help-33

[knotty trellis]

im asking for what values of t can we find a spot for car 2?

we know that t must be between 0 and 8 for car 1 to even fit there

but what about car 2?

For what values of t will there be space for it?

[slender comet]

8?

[knotty trellis]

Yes, for t = 8 there definitely is space for it

[slender comet]

yea

[knotty trellis]

but what are all the possible values of t?

[slender comet]

13

[knotty trellis]

there are whole intervals of them

[knotty trellis]

T cant be 13

t must be between 0 and 8

can you open this graph?

[slender comet]

yea

[knotty trellis]

https://www.desmos.com/calculator/23yxk5hxre

sorry, wrong one

as you can see t can be at most 8

[slender comet]

im so stupid

yea

the first car has a wiggle room of 6t

[knotty trellis]

okay, so now try finding the interval(s) of all possible t, such that there is still a place for car2 to be parked

wdym by wiggle room?

[slender comet]

well the first car can park within the first 3 meters, or within the last 3 meters, and the second car can still fit in

[slender comet]

8/13?

[knotty trellis]

Why 3?

[slender comet]

because it leaves 5 meters on the other side

[knotty trellis]

oh i see what you mean

[slender comet]

I think thats how they got the answer

6/8

[knotty trellis]

the answer is not 8/13 though

[slender comet]

no its 6/8

[knotty trellis]

yeah

thats because t must be either between 0 and 3 or between 5 and 8

[slender comet]

the first car right?

[knotty trellis]

yep

[slender comet]

right

man

[knotty trellis]

if its between 0 and 3, the 2nd car can be parked on the right side

if its between 5 and 8, we can park another car on the left side

[slender comet]

yea

my head hurts

I dont like probabilities

appreciate the help

.close

[marsh citrus]

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[marsh citrus]

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[nova rivet]

You have the numbers 2, 3, 6, 42 and 27. You can only use one +, one -, one × and one ÷ and you have to use all the numbers, so that the math expression totals to 0

[nova rivet]

it works with
(42 ÷ 6) - (27 ÷ 3) + 2 = 0 but it uses two ÷

and no ×

is this possible?

[nova rivet]

<@&286206848099549185> is this possible?

[hollow glen]

yes

try to get 0 with a subset of these numbers

[odd crest]

yes

i got one

a solution

it is possible

[hollow glen]

and no multiplication

[odd crest]

that works as well

i got one using all numbers and all operators once

so it is possible

[nova rivet]

do you maybe remember the answer?

[odd crest]

i have the answer written down

but i can't just give it

[nova rivet]

bet

[hollow glen]

as I said, can you get 0 with a subset?

[odd crest]

you're correct by saying 42/6

you can do that too

[nova rivet]

so only using some of those numbers?

[hollow glen]

yeah

[nova rivet]

no i think i have to use all of them

[hollow glen]

that will work towards a solution

[odd crest]

0 multiplied by the last number gives 0

[nova rivet]

but you can only use those numbers

0 is only the answer

[odd crest]

if you can find 0 by using only some of the numbers and no multiplication

you can multiply the rest of the numbers by that 0

to get 0

did you find a solution using that rbit?

[hollow glen]

let's say the numbers are 2, 5, 7, 102, 306, then you can just do (7-(2+5))*(306/102)

[odd crest]

i found a solution using that method as well

[hollow glen]

it could be anything really in place of 102 and 306

[nova rivet]

so like (y)*(0) but i need to get something to 0 and a y without multiplying

[hollow glen]

why not 732502 and 83991013

[odd crest]

yes

the y is the least interesting because it can be any combination of numbers and operators

as long as you can find 0 without multiplication

[nova rivet]

yes

ill try

(42+27)*(6/3 - 2 ) = 0

this may be it

[odd crest]

yeah

that's one of the solutions

[hollow glen]

yeah theres more it seems

[odd crest]

you can swap the 2 and 3 to get another

[hollow glen]

42÷6 was suggested too

[odd crest]

and you can find one without this trick by doing 42 / 6

to get 7

and then using 7, 2, 27, 3, +, -, x to get 0 as well

[nova rivet]

27-(7+2)×3

and then
27-(((42÷6)+2)×3)

27-(42÷6+2)×3

actually

[odd crest]

you got the solution i found :D

[nova rivet]

or i you can just divide 0 by anything to get 0

(6-3*2)÷(42+27)

could also work

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[static geode]

I am supposed to do this by completing the square.

