help-0

[stable cove]

Whereas the additive approach doesn't, so not only do we accumulate loss step by step

But the higher we get, the more we lose each step

[leaden zenith]

And how can i use this to calculate how much is the real ammount i get?

[stable cove]

What are you calling the "real amount" here?

[leaden zenith]

The ammount that have effect on each ¨level¨

Like if i im on level 7 the real ammount i get of % it would be lower than the other one

But how much is that ammount?

[stable cove]

This one, then

[leaden zenith]

Niceee tysm

[stable cove]

This one tells you, on levelling from 7 to 8, how much you are losing by adding instead of multiplying

[leaden zenith]

Let me try to do it on the excel to see if everything goes smoothly

[stable cove]

The other one tells you how much you've lost since level 1

[leaden zenith]

OHHH

I understand now the difference between each one

Then to know how much i would have on that step i should only just substract that from the original ammount (in this case 10,38%)

[stable cove]

So 10.68% minus this, if I'm not mistaken

[stable cove]

Yep

Oh, actually, n-1 on the dividend too

[leaden zenith]

Niceee this excel will be so sexy when i finish it tysm that helped a lot i highly doubt i would reach this far on my own

[stable cove]

If you want it as a percentage of your level 7 value, not your level 8 value

[leaden zenith]

You mean having this [1+(n-1)0.1068] in the dividend too?

[stable cove]

Actually

If what you want to figure out is the real percentage

The easier way to do that is just:

$\dfrac{0.1068}{1+n*0.1068}$

[ocean seal]

A miscellaneous Fern

[stable cove]

Where n is your current level

[leaden zenith]

Thats it? :0

Dayum that was less complicated than i thought it would finish

I need to chain it with the data i have on the excel but i think with this i have it enough

Then if i want to know how much % i have at the end i can do a summatory of all the numbers?

[stable cove]

How do you mean?

[leaden zenith]

Imagine i want to know how much i have from level 6 to 8 i would just need to take the numbers used from 6 to 8 and summ them? 🤔

[stable cove]

No, they're percentages of different totals.

[leaden zenith]

I doubt that

Okay

[stable cove]

The number you get when you calculate it for n=6 is the percent of your level 6 stat by which your souls increase when going to 7

[leaden zenith]

Then the n is the level minus one

[stable cove]

N is "current" level, so to speak.

If you want to see the % increase when going from Lvl 6 to Lvl 7, n is 6

oh wait actually

is lvl 1 100% or 100.68%?

[leaden zenith]

Lvl 0 is 110'68% TT

[stable cove]

Oh.

[leaden zenith]

Yes xD but dw i can do magic with excel and make it like lvl 1

[stable cove]

In that case, n+1

[leaden zenith]

=0,1068/(1+(K4)*0,1068)

This is the one that works for me right now being K4 ¨n¨

And n being 0

The () was bec there was a N-1

[stable cove]

Should be K4+1

[leaden zenith]

Well if k4 is 0 that would make it 0,1068/1

It would be because ¨our 0¨ its 0,1068 as we start counting from there?

[stable cove]

If you do it like that, it will give you the increase given by the level of the same row

If you make it K4+1, it will give you the increase given by a levelup (aka the next one)

[leaden zenith]

Wait a second let me put it in excel

Thats what i get with
=0,1068/(1+0*0,1068)

[stable cove]

Yes.

So right now, the right column tells you "If I am on the previous level, and I go to this one, what bonus do I get?"

So the % you get from going from level 0 to level 1 is 0,0964944...

[leaden zenith]

I just saw it was wrong

That one was the true one

Then if i want to know how much i gain from n to m being m desired level for example

I just found out the 13 was wrong lol

Fixed it

If i want to go for another one then i should just summ the ones around that number right?

[stable cove]

No.

[leaden zenith]

Like from lvl 6 to lvl 8 i summ the 6th 7th and 8th?

[stable cove]

$\dfrac{1+0.1068(m-n)}{1+0.1068(n+1)}$

[ocean seal]

A miscellaneous Fern

[leaden zenith]

And remember it starts on 0 TT

Not from 1

[stable cove]

Yep

[leaden zenith]

Okay let me check

Im back

Had problems with keyboard TT

Ehhh

Let me search more there's something or curious or wrong

Aaaa

That 32,04% the raw one

Better to say it givs me 1+0'1038*3

If you level up 3 levels

I love maths

=(1+0,1068(5-10))/1+0,1068(5+1) thie is literally 110'68

I dont know how that works but is funny

You there? D:

[leaden zenith]

<@&286206848099549185>

I still need help TT

If someone comes new ask me if you want to know what we were doing

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[leaden zenith]

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[upper veldt]

Apparently this simplifies down to 17, but idk how.

I know u can verify by saying like 17 = that then equating them

[tacit arch]

Rationalize the denominator

[upper veldt]

What does that mean

[sterile trench]

multiply and divide with root 290 + 17

switching the signs

[upper veldt]

Okay and what made u know that was what to do? The sqrt binomial on the denominator?

[tacit arch]

Yes

[upper veldt]

Ok ty mr riemann

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[broken pivot]

What is du

Not /; du = 2x dx

No, because you havent fully implemented du

Ok can you use this to solve for dx

$$\dd u = 2x \dd x$$

[ocean seal]

King Leo

[broken pivot]

Now solve for dx

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[broken pivot]

Use the least common multiple

[thorn igloo]

how

[sterile trench]

#help-17

#help-17 message

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[abstract oasis]

see which numbers are both multiples of 3 and 4, so 12. Every 12th day they are on the same day so this will only happen twice in a space of 31 days 12, 24 then 36 is out of the range

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[glossy obsidian]

hello, could someone maybe explain me the function of the confidence interval, aswell as the forecast interval

[tacit arch]

Can you give a more concrete example where you're reading this

[glossy obsidian]

as far as i know this is the general formula, but sometimes the z is replaced by a t

i dont understand why

or is it kind of the same?

