[visual linden]

,calc integrate sec^(5)x

[flat frigate]

The following error occured while calculating:
Error: Undefined symbol integrate

[visual linden]

Alright I give up, manually it looks fine, let's just check it on integral calculator to confirm

[visual linden]

where's the 8 coming from, yeah I think there might be some calculation mistake we missed

[twilit cargo]

i think it fell apart when i was limited by space on the bottom and tried to force it

imma start again from the middle piece

[visual linden]

yeah maybe we had to multiply by 3/4

[twilit cargo]

up till was it good?

[twilit cargo]

^ full img

[visual linden]

yeah I'd say looks fine

oh wait

dude

you found I

wait wait, you found I, that is the integral of sec^5x

you had to find 3I/4

[twilit cargo]

yeah thats the part confusing my

how do i make it equal to the integral?

do i take it as 3/4 I = ____what?

or another way?

[visual linden]

okay so here comes the way I do it

now, from a new line, write I = integral of sec^3x

then solve for it

and later on substitute it as I back into 3I/4

[twilit cargo]

ok doing it rn

this^

[visual linden]

yes

now substitute this I into 3I/4 you had in the original integral

[twilit cargo]

oHH soo nice

thanks for the clean understanding of how u do it,

really helped me

appreciate that

[visual linden]

no problem man

[twilit cargo]

.close

[safe radish]

Channel closed

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[safe radish]

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[safe radish]

Channel closed

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[fair holly]

how to factorise it

i dont understnad after finding the like term

[open wedge]

Factorize what?

[fair holly]

after y derivative

[open wedge]

4(x^2 + 2)?

[fair holly]

yes

i dont understand how to factorise

[open wedge]

You can't factorize that any further

[fair holly]

no cuz there are like terms

what do i do after

[open wedge]

You can't factorize this any further

[distant wolf]

Damn

[fair holly]

no i know that

bro look at the working out

you can write it simpler thats what im asking how to do

[open wedge]

Just expand each parentheses out

[fair holly]

bro

what do i do after this step

is what im asking

[open wedge]

.

[fair holly]

bro are u listening

[open wedge]

Just expand the PARENTHESES out

🙏

[fair holly]

ok how

[open wedge]

Like for 4(x^2 + 2)(x^2 + 2)

Expand the (x^2 + 2)(x^2 + 2) first using the distributive property

[distant wolf]

Bro use the product rule

[fair holly]

omg

i already did

this is after

buddy

[distant wolf]

Aha

[open wedge]

I think he used that

[distant wolf]

Then

[open wedge]

Ts what's y' is

[fair holly]

from that

how do you get to this

[safe radish]

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[open wedge]

!nosols

[open wedge]

.close

[safe radish]

Channel closed

Closed by @open wedge

Use .reopen if this was a mistake.

[distant wolf]

What?

Is his doubt cleared

[visual linden]

well the channel was closed to being with, so it just became unavailable after one point

[open wedge]

Don't post solutions so he can copy

[distant wolf]

I see

He isn't understanding

Iam sorry

[open wedge]

No worries

[open wedge]

Damn the factoid doesn't work when i open the channel 🥀

Someone put an else if statement somewhere in their code...

[safe radish]

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[small rampart]

What is the name of this theorem and what does it say?

[small rampart]

I translated it from my language, it was listed there as Cantor's Principle but I couldn't find this by searching that

[delicate shore]

cantor intersection theorem

[small rampart]

Also I don't understand what 2 says

[delicate shore]

sometimes known as "nested intervals property"

or a variety of other names

[delicate shore]

it says that the intervals are descending, ie they are nested

[small rampart]

Oh those are sub-intervals?

[delicate shore]

mhm

[small rampart]

So like [a_n,b_n] ⊂...⊂ [a_2,b_2]⊂[a_1,b_1]

[empty horizon]

yo can you help me i have my maths exam on 25 any tips?

[delicate shore]

yes

[delicate shore]

Wrong channel, see #❓how-to-get-help

[empty horizon]

thanks mate

[small rampart]

The version on wikipedia only said that the intersection is not empty, but doesn't mention the unique element

[cloud hound]

I'm surprised in our book they call it the nested interval theorem when it has an actual name

[small rampart]

I'm Vietnamese and our books tend to mention the author of theorems

Like AM-GM is Cauchy's inequality

And Cauchy-Schwarz is Cauchy-Bunyakovsky-Schwarz because he also independently discovered it

[flat frigate]

Nerdyasianguy

[cloud hound]

ik in europe they like to censor names of non european discoverers but I thouhgt cantor was european

[small rampart]

And you'd be correct, he's German

[cloud hound]

how odd

[delicate shore]

the condition that diam C_n -> 0 gives the result

you basically get it for free in metric spaces

[small rampart]

What's diam?

[cloud hound]

a is unique as stated in the theorem

looking it up in google it looks like cantor and nested interval are different

cantor is more general

nested inverval is for R

the Nested Interval Theorem applies specifically to closed, bounded intervals in R, while Cantor’s Theorem is a generalization applying to compact sets within metric or topological spaces.

[cloud hound]

diameter I assume

[small rampart]

I didn't know sequences have diameters

[cloud hound]

sets in metric spaces, I believe

[delicate shore]

it's the diamater of the interval

diameter

ie b_n - a_n

[small rampart]

Oh this is true because it's \subset and not \subseteq

Right?

So the intervals get smaller an infinite amount of times, so the diameter approaches 0

[delicate shore]

ok honestly I don't know why they include 3 as part of the theorem statement

the key part is that it is nonempty

if you assume 3 then obviously it can only be a singleton, so that is a trivial corollary of the "main result"

[small rampart]

Does 3 make them bounded?