[static geode]

This is the one I came up with

Any way to improve it?

Should I have come up with another square or should I try to find the lowest value with this one?

I see there are (y+1) and (y-1) involved but I don't see a way to do something with this knowledge

Oh, thanks for your silence. I found the answer myself. It is even more precious.

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[slender comet]

Den stokastiska variabeln $\xi \in N(0,1)$ Bestäm talet a så att a) $P(\xi > a) = 0.001$, b) $P(\xi > a) = 0.999$, c) $P(|\xi| < a) = 0.95$, d) $P(\xi > a) = 0.05$. Can someone help me understand how to solve these?

[elfin berry]

Totalani

[marsh citrus]

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[slender comet]

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[still temple]

Saw a youtube short of 70%÷14%. And the comments were having a debate if the answer was 5% or 500%. What’s the answer?

[odd orchid]

you can see the % sign as a fraction, 1/100

so (70/100)/(14/100) is 5

or 500/100 or 500%

[still temple]

Yeah, that’s what I thought. Most comments said the answer is 5%. So I was confused

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[modern tiger]

Need help integrating

tan^8(x)sec^2(x)

[odd orchid]

try a subsituion

[novel juniper]

hint: Write tan in terms of sec

[odd orchid]

i think theres a much simpler way that that

[novel juniper]

ooh, yeah

tehre is

[modern tiger]

I’m still a bit confused, where would I substitute here?

[thin shell]

consider the following: the derivative of tan is sec^2x

[odd orchid]

u = tan x

[modern tiger]

Ohhhhhhh

i cant believe i got stuck on this one

Answer is 1/9tan^9x right?

[odd orchid]

(1/9) but yes

[thin shell]

weird order of elements, but if you mean tan^9x/ 9 then yes

[modern tiger]

Yes, thank you!!

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[faint anchor]

Hello , is there anyway to solve it without dx/dy ?

iam stuck on this step
1-cos+cos^2/x^4

[limber hearth]

Without dy/dx you mean without l'hôpital?

[faint anchor]

yeah

[sullen dagger]

Since x -> 0, you can use the small angle approximation for cos (i.e. take the first 2 non-zero terms of its Taylor expansion, and drop the rest)

[spark otter]

it's not cos(x)(1-cos(x))

it's cos(1-cos(x))

[proud ice]

Do you mean $$\lim_{x\to 0}\frac{1-\cos x(1-\cos x)}{x^4}$$
or $$\lim_{x\to 0}\frac{1-\cos(1-\cos x)}{x^4}$$

[elfin berry]

SWR

[spark otter]

then '1 - cos + cos^2' is wrong

[proud ice]

This is what I thought at first, but his working is making me think that perhaps he just omits the variables for cosine (which is very egregious and error prone if true)

[proud ice]

yikes

[faint anchor]

oh

[spark otter]

anyways, you can use that 1-cos(t) is equivalent to t^2/2 when t-> 0 as suggested before

[faint anchor]

only trig identities is allowed

[spark otter]

start by showing that (1-cos(t))/t^2 -> 1/2 when t-> 0 then

hint: cos(t) = cos(2 * t/2)

[faint anchor]

2sin^2(x/2)/x^2

2(sinx/2 / x ) ^2
2(1/4)
=1/2

[faint anchor]

i stopped here

[spark otter]

That's not what I would have done

now that you found that (1-cos(t))/t^2 -> 1/2 when t-> 0

why not use it with t = 1-cos(x)

[faint anchor]

i didnt get it
can u please explain it in mathmatical solution ?

[spark otter]

(1-cos(1-cos(x)))/(1-cos(x))^2 -> 1/2

when x -> 0

so now

use this

[spark otter]

to find something about this limit

[marsh citrus]

@faint anchor Has your question been resolved?

[marsh citrus]

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[rocky sedge]

hemlo

[rocky sedge]

how to integrate $4\int u^4 \cdot e^{-u^4} du$

[elfin berry]

Luh Roub

[devout mauve]

you dont

[rocky sedge]

pls

for the love of god

[devout mauve]

!original

[marsh citrus]

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.

