help-23

[trail fjord]

Could someone help me figure the steps out and explain why it’s done like that step by step pls

[dapper grail]

What do you mean by removing the parenthesis?

If you meant that as in distributing the coefficient then yes, you would be correct

[trail fjord]

But what happens to the 6?

[dapper grail]

im assuming this is the question?

it's kind of blurry

[trail fjord]

Yup my bad

I think we have to do something to the 6 first, no?

[dapper grail]

No, you dont have to do anything with that yet

I mean you could depends on how you want to approach the problem

You could multiply everything by 6 do get rid of that denominator if you wish

Or

You could distrbute the fraction first

[trail fjord]

what’s easier

I mean Ik I have to do them both eventually

[dapper grail]

I believe that multiplying the entire thing by 6 would help you out

[quasi bison]

definitely multiplying boht sides by 6 is easier

[trail fjord]

yea that’s what I was thinking

[quasi bison]

if only because it leaves less room for error

[dapper grail]

[trail fjord]

So it’d be 5(x-4)=3(x-2) right

[dapper grail]

[trail fjord]

ah alr Ik what to do from here thanks

.close

[safe radish]

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[unborn wigeon]

can someone please explain this claim?

why do we need the exopntent of 2 and p-1 to be coprime?

<@&286206848099549185>

[fiery skiff]

if $\gcd(n,138) > 1$, then that means $2^n$ will "skip over" at least half the terms/some terms

[flat frigate]

Dee3Cay

[fiery skiff]

primitive root means when iterating along it's power (power 1, 2,3 so on), it should go through all values modulo 139 (except 0) once

this is just a rough idea but

[safe radish]

@unborn wigeon Has your question been resolved?

[safe radish]

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[polar grove]

Can anyone help me with this

[polar grove]

Shorten this as much as possible

[frigid locust]

a+a = 2a
b+b = 2b

(2a)^3 . (2b)^3 = ??

[polar grove]

So this is the answer?

[safe radish]

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[grim plover]

$$f(x) = \begin{cases}
x+1&x \text{ is odd}\
3x&x \text{ is even}\
\end{cases}$$

[flat frigate]

ItzKraken

[grim plover]

let $a_{i+1} = f(a_i)$

[flat frigate]

ItzKraken

[grim plover]

could someone guide me how to find the solution to the recurrence relation?

I think I found something similar (https://math.stackexchange.com/questions/3720790/solving-the-piecewise-recurrence-f-n-f-n-1f-n-2-for-f-n-1-even-and) but I have no idea what they does after that O thing. Is it big O notation? If yes how is that useful in this context? If no, does it refer to parity?

[devout shale]

Collatz’s disabled cousin

[grim plover]

what gave it away?

[quasi bison]

the 3x+1 shit

[grim plover]

I only wrote 3x

well in anycase

do you guys have any links to resources?

[grim plover]

he also does some matrix crap at the end, no idea what that is

[quasi bison]

this looks kinda ugly ngl

[grim plover]

wdym

[quasi bison]

i think your recurrence is gonna be hard if not impossible to obtain an explicit formula for

or wait hold on.

[grim plover]

okay

[quasi bison]

the sequence never has an odd number after an even

so if a1 is even only the even condition will ever trigger and it's just a geometric seq

if a1 is odd it'll be geometric starting from the 2nd term onwards instead

[grim plover]

what?

lemme process that

[quasi bison]

f(x) is even nmw

[quiet juniper]

yeah I thought I was tripping cus it looked that simple to me

[grim plover]

wdym, we have 3 after 2 no?

[quasi bison]

no we don't

2 is even

2 becomes 6

[grim plover]

oh i forgot to mention, let $a_0 = 900$

[flat frigate]

ItzKraken

[peak estuary]

can you please calculate a1,a2 and a3 for us

[grim plover]

aight

a1 would be 2700, a2 would be 8100, a3 would be 24300

yeah if u start with even u never got odd

[peak estuary]

yes

what if you started with a0=899?

aka odd?

[quasi bison]

!xyproblem

[grim plover]

wdym?

[grim plover]

hmm well then it becomes even and stays even

[peak estuary]

are you sure this is the problem you want to ask about or is this some sort of intermediate step in a different problem you actually want to solve

[quasi bison]

http://xyproblem.info/

[peak estuary]

i thought we had a command for that

!xy

hmm nope

[grim plover]

well no, am just asking how to solve a piecewise recurrence relation in general

[devout shale]

I’d be surprised if there is a general method

[grim plover]

well if $a_0$ is even,
then this becomes $a_{i+1} = a_i + 1$

[peak estuary]

if we knew then we could solve collatz

[flat frigate]

ItzKraken

[grim plover]

...

[devout shale]

It’s all collatz

Always has been

[grim plover]

ah okay

[grim plover]

can u explain atleast what they did in this MSE thread?

[peak estuary]

at a quick glance, they notice that the parities repeat

[toxic stratus]

they analysed the recurrence and realised that it would always be odd, odd, even, and thereby the piecewise could be turned into not piecewise

[peak estuary]

so they consider one such period as "one step"

which can be nicely expressed using linear algebra

[grim plover]

could I have realized this fact and solved it using some other method? like characteristic root technique or smthing

[toxic stratus]

same strategy that was applied for your dodgy recurrence relation

[grim plover]

what if we dont find such a pattern? then it becomes a collatz-like problem?

[peak estuary]

at least it immediately becomes much harder

[grim plover]

okay

[safe radish]

@grim plover Has your question been resolved?