The 2 sequences i meant

[delicate shore]

not really since all the intervals are contained in [a1, b1]

and a_n is increasing, b_n is decreasing

[small rampart]

Or both are increasing but b_n converges quicker?

[cloud hound]

3 is common for nested interval theorem, we have that both sequences converge but they may converge to not the same number

$a_n = - \frac{1}{n} $ and $b_n = 1 + \frac{1}{n}$ I think satisfy conditions 1 and 2

[flat frigate]

Coolempire93

[delicate shore]

yes of course I just find it somewhat distracting because it's not the meat of the proof

[small rampart]

Is this statement "stronger"?
(Like is it more general?)

[cloud hound]

In our book they take about equal sections (with monotone convergence), so that's interesting maybe because we are missing some results that make it trivial

[small rampart]

Also the uniqueness of the element of the intersection also holds here right?

[cloud hound]

this appears to be the same thing as before but without condition 3

[small rampart]

So the conclusion is exactly the same without 3?

[cloud hound]

it yields that the intersection of these is nonzero (we know it is [0,1])

not a unique element because 3 is gone

[delicate shore]

If it's nonempty, then it has to be a singleton because the diameter of the intersection is dominated by the diameters of all the sets in the sequence, which goes to 0

[cloud hound]

oh I see

interesting
and a doubleton is a set with 2

[small rampart]

Singleton means a set with exactly 1 element?

[delicate shore]

that's why i think 3 is somewhat distracting and it suffices to prove the intersection is nonempty while assuming 1 and 2

[small rampart]

Thanks y'all

.close

[safe radish]

Channel closed

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[safe radish]

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[small rampart]

Does lim_{n→∞} (x_n) = l imply that lim_{n→∞} (M_{k,n}) = l where M_{k,n} is the k-power mean of the first n terms of (x_n)?

[flat frigate]

Nerdyasianguy

Does $lim_{n \rightarrow \infty} (x_n) = l$ imply that $lim_{n\rightarrow \infty} (M_{k,n}) = l$ where $M_{k,n}$ is the $k$-power mean of the first $n$ terms of $(x_n)$?

[spring glacier]

can you show the sequence is cesaro summable? if so, it follows by choosing the right continuous functions

[flat frigate]

Nerdyasianguy

[small rampart]

Here's where i'm unsure, for example where k=2 and some a_n are negative then a^k is not defined

[toxic stratus]

the case when k=1, so the usual arithmetic mean, is a commonly stated theorem

the others follow by considering the sequence (x_n^k) and applying the case when k=1

[small rampart]

Yes I am aware, it is the motivation of this question

[small rampart]

I can also work around this with the Power Mean inequality and Squeeze Theorem right?

[toxic stratus]

M_n,k(x_n)^k = M_n,1(x_n^k)

oh you called your original sequence x

[small rampart]

Yes, x_n is a sequence of real numbers

[toxic stratus]

anyway take the limit of the RHS

then ^1/k of that is your desired limit

[small rampart]

Is this missing brackets? I can't read this

[small rampart]

I meant this

[toxic stratus]

uh

it can use brackets for more clarity but im typing on a phone so its a bit awkward

[small rampart]

Same

[toxic stratus]

,, \lim_{n\to\infty} \f {x_1^k + \dots + x_n^k} n = l^k

[toxic stratus]

this follows from the fact that the sequence (x_n^k) has limit l^k

now take ^1/k

[small rampart]

My concern was that ^(1/k) might not always be defined

[toxic stratus]

such as?

[small rampart]

Like you can't take ^(1/2) with negative numbers

[toxic stratus]

well

if you have negative numbers you're kinda screwed from the get go anyway

[small rampart]

Why so?

[toxic stratus]

how do you propose we calculate the quadratic mean if some of the intermediate results are negative

well

your numbers are real so that's actually not a problem i suppose

but how will you calculate some other power mean

like say the πth power mean

what is (-1)^π?

[small rampart]

Using pi as a variable

Oh wait no

It's not a variable

[small rampart]

Uhmm...

I guess you could take the limit of a sequence that converges to pi

[toxic stratus]

(-1)^x is not exactly a continuous function

if you want real outputs, good luck i guess

if you're okay with complex outputs, then you gotta mess with complex logs

[small rampart]

What if I only consider integers k?

[toxic stratus]

then there can be no problem

[small rampart]

With k=2 there's a square root

[toxic stratus]

for odd k, the kth root is always defined

for even k, the "inside" is non-negative regardless

[small rampart]

Oh that's trivial

[toxic stratus]

blud

[small rampart]

If k is a real number and x_n ≥ 0 forall n ∈ ℕ then it still works?

[toxic stratus]

yeah

unless uh

k<0 and x_n = 0

[small rampart]

Nice, thanks it works for x_n>0 then

.close

[safe radish]

Channel closed

Closed by @small rampart

Use .reopen if this was a mistake.

[toxic stratus]

maybe be careful when k=0

the definition is different

[small rampart]

The power mean when k=0 is the geometric mean

[toxic stratus]

yeah

[small rampart]

And that's correct too

[toxic stratus]

but that's a separate result

you need some sign assumptions on x_n i assume

[small rampart]

It's correct for x_n > 0

[small rampart]

Similiar technique to this but with $a_n = log(\prod_{i=1}^n x_i)$

[delicate shore]

sssumptions

[toxic stratus]

🐍

[flat frigate]

Nerdyasianguy

[toxic stratus]

you need x_n > 0 to do the log thing but it should hold true in slightly more generality

anyway

[safe radish]

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[graceful lichen]

.

