help-36

[crystal lion]

im not too sure how to get the second solution of -1/2 ln2, ive tried simplifying and using the log rules for the part in red writing but i keep getting 1/2 ln2

[blissful meadow]

The value you get in the red part you wrote isn't defined in the real numbers since you have ln(**-**sqrt(2)/2)

[crystal lion]

yeah

im not sure how else to manipulate the form of it

[blissful meadow]

Well cosh is even, so its inverse is only defined on half the domain.

It's kind of like solving $x^2 = c$, you need to take $\pm$ the solution you end up with

[soft zealot]

Azyrashacorki

[crystal lion]

so there's only one solution?

[blissful meadow]

No what I mean is that your arccosh (your log formula) only spits out the answer which is >= 0, but there is another answer <= 0

[crystal lion]

ohhh

[blissful meadow]

So in other words, $arccosh(cosh(x)) = \abs{x}$

[soft zealot]

Azyrashacorki

[blissful meadow]

So if you solve for $\cosh(x) = c$, then the answer (provided $c \ge 1$) is $x=\pm arccosh(c)$

[soft zealot]

Azyrashacorki

[crystal lion]

so u could just simply make the second solution be the negative version of the first solution?

[blissful meadow]

Yes

[crystal lion]

okk thxx

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[hard blaze]

during the last two lines can someone explain to me why p and q switched signs?

[hard blaze]

did the book make a mistake or is there something I'm not seeing?

oh wait im dumb they multiplied everything by -1

[vital crag]

yes

[proper dagger]

mhm

[hard blaze]

i swear i started at the answer sheet for 5 minutes and couldn't figure it out

as soon as i post it here i figure it out smh 🤦‍♂️

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[ruby crypt]

Infimum = 1, Supremum = e. Am I missing something?

[ruby crypt]

I just need clarification

[odd rivet]

lets say the set would be just (0,1)

do you know what the supremum would be then?

[ruby crypt]

basically its like upper bound right?

smallest upper bound

yea its e no?

[halcyon vine]

the math ping is too big

[odd rivet]

yes

[odd rivet]

yes

[ruby crypt]

ok, so my answer is correct right?

[slate tiger]

wtf happened to the serv icon

[safe cedar]

lmaooo

[slate tiger]

no its the icon ig

[odd rivet]

wrong channel guys

[ruby crypt]

lol

[slate tiger]

MB

[hybrid heath]

[ruby crypt]

@odd rivet what did I miss?

[kind cove]

guys...did the server face change in ur dc too?

like theres a big one written over there?

[odd rivet]

so do you know what you need to change if you intersect the set with the rational numbers?

[ruby crypt]

y'all talking here like its #general

[odd rivet]

keep in mind that exp(sqrt(x)) is a continues function

[true pawn]

april fools

[ruby crypt]

wait I just realized rational is not same as real number

[true pawn]

lol

the icon changed again

[kind cove]

ow silly me

[hybrid heath]

!redir

[final saddle]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[true pawn]

im sos orry

[ruby crypt]

rational is the number that can be represented as m/n with m,n being integers right? @odd rivet

[odd rivet]

yes

[ruby crypt]

ok

uh

[hybrid heath]

(And n is non zero)

[ruby crypt]

yes

[rapid crypt]

What subject does this require to solve this , id like to learn it

[ruby crypt]

9/10, 10/11, 11/12, 12/13, ... 999/1000

[ruby crypt]

math analysis

[hybrid heath]

Real analysis you mean?

[ruby crypt]

no

its like introductory lesson before real

im lost @odd rivet , won't the limit tend to e at x = 1

how ddoes it change anything

[hybrid heath]

I'm not tobi, but yes it should

[ruby crypt]

yea idk if he is complicating things or what, but i gotta listen first

[odd rivet]

I just want to make sure you understand the question correctly

[ruby crypt]

and there is probably a reason why its intersection with Q and not R

[ruby crypt]

a

I understood no worries

[odd rivet]

the thing is the supremum

is only accurate up to an epsilon

for continues functions

[ruby crypt]

the same epsilon from formal def right?

wait its that deep??

[odd rivet]

yes to formally understand it you need to think about epsilons

and that Q is dense in R

and that you have a continues function here

[ruby crypt]

yes

notice that it is part a, part b is absolute hell lol

or I just do not have the energy to solve at the moment

I will go sleep

thanks guys

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[halcyon vine]

ofc ik

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[runic phoenix]

Find a number 𝑡 such that
(3, 1, 4), (2, −3, 5), (5, 9, 𝑡)
is not linearly independent in 𝐑3

[runic phoenix]

Okay

[abstract bramble]

make matrix?

[runic phoenix]

no

I can't work with that

I am using LADR

I am unsure on whether the linear dependence lemma works

for this

I am sure it is a nice way to solve it

I just do not know how to solve it

[candid hull]

which is ?

[runic phoenix]

I am thinking on rewriting the last vector with the other vectors

I do not know how to solve this

I could not solve this yesterday

I have not taken any computational courses

[abstract bramble]

what does it mean for the three vectors to not be linearly independent

[runic phoenix]

that I can get their sum to be 0 with scalars not all zero

That is not a problem

Then it is linearly dependent

yeah

[abstract bramble]

you would have
aA + bB + cC = 0
so aA + bB = -cC
but they are nonzero scalars so divide by -c
you still get something like
aA + bB = C

[runic phoenix]

I am thinking of making $$\forall a_1, a_2, a_3 \in \mathbb{F},$$ not all zero,
$$a_1(3,1,4) + a_2(2,-3,5) + a_3(5,9,t) = (0,0,0)$$

[abstract bramble]

Hmm why need a3 though

[candid hull]

you don't know whether c can be nonzero or not

it's at least one of them nonzero

not necessarily all

[quiet carbon]

guys any one have 6month + weechat account?

