help-42

[tidal condor]

i thnk this good too

do student?

[heavy mantle]

Yes

[tidal condor]

I was a student once, too.

[heavy mantle]

Not math student

[tidal condor]

oh

so what?

your subjeckt\

[heavy mantle]

I am about to join engineering

[tidal condor]

I am an design engineer

[heavy mantle]

I see

[heavy mantle]

Like machines??

[tidal condor]

I am engaged in drawing up drawings in the field of advanced technologies

[heavy mantle]

I see

See you later then ..👋

[tidal condor]

bye

[calm coral]

@brittle basin Has your question been resolved?

[tidal condor]

i think it was

[calm coral]

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[frail folio]

hello

[frail folio]

so

im doing transformations basically

i dont understand what it means they are constants and how to find them.

[left wren]

plug in the values

you are given in the graph

[frail folio]

so like

[left wren]

when x = 0 y = -3, x = 3𝜋/4 y = 2

and solve for a,b and c

also the maximum magnitude of the sinusoidal is 3

so that means a = 3

[frail folio]

yeah i thought so for a

[left wren]

or -3

one of those two

c = -1 cause the graph is shifted by 1 unit downwards

[frail folio]

okay so

you sure? c is supposed to be the central axis

[left wren]

c shifts the graph up or down

[tidal matrix]

well right off the bat you can see that c = -3, it was shifted downwards by 3

[left wren]

if c was 0 the sinusoidal would go 3 units up and 3 units down at its peaks

[frail folio]

the other guy said -1

im confused to what to listen to

[left wren]

it is -1

[heavy mantle]

How can u confirm it's -3?

[left wren]

the magnitude of peaks is decided by the coefficient of sin

[tidal matrix]

nevermind yeah it's -1, i didn't take a and b into consideration

[frail folio]

c is supposed to be central axis, a is amplitude and in the parenthesis is supposed to be the period

as its transformations

[left wren]

i am not sure what you mean by central axis

[frail folio]

example

say the top one is

2

and the bottom is -2

(y axis)

[left wren]

yes

[frail folio]

the midpoint/central axis is 0

[left wren]

oh you mean the middle y value

between the two

[frail folio]

yeah we call it central axis at my school for some reason

[left wren]

ye so for this it has to be -1

(2-4)/2

it is the mid point

[frail folio]

whats the formula

thats the one i was looking for

[left wren]

(ymax+ymin)/2

[frail folio]

thank you

what does the yman-ymin/2 solve btw

as i remmeber it did something

im writing these down to remember them

oh amplitude

mb

[frail folio]

3sin(x+b) -1

i did a & c

[heavy mantle]

You got b?

[frail folio]

nope and i did c wrong i just realised lemme fix it

[heavy mantle]

Put x=0 and get b

[frail folio]

wait so

3sin(b)-1=0

and move -1 and divide

?

[heavy mantle]

=0?

Isn't it -3?

[frail folio]

no i said i made mistake

its -1

i corrected it above as well

c = -1

[heavy mantle]

No I mean at x=0 the value is -3 right?

[frail folio]

yes

[heavy mantle]

You equated it here to 0

[heavy mantle]

Here

[frail folio]

how can i solve then for

x+b

[heavy mantle]

3sin(b) - 1= -3

b= -sin^-1(2/3)

[frail folio]

ohh i see

ty

[heavy mantle]

Is it right . You know the answer?

[frail folio]

what the...

[frail folio]

yeah i mean it makes sense

thank god i just realised this problem is with horizontal shift

the ones on the test tommorow are with period which is simple its just 2π/b

[calm coral]

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[calm coral]

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[vital chasm]

am i tripping on taking mod m on both sides means n is a multiple of m

[neon shadow]

where are they taking mod?

[vital chasm]

im taking mod

[neon shadow]

if n^3 is a multiple of m you can do that

[vital chasm]

why if

isnt that implied

taking mod m both sides nets us n^3 is congruent to -n^3 mod m

[neon shadow]

why are you taking mod tho?

[vital chasm]

because i can

[neon shadow]

then yea, if you take mod on both sides, then it implies n^3 is a multiple of m

[vital chasm]

m*

[vital chasm]

but then these are the solutions

how is that possible

[neon shadow]

yea

[vital chasm]

ok nvm

im tripping

.close

[calm coral]

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[neon shadow]

you dont need mod at all i guess?

[left wren]

ye

[vital chasm]

no n^3 is cong to 0 mod m doesnt imply n is 0 mod m

[left wren]

you can just expand

terms will cancel

[neon shadow]

you are reallly overthinking it

just expand both sides

[calm coral]

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[spice venture]

i need help in this question

this is my work so far

(can start from scratch if required to)

[left wren]

nah you are doing alright

[spice venture]

ohk so what do i do from here?

[formal tulip]

Hmm I have a different solution

[spice venture]

yeah?

[formal tulip]

Assume a solution of the form r^x

[spice venture]

ohk is that because of the range given

[formal tulip]

You will get a quadratic equation with complex roots that will break down into a sin and cos term hopefully

[formal tulip]

it's because of how the equation looks

[spice venture]

how is ur intuition that good 😭

also i didnt understand..after assuming it how do i get the quadratic eqn?

[left wren]

but isnt r^x a non-periodic function

[formal tulip]

$r^{x - 1} + r^{x + 1} = \sqrt{3}r^x$

[potent lotus]

jewels!

[formal tulip]

@left wren

[left wren]

oh

[formal tulip]

So this turns into $r^2 - r\sqrt 3 + 1 = 0$

[spice venture]

oh alright let me try this

[potent lotus]

jewels!

[formal tulip]

So you get $r = \frac{\sqrt{3} \pm i}{2}$

[potent lotus]

jewels!

[spice venture]

yeah i got till there

[formal tulip]

So this is e^(i*pi/6)

So take $f(x) = A \sin \left ( \frac {\pi x}6 \right ) + B \cos \left ( \frac {\pi x}6 \right )$

[potent lotus]

jewels!