[nimble wyvern]

what's the problem?

[graceful lichen]

wait i have typed wrong latex

let me fix it

[nimble wyvern]

e

[prisma karma]

put it in dollar signs

[graceful lichen]

$ \text{For three events } A,B,C,\ \text{it is given that} $
$ P(\text{exactly one of }A\text{ or }B\text{ occurs}) = p $
$ P(\text{exactly one of }B\text{ or }C\text{ occurs}) = p $
$ P(\text{exactly one of }A\text{ or }C\text{ occurs}) = p $
$ P(A \cap B \cap C) = p^2 $
$ \text{If } A,B,C \text{ are exhaustive, find } p $

[nimble wyvern]

ee

[graceful lichen]

$ \text{For three events } A,B,C,\ \text{it is given that} $
$ P(\text{exactly one of }A\text{ or }B\text{ occurs}) = p $
$ P(\text{exactly one of }B\text{ or }C\text{ occurs}) = p $
$ P(\text{exactly one of }A\text{ or }C\text{ occurs}) = p $
$ P(A \cap B \cap C) = p^2 $
$ \text{If } A,B,C \text{ are exhaustive, find } p $

[nimble wyvern]

eee

[graceful lichen]

i will not ask question now

lol

[prisma karma]

yeah the bot is tripping ig

[graceful lichen]

i am closing

[prisma karma]

$a$

[flat frigate]

qwertytrewq

[nimble wyvern]

$\text{For three events } A,B,C,\ \text{it is given that}$
$P(\text{exactly one of }A\text{ or }B\text{ occurs}) = p$
$P(\text{exactly one of }B\text{ or }C\text{ occurs}) = p$
$P(\text{exactly one of }A\text{ or }C\text{ occurs}) = p$
$P(A \cap B \cap C) = p^2$
$\text{If } A,B,C \text{ are exhaustive, find } p$

[flat frigate]

Itsuki

[mossy lotus]

no spaces near the $s

[nimble wyvern]

yes

[prisma karma]

$ a $

fr? bruh

[nimble wyvern]

"keep it close to the moneh"

[mossy lotus]

draw a venn diagram corresponding to each of these so you can see the situation clearly

[graceful lichen]

no i had to ask something else

[nimble wyvern]

bruh

🤦‍♂️

[graceful lichen]

Que: For 3 events A,B,C
P(exactly one of A or B occurs) = p
P(exactly one of B or C occurs) = p
P(exactly one of A or C occurs) = p
P( all 3 occur together) = p^2
find value of p if all the events are exhaustive.

This is question

[flat frigate]

BlackidoZΣ

[graceful lichen]

this is my solution

what to do next now

[mossy lotus]

clearly P(AUBUC) = 1 coz thats exhaustive

you want intersection to be p^2

$P(A\cap B\cap C) = p^2$

[flat frigate]

Bacter14Fr0g

[graceful lichen]

definition of exh events is that Union of all events = Sample Space

right

[mossy lotus]

yes

[graceful lichen]

it means here S = 1

[mossy lotus]

mhm

[graceful lichen]

how

[mossy lotus]

can you clarify what S = 1 means

I was expecting you to explicitly state what that means

[graceful lichen]

here

[mossy lotus]

ok

then wdym how?

[graceful lichen]

I didn't understand why P(AUBUC) = 1

[mossy lotus]

the probability of an event being one of all possible events is always true

[graceful lichen]

okay

[mossy lotus]

say you have a coin thats getting tossed, A = heads, B = tails and C = edge

ofc thats exhaustive

[graceful lichen]

got it

[mossy lotus]

and P(AUBUC) = 1

[graceful lichen]

okay got it

what's next now

[mossy lotus]

.

[graceful lichen]

3p/2 + p^2 = 1

[mossy lotus]

yep

[graceful lichen]

ez now

thx

.close

[safe radish]

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[desert pasture]

really confused here, isn't the adjoined column supposed to be the constants

[mortal sandal]

that doesn't even seem like it should be an actual adjoined column

[desert pasture]

should be part of the matrix, right

[mortal sandal]

that's what I'm thinking

[desert pasture]

cool

thanks

[mortal sandal]

Idk why it's separated off

[desert pasture]

and are b,c,d supposed to be for arbitrary numbers or wrt (a)

.close

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[worn geyser]

Hi i need help with the task below.

The task
"The graph of a sine function of the form f(x) = Asin(cx) +d intersects the positive part of the x-axis at the x-values 1, 5, 13, 17 and so on. The graph intersects the y-axis at the y-value 3."

(a): What are the period, the midline, and the horizontal shift along the midline?

[worn geyser]

Thanks in advance!

I have found the period

Which is 12

p = 12

But now i am not sure how to find the midline...

[edgy breach]

you can just make some simultaneous equations

since theres no horizontal phase shift and its a sin function you can find d with the y intercept

[worn geyser]

Hmm okai

[edgy breach]

when x is 0 sin(cx) is 0

[worn geyser]

Ohh so there are no horizontal phase shift because there is no phi in the expression of the function?

[edgy breach]

yeah theres nothing added to cx

[worn geyser]

Ohh okai

Understood

So the function would start at x = 0?

Or no

That doesnt make sense

[worn geyser]

Ohh okai

[worn geyser]

But how do i set up two equations in order to find the midline?

<@&286206848099549185>

[worn geyser]

Yess

[safe radish]

@worn geyser Has your question been resolved?

[worn geyser]

Noo

But the horizontal shift along the midline is 0, right?