[candid hull]

also just setting up the system like andre did, it's exactly the same thing as setting up a matrix like you suggested

=(0,0,0) tho

then solve for a1 a2 a3

[runic phoenix]

okay

sure

[soft zealot]

andre

[candid hull]

and hopefully there's a t for which you get solutions other than a1=a2=a3=0

this is what you're looking for

[runic phoenix]

I was stuck at this bit yesterday

my intuition is that I can just get scalars that span the last vector since I can just solve for t at the end

and that is exactly the problem; I don't know how to

$$
3a_1 + 2a_2 = 5 a_3 \implies a_1 = \frac{2a_2 - 5a_3}{3} $$

[soft zealot]

andre

[runic phoenix]

Am I in a good path here?

[candid hull]

you've never done gaussian elimination then ?

[runic phoenix]

no

[candid hull]

if you say you never had a computational class

[runic phoenix]

sadly

[candid hull]

you better remedy that then

[runic phoenix]

well

okay

[candid hull]

it's like the algorithm of intro lin alg

[runic phoenix]

I guess I will work with that

thanks

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[coarse mantle]

$\oint_L{(xy+x+y)dx + (xy+x-y)dy}$, $L: x^2 + y^2 = 4x$

[soft zealot]

Bakoles

[drowsy epoch]

What's your question?

[coarse mantle]

this is what i did so far

[coarse mantle]

i need help?

[drowsy epoch]

i see

i would have used a diff sub

[coarse mantle]

hm

[coarse mantle]

wdym

[coarse mantle]

huh

that's

not polar coordinates

[drowsy epoch]

what is it then

[coarse mantle]

it's parametric

[hallow yarrow]

1+1 = 3

[coarse mantle]

polar has rho and phi

[drowsy epoch]

ok

use green's theorem gng

[vital crag]

<@&268886789983436800> troll

[hushed sequoia]

@hallow yarrow please don't use help channels to troll.

[tidal beacon]

not the place for such idle meanderings

[drowsy epoch]

yeah this isnt modular arithmetic anyway rn

[coarse mantle]

how do i find the bounds for rho

[drowsy epoch]

if you use my sub what is rho at most

[coarse mantle]

2

[drowsy epoch]

ye

[coarse mantle]

idk

i mean

isn't rho supposed to be a function

[drowsy epoch]

rho=4 is still a function

it's the boundary of your disk

[coarse mantle]

ok

[drowsy epoch]

you can also do default polar sub and get something in terms of cos(theta) but thats not so elegant

[coarse mantle]

something is wrong

[drowsy epoch]

So you used the standard sub

[coarse mantle]

yeah

i think phi might be on the wrong interval

[drowsy epoch]

no it seems correct

[coarse mantle]

hmm

[obtuse grail]

check your phi

[coarse mantle]

yeah

[drowsy epoch]

you forgot the jacobian it seems

[coarse mantle]

ohhh

yup

i did

[drowsy epoch]

phi should be correct

[coarse mantle]

thanks

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[hot pawn]

I would appreciate help on this question! Not the answer, but I don't know how to go about solving it. The first time I answered, I got it wrong. I answered A'=(2,4) B'=(1,1) and C'=(4,2). I kinda was just guessing.

[drowsy linden]

Well do you know the definition of vertices?

like do you need an explanation or just a clear answer?

[muted prairie]

your answer is dilated with respect to the origin, not with respect to D

[lofty goblet]

you are translating it so that D becomes the origin, scaling, then translating back

[muted prairie]

what's with the weird pictures

[hot pawn]

Where the two points/lines meet?

[drowsy linden]

Yes

[hot pawn]

Explanation

[drowsy linden]

The answer is doesn't involve D tho

So now try solving

Ill explain in a moment after you attempt it

[hot pawn]

I'm just not sure how to dialate with D and not the origin. I was taught this at the very beginning of the unit and theres been a lot of stuff so i forgot how to do it lel..

[muted prairie]

if you move the origin to D, what do the coordinates become?

[hot pawn]

im still confused

[hot pawn]

O

[drowsy linden]

Yea that lol

You beat me to it

[muted prairie]

the coordinates of A,B,C

[hot pawn]

b is 4,4 a is 6,10 c is 10,6?

[muted prairie]

yeah

[drowsy linden]

Yes

It's tricky to understand the question but its simple in practice

Do a few more and you'll be clearing these questions no problem

[hot pawn]

soo if i dilate it by 1/2 it would become 2,2 3,5 and 5,3 but those werent in the answer choices iirc

[muted prairie]

because that's when D is the origin

then you have to rememeber to switch the origin back

[hot pawn]

OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOH wait a minute

[drowsy linden]

You are talking D as the beginning point. You gotta reverse it

[hot pawn]

yeah nvm i thought i got it but no

uhhhmm

lemme think

[drowsy linden]

Move back to origin after

[hot pawn]

wouldnt that just give me the same answer

im so lost

i dont think i understand what you guys are saying

uuuuuuuufgjhdsfgjhglirghlkdjcnpojhrf

looked thru our notes and my teacher did not teach this ever lol no ownder idk how to do it

[hot pawn]

after which step? after i get the 4,4 6,10 10,6 coordinates?

is B (0,0)?

because the coordinates from D are 4,4 but it a dilation of 1/2 and if you count 2,2 from D it lands on 0

or am i completely making that uo

up

<@&286206848099549185>

[south dirge]

once you got this,

these coords were with respect to D as origin

well now when switch out origin back, they would change too right?

