help-23

[digital sparrow]

Or intersection

[plucky elk]

fluxion knows what union means

and intersection

[digital sparrow]

Hope so

[compact flame]

union and intersection are totally different operations

[digital sparrow]

Did you know what these meant?

[compact flame]

of course

U -> union

upside down -> intersection

[digital sparrow]

?

[plucky elk]

none of what you're saying makes this true

[digital sparrow]

https://images.app.goo.gl/JL6P6tQx7tH8Jtxy6

[compact flame]

I think by replacing

they just mean removing

let's take it like this

[digital sparrow]

Yes mb

@compact flame PYQ?

[compact flame]

no

[digital sparrow]

Main ya Adv?

[compact flame]

i dunno

it's a question

[digital sparrow]

Kk

[compact flame]

that's all i care abt rlly

OH HOLD ON

HOLD ON

I think by replacing they mean that the set A is restored again

So A = {1, 2, 3, 4, 5, 6}, P1 = {1, 2, 3}

then P2 = {2, 3, 4}

[digital sparrow]

Yeah makes sense

[compact flame]

or P2 = {4, 6, 1}

like that

@plucky elk

[plucky elk]

When did A get restored in your example?

[compact flame]

I didn't remove any of the elements taken inside P1

they can still be selected in P2

and so on

[plucky elk]

maybe try that in the simpler case with A= {1,2,3}, and \cup P_i = A, e.g. P_1={1,2} and P_2 = {3} to see if (d) is the right

[compact flame]

A and D both are correct

but i'll check with ur example

[plucky elk]

yea i just mean to do the counting in that case to see if you get (2^m -1 )^3

[compact flame]

i think i fucked up somewhere

oh wait

nvm yes I fcked up

hold on

[plucky elk]

oh god that's 27 with just m=2

[compact flame]

yeah

@plucky elk I do have the solution to this problem

i haven't looked at it

[compact flame]

27? There's 7 possible proper subsets of A when n = 3

from the result 2^n - 1

[safe radish]

@compact flame Has your question been resolved?

[safe radish]

@compact flame Has your question been resolved?

[jade magnet]

for (d) you can consider an element a in A and for the union condition to be satisfied a should be in atleast 1 of those subsets P_i

and we have n ways to choose a aswell

the logic is similar for (a)

i dont understand the replacing aspect of the question

[safe radish]

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[compact flame]

Tyyy I get it now

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[astral sparrow]

Can i have help with this prblem

[modest hazel]

whats f(4)?

you might have to draw some number to the graph

that would be easier

[astral sparrow]

like you mean adding more numbers to the graph

[lean otter]

draw the origin

and then number it like its a graph

so you can easily see x= 4

or you can just look at it, it isnt that difficult since theres x=4 and x = 1, better to number for a difficult question tho

[astral sparrow]

im a bit confused if im being honnest

i dont really know what to do with that information

[safe radish]

@astral sparrow Has your question been resolved?

[prisma perch]

pretend that you're a function

like really manifest your functioninity

so if you are g and you are walking along that big thick line

and look at the y axis when you hit x = 4

what y do you see

👼 👼 👼

[astral sparrow]

i ran out of time for my test unfortunately thanks for the help but there's nothing else i can do

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[safe radish]

@mossy ridge Has your question been resolved?

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[lean otter]

i just need clarification if this the answer is 1, and if not pls explain

[tiny trench]

How d u get 1 pls explain in short

[lean otter]

L hospitals rule

[lofty pendant]

can i

[lean otter]

what?

[lofty pendant]

can i solve it

[lean otter]

sure

[lofty pendant]

alright

we can apply the L hospital's rule

this is an intermediate form

[lean otter]

thats what i did

[lofty pendant]

0/0

[lofty pendant]

aight

let me finish

the numerator is ex + e-x -2

now we differentiate it term by term

it should look something likethis

then

the denominator is x tan x

we use the product rule

for differentiation

for

x and tanx

like

:

now we applythe Lhospital rule

then we evaluate

as X →0

ex-e-x →0 because e0 equals 1, e-0 equals 1

tan0 equals 0

and sec2 0 equals 1

therefore

the limit is zero.

[lean otter]

i got 1

[lofty pendant]

hm

show work

[worthy estuary]

this is indeterminate again

[lean otter]

yes

was gonna say that

[lofty pendant]

o

right

[lean otter]

ill type it since i lost the paper

[lofty pendant]

i think it's 1

[lean otter]

1: rearrange terms

2: eval terms by substituting 0 for x

[lofty pendant]

when we differentiate numerator

[lean otter]

simplify: 0/0 undefined

apply l hospital

[lofty pendant]

hm\

WAIT

igot it

[lean otter]

take derivative of numerator and denominator

substitute 0 for x

simplify

then i got 1

[lofty pendant]

when we differentiate tanx + xsec2 x

[lean otter]

i didnt use differentiation

[lofty pendant]

o

u substituted?