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[fleet pine]

can someone explain this step?

[fleet pine]

i think cross multiplying, but idk how you get to that answer

[lean otter]

@fleet pine

[fleet pine]

oh thank you!

.close

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[short sparrow]

Consider the function $f:\mathbb{N} \longrightarrow \mathbb{N}$ with
$$
f(x) = \begin{cases}
3(x-1) & \text{ if }x\text{ is odd,}\
\frac{x}{2} & \text{ if }x\text{ is even but not divisible by 6,}\
x+3 & \text{ if }x\text{ is divisible by 6.}
\end{cases}
$$
Examine $f$ on injectivity, surjectivity, and bijectivity.

[flat frigate]

Levens

[short sparrow]

now i know the definitions of the properties, but how do i examine the function and find out if it is injective/surjective? Do I split it up?

[mortal ingot]

Well, you know the def of e.g. injectivity?

[short sparrow]

yeah, f is injective if x =y follows from f(x) = f(y)

And vice versa

[mortal ingot]

Vice versa? That won't be needed

[short sparrow]

but its true lol

[mortal ingot]

Well yea

[raven heart]

it's true, it's just what it means to be a function

[short sparrow]

yeah

[mortal ingot]

Anyway so assume f(x) = f(y)

And then see if you can get x = y

[short sparrow]

but how can i do that if there are 3 cases for f

[mortal ingot]

Case-by-case 💀

[short sparrow]

hey no mean skull

should i write 2y+1 for uneven x?

[mortal ingot]

You may, also you may not

Perhaps Try yourself to find out

[short sparrow]

probably wouldnt matter right

[split ether]

It can be helpful

[short sparrow]

ok ill give it a go

[split ether]

First show that in each case the remainder of output when divided by 6 is different

[short sparrow]

i dont get it

[split ether]

If x is odd, what's the remainder of 3(x - 1) when divided by 6? (Use the fact that there exists an integer y with x = 2y + 1)

[short sparrow]

oh the remainder is y

[mortal ingot]

Huh

[split ether]

You mean the quotient?

[short sparrow]

sry what does remainder mean english isnt my first language lmao

[mortal ingot]

Ah, I see

[split ether]

E.g., remainder of 57 when divided by 25 is 7 because 57 = 2 * 25 + 7

[short sparrow]

oh

then there wouldnt be a remainder?

0?

[split ether]

Yes

Now consider the other cases

[short sparrow]

how can i present x as a number thats even but not divisble by 6..

[mortal ingot]

You do not need to present it

[quasi bison]

what is your first language

[short sparrow]

for x/2 the output would be y again if we say x=2y... but how would that help us

[short sparrow]

german

[quasi bison]

Teilungsrest

[short sparrow]

yeah figured.. thanks!

[split ether]

First, x is even, so x = 2y for some integer, now you need y not to be divisible by 3, so let y = 3k + 1 or y = 3k + 2 for some k

[short sparrow]

ah yeah cool

but whats the mission here.. how does it help

[quasi bison]

@split ether do you think you might be overcomplicating stuff

[split ether]

Don't know, this is how I'd do it

Once we show that the remainder is different in each case, we will be able to say that if f(x) = f(y), then we have either 3(x - 1) = 3(y - 1), x/2 = y/2 or x + 3 = y + 3

[quasi bison]

6k+0 -> 6k+3
6k+1 -> 18k
6k+2 -> 3k+1
6k+3 -> 18k+6
6k+4 -> 3k+2
6k+5 -> 18k+12

is how i would do it

[split ether]

Ah, fair enough

That's better yeah

[short sparrow]

ooooo

cool

[split ether]

And just show that in each of the situations you have x = y

[quasi bison]

btw does N start at 1 or at 0

[short sparrow]

but then youd have to do the first 6 natural numbers manually right

[short sparrow]

we start at 0

[quasi bison]

right yeah you would need to check the low ends first

so yeah f(0) through f(5) manually

[short sparrow]

i mean with this it also show surjectivity right, since they all have an output

thats in N

[quasi bison]

not quite no

[short sparrow]

oh yeah oops nvm

[quasi bison]

you can show INjectivity by looking at what residue classes mod 18 are covered by each case

[short sparrow]

surjectivity would by that for all n in N there exists a x in N with f(x) = n

[quasi bison]

yes

and that is why you need to check the first fivee

so that you know which output is the lowest possible for each case

[short sparrow]

?

[quasi bison]

f(0) = 3, and this is the lowest output that you can get from f(6k+0)

likewise the lowest output possible from f(6k+1) is f(1) = 18

[mortal ingot]

Well, that is assuming the function is increasing within each residue class (which it is).

[quasi bison]

and so on for the other cases

[quasi bison]

the restriction thereof to any residue class mod 6 obviously is.

and you can use the formulas i wrote down to prove it. it is not hard.

[mortal ingot]

(But Levens, did you learn some about residue class?)

[short sparrow]

like e.g. for 6, the numbers 11 and 17 would be in the residue class 5?

[mortal ingot]

Ah I was worried you could be confused, don't mind it

[short sparrow]

you might be onto something though, cause i never really used them before

i just know them from seeing them in the internet here and there

[mortal ingot]

Yea, you can partition the natural numbers into residue classes - that is all you need to know methinks

[short sparrow]

but why would it be important to know the lowest output for each case?

that kind of confuses me

[quasi bison]

so that you know the lowest output of f globally

and if it isn't 0 you know it isn't surjective

(because f will miss 0)

[short sparrow]

okay but there is an output thats 0

[quasi bison]

is there?