[hot pawn]

uhh ys

yes**

[south dirge]

Right well then you just need to shift them back!!

Oh i see where your confused

You agruement would be that they just come back as when they were not reversed due to D and dilation?

[hot pawn]

aaargh im just not seeing how my answer is going to change because when i try to do it the way someone tells me i get (2,2) or (4,4) for B

i just dont know what im doing

ive been doing schoolwork for 8 hours lol my brain is fried

Could you maybe explain to me how to get the B' coordinate? like the steps

if thats allowed im not sure if it is

[hot pawn]

yes this

because if i count to 4,4 for b after i do the dilation it would just be 2,2 again

i feel like im going in circles

[south dirge]

Sorry, was lil busy

[hot pawn]

its alright im just very stuck

[south dirge]

One sec am helping somelse really sorry,
you could try another helper in the time being

[hot pawn]

its okay!! no rush, im the one asking for help after all

i need to do something real quick anyways

[south dirge]

Hey sorry about that

[south dirge]

b is 4,4 a is 6,10 c is 10,6?

This 👆 was correct!

[south dirge]

correct!

[south dirge]

No see, thats where your going wrong, you are not dilating the entire coordinate system, only the traingles point ABC

[south dirge]

So after this step, all you gotta do is shift the D(new origin) back to the old origin.

So in short, just subtract 2,2 to each point (ABC)

[south dirge]

similar method applied as what you did here too!

[final saddle]

@hot pawn Has your question been resolved?

[urban ice]

need help in math

[vital crag]

#❓how-to-get-help

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[sour fiber]

Ik what x is

I also know

There’s a correlation between x and y that I’ve forgot

Or y and 65

X is 130

[rugged stone]

u got x right

[sour fiber]

How do I get y

[rugged stone]

u have to use ABCD

[sour fiber]

Ohhh

Im still a bit lost I know know d

[rugged stone]

ABCD is a cyclic quadrilateral

[scarlet mango]

whats the problem

what do i find

[sour fiber]

Idk how to get y

[scarlet mango]

whats the problem translated to english

[lofty goblet]

use properties of this

[rugged stone]

what is do you know about it s opposite angles

[sour fiber]

Find x and y

[scarlet mango]

they're supplementary

but what am i supposed to do

how do i translate problem to english

[sour fiber]

The question is

Find x and y

[rugged stone]

use that

[sour fiber]

That’s all

Can u explain cus I don’t use englihs terms

[rugged stone]

what language u speak

[scarlet mango]

y is 130 as well?

[scarlet mango]

english

[rugged stone]

no

[scarlet mango]

it looks obtuse but apparently its 50

[sour fiber]

I thought it was so I checked and it’s not

[sour fiber]

Swedish

[scarlet mango]

is it 50

[sour fiber]

Its 115

[scarlet mango]

wtf

how??

[sour fiber]

Idk

That’s what the answer sheet said

Im tryna figure out how

[scarlet mango]

...

[rugged stone]

so the ADC is opposite to ABC

[kindred mortar]

is M the center of the circle ?

[rugged stone]

then 65+y =180

[sour fiber]

Yep

I don’t get itttt

[rugged stone]

that is a priority of cyklisk fyrhörning

[sour fiber]

A figure with four corners

Oh I think I’ve heard it before

Just forgot it

What’s the correlation between the opposite corners

[rugged stone]

Motsatta vinklar i en cyklisk fyrhörning summerar till 180∘.

[sour fiber]

Ohhhh

OH

OHHHHHH

thank you

[rugged stone]

you are welcome

[sour fiber]

I remember I’ve heard that I just forgot it

[sour fiber]

The opposite corners make 180 together so 180- the corner we know which is 65

Anyway thanks again

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[sour fiber]

So i basically only know that one corner is 90

[whole halo]

is that all they tell you on this problem?

as it is, you dont have enough to figure this out

[sour fiber]

Yep they say deside all the angles

[whole halo]

well here's one idea you can try

can you figure this one out from x?

[sour fiber]

I tried doing this hold on

[whole halo]

then you have that, not very useful but it's there

[south dirge]

!xy

[final saddle]

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

[sour fiber]

I got a rlly big decimal though

[south dirge]

that doesnt seem right

how did you get here?

[sour fiber]

I thought opposite angles in a figure with four corners make 180*

[lofty goblet]

eh i got a value of x

idk if im wrong or smt

[south dirge]

nope idts thats true for all quardilaterals

[south dirge]

what did you apply?

[sour fiber]

Wdym

[south dirge]

well take a trapezium for example

[lofty goblet]

draw line from centre to that other point

isosceles triangles

[whole halo]

idts means "i dont think so"

[south dirge]

they arent isosceles

[lofty goblet]

but its all same radius

[south dirge]

um sry? did i say smth wrong😭

[drowsy epoch]

idts

[sour fiber]

I figured the radius needs to have smt to do with it

[whole halo]

replacing idts with I dont think so in your message gives "nope I dont think so thats true for all quadrilaterals" which is the opposite of what you intended

it's what it's

[bold turtle]

Opposite angles in a cyclic quadrilateral sum to 180 degrees

[rugged stone]

so this is 135 according to the rule u have it and the right angle ends on the sam point on the circle so $frac{360-90}/{2}$

[soft zealot]

NoX

[sour fiber]

Ohh thank you

[whole halo]

someone else go convince nox thats not true

Im going to do something else

[south dirge]

-# Oh i see, so in mathematical terms i just put two negatives

[lofty goblet]

is M even the centre

[sour fiber]

It is

[whole halo]

thats not correct either

[south dirge]

Yes it is

[south dirge]

-# okay just leave me alone😭😭😭

[sour fiber]

Ohh

[rugged stone]

why is that?