<@&286206848099549185>

Hope its ok i pinged

[open wedge]

!15min

[safe radish]

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

[open wedge]

please don't ping twice

[worn geyser]

Ohh okai

[safe radish]

@worn geyser Has your question been resolved?

[safe radish]

@worn geyser Has your question been resolved?

[simple galleon]

[the graph] intersects the positive part of the x-axis at the x-values 1, 5, 13, [and] 17
if it's intersecting the x-axis then f(x)=A sin(cx)+d=0
so f(1)=0, f(5)=0, f(13)=0, and f(17)=0

those are simultaneous equations (but you'll probably want to put 1 in f)

[worn geyser]

Oh i thought i could find d (midline) by doing this: 3 = A (sin c*0) + d

Which would be 3

d = 3

But that’s incorrect

But why is my method wrong?

[simple galleon]

idk, what did you use for A and d?

wait

nvm

[worn geyser]

Well i need to find d

Which is the midline i believe

[simple galleon]

yeah

[worn geyser]

And A becomes zero

When its being multiplied by sin(0)

Because sin(0) = 0

[simple galleon]

if its not 3 then the question must be wrong

(or c isn't a constant but i doubt it)

[worn geyser]

Hmm i am not sure

The conclusion says midline = 6

But i am not sure how

[worn geyser]

But is d = 3 by using this method as well?

[simple galleon]

i tbh dont know, i don't understand what oakley was trying to do

[worn geyser]

Me neither 😭😭😭

[simple galleon]

and i tried using desmos and it gave me this bs

which makes zero sense but desmos says it works

[simple galleon]

are you entirely sure that the x-values are 1, 5, 13, 17, ...?

oh i see a pattern nvm

[worn geyser]

Yess

This is the task ^

[brave wolf]

it does work

but you're prolly supposed to do it without desmos

[simple galleon]

well yeah up to some precision im assuming

theres probably a transcendental number for b that im missing

[brave wolf]

yeah, its gonna have pi in it

[simple galleon]

OHHH

[brave wolf]

are you done with a) already?

[simple galleon]

@brave wolf

[brave wolf]

oh wait you arent the OP

nvm

[brave wolf]

it's not 6, it's 3. 6 is the amplitude

[simple galleon]

that would make sense yes

[worn geyser]

Yes 3 makes sense

[safe radish]

@worn geyser Has your question been resolved?

[worn geyser]

Okei so d = 3

But now i am not sure how to find the horizontal shift along the midline

[worn geyser]

Do you know?

So the horizontal shift along the midline means how far the entire "wave" or graph is moved left or right, measured along the midline of the graph

midline = d = 3

Phi, φ, controls the horizontal shift

But there is no phi in the given function...

[brisk imp]

Amplitude is 6 as stated above

the equilibrium line is 3

Since one full cycle intersects the x axis 2 times

The period is 13-1 = 12

c = 2pi / period

= 2pi/12

= pi/6

since sin pi/6 is 0.5, 0.5A + 3 = 0

so A = -6

our final function is

-6sin((pi/6) * x) + 3

@worn geyser

[worn geyser]

But how do we know that the amplitude = 6?

[worn geyser]

Yes

[worn geyser]

Yes

[simple galleon]

do you understand how to get everything else

[safe radish]

@worn geyser Has your question been resolved?

[safe radish]

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[cerulean socket]

Can someone check my proof, and verify that it is well explained and correct?

[cerulean socket]

Given $f,g$ integrable on $[a,b]$, It is required to prove that,
[
\int_{a}^{b} f(x),dx\ + \int_{a}^{b} g(x),dx\ = \int_{a}^{b} (f(x)+g(x)),dx
]

[safe radish]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[flat frigate]

Arnob | Sukuna Enjoyer

[cerulean socket]

$f,g$ are both integrable so, $\exists P_n, Q_n$,
\begin{align*}
\lim_{n\to\infty} U(f,P_n) &= \lim_{n\to\infty} L(f,P_n) &= \int_{a}^{b} f(x),dx\ \
\lim_{n\to\infty} L(g,P_n) &= \lim_{n\to\infty} U(g,P_n) &= \int_{a}^{b} g(x),dx
\end{align*}
Let, for $n \in N$
[
M_n = P_n \cup Q_n
]
So, $M_n$ is a refinement.
Now,
[
L(f,P_n) \leq L(f,M_n) \leq U(f,M_n) \leq U(f,P_n)
]
Taking $n\to\infty$ on all sides,
[
\int_{a}^{b} f(x),dx\ \leq \lim_{n\to\infty} L(f,M_n) \leq \lim_{n\to\infty} U(f,M_n) \leq \int_{a}^{b} f(x),dx
]
So,
[
\lim_{n\to\infty} U(f,M_n) = \lim_{n\to\infty} L(f,M_n) = \int_{a}^{b} f(x),dx
]
Simillarly,
[
\lim_{n\to\infty} U(g,M_n) = \lim_{n\to\infty} L(g,M_n) = \int_{a}^{b} g(x),dx
]
Again, WLOG, $M_n = {x_0,x_1,x_2, \cdots, x_n}$
Now, let,
[
M(f,i)={f(x): x \in [x_i,x_{i-1}]}
]
Now, for $i \in {1,2,3,\cdots, n}$, because ${f(x)+g(x): x \in [x_{i-1},x_i]} \subseteq {f(x)+g(y): x,y \in [x_{i-1},x_i]}$,
\begin{align*}
\sup(M(f,i)) + sup(M(g,i)) \
&= \sup({f(x)+g(y): x,y \in [x_i,x_{i-1}]}) \
& \geq \sup({f(x)+g(x): x \in [x_i, x_{i-1}]}) \
&= sup(M(f+g,i))
\end{align*}