[hushed locust]

some confidence intervals are based on the normal distribution (using z) and some are based on the student t-distribution (using t)

[glossy obsidian]

ahh

[hushed locust]

note that the t-interval can be done using the sample standard deviation (s), whereas strictly speaking the z-interval should be done with population standard deviation (sigma)

the t-distribution can also be more accurate for small samples

[glossy obsidian]

okay i see

and whats the real difference between the confidence interval and the forcast interval?

i just can't really understand the difference

i just know that the forecast interval gives you a wider range of confidence or smth like that

idk😅

[hushed locust]

(assuming we are using intervals for the mean)
in a confidence interval, we are trying to find a reasonable range for what the population mean is based on the sample statistics.
in a forecast interval, we are trying to find a reasonable range for a single data point which could be observed in the future

so let's say you have a farm growing tomatoes, and you weigh a boxful of tomatoes, and take their average weight (sample mean)

• the 90% confidence interval would tell you a range with which you are 90% confident the average weight of the entire field will be (population mean)
• the 90% confidence forecast/prediction interval gives you a range where you are 90% confident that any individual tomato picked from the field will have a weight in this range (this is a bigger range because the tomatoes can have weights significantly different from the population mean)

[glossy obsidian]

thank you very much for that information

The forecast interval does not indicate where the true mean lies, but rather where the value of a single new observation from the population lies with a certain probability

what do you think of my deffinition?

is there maybe something very important i left out?

@hushed locust ?

@tacit arch ?

just if someone could have a quick look at my definition?

.close

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[lilac ibex]

help me

PLASE

<@&286206848099549185>

[alpine sable]

help me #help-15

[lilac ibex]

help me please

[lilac ibex]

go away

u have your own spot

@broken pivot can u help

[alpine sable]

u went to my territory therefore i can go into urs

[split mantle]

If you we're given x = 4.5, would you sell four muffins, and half a muffin?

[lilac ibex]

you'll hear back from my lawyers

ima sue you

[lilac ibex]

ohhhhh

BRO

ur the goat

[midnight wharf]

it's a party here

[lilac ibex]

genius

[broken pivot]

Dont get banned from this server just to make a point

[alpine sable]

first figure out how to do that math lil bro

[frozen lodge]

lol

[lilac ibex]

i got more than u

[alpine sable]

good point i need this server

[lilac ibex]

lets go band 4 band rn

[lilac ibex]

ok discord mod

[buoyant saddle]

lol

[broken pivot]

This guy is 100% trolling

searchfrom: abd3327 help

[lilac ibex]

The eagle in comin'... You'd better start runnin'... His blood is gushin'... Stunnin' and gunnin'...

ok leo

"king" leo

[broken pivot]

Ping mods?

[lilac ibex]

bros so mad rn

ur not even a mod

get outta here

glorified unemployed

[jagged imp]

19/43 messages contain the word help 😭

[broken pivot]

I guess im mad

Damian i think im mad

😡

[lilac ibex]

king leo more like peasant leo

[broken pivot]

See

Im so mad

[lilac ibex]

get outta here ur not a mod

bros a mall cop

[alpine sable]

abd why r u risking ur chances

[lilac ibex]

boy u aint no MOD

[alpine sable]

this server is like a blessing

stop trolling

[broken pivot]

Facts

[lilac ibex]

im not im locked in

i just got into umich EA

[broken pivot]

I joined here for the sole purpose of getting my friend unbanned lol

[lilac ibex]

u can thikn whatever u think lol

[hot drift]

This doesn't look like math

[lilac ibex]

leo just got mad that he's not a mod

[kind trellis]

What don't you understand about the question @lilac ibex

[lilac ibex]

no i got it now

[broken pivot]

I dont use discord enough to warrant being a server moderator

[lilac ibex]

@split mantle is the goat

[kind trellis]

Then why were you typing in every channel lmao

[lilac ibex]

that was b4 he answered

[lilac ibex]

u aint no mod

😹

[broken pivot]

Thats basically what i said but ok

[mental python]

!done

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If you are done with this channel, please mark your problem as solved by typing .close

[alpine sable]

@broken pivot what do i do 🫠

[broken pivot]

Idk 😭

[alpine sable]

What do u do in moments of crisis

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[solemn wyvern]

when subtracting vectors, does angle become inverted? (off by 180 degrees?) does magnitude become negative?

[broken pivot]

Either the magnitude is negated, or the angle is inverted

You must pick one or the other to subtract the vector

[solemn wyvern]

so i can actually pick and choose?

https://tenor.com/view/limbus-company-limbus-don-quixote-excited-gif-27610057

[solemn wyvern]

also, in terms of "multiplying" vectors as seen above (2A, 3B, 2C etc), is the magnitude the only thing multiplied?

[celest night]

mhm, essentially

[north rover]

If you have done polar coordinates, it is exactly the same as this.

[broken pivot]

Yes

[north rover]

Yes.

The war of scalars vs. vectors.

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[thorn flame]

Hello, can anyone tell me if, and if so how, this expression can be simplified? I am not used to working with sums so I’m not sure. (EDIT apologies please see next image for actual expression, this one has a typo in)

[safe meadow]

is that (xi)*(yi)

[thorn flame]

it might not help but, the expression is for the general constant coefficient of a linear fitted model derived by minimising chi squared. (y is the measured quantity, with uncertainty sigma, measured relative to x which is known exactly, with k data points)

[thorn flame]

it is x_i * y_i

[safe meadow]

oh

[thorn flame]

sorry my subscripts are not the best

[safe meadow]

could they be simplified

are they the same

we could write

(the expresion)^2

[thorn flame]

sorry you’re completely right, I made an error when copying the expression it should be:

Again, I have no real basis to suggest it does simplify other than it looking at the patterns within the expression. I just don’t really have experience working with sums so I’m unsure if there are ways to manipulate them.

<@&286206848099549185> question relating to simplifying an expression with sums.

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[late slate]

Hey, can someone help me with this question? I dont know how to work with this structure

[wanton wigeon]

HI

[broken pivot]

Do you have a math question

[wanton wigeon]

yes

[broken pivot]

Send in an unoccupied channel

[wanton wigeon]

im just new here in discord so idk whats that

[broken pivot]

#help-24

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@late slate Has your question been resolved?

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@late slate Has your question been resolved?

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@late slate Has your question been resolved?

[keen plinth]

have you tried using the norm?

[late slate]

What do you mean by norm?

Like absolute value?