[short sparrow]

yeah for x=1

[quasi bison]

ah

ok yeah you're right

hm

still though

would be good to map out all of its outputs

actually maybe we'll prove it is surjective that way

[short sparrow]

well
f(0) = 3
f(1) = 0
f(2) = 1
f(3) = 6
f(4) = 2
f(5) = 12
...and then your list that generalises the further terms:

6k+0 -> 6k+3
6k+1 -> 18k
6k+2 -> 3k+1
6k+3 -> 18k+6
6k+4 -> 3k+2
6k+5 -> 18k+12

[safe radish]

@short sparrow Has your question been resolved?

[mortal ingot]

Hmm, are you still having problems?

I thought you solved it

[short sparrow]

yeah i kind of still dont get how it shows injectivity/surjectivity

esp surjectivity

[mortal ingot]

Injectivity as well?

What part are you having problem with regarding injectivity?

[short sparrow]

do i show f(x) = f(y) ==> x=y for all 12 cases? like idk.. somethings missing in my head lol

and with all 12 cases i mean f(0),...,f(5) and then 6k+0,...,6k+5

[mortal ingot]

Well you can simplify it a bit.

If you assume f(x) = f(y), the two should be in the same residue class, right?

[short sparrow]

yes

wait..

oh what you were trying to say before?

[mortal ingot]

Wdym "trying to say"?

[mortal ingot]

Anyway, assuming you understood this, you would see that you may compare the residue classes first.

[short sparrow]

hm.. ok then what about surjectivity

[mortal ingot]

For surjectivity, you first consider partition of natural numbers into residue classes.

Then, for each residue class, you can see what kind of x you need for f(x) to be in the class

[short sparrow]

basically this right

6k+0 -> 6k+3
6k+1 -> 18k
6k+2 -> 3k+1
6k+3 -> 18k+6
6k+4 -> 3k+2
6k+5 -> 18k+12

[mortal ingot]

Well, the reverse

Consider residue class of y = f(x).

[safe radish]

@short sparrow Has your question been resolved?

[safe radish]

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[velvet hearth]

im going through a proof

[velvet hearth]

kind of confused tbh

i follow it up to the since part

the implications should be the other way round no?>

and then the next part just seems totally fucked

<@&286206848099549185>

this is the whole proof

[safe radish]

@velvet hearth Has your question been resolved?

[safe radish]

@velvet hearth Has your question been resolved?

[safe radish]

@velvet hearth Has your question been resolved?

[safe radish]

@velvet hearth Has your question been resolved?

[cold relic]

Who gave you this proof?

$gcd(6r+1, 6^2r^2+5)=6$

[flat frigate]

FirstNameLastName

[cold relic]

This is just straight up false

6 doesn’t divide 6r+1

Whoever wrote this did not make sure it’s error free, and I therefore wouldn’t assume that no other errors were made

That aside though it seems the proof itself is correct

@velvet hearth I think your channel is gonna be closed soon btw

Yeah that proof has a lot of errors and unproven assumptions baked into it

To fulfill the proof you should be aware that gcd can neither be 2 nor 3, because they can’t divide 6r+1

[safe radish]

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[crisp gyro]

I got the correct answer but how do I put my answer in the form they're asking in?

[junior smelt]

Multiply this out by 2x^2 + 2

[crisp gyro]

Is the RHS = 0 tho?

[crisp gyro]

Is it = to 0 or 2x^2 +2?

[junior smelt]

You had on the top line that $\frac{8x^3 + 2x^2 + 5}{2x^2 + 2} = \frac{-8x + 3}{2x^2 + 2} + 4x + 1$, and the suggestion to multiply both sides by $2x^2 + 2$ changes your right hand side to be what you wrote (but with appropriate bracketing around the $2x^2 + 2$)

[flat frigate]

@junior smelt

[junior smelt]

Then the left hand side but becomes 8x^3 + 2x^2 + 5

[crisp gyro]

Tysm

[pulsar grail]

basically find natural number n so that...

[safe radish]

@crisp gyro Has your question been resolved?

[safe radish]

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[shrewd topaz]

i keep getting this wrong and idk why

(x-3+2i)(x-3-2i)(x^2+2x+1)

(x^2+6x+9)(4)
(4x^2+24x+36)(x^2+2x+1)

4x^4+8x^3+4x^4+24x^3+48x^2+24x+36x^2+72x+36

[final halo]

where did that (4) come from?

[shrewd topaz]

2i * -2i

-4i^2

-4(-1)

[final halo]

yeah but it wouldnt be multiplying the whole polynomial

[shrewd topaz]

i have two x- 3 so i did (x-3)^2

and -2i and 2i left over

[glacial cairn]

Where did the x^2+2x+1 come from

[shrewd topaz]

(x+1)^2

[glacial cairn]

If you want a zero of 1, shouldn't it be (x-1)^2?

[shrewd topaz]

i did the perfect square thing

[shrewd topaz]

oh u right

i guess that messed with everything?

[final halo]

are you able to expand (x-a)(x-b)? if so then do that and then plug in 3+2i for a and 3-2i for b

[shrewd topaz]

(x-3+2i)(x-3-2i)
this too maybe? (x-3)^2 (4) made sense to me

[final halo]

you cant split it up like you're trying to do

its not (x-3)2i*(x-3)(-2i)

[shrewd topaz]

well then what is it

[final halo]

its this

like its just (x-something)(x-something else) and you know how to expand that

[shrewd topaz]

so this (x-3+2i)(x-3-2i)

or (x-(3-2i))(x-(3+2i))

[final halo]

same thing bc theyre conjugate

[shrewd topaz]

well is there a shorter way to do that

can you answer this question in 3 minutes?

takes me forever

[final halo]

well yeah but only because i've done this kind of stuff alot for a while

[shrewd topaz]

damn so skill issue i guess

ight well thanks

.close

[safe radish]

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[worn scroll]

Hi

[worn scroll]

B23

I know how to get the range.