[lean otter]

i forgor

just continue

[lean otter]

sounded rude sorry

.close

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[lofty pendant]

Ya I have to do group study with my friends now when I'll be free I'll finish it

Rlly sorry

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[lean otter]

who is taking amc10 this year

[frozen marlin]

???

that's not really a question you ask here

if you need help with amc try #competition-math

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[austere turret]

claim

[austere turret]

if "4.X people" fail an exam, how many people failed the exam?

is it a case where you should always round up/down

or do you round to what is closer

[lean otter]

depends on the quesiton

[austere turret]

the question asked how many people fail on average

answer is 4.8

[lean otter]

Since you can't have a fraction of a person in practical terms, you typically round to the nearest whole number, depending on the context. However, in some cases, especially in statistical or analytical contexts, you might keep the decimal to indicate the average or estimated number of people affected.

I would probably say

round up

to 5

that's what I would do personally

[austere turret]

what if it was 4.4

still up?

[lean otter]

then I would round down

[austere turret]

looks like this question is really subjective

[lean otter]

Actually it really really depends on the full question

can you post the full quesiton?

I'll try to judge it

[austere turret]

if 100 students took the test, how many fail

yeah that's it

it's subjective how you round the number

[lean otter]

make it 5

just apply normal rounding rules

that's what I would say

[austere turret]

aight bet

ty for advice

[lean otter]

happy to help

note that my answer is not definite

it may different based on who you ask

[austere turret]

yeah I understand that

[lean otter]

some might prefer rounding down

its just me

I would round up

[austere turret]

let me ask you something

[lean otter]

cause its

a) a lot closer to 5

b) you can't have 0.8 of a person

[austere turret]

if I tell you that I have money to buy 4.8 printers

the answer will always be 4 right?

[lean otter]

definitely

[austere turret]

this one is clear

the student one isn't

unfortunately

[lean otter]

yeah

language and math

don't always go hand in hand

[austere turret]

I feel like since it just says "on average" they should be accepting both 4 and 5

regardless ty for helping

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[lean otter]

couldn't agreee more

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[tawdry salmon]

About the middle part

wouldn’t 1/infty go to zero?

[undone burrow]

are you referring to the 1/x in the numerator or the whole thing?

[tawdry salmon]

its equal tot his right

[frozen marlin]

yeah

that's literally what they've written

use

[flat frigate]

ren

[frozen marlin]

these

specifically 3

[tawdry salmon]

alright got it ty

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[toxic ravine]

whats the difference beetwin these two

[toxic ravine]

whats the both means

[digital sparrow]

bitwise operators?

[raven heart]

or gcd / lcm

how can we know

[toxic ravine]

about divisiblity

a ∧ 2a+1

a ∨ 2a+1

whats these means

[lean otter]

honestly

not much

Must be some notation specific to your class

Please send context

[toxic ravine]

it's a question :

calculate for any non-zero natural integer n

a ∧ 2a+1

and a ∨ 2a+1

[lean otter]

huh

Any files that might explain what these notation mean?

I mean surely your teacher gave you something that says exactly what these symbols are for

[raven heart]

∧/ ∨ as gcd/lcm is really not unheard of

lapiwin

[lean otter]

Oh?

[raven heart]

especially the first

[split ether]

The notation makes sense tho, in relation theory those denote sup and inf

[raven heart]

@toxic ravine

[toxic ravine]

gcd

[split ether]

If the relation is division, that's what gcd and lcm are

[toxic ravine]

ohh ok dude

thaks for helpin <3

i understanded

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[fresh orbit]

theres a question which i have been wondering: is there an equation of a locus which connects the minima of all curves in the form x^2+ax-a?

[hard crest]

well the minimum of a quadratic ax²+bx+c is achieved at -b/2a

[fresh orbit]

but what if i have like x^2+3x-3, x^2+4x-4, x^2+5x-5, ….., etc., is there an equation which connects the minima of all these curves?

[dense pulsar]

C-b^2/4a i think

[hard crest]

well let's try to find one

what's the minimum of your expression?

it's achieved at x = -a/2

[fresh orbit]

yep

[hard crest]

and what is its value?

[fresh orbit]

a^2/4+a^2/2-a

[hard crest]

okay, so if x = -a/2 then y = that thing^

can you turn that into y being an equation in terms of x?

try to get rid of a

[fresh orbit]

ohhhhh

i got -x^2+2x

okay thanks!