[keen plinth]

no the norm on Z[sqrt(-5)]

,, N : \Z[\sqrt{-5}] \to \Z, \quad N(a + b\sqrt{-5}) = a^2 + 5b^2

[late slate]

Hmm.. I think we didnt introduce this concept in the lecture

[keen plinth]

this is the standard way of finding units

[late slate]

Is the norm like the euclidean function?

for euclidean domains

[keen plinth]

kinda

but its not the same

here the norm is multiplicative

N(αβ) = N(α)N(β)

it tells you that in order to be a unit in Z[sqrt(-5)], the norm must be a unit in Z

[late slate]

We generally did not discuss a way to find units

[keen plinth]

strange

[keen plinth]

you can verify this property pretty easily since N is just the square of the complex modulus

if α and β are inverses then N(α)N(β) = N(1) = 1

so you quickly conclude that N(α) = 1 (-1 isn't possible since N is always >=0)

[fleet ether]

anyone who can help with a physics problem help 8 pls

[late slate]

.close

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[carmine topaz]

i dont get why the answer is (E) i know its gotta be either (C) or (E) because of the number of possible x intercepts but why does it have to be (E)?

[proud obsidian]

by definition of derivative, which is the rate of change you see that when graph E < 0, h(x) is decreasing

[fierce cipher]

because:

1. h(x) has 2 extrema so h'(x) = 0 needs 2 distinct roots (which you can understand)
2. when x -> infinity, h(x) tends to infinity, so h'(x) tends to infinity as well
3. h(x) is a cubic, when differentiated, it becomes a quadratic (which you also understand)

[carmine topaz]

the solution says something like relative maximum but i dont get it

[fierce cipher]

just think simple:

• if h(x) -> +inf when x -> +inf, what happens to h'(x) when x -> +inf

[carmine topaz]

infinity? or zero? i dont know 😭

[fierce cipher]

... of course it's +inf

how can it tend to 0

[carmine topaz]

my brain aint braining

[carmine topaz]

i didn't sleep well

[fierce cipher]

go sleep now

[carmine topaz]

umm

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[carmine topaz]

if a curve has horizontal or vertical tangents at a certain x value is it differentiable (smooth)?

[carmine topaz]

graph like this one...?

[junior vigil]

it is not differentiable where it has a vertical tangent yes

[carmine topaz]

ooh is like not smooth...?

oh

[junior vigil]

this is more obvious because it is not continous here :3

[carmine topaz]

every vertical tangent is not differentiable right?

[junior vigil]

yes

[carmine topaz]

and at the point when a graph crosses 0 on x-axis its also not differentiable at that exact value right?

[junior vigil]

or -inf

[carmine topaz]

inf and -inf is like an asymptote thing right?

[ocean fossil]

/0

[junior vigil]

no you're right its technically infinity

limit-wise

[carmine topaz]

thanks yall for helping!

.close

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[tepid scroll]

can someone show me how the derivative of a matrix with respect to itself is somehow identity?

[tepid scroll]

for refrence I am doing part b of this question

[vale crag]

where do they say that ? can you highlight what you have trouble with ?

[tepid scroll]

Well it doesn't but how would we remain with BA?

So it has to turn to identity?

[vale crag]

so this equality is what you don't get right ?

@tepid scroll

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@tepid scroll Has your question been resolved?

[tepid scroll]

Sorry for the late reply, yes @vale crag

[vale crag]

ok yea I'll just change the indices in the partial derivatives so that it's a bit less confusing [cause the i,j in the partial derivative aren't the same as the i,j in the sums]

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[vale crag]

.reopen

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✅

[vale crag]

so ok we wanna find $$\pdv{}{X_{pq}} \sum_k \sum_i \sum_j A_{ki}X_{ij}B_{jk}$$

[ocean seal]

aPlatypus

[vale crag]

the derivative is a linear operation so you can just insert/distribute it inside the three sums $$\sum_k \sum_i \sum_j \pdv{}{X_{pq}}A_{ki}X_{ij}B_{jk}$$

[ocean seal]

aPlatypus

[vale crag]

now the entries in A and B are just constants wrt the entries of X so you can pull them out

$$\sum_k \sum_i \sum_j A_{ki}B_{jk}\pdv{}{X_{pq}}X_{ij}$$

[ocean seal]

aPlatypus

[vale crag]

@tepid scroll still here ?

now we gotta figure out what that $\pdv{}{X_{pq}}X_{ij}$ is

[tepid scroll]

Just got here ill read wait

[ocean seal]

aPlatypus

[tepid scroll]

But but they are matrices

AXB is not equal to ABX

[vale crag]

A_ki, B_jk X_ij are just numbers inside the respective matrices

[tepid scroll]

Oh yeah true sorry

[vale crag]

right so do you have any idea about that now ?

@tepid scroll

[tepid scroll]

Umm okay so

It's 1 if ij= pq

Else it's 0

[vale crag]

yeah

[tepid scroll]

So like we have 12 matrices

With one 1 rest zeros

Such that if you add them all up we will have a three by four matrix of all ones

Obv the addition has nothing to do here I'm just trying to explain what I think the partial derivative matrix looks like

[vale crag]

yeah dX/dX_pq or something

but here it's just one number yes

ok so $\pdv{}{X_{pq}}X_{ij} = 1_{(i,j) = (p,q)}$ 

[ocean seal]

aPlatypus

[tepid scroll]

Yee

[vale crag]

so there's a ton of terms in these sums that become 0 now

unless i=p, j=q

[tepid scroll]

True

[vale crag]

that's how you end up with what they have in your book

the double sums get reduced to one term only

[tepid scroll]

Wait I don't follow

[vale crag]

the only terms left in the i,j sums are those with i,j=p,q

[tepid scroll]

Right right

[vale crag]

all the other terms are killed by the derivative thing

[tepid scroll]

Right

[vale crag]

so you end up with $$\sum_k \sum_i \sum_j A_{ki}B_{jk}\pdv{}{X_{pq}}X_{ij}$$
$$=\sum_k A_{kp}B_{qk}$$

[ocean seal]

aPlatypus

[tepid scroll]

But still first off in the second lady equation how does that sum equal (BA)ij

[vale crag]

well it's exactly the equation of a matrix product

$$=\sum_k B_{qk}A_{kp}$$

[ocean seal]

aPlatypus

[vale crag]

that's (BA)_qp

[tepid scroll]

Oh shiii

Thank you so much Mr platypus

With a deer pfp

[vale crag]

hehe

yw

[tepid scroll]

.close

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[willow stone]

Hey guys, is that what I wrote correct? Like I got 2 type of solutions, so I need to solve for yp1 and yp2 separately?