Not sure about basis.

[hearty egret]

do you want a basis for the range ?

[worn scroll]

I am not sure how they got the basis Null(L)

everything else i get

[hearty egret]

so if the range has dimension 2 ..then you have only to search one vector that goes to zero right ?

?

[worn scroll]

uh

no im not sure about that

how would the matrix look

thats all i need to know

augmented

[hearty egret]

do you know that dim ker+ dim Im= dim vector space=3 (for your case)

[worn scroll]

no

i know dim

i just need the matrix really

bc its straight forward from there and i can see how u got it just by looking at it

[hearty egret]

you cannot everytime compute the matrix

[worn scroll]

how did they get Null(L) is [2;-1;1]?

[hearty egret]

it takes long..

[worn scroll]

could u show the calculations

[hearty egret]

can you find a vector that goes to zero ?

[worn scroll]

wait

so the matrix would look like this

k wiat

[worn scroll]

: 1 2 0 0
0 1 1 0
-1 0 2 0
1 2 0 0

RREF:
1 0 -2 0
0 1 1 0
0 0 0 0
0 0 0 0

[hearty egret]

mh...

this is the matrix that represent the matrix ?

[worn scroll]

yea

x1 + 2x2 −x1 + 2x3
x2 + x3 x1 + 2x2

[hearty egret]

its impossible ...why there is 4 column ?

[raven heart]

yeah the transformation goes from R^3 to (essentially) R^4

[hearty egret]

the starting space is R^3

[worn scroll]

thats what im asking

how do i put this into a matrix

[raven heart]

so you found that matrix yourself?

[hearty egret]

delete last column (?)

[worn scroll]

yea

[raven heart]

it's an augmented matrix

to solve L(x) = 0

[worn scroll]

yea

[raven heart]

yea seems fine

[worn scroll]

so

: 1 2 0 0
0 1 1 0
-1 0 2 0

[hearty egret]

however its right the 4x3 matrix inside your 4x4 matrix

the first one

[worn scroll]

bro...

@hearty egret im sorry but i cannot uinderstand u..

[hearty egret]

this matrix represents your linear function

[worn scroll]

i got the answer already

all i needed was the matrix but thanks

.close

[safe radish]

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[hearty egret]

you can go faster without matrix computation

[safe radish]

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[granite cape]

Hey guys could someone help me out with this problem

[granite cape]

I know I have to long divide, although the remainder I get is 2x is this correct?

[acoustic goblet]

Do a polynomial division, youll be left with only easy to integrate addends

[granite cape]

Okay I've just finished that question, although I have one more question.

For integration word problems such as this:

How do I know whether to seperate variables and integrate or otherwise?

These two questions,

How would the approach to each question differ?

[safe radish]

@granite cape Has your question been resolved?

[granite cape]

<@&286206848099549185>

[safe radish]

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[unreal kindle]

hello

[unreal kindle]

how do i know the sign of cos(2nx) - sin(2nx)

when $n \in \mathbb{N}$

[final halo]

what is n?

[unreal kindle]

and x is a positive real number

[obsidian oracle]

You can show that cos(2nx) - sin(2nx) = sqrt(2)cos(2nx + pi/4)

[flat frigate]

lilisworld

[obsidian oracle]

then the sign of that will only depend on the sign of cos(2nx + pi/4)

[unreal kindle]

ok but actually i don't think that's how i'm supposed to do sorry but i don't think i need to find the derivative

huh

that is the og question

[obsidian oracle]

and you're tasked to find?

[unreal kindle]

is fn uniformly convergent on R+

and i tried to find the sign of the derivative

the cos sin thing

[obsidian oracle]

you know the pointwise limit right?

[unreal kindle]

it's 0?

[obsidian oracle]

yes

so to show uniform convergence

you need to find ||fn||

[unreal kindle]

i usually differentiate

to find the sup

[obsidian oracle]

yes, but the dif gives ne^(-nx)(2cos(2nx) - sin(2nx))

you only need to find when the derivative = 0

those will be your extrema

[unreal kindle]

ok

but i don't know when it's 0

uhm

[unreal kindle]

how do i solve derivative=0?

[obsidian oracle]

tan(2nx) = 1/2

what interests us is the first extrema, because we know THIS will be the biggest fn will ever get

so we're solving on 2nx in [0,pi/2[ and so x = arctan(1/2)/(2n)

so yes requires a bit of proof but this has to do with the fact that e^(-nx) is decreasing and sin(2nx) is periodic and bounded by 1, so the max is the first extrema we ever cross

[unreal kindle]

is there another way that doesnt involve arctan and all of those things?

[obsidian oracle]

yes, you can bound fn by another, easier function which will be easier to bound

[unreal kindle]

ok can you give me a hint

[obsidian oracle]

but here I don't think it's possible

as I told you I think fn cannot be well bounded here

[obsidian oracle]

if you compute fn(x) for this x

you will find a constant

So norm of fn is constant and doesn't converge to 0

Another way is simply to find a sequence xn such that fn(xn) never converges to 0

the sequence xn = 1/n easily settles it

fn(xn) = sin(2)/e

[unreal kindle]

oh yes

ok thanks

.close

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[pulsar vine]

How do you calculate the last one?