[spice venture]

so our function is
f(x) = e^x*(ixpi/6)

[spice venture]

oh

[formal tulip]

the specifics aren't really important now

[spice venture]

so period is 12

[formal tulip]

yeah

[spice venture]

oh man

i thought of assuming the function as well

but then i was like how would i proceed from there and then stopped

thank you

.close

[calm coral]

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[spice venture]

@left wren thx bro

[left wren]

i didnt really help with that

[spice venture]

its fine u still responded

[calm coral]

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[calm coral]

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[calm coral]

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[glad parrot]

!msgdelete

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[lone sparrow]

Hi, just wanted to ask if this is correct

Confused with “at the point c(10)” like it doesn’t expect me to plug 10 into sin or cos right?

[stoic oyster]

the point c(10) is for t=10 yes

[calm coral]

@lone sparrow Has your question been resolved?

[lone sparrow]

so is this answer wrong?

[stoic oyster]

what answer

I'm saying your assumption is correct

@lone sparrow

[lone sparrow]

65/2197

[stoic oyster]

I don't see anything wrong really

[lone sparrow]

o

so i just ignore that fact in the computation?

[stoic oyster]

what fact ?

[lone sparrow]

that t=10

like if i were to compute a tangent plane at a given point, id plug that in

[stoic oyster]

curvature just doesn't depend on time here

[lone sparrow]

o rlly

[stoic oyster]

it's the same whatever t

that curve c(t) is an helix so the curvature being constant sounds alright

[lone sparrow]

ooo

oki

tnx

.close

[calm coral]

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[restive harbor]

Can someone help me with 8. B)

I wa stuck for a while

this is what I have done

[restive harbor]

idk why its sending a file

[next hound]

required for a perpendicular bisector:

1. perpendicular
2. bisector

do you know how to show either?

sorry I can't open the HEIC

[restive harbor]

yea what i did was correct ?

I found the gradient

of the perp bisector

[next hound]

okay that's the first step

did you find the gradient of AB?

[restive harbor]

yes

m = -3

[next hound]

okay, based on the gradients are they perpendicular?

[restive harbor]

yea

[next hound]

if so, we have the perpendicular down and just need the bisector part

[restive harbor]

whats bisector part?

I thought it was called perpendicular biscector itself

[next hound]

prove that the line they gave you bisects (splits in half) AB

[restive harbor]

oh

so the perpendicular has to cut through ab half equally

[next hound]

otherwise it's just a line perpendicular to AB and not a perpendicular bisector

[next hound]

yes, but there's a simple way to show that it does

think about how you would do it, then read this: ||Show that M is on the line||

[restive harbor]

utiliise the Mid point?

[next hound]

yes!

[green crag]

No I can’t help you

[drifting seal]

,av sizzly0376

[potent lotus]

sizzly0376's Avatar

Click here to view the image.

[restive harbor]

what do i do then?

Idk how to show M is on the line

[next hound]

well you need to know that the perpendicular bisector goes through the midpoint

[restive harbor]

yea

[next hound]

is (2,1) on the line y=3x?

as an example ^

[restive harbor]

Well the M is one point, but I need another point

[next hound]

no

you need two points to define a line, but you can check if a point is on a line a different way

||plug the point into the equation and see if it's true||

[restive harbor]

@next hound this is what I have done

to find the equation of the perp bisector

[next hound]

yeah that is the correct equation

and it's the same as the one they asked you to show (if you do a few steps)

[restive harbor]

Yeah

[restive harbor]

Do I need to rearrange this equation so it matches x-3y+9=0?

but I dont think that will work

[next hound]

yes

it would work, but I don't think it was the intended way to solve the problem

[restive harbor]

it wont work because it would be then 1/3x -y +3 = 0

which isnt the same as x-3y+9=0?

[next hound]

multiply everything (both sides) by 3

then it's the same

[restive harbor]

ohhh

so i got x-3y+9=0

alr that makes sense now, thank you

[calm coral]

@restive harbor Has your question been resolved?

[restive harbor]

@next hound I have another question for 8. c)

[next hound]

ok

[restive harbor]

am i on the right track

since c lies on the y axis, i know x coordinate of c is 0

[next hound]

I don't think so

why did you calculate the distance between A and B?

you want to make the distance between C and A equal to the distance between C and B

[restive harbor]

it says it is equisdistant from a and b

[next hound]

yes, read my message directly above for what that means

the distance between A and B is irrelevant

[restive harbor]

alr I get it

[tidal matrix]

a) M = (x1+x2/2, y1+y2/2)
b) Show m1m2 = -1, and then use y - (y1+y2/2) = m2(x-(x1+x2/2))
c) x = 0, d(AC) = d(BC) so therefore d^2(AC) = d^2(BC) or in other words, (2-x)^2 + (2 - y)^2 = (1-x)^2 + (5-y)^2, from there plug in x, find y, coord of C is (0,y)
d) to find the coords, do system of equations
y = 5 and x - 3y + 9 = 0, so basically just plug in y = 5 and solve for x, and that's ur coords -> D(x,5)

[calm coral]

@restive harbor Has your question been resolved?

[calm coral]

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[steady nebula]

can anyone help? i dont know what to do here

[calm coral]

@steady nebula Has your question been resolved?

[calm coral]

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[agile jetty]

Quick question, If I'm solving a lagrange multiplier system and I get only one point how do I know if it's a minimum or maximum point?

[agile jetty]

Will I have to resort to using hessian matrix?

[sharp narwhal]

u could compare with other values

that ur trying to minimize/maximize

i think lagrange multiplier gives a necessary condition

[agile jetty]

Actually I just noticed something, I think in the context of the function I'm maximizing/minimizing I can say that it doesn't have a maximum

the function is f(x, y)=x²+2y²

the restriction is 3x+y-1=0

[agile jetty]

So this function doesn't have a maximum point (just from looking at it)

And the constraint isn't bounded

So for those reasons it shouldn't have a maximum point right

[warm warren]

What do you mean the constraint isn’t bounded

[agile jetty]

3x+y-1=0 isn't bounded

it's a line

[warm warren]

But you can’t take every point on the line you need it to intersect with f

[agile jetty]

sorry I don't understand

Do I need it to intersect with f?