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[celest marsh]

hi

[celest marsh]

A box of spheres has 8 rows with 5 spheres each, how many spheres are there in each box?

[indigo valve]

40?

[fleet wadi]

how many spheres are there in 1 row?

[celest marsh]

has 8 rows

[fleet wadi]

now repeat that number 8 times

[celest marsh]

like 8+8+8 etc???

[fleet wadi]

you could say so

but theres already an operation for repeated sums

(multiplication)

you have 5, 8 times

8 times 5

8x5

[celest marsh]

okey

are 40

[fleet wadi]

yes

[ocean crypt]

@celest marsh you're taking up all the rooms 😡

[celest marsh]

I live in a building that has 9 floors and there are 4 apartments in each one, how many apartments are there in total?

.close

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[waxen lichen]

How do i start solving this
(Z-2+i)²=9i

[thin bridge]

express Z as (x+iy)

[waxen lichen]

How do i get it out of the ()

[versed pendant]

root both side thus, Z-2+i=±3sqrt(i)

[waxen lichen]

Thanks i shouldve known that one

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[thin bridge]

@waxen lichen @versed pendant sqrt of 9i is wrong

whoops sry misread that

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[tough elbow]

can anyone explain why this is considered wrong?

i also tried 8113 because we're talking about a real case

so bacteria wouldn't really count the decimals

but it is still wrong

[sleek plank]

db/dt = 3.079e^1.8t

find b(2)

[tough elbow]

so this step process isn’t correct way of going about this ?

[sleek plank]

nope

[tough elbow]

how would i know to integrate the problem vs the above way

[sleek plank]

it specifically state that the rate of increase = 3.079e^1.8t

rate of increase implies derivative

if it had said "the colony's size is modeled by 3.079e^1.8t", then all you would do is plug in 2 for t

[tough elbow]

would i have to use u substitution for this problem?

[sleek plank]

uhh yeah

although you can get good enough to do it in your head

[silver glade]

so it should be like || 133 || right? (sorry idrk calc so i might've got it wrong)

[tough elbow]

would you round it ?^

[sleek plank]

for that method yea, but that's the incorrect method

[sleek plank]

you should round down i'd say

[tough elbow]

makes sense

[ember canopy]

Savir?

[sleek plank]

can turn 0.5 a bacteria into 1

[silver glade]

Bro why is ryan sharma here 😭

[tough elbow]

who?

[silver glade]

nvm nvm

[tough elbow]

so would i still be able to plug in my numbers for time t=2 hours ?

[silver glade]

wait is the right answer not 132

or 133

[sleek plank]

after you've integrated

[sleek plank]

no

[silver glade]

i found the derivative

oh

[sleek plank]

please ask questions about this in your own help[ channel

[tough elbow]

and when you integrate that is also considered taking the antiderivative?

[sleek plank]

yes

you would do this process called separation of variables

[simple socket]

ur handwriting eats

[silver glade]

fr

[tough elbow]

lol thank you

after we plug in the value for 2 hours, do we plug in the constant ?

I’m a bit new to separation of variables ngl

[sleek plank]

after integrating, you get b = 1.711e^18.t + C

now you plug in t=0 and b=72 for the initial condition

and then solve for C

[tough elbow]

we have c here because there are no specific points were given ?

[sleek plank]

they did however

they gave (0, 72)

[tough elbow]

oh okay

so then why do we have C if we’re given those points

[sleek plank]

because we need to find b(2)

and if we try to do that, we get b(2) = 1.711e^3.6 + C

and we need a number for b(2)

so solving for C lets get a straight number for b(2)

[tough elbow]

okay i think i get it thank you !

[sleek plank]

!done

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[waxen lichen]

.close

[silver glade]

bruh this problem is so confusing
I got C=70.289 and i keep getting 132 😭

do i need to ask in another channel

if i have a q

sorry im new to this server

[sleek plank]

yes, make your own help channel if you want help

#❓how-to-get-help

[silver glade]

ty

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[polar pike]

what is cot(11pi/6)

[unique cradle]

do you know what contangent is in terms of sines and cosines?

[polar pike]

cos(11pi/6)/sin(11pi/6)

[unique cradle]

yes

and cos(11π/6) and sin(11π/6) should be known values

do you know what they are?

[polar pike]

cos(11pi/6) = sqrt(3)/2 and sin(11pi/6) = -1/2?

[unique cradle]

yes

so (√3/2)/(-1/2) = ?

[polar pike]

-sqrt(3)

thanks

[unique cradle]

np

[polar pike]

.close

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[cloud kelp]

i got ans to (i) i dont get how to get (ii)

[fathom jewel]

@cloud kelp

[cloud kelp]

the answer is √27 at top rest is fine

[flat frigate]

bacc (unhelpful)

[fathom jewel]

The least value of x is to find the absolute minimum

[cloud kelp]

so maximum of cos2 theta?