[rocky sedge]

okax

okay

The original problem is

$\frac{1}{4}!$

[elfin berry]

Luh Roub

[rocky sedge]

parenthesis ig

[devout mauve]

you can numerically solve the integral

but I dont think there is a closed form

[rocky sedge]

i want to introduce the pi function, respectfully

[devout mauve]

(at least WA doesnt give it)

[rocky sedge]

what does numerically even mean fr

[devout mauve]

approximate it

[rocky sedge]

no

i need pi function

[limber hearth]

,w factorial of (1/4)

[elfin berry]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by yakubros

[rocky sedge]

omg

i wouldnt have asked if i wanted to know it asap

I wanna do it w pi function

[rocky sedge]

Hence this question

[fervent rampart]

Simple closed-form expressions of this type do not appear to exist for Gamma(1/n) for n a positive integer n>2.

[rocky sedge]

bro

just help me integrate that mf

[devout mauve]

"just help me integrate something which no one has done so far"

[rocky sedge]

Its been done before wym

[fervent rampart]

if you have an answer which you are aiming for then can you share it?

[rocky sedge]

uh im ideally aiming for this

And so far

Ive introduced the pi function

substituted everything in

but the integrating part

Idk ab that

need the di method and no idea how to apply it here

[main idol]

di method?

[rocky sedge]

yea

[devout mauve]

a method to write down ibp

[dry prawn]

ibp wearing glasses and a moustache

[main idol]

di, chi's successor

[mellow crag]

as was mentioned, there is no nice closed form

that means that you can't write down some expression involving i.e. e, pi, sin/cos etc. that gives you the answer

you can only approximate what the integral equals

[rocky sedge]

uh

0.5! = √π/2

[devout mauve]

and?

[rocky sedge]

so i dont see why 0.25! wouldnt have one

[devout mauve]

different number, different thing

[mellow crag]

i.e. sin(pi/2), sin(pi/4) etc. all have nice closed forms

sin(1) doesn't

[devout mauve]

some numbers just dont have a nice form

[rocky sedge]

hmm

makes sense then

how do i tell which one does and which one doesnt have a closed form

[fervent rampart]

gamma(1/2) has a nice expression but gamma(1/3), gamma(1/4) etc do not

[mellow crag]

plug it into wolframalpha, if wolframalpha gives you the option for a closed form then a closed form exists

if it doesn't have a closed form it probably doesn't exist lol

but anyway in general, most integrals you write down do not have closed forms

[fervent rampart]

because gamma(1/2) can be related to the gaussian integral

[mellow crag]

in some sense, if someone gave u a random mess of functions, 99/100 it won't have a closed form for the integral

so this means ones that do have a nice closed form are special

[rocky sedge]

indeed

[mellow crag]

i think like school etc. can skew someone's perspective on the number of functions we can integrate and get a closed form for

ofc in school, they will only ask you things you can integrate

[rocky sedge]

alright thanks lads😔

[mellow crag]

but most of the time, there just doesn't exist a closed form

[mellow crag]

nw

[rocky sedge]

wait

holon

i still wanna integrate that🙏

f that closed form

I want 0.9 to appear at the end of my calculations ong

[mellow crag]

there's lots of ways of numerically integrating a function

i.e. trapezium rule etc.

unfortunately i'm not really that knowledgable on numerical methods so someone might be able to suggest better methods

[rocky sedge]

idk about all of that fr

i jus came asking for the uh

di

[fervent rampart]

integration by parts will not get you anywhere on this integral. you're welcome to try it and find out why

[rocky sedge]

.close

[marsh citrus]

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[native garden]

Need help understanding proofs in discrete math of the types

1. Direct
2. Contrapositive
3. Biconditional
4. Proof by induction
5. Contradiction

I have some background knowledge on it already, but I still struggle with writing good and rigorous proofs. I struggle specifically with knowing which situation is optimal for each type. Like identifying what might be easier for me try (eg a direct proof vs a proof by contrapositive)

id like to figure out a more structured way of approaching proofs

[marsh citrus]

@native garden Has your question been resolved?