[warm warren]

Oh wait I see what you mean

Yeah you’re right

[agile jetty]

I see

Thanks

.close

[calm coral]

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[waxen folio]

Help

[rough blade]

!da2a

[calm coral]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[waxen folio]

@agile jetty

[serene talon]

<@&268886789983436800>

[agile jetty]

The hell

[wide chasm]

Hi! Welcome to mathcord. We cannot allow people to practise academic dishonesty on this server. Sorry!

[agile jetty]

Why did you ping me

[wide chasm]

This user has been timed out.

.close

[calm coral]

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[leaden pendant]

parent function

[leaden pendant]

<@&286206848099549185>

[bold hatch]

Is there a sign missing here?

[leaden pendant]

uh i didnt realize that

do anyother one except for that

[leaden pendant]

yea i think so

<@&286206848099549185> can someone help me

[slender gust]

y=f(x-1)+3 ---> you are plotting the parent function offset to the right (x axis) by one and offseting +3 unit upwards on y-axis

[sand compass]

I think the sign missing is an = as in your starting X is 1

[leaden pendant]

yea f is broken ingore it

[sand compass]

lol

[leaden pendant]

my teahcer mesed up

[leaden pendant]

i know but how to draw i know the rules

bro can someone ehlp em

<@&286206848099549185>

<@&286206848099549185>

can someone help me

[scenic grail]

Easiest way if u already know where u want to go

Is to draw the parent function first

and then translate each significant point

[leaden pendant]

thats done for me

[scenic grail]

Great

[leaden pendant]

i just dont know how to draw the trnaformed grpah

even though i memorized all the rules

bruh

@scenic grail

[scenic grail]

Start with the most important points

Let's try from left to right on the first q

[leaden pendant]

idk what those are

the first one is broken

[scenic grail]

(-2,4)

[leaden pendant]

dont do f

[scenic grail]

k second 1 then lol

Any important point will be where its easy for u to identify the function

(-2,4) on the parent function is important cus its the end of the function on the left side

So u can translate that one first

Did u get that

[leaden pendant]

naw

[scenic grail]

Try translating the point (-2,4) according to what qg asks

What new point do u get

[leaden pendant]

the -1,7

@scenic grail

[ancient grotto]

Yeah.

[leaden pendant]

but how would i draw it i dont have a drawing tool

[ancient grotto]

But I would follow this.

[ancient grotto]

Shift the graph to the right one unit and up 3.

[leaden pendant]

i cant draw this genuis

otherwise i dont need your help like can you draw it

[ancient grotto]

So are you asking for someone to draw it for you?

[leaden pendant]

no but i cant do it myslef

it doent work

[ancient grotto]

What exactly are you not understanding?

[leaden pendant]

i cant draw it

[balmy bane]

(There's an amazing tool that mankind has had for years, maybe even decades, specifcally for drawing; you might of heard of it - a pencil)

[leaden pendant]

buddy i dont ahve one next to me

other wise i would have done it ages ago

[still kindle]

Are u on cellphone or pc?

[leaden pendant]

pc

[balmy bane]

(there are graphing calculators)

[ancient grotto]

That explains it.

[still kindle]

Use paint or something like that

Or desmos

[balmy bane]

(but beyond that idk what else I can say)

[ancient grotto]

Why were you attempting this without having a utensil to actually graph with?

[leaden pendant]

becuase its online

idk

[balmy bane]

You should still have a pen(cil) and paper to hand

Maths is a hands-on subject

[leaden pendant]

naw but my whole summer course is online

[leaden pendant]

i use my head

[balmy bane]

It doesn't matter

[leaden pendant]

proving is too hard\

[balmy bane]

Online art classes are useless if you then don't proceed to do any art yourself

Maths is no different

Geometry (and by proxy function graphing) even more so no different

[leaden pendant]

?

[manic oracle]

Alternatively use paint tool on computer

Doing math without a visual doesn't help train intuition as well as with a visual

[balmy bane]

This is what I was getting at

[manic oracle]

ASCII art it if you are desparate for a drawing tool

[balmy bane]

It doesn't suffice to read without practice

[balmy bane]

("desperate*" ik but still) damn imagine using ASCII art in an exam as a last-minute resort

[leaden pendant]

an exam what the hell is an exam

[manic oracle]

Other approaches:

• Go to the beach and write in the sand
• Rearrange your next meal into the diagram
• Take strands of your hair and arrange them into the diagram

[leaden pendant]

?

.close

[calm coral]

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[blazing coyote]

My original idea was to let $A_1 \cap A_2 = t+U; t\in V, U \subseteq V$

[potent lotus]

wai

[blazing coyote]

and Then I show this?

or it this too general

[ancient thistle]

but you don't know that to be true

[unique jackal]

you're trying to prove that A_1 \cap A_2 is the translate of some subspace of V

you can't just assume it

[blazing coyote]

hmm, okay

What I also thought of was

let $x \in U_1; y\in U_2$. We then have for any element in the intersection $x+v=w+y$

[potent lotus]

wai

[unique jackal]

what is v? what is w?

also, that's not how you show equality

[blazing coyote]

from the question, some vectors in V

[blazing coyote]

wydm, I'm just saying for any vector in the intersection, this equality is true

[ancient thistle]

thats a weird way to phrase it

[unique jackal]

you need to show that either $A_1 \cap A_2 = \emptyset$, or $A_1 \cap A_2 = v + U$ for some subspace $U \subseteq V$

[potent lotus]

higher!

