help-10

[balmy mortar]

$$(ab)^k = a^kb^k$$

[warm shale]

Shuri2060

[balmy mortar]

ok?

This is only possible since G is abelian

[wraith gazelle]

Yes

[balmy mortar]

So now, what is the order of ab?

Remember gcd(a, b) = 1

[wraith gazelle]

It's mn

[balmy mortar]

yes.

What does this tell you about G?

[wraith gazelle]

The elements in G commute?

[balmy mortar]

You already knew that. G is abelian.

However, you have just found an element in G which has order mn, the size of G.

|G| = mn. If G contains an element of order mn, that means it is cyclic

[wraith gazelle]

Oh

[balmy mortar]

So now you know ab is a generator

G = <ab>

[wraith gazelle]

Sorry, I've been very distracted by everyone around me. Finally quiet🥺

[balmy mortar]

is ok

So now we want an isomorphism between <ab> and <a> x <b>

make sense?

[wraith gazelle]

yes

[balmy mortar]

Let's go with this example

when order of a = 2

order of b = 3

So mn = 6

So order of ab = 6 in G

We need to try to construct an isomorphism.

What seems something sensible if we want to succeed?

Just think about mapping 1 element for now

[wraith gazelle]

We're trying to show G is an isomorphism?

[balmy mortar]

We want to create an isomorphism

between G and <a> x <b>

To do that we need to think of a mapping

[balmy mortar]

Try thinking whats a good idea to try

things often turn out the way you expect them to in group theory

[wraith gazelle]

I'm unsure of what you mean by mapping

[balmy mortar]

A function

might be referred to that or a map?

Because you have something like this

Elements from the domain 'map' to the codomain

1 is 'mapped' to 5 in my picture

For this, I am suggesting you think of a 'map' from elements of G to here

but don't try to map everything at once

Just 1 element at a time

So right now, maybe it's easier to map from G to <a> x <b>

There are actually only a few elements of G we know (explicitly)

[wraith gazelle]

So mapping from these elements to one another?

[balmy mortar]

No, from group G

to this group <a> x <b>

What's an element in group G you know?

any

[wraith gazelle]

a

[balmy mortar]

sure.

So what might be a good idea to map a to

in here

[wraith gazelle]

(a,e)?

[balmy mortar]

ok

so lets call our function f

f(a) = (a, e)

Another element from G?

[wraith gazelle]

b

[balmy mortar]

f(b) = ?

Just take a guess

if it doesn't work we can try again

[wraith gazelle]

(e,b)

[balmy mortar]

ok

so

f(a) = (a, e)
f(b) = (e, b)

Then it turns out this actually locks down our function

You chose where a and b map to

everything else must follow

if we want it to be an isomorphism

f(ab) = f(a)f(b) = (a, e)(e, b) = (a, b)

Good so far?

[wraith gazelle]

Mhm

[balmy mortar]

Then you can also check

a^2

b^2

ab^2, etc.

I will tell you now what you chose works

But for the original question, we don't know if m and n are 2 and 3

So we need to prove this idea works in general

$$\langle a\rangle \times \langle b\rangle= {(a, b):a\in \langle a\rangle, b\in \langle b\rangle}$$

[warm shale]

Shuri2060

[balmy mortar]

$$ord(a) = m$$
$$ord(b) = n$$

[warm shale]

Shuri2060

[balmy mortar]

So try to do the same thing in this case, and in particular you need to prove what you construct is an isomorphism

That doesn't sound easy, but we can also use something we found our earlier

$ab$ generates G because its order is $mn$

[warm shale]

Shuri2060

[wraith gazelle]

And we want to show <ab> is isomorphism with <a>×<b>?

[balmy mortar]

yes exactly

[wraith gazelle]

Well, we got ab from that question, what about the other question it hints at? Where does this go into play?

[balmy mortar]

I think you used them already

To show G is cyclic

First try to construct the function $f$

[warm shale]

Shuri2060

[balmy mortar]

I think it will be easier to map from <a> x <b> to G instead

[wraith gazelle]

Can we do that? Wouldn't it be assuming what we are proving?

[balmy mortar]

no.

We construct f

[wraith gazelle]

Or to just get an idea?

[balmy mortar]

then prove it is an isomorphism

[wraith gazelle]

Oh

[balmy mortar]

$$f : \langle a\rangle\times \langle b\rangle$$

[warm shale]

Shuri2060

[balmy mortar]

$$f((x, y)) = ???$$

[warm shale]

Shuri2060

[balmy mortar]

Then define where (x, y) must map to

for any (x, y) in <a> x <b>

A hint is --- you can write any generic (x, y) in a certain way

which makes defining f much easier.

Try it 🙂

[wraith gazelle]

I'm not sure what you meant by this

@balmy mortar Shuri I'm still very confused

[obtuse pebble]

@wraith gazelle Has your question been resolved?

[balmy mortar]

@wraith gazelle

$={(e, e), (a, e), (a^2, e), \cdots, (a^m, e), \
(e, b), (a, b), (a^2, b), \cdots, (a^m, b), \
(e, b^2), (a, b^2), (a^2, b^2), \cdots, (a^m, b^2),\
\vdots,\
(e, b^n), (a, b^n), (a^2, b^n), \cdots, (a^m, b^n)}$

ugly, one sec.

[warm shale]

Shuri2060

[wraith gazelle]

f((x,y)) = f(x^m)f(x^n) can we use order?