[flat frigate]

Arnob | Sukuna Enjoyer

[cerulean socket]

Simillarly,
\begin{align*}
\inf(M(f,i)) + \inf(M(g,i)) \
&= \inf({f(x)+g(y): x,y \in [x_i,x_{i-1}]}) \
& \leq \inf({f(x)+g(x): x,y \in [x_i, x_{i-1}]}) \
&= \inf(M(f+g,i))
\end{align*}
For the upper sums,
\begin{align*}
U(f,M_n) + U(g,M_n) &= \sum_{i=1}^{n} (\sup(M(f,i))+\sup(M(g,i)))(x_i-x_{i-1}) \
&\geq \sum_{i=1}^{n} (\sup (M(f+g,i)))(x_i-x_{i-1}) \
&= U(f+g,M_n)
\end{align*}
Simillarly, for lower sums, it can be proved that,
\begin{align*}
L(f,M_n) + L(g,M_n) \leq L(f+g,M_n)
\end{align*}
Therefore,
[
L(f,M_n)+L(g,M_n) \leq L(f+g,M_n) \leq U(f+g,M_n) \leq U(f,M_n) + U(g,M_n)
]
Taking limit on all sides,
[
\int_{a}^{b} f(x),dx\ + \int_{a}^{b} g(x),dx\ \leq \lim_{n\to\infty} L(f+g,M_n) \leq \lim_{n\to\infty} U(f+g,M_n) \leq \int_{a}^{b} f(x),dx\ + \int_{a}^{b} g(x),dx
]
By sequential definition of integral, $f+g$ is integrable and equals,
[
\lim_{n\to\infty} L(f+g,M_n) = \lim_{n\to\infty} U(f+g,M_n) = \int_{a}^{b} (f(x)+g(x)),dx\ = \int_{a}^{b} f(x),dx\ + \int_{a}^{b} g(x),dx
]
This is what was to be shown.

[flat frigate]

Arnob | Sukuna Enjoyer

[cerulean socket]

Had to split it into two parts because I don't have nitro

It is darboux integral btw

[safe radish]

@cerulean socket Has your question been resolved?

[cerulean socket]

Edit: There were a few typos I made on second revision. On second line,

$\lim_{n\to\infty} L(g,Q_n) &= \lim_{n\to\infty} U(g,Q_n) &= \int_{a}^{b} g(x),dx$

[flat frigate]

Arnob | Sukuna Enjoyer
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[cerulean socket]

Also, definition of M(f,i)

$M(f,i)={f(x): x \in [x_{i-1},x_i]}$

[flat frigate]

Arnob | Sukuna Enjoyer

[cerulean socket]

But other than that, it looks correct to me

[cerulean socket]

<@&286206848099549185>

[mighty saddle]

Yeah?

[cerulean socket]

^

Okay, since no one is replying, I guess I'll ask my professor tomorrow when he is free.

Thanks anyways!

.close

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[maiden schooner]

For A do we just do a factor of 4 with pi r^2

and radius how to do mathematically work this out

[buoyant shadow]

no

[mighty mango]

the area of each circle represents the population. there's a ratio between the population of A and B in given information.

read (a) again.

[buoyant shadow]

that's not how you A

[mighty mango]

this will be simple once you've concluded (a)

[maiden schooner]

how would we express the ratio 4:1???

[mighty mango]

population of A = 4 * population of B
A:B = 4:1 indeed.

what does that say about the ratio for the areas?

[maiden schooner]

we wouldnt just write 4:1 tho

cause its asking area to area

[mighty mango]

its asking for the ratio of the two areas.

[queen pine]

So lets say the area of B is 10 units^2
then A must be 40 units ^2
therefore the ratio is 40:10
thus 4:1
hopefully this makes sense

[maiden schooner]

so the answer is jsut 4:1

[queen pine]

yes

[maiden schooner]

for 1 mark

[mighty mango]

for (a).

[queen pine]

ratios are represented with [number]:[number]

[maiden schooner]

any clues to how the radius may be affected

or is it the same since pi r ^2

[queen pine]

You can let x be the radius of the circle A and r be the radius of B (you can choose what ever variables you choose).

From this try to construct an equation that is correct

then solve for x in terms of r

(or what ever variables u chose)

[mighty mango]

calling radius of a ---> r_a and radius of b --> r_b

you just concluded that the area of A = 4 * area of B

can you translate the previous sentence into an equation?

oops, I repeated what creamy custurd said.

[maiden schooner]

not rlly sure what equation to construct

nor show

[mighty mango]

the equation can be construced from the ratios of the areas.

[maiden schooner]

4pi r^2:pi r^2????

then cancel them out

ur left with 4:1

then with radius

oh wait

[mighty mango]

that's not an equation tho

area of A = 4 * area of B

area of any circle = pi r^2, sub that into area of A and B (just name their radiuses differently.)

[maiden schooner]

ohhhh

ok that makes sense

so b) would be

how would we express b) as an answer

give me a min

.close

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[honest perch]

!status

[safe radish]

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

[zealous ingot]

how far along ru

[safe radish]

@granite siren Has your question been resolved?