[fathom jewel]

yep

[cloud kelp]

so we want 1 ?

so theta will be what value

[fathom jewel]

theta = 0 for example yes

but it's periodic

so it doesn't matter

[cloud kelp]

okk thank u helpful brother

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[twilit badger]

I have some personal issue and skipped one of my math class, thing is they dont provide any online files and now Im having trouble to understand the review sheet

[rustic goblet]

sure, which part are you having trouble with?

[twilit badger]

pretty much reading the graph in general, I tried to watch video online but still confuse

I understand the little - and + on the right side of number, but thats pretty much it lol

[rustic goblet]

so you're wondering about limits?

we read $\lim_{x \to -3^{-}} g(x)$ as "the limit of $g(x)$ as $x$ approaches 3 from the left"

[flat frigate]

higher!

[twilit badger]

alright

[rustic goblet]

it is the value that g(x) gets closer and closer to as x gets closer and closer to 3, from the left

it doesn't necessarily have to equal g(3)

[twilit badger]

so on the part a. it should be -2 since as x approach -3 from left?

[rustic goblet]

what about as x approaches -3 from the right?

[twilit badger]

0? im not quite sure why is there close circle and open circle

[rustic goblet]

the open circle means there's a hole

it means that the function doesn't actual take on that value at x = -3

[rustic goblet]

but yes, it is 0

good job!

[twilit badger]

[rustic goblet]

now, what about $\lim_{x \to -3} g(x)$?

[flat frigate]

higher!

[rustic goblet]

[twilit badger]

yeah this is the part I feel lost, it translate to x approach infinite close to -3, but then there is two line referring to -3

[rustic goblet]

exactly

[twilit badger]

and if I know something about math, it is having two answer means both is wrong lol

[rustic goblet]

as x approaches -3 from the left, g(x) gets closer and closer to -2. but as x approaches -3 from the right, g(x) gets closer and closer to 0

these left and right hand limits do not agree

so the overall limit as x approaches -3 does not exist

[rustic goblet]

pretty much lmfao

[twilit badger]

XD

guess that works

[rustic goblet]

what's g(-3)?

[twilit badger]

0? or DNA since its a open circle

DNE*, god I really shouldnt study biology

[rustic goblet]

oh no

[twilit badger]

[rustic goblet]

g(x) is defined at x = -3

there is a value it's been assigned

the open circle just means that the function doesn't ever take on the value of the hole, but that doesn't mean it takes on no value at all

[twilit badger]

then how is it not 0?

[rustic goblet]

the filled in circle means the function does take on said value

[rustic goblet]

because there's a hole there

[twilit badger]

so its not exist?

[rustic goblet]

it exists! >.<

it's just not 0

[twilit badger]

HMMM

[rustic goblet]

the red line is solid all the way to -3; the blue line is solid up to, but not including, -3

[twilit badger]

-2?

[rustic goblet]

yeah

well done!

[twilit badger]

this is weird but I think im getting the hang of it

[rustic goblet]

can you tell me how to do part b), now?

the Helpfuls lurking this channel (I know you're all there!) are going to have a field day with c)

but let's do b)

[twilit badger]

a.2 b.1 c.3 d.DNE e.infinite, I think im wrong with the 3 and DNE :(

[rustic goblet]

a)
b)
c)
d)
e)

d) depends on who you ask, ig. some would say DNE, others would say infinity

it just depends on if you count infinity as a valid limit

but e) is definitely not correct

as x tends to infinity, you can see that the graph of g(x) leveled off

[twilit badger]

hmmmm, as X approach pos infinite, shouldnt the g(x) also be?

[rustic goblet]

not necessarily

for example, if $f(x) = 1$, then $\lim_{x \to \infty} f(x) = 1$, not $\infty$

[flat frigate]

higher!

[rustic goblet]

in that case, the graph of f(x) is a flat line, which doesn't grow to infinity as x goes to infinity

in your case, you have something similar

the graph of g(x) plateaus and doesn't get above a certain value, even as x gets larger and larger

that value is $\lim_{x \to \infty} g(x)$. what is it?

[twilit badger]

so the answer is 4?

[flat frigate]

higher!

[rustic goblet]

yeah

now... c)

agh

@hard crest @solar hazel are you two still watching

do you two want to deal with c

[solar hazel]

i wasn't watching at all

[rustic goblet]

ic

@twilit badger give c) some thought

where is the graph of g(x) discontinuous?

[twilit badger]

at -3, 0, and 2?