[balmy mortar]

almost

Look how I have written <a> x <b>

it is almost like a matrix

All of the elements in there can be written in a specific form

[wraith gazelle]

You wrote it as (a,b)

[balmy mortar]

everything in <a>

can be written a^k for some k

everything in <b> can be written b^j for some j

hint?

[wraith gazelle]

(a^k,b^j)

[balmy mortar]

thats what i meant

[wraith gazelle]

Yeah, just clarifying that bit

I'm just not sure what you meant in that hint

[obtuse pebble]

@wraith gazelle Has your question been resolved?

[balmy mortar]

?

[balmy mortar]

(a^k,b^j)

f((a^k,b^j))

You define where this maps to for every k and j

to define the function.

[wraith gazelle]

So in this case with the problem, can we say f((a^m,b^n))?

[balmy mortar]

m and n are constants.

a^m = e

b^n = e

You need to specify where the function maps to for every k and j

f((a^k,b^j))

[wraith gazelle]

So we can't have that, which is why we are using x,y with k,j

[balmy mortar]

not x, y

a and b

[wraith gazelle]

Well a,b

[balmy mortar]

You should do the same as in the previous example

where should we send (a, e) to?

f((a^1,b^0)) = ???

f((a^1,b^0)) = f((a, e)) = ???

[wraith gazelle]

(e,b)

[balmy mortar]

no ???

let's get this clear

We need to create a bijection from G to <a> x <b>

The way I've written it, f maps starting from <a> <b>

[wraith gazelle]

Ohh sorry, I thought we were using the example since we stated it

[balmy mortar]

$$f: \langle a\rangle\times\langle b\rangle\to G$$

[warm shale]

Shuri2060

[balmy mortar]

yh. I think it's probably easier to think the other way round

though for the general case

$$f((a^1,b^0)) = f((a, e)) = ???$$

[warm shale]

Shuri2060

[balmy mortar]

So this must map to element in G

[wraith gazelle]

Which is a?

[balmy mortar]

$$f((a^1,b^0)) = f((a, e)) = a = ae = a^1b^0$$

[warm shale]

Shuri2060

[balmy mortar]

I think you should be able to figure out what f should be

Hint is there

[wraith gazelle]

f((a^k,b^j)) = a^kb^j

[wraith gazelle]

.close

[obtuse pebble]

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[spark talon]

hello i don't seem to understand where my working went wrong

[spark talon]

according to the markscheme the answer is supposed to be 1441π/20

by the way this is for question (a)

[obtuse pebble]

@spark talon Has your question been resolved?

[obtuse pebble]

@spark talon Has your question been resolved?

[tardy epoch]

@raven spire , @inland ginkgo and I did almost an exact problem in #calculus

[inland ginkgo]

yea thats similar

[spark talon]

Alright I will check

[obtuse pebble]

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[timid silo]

How do I express this surd with a rational denom

[wet drift]

Rationalize

[timid silo]

🧠

Great idea

Why haven't I thought of that

Please expand on your idea

[wet drift]

$\frac{1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1}$

[warm shale]

BrownMunde

[timid silo]

Sounds rational

Cheers for the help

.close

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[weak wind]

any ideas how should I go about solving this?

[obtuse pebble]

@weak wind Has your question been resolved?

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[timid silo]

My answer is coming wrong

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[wary vigil]

what did you try

[obtuse pebble]

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[ebon axle]

Hey everyone, I have a question and i'm no good at matths so i wanted some input
so imagine you've got a grid tile surface like this:

[ebon axle]

you can deform the 4 corners of the tile up or down to change the shape of the tile

my question is: is it possible to deform a whole line of tiles so that each tile surface becomes coplanar, but still keep the corners and edges meeting the neighbouring tiles?

i suspect it might be possible for a single straight line of width 1, but what about a corner L, or a line of tiles with width larger than 1

[wary vigil]

i'm not sure what you're asking

[ebon axle]

let me try to draw something

so on a given tile you can move the corner points up and down to change the shape

if you move only one the tile surface is no longer coplanar

(or you could make a tile that's not coplanar be coplanar again)

i want to know if it's possible to make tiles coplanar while also still having their corners and edges meeting those of neighbouring tiles

[wary vigil]

i think so

you'd have to move more than 1 corner to do that

but using this method you could make them all lie in the same plane

which would be the condition you're asking for

[royal solar]

so you want z to be a wave function of x and y?

[ebon axle]

if i had a rectangle region of multiple tiles, and i tried to make them all coplanar, is that possible?

[wary vigil]

is that what you're asking?

[ebon axle]

ok so they'd all be forced into a straight line 🤔

[wary vigil]

i think it has to do with curvature

you would only be able to have curvature in one direction

otherwise you'd get some rectangles that do not have coplanar vertices

i made them triangles to better show when a non coplanar rectangle is existing

[royal solar]

https://www.geogebra.org/calculator
use the 3d calculator setting
type z = sin(x)-sin(y)

tell me if that’s what you’re looking for

@ebon axle

[ebon axle]

i think from that demo i can tell what i imagined isn't possible

@wary vigil thank you :)

[royal solar]

demo?

[ebon axle]

demonstration

[royal solar]

yeah no, wait, i’ll read the rest of the messages. apologies for rushing ahead

[ebon axle]

ok so this fact makes my task a little more complex

[ebon axle]

so next i am looking for a way to determine the height of an arbitrary point on the tile surface

this is easy when the tile is coplanar because i can interpolate by distance between two corners

but it seems that when the tile's not coplanar, the slope will be different in each of the four 'quadrants'

[wary vigil]

i don't think you can find any arbitrary point on the tiles if they are not coplanar

because there are 2 ways to cut up the quad into triangles

and infinite ways of constructing a surface from it

without straight lines that is

you'd need to necessarily use triangles

cuz they are always coplanar

[ebon axle]

ohh ok

so for each triangle

wait

so i'd need the highest point and lowest point on each triangle

and the 'height' is the interpolation based on distance from the lowest to highest point

does that sound correct?