[granite siren]

ive gotten nowhere

[granite siren]

i used leibniz and somehow tried eliminating those ugly integrals but in vain

[zealous ingot]

that even function bit is very important

[granite siren]

hmmm

f(x) = f(-x)

[zealous ingot]

f(x) = f'(x) + 2xg(x) -x right

[granite siren]

how can i use that

[granite siren]

oh yeah

i didnt see that

[zealous ingot]

if f(x) is even, f'(x) must be odd

[granite siren]

yes

[zealous ingot]

so rearrange as f(x) - f'(x) +x = 2xg(x)

differentiate to get rid of that integral thing

you'll get g'(x) as something along the lines of f(a)f(x)/(a^2 x^2)

[granite siren]

wont it reappear

2x.g'(x) + 2.g(x)

[granite siren]

oh i can substitue this in place of g(x)

[zealous ingot]

exactly

[granite siren]

ok lemme try

ill get back to u in a few

absoulte mess

all cause of 'a', when im differntiating g(x), the term f(a) appears

[zealous ingot]

mm yea true

if this is a jee problem, like I suspect

your best bet is usually to find a pattern and try thinking of a suitable form for f(t)

[granite siren]

jee advanced

[solar hazel]

[zealous ingot]

send method

[solar hazel]

JEE advanced is trivial

the problems are easy and they give you way too much time

[granite siren]

4/10 ragebait

[zealous ingot]

ok bro 😂

[zealous ingot]

they only ever use 2 forms there btw

[solar hazel]

idk i thought it was better than 4/10 ragebait

at least 7/10

[zealous ingot]

$\alpha e^{\beta t}$ or $\alpha t^n$

[flat frigate]

donkey

[granite siren]

2 forms?

[granite siren]

nah u gotta make it less obvious for a 7/10

[zealous ingot]

yeah like most of the times u can js guess the structure for f(t)

[granite siren]

yeah usually its smth very simple

[zealous ingot]

this one is too 💀

[solar hazel]

yea

[zealous ingot]

Ct^2

C is prolly 1/2

[granite siren]

no amount of guessing is going to work in this

[zealous ingot]

ok fine

use the f(x) = f'(x) + 2xg(x) - x

since it must be even, the odd parts must cancel out

2xg(x) might have some special relationship, but idts

[zealous ingot]

so by this logic f'(x) = x

[granite siren]

ok let me check

thats the most generic f(x)

but ok

[zealous ingot]

f' = x, not f

[granite siren]

yeah

i mean yeah u can also see it in the g(x) term

f(t)/t^2

[zealous ingot]

mhm yea

[granite siren]

youre right lmfao

[zealous ingot]

im not quite satisfied with the robustness of this thing, but yeah 😭

[granite siren]

if it works it works

now its just the area resolved by y=x and y=x^2

bruh

[zealous ingot]

💀

[granite siren]

welp

that was a colossal waste

[zealous ingot]

x^2/2 no

[granite siren]

or y=2x

[zealous ingot]

happens all the time

[granite siren]

i put f(x) as x^2

[zealous ingot]

nah nah

[granite siren]

oh yeah

[zealous ingot]

f'(x) being x means f(x) must be x^2/2 + c

[granite siren]

youre right

[zealous ingot]

c=0 cuz f is even

[granite siren]

yeah mb

[zealous ingot]

you're good

[granite siren]

thanks man'

youre a jee guy too?

[zealous ingot]

yeah

not fond of chem

[granite siren]

how was jan

[granite siren]

same

[zealous ingot]

ah it was a bit of a messup

got 19 in chem so yeah 💀

overall 97.91%ile

[granite siren]

damn

[zealous ingot]

how was urs

[granite siren]

bad

i got 93.1

absolutely bombed the paper

[zealous ingot]

april attempt bro

[granite siren]

yes

youre a dropper?

[zealous ingot]

nope, first year

[granite siren]

oh youre in 1st year and u knew this much calculus?

geez

wait what

[zealous ingot]

like first year of jee

12th grade

[granite siren]

ohhhh

[zealous ingot]

I was into competition math since early hs so yeah

[granite siren]

i see

IMO and stuff?

[zealous ingot]

partly

oh btw, you want a differential eqn?

found a nice one in allen tests from yesterday

[granite siren]

give

ill solve it in the evening

i really need to speedrun chem

[zealous ingot]

us 💀

[brave slate]

y=xt then it becomes a LDE?

[zealous ingot]

nope

that'd work if it was homogenous

[brave slate]

wait lemme try it again

well its my first time seeing jeetards here

[zealous ingot]

there's a lot 💀

[brave slate]

hmm. What's the idea behind this problem

[brave slate]

really?

X+y=t?

[zealous ingot]

nope

[dusky totem]

Nice y’all are talking about the JEE too

I’m focusing for advanced 🙏

[brave slate]

Got it

This is dy/dx of y/x^2

if we rearrange a bit

[dusky totem]

Brilliant!

[brave slate]

y=x^2 tan(x+c)

[dusky totem]

I see it

[brave slate]

But I took too much time. Guess i have been a bit lazy after first attempt

[dusky totem]

You’re a JEE guy too?

How much

[brave slate]

I fucked up jan attempt

[brave slate]

yours?

[dusky totem]

233 marks

This is probably the wrong channel to talk abt this stuff

[brave slate]

I got 99.3 only

[open wedge]

oof

[brave slate]

yeah why isn't op here

[open wedge]

i don't even know what JEE is 😭

[brave slate]

its coz 1 divided zero is infinity

@granite siren close the channel if your doubt's solved

[open wedge]

39th bait

[safe radish]

@granite siren Has your question been resolved?