[rustic goblet]

yes, as much as it hurts me to say

[twilit badger]

my logic is they have a disconnetion and a new line there

huh

[rustic goblet]

sure, that logic is okay for this question

[rustic goblet]

so this is correct

the reason it bothers me (and many others!) to say that 0 is a discontinuity point is because g(x) isn't even defined at 0

but this is smth I don't think you should worry about

so I'll let it go

0 is a discontinuity for our purposes

[twilit badger]

[rustic goblet]

you're done!

congratulations!

[twilit badger]

yay

[rustic goblet]

[twilit badger]

now I can move to next part of the suffering :D

[rustic goblet]

[hard crest]

no

[rustic goblet]

fair enough

I did it

[twilit badger]

Im gonna try to hit myself with some practice and video to see if I can deal, before I come back here defeated

thank you @rustic goblet

[rustic goblet]

okie dokie

[twilit badger]

[rustic goblet]

sounds good to me

[rustic goblet]

no problem!

I hope you have a great day!

[twilit badger]

you too!

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[swift owl]

would this work?

[junior smelt]

With the exception of this one, yes to the rest

[swift owl]

thanks, should i sub for s?

nah that wouldn't make sense

[junior smelt]

Not s, no think about it a little bit

[rustic goblet]

dang it cat bit

you beat me to it

[swift owl]

wouldn't the same case be for f then?

[rustic goblet]

[swift owl]

lmao the cats are cute

[junior smelt]

No, it works for (f), you get the integral becoming that of 1/(2u) wrt u if I can mentally do them

[rustic goblet]

for g), maybe it'll be useful to remember that $\int \frac{1}{1 + u^2} du = \arctan(u)$?

[junior smelt]

I was thinking about phrasing my post above to say "think a bit @rustic goblet " as well

[flat frigate]

higher!

[rustic goblet]

it's good to recognize that this is a known antiderivative, because it'll help you decide what to substitute

[rustic goblet]

higher!

[swift owl]

i see thank you!

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[narrow oak]

I need to find the derivative of the function

and this is my work (from the red circle continued) but I (think? that I) got it wrong but idk how

I multiplied out the square in the begenning so that I didn't have to deal with it later

and so that I could just do quotient rule

[exotic bolt]

i cant read that

zoom in

might be easier

if you didnt already

to set the inside as u

so u have this

then just find the deriative of the inside function by itself

[exotic bolt]

dont do that

ur wasting ur time

by distrbuting

and making it a hell for later

[narrow oak]

oh

okok i will try that and come back to lyk tyy!

:( im still wrong

[exotic bolt]

show work

[narrow oak]

The last thing is my answer^^

but this way was much less painful

and i can see the light at the end of the tunnel maybe

But where did i go wrong 😞

[exotic bolt]

u cant do that

[narrow oak]

aw man

Ur right

AAAAAAAAHHHHGGGGGGGDHDHD

Ok thank u so much i will try again

[exotic bolt]

do u have to simplify

[narrow oak]

erm this is the answer the book has

is this simplified?

[exotic bolt]

lol thats an interesting way to simplify

i would just leave this but

[narrow oak]

Wait im dumb

I sont need to multiply everything out jist to re-simplify

this is revolutionary

i swear im not this dumb

Ok THANK YOU SO MUCH!!!

I GOT IT

THAAAANKSKS

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[quiet stump]

Hi

[quiet stump]

1 digit numbers are 2,4,6,8

Second digit numbers are 0,2,4,6,8

3rd digit numbers the same as second

So the possible totally even numbers so far are 124 right?

I am not sure how to proceed

[young nexus]

write a totally even number by replacing 0 -> 0, 2 -> 1, 4 -> 2, 6 -> 3, 8 -> 4, and then interpret the resulting number in a base 5 number system.

for example:
40 -> 20, 20 interpreted in base 5 -> 2 times 5 + 0 = 10.

[quiet stump]

This is too difficult to me

But I appreciate your reply

The answer is 132 I think

[young nexus]

what do you get for 2024 if you do this replacement? write a totally even number by replacing 0 -> 0, 2 -> 1, 4 -> 2, 6 -> 3, 8 -> 4,

[quiet stump]

I don’t know sir

[young nexus]

ok, i see.

[lime dust]

What is the totally even number right before 2000? @quiet stump

[quiet stump]

888

Got it

+13 numbers

[lime dust]

What do u mean

+13

[quiet stump]

124+13 numbers

13 four digit up until 2024

[lime dust]

U sure?

[quiet stump]

Hmmm

[lime dust]

I only count 8

[quiet stump]

Oh yes

[young nexus]

124+13 = 137 which isnt a possible solution.