[obtuse pebble]

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[fickle falcon]

im stuck

[chilly patrol]

lol

[fickle falcon]

i dunno know how to find f(-1)

[raven spire]

It's probably: The remainder when f(x) is divided by (x - 1) and (x + 1) are -1 and +1 respectively.

[fickle falcon]

my teacher keep saying the question is not wrong and i panic

[raven spire]

Or you might've been given something additional info to help you here... can you click a pic of the question?

[fickle falcon]

thats the question already🥲

is it even possible to do it....

[raven spire]

I mean: consider $f(x) = (x-1)(x+2)Q(x) -\frac{2}{3} x - \frac{1}{3}$

It satisfies the condition you gave but you can't really figure what the remainder is gonna be here

[warm shale]

Ansh

[fickle falcon]

i did told my teacher this

i found the ans by using x-1 x+2

and the ans is excatly the one i found

but idk how to transfer x+2 to x+1

[raven spire]

then maybe 🤦‍♂️ they meant, what is the remainder when f(x) is divided by (x - 1)(x + 2)

[fickle falcon]

dang

imma ask him one more time

thx a lot

.close

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[humble umbra]

So I'm compressing the graph y=x^2 to sketch the graph of f(x)=1/2x^2 and the example is telling me to do this by compressing the graph of y=x^2 by a factor of 1/2 but I don't know what that means

[haughty coyote]

The point (2,4) belongs to the graph of y=x^2. The point (2,2) is in the graph of y=x^2 / 2 as a consequence, because x is unchanged but y is halved

[humble umbra]

I think I wasn't clear. I want to know what "by a factor of 1/2" means here. If you already answered I'm sorry I just don't' understand yet. IF you have a video or article I could read on this that would be great too so you don't have to waste your time. I can post the question if needed.

[raven spire]

https://youtu.be/Tmdrjs9xufc

Ik the "Organic Chemistry" title for a math video is sketchy butttt

he's goodddd 👀

[haughty coyote]

"by a factor of 1/2" is what is done to y. y is multiplied by a factor of 1/2, i.e. if (x, y) was in the graph, the new graph contains (x, y/2): y has been multiplied by 1/2

[humble umbra]

OKAY

WOW THANK YOU BOTH

❤️

[humble umbra]

So just to make sure I have this correct, if I were to use the key points I originally created, could I just multiply the y's by 1/2??

[haughty coyote]

Yes

[humble umbra]

WHAT A BRILLIANT DAY. THANK YOU

[haughty coyote]

You bring every point halfway closer to the x-axis

[humble umbra]

Okay. How do I close this?

or do I just leave it

[haughty coyote]

.close

[humble umbra]

.close

[obtuse pebble]

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[obtuse pebble]

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[inland thicket]

For $120 you could buy 2 chocolates more than after they had gotten 25% more expencive, how much chocolates could you buy before it became more expencive?

[inland thicket]

my teacher solved this but its hard to understand

she is weird

I can show how she solved it but It would be nice if someone just like said yo u get this then put there like that n ting

its those with 2 equations and 1 gets inserted

like first thing she did is x*y=120 right i get that

after that she did the part i dont get:

x+25%x=125%*x=1.25x (price of 1 chocolate)

like this is how i written this from blackboard, im questioning if I written it wrong or like whas goin on

[timid silo]

basically how this works is that she is using percentages

you understand that 100% of something just means it's equal to itself right?

so what you can rewrite it as:

100%x+25%x=125%x

make sense so far?

@inland thicket

[obtuse pebble]

@inland thicket Has your question been resolved?

[inland thicket]

yo

i was eating shiii

oh shit ye

[obtuse pebble]

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[brittle swan]

calculator said ‘0’

where did i go wrong

my idea was to make x = 5 the y-axis by shifting both functions left by 5

[spiral maple]

your bounds are x bounds

not y bounds

[brittle swan]

are they not

[spiral maple]

(3,1)

x=3, y=1

you're summing a bunch of anuli

[brittle swan]

oh

.close

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[cyan heron]

I’m back with the iteration shit

[sage geode]

Just use the formula to find x1, then x2 and finally x3?

[cyan heron]

You just press the equals 3 times right

On the calculator

[sage geode]

Not really

The calculator would need to know what x2 is for that

[cyan heron]

Or do you get the answer for x1 then plug that back in?

[sage geode]

In this case to know x_n you need to know the previous term

To know x3, you need to know x2

To know x2, you need to know x1

And to know x1, you need to know x0, which is given

[cyan heron]

So once I have x1 using the formula

Do I substitute x1 into the xs in the formula to find x2

[sage geode]

Yes

[cyan heron]

Do I round x1 to 3 dp the plug back in?

[sage geode]

I think you'd need to simplify the expression

[cyan heron]

I’m to confused for this 😭

Omg I get it now

Sorry man

Thanks for the help

[sage geode]

Np

Feel free to .close

[obtuse pebble]

@cyan heron Has your question been resolved?

[obtuse pebble]

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[coral sonnet]

.reopen

[coral sonnet]

I can't understand the question at the first place

ya, but it just says chooses x to maximize we dont have to do that

wait

is it not xp - f(x)w?

they are selling at p per unit so revenue is x times p?

ok thanks

.close

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[lunar oracle]

Curve has equation y=x^2 -6kx +9k^2, PROVE that the x axis is a tangent to the curve for all values of K

[timid silo]

for every value of k, prove the tangent at the x value k is the axis line?