[zealous ingot]

its this tiny local exam

dw abt it

[open wedge]

ah okay

[proud tree]

-# yeah def not a nationwide examination

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[granite siren]

aight

[granite siren]

.close

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[dusky totem]

💀💀💀

[open wedge]

LMAOO

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[lean otter]

can someone help me with this question? here is the translation:

Consider the function f defined for all real numbers x other than -1 by
f(x) = (look the picture)
Which of the following sets corresponds to the domain of definition of the function f o f where the symbol o denotes the composition of functions?

during the simulation i will only have 3 mins for answering each questions and calculators are forbidden.
if anyone could be able to explain me how to answer those kind of questions that would help me a lot!
im very lost here

[stoic dune]

If you're desperate, one way to do this question is to actually calculate f(f(x)) and check its domain

The smarter way is to realize that, whatever value of u such that f(u) = -1, must not be in the domain of f(f(x))

Because:
f(f(u)) = f(-1)
can't be done

[lean otter]

OH that actually helps a lot! i will do that thank u so much for ur help

.close

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[stoic dune]

Whoops I meant -1. See the edit.

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[jade viper]

I am currently doing a geometry problem and I'm almost done with it just need a bit of the last part

[open wedge]

!da2a

[safe radish]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[open wedge]

just send the question right away

[jade viper]

Right now I need TA^2 = TZ.TF

T is an arbitary point on (BOC)

[open wedge]

toán chuyên đúng ko

[jade viper]

oh?

hi

[open wedge]

does the question give any other info?

[jade viper]

I'm typing it out

[open wedge]

!xy

[safe radish]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[jade viper]

ok

[open wedge]

sending a picture would be better

[jade viper]

radical axis*

also AX intersects (XYZ) at a second point E, TZ intersects (BOC) at a second point F

And I got it to proving TA^2 = TF*TZ

Which is same as proving TA is the tangent of (AZF) but I have no idea right now

[safe radish]

@jade viper Has your question been resolved?

[safe radish]

@jade viper Has your question been resolved?

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@jade viper Has your question been resolved?

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@jade viper Has your question been resolved?

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[glossy dune]

@safe radish hey

[glossy dune]

@safe radish do you solve problems?

[tulip sundial]

no it doesnt

read #❓how-to-get-help

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@glossy dune Has your question been resolved?

[plucky elk]

.close

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[tawdry marlin]

can someone help how are u supposed to 5c)

[plucky elk]

Did you learn what it means for a function to be odd or even

[tawdry marlin]

uhhhhh i think so

like ik the odd functions have point symmetry and the even ones have line symmetry

[warm copper]

I think graph b / equation 4iii is the easiest to start with because you can easily see that it has symmetry across the y-axis, like a mirror image

if you can figure out what that means mathematically then you can verify it easily

[tawdry marlin]

ok wait

i get that but im just confused on what it means to verify algebraically

esp for graph a) and c)

[velvet nebula]

What symmetry do you think it has?

Try phrasing your answer in terms of "if (x, y) is on the graph, then some other point (..., ...) is on the graph"

[tawdry marlin]

i think graph a and c have point symmetry?

so if u were to verify would u like plug in a point from the graph or smth

[plucky elk]

Were you given definitions in terms of a function

Like f(-x) = ...

[tawdry marlin]

uhhhhh

[warm copper]

that would work but you can also do it more abstractly

[tawdry marlin]

i dont think so

but i saw smth ab that when i searched my question up

[warm copper]

again using graph b as an example, a point is (x, y) and it's mirror is (-x, y)

[tawdry marlin]

ohhhh

ok but how ru supposed to do it for graph c

[warm copper]

basically the same thing, take a point's coordinates and think about it's mirror's coordinates

[tawdry marlin]

i see

but like earlier i tried doing that

and like the y values are different when u do that method

for graph c

bc everything is shifted up one

[warm copper]

you mean the value is different, not the sign?

[tawdry marlin]

wait wdym

[warm copper]

graphs a and c are clearly not symmetric around the y axis (even)

the other possibility is that they are, like you said, point symmetric

if you took a y value at x and compared it with the y-value at -x, you'd get the same magnitude but with the sign flipped

or simply (x, y) -> (-x, -y)

[tawdry marlin]

wait

but i tried doing that and the y values were not like the same value

[warm copper]

then it isnt symmetric

[tawdry marlin]

wait so for it to be symmetric does it have to pass through (0,0)

or something

bc i think the y values are not the same bc the graph only passes through (0,1)

[warm copper]

im not sure exactly what that means but even symmetry is thought of as reflecting on the y-axis and defined as f(-x) = f(x)

odd symmetry is putting your finger on the origin and rotating it by 180deg or f(-x) = -f(x)

[tawdry marlin]

ohhh okay

[warm copper]

if you can come up with a counterexample to both of those then it isnt symmetric

[tawdry marlin]

okay thanks 🙏

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[austere lark]

I was thinking 4

But I’m thinking wrong aren’t I

[glacial cairn]

4 isn't one of the choices...?

[heavy glade]

how can it be 4? its clearly larger than 4

[buoyant shadow]

i can't solve it in my head 😭

disqualified from using tp

[heavy glade]

pi*r^2 = (6^2-2^2)pi/2 + 4 pi

[glacial cairn]

sqrt((12^2-4^2)/2+4^2) isn't that hard

[heavy glade]

root 12?

not even an option

[austere goblet]

I would like to point everyone's attention to the phrase closest to just in case.

[buoyant shadow]

72−8+16 = 80

so 9

[heavy glade]

how come

naww it's root(20) for sure, unless im misunderstanding the question

[glacial cairn]

You're definitely misunderstanding since sqrt(20) is not an option

[shy dune]

find the initial paper area and / 2 isnt it?