[pulsar vine]

It should be 2. something

[frigid prairie]

Can you use a graphing calculator for this?

[pulsar vine]

?

I think so

[glacial cairn]

The only relevant part is x - 35/x^2 + 5 = 0, right?

[pulsar vine]

Yes that is the derrivative

Just need to find where the derrivative is 0 using a calulator

[frigid prairie]

Well in the text above it says you should put the derivative equation into a graphical calculator and then find where Cbar’(x)=0

So where the line intersects the x-axis.

I have no idea what kind of graphical calculator you use, but there might be a way to press a ‘root’ button or an ‘xcal’ button once in the graph screen. This can find the exact value for which the derivative is 0

[pulsar vine]

Ohh I figured it out

You just do 0=x-35/x^2+5

[frigid prairie]

That can work too

[pulsar vine]

Okay thanks

[frigid prairie]

Anyways the answer should be 5.9 something

[pulsar vine]

no it was 2.204

[glacial cairn]

Yes it's 2.204...

[frigid prairie]

Yep you’re right, sorry I mistook a plus for a minus

[pulsar vine]

.close

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[latent wedge]

how do i find two square roots of z?

[safe radish]

@latent wedge Has your question been resolved?

[latent wedge]

<@&286206848099549185>

[safe radish]

@latent wedge Has your question been resolved?

[fair hound]

if w is a square root of z, w² = z
so |w| = sqrt(|z|)
and for the arg, you have two choices since there are 2 roots, but both choices should lead to 2arg(w) = arg(z)+2kpi

here arg(z) = -3pi/4 and you can write it 5pi/4 too
so for arg(w) you can choose 5pi/8 or -3pi/8 which is the same arg as 13pi/8

[latent wedge]

wouldn't 5pi/4 not be part of the domain though?

[fair hound]

5pi/4 is between 0 and 2pi

that would be the main arg for z

[latent wedge]

ohhhhh

is that the arg and not the Arg

i kinda see that

[latent wedge]

is this the formula?

[fair hound]

yes

[latent wedge]

👍

[fair hound]

you can also solve it algebraically

[latent wedge]

which way is easier?

[fair hound]

the formula is more direct, it's linked to the exponential form which makes it really easy to understand

[latent wedge]

n is

n is the amount of square roots we want to find?

i said that wrong

n is the amount of roots we want to find?

is that right? im not sure

[fair hound]

when you want to find the nth roots of z

it's equivalent to solving w^n = z for w

which always yield n solutions since w^n-z is a polynomial of degree n in w

then the formula you posted gives the n solutions

[latent wedge]

im just not sure if i have all the values to plug in to the formula

do i?

i have z

r

theta

[fair hound]

z is the number you want to find the roots of

r is |z|

theta is arg(z)

[latent wedge]

oh shoot

i thought z would be

-1 -i

no?

[fair hound]

yes

[latent wedge]

okay

i got r

yeah i can plug those in right

and the only value i wouldn't have is

what, 2?

or whoops

n?

[fair hound]

n is 2

since you want to find square roots

[latent wedge]

because you want to find two

gotcha

[latent wedge]

so

n = 2
z = -1 -i
r = sqrt(2)
theta = -3pi/4

did i do that right

and i can just plug those in?

[fair hound]

yes

[latent wedge]

1 sec

sqrt(sqrt(2)) -> 4sqrt(2)

[fair hound]

note that in the case of square roots
if w² = z
then (-w)² = w² = z
so the formula is just telling you that there are two roots with one that is minus the other

the root for k = 1 is minus the root for k = 0

[latent wedge]

would i need to change anything if i were to plug anything in because of this?

[fair hound]

no

[latent wedge]

what would k be again? and are you saying in this instance it's 0?

[fair hound]

k is a thing that goes through [0, n-1] to give you all n roots

but since we have n = 2 bc we're talking about square roots

there are only two roots

one which is given when k = 0

and the other when k = 1

[latent wedge]

ohhhh

[fair hound]

with the k = 1 one being minus the k = 0 one

[latent wedge]

im pretty lost on that, so id plug in what for k?

sorry

[fair hound]

k is a range of values

you have a root when k = 0

and one when k = 1

[latent wedge]

so i write the formula twice, one for when k = 0 and one for when k = 1

[fair hound]

yes

[latent wedge]

got it

[fair hound]

alternatively, the roots are the one with k = 0, and minus this one

but in a general case, it's easier to think efficiently more than trying to remember the formula
you have z = -1-i = sqrt(2)(cos(-3pi/4)+isin(-3pi/4))
so one root is sqrt(sqrt(2))(cos(-3pi/8)+isin(-3pi/8))

[latent wedge]

im just a little confused on that so are you able to dumb that down and help me easily get used to understanding it more

[fair hound]

and the other is -sqrt(sqrt(2))(cos(-3pi/8)+isin(-3pi/8))

when you have a degree 2 polynomial like

x²-2

and you want to find roots

you solve x² = 2

the solutions are the two square roots of 2

one is sqrt(2)

the other is -sqrt(2)

bc (-x)² = x²

here, the same thing happens for w² = -1-i

there is a first root, sqrt(sqrt(2))(cos(-3pi/8)+isin(-3pi/8))

[lean otter]

For f(x)=1/(x+3) , g(x)=1/(x-9) the domain of f(g(x)) would be (-inf,-3)U(-3,9)U(9,inf) right?