[lunar oracle]

Yeah basically

I just need to prove the turning point always touches it right

I’m not sure

[spiral maple]

yes

prove the vertex always lies on the x axis

[timid silo]

wait i think they mean the tangent at the vertex

[spiral maple]

well duh

how would you have the x axis be tangent anywhere else?

[spiral maple]

yeah

[lunar oracle]

So is that dy/dx = 0

[spiral maple]

calc isn't needed

[lunar oracle]

Oh

[timid silo]

well on a prabola the slope of the tangent at the vertex is alwayse 0

ie the tangent at the vertex is a flat line

[spiral maple]

it suffices to show the vertex is always (x,0), so write the quadratic in vertex form

[timid silo]

yeah what i was about to say lol

[lunar oracle]

What do you mean by vertex form?

[timid silo]

if the tangent at the vertex is always flat, then if the vertex is alwayse on the x axis it will be the same line

[spiral maple]

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

[warm shale]

Mosh

[spiral maple]

The form that lets you read off the vertex.

[lunar oracle]

Ohhhhh I see, I never knew it could be used this way

[spiral maple]

...

[lunar oracle]

😦

[spiral maple]

you didn't know vertex form had something to do with the vertex?

[lunar oracle]

yeah cos they didn’t teach it to me that way

They just told me to use it

[timid silo]

hay man, maybe he just never made that connection

[lunar oracle]

Xd

[timid silo]

i miss the point all the time and then 10 min later im like "OOOHHHHH"

[lunar oracle]

They called it like the square form or something

[timid silo]

lol

[lunar oracle]

👹 who the hell comes up with this shit

Anyway thank you guys :>)

[timid silo]

no problem, come back (or give me an @) if you get stuck somewhere along the way

[obtuse pebble]

@lunar oracle Has your question been resolved?

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[stable helm]