[lone heart]

@willow stone Has your question been resolved?

[split mantle]

You find the yp1 and yp2 separately, but apply them in yp=yp1+yp2.

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[shrewd flint]

i need a help with my math homework its too hard for me

[forest marsh]

Sure

[shrewd flint]

could you tell me the results for it when i will send you the picture of the math assignment?

[forest marsh]

I can help you understand and give hints but i won't gives the answer freely

<@&268886789983436800>

[shrewd flint]

i dont know how to understand it

my teacher is very bad at teaching

[forest marsh]

You have to use the foil method for thus

<@&268886789983436800> please

[shrewd flint]

what is the foil method iam from czechia

[forest marsh]

He is deleting post gif

Thanks

[forest marsh]

Let me send you the image

[night geyser]

@tight pier don't thumb up react to mod pings

[shrewd flint]

i need help with this i accidently skipped that

[forest marsh]

You have to cross develop it like (a+d)(b+c) = ab + bc + db + dc

[shrewd flint]

can you make it in paint and explain it somehow there?

[tight pier]

mb

[shrewd flint]

i think i understand it now thanks

how do i multiply it when there is one more a the end?

[tight pier]

You can factor a

Or make it one term

[lone heart]

@shrewd flint Has your question been resolved?

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[rancid pilot]

determine the equation of the line q, which passes through the given point and is perpendicular to the line p:

p: 1.5x + 2y + 3 = 0 , R(-2, -3)

= 4x−3y−1=0

can anyone check if its right

[wanton peak]

Seem like it

Yep

Just do the math myself

Its correct

@rancid pilot

👍

[rancid pilot]

thanks

.close

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[tropic sequoia]

Can anyone help with quadratic formulas?

[broken pivot]

Try to make a substitution $u = x^2$

[ocean seal]

King Leo

[broken pivot]

Wait which problem are you working on

[tropic sequoia]

both of them

but a first

[broken pivot]

What do you not understand about a

It seems to me like an (almost) direct application of the quadratic formula

[tropic sequoia]

I'm not good at the quadratic formula, that's the issue

[broken pivot]

First, for $(2x - 1) \cdot \qty(x^2 - 4x + 3) = 0$, do you agree that:
$$\begin{cases}
x^2 - 4x + 3 = 0 \

2x - 1 = 0 \end{cases}$$

[ocean seal]

King Leo

[tropic sequoia]

yes

one, or both, of the parts should equal 0

[broken pivot]

Can you identify $a$, $b$, and $c$ in $x^2 - 4x + 3 = ax^2 + bx + c$

[ocean seal]

King Leo

[tropic sequoia]

no

[broken pivot]

Ok what if i write it like this:
$$1x^2 - 4x + 3 = ax^2 + bx + c$$

[ocean seal]

King Leo

[tropic sequoia]

oh like that, yeah

I can

[broken pivot]

Ok find their values and ping me

[tropic sequoia]

a=1, b=-4, c=3 @broken pivot

[broken pivot]

Now just apply:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

[ocean seal]

King Leo

[broken pivot]

Texit died

[tropic sequoia]

+=3 and -=1

[wanton peak]

?

Weird way to write pos/neg value

[tropic sequoia]

my bad

[wanton peak]

Huh

Recheck ur result

Oh

[broken pivot]

He got it right

[wanton peak]

I get it

• • stand for the +- case

[broken pivot]

Did you think he said x = +3, x =-1

[wanton peak]

Not pos/neg

[wanton peak]

Yeah I did

lol

[broken pivot]

$$\begin{cases}
x^2 - 4x + 3 = 0 \implies x = -1, 3 \

2x - 1 = 0 \implies x = \frac 12 \end{cases}$$
$$(2x - 1) \cdot \qty(x^2 - 4x + 3) = 0 \to x = ?$$

Now solve the bottom line (this should be easy)

[tropic sequoia]

x=1/2

yes

[ocean seal]

King Leo

[tropic sequoia]

x=-1,3 or x=1/2?

[wanton peak]

Nope

And

[broken pivot]

Dont use .

[tropic sequoia]

x=1/2 and x=1,3?

[broken pivot]

What happened to x = -1

[onyx swallow]

yeah should be a 1 there instead of -1

[broken pivot]

Whoops

[lone heart]

@tropic sequoia Has your question been resolved?

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• Be polite and have a nice day!

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[odd reef]

Hello, I know I asked this question already however I was unable to respond, before the session timed out

[odd reef]

But here is link to message #help-18 message

So basically what I’m understanding is that we just show set equality?

That there exists an infinitely countable union of open rectangles that equals any open subset?

[wary fable]

yes

[odd reef]

Ah I see. And here why can’t we use a mapping from integers? Is it because the subset might not have integers in it?

[wary fable]

Yes

use the density of rationals

[odd reef]

Wait, do I use the fact of open balls, wait nvm… need to use open rectangle

I’m thinking of constructing an open rectangle around each rational

Or like map each open rectangle to each rational

But I think I need to show that each real in the subset is contained in one of those rectangles

[odd reef]

I think I’m kinda lost on the “size” of each rectangle though.

Because we need it to match exactly the subset

[clear sky]

What's an open rectangle here

[wary fable]

covered by R^n

[pallid scarab]

a cartesian product of open intervals

[clear sky]

So is it just the idea of a rectangle in R^2 extended to R^n or something

[wary fable]

use the fact that the set of points in R^n with rational coords is countable to construct the collection of countable open rectangles

[odd reef]

But I need to make sure that each rectangle does not “extend out” of the subset right?

[wary fable]

consider an open set U subset of R^n

[clear sky]

Sorry for interrupting but I'm curious, can someone explain what's going on in this question

[wary fable]

oke rectangle make the standard topology for R^n right?

so instead of working with open sets, we work on open rectangles

[clear sky]

Open subset of R^n means that the boundary is also within the subset, yes?

Ok maybe I'll read myself on wiki, TQ goëtia

[wary fable]

no?

the defining property of an open set is that it excludes its boundary

[odd reef]

Do I perhaps need to consider an open ball when I do this?