[heavy glade]

yep

[shy dune]

sry if u cant see

so the new area* is pi 32 / 2 = pi 16

then u just have to find the diameter (by finding radius)

[heavy glade]

16 pi is the radius?

[shy dune]

no sry type 😭

[heavy glade]

and we also need to account for the area of the "tube"

[shy dune]

does op have more working out?

[heavy glade]

OP hasnt been here for half an hour....

[shy dune]

oh...

[heavy glade]

but my solution isnt even one of the options.

[buoyant shadow]

it's 2 root 20

you got radius instead of diameter

[heavy glade]

so 9 is the solution like you said. i guess i messed something up

[shy dune]

the reason the answer isnt 8 btw is because intuitively the paper roll is cut in half, but its not the radius that is cut in half, it is the volume which doesnt have a linear relationship with the radius/diameter.

[austere lark]

uh

I’m not getting it

Roll area - tube area?

[buoyant shadow]

yes

gives paper area

[austere lark]

Then what do I do with paper area

[buoyant shadow]

divide by 2

add roll area back

[austere lark]

but what does that do

[buoyant shadow]

area of roll that lost half its paper

[austere lark]

do you mean add tube area back?

[buoyant shadow]

yeah sorry

[austere lark]

Then to find radius I divide by pi then find square root?

And then x2

[buoyant shadow]

right

[austere lark]

So (144pi-16pi)/2+16pi = area of roll that lost half paper

Then sqrt(answer) pi))2

I got 42

[buoyant shadow]

no that's using diameter as radius

(36pi-4pi)/2+4pi

2 sqrt 20
sqrt 80
≈9

[heavy glade]

bruh so the question asks for diameter? 😭

like imagine writing up this question such a jerk

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[safe radish]

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[north fulcrum]

four fair dice are thrown, find the probability that 5 is the largest number and 3 is the smallest

which of these methods is right?

[mortal sandal]

Neither

The first die is not required to be a 3, nor the second a 5

For example, 4,4,5,3

The second method also fails to differentiate the order the dice are rolled in, but even more so

[solar hazel]

amen dreyuk

[buoyant shadow]

wow that's complicated

[solar hazel]

you could always just count up the permutations of each of these and add them up

3 5 4 4
3 5 5 4
3 5 5 5

3 3 5 4
3 3 5 5

3 3 3 5

pretty fast since the lists are small (and some of the lists are isomorphic)

but there is a better way maybe

[buoyant shadow]

like
range 3−4 has 32 outcomes
range 4−5 has 32 outcomes

so 63 outcomes that don't do both

😐

[solar hazel]

😐

[mortal sandal]

3-4 has 32 outcomes?

[buoyant shadow]

oh i thought 5 dice

31 outcomes that don't do the thing
subtract from 81

i think

[solar hazel]

yea actually pie sounds good

[buoyant shadow]

sounds very wrong somehow

[solar hazel]

[buoyant shadow]

12+12+4+12+6+4

also 50

[solar hazel]

,calc 3^4 - 2^5 + 1

[flat frigate]

Result:

50

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[pseudo compass]

Please give me some tips on rational number, o keep making errors and silly mistakes.

[pseudo compass]

Eight grade.

[manic patrol]

may I know what kind of questions you are making errors or silly mistakes on? do you have examples of said questions and mistakes? we could look at those and then see what might be going on.

[frozen marlin]

!xy

[safe radish]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[frozen marlin]

do you have perhaps specific problems

if its a general thing i suggest #study-discussion

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[safe radish]

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[small rampart]

I'm learning how to solve different kinds of linear difference equations but the methods don't seem intuitive, which makes it very hard to grasp. Someone please help me build my intuition on the motivation behind the methods of solving those

[twilit juniper]

Hello i need help doing this problem.

calculate the surface area of ​​a regular four-sided prism if D=42 and the side diagonal d=30

[small rampart]

!occupied

[safe radish]

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).

[twilit juniper]

Oh sorry my wifi is not that good rn

[flat frigate]

Nerdyasianguy

[small rampart]

I don't understand where of the method of using the characteristic polynomial and stuff come from

Especially the part where you convert complex roots into trig

[safe radish]

@small rampart Has your question been resolved?

[small rampart]

Any video/file/book recommendations on the topic would be appreciated too

[safe radish]

@small rampart Has your question been resolved?

[small rampart]

.close

[safe radish]

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[north fulcrum]

im reviving this

[north fulcrum]

i got the ans of 50 from doing this

but is there a second method?

(its also required by the worksheet)

[winter whale]

always bet on mathematics

[north fulcrum]

the permutation method showed me that every case has a different count, so at the end i have to break it down to this level

[brave wolf]

Well, you will have to count those permutations in some way. In this case listing them is simplest.

You could probably do something with inclusion exclusion though
||3^4 is number of permutations with 3,4,5
2^4 is number of permutations with 3, 4
2^4 is number of permutations with 5, 4
1^4 is number of permutations with 4

3^4 - 2*2^4 + 1^4||

[north fulcrum]

ohh

[flat frigate]

The following error occured while calculating:
Error: Value expected (char 1)

[brave wolf]

maybe the incl excl is simpler actually

i thought it would be more difficult than that, but somehow it worked lol

[north fulcrum]

so if theres no 3/4/5, exclude it, if theres just 3/4/5, include them back

[brave wolf]

not what i meant

Start by counting all rolls with only the numers 3,4,5

that overcounts though, because that also counts throws like 3,4,4,3 (max isnt 5 here)

so you subtract those that miss either 3 or 5

that undercounts, because you subtract those that have just 4 twice

so you add back the one permutation which has only 4's

[north fulcrum]

ok i got it

thats all thx

.close