MISSISSIPPI in which there are no consecutive symbols S?
E.g. SIISIMSIPPS is ok, but MISSISSIPPI, ISIPIMSIPSS are not.```
Is 70 correct answer here? (8 chooses 4) ?





```Suppose we want to make strings containing n distinct symbols (different from
S and each occurring exactly once) and k occurrences of S which are not consecutive, where
n ≥ k ≥ 1 (so the length of such strings is n + k). How many such strings can we make?```


is the correct answer (n+1 chooses k) ?

[stable helm]

S and each occurring exactly once) and k occurrences of S which are not consecutive, where
n ≥ k ≥ 1 (so the length of such strings is n + k). How many such strings can we make?

is the correct answer (n+1 chooses k) ?

[obtuse pebble]

@stable helm Has your question been resolved?

[stable helm]

@obtuse pebble No

<@&286206848099549185>

[slate zephyr]

@stable helm How about calculating it via

words without consequtive s = # words - # word with consequtive s

[scarlet gale]

It's easier to first place the non Ss, then to place the Ss.

Like maybe you'll get xMxIxIxIxIxPxPx when you place the letters other than S in one way.

[slate zephyr]

How did you ge those numbers?

words should

[stable helm]

okay. #word = 34650

[scarlet gale]

And then you have 8 positions to put 4 Ss.

[stable helm]

yeah that's how i got 70.. (8 chooses 4) is that correct then?

[scarlet gale]

It's the way to place the Ss.

But you need to place the other letters in all the different orderings before the Ss as well.

The trick for that is you have 7 other letters.

So, 7! on top.

Then you have 1 M, 4 Is, 2 Ps, so 7!/(1! 4! 2!).

So, altogether 7!/(1! 4! 2!) 8C4.

[stable helm]

7!/(1! 4! 2!) the relation here is multification? 8C4?

[scarlet gale]

7!/(1! 4! 2!) is the number of ways to rearrange MIIIPPI.

You place those first.

[slate zephyr]

,wolf Multinomial[1,4,4,2]

[warm shale]

Results provided by WolframAlpha

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Requested by M8732#6402

[scarlet gale]

No, the multinomial there won't work, since it doesn't count Ss next to each other.

Or it does count them, rather.

[slate zephyr]

it's just number of any words

wanted to verify the number he got above

as for ss next to each other I think you can just threat "ss" as one symbol

thus 2 single s to place and one double s

,wolf Multinomial[1,1,2,4,2]

[warm shale]

Results provided by WolframAlpha

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Requested by M8732#6402

[slate zephyr]

wait no huh

[stable helm]

but there are 4 S's.. if i treat them as one there's still a chance that the other 2 will be consecutive

[slate zephyr]

I think that is fine though, we are just interested in the number of words

with consecutive ss.

[stable helm]

this makes sense to me .. so 34650 - 70 = 34580 is the answer? i get the 70 as @scarlet gale suggested

[slate zephyr]

How do you get 70? That seems veeery few.

let's say you put the ss in the first position, then you have still 9 arbitrary ways to put all the rest

no way that's 70

[stable helm]

it does. but i got it this way - @scarlet gale : "Like maybe you'll get xMxIxIxIxIxPxPx when you place the letters other than S in one way."

[scarlet gale]

See how I arranged the non S letters there?

See how there are 8 xs in which to place Ss?

And you have 4 Ss.

[stable helm]

exactly

[scarlet gale]

So, 8C4 is the ways to arrange the Ss after the rest.

[stable helm]

8 postion to put 4 S's

[scarlet gale]

Subtracting that doesn't make sense.

You multiply it with the number of ways to arrange the non S letters since they're happening simultaneously.

[stable helm]

oh yeah true. we need to find the arrangement that doesn't have consecutive S's.. so why subtracting !!

[slate zephyr]

Oky I see why my second query yields something to large

it's because it counts positions doubly since you can place either double ss or 2 single s.

[scarlet gale]

Verified it with computer.

[slate zephyr]

verify what

[scarlet gale]

The answer.

It's 7!/(1! 4! 2!) 8C4

import Data.List  (isInfixOf, nub, permutations, tails)
import qualified Data.Set  as Set (filter, fromList, size)

main :: IO ()
main = print . Set.size . Set.filter (not . isInfixOf "SS") . Set.fromList . permutations $ "MISSISSIPPI"

Take the permutations of MISSISSIPPI, put them all in a set, keep only the ones with no neighboring Ss, count how many elements are in the set.

Gives 7350 after checking everything.

[slate zephyr]

I was gonna calculate it a different way without the 70.

I thought he meant 70 as a step in my solution strategy.

[scarlet gale]

,w 7!/(1! 4! 2!) Binomial[8, 4]

[warm shale]

Results provided by WolframAlpha

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[stable helm]

okay. it's starting to make sense.

[stable helm]

how about this question?

[stable helm]

<@&286206848099549185>

[obtuse pebble]

@stable helm Has your question been resolved?

[stable helm]

<@&286206848099549185>

[scarlet gale]

S and each occurring exactly once) and k occurrences of S which are not consecutive, where
n ≥ k ≥ 1 (so the length of such strings is n + k). How many such strings can we make?

OK, so place the distinct symbols in a particular order. How many ways are there of doing that?

@stable helm

[stable helm]

n!

[scarlet gale]

OK, so let's say you have the distinct symbols A, B, C.

And one ordering is BAC.

Where can you place the Ss?

[stable helm]

in 4 places xBxAxCx

[scarlet gale]

OK, so how many is that in terms of n?

[stable helm]

n+1

[scarlet gale]

And how many of the places are we choosing?

[stable helm]

n+k

[scarlet gale]

No, for the Ss.

We already placed the distinct letters.

[stable helm]

for S, we're choosing k

[scarlet gale]

OK, so first we do n! arrangements, then we choose (n + 1)Ck.

So, n! (n + 1)Ck.

Do you see why that works?

[stable helm]

okay because first we find out how many ways we can arrange the distinct letters which is n! and when we incorporate S into these strings we have k places to plug in S.. which is n+1 C k
Since they are related and happens together , we've got n! (n+1) C k

[scarlet gale]

Yes, also you can see it another way. First, we place the distinct letters:```
xAxBxCx
xAxCxBx
xBxAxCx
xBxCxAx
xCxAxBx
xCxBxAx

Let's say k = 2.

So, there will be 4C2 = 6 variations for each row there.