[blazing coyote]

yes

[ancient thistle]

like what you've written here is not logically correct

x + v might not equal w + y for arbitrary elements x in U_1 and y in U_2

[elder pawn]

v already used

[ancient thistle]

and they might not live in the intersection

[unique jackal]

oops

point remains though

[elder pawn]

$A_1\cap A_2=x+U$ for some $x,U$

[potent lotus]

ロケットジャンプ

[blazing coyote]

Lemme think of another proof then

[ancient thistle]

the correct phrasing is: for any $a \in A_1 \cap A_2$, there exist $x \in U_1$, $y \in U_2$ such that [ a = x + v = w + y ]

[blazing coyote]

yes, that's what I meant

[ancient thistle]

so you should say what you mean

[blazing coyote]

Is this a good place to start

[ancient thistle]

well thats kinda just the definition of the intersection A_1 \cap A_2

so you can't not use it

[blazing coyote]

hmm

can I have a hint

[ancient thistle]

if A_1 \cap A_2 is empty, theres nothing to do

so you should just assume otherwise

say t \in A_1 \cap A_2

use that to "unify" the definitions of A_1 and A_2

[blazing coyote]

What I got so far, $\exists x \in U_1 , y\in U_2$ st $v+x=t; w+y=t$ . Adding them, $v+x+w+y=2t$

[potent lotus]

wai

[blazing coyote]

which doesn't really help

[ancient thistle]

what's the relationship between t + U_1 and A_1?

[blazing coyote]

t+u \in A_1

[ancient thistle]

okay but how exactly are t + U_1 and A_1 related

[blazing coyote]

t+U_1=A_1

[ancient thistle]

_1

what about t + U_2 and A_2?

[blazing coyote]

t+U_2=A_2

[ancient thistle]

okay so with that

you can rephrase the question

Suppose $A_1 = t + U_1$ and $A_2 = t + U_2$ for some $t \in V$ and some subspaces $U_1, U_2$ of $V$. Prove that the intersection $A_1 \cap A_2$ is a translate of some subspace of $V$.

[ancient thistle]

can you guess which translate this should be?

[blazing coyote]

t

[ancient thistle]

no thats not a translate

,, A_1 \cap A_2 = (t + U_1) \cap (t + U_2) = {???}

[ancient thistle]

what goes in the place of ???

[blazing coyote]

I have no idea, shouldn't I find that first

[ancient thistle]

hmm

well you should be able to guess without having done the proof

like take the trivial case for example, when t = 0

,, A_1 \cap A_2 = (0 + U_1) \cap (0 + U_2) = {???}

[ancient thistle]

what goes in the place of ???

[blazing coyote]

then it's U_1 \cap U_2

[ancient thistle]

so what should it be when t isnt 0

[blazing coyote]

t+ U_1 \cap U_2

[ancient thistle]

yes

so you should try proving that

like this should be the picture you have in your mind

[blazing coyote]

But say I didn't know that, I'd want to be able to explcitly find it , no?

[ancient thistle]

well

if you didn't know that

i don't have high hopes for you being able to solve the problem

[blazing coyote]

Okie, lemme try to prove it then

so now I just prove each set is a subset of the other?

[ancient thistle]

its a possible direction

[blazing coyote]

Let $\alpha \in A_1 \cap A_2$. Then $\alpha \in t+U_1 \land \alpha \in t+U_2$. Thus $\alpha \in t+ U_1 \cap U_2$. Therefore $A_1 \cap A_2 \subseteq t+( U_1 \cap U_2)$

[potent lotus]

wai

[ancient thistle]

Thus ...
why

thats the thing you're trying to prove in the first place

justify

[blazing coyote]

Let $\alpha \in A_1 \cap A_2$. Then $\alpha \in t+U_1 \land \alpha \in t+U_2$. Let $ x \in U_1; y \in U_2$. Then $t+U_1 = {t+x \mid x \in U_1}$. Similarly $t+U_2 {t+y \mid y \in U_2}. Let $z \in t+ U_1 \cap U_2. so ${t+z \mid z \in U_1 \cap U_2}$Thus $\alpha \in t+ U_1 \cap U_2$. Therefore $A_1 \cap A_2 \subseteq t+( U_1 \cap U_2)$
\
Let $\beta \in t + U_1 \cap U_2$. Then $ \beta. \in t+U_1 \land t+U_2$. Therefore $ \beta \in A_1 \land \beta \in A_2$ so $ \beta \in A_1 \cap A_2$

[potent lotus]

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

[ancient thistle]

whats with the "Let x \in U_1" and "Let y \in U_2"

here as well

you can't just state the result you're trying to prove the prove the result you're trying to prove

you need to justify

[blazing coyote]

That's by defn

right

[ancient thistle]

i mean clearly not

[blazing coyote]

wait, lemme rephrase myself

$\beta \in t+ U_1 \cap U_2$. $t+U_1 \cap U_2 = {t+z \mid z \in U_1 \cap U_2}.$

[potent lotus]

wai

[blazing coyote]

$t+U_1 = {t+ x \mid x \in U_1}$. $t+U_2 ={t+y \mid y \in U_2}$. $t+ U_1 \cap U_2 ={t+z \mid z \in U_1 \land z \in U_2}$

[potent lotus]

wai

[ancient thistle]

just take an element and do something with it

$\beta \in t + U_1 \cap U_2$ means $\beta = t + u$ where $u \in U_1 \cap U_2$, i.e., $u \in U_1$ and $u \in U_2$

[ancient thistle]

so $\beta \in t + U_1$ and $ \beta \in t + U_2$

[ancient thistle]

you don't have to contort your proof to write out sets over and over again

[blazing coyote]

oh

yea, makes sense

thanks

[ancient thistle]

theres also the other direction

you should try writing it without faffing around with set notation

[blazing coyote]

That is $\alpha \in t+U_1 ; \alpha \in t+U_2 \implies \alpha \in t+( U_1 \cap U_2)$

[potent lotus]

wai

[ancient thistle]

yes you need to prove that

[blazing coyote]

Let $\alpha \in t+U_1 ; \alpha \in t+U_2$. Then $\alpha -t=x ; x \in A_1; \alpha - t=y;y\in A_2$. Thus $ \alpha -t \in U_1 \cap U_2$. So $\alpha-t+t= \alpha \in t + U_1 \cap U_2$

[potent lotus]

wai

[ancient thistle]

seems good

[blazing coyote]

Cool. Then I'm done with this

Thanks!