[quiet stump]

I am counting the 1 lol

[lime dust]

Totally

2010 doesnt count

[quiet stump]

My head hurts

Yeah so there are 8

Thank you

[lime dust]

!done

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[quiet stump]

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[lean otter]

2-|x|<=|x-2|

lets stipulate del <= 1/2
|x-3| < 1/2 ( 1 don't work as x-2 = 0 so 1/2)
3 - |x| <= |x-3| < 1/2
5/2 < |x|
minimum is 5/2 so
2-|x|<=|x-2|
2-5/2 < |x-2|
-1/2 < |x - 2|

well lmao |x-2| minimum coming negative (-1/2)

but i should be 1/2

idk wtf i did

help!!!

[long oracle]

If you know |x| > 5/2, then -|x| < - 5/2, so you can only say that 2 - |x| < 2 - 5/2 but not that this is less than |x-2| (which it is in this case and is not helpful)

[lean otter]

why the
2 - |x| < 2 - 5/2 ? 5/2 is the minimum of |x| isn't the 2 - |x| > 2 - 5/2 should be the case?

[long oracle]

Alternatively, I think it might be better if you try and do something with |x-3| and |x-2| together with the reverse triangle inequality,
e.g. 1 - |x-2| <= |1-(x-2)| = |x-3|

[long oracle]

No

[lean otter]

OH IM

YEAH

5/2 is less that means

2-|x|

will be MORE

YEAH

SORRY

nvm

that

[long oracle]

As I said if you know that |x| > 5/2 then multiply by -1 it will be reversed

No worries!

[lean otter]

ohhhh together?

[long oracle]

Yeah, see if this is helpful

[lean otter]

oh lmfao i actaully know this why the hell i didn't use that xd
|x-3| - 1 <= |x-3-(-1)|

but the problem is we only the upper bound of |x-3|
|x-2| - 1 <= |x-2-1|
1- |x-2| <= |x-2-1|
1 - |x-2| <= |x-3| < 1/2

oh yes that works good

1/2 <= |x-2|

!!!

also makes sense why my approach

didn't work

[lean otter]

i acutally have to like

find upper bound of

|x|

[long oracle]

Yeah so it would be different if you had an upper bound, say it being 3/2 somehow. Then you’d get the same lowebound for |x-2|

[lean otter]

makes so much sense

ngl i think my understanding

increased by 10x

after this convo

thank you so much!!!!

❣️

[long oracle]

You’re welcome!

[lean otter]

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[lean otter]

sorry for pinging but
5/2 <= |x|
|x| - 2 <= |x-2|
5/2 -2 < |x| -2 <= |x-2|
1/2 < |x| -2 <= |x-2| ?

[lean otter]

yes going through 24 epsilon delta proofs

xd

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[blissful flicker]

How is the answer b

[blissful flicker]

I know according to the extended pigeonhole principle, this should be ceil(n/k) but then how is ceil(n/k) the same as (n-1)/m + 1

[light shoal]

how does k relate to this

do you mean ceil(n/m)?

[blissful flicker]

yes

[light shoal]

these all seem wrong tbh

unless i'm missing something

what if say n=10 and m=2, then one possibility is that each hole gets n/m = 5 pigeons

the formula for (b) would give you (10-1)/2 + 1 = 5.5, which is false

[magic junco]

My thoughts:
distributes all the pigeons to the holes equally. If the numbers of pigeons is not divisible to the quantities of holes, the answer would be the quotient + 1.
On the other hand, if they are divisible, the answer should be quotient.
If the amount of holes is greater than pigeon, the answer should be 1

[blissful flicker]

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[magic junco]

Do you understand?

[magic junco]

Agree, they all look incorrect

[blissful flicker]

yes

[long oracle]

That works!

[lean otter]

ohhh!

[long oracle]

(sorry for intervening in this channel)

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[lean otter]

yay! but wait what if u put maximum value of |x| in
|x| - 2 <= |x-2|

[lean otter]

how is that gonna work

[long oracle]

Like |x| - 2 <= m - 2, where m is max?

Then you wouldn’t be able to say anything (directly) in relation to |x-2|

[lean otter]

ohhhh

we just don't know where should M-2 will lie respect

to |x-2|

right

[long oracle]

I think it might help if you do something concretely with numbers, suppose you had 1 <= 2

Then we know that 1 <= 10

But 1 <= 10 <= 2 is not true

[lean otter]

yess

that makes perfect sense mapping

my problem on this

yay nice i think all my doubts went away

thank you again!! 🙏🏻

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[lean otter]

is it gonna be like
|x| - a <= |x-2|
|x| - a <= m-2
but can we do
0 <= m-2 - |x-2|

|x-2| <= m-2

.reopen

[safe radish]

✅

[long oracle]

How?