xAxBxCx

SASBxCx
SAxBSCx
SAxBxCS
xASBSCx
xASBxCS
xAxBSCS

That row turned into 6 final rows.

And so do all the other 3! - 1 rows.

And so, since we have 3! things that turn into 4C2 final rows each, that's 3! × 4C2.

[stable helm]

I'm not sure i understand this. but i understood the previous way

[scarlet gale]

Well, you know that you get an arrangement of the ABC, right?

[stable helm]

yes

[scarlet gale]

And then you place the 2 Ss.

There are 4C2 ways of placing the 2 Ss, so if your arrangement is ACB, you get SASBC, SABSC, SABCS, ASBSC, ASBCS, ABSCS.

So, your 1 ACB arrangement becomes 6 arrangements with the Ss filled in.

Does that make sense?

[stable helm]

yes.. so i just have to multiply with other arrangement of ACB and get final result

[scarlet gale]

Yes.

3! arrangements before Ss are added in, and each one turns into 4C2 final ones after the Ss are filled in, so 3! × 4C2.

It's basically the idea of why you multiply.

Each of the 3! is worth 4C2 final arrangements.

[stable helm]

yeah. thank you so much! it makes sense now

[scarlet gale]

No problem.

[stable helm]

.close

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[obtuse pebble]

@bright plinth Has your question been resolved?

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[brisk heron]

Can someone help me

[obtuse pebble]

@brisk heron Has your question been resolved?

[brisk heron]

[obtuse pebble]

@brisk heron Has your question been resolved?

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[pure locust]

Hi! I'm trying to disprove the following by contradiction.
$$c\sqrt{n} \leq log_2^3(n), \forall n \geq n_0$$

[warm shale]

Phascolarctos

[pure locust]

My professor has told me that I cannot use limits or derivatives, but is this even possible? I've been trying to wrack my brain around this for several hours 😭

I've tried using properties such as $$log(n)<log^2(n)<log^3(n)$$

[warm shale]

Phascolarctos

[pure locust]

$$log(n)<n$$

[warm shale]

Phascolarctos

[brave bramble]

Well, the contradiction would be to assume that
c√n ≤ log2(n)³
Is true for all n ≥ n0, and prove something obviously false.

[pure locust]

I've been doing that, but I can't quite seem to get it to line up properly.

The log and that pesky cube are especially hard to manipulate

[brave bramble]

You can manipulate this into
2^(³√[c√n]) ≤ n
But this doesn't seem to provide anything obvious

What class is this? What theorems are you working with?

[pure locust]

I hadn't actually thought of doing that! I guess I just thought that cube-root would be difficult to manipulate, but maybe I'll try playing around with this. But off the bat, I would agree with you 😛
It is for a Data Structures & Algorithms course, in regards to proving whether a function f(n) is is O(g(n)), Omega(g(n)) or Theta(g(n)).

We don't really have many theorems to play with unfortunately. Just algebraic manipulation and proof techniques.

the bummer is that if you just write it as $$\frac{log_2^3(n)}{\sqrt{n}}$$ and take the limit as n goes to infinity, you get $c<0$, but the conditions of the problem state that $c>0$, so boom! Contradiction.

[warm shale]

Phascolarctos

[pure locust]

but apparently we can't do that 😦

Anyways, thanks for your help @brave bramble !

.close

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[low fossil]

I would like to ask part b plz

The p=k+1 of the induction

I used the assumption already

Now I want to conclude the rightmost term is less than or equals to RHS, with a proper reason

I wonder if n$\sum_{i=1}^n x_i^{k+1}$ $\sum_{i=1}^n x_i^{k+2} $ is less than or equal to $n^2$$(\sum_{i=1}^n x_i^{k+2} )^2$

[warm shale]

Trenton

[low fossil]

<@&286206848099549185>

.close

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[delicate flint]

Take a pack of playing cards and discard the picture cards, that is the jack, queen and king, of each suit. Shuffle the cards and deal out 10 of them. Calculate the mean score by adding together the 10 face values and dividing by 10. Put the 10 cards back in the pack and shuffle again. Deal out a further 10 cards and calculate the mean score again. Repeat this to give 100 observations, each of-which is the mean score of 10 cards. Calculate the mean and the variance of these scores. Compare these values with the ones you would expect from the theoretical results.

[obtuse pebble]

@delicate flint Has your question been resolved?

[delicate flint]

<@&286206848099549185>

[crimson berry]

so... it actually wants you to run the experiment huh

did you do it?

[obtuse pebble]

@delicate flint Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

[little lily]

.close

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[long root]

Would any one be able to help me out on question 5 please

It is a question about polynomials

<@&286206848099549185>

[crimson berry]

Are you deleting the ping and reporting it every few minutes?

Don't do that, just be patient

[long root]

Sorry

[obtuse pebble]

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[rough bough]

Shouldn’t it be 4 instead of the 2

This is my third time asking, please help, I have an examination tomorrow.

[warm shale]

Ansh

For (ii), $\alpha = 2e^{i \frac{2\pi}{3}}$ I suppose

[raven spire]

where do you it should be 4?

[raven spire]

$\implies (\alpha + \alpha^2)(\alpha - \alpha^2) =\alpha^2 - 8\alpha$

[warm shale]

Ansh

[rough bough]

@raven spire this is from finding the cube roots from part (a)

Which has a mod 2 for each of the cube roots

But then it squared itself so I believe it should be 4e^i(2pi/3)

[raven spire]

huh?

$\alpha= 2e^{i \frac{2\pi}{3}} \implies \alpha^2 = 4e^{i \frac{4\pi}{3}}$ no?

[warm shale]

Ansh

[rough bough]

Yeah

So this is wrong right?

@raven spire

[raven spire]

Sorry, yep it's wrong

actually

maybe the arg(alpha) in (0, pi/2) has sth to do here and maybe the alpha you got isn't correct?

Aaah got what's going on here lmao

$\alpha = 2e^{-\frac{2\pi i}{3}}$ here

[warm shale]

Ansh

[raven spire]

@rough bough

SO your equation would be: \ $x^2 - (4e^{-\frac{2\pi i}{3}}) + ( 4e^{-\frac{4\pi i}{3}} - e^{-\frac{2\pi i}{3}}) = 0$

[warm shale]

Ansh

[rough bough]

Okay thanks and there is another quite challenging problem

This is my working, wondering if it is correct or wrong?

Especially for part (I)