[wary fable]

ofc

[pallid scarab]

you're mixing up with closed

[clear sky]

Oh true

[wary fable]

and use the fact that $\mathbb{Q}^n$ is countable and dense in $\mathbb{R}^n$ @odd reef

[ocean seal]

Goëtia

[odd reef]

I see hmm. So then the size of each rectangle should be the radius of the open ball… or wait 2 times the radius of the open ball?

[wary fable]

so for any point x in the open set U there is an epsilon > 0 there is an open ball containing x, and we can use density to say there exists a point $q \in \mathbb{Q}^n$ arbitrarly close to x

[ocean seal]

Goëtia

[wary fable]

and then we can construct an open rectangle center at this new rational point

[odd reef]

Ah okay. As long as the yes as long as the open rectangle is “smaller” or contained in this open ball

And contains x

[odd reef]

Yeah, wait, but then doesn’t this create a one to one mapping between each open ball in the subset and the open rectangles?

But the open balls are one to one mapping with reals

[wary fable]

one to one?

open rectangles with rational endpoints does not create a one-to-one mapping between open balls and rectangles, it establishes a covering relationship

Each point in U is covered by at least one rectangle

Multiple rectangles may overlap and cover the same region

[odd reef]

Oh but aren’t we creating an open rectangle for each open ball?

Since the “size” of the open rectangle is determined by the open bal

[wary fable]

we are so where is the problem?

[odd reef]

But isn’t there a one to one mapping beteeen open balls and each point in the subset?

[wary fable]

no

why are you thinkin about mappings here, that isnt helpful

[odd reef]

I mean I’m a bit confused on how we are constructing this open rectangle

I understand the part where each rectangle corresponds to a rational point

However isn’t the size of the open rectangle based upon each open ball?

[wary fable]

Consider $x \in U : x = (x_1,..,x_i,.., x_n)$ using density we can find $$a_i, b_i \in \mathbb{Q}: a_i < x_i < b_i \land b_i - a_i < \varepsilon:$$

[odd reef]

Yes, but how do we know that b_i is contained in the subset U?

Because if b_I say extends far out

We will have a real between x_i and b_i

That could be “outside” the set of U

[wary fable]

no

look at the conditions i set

[ocean seal]

Goëtia

[odd reef]

And here, epsilon is the radius of the open ball at b_i/a_i?

[wary fable]

epsilon is the radius of the ball in the neighborhood of x

[odd reef]

Oh wait wha.. so it’s the ball centered at x?

[odd reef]

But why dojt we take the ball centered at b_i?

[wary fable]

why?

[odd reef]

Because then each rectangle depends on the open ball right

[wary fable]

no

[odd reef]

But there are uncountable amount of open ball

[odd reef]

But aren’t we taking b-a to be less than radius of open ball at x?

[pallid scarab]

this won't suffice if you need the ball of radius epsilon to contain the point

[wary fable]

mhm?

[pallid scarab]

you need sqrt(n)(b-a) < epsilon

[wary fable]

yes

well might aswell give him the whole proof now

[odd reef]

Huh

[wary fable]

raf will take it from here

[odd reef]

Oh ok thanks for the help so far.

[pallid scarab]

ok so

taking a random x in U

we know there is a ball centered on x contained in U

because U is open

[odd reef]

Yes

[pallid scarab]

so B(x,r) contained in U

[odd reef]

Yes

And we can make a rectangle contained in B(x,r)?

[pallid scarab]

yes, with rational coordinates, here's how

writing x = (x1,...,xn)

for each xi

(xi - r/sqrt(n), xi) has some rational, select one and call it ai

because Q is dense in R

and (xi, xi + r/sqrt(n)) also has a rational

select one and call it bi

[odd reef]

Yes

[pallid scarab]

now what can we say about the rectangle (a1,b1) x ...

[odd reef]

It must contain x_i for all i and it is contained in B(x,r)?

[odd reef]

But this creates uncountable many rectangles right? Because it is based upon each ball right

[wary fable]

no

(a1,b1) is in Q

Q^n is countable

[odd reef]

ig this is the part im confused on. Because we create the rectangle based upon the radius of the ball with center x

but there are uncountbaly inifnite many balls since uncoutnably infintiy many real x's in U

[wary fable]

we didnt construct the rational endpoints interval (ai, bi) from open balls

[odd reef]

yeah, didnt we require that bi - ai < e where e is radius of the ball centered at x?

[pallid scarab]

it's enough to have b_i - x_i < epsilon/sqrt(n) and x_i - a_i < epsilon/sqrt(n)

but taking b_i-a_i < epsilon/sqrt(n) works too

[odd reef]

but isnt this dependent on epsilon? but epsilon is differentn for each open ball right

[wary fable]

he only worked with one open ball the one containing x

[odd reef]

but we do this for every open ball containing x for all x in the subset right

[wary fable]

what?

[odd reef]

like we are constructing a rectnagle for all x in U based upon the open ball centered at x

[wary fable]

then?

[odd reef]

but subset U is uncountable right

[wary fable]

yes, but the rational rectangle is countable

[pallid scarab]

but the number of boxes that contain them is countable

because each box has rational coordinates

[odd reef]

so then set of openballs is countable

[wary fable]

no

[pallid scarab]

no

[odd reef]

sry uncoutnable

i mean set of open balls is uncountable

[pallid scarab]

the boxes themselves contain an uncountable amount of elements

[odd reef]

but we construct the rectangle based upon each openball right

[pallid scarab]

but the number of boxes is countable

[odd reef]

that is there is one to one map between the set of openballs and the set of open rectangles

[pallid scarab]

but some rectangles will be used for multiple x

[odd reef]

but the way we construct it, it is not so right

[pallid scarab]

it is implied actually

[odd reef]

wait how?

[wary fable]

we are just using the radius of the open ball to make the rectangles inside the ball, we dont use the openess of the ball or anything on the rectangles

[odd reef]

yes, but each ball can have different radius. and we are using all balls that have centers in tehs subset right?