[fair hound]

so the other is just -sqrt(sqrt(2))(cos(-3pi/8)+isin(-3pi/8))

[latent wedge]

!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).

[latent wedge]

sqrt(sqrt(2)) is just 4sqrt(2)

but i just wanted to plug them in to see if i got those right

[fair hound]

you can just simplify in the cos and sin now

[latent wedge]

-1/sqrt(2)

both cos and sin

[fair hound]

huh

I don't really know what you mean by that

[latent wedge]

is it not

like

[fair hound]

in each cos and sin there is a fraction

[latent wedge]

special triangles

[fair hound]

with a sum in numerator

[latent wedge]

would that not be like

[fair hound]

just simplify it

[latent wedge]

i made a special triangle with -3pi/4 as the angle and got -1/sqrt(2) where did i go wrong

[fair hound]

you went wrong bc there is no cos(-3pi/4) anywhere

cos((-3pi/4)/2) and cos(-3pi/4) are not the same thing

in each cos and sine you have fractions

simplify that

[latent wedge]

oh shit whoops

cos(-3pi/4/2)

id multiply by 2 right?

would i multiply the two in the front of cos or the numerator inside the cos

[fair hound]

(-3pi/4)/2 = -3pi/8

[latent wedge]

thanks

where would i go from there though?

given that i cant use like calculators or anything

[fair hound]

to go further, you need to know cos and sin for the known angles

then you can use double angle formulas for example

so that you can deduce what cos and sin of -3pi/8 are

[latent wedge]

sin/cos of -3pi/8

i have no idea what that owuld be

[fair hound]

cos(-3pi/8) is sqrt((2-sqrt(2))/4)

[latent wedge]

can u get the bot to visualize

[fair hound]

$cos(\frac{-3\pi}{8}) = \sqrt{\frac{2-\sqrt{2}}{4}}$

[latent wedge]

ty

[flat frigate]

Mélo

[latent wedge]

im curious to know if there's a way to get that w/o a calculator easily?

as well as with sin

[fair hound]

I told you, double angle formula

[latent wedge]

my bad i didn't notice

was this what you used?

[fair hound]

yes

and it's a + bc theta is between -pi/2 and pi/2

[latent wedge]

my friend used some sort of exponential form on this question

i don't know if that would be easier or not than needing to do this

[fair hound]

for complex numbers, we usually say there are 3 forms

the algebraic one, you're trying to find

the trigonometric one, which you've already found

and the exponential one, which is essentially the same as trig one, bc e^(itheta) = cos(theta)+isin(theta)

[safe radish]

@latent wedge Has your question been resolved?

[safe radish]

@latent wedge Has your question been resolved?

[safe radish]

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[young flume]

I don't get what they do after getting y=x

I was thinking we could rather apply triangle law of vector addition

But for that I don't know what to put in y=x

<@&286206848099549185>

[safe radish]

@young flume Has your question been resolved?

[young flume]

<@&286206848099549185> 😭

[radiant bough]

What

[young flume]

.

Do you have a clue?

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[fathom lake]

i need help in 2.

we have x,y,z in [0;π] and x+y+z=π

we might use complex numbers

[young flume]

.reopen

[fathom lake]

<@&286206848099549185>

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@fathom lake Has your question been resolved?

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[fathom lake]

i need help in 2.
we have x,y,z in [0;π] and x+y+z=π
we might use complex numbers

[tiny wraith]

Look at S as a quadratic equation in terms of cos(x)

Where its second coefficient is between -1 and 1

[fathom lake]

ok

yeah it worked

thx

.close

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[tardy copper]

if a tasks says "if y is true then prove that x is true", can i do the counter and assume that x is true and prove that y is true from it?

[dull sequoia]

you want to prove the implication "if y is true then x is true"

an equivalent statement is "if x is not true then y is not true"

proving that "if x is true then y is true" is the converse and not the same statement

[tardy copper]

ah okay i get it

[dull sequoia]

example:
If it rains, then the floor is wet
you may not be able to show that if the floor is wet then it rained

[tardy copper]

got it

[dull sequoia]

there could be many other sources to make the floor wet

[tardy copper]

thnx

.close

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[mighty wadi]

What am I doing wrong here

[main mural]

x cannot be 0

[dapper venture]

also x might be negative

[main mural]

and that

consider 2 cases, x>0 and x<0

and combine them afterwards

[mighty wadi]

this is the part where get confused

[dull sequoia]

well you didn't even start with the question

[main mural]

you forgot to consider the 1/x < 0 part

which is only true for negative numbers, thus x<0

[mighty wadi]

but how i see i got the inequality in the wrong way

[mighty wadi]

Yeah, how I saw they do this, but why?

[main mural]

step 1->step 2 is wrong

because x<0

[mighty wadi]

I cant multiply both sides with X?

[dapper venture]

you know when multiplying/dividing negative numbers, we need to flip the inequality

[main mural]

if you multiply by it on both sides, you have to flip the inequality sign

[dapper venture]

and it is also true when you multiply/divide x

[main mural]

1 < 2 but multiplying by -3 gives -3 > -6

[mighty wadi]

Ahh and because x is a negative number in 1/x < 0 i should also flip it?