I mean ii

Part ii i think there exist two solutions

[raven spire]

honestly, there's 3

[rough bough]

What where is the third?

[raven spire]

oiii =_+ rotation is multiplication by a complex number? idkkk what you did for (i)

[rough bough]

What i did from (ii) is using the property of the lines joining from the origin to the circumference is double the angle from the angle created from the line from the circumference joined to a single point

[raven spire]

I thinkkk I get what you did with i. Although I don't get how you arrived at your conclusion 🤦‍♂️

As for (ii), there's RS parallel to PQ, PS parallel to QR and SQ parallel to PR

[rough bough]

Is my solution for part(ii) okay?

This is only 2 marks, and that’s too much work

[raven spire]

I think z2 + z3 - z1 only gives you one of the possible choices for S, so.. if your question's good with only one S, then okie

[rough bough]

So my answer does give one possible solution, or am I just stupid? And it doesn’t

[obtuse pebble]

@rough bough Has your question been resolved?

[raven spire]

yeah only one possible soln

consider z2 + z3 - z1, z2 + z1 - z3, z1 + z3 - z2

[obtuse pebble]

@rough bough Has your question been resolved?

[rough bough]

I asked someone from my house, and he said that it is PQRS and that it should be in the fourth quadrant

[obtuse pebble]

@rough bough Has your question been resolved?

[raven spire]

Yep, with that reasoning, it's restricted to only one solution

which is still not the one you found

your point lies still in the third quadrant

[crude cloud]

Rotation theorem will work here?

Final vec/ initial vec= length of final vec/ length of initial vec×e^itheta

Isn't that a mains question?

[raven spire]

they got the first part :o

the second part tho, they're confusing the quadrants

[crude cloud]

Quadrants? Wait a sec

[raven spire]

the choice of S for PQRS parallelogram. Boruto figured that S must be in 4th quad but not clear about why their computation spits out S in 3rd quad

[crude cloud]

Dude i think is u nedd to find out Z2 and z3

For 2nd part

[raven spire]

._. Isn't z_2 and z_3 already figured in (i) lol

[crude cloud]

Ok ig

Yeah it's given lol

It's there in question only 🤣

So covert it into Cartesian equation and solve the slopes of a parallelogram

[raven spire]

watttt? You can directly write the third point lmao

[raven spire]

@rough bough the 4th point will be $z_3 + (z_1 - z_2)$ and not $z_3 + (z_2 - z_1)$

[warm shale]

Ansh

[obtuse pebble]

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[tender nest]

What are the total number of 12 letter words that contain 3 'A's and 2'E's?

[brittle swan]

define 'word'

[crude cloud]

Dude we can make words from all the alphabet?

[brittle swan]

sufosumosdumfosdmf <- is this a word?

[crude cloud]

@brittle swan no need of meaning ig

[brittle swan]

i need to make sure by word he means 'any combination of letters'

[crude cloud]

My answer is 26c12/3!×2!

[sudden fox]

26c12 assumes that its all unique letters right?

i think its 26^7/3!2!

[crude cloud]

Why?

[tender nest]

all the words possible

with or without meaning

[brittle swan]

???

oh

[tender nest]

yeah exactly any combination of letters

[brittle swan]

you know combinations right?

computing n choose k

[tender nest]

yeah i do

[crude cloud]

@sudden fox in this case u r saying repitition is allowed for all alphabets

[brittle swan]

but you don't know how to approach this?

[agile owl]

What you all talking bout

[tender nest]

but im confused with how to calculate the question

[sudden fox]

wait is

zzzzzzzzzzzz

a possible word

assuming thats 12 Zs

[crude cloud]

But there's no 3a or 2 e?

[brittle swan]

so you're asking for the number of strings such that:

• contains 3 'A's
• contains 2 'E's
• is 12 in length

[tender nest]

like... i dont know if i should do permutation for those 5 letters or do permutation for all or do multiplhying with all

[tender nest]

yeah

[sudden fox]

that's the total

now you would want to then change that to yk

[crude cloud]

It' should contain 3a and 2e

[brittle swan]

try considering (conditions 1 and 3) and (conditions 2 and 3)

[sudden fox]

yes BUT im not doing the conditions yet

i wanna know

[crude cloud]

?so how can it be z 12 times?

[brittle swan]

and then combine them

[sudden fox]

how many possible words you can form then do complimentary counting

[crude cloud]

Nope

We have 12 places i think the answer is 3c1 2c1 24c10/2!×3!

[sudden fox]

actually

24^7 x [ways you can mess up 3As and 2Es]

so now second term is 5!/3!2!

[brittle swan]

lol

[crude cloud]

U r taking power 7 cuz a and e aren't same?

[brittle swan]

just consider the possible placements

of a and e

the rest of the spaces can be anything but a or e

[crude cloud]

Yeah but ita very long

[brittle swan]

wdym

[crude cloud]

Yeah I didn't think of it

Then it'll be 12c2×24c10?

@brittle swan uhm makes sense

Yeah I got it ig 12c3×9c2×24^7

[brittle swan]

dont spoil

[obtuse pebble]

@tender nest Has your question been resolved?

[tender nest]

is it (12C2) * (10C3) * (26)^7?

[crude cloud]

@tender nest I'm getting that it can be wrong

[tender nest]

yeah i got that too and a friend of mine got that answer as well

for some reason i have a bad feeeling about this answer

cuz the last time i used this method of "eliminating" the requirements it didnt end up correct

[crude cloud]

Oh ok

[tender nest]

but meh ig since the 3 of us got the same answer it must be correct

thanks :)

.close

[obtuse pebble]

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[opal swallow]

Hi, I need help with 8b

[opal swallow]

And 5a

[deft hazel]

bruh if

CO and OB are radii to the circle

then COB is an isosceles triangle

since they share the same chord

so find COD

then yeah

apply the alternate segment theorem for part a of q 5

i mean you've written it there

[obtuse pebble]

@opal swallow Has your question been resolved?

[timid silo]

OBC isn't 25 deg

[opal swallow]

Isn’t it isosceles tho

[timid silo]

Umm

look carefully

it is isosceles

[opal swallow]

There’s D

[deft hazel]

what

[opal swallow]

Wheres***

Oh

I WAS LOOKING AT THE WRONG DIAGRAM

[opal swallow]

Wait but it isn’t 25

I still can’t find OCB😭

[timid silo]

65 degree...

Connect BD

[opal swallow]