[ancient thistle]

pretty much

[blazing coyote]

.close

[calm coral]

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[calm coral]

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[blazing coyote]

(b)
I was thinking ${(1,0,\dots)+ U}, {(0,1,\dots) +U}....$ could be a basis

[potent lotus]

wai

[velvet osprey]

nope!

(1,0,...)+U and (0,1,...)+U are the exact same vector in F^infty/U.

[blazing coyote]

hmm

[velvet osprey]

you need to exhibit a family of vectors in F^∞ which satisfy the following properties:

• they are linearly independent (in the sense of any finite subfamily being bonafide LI)
• no two of them differ in only a finite number of positions (i.e. no two differ by an element of U)

[blazing coyote]

the set containing the set of these vectors aren't the same though, right

[velvet osprey]

there are a bunch of different ways to do it but they do require being comfortable with describing stuff in words precisely

[velvet osprey]

sorry what

[blazing coyote]

${(1,0,\dots)+U}, {(0,1,\dots)+U}$

[potent lotus]

wai

[velvet osprey]

these are the same set

but also you're off base

$e_1 + U$ and $e_2 + U$ are the same thing.

[potent lotus]

Ann

[blazing coyote]

got it

[velvet osprey]

i'm using $e_k$ here to mean the sequence with a 1 at the $k$'th position and 0s everywhere else

[potent lotus]

Ann

[velvet osprey]

even worse $e_1 + U = e_2 + U = U$ because $e_1$ and $e_2$ themselves both lie in $U$.

[potent lotus]

Ann

[blazing coyote]

Right

[velvet osprey]

so in F^∞/U these would both be 0

[civic dirge]

The way I like to think of quotients:
two vectors in V/U are the same if you can get from one to the other by adding some element in U

(Where addition is done in V but forget about that part)

[blazing coyote]

hmm, so constructing the basis would be a bit challenging here

[velvet osprey]

you do not need a basis

merely an infinite LI family

[blazing coyote]

hmm,so what i want is a list of linearly independent sets of the form ${v+U \mid v \in F^{\infty}}$

right

[potent lotus]

wai

[dire gulch]

wouldnt it be easier to show that F^inf/U is not finite dimensional? as in for any finite set of vectors in F^inf, show there is an element of F^inf not in the set of vectors + U

[civic dirge]

That's very similar. That is, use your generation method to generate a linearly independent set

Or, if you have a lin ind set, then you probably used a generation method to get them

[blazing coyote]

You're talking to me?

[civic dirge]

No

So you'll need to change each vector in infinitely many components, to generate more @blazing coyote

[dire gulch]

we can do this case by contradiction tho, so we dont need to give a generation method

[blazing coyote]

Yea, that was my next thought, but not sure how to formalise it

[civic dirge]

Ah like, we can just show such a list exists, but don't actually construct any vectors? Yeah that would work and not give a set

[blazing coyote]

like in the first basis element, all elements in v are 1, in the second all but one elements are 1 and so on

[civic dirge]

We can construct an infinite LI set. I did think of one, it's not too complicated.

[blazing coyote]

would this work

[civic dirge]

Yeah I think it does!

[blazing coyote]

${(1,1,\dots)+U},{(0,1,1,\dots)+U},\dots}$

[potent lotus]

wai

[civic dirge]

Then two 0's, then three 0's, etc?

[blazing coyote]

yes

Now to prove linear independence

[blazing coyote]

so here the zero vector would be {0} right

[dire gulch]

all of these are the same

they all only differ in finite elements

[blazing coyote]

Well, none of them are in U

!occupied

[calm coral]

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).

[dire gulch]

yes, but they are still the same

[blazing coyote]

hmm, but a basis must exist

[elder pawn]

recall anns tip

you need to exhibit a family of vectors in F^∞ which satisfy the following properties:
they are linearly independent (in the sense of any finite subfamily being bonafide LI),
no two of them differ in only a finite number of positions (i.e. no two differ by an element of U)

[blazing coyote]

ooh, okay

so I need something like all positions which are multiples of say 2 are 1

then all positions that are multiples of 3 but not 2

[dire gulch]

you can just say all positions that are multiple of 3, of 3 but not 2 doesnt change the LI

[blazing coyote]

fair yes

let $e_n$ be that vector in $F^{\infty}$ with $1$ at all positions that are multiples of $n$. We then haave the basis ${e_n+U}; n \in \N$

does this work

[potent lotus]

wai

[blazing coyote]

Does this work

[dire gulch]

i think it does, but i found it easier to prove by choosing specific n instead of all n

[blazing coyote]

Thanks!

Lemme just verify linear independence now

[warm warren]

you can't call it a basis unless you show that it is one

[blazing coyote]

Okay, in that case an infinite list

so I have to show $\left {\sum_{i=1}^{\infty} \lambda_i e_i + U \right}\neq {0}$ for any values of lambda

[potent lotus]

wai

[blazing coyote]

Can I have a hint

[lyric ravine]

what?

you have to show every finite linear combination equalling 0 has all \lambda_i = 0

[blazing coyote]

so I sum up to arbitrary n here?

[warm warren]

you can't use the symbol infinity in just LA

you don't have a concept of limits

[warm warren]

so you're looking at every finite subset of this list to be LI

[blazing coyote]

yes

[lyric ravine]

even if you did, the definition of basis says that you need to check every finite subset

[blazing coyote]

ok so as @dire gulch suggested, a proof by contradiction is probably easier

[blazing coyote]

Couldn't we use induction?