[south dirge]

oh you meant those two points

[lofty goblet]

yeah

mb

[south dirge]

Yeah that seems solved then!!

[sour fiber]

Huh

I am still confused

[south dirge]

Well ill leave it to kiwie then

Since he solved it

[lofty goblet]

😟

[sour fiber]

Kiwi help

[bold turtle]

Looking at that annotated diagram, can you see how introducing the line MB gives us two isosceles triangles?

[sour zealot]

you have isosceles triangles now, since the radii of a circle are the same

[sour fiber]

What is isosceles

I don’t use englihs terms

[lofty goblet]

triangle with 2 lengths that are the same

[rugged stone]

du har likbenta trianglar nu, eftersom radierna för en cirkel är desamma

[sour fiber]

Oooooohhhh

OH

YES

I see it now

And how does that help me..

Omg should I give up

[whole halo]

it turns out theres nothing immediately interesting happening anywhere else

I am going to say

[bold turtle]

Can you deduce what the angles around B are?

[whole halo]

yall are missing an easy way to find this missing angle

[rugged stone]

deduce = härleda

[sour fiber]

Lol

Thanks

[lofty goblet]

just MAB+MCB

[sour fiber]

Idk tbh

[whole halo]

you need to consider the "cyclic" version of this quadrilateral

you mentioned it earlier
if you were to move point M but keep A, B, C fixed, you can prove it that way

[sour fiber]

Remove M?

[south dirge]

i sse your point being, but wouldnt that be more lenghty

[sour zealot]

oh yes mtt is correct

[lofty goblet]

yeah

that workstoo

[whole halo]

it is a grand total of 2 steps

[south dirge]

i know but circles properties arent that well easy to grasp

[whole halo]

both rules are commonly taught

so no

[south dirge]

-# Kay go ahead then🙏

[whole halo]

you first have to use one thats taught as one of the "first" circle properties

[bold turtle]

Move - not remove
Move the point M onto the circumference (call this point M'), and keep A, B and C where they are

[whole halo]

then after that, use that cyclic thing youve heard about to turn this angle M to angle B

[sour fiber]

Where am i moving it

[bold turtle]

What could you say about angle AM'C (compared to angle AMC)?

[whole halo]

oh right, "circumference" = "edge of the circle"

[bold turtle]

"omkrets"

[sour fiber]

Ohhhhh

[sour zealot]

wait yall are talking about ||inscribed angle thm|| right

[whole halo]

lower-left works

[sour fiber]

So if I move it

[whole halo]

yep

[marsh gorge]

Who needs help

[lofty goblet]

yeah ||1/2*270 for ABC||

[bold turtle]

oh I wonder given the channel name, truly a difficult question /s

[sour zealot]

[sour fiber]

Where I put the arrow?

[bold turtle]

^

[marsh gorge]

@sour fiber how can I help

[sour zealot]

well it technically can be any one point on the circle

[bold turtle]

(meaning I need to draw in AM' and CM')

[sour zealot]

but yeah, what you drew works

[whole halo]

yes

[whole halo]

no, it has to be on major arc AC

[sour fiber]

What happens if u move it thre

[sour zealot]

oh yeah 💀 any point that forms a cyclic quad with vertices MABC

if you name it clockwise

[whole halo]

waes I think you should take it from here

you actually speak the language

[sour zealot]

can you find a way to now relate the original AMC and the new AMC?

[whole halo]

(unless you just google translated the word in which case nvm)

[sour fiber]

Maybe that the opposite angles Make 180

[bold turtle]

I don't

I did just GT it

[whole halo]

dammit

[rugged stone]

@whole halo while your method is right i don't know why reject my method and they both gives the samethink + mine is faster

[bold turtle]

Incidentally - @sour fiber we're referring to this, if that jogs your memory:

[drowsy epoch]

-# congratz on 🟨

[sour zealot]

hmm, another property? have you heard about inscribed angle theorem before?

[sour fiber]

Ohhh but I don’t see one of those in this

[whole halo]

sorry, I cant understand a word you said when you wrote down "(360 - 90)/2" which also isnt exactly correct either

[sour zealot]

im pretty sure helping in channels decreases your virtue 💀 i got yellow from sending a message here, hahha

[bold turtle]

We've built one - look at angle AMC and AM'C (where M' is that new point we've been making)

[sour fiber]

Have we?

Wait y mean

When we moved m

[whole halo]

M' is the new M

[sour fiber]

Does that jus make a longer square

[lofty goblet]

i think they mean major arc AC becomes 360-AMC because B is on minor arc AC

[sour fiber]

We have two m

[bold turtle]

Well

The original centre, which is M

[rugged stone]

the reflex angle at M angle, at the center should be 360-90 then it is the same from here just devide by two same, same but different

[bold turtle]

And the extra point we've made, which (at least I am, but) we're calling M**'**

[sour fiber]

New m is 180? Since old m is 90

Or

Nvm

[bold turtle]

Other way around

[sour fiber]

45

[whole halo]

yea thats better explanation

[sour fiber]

Mb

[bold turtle]

ye

[whole halo]

thats actually really good, thats one step instead of two

we can show that one later on after this one

[bold turtle]

Now look at the quadrilateral M'ABC

[sour fiber]

Yes

[bold turtle]

All points lie on a circle, so what have we got here?