OHHH

Wait if we connect how did you get 75?

[timid silo]

Wait sorry should be 65

Double angle theorem

[raven spire]

BRUHHHHHHHHHHHHh

You're not even supposed to find OCB

Find BCD

Not BCO

🤦‍♂️

[timid silo]

emotional damage +1000

anyways

[opal swallow]

i

i-

[raven spire]

Also, isn't CAB just PBC? secant tangent theoreM?

[opal swallow]

https://tenor.com/view/kekw-kek-bttv-twitch-emote-gif-15123134

[raven spire]

.

[opal swallow]

i found 5a

[raven spire]

BCD is so much easy to find as well

Dew itt

[opal swallow]

WAIT HOW DO WE FIND BCD

MY BRAIN CANT DO CIRCLES

[raven spire]

2BCD = BODDDD

double ankle theorem

[deft hazel]

then after that you use the double knee theorem because what's a leg without a knee amirite i'm truly the funniest man alive

[opal swallow]

WAIT BUT WE DIDNT COVER THIS

IN CLASS

[deft hazel]

yes you cover analytical trigonometry a bit later

[raven spire]

welllllllll binary fission is covered in biology class soooooo obv you wouldn't have heard about double ankle theorem

[opal swallow]

wait its actually called double ankle

wait nvm

[raven spire]

||not like by double angle theorem we mean: angle subtended by the chord at the center is double the angle subtended at the corresponding arc||

||which was pretty visible when I wrote 2BCD = BOD||

[opal swallow]

oh.

oh

I CANT FIND T

WAIT I FOUND IT

ok tysm everyone

ganyu pfp helping me with all my hw

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[proper apex]

Can someone explain why the answer is the eighth root of 128?

[timid silo]

sqrt(x) = x^(1/2)

[proper apex]

yes

[timid silo]

So you just trasform all sqrts into power

[proper apex]

did you see the picture

it's literally what i did

[timid silo]

Hmmm I think your answer is correct??

I plugged it in a calculator

[proper apex]

bruh

[timid silo]

You can try the answer should be 8th root of 2

[proper apex]

💀

the book was very clear to say the answer is D

is it wrong then

https://media.discordapp.net/attachments/733739272217034898/859566012293447700/bimgus.gif

WAIT

HAHASSADGAJSGDJAWUTGYDIW/UARGKAWUYG

oh my god

sigh

it was like 2 am

im so dumb

jesus

[long sinew]

Bruh

[proper apex]

Solved it :)

uh thanks for the help lol

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[proper apex]

If this equals 4 -x, find the value of x.

[proper apex]

These are the possible answers.

[timid silo]

What was your thought process?

What did you do first?

[proper apex]

(a - b)^2 = a^2 - 2ab + b^

this

i don't really know what to do or if what i did was correct

[timid silo]

Good

[proper apex]

it's the second line in the picture lol

[timid silo]

It’s correct

Cool

You then combined the fractions

[proper apex]

Yes

oh wait the last step looks very similar to B

[timid silo]

Everything seems correct

[proper apex]

o

[timid silo]

Yes

[proper apex]

O

i think i got it

[timid silo]

In fact they are the same

[proper apex]

so you take 10 from 60, leaving 6, and multiply it by the square root of 1/10

so it turns into 100 * 1/10

10

[timid silo]

Yes

[proper apex]

https://tenor.com/view/halo-halo-infinite-the-weapon-cortana-ai-gif-24406695

[proper apex]

ignore the dialogue

LOL

thank you very much :)

[timid silo]

Np

[proper apex]

.close

[obtuse pebble]

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[proper apex]

hold up

[brittle swan]

?

[proper apex]

uhhhh

so the last two steps

shouldn't it be 7/8 * 1/2?

[brittle swan]

which to which

[proper apex]

do you just replace whatever root there is with the new denominator?

[brittle swan]

2nd last to last?

[proper apex]

yes

[brittle swan]

√ = ^(1/2)

do you know that

[proper apex]

of course

look

[brittle swan]

alright

[proper apex]

this is what's confusing me

[brittle swan]

they wrote √(2^(7/8))

which isn’t right

[proper apex]

huh

i mean

that's what i did

[proper apex]

here

i was told it was right

and my book says the answer is 8√128

[proper apex]

oh uh

[nocturne minnow]

If you were adding exponents for $2^1 \cdot 2^{\frac{3}{4}}$

That's not 7/8

[proper apex]

,

shittttt

wait

where is there addition in the picture

[warm shale]

dldh06

[proper apex]

isn't that 2^(7/8)

[nocturne minnow]

No

1+3/4 doesn't equal 7/8

[proper apex]

O

OH

WHAT

ITS

7/4

[brittle swan]

lol

[proper apex]

https://media.discordapp.net/attachments/778043543515037776/861306390861316136/wdwdad.gif

what's wrong with me T-T

[proper apex]

so is the concept behind this correct?

it's just that

i should've gotten 4

and then multiplied it by 2

to get 8

[nocturne minnow]

Conceptually it's right

[proper apex]

oh good

[nocturne minnow]

Mathematically wrong

[proper apex]

yeah

so like this?

[nocturne minnow]

Yeah

[proper apex]

ok nice

i should really get more sleep

[nocturne minnow]

,w sqrt(2sqrt(2sqrt(2)))

[warm shale]

Results provided by WolframAlpha

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Requested by dldh06#2659

[brittle swan]

we should all get sleep before math

last night at 2:30 i struggled finding some infantile mistake

[nocturne minnow]

,w 2^(7/8)

[warm shale]

Results provided by WolframAlpha

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Requested by dldh06#2659

[nocturne minnow]

Math checks

[proper apex]

lol same here

[proper apex]

neat

thanks guys

imma close this now

https://media.discordapp.net/attachments/733739272217034898/859566012293447700/bimgus.gif

.close

[obtuse pebble]

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[trim island]

Why is e^2x differential shown like this on a graph

[trim island]

And why doesnt this disprove that the differential of e^2x is not 2e^2x

[obtuse pebble]

@trim island Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[trim island]

.reopen

[obtuse pebble]

✅

[timid silo]

temporary

What is your issue with the graph?

[trim island]

they do not touch at the minimum/maximum points

[timid silo]

there is no maximumm

and no minimum either actually, only an asymptote

also why does the derivative have to have this property?

[trim island]

ah i see

[timid silo]

the differential is the rate of change at a point

[trim island]

because isnt the function of a derivative to find the min/maxi

[trim island]

ah okay

[timid silo]

it can help, yes.