[wary fable]

what ball r u talkin bout

[odd reef]

ball centered at x

[wary fable]

"all balls"

what balls u talkin bout

[odd reef]

like for all x in U, the open ball with center x

[wary fable]

oke then what

[odd reef]

but we are constructing an open rectangle with each of those open balls right?

like each open ball is mapped to a specific rectangle (due to radius)

[wary fable]

define the other balls

[wary fable]

no

we dont use that

[odd reef]

then what is raf doing here?

like isnt "r" here dependnt on the ball?

[wary fable]

tellin u how did the existence of the rational endpoint interval arised

[odd reef]

yes, but the exsitnence of that endpoint is based upon the radius of the ball centered at x there

but we do this for each x/each radius

[wary fable]

no

we did divide by sqrt(n)

so we only depend on the first ball

suppose you have a segment of length 1, you want to partion it into the union of disjoint open rational intervals of same length

how you would do it? @odd reef

[odd reef]

i would divide by 1/n where n is the number of intervals

[wary fable]

correct

so how would each interval look like?

give me an explicit form

[odd reef]

(0,1/n) U(1/n, 2/n), .... (n-1/n, 1)

[wary fable]

there you go

is there a 1-1 map here?

i mean regardless of the length of each rational interval

as long as it is contained by the line ur good

[odd reef]

yes.... wait but here we didnt use the fact of open ball

im saying because we use the open ball fact

we do this for every open ball contained in U

[wary fable]

the only reason we used the open ball, is because we are working with an open set

[odd reef]

we construct an open rectangle from every open ball in U

[odd reef]

but to also decide the "size" of the open rectangle right

[wary fable]

we dont care about the size as long as it is contained by the open set

[odd reef]

are you saying we just choose an arbitrary real number in the interval?

and then we take the ball from there?

[wary fable]

as long as ur not leaving the set

ur good

[odd reef]

thats the thing am kinda stuck on. Because im confused on how we can make it so it doenst elave the set

without making it uncountable

[wary fable]

ok to not waste my time , look at the proof : Q^n is countable

spend an hour studying the proof

come back later

[odd reef]

yeah, i think I know how Q^n is countable

since Q is countable

thus Q x Q is countbale .... and by induction Q^n is countable

[wary fable]

we just told ya

each subinterval is of the size epsilon/sqrt(n)

[odd reef]

yes, but that epislon depends on the open ball at x

[wary fable]

so what?

[odd reef]

but there are uncountable many open balls

like the set of open balls contianed in U is uncountable

but we are constructing an open rectangle based upon an x and open ball

[wary fable]

we dropped the open ball we took just its size

[wary fable]

take 1 open ball and make it look like the union of countable intervals in Q

[odd reef]

wait.... are you saying we take each open interval (each is mapped to a rational number). Then we know by density of reals in ratioanls that there exists a real number between a_i and b_i. Then based upon that real number,we take the ball centered there. Then we make sure a_i, b_i lies within that ball?

[wary fable]

ye

[odd reef]

ohhh, i see. But dont we first "decide" a_i, b_i before we choose the real?

[wary fable]

no

we choose the real

[odd reef]

i thought we choose b_i (the rational first)

then choose a real

[wary fable]

x = (x0, ..., xi , .., xn)

[odd reef]

that way, we ensure it is not based upon uncountability of reals and more so on countability of ratioanls

[wary fable]

u find ai bi converging to xi

[odd reef]

but then we are choosing based upon real.... no?

which is not what we want

[wary fable]

good luck

[pallid scarab]

it doesn't matter

we're saying

for every x, there is a rational coordinate box that contains it

the way we know there are a countable amount of them comes from a different result

[odd reef]

i see ohhh okay.. i think im kinda getting it?

[odd reef]

so its less about mapping an x to a ratioanl coordinate but rather rational coordiantes are countable, and each x is in a ratioanl corodiante box

[wary fable]

yeah i told u , thinking of mappings is useless

[odd reef]

oh ok

ig its kinda hard me to wrap my head around how we create a box for every real

because that would imply each "real" gets its own box

[wary fable]

do relatively easier problems, dont jump to this

[pallid scarab]

but multiple reals^n can get sent to the same box

anyways

the way we know there's a countable amount of them

is:

[wary fable]

genre m saoul c mec

[pallid scarab]

let $\mathcal B = {B, B \text{ is a box with rational coordinates contained in } U}$

[ocean seal]

rafilou is not not born in 2003

[odd reef]

this is my homeowrk and its only problem 2

[wary fable]

go back to the lecture, watch u2b, tons of resources online before doing this stuff

[pallid scarab]

$\mathcal B = {B = (a_1,b_1) \times ... \times (a_n,b_n)\mid a_i,b_i\in \bQ, B\subseteq U}$

[ocean seal]

rafilou is not not born in 2003

[odd reef]

okay thanks for help, ill try to keep on tackling it a bit more

[pallid scarab]

if you want to try out latex bot go to #bot-testing

[wary fable]

lmfao there is no more tackling u already got spoonfed the solution

[pallid scarab]

anyways this set is countable

$U = \bigcup_{B\in \mathcal B}B$

[ocean seal]

rafilou is not not born in 2003

[pallid scarab]

we know each B is a subset of U, so the right hand side is a subset of U, obviously

and for each x in U, we proved there was a box B that contained it

so each x in U, is also in the union on the right hand side

U subset RHS

conclusion U = the union

U = a countable union of open rectangles

[wary fable]

you want to do tackling? consider $Q = \prod_{i=1}^n [\frac{k_i}{2^n}; \frac{1+k_i}{2^n})$ use this to cover $\mathbb{R}^n$ and make a delicate proof, this is a good exercise if you think you understood what was digested here

@odd reef

[ocean seal]

Goëtia

[odd reef]

okay, ill try

[lone heart]

@odd reef Has your question been resolved?

[lone heart]

@odd reef Has your question been resolved?

[serene ridge]

if JIH and HUT are congruent triangles, then what does the side HJ equal?

[lone heart]

@odd reef Has your question been resolved?

[lone heart]

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[stuck eagle]

Is this right?