[sour fiber]

180-45

Is it 135

[bold turtle]

For which angle, to check?

[sour fiber]

B

[bold turtle]

Yeeee

[whole halo]

very nice

[sour fiber]

Yayyyy

[whole halo]

now so far we've done it with 180 - 90/2

[bold turtle]

Now we have all four angles of our original quadrilateral MABC - now time for the algebra

[sour fiber]

Yes

[whole halo]

oh right

[bold turtle]

This should hopefully be the smooth-sailing part

[sour fiber]

So

Wait

Lemme try

This right?

Flip it dammit

I forgot 90

[lofty goblet]

,rccw

[sour fiber]

Throw that in there pretend it’s there

[marsh gorge]

You can check your answer by solving for x

[bold turtle]

...?

[marsh gorge]

My fave thing to do on here is to click on people profiles and listen to the songs they listen to, Like Ve's

[bold turtle]

Solving for x IS what the question's asking for

[sour fiber]

I got x to 16

15

[marsh gorge]

Ohh I didn't read the question i assumed that's the answer

[sour fiber]

OHHHH

Omg that wasn’t that hard

[lofty goblet]

good job

[sour fiber]

YAY

[marsh gorge]

So u don't need help anymore?

[sour zealot]

congrats! c:

[sour fiber]

TYYY

[sour fiber]

Nop

[bold turtle]

Wait

[marsh gorge]

Check ur dms

[bold turtle]

You sure?

@sour fiber Did you sum to 260 by any chance?

[sour fiber]

Wdym

I haven’t done the like corners yet I only jus got x lemme check if it’s correct

It’s right

Dw

I checked answer

[marsh gorge]

@sour fiber i sent u a dm, check, really important

[sour fiber]

Thanks moey but I’m good

[bold turtle]

oh wait, yh, you did say you were missing the 90 in the thing

[sour fiber]

Thanks for the concern though

[bold turtle]

nvm, you're good

[marsh gorge]

NP

ok if u dont need anymore help imma go

have a good day and dont forget to do .close when youre done

[sour fiber]

U tooo

.close

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[rain agate]

Hi, how do you simplify -i over 2i?

[royal gust]

You're probably expecting this, but i/i = 1

[rain agate]

I got it wrong chat

[royal gust]

I actually liked the suggestion that was deleted. Multiplying by i/i is an interesting thing you can do here

[royal gust]

Note I am not answering your original question here, but a related idea

[rain agate]

Wait was I supposed to put in 1? 😭

[karmic glen]

oh I thought it was silly haha

[rain agate]

Okay guys, how do ya simplify 5i/-4i?

[royal gust]

Similar idea. i/i = 1

So we're left with 5/(-4) = -5/4

[rain agate]

Ohh okay, thanks

How do ya simplify 4/3i? There's no other i

[chrome heart]

what is 1/i

how can u simplify that

[karmic glen]

ooh multiply by i/i

[rain agate]

just i?

[bold turtle]

Careful

If you're multiplying top and bottom by i, the top becomes i, sure - but what does the bottom become?

-# Use \ to escape Discord formatting an initial - as a bullet point, if that ends up happening

[rain agate]

Wouldn't the top be 4i?

[bold turtle]

@rain agate I believe we were first addressing this

-# We need to make sure this is down correctly before tackling the 4/3i one

[rain agate]

i/i?

[bold turtle]

So, as in, if we want to realise 1/i, we can multiply top and bottom by i

(cf. rationalising fractions with surds in the denominator)

[rain agate]

What are surds?

[bold turtle]

hoo boy

https://www.bbc.co.uk/bitesize/guides/zg6vcj6/revision/1 (I'm hoping this can be accessed)

But you typically should have come across these before dealing with complex numbers, fwiw

[rain agate]

My teacher never taught me abt surds

[final saddle]

@rain agate Has your question been resolved?

[rain agate]

.close

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[chrome ermine]

int of ln(4-x^2) dx from 0 to 1

[drowsy epoch]

Can you factor 4-x^2

[lofty goblet]

Yes I can

[versed crater]

What have you really tried

Sometimes it’s good to just try the obvious thing and see where that leads you

[lofty goblet]

after doing this try thinking about what log property you can use

[thorny otter]

what is ln(0), beautifulsoap?

[chrome ermine]

i didnt do that

but i can

[drowsy epoch]

Do that and then this

[chrome ermine]

2-x * 2+x

then you set that to like ln (2-x) + ln(2+x)?

[drowsy epoch]

yes with parentheses please

[chrome ermine]

i started by going directly into ibp

is that wrong

[merry trench]

-inf

[drowsy epoch]

I prefer the word counterproductive rather

[chrome ermine]

ok

[final saddle]

@chrome ermine Has your question been resolved?

[final saddle]

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[final saddle]

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[clever carbon]

Ummm hlo

[clever carbon]

Anyone here?

I have a doubt

[sturdy flax]

!da2a, please. just send your question!

[final saddle]

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

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

[crystal pawn]

@clever carbon do !close here

[clever carbon]

!close

[sturdy flax]

.close, not !close.