Again, m could really be anything bigger than |x|

Even if it’s specially the max we can’t say for certain this was true

[lean otter]

oh yeah

[lean otter]

ah

[long oracle]

Is there a specific reason you’re attempting this?

Your previous way of going about it was solid

[lean otter]

oh its just xd

[lean otter]

is it cuz

m-2 - |x-2|

we don't know

positive or negative

[long oracle]

I mean yeah, it could be anything

However when you subtract two inequalities that’s now how it’s done

I’m assuming that’s what you did

[lean otter]

yeah

i think i assumed stuff

[long oracle]

Because you can say something, but it’s probably not useful

But before we sidetrack again, was there a reason you abandoned your old way of achieving the bound? If anything that’s the preferred way of doing it

[lean otter]

ahh well i think im just overthinking about this and i should just move on fr

[long oracle]

It’s good that you’re experimenting, epsilon delta proofs like these as you may have noticed is just knowledge about inequalities

[lean otter]

yess

[long oracle]

So keep at it with the inequalities, I think if you take the time to study inequalities in general you’ll thank yourself in the future!

[lean otter]

is there any

book

i can read

[long oracle]

That’s a good question, I know that there is one but i don’t remember if its overkill but it might not hurt, let me look it up

The Cauchy Schwarz Master Class

This book is specifically about inequalities

[lean otter]

thank you!!! i might start reading it

[long oracle]

But it may be overkill

[lean otter]

ah

[long oracle]

But I guess it doesn’t hurt to try! In any case, I think most intro books should have something about inequalities

[lean otter]

true

[lean otter]

🙏🏻 thanks for the help will look up the book

closing the post

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[rare thistle]

Given triangle ABC. A secant intersects AC at F, intersects AB at E and intersects BC at D. Prove that the orthocenters of the three triangles ABC, BDE and CDF are collinear. Prove this in affine coordinate system

[rare thistle]

G,H,I are orthocenters

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[rare thistle]

<@&286206848099549185>

[rare thistle]

<@&286206848099549185>

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@rare thistle Has your question been resolved?

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[sacred mountain]

, rotate

[sacred mountain]

Can anyone explain

[plucky elk]

Did you write the probability as a fraction in terms of counts of marbles

[sacred mountain]

I didn’t do anything because I’m a bit lost

U there

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@sacred mountain Has your question been resolved?

[digital sparrow]

lets say the number of silver marbles is x

what is the total number of marbles?

@sacred mountain

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[soft blade]

so uhhh, the first one I put -64 as the answer and its wrong

[soft blade]

is it not evaluating?

[thin bridge]

can you show your work

[soft blade]

6(4+0.1)-7 - 6(4)-7

all over 0.1

im right so far?

[thin bridge]

you didn't distrubte the - sign to the f(4) correctly

[soft blade]

OH

sorry I didnt respond

[soft blade]

that doesnt change anything?

hold on

so the first part is 17.6

then the second part is just 6(4) -7

right?

[safe radish]

@soft blade Has your question been resolved?

[thin bridge]

and then subtract properly

[soft blade]

like

0.6

[thin bridge]

for the numerator yes

[soft blade]

0.1

denominator

so like 6

[thin bridge]

yes

[soft blade]

alr

6 for the second one

bruh its all 6

thanks

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[remote lily]

In the art of electronics there is this constant A, it stated later that A is a constant based on the initial conditions of the RC circuit. Which is -Vf or -Vsupply. What I cannot understand is how they got this constant, by my logic it should be 0 when time = 0. I'm failing to understand how the initial voltage across the capacitor is -Vf.

[upper hatch]

well when you first connect the battery you definitely have some sort of voltage, so why would Vout be 0?

when you initially connect the battery, you only have the battery's voltage going out towards the rest of the circuit at Vout

[remote lily]

Initially there is no charge on the capacitor so almost all the current is dissipated by the resistor, which gives you practically supply voltage across the resistor initially

[upper hatch]

right

so maybe A is Vf?

[remote lily]

Well no its -Vf according to the equation given by the book. And it is correct as I've tried it with values and an equivalent equation.

[upper hatch]

yes thats what i meant, i thought the original equation itself had the - in it

[upper hatch]

for t=0, did you calculate the Vout?

[remote lily]

Here's the rest of the excerpt

[upper hatch]

if A=-Vf?

[remote lily]

Vout = 0 as its essentially measuring across the capacitor.

And that's why I'm so confused

[upper hatch]

so the physics adds up for you, why Vout=0, but the math isnt?

[remote lily]

Essentially yes

And I'd like to understand why

[upper hatch]

Vout(0) = Vf - Vf* e^0/RC = Vf - Vf *1 = Vf- Vf = 0

remember that e^0 is equal to 1

[remote lily]

Yes that makes sense but how do you derive the A?