[dapper venture]

exactly

[main mural]

mhm

[mighty wadi]

that makes sence, thanks

oh and that why we consider 2 cases when x is involved

[main mural]

yes

[main mural]

but you don't have to because this is what your problem is saying

1/x < 0

[mighty wadi]

Yeah, I understand

Thanks

.close

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[trail lintel]

this might be a stretch but would anyone know how to find the equilibrium points of a non linear system and linerise it

[trail lintel]

https://cdn.discordapp.com/attachments/760227128020041818/1165850584667144302/image.png?ex=654859f7&is=6535e4f7&hm=57d7ee02b9fb3835c36d288cb24786046ca334403b3ee620b2369d82ddeca239&
https://cdn.discordapp.com/attachments/760227128020041818/1165850629340672081/image.png?ex=65485a02&is=6535e502&hm=630f14f40b0def8d3ef0b246e05e07fbaa253c768efcbe263bcff4fac9ad1158&

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@trail lintel Has your question been resolved?

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@trail lintel Has your question been resolved?

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@trail lintel Has your question been resolved?

[lean otter]

....

That question is out of my senses

[safe radish]

@trail lintel Has your question been resolved?

[safe radish]

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[rustic goblet]

!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

[rustic goblet]

2

So the condition is that w1 + w2 + 2w3 + 2w4 = 0, I believe

I want to show that w1 + w2 + 2w3 + 2w4 = 0 <=> there exists a linear transformation st. T(v_i) = w_i

i have shown the backwards implication

but I don't know how to show the forwards implication

any help is appreciated

<@&286206848099549185>

[steep lily]

you can define a linear map by what it does to a basis

here, you have an extra element

so what you'll need to double check is that when you extend by linearity, you end up doing the right thing to the extra element

e.g. you can define $T\colon V\to W$ by $$T(\lambda_1v_1+\lambda_2v_2+\lambda_3v_3)=\lambda_1w_1+\lambda_2w_2+\lambda_3w_3$$

which is well defined and linear because v1,v2,v3 is a basis

[flat frigate]

Edward II

[steep lily]

which is well defined and linear because v1,v2,v3 is a basis

[rustic goblet]

okay..

[steep lily]

and then the only remaining property you need to show is $T(v_4)=w_4$

[flat frigate]

Edward II

[rustic goblet]

how does this show the implication?

[rustic goblet]

the forwards one, that is

[steep lily]

you're going to use that property in showing this

because it clearly won't be true in general, as there's no way to getting a w4 when the definition of T only involves w1, w2, and w3

[rustic goblet]

I see

[safe radish]

@rustic goblet Has your question been resolved?

[safe radish]

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[wicked basin]

Hello, I'm unsure of how to start this question?

I need assistance with solving it please and ty

[quiet juniper]

same way u do it in decimal but with binary

Long divisioooooooon

Instead of doing long division with 5 and 7 tho

[wicked basin]

but what is that way😭its hard to visualize not doing it the normal way

[quiet juniper]

U do it with their binary representations

Well try doing it in decimal first

Then try converting 5 and 7 into binary

And try generalizing what you did in decimal

ngl doing this made me realize I forgot how to substract without a calculator

but if I cna rederive it

So can u

[wicked basin]

like this?

[quiet juniper]

Cmon u don't need me to check decimal long division :p

[wicked basin]

no i understand this

but like

how binary

[quiet juniper]

Do that but turn 7 and 5 into binary

[wicked basin]

binary is 0 and 1 correct

[quiet juniper]

Ye

[wicked basin]

how can i just turn 5 and 7 into 0 and 1

subtract 4 and 6 ??

[quiet juniper]

Um

How did you get this question

no offense but if you didn't know what binary meant this might be harder than I thought 🙃

[wicked basin]

yeah none taken, i came here for that exact reason

i’m unsure of how to go about this

[quiet juniper]

Is this for a class or what

[wicked basin]

yes

we touching on binary for first time

[quiet juniper]

Did u skip class or had to miss or what

[wicked basin]

^

[quiet juniper]

Yeah but to not even know what binary means

[wicked basin]

what is it that i am missing

[wicked basin]

it’s 0 and 1

[quiet juniper]

Feel like they would've taught you that

How long frkm now is the homework due? If it's a while maybe best to just wait til you cover more material

[wicked basin]

midnight

[quiet juniper]

what

[wicked basin]

it’s 7:05 pm

[quiet juniper]

OK r u sure u didn't skip or miss class

[wicked basin]

yes i’ve attended all of them

it’s just a new concept

[quiet juniper]

So weird

[wicked basin]

that’s y i’m untrained

[quiet juniper]

well

I mean

They might as well have taught you nothing in class cus imma have to teach you everything 🙃

you know how decimal representation works at least?

like sum of digits between 0 and 9 times powers of 10

[wicked basin]

is that like saying 0.123 = 1/10 + 2/100 + 3/1000

[quiet juniper]

Yeah

[wicked basin]

yes i know

[quiet juniper]

do that but with power of 2

And the digits are 0 and 1

[wicked basin]

by what you’re saying this is by power of 1 right

[quiet juniper]

no

Power of 2

[wicked basin]

10*

[quiet juniper]

Oh yeah

[wicked basin]

1/10^1 and 2/10^2 and so on

okay

uh

[quiet juniper]

for instance 10 = 1*2^1 + 0* 2^0 = 2

11 = 3

[wicked basin]

0.789 =

bruh🤦🏾‍♀️okay wait hold on

0.789=

7/2^1 + 8/2^2 + 9/2^3

???

wait

bruh i’m lost how do i get 0 and 1 out of .789 with powers of 2🤦🏾‍♀️🤦🏾‍♀️🤦🏾‍♀️🤦🏾‍♀️🤦🏾‍♀️

[quiet juniper]

hmm honestly I don't know the algorithm for how to convert a number with decimal points to binary off the top of my head

But you just need to convert 5 and 7

[delicate sphinx]

Nah you don't need that

You can use the decimal equivalent of 5/7 and convert to binary

[wicked basin]

how so?