[crystal pawn]

oops

.close

mb

😂

[clever carbon]

.vlose

.close

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[crystal pawn]

im used to ! lol

[clever carbon]

Ok yo so if I want to create a grafh lets say like a infinity symbol can I create it on my own?

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[still cradle]

I'm trying to solve for ∠B. This is what I have gotten so far?

[still cradle]

oh and ∠DEA is 100°

[quasi rapids]

BDA is 130 btw

[still cradle]

whoops, thank you for catching that

[quasi rapids]

ah this is a sine rule problem i think

[still cradle]

Where AE/sin50 = DE/sin30?

[quasi rapids]

AD/sin(100deg) = AE/sin(50deg)

[still cradle]

Thank you!

[molten forum]

if we are never given the length of the lines it's impossible to solve it

yes

because ABD and BAD can be anything

ANYTHING

and the 130 degrees will stay the same

[south dirge]

Are you sure?

[still cradle]

Okay so ∠CEA = 80° and ∠ECA = 80°, so it's an isosceles triangle where AE = AC
∠DAC = 50° and ∠CDA = 50° so it's also an isosceles triangle where AC = DC

So AE = AC = DC

[molten forum]

ok you extend the length of BD, will it affect the 130 degrees on BDA?

the answer is no

[quasi rapids]

so confident yet so incorrect

[south dirge]

-# Did you just contradict yourself?

[molten forum]

BD = CD

nvm

[still cradle]

BD = CE and DC = DE + CE

[molten forum]

I'm blind

shit

[still cradle]

And DC = AC

So AC = DE + BD

[final saddle]

@still cradle Has your question been resolved?

[still cradle]

Okay DC = DE + CE
And if we substitute CE for BD we get CD = BD + DE = BE
And AE = CD and CD = BE which means AE = BE

And that must mean ABE is an isosceles triangle

So ∠ABE and ∠BAE are equal

SO 180 = 130 + 2B - 30

[south dirge]

isnt AC = CD

[quasi rapids]

yup thats correct

[south dirge]

oh nvm

[quasi rapids]

now u only need angle AEB

which is 100deg

and since ABE is isosceles, the answer is just 40deg

[still cradle]

2B = 80 which means B is 40 degrees

[still cradle]

YIPPIEEE

[quasi rapids]

YAYYYY

[still cradle]

thanks yall

so much

[quasi rapids]

no need you did most of it on your own!

[still cradle]

you helped guide me and motivated me 🙏🏽

.close

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[rugged merlin]

,tex So for Fourier analysis, there are two transforms you consider. The Fourier Transform and the Inverse Fourier Transform:
\e{align}{
x(t) &= \int_{-\infty}^\infty X(f)e^{j2\pi ft}\dd t \
X(f) &= \int_{-\infty}^\infty x(t) e^{-j2\pi ft } \dd t
}
I have an understanding of (1) in the sense of it saying that complex exponential signals can form a basis to describe any (most) functions, but I can't wrap my mind around of (2), especially the negative sign in the power of the exponential. Can we see it as the inner product of two functions
[
(f,g) = \int f(t)\overline{g(t)}\dd t
]
which makes the negative sign appear?

[royal gust]

(2) is the inverse of (1)

Your explanation of (1) aligns with mine for the Fourier series but doesn't really describe the Fourier transform.

[desert mantle]

well the transform is just the continuous version of the series

so as far as intuition goes, it kinda translates

[royal gust]

Okay fair!
Is there a similar intuition for (2)? I actually don't know it

[desert mantle]

for the fourier series you have $\hat f(n) = \int f(x) e^{-nx} dx$ modulo some constants which I dont wanna bother with and then $f(x)=\sum_n \hat f(n) e^{nx}$. so its kinda a continuous version of that I suppose?

[soft zealot]

Denascite

[desert mantle]

whoops forgot the i in the exponent. but you get the idea

but of course the inner product is a good observation and the fourier transform should also be viewed as an operator on some L^2 space

[final saddle]

@rugged merlin Has your question been resolved?

[rugged merlin]

aight thanks

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[gentle zephyr]

\textbf{Exercise 3.} Determine, for each $n \in \mathbb{N}$, the remainder of the division by $50$ of
[
3(-49)^n + 151^{3n+1} + \sum_{k=0}^{n} 2^k (2k)!
]

[soft zealot]

Renato

[stone wagon]

what have you tried?

firstly, what do the first two terms become mod 50?

[gentle zephyr]

3 and 1

[stone wagon]

ok nice

[gentle zephyr]

then what

[stone wagon]

compute a couple of terms out and see if you notice a pattern

try for n=0, n=1, etc.

[gentle zephyr]

n = 0 is 1

[stone wagon]

or actually notice what happens at n=5

what is 2^5 (2*5)! mod 50

[gentle zephyr]

who knows

[stone wagon]

well 10! = 10*...*5*...*1

[gentle zephyr]

ok so n >= 5 is 0

[stone wagon]

yup

the rest you just compute ig

i don't think there's a trick

the only trick i suggest is doing everything mod 50

[gentle zephyr]

dude shit seemed intimidating but it actually wasn't that big of a deal if you notice n >= 5 is 0

[stone wagon]

that's how they get you

[gentle zephyr]

i appreciate the help

[stone wagon]

ofc

[gentle zephyr]

.solved

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[worldly spruce]

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[rough nimbus]

What went wrong? What am I lacking here?