[warm warren]

if N is infinite then the power set of N is also infinite

you can't do induction on infinite things

also you don't really have a nice strict ordering on P(N)

[lyric ravine]

if $B$ is the set you are claiming to be a basis, then you need to show that:\
1)it is spanning\
2) $$\forall S\subseteq B: |S| < \infty(\sum_{b\in S}\lambda_b b = 0 \implies \lambda_b = 0 \forall b\in S)$$

[potent lotus]

Herbert

[lyric ravine]

this is the definition of a basis

[ancient thistle]

it suffices for lin indep though

[lyric ravine]

yeah, you need to justify that: becuase subsets of lin indep subsets are lin indep

ah okay so i guess this is strictly wrong then

[ancient thistle]

you won't find a basis for F^infty/U

[blazing coyote]

Okay, so I proceed with this?

[ancient thistle]

don't attempt this

[blazing coyote]

So I proceed by contradiction?

[ancient thistle]

no, show your chosen list is LI

it won't be spanning but that's not necessary

[blazing coyote]

for a finite number n?

[ancient thistle]

?

what is n

[blazing coyote]

Finite number of terms

[dire gulch]

that was wrong

instead of talking all n take a nice subset of all n, that should make this easier

[ancient thistle]

use the result of this exercise

you've seemingly already picked a sequence

[calm coral]

@blazing coyote Has your question been resolved?

[blazing coyote]

okay, got it I think

thanks

[calm coral]

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[blazing coyote]

have to eat now

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[idle geode]

You were playing a game JToH when you were the victim of a mug by a mugger. The neighborhood you live in can be represented as a simple undirected connected bipartite graph. Your house is at vertex 1, and both you and the mugger are there.

You then take turns playing the following game. The mugger goes first, choosing an adjacent house v from his current house u and burning the edge (u,v).

Then, you move to an adjacent house and keep the edge intact.

The goal is for you to get to house n to corner the mugger. Devise an algorithm to determine the answer to this question efficiently, assuming optimal play from the mugger.

[calm coral]

@idle geode Has your question been resolved?

[calm coral]

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[scenic briar]

What in the world am I getting wrong here

[scenic briar]

That's my full answer

I keep doing this question again and again, and this is my answer

[nimble harbor]

${-(2x+1)(x+2) = -(2x^2 + 5x + 2) = -2x^2 - 5x - 2}$

[potent lotus]

k

[unkempt drift]

also your denominator should just be $(x + 2)(x - 2) = x^2 - 4$

[potent lotus]

south

[unkempt drift]

if you do $(x + 2)(x - 2) - (x - 2)(x + 2)$ that's just zero

[potent lotus]

south

[scenic briar]

My denominator is fine according to the makr scheme

Mark*

[unkempt drift]

the way you achieved it is not

but anyway just remember that for next time

[scenic briar]

I just care abt my numinator

[unkempt drift]

the real issue is this

missing brackets

[scenic briar]

Oh, brackets then change of signs?

Shittt

[unkempt drift]

yes!

[scenic briar]

Alright thanks sm

[unkempt drift]

so always be careful with subtraction

it's like how 10 - (5 + 3) = 10 - 5 - 3

[scenic briar]

Yeah I understand now

[unkempt drift]

you always should use brackets after a minus sign

cool np

[scenic briar]

👍👍👍

.close

[calm coral]

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[calm coral]

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[ashen imp]

yo can anyone pls assist me with part ii and iii i genuinely dont know what to do do i diff again or some or like?

[unkempt drift]

how did you do part i?

[ashen imp]

i let x = x0 as thats the point of intersection then equated and then diffed both sides

[unkempt drift]

oh that's a nice method

[ashen imp]

oh is that not how i was meant to do it?

[unkempt drift]

ah then it must be using $(1 - \cos^2 (x_0))^{1/2} = \sin x_0$ then

[potent lotus]

south

[unkempt drift]

I didn't think of that haha

it's a compliment, take it

[ashen imp]

ah ok fair (i may have used other sources)

[unkempt drift]

you just need to eliminate the possibility $\sin(x_0) = \sin(x_0 - \alpha)$ then

[potent lotus]

south

[ashen imp]

sorry im not following

so u square rooted and sqaured and made the sin into 1-cos

so its root sin^2(xo) = sin x0

wait nah i donrt get it sorry lmao

[unkempt drift]

so $1 - \cos^2 x_0 = 1 - \cos^2 (x_0 - \alpha) \implies \sin^2 x_0 = \sin^2 (x_0 - \alpha)$ right

[potent lotus]

south

[unkempt drift]

now if you take the square root

you have $\sin x_0 = \pm \sin(x_0 - \alpha)$

[potent lotus]

south

[unkempt drift]

you want to eliminate the case with the positive sign

[ashen imp]

hmm ok i get those parts but how does the 1-cos^2(x0) = 1-cos^2(x0-a)

oh wait

my faukt

i see

one min

hmm do we ignore the positive case due to the domains they gave?

[unkempt drift]

specifically the domain of alpha

[unkempt drift]

cause that would imply that sin is periodic with period alpha

[ashen imp]

so alpah is > 0 and less than pi but thats still in the first two quadrants and sin is +ve in those

[unkempt drift]

yes, the period of sin is 2pi, but alpha can't possibly be 2pi

[ashen imp]

oh ok

so...

since xo - alpha is the period of that equation, and sin's period is 2pi u have to add a -ve...

becuase...

uhm

errrr

sorry im kinda slow at trig graphs and stuff

[unkempt drift]

$\sin x_0 = \sin(x_0 - a)$ implies that $\alpha$ is the period

[potent lotus]

south

[unkempt drift]

that's the definition of a period, so if the period of a function is T, then $f(x) = f(x - T)$

$f(x) = f(x + T) = f(x + 2T)$ and so on also

[potent lotus]

south

[ashen imp]

hmm ok so alpha is the period, which is within the domain of 0 and pi but sin's period is 2pi , i dont see whereh the negative comes from is it coz we need alpha + xo or something or

[unkempt drift]

so f(x - T) means you translate the graph of f(x) right by T units

so if f(x) = f(x - T), the y-value doesn't change if you move the point (x, f(x)) by T units to the right

[ashen imp]

mhm

[unkempt drift]

wait sorry but we only know this for x0, hmmm

there actually is a value of alpha that isn't 2pi
it's the reflection across x = pi/2, wait

[calm coral]

@ashen imp Has your question been resolved?