[lavish cave]

,calc 2(-2)^2+3(-2)-2

[ocean seal]

Result:

0

[lavish cave]

yes, that's correct

[lavish cave]

.close

[lone heart]

Channel closed

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[lone heart]

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• Show your work, and if possible, explain where you are stuck.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[still plume]

Hii

[still plume]

I need a suggestion should I buy a scientist calculator for Matrix and vector or I should learn to solve them

[lunar saddle]

Wolfram alpha

[native cloud]

Ask your school if they allow calcs on tests

[still plume]

My previous scientific calculator don't have matrix and vector features

[still plume]

My school does allow it

[native cloud]

Ti-Inspire is pretty good imo

[still plume]

But I feel like it's wasting my money and my other part is like no it's better to have a calculator

I'm indecisive

[native cloud]

So you're looking for an affordable and good calc, right?

[still plume]

Idk... I'm just confused right

Like can I solve the sum without calculator??

[native cloud]

But like if it's for like general self studying, I would say symbolabs is pretty good

[still plume]

What you think

It's just martix and vector

[native cloud]

I mean you can pretty much use your phone calc to calculate computations of matrices

[still plume]

Yeah right

But in examination

[native cloud]

https://satsuite.collegeboard.org/sat/what-to-bring-do/calculator-policy

Honestly one of these

[still plume]

Hmm exactly in my country we have different policy regarding calculator

[lone heart]

@still plume Has your question been resolved?

[lone heart]

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[final badge]

How do I find the correct domain for this?

[tough relic]

3x³+11x²≥0

[broken pivot]

Do you know the restriction for a square root

[final badge]

nvm i see now lol.

It wanted [ , )

[fierce cipher]

but the domain is still not 4.681 😭

[final badge]

ik

originally had -3.667 but with ( , )

Thanks!

.close

[lone heart]

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[turbid field]

All of them are correct. Right?

[turbid field]

<@&286206848099549185>

[turbid field]

,tex $\vec{AC}$\ + $\vec{BD}$\ = $\vec{AB}$\ + $\vec{BC}$\ + $\vec{BD}$\ = $\vec{AB}$\ + 2 * $\vec{BJ}$\ = $\vec{AJ}$\ + $\vec{BJ}$\ = 2 * $\vec{IJ}$\

[ocean seal]

h

[turbid field]

,tex $\vec{AD}$\ + $\vec{BC}$\ = $\vec{AB}$\ + $\vec{BD}$\ + $\vec{BC}$\ = $\vec{AB}$\ + 2 * $\vec{BJ}$\ = $\vec{AJ}$\ + $\vec{BJ}$\ = 2 * $\vec{IJ}$\

[ocean seal]

h

[turbid field]

the last two equations are the same with $\vec{AC}$\ + $\vec{BD}$\

[ocean seal]

h

[lone heart]

@turbid field Has your question been resolved?

[lone heart]

@turbid field Has your question been resolved?

[lone heart]

@turbid field Has your question been resolved?

[turbid field]

.close

[lone heart]

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[exotic canopy]

no clue how to prove that. here are the definitions we use

i guess we use contradiction and start by assuming f is unbounded

=> there exists a sequence x_n such that |f(x_n)| gets arbitrarily large

in fact by continuity there exist points such that $|f(x_n)| = n$

[ocean seal]

artemetra

[exotic canopy]

idk if we need the absolute value

[exotic canopy]

the reason for this is cuz the prof did something like that last lecture lol

[slim belfry]

Choose epsilon to be 1

[exotic canopy]

ok

i was about to say that ahah

let me think

[slim belfry]

On, and start with a base point at the origin

f(0) equals something, but wlog f(0)=0

[exotic canopy]

sure

[slim belfry]

If g(x) = f(x)-f(0) is bounded, then f(x) is bounded too

[exotic canopy]

$\implies \exists \delta > 0$ s.t. $\forall x,y\in B$ (??), $||x-y|| < \delta \implies |f(y)- f(x)| < 1$

[ocean seal]

artemetra

[exotic canopy]

aha i see, and we are considering the contrapositive of that

[exotic canopy]

do we pick B or some other subset tho

[slim belfry]

All of B

We will show directly that any point in B is bounded by... something

[exotic canopy]

i'll try and come back in a moment

[slim belfry]

I can tell you what that something is if you want.

[exotic canopy]

hm

yeah no idk

[slim belfry]

Let x be any point in B

And consider the line segment from the origin to x

[exotic canopy]

oh

and we consider the function on $p(t) = f(t\mathbf{x})$, $t \in [0, 1]$

[ocean seal]

artemetra

[exotic canopy]

and p(t) is continuous on a compact set

which makes it bounded

this feels.. off but what's wrong

[slim belfry]

The fact that that set it compact is misleading here

Because we started by fixing an x in B

[exotic canopy]

yes

[slim belfry]

And B was not compact.

[exotic canopy]

indeed

[slim belfry]

So if we get different bounds for each x...

[exotic canopy]

yeaaa

i see

[slim belfry]

What does uniform continuity mean to you intuitively?

[exotic canopy]

one delta works for an entire interval

well, "interval"..

subset

[slim belfry]

For our purposes in this question, uniform continuity means there's a maximum slope

[exotic canopy]

yes

[slim belfry]

So if x is within 1 of the origin, then the largest f(x) can be is f(0) plus the maximum slope

[exotic canopy]

OOOOHHHH

[slim belfry]

(do note that uniform continuity does not imply maximum slope in general, or even here, but this idea is fine for what we want to do with it)

[exotic canopy]

i'll try this again

[exotic canopy]

is my p(t) idea good tho?

like apply that max slope idea on that one variable function

[slim belfry]

You could do that. It's not wrong, but I wouldn't bother with it

[exotic canopy]

so we need to show first that $\frac{f(y)-f(x)}{||x-y||}$ is bounded

[ocean seal]

artemetra

[slim belfry]

You have a delta that works for epsilon=1.

[exotic canopy]

indeed

[slim belfry]

So you know that B(delta/2) around the origin is bounded

Bounded by 1, in fact

(gonna use delta/2 so I don't have to worry about strict or non strict inequalities)

[exotic canopy]

the slope, right?

[slim belfry]

Instead of a line from origin to x, consider a finite sequence of equally spaced points (which are indeed along a straight line).

[exotic canopy]

or |f(y)-f(x)|

yeah it's the latter

silly question

[exotic canopy]

is x in B_delta/2 or in B in general

[slim belfry]

B in general

[exotic canopy]

i know i need to do something like $\frac{f(a_{i+1}) - f(a_i)}{||a_{i+1} - a_i||}$

[ocean seal]

artemetra

[exotic canopy]

OH and we can pick a_i+1 - a_i to be less than delta

right?

damn this is really pretty

[exotic canopy]

and we can use this now?