[rough nimbus]

For correction, in last step
3/4(sin10 + sin70 - 1)

[final saddle]

@rough nimbus Has your question been resolved?

[rough nimbus]

<@&286206848099549185>

[strange pelican]

hi

[strange pelican]

you messed here

,, \sin \theta.\cos (\frac{4\pi}3) = \sin \theta.\cos(\pi + \frac{\pi}3) \
= \sin \theta(-\cos(\frac{\pi}3) )

[soft zealot]

professor paradox

[rough nimbus]

AH- thanks for the help Prof 😭

itshould be 0 right?

the ans

[strange pelican]

yes

np

[rough nimbus]

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[warm python]

Need (b) and (c) checked

[warm python]

Let $x$ be rational. We then have $\frac{1}{q}≥ \frac{\varepsilon}{2}>0 \iff \frac{2}{\varepsilon}≥q>0$. Which is finite. At the same time we have $p≤q$. We thus find $D_{\frac{\varepsilon}{2}}$ is finite

[loud sundial]

Forgive me if I'm being thick, but what?

[warm python]

FUCK

[loud sundial]

[soft zealot]

Wai

[loud sundial]

Assuming that x=p/q in lowest terms, this is fine. Maybe briefly specify why you don't care about irrationals, but this should be fine.

ik you're probably just sketching things

[warm python]

thanks

[loud sundial]

[warm python]

I'll work on (c)

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[warm python]

.reopen

[final saddle]

✅ Original question: #help-36 message

[warm python]

oh, that's easy

we make our partition of those points in $D_{\varepsilon/2}$ then the upper sum is less than $\varepsilon/2$

[soft zealot]

Wai

[warm python]

.close

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[warm python]

It converges pointwise to $0$.
I suspect it's not uniformly convergent on $(0,

[soft zealot]

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

[warm python]

I accidentally sent it

was typing out the entire thing

[signal vector]

Not quite sure about it converging pointwise to 0. For fixed x,
\begin{align*}
\lim_{n\rightarrow\infty}\left(\frac{nx}{1+nx^2}\right)&=\lim_{n\rightarrow\infty}\left(\frac{1}{1/nx + x}\right)\
&=\frac{1}{x}
\end{align*}

[soft zealot]

what

[warm python]

huh

right, okay

thanks

hmm, now to show it's not uniformly convergent at 1

[signal vector]

Uniform convergence isn't a property of a function at a point

[warm python]

I know I meant it breaks down at 1

okay, got it

thanks

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[grim frigate]

hello is anyone here?

[rugged merlin]

yes

[quasi rapids]

hi!

[grim frigate]

I'm going to ask if I graphed something correctly

[leaden moon]

what's your question?

[leaden moon]

!da2a

[final saddle]

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

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

[leaden moon]

ask right away

[quasi rapids]

woohoo da2aja

[grim frigate]

actually nevermind I think I have it correct

.close

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[grim frigate]

o u h

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[warm python]

Each $f_n$ is continuous at $0$ as by the archemdian property there's $m$ such that $\frac{1}{m}< \frac{1}{n}$ giving us an interval in which the function can be brought arbitraily close to $0$.

[soft zealot]

Wai

[rugged merlin]

sure

what about the rest

[warm python]

It doesn't converge uniformly

[rugged merlin]

yes

why

[warm python]

for fixed $n$ you can only get at most $\frac{1}{n}$ close

[soft zealot]

Wai

[final saddle]

@warm python Has your question been resolved?

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[sturdy flax]

<@&268886789983436800>

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[warm python]

\begin{enumerate}
\item Define $f_n=x^n$ where $x \in [0,1]$ , each $f_n$ is unioformly continuous, but $f$ is not continuous , let alone uniformly.
\item Define $f_n(x)=x ; x \in [0,n]$ , $0$ eveywhere else.
\item Define $f_n(x) = \frac{1}{x} \forall x \in [0,n]$ and $0$ elsewhere
\end{enumerate}

[soft zealot]

Wai

[warm python]

for (d) I can think of quite a few examples

but (e) seems to be tricky

hmm, I;ll basically want the discontinuities to cluser around every irrational

but that won't work

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[warm python]

really silly, but could someone pleas explain this simplification?

[warm python]

from the sum to the next line

[sonic crystal]

wai

[warm python]

hmm?

Hmm, I see it now I think

cool

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[twin pivot]

What does it mean to write ds/dt?

[twin pivot]

Which would be v(t)

[iron mist]

Derivative of s with respect to t, aka instantaneous rate of change of s with respect to t

[twin pivot]

Why does it mean how fast s changes?

[iron mist]

Yes

[twin pivot]

Why?

[iron mist]

Question: have you taken a physics class

Or any class dealing with kinematics or mechanics

[twin pivot]

Isn't the derivative the value to which the limit tends?