[calm coral]

@ashen imp Has your question been resolved?

[calm coral]

@ashen imp Has your question been resolved?

[calm coral]

@ashen imp Has your question been resolved?

[calm coral]

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[mossy cipher]

Prove that sin(x) + cos(x) >=1 for x ∈ [0, pi/2].

[spice storm]

[kind axle]

Multiply and divide by √2

and continue

[spice storm]

Derivatives

[kind axle]

Ew

[spice storm]

Ew but better

[mossy cipher]

cos(x) = root(1 - sin^2(x))
sin(x) + root(1 - sin^2 (x)) = K
sin(x) goes from 0 to 1.

Let sin(x) = p.
K = p + root(1 - p^2)
p ∈ [0, 1]

p^2 + (root(1-p^2))^2 = 1
p^2 + root(1-p^2) ^2 + 2 p root(1-p^2) = 1 + p root(1 - p^2)
(p + root(1-p^2)) = root(1 + p root(1-p^2))
p and root(1-p^2) are positive because p ∈ [0, 1].
So root(1 + p root(1-p^2)) >= 1
(p + root(1-p^2)) >= 1
sin(x) + root(1-sin^2(x)) >= 1
sin(x) + root(cos^2(x)) >= 1
sin(x) + cos(x) >= 1

[spice storm]

I see Philosophy driven solution

[mossy cipher]

Is this correct?

[desert lichen]

If you're completing the square, then shouldn't the the coefficient of the p root term on the RHS be 2?

Tbh, for a question like this my first instinct would be to try to express the sum as a single triogonometric function (which is rather easy here)

[mossy cipher]

Yup. That doesn't change anything.

[mossy cipher]

Didn't I do same the same?

[desert lichen]

I suppose but I meant more in a sense where we can avoid roots and stuff, just play with the amplitude and angle

[calm coral]

@mossy cipher Has your question been resolved?

[calm coral]

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[brittle sail]

1-2x+3x^2-4x^3+5x^4-...

[brittle sail]

I need to find the series sum and convergens

So my first thought was it looks like derivative of e^x?

[unkempt drift]

try integrating the series

[unkempt drift]

not the right one here

[brittle sail]

How can I integrate it?

[blazing coyote]

When x is negative, this is equivalent to $\sum_{i=1}^{\infty} (i+1)x^{i}$

[potent lotus]

wai

[blazing coyote]

Now you can use perhaps the root test to check for convergence

[brittle sail]

But my question is can we find the sum of this series?

[unkempt drift]

power rule for integrals....

well just integrate 1 dx, -2x dx, 3x^2 dx separately

see what you get

[brittle sail]

x-x^2+x^3-x^4

[drifting seal]

$\sum_{n = 1}^{\infty} (-1)^{n - 1}n x^{n - 1}$

what is this the derivative of

[blazing coyote]

again

[unkempt drift]

yeah and do you see how that's a geometric series

[ancient grotto]

Mayhaps you can put this in a geometric series form?

[brittle sail]

It is but terms are alternative sign

[drifting seal]

and

[unkempt drift]

yeah so the common ratio will be negative

[ancient grotto]

So what?

[drifting seal]

^

[ancient grotto]

Just slap a (-1)^n in front

[potent lotus]

knief

[brittle sail]

No

[drifting seal]

word

[brittle sail]

if we can do this directly then why asked me to do integration

[ancient grotto]

Because its clever

[brittle sail]

.

[ancient grotto]

It makes you think 🗣️ (closed form is easier to find)

[brittle sail]

What is the purpose of this integration

[drifting seal]

to find the sum bruh

[brittle sail]

x-x^2+x^3-x^4

[ancient grotto]

^

[drifting seal]

^

[brittle sail]

I am asking this

[ancient grotto]

You can easily just use the geometric series here, then differentiate your result

[drifting seal]

$\dv{x} \left(\sum_{n = 1}^{\infty} (-1)^{n - 1} x^n\right) = \sum_{n = 1}^{\infty} (-1)^{n - 1} n x^{n - 1}$

[potent lotus]

knief

[brittle sail]

And how we will get sum of it which is asking 4th time

[drifting seal]

the left series has a nice closed form

a/(1-r)

[brittle sail]

I know the sum but I don't know when it has (-1)^(n-1)

[drifting seal]

$\sum_{n = 1}^{\infty} -(-x)^n$

[potent lotus]

knief

[drifting seal]

left side before the derivative

[brittle sail]

You are not understanding my problem

[drifting seal]

no you’re not understanding

[brittle sail]

Do you think i don't know the sum of x+x^2+x^3

[drifting seal]

i moved the (-1)^n and x^n into one

[brittle sail]

What one?

[drifting seal]

ratio is -x

[drifting seal]

here

on the left

[brittle sail]

So we can take it out of sum?

[drifting seal]

$(-1)^{n - 1}x^n = (-1)(-1)^n x^n = (-1)(-x)^n$

[potent lotus]

knief

[drifting seal]

^

r = -x

[drifting seal]

no

[brittle sail]

You are explaining wrong algebra to me

[drifting seal]

right..

[brittle sail]

I know such things

[drifting seal]

🤣

wrong algebra

point out my mistake please

[brittle sail]

Bro i didn't say you are wrong

I am saying you are pointing out wrong things

[drifting seal]

what does this mean then

[brittle sail]

I am not stuck there

[drifting seal]

where are you stuck?

[brittle sail]

Good question

x-x^2+x^3

so sum will be here

ratio is -x

[drifting seal]

^

mhm

[brittle sail]

x/(1+x)

[drifting seal]

yep

[brittle sail]

okay so what we will do next?

With this sum?