[quiet juniper]

OK I mean

"if you don't know the algorithm you just need to convert 5 and 7"

[delicate sphinx]

It's easier than you think

[quiet juniper]

bet

[delicate sphinx]

The decimal equivalent of 5/7 is 0.7142857....

[quiet juniper]

Oh

You just convert the repeating part

[delicate sphinx]

To convert a decimal number to binary, double the number, then the number left of the decimal point is the carry bit

Then you ignore the unit value and focus on the decimal part

Here's an example

0.188
You double it and get 0.376 that 0 in the units is the first binary digit

[quiet juniper]

kool

[wicked basin]

i agrée

very kool

tyvm mr captain nova 22🙏🏾

[delicate sphinx]

Then take 0.376 double it and get 0.752 the 0 in the units is the next digit, etc. If it's a 1 then it's a 1 in the binary representation

[wicked basin]

i grasp this

and also thank you mr 992 qqoloy

[quiet juniper]

np

[delicate sphinx]

Definitely more nice and simple then converting 5 and 7 into binary

[wicked basin]

!close

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[fresh tree]

what am i doing wrong

[fresh tree]

45/64 is somehow wrong

ET= | 3/4 - 3/64 |

this makes 0 sense

<@&286206848099549185>

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[marble goblet]

what is this?

what kind of math

[fresh tree]

calc 2

[marble goblet]

I dont understand a single thing of what its asking

k

very nice how did u do in calc 1?

[fresh tree]

good

[marble goblet]

im slowly making my way up the ladder of math. currently in intermediate alg, going to take college alg next sem

looking at this, I have no idea what anything means what so ever. whats Et?

[fresh tree]

you honestly shouldnt go to college for algebra

i would watch online courses up to calc

save bread

[fresh tree]

error

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[fresh tree]

<@&286206848099549185>

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[unborn wigeon]

wait why can we do this!?!?

also why do we need to consider even power x^82!?!?

[quasi bison]

i think they meant (x^139)^82 ...

x^138 = 1 so x^139 = x

wait

this is sus actually

[unborn wigeon]

why do I even need to consider the 83=82+1

[quasi bison]

yeah i dont know anymore either

[unborn wigeon]

how are you meant to do this

wouldn't I need to find a primitive root first?

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[tough canopy]

Hello , can someone help me with the parameters of the asymptotes?

[quasi bison]

,rccw

[tough canopy]

🥹

<@&286206848099549185>

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[wooden ridge]

I don’t understand why the answer is 3! x 4!=144 rather than 2 x 3!x4! = 288 as either a boy or girl can start?

[quasi bison]

nope, only a girl can start

GBGBGBG

[wooden ridge]

Oh, I’m so dumb 🤦‍♂️

That never occurred to me

That it would be GG at the end of a boy starts

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[lean otter]

Solving for y, which one of these is correct?

[lean otter]

first

Ight thankyou

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[mighty wadi]

Does someone know why 0 itself is not valid?

[main mural]

show the original problem

[mighty wadi]

we need to find y and the valid interval of y

[lean otter]

seems like it was a mistake

I'd say 0 should be included

[mighty wadi]

Yeah, right
I understand that negative values should be excluded but 0

[quasi bison]

btw "yeah, right" (specifically those words specifically like that) is often a sarcastic phrase

[mighty wadi]

No I meant it like YEAAAAHHHHH!!!! is that true?

Because maybe we are overseeing something

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[silk token]

i asked before the best way to get the determinant, so i should reduce by rows the matrix and then use Laplace, but when i do row reduction, the matrix may not have the same determinant, so is this method only useful to know if the determinant of the original matrix is 0 or different from 0, or there's a row reduced matrix that has the same determinant as the first one

[raven heart]

not all row operations conserve the determinant that's true

1. "Ri <- Ri + k*Rj", that doesn't change the determinant

2. swapping rows multiplies the det by -1

3. "Rj <- k*Rj", that multiplies the det by k

you just have to keep track of which operations you take change the determinant

and then go through your row operations in reverse

"the determinant I get at the final step is 27, so the det at the step just before is ...etc"

@silk token

[silk token]

oh thank you so much!

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[exotic owl]

I plugged in x to find y = 1/2

then I found the slope of 1

through derivative

but then i did

y-1/2 = 1(x-pi/4)

i dont get i

[spice grove]

How'd you get the slope again?

Can you show all your steps for that?

[exotic owl]

i got my slope from

(sinx) (-sinx) + (cosx)(cosx)

-sin^2x + cos^2x

[spice grove]

Great, how's that 1?

[exotic owl]

oh shoot

wait

can it be

(wait no nvm

is it just 0

[spice grove]

yeah

[exotic owl]

ohh

thx

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[glacial flicker]

is b correct?

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[hallow forge]

slept midway through math class now idk how to solve this

[hallow forge]

need some help

maybe explain it as well

im guessing CEB would also be 80 degrees?

[knotty coyote]

CEB would be 180 right

thats 3 angles

[hallow forge]

how would it be?

180's a semicircle

[knotty coyote]

I’m trying to remeber

yeah a semicircle is 180 degrees

but a triangle is made up of 3 angles that will always equal 180 degrees

CE&B are 3 different angles