[twin pivot]

Yes

[iron mist]

It is defined as a particular limit

[iron mist]

So yeah, if you have, then you probably know the formula for average speed

[twin pivot]

Yes

∆s/∆t

[iron mist]

Okay

[twin pivot]

I don't understand why it represents how much s changes

[iron mist]

But a natural question is: how can we find the instantaneous speed. That is, the speed you’re traveling at an instant

[twin pivot]

Time cannot be 0 otherwise there is no speed

So they find the values in that very small interval ∆t

[iron mist]

Yes

As Δt gets really really small

[twin pivot]

Yes

[iron mist]

What does ∆s/∆t approach

So maybe let’s say t1 and t2

[twin pivot]

The value from right and left must be equal to exist

Since the intervals are very small, ds/dt comes out

[iron mist]

Yes

In fact, $v(t)=\lim_{h\rightarrow0}\frac{s(t+h)-s(t)}{h}$

[soft zealot]

CST (reply ping for help)

[twin pivot]

yes but this thing tells me the value I'm getting close to, it doesn't tell me how quickly it changes

[iron mist]

What do you mean how quickly it changes

What’s it

Because if “it” is the position, then recall that the derivative is the instantaneous rate of change because it’s what the average rate of change approaches as the time difference approaches 0

[proper dagger]

<@&268886789983436800>

[twin pivot]

<@&268886789983436800>

[iron mist]

…anyway…

[twin pivot]

Perhaps a practical example would be clearer

[iron mist]

Okay

Let’s let a position function $s(t)=t^2+4t+4$

[soft zealot]

CST (reply ping for help)

[twin pivot]

Yes

v(t)=2t+4

[iron mist]

Find the average speed between t=1 and t=2

[twin pivot]

(16-9)/(2-1)= 7

[iron mist]

Ok

Can you do say t=1 and t=1.1

[twin pivot]

7/0.1=70

Oh no

I have to recalculate too

S(t)

its 6.1

average speed Is 6.1

[iron mist]

Alright

Now t=1 and t=1.01

[twin pivot]

496?

[iron mist]

Too big

[twin pivot]

6.01

[iron mist]

Okay

Can you see where we’re going with this?

[twin pivot]

∆t->0

[iron mist]

Yes, and…

[twin pivot]

The speed at time 1?

Which would be about 6

[iron mist]

Yes

The derivative is the instantaneous rate of change specifically because it is the limit of the average rate of change

Graphically the rate of change can be represented as the slope of a line, and as the difference goes to 0, the line becomes more and more tangent to the graph

And so the slope of the tangent line is the instantaneous rate of change

[twin pivot]

So in our case v(t)=2t+4

[iron mist]

Ye

[twin pivot]

So the tangent line to s(t) at the point t=1 has a slope of 6?

[iron mist]

Yep

in the realm of kinematics, average velocity is the average rate of change of position with respect to time, hence instantaneous velocity is the instantaneous rate of change of position with respect to time, i.e. derivative of s with respect to time

(Speed is strictly nonnegative which is why velocity is the preferred term, since v(t) can be negative)

[twin pivot]

Instead at t=1.01 v=6.02

So the speed is increasing

And what can be said about s(t) instead?

From t=1 to t=1.01?

[iron mist]

You can’t say the speed is increasing using the velocity function

Because we’re measuring the rate of change of position, not of speed

[twin pivot]

Since v is increasing there is an acceleration

[iron mist]

Mmm yes but acceleration is a different function

[twin pivot]

I don't understand what v(t) tells me about s(t)

I can only say that if v(t) is >0 on an interval then s(t) is increasing on that interval

I understand

Thanks !!!

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[shy grove]

Guys why cant i use area = base * height in this letter c?

[blissful meadow]

This triangle does have the right ratios to represent cos(theta), but it's not the triangle with the height as its other leg

You need the green height here.

And in particular, you don't know the length of the leg at the bottom.. it's not 2, and the hypotenuse is not sqrt(62) as it's the length of BA.

[shy grove]

BA is not the hypotenuse?

[blissful meadow]

BA is the hypotenuse. What I'm saying is that its length isn't sqrt(62)

[shy grove]

Why is it not the length?

[blissful meadow]

What is |BA|?

You computed it above.

[shy grove]

3

[blissful meadow]

Right, so it's not sqrt(62)

[shy grove]

Ooooohh

I seee

Okk thanku so much

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[ruby crypt]

part (b)

[ruby crypt]

im confused

nvm I got it

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[thin jay]

ive seen that 1/89 equals the fibbonacci sequence shifted by 10 at each decimal place
i understand how it works fundamentally, but im not good enough at math to be able to reach the "compact" form

i also really want to know if there is any cool 1/x form for the fibonacchi shifted by 100 or 1000, (mostly because there are only 7 digits who are smaller than 10 in the sequence, then it gets all messed up)

(i added a cool explanation i found on reddit :p)

[thin jay]

i also found another video explaining this, it mostly gets to the first step and then sets S to 1/10 (for some reason i dont understand)
where does this 1/10 come from???

[rugged merlin]

so if i understand what you are asking correctly, you want to 1- understand how the image gets to 1/89 2- the version of the solution with 2-digit and 3-digit shifts?

[thin jay]

yep!!!

[rugged merlin]

hmm ok

[thin jay]

my main issue rn is with the infinite sum
how do you get things in and out of it?!?!?

isnt it supposed to be infinite 😭

[rugged merlin]

can you elaborate what you mean by highlighting it in your image?

[thin jay]

okey

ill try

i dont understand why and how the guy added 2 extra terms on the infinite sum and what are the Sses doing

i also dont understand most of the forth and fifth line 😭

i can kind of imagine the Sses are cancelling out with the extra terms, but, how do you even come up with them???

[rugged merlin]

ok so like

for a sum $S= \sum_{n=0}^\infty a_n$, we can do the following generally:
[
c\2S=\sum_{n=0}^\infty c\2a_n
]

can you accept this?

[desert mantle]

if we are ignoring convergence issues then you might aswell just imagine that the sum is finite

say up to 100

and it can often help to write out the first few terms of the sum notation

then it will also be much clearer why all those 1/10's appear