That's the problem I'm having, where on earth does A come from

[upper hatch]

by the initial condition

you yourself said that for t=0, the Vout should be 0

so, Vout(0) = Vf+A * e^0/RC. but Vout(0) = 0 so Vf+A * e^0 = 0 => Vf + A = 0 => A=-Vf

[remote lily]

I think I get it now, the wording was really confusing

Essentially you solve for A using the initial conditions of the circuit =0v

[upper hatch]

exactly

[remote lily]

That could really have been worded better and have more work shown but atleast I get it now

The way it's worded in the book makes it seem as if the v=0 at t=0 is directly used to solve for A

Which makes no sense at all

Thanks for the help

.close

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[reef junco]

how do i do 7b

working out and answer for 7a

its all right

[strong token]

u need help with b?

[reef junco]

yes

idk how to start it

nvm

i got it

i was being stupid

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[scarlet comet]

@white umbra #help-14 message do you know how to do the manipulation?

[scarlet comet]

I am stuck trying to manipulate the values for b and c

I am at a[x1; w1] + (1-a) [x2; w2] rn

and i want to show that this is also in E_f

[short topaz]

sps f is convex

and let (x_1, w_1) and (x_2, w_2) be in E_f

we want to show

$(\lambda x_1 + (1-\lambda) x_2, \lambda w_1 + (1-\lambda) w_2) \in E_f$

[flat frigate]

LY

[short topaz]

write down what this means

and then basically it just falls out from ur definitions basically

[scarlet comet]

Thats the same as mine

Now I'm stuck

[short topaz]

so as in what does this mean?

[scarlet comet]

no idea

[short topaz]

what's the definition of E_f?

[scarlet comet]

I think we need satisfy one more condition

f(x) <= w

but the rest are fine

agreed?

[scarlet comet]

This*

[short topaz]

yes but written out, we wanna show

[scarlet comet]

since the first coordinate is in R^n and the second coordinate is R

[short topaz]

$$f(\lambda x_1 + (1-\lambda) x_2) \leq \lambda w_1 + (1-\lambda) w_2$$

[flat frigate]

LY

[scarlet comet]

So i think we just substitute it into the expression?

[scarlet comet]

I wrote this expression too now, how do I show that it's true

I don't know how to use the convex function definition to show it

[short topaz]

so start with the LHS

we know f is convex, what does that tell us?

[scarlet comet]

<= a f(x) + (1-a) f(y)

[short topaz]

yeah ok

and f(x) <= w_1 right?

so we're done

[scarlet comet]

Omg

[scarlet comet]

How do we do the other way?

wts f(ax + (1-a)y <= af(x) + (1-a)f(y)

[safe radish]

@scarlet comet Has your question been resolved?

[scarlet comet]

@short topaz

[short topaz]

we've shown in part a that x,f(x) is in the set for all x

[scarlet comet]

What does that tell us?

[short topaz]

let's try plugging in x,f(x) and y,f(y) into the condition for E_f

so as in we know

$\lambda (x, f(x)) + (1 - \lambda) (y,f(y)) \in E_f$

[flat frigate]

LY

[short topaz]

what does this condition actually mean?

[scarlet comet]

Not sure

It is different dimension

+1

[short topaz]

so as in

$f(\lambda x + (1-\lambda) y) \leq \lambda f(x) + (1 - \lambda) f(y)$

[flat frigate]

LY

[short topaz]

this is the condition for convexity

so we're done

[scarlet comet]

Can you explain more how we get this result

[short topaz]

literally from this

[short topaz]

as in that's true cus E_f is convex

so plug in what it means for something to be in E_f

[short topaz]

and u end up with this

[scarlet comet]

What exactly is done here

|

[short topaz]

so from part a

$(x, f(x))$ and $(y, f(y))$ are elements of $E_f$, right?

[flat frigate]

LY

[short topaz]

so then E_f is convex by assumption

so $\lambda (x, f(x)) + (1 - \lambda) (y, f(y)) \in E_f$ right?

[flat frigate]

LY

[short topaz]

so that means

$(\lambda x + (1-\lambda)y, \lambda f(x) + (1-\lambda) f(y)) \in E_f$

[flat frigate]

LY

[short topaz]

(just expand out this)

right?

[short topaz]

so then because $(\lambda x + (1-\lambda)y, \lambda f(x) + (1-\lambda) f(y)) \in E_f$

[flat frigate]

LY

[short topaz]

if it's in E_f it means f(x) <= w

so

$f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda) f(y)$

[flat frigate]

LY

[scarlet comet]

Tysm

I forgot I can expand it

[short topaz]

nw

[scarlet comet]

.close

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