[drifting seal]

take the derivative

[brittle sail]

Fine

[unkempt drift]

we integrated before, so now you have to differentiate back

[brittle sail]

Yes right

[drifting seal]

^

we have d/dx (x/(1 + x)) = series we want

[brittle sail]

1/(1+x)^2

Hmm i got it now

(1+x)^-2 yes it is same by binomial expansion

[unkempt drift]

yep and one more thing, radius of convergence = ?

[drifting seal]

🔥

[unkempt drift]

nice

[brittle sail]

so here it will converges only when |x|<1

[brittle sail]

sure

[unkempt drift]

yay you got it!

[brittle sail]

For radius of convergence what will be an?

[drifting seal]

wdym?

you mean like

$\sum a_n x^n$

[potent lotus]

knief

[brittle sail]

yes

so here an will be (-1)^n-1(n+1)

[drifting seal]

just the coefficient on the right here

[drifting seal]

well that isn’t quite the original series is it?

[brittle sail]

Yes radius is1

n=0 to infinity

(-1)^n

Okay so radius is 1 now should I have to check it over x=1

1-2+3-4-5

[drifting seal]

$\sum_{n = 0}^{\infty} (-1)^n (n + 1) x^n$

[brittle sail]

this is wrong

first term is not -x

[potent lotus]

knief

[brittle sail]

Good

[brittle sail]

...

[brittle sail]

.

[drifting seal]

yea let x = 1 and x = -1

[brittle sail]

I see

[brittle sail]

For x=1

x=-1 i got

-1-2-3-4-5...

[drifting seal]

hmm x = -1 you should just get the sum of n + 1

1 + 2 + 3 + …

= -1/12

[brittle sail]

Yes

n(n+1)/2

[drifting seal]

well

[brittle sail]

What?

[drifting seal]

it goes to infinity

[brittle sail]

yes

1

[drifting seal]

it’s a joke (sort of)

[brave girder]

Is that my goat knief? 🔥🐐

[drifting seal]

,w zeta(-1)

[potent lotus]

Results provided by WolframAlpha

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Requested by kniefy

[drifting seal]

maybe

[brittle sail]

I don't think we need zeta here?

[drifting seal]

if you let x = -1 the (-1)^n and (-1)^n becomes 1

so you have the sum from n = 0 to inf of n + 1

or just the sum of the naturals

which diverges

[drifting seal]

don’t worry about this lol

[brittle sail]

.close

[calm coral]

Channel closed

Closed by @brittle sail

Use .reopen if this was a mistake.

[brittle sail]

.reopen

[calm coral]

✅

[brittle sail]

At x=1 it will be like

1-2+3-4+5

Which is oscillation?

[drifting seal]

divergent

[brittle sail]

Can I say oscillation in finately

@drifting seal

[drifting seal]

what does that mean

oh infinitely?

well that misses the point right

[brittle sail]

Which one?

[drifting seal]

we want to know if it converges or diverges

saying it "oscillates" isn’t precise

could be taken to mean it alternates

which tells us nothing about convergence

it diverges by the nth term test (limit of terms is nonzero)

[brittle sail]

Thanks

.close

[calm coral]

Channel closed

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[calm coral]

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[blazing coyote]

What I did was re-write this as $v_m + \sum_{i=1}^{m-1} \lambda_i(v_i -v_m)$

[potent lotus]

wai

[blazing coyote]

This is obtained by setting $\lambda_m=.1-\sum_{i=1}^{m-1} \lambda_i$

[potent lotus]

wai

[blazing coyote]

We this is a translate of the subspace generated by $\sum{i=1}^{m-1} \lambda_i(v_i -v_m)$

[potent lotus]

wai

[velvet osprey]

.1?

why did 0.1 (one tenth) come into this?

[velvet osprey]

your notation is off i think.

[blazing coyote]

typo

yea

[velvet osprey]

maybe you meant that A-v_m is generated by the vectors v_i-v_m for i from 1 to m-1?

[blazing coyote]

Let me retype it

[velvet osprey]

but then you confused "generated by" with "consists of"

[blazing coyote]

This is a translate of the subspace spannedby $\lambda_i (v_i -v_m); 1≤i≤m-1;\lambda_i \in F$

Well, that's a clunky way to express it

[potent lotus]

wai

[velvet osprey]

you should have dropped the lambdas

this is like saying R^3 is generated by xi, yj and zk for x, y, z in R instead of saying it's just generated by i, j and k

[blazing coyote]

my bad

yes

okay, I think I'm done

Thanks!

.close

[calm coral]

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[blazing coyote]

<@&268886789983436800>

[calm coral]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[terse shore]

how do u make this jump

[topaz raft]

distribute sqrt(u)

[velvet osprey]

the two halves were pulled out in front

and that

[terse shore]

u can just bring out the half from du??

[velvet osprey]

yes, it is still just a 1/2 multiplier

[terse shore]

.close

[calm coral]

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[glacial umbra]

When building a high-rise building, you first need to dig a large hole to pour a foundation. A person stands 2.4 meters from the edge and sees just the far corner of the hole. How deep is the hole? The distance between the person's eyes and the ground is 1.8 meters.

[glacial umbra]

I need help

[keen flare]

it also helps to write out the 1.8 m measurement

[rocky flower]

Do you know similar triangles?

[glacial umbra]

yes

same shape but different sizes

[rocky flower]

Okay, you will have to use it here to determine the depth of the hole

[rocky flower]

Yes, but you have to be careful because they could have the same size and still be called similar. You mean “not necessarily” the same size

[glacial umbra]

yes

[rocky flower]

Okay do you see that there are two triangles you can make? What do you notice about their angles?

[glacial umbra]

90 degrees

[jolly pilot]

uhh, the 1.8m distance mentioned in the problem is not what you have marked in this figure

[glacial umbra]

yes does it represent the depth right

so 2,4/28=1,8/x or am i wrong?

[jolly pilot]

its the (almost) height of the person

[jolly pilot]

yea, the equation is right

[glacial umbra]

so irs 21