[slim belfry]

I wouldn't actually use it directly. I'd only think of it to plan a strategy. Once you have the plan laid out, it's abandon it.

[exotic canopy]

okay

[exotic canopy]

so we know that there exists a maximum out of all of these slopes tho

[slim belfry]

If each a_i is within delta of the last, then f(a_i) is at most larger by 1 than the previous

[exotic canopy]

ahaaaa

[slim belfry]

And it's a bounded number of steps to any x

[exotic canopy]

because each sequence is finite right

[slim belfry]

More than that, the same bound applies to every x

[exotic canopy]

and that's it?

idk when the proof is done haha

[slim belfry]

Kind of. Can you put it all together and write the explicit bound?

[exotic canopy]

yes, moment please

[lone heart]

@exotic canopy Has your question been resolved?

[exotic canopy]

$|f(x) - f(0)| = |f(a_n) - f(a_{n-1}) + f(a_{n-1}) - f(a_{n-1}) + \cdots + f(a_1) - f(a_0)|$\
$\leq |f(a_n) - f(a_{n-1})| + |f(a_{n-1}) - f(a_{n-2})| + \cdots + |f(a_1) - f(a_0)|$
$\leq n$

[ocean seal]

artemetra

[exotic canopy]

my friend showed me this

man this is a nice way to do it

[exotic canopy]

does that suffice

[slim belfry]

It's very good, but you still need to bound n

[exotic canopy]

oh but $\frac{1}{n} \leq \delta$

[ocean seal]

artemetra

[slim belfry]

With the condition that each consecutive a_i is close to the previous

That doesn't bound n.

[exotic canopy]

wait hm

[exotic canopy]

yes

[slim belfry]

You have a total distance of at most 1, and it's chopped into pieces of length...

[exotic canopy]

1/n (??)

oh no

x/n

right?

||x||/n

sorry i don't get it :(

[slim belfry]

None of those

The terms in the sequence need to be close enough that you can apply the epsilon delta

They need to be less than delta apart

[exotic canopy]

yes

[slim belfry]

I'd choose delta/2

So how many pieces are there?

[exotic canopy]

2/delta ?

[exotic canopy]

interesting.. he did it on the board and TA didn't notice that part

[slim belfry]

I'm just realising that is probably better to write 2/delta +1

For rounding at the end

[exotic canopy]

final answer: $\frac{2}{\delta} + 69420$

[ocean seal]

artemetra

[exotic canopy]

phew okay

[slim belfry]

There are ways to avoid it, but I think they're all messier anyway

[exotic canopy]

but what he did is he just picked the smallest k s.t. $\frac{1}{k} \leq \delta$

[ocean seal]

artemetra

[slim belfry]

It's that your choice for f(0)? 🤪

[slim belfry]

Yeah, that works too

[exotic canopy]

ok

let me now write the whole thing

Fix $\epsilon = 1 \implies \exists \delta$ s.t. $\forall x,y \in B$, $||x-y|| < \delta \implies |f(x)- f(y)| < \epsilon$. \
Pick smallest $k$ s.t. $\frac{1}{k} \leq \delta$ and pick an arbitrary point in $x \in B$. Make a sequence $a_0 = 0, a_j = \frac{j}{k} x$.\
$\implies ||a_{j+1} - a_{j}|| = |\frac{j+1}{k} - \frac{j}{k}| ||x|| < \frac{1}{k} \leq \delta$\
$\implies ||f(x)-f(0)|| = |f(a_j) - f(a_{j-1}) + \cdots + f(a_1) - f(a_0)|$\
$\leq |f(a_j) - f(a_{j-1})| + |f(a_{j-1}) - f(a_{j-2})| + \cdots + |f(a_1) - f(a_0)|$\
$\leq j \leq k$ which is the same for all $x\in B$

@slim belfry any holes in the proof?

oh and one more thing

[ocean seal]

artemetra

[exotic canopy]

$\implies |f(x)| \leq k + |f(0)|$

forall x

[ocean seal]

artemetra

[exotic canopy]

and we picked k to be the smallest possible one

[slim belfry]

One tiny mistake

[exotic canopy]

oh right it should be k

man this felt unreasonably tough

[slim belfry]

That definition of a_j is so slick I thought it was a mistake. I would have done this without specifying a_j, only describing its requirements, but what you did is better.

[exotic canopy]

thanks

[exotic canopy]

all of this was really helpful

i need to brush up on my epsilonics

thank you so much!!!!!!!

.close

[lone heart]

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[waxen jacinth]

What is the sum of the last two digits of the quotient? (10^50)/((10^25)+7). note that 10^25 + 7 is in the denominator

[mortal trellis]

$\frac{10^{50}}{10^{25}+7}$ ?

[ocean seal]

Denascite

[mortal trellis]

thats not an integer. do you mean if you round that down?

[waxen jacinth]

I don't know actually. I got this in one exam. It says quotient. So I guess we can have remainder too

[mortal trellis]

clearly the result will be quite close to 10^25. you could for example set it as 10^25-x for some number x

and then solve the equation for x

you'll get some approximation and then you can round

but I dont know if that is what you are supposed to do

[waxen jacinth]

I didn't quite understand

[mortal trellis]

$\frac{10^{50}}{10^{25}+7} = 10^{25}-x$ for some (compared to $10^{25}$) small $x$

[ocean seal]

Denascite

[mortal trellis]

you can rearrange a bit and get an expression for x from which you can easily tell an approximate value of it

[lone heart]

@waxen jacinth Has your question been resolved?

[waxen jacinth]

I think it doesn't work. It keeps coming back to a complex equation

[mortal trellis]

you can get something involving $\frac{10^{25}}{10^{25}+7}$. which is essentially 1. now we can feel much more comfortable rounding like that

[ocean seal]

Denascite

[waxen jacinth]

I mean it says we need a quotient. So, If we round it, we can get the first few digits correct. But the question asks about the last digits

[waxen jacinth]

I don't understand that actually

[mortal trellis]

if you solve the equation you get $x=7\cdot\frac{10^{25}}{10^{25}+7}\approx 7\cdot 1 = 7$

[ocean seal]

Denascite