[idle ridge]

I don't think there's a general strategy here, maybe it'd be easier to give you help if you gave a specific problem where you were having trouble figuring out which type of proof to try?

[native garden]

The thing is there’s no specific problem it’s just

Okay

Firstly here are the course definitions

Here’s an example

Let n be an integer. Prove that if n^3 + 5n is divisible by 6, then n is divisible by 2 or 3 or both

I’ll explain how I’m thinking about the problem and how I would approach it

So we have a few definitions to work with here

We know n exists in Z

We can say that n^3 + 5n = 6m where m is some integer

And now we need to prove that n is divisible by 2 or 3

Well we rewrote n^3 + 5n as 6m

We can set 6m = 3 and solve for m

Set 6m = 2 and solve for m

But I still don’t think this gets to the heart of showing that the proof works in general

And at this point now I’m lost and not sure where to go

[idle ridge]

For modular arithmetic stuff, when you're thinking about divisibility, I think it's often helpful to argue by cases

Here you could consider six different cases

[native garden]

They never taught us proof by cases!

🥹🥹🥹

I’m going to cry

[idle ridge]

It's okay, it's not a difficult concept

[native garden]

I literally feel like I can’t get this no matter what

[idle ridge]

The idea is like this

You just consider all the different possibilities for what n could be

And then prove the statement for all of them

Have you learned about modular arithmetic yet? Like n = 0 (mod 6)

[native garden]

I sort of know it but not really

It wasn’t taught to us exactly

I just know from programming what the mod of a number will be

[idle ridge]

alright well let's write it like this then

n could either be a multiple of 6

so n = 6k

or it could be a multiple of 6 plus 1, like n = 6k+1

or n = 6k+2 or n = 6k+3 or n = 6k+4 or n = 6k+5

and those are all the possible cases for what n could be

does that make sense?

[native garden]

Maybe?

I get n = 6k

I am not sure on n being a multiple of 6k + 1

Because then this doesn’t cleanly go into what we wanna prove anymore no?

We wanna prove it’s divisible by 6

[idle ridge]

we want to prove what's divisible by 6?

[native garden]

Wait no

Ugh

We want to say that If It’s divisible by 6 then it is also divisible by 2 or 3

[idle ridge]

that n is divisible by 2 or 3 right

[native garden]

Yeah

So if we represent it as 6m

Or 6k

Maybe we can prove it’s divisible by 2 or 3 by just dividing it by those values and showing there’s no remainder?

[idle ridge]

How do you know if it's gonna be divisible by 2 or by 3

If you're dividing by one of them

[native garden]

Ig I don’t

Ugh

Well

Wait

What if I

Take 6k and divide it by 3 to get 2. Then we know for a fact it’s divisible by 2

And then divide it by 2 to get 3

Maybe I’m mixing this up

Ugh

[marsh citrus]

@native garden Has your question been resolved?

[marsh citrus]

@native garden Has your question been resolved?

[idle ridge]

If you're having trouble with a direct proof, sometimes that's a good sign to try a contrapositive proof instead

What would be the contrapositive of this statement?

(In general, there can be plenty of ways to approach a proof, not just one correct answer)

[native garden]

I just don’t get why I’m having trouble

I don’t get what I don’t get

[native garden]

The contrapositive would be

If n is not divisible by 2 or 3 then n^3 + 5n is not divisible by 6

[idle ridge]

yup

so now judging based on the cases we said earlier with n = 6k, n = 6k+1, etc.

which ones are possible if n is not divisible by 2 or 3?

[marsh citrus]

@native garden Has your question been resolved?

[marsh citrus]

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[past comet]

I am doing the part b of this question, and the answer said using the -b/2a method to find the midpoint coordinates. I don't really understand how this -b/2a works and I also remeber there is another few similar rules of these types of equations. Can anyone explain it to me please?

[weary karma]

The -b/2a here will give the midpoint of the function and the absolute minimum

It is because the quadratic will form a parabola

And the cordinates of its vertex are -b/2a, -D/4a

[past comet]

But this is a circle equation

Dont only normal quadratic have a vertex?

[weary karma]

The circle equation should have y^2 coeff too

The last equation is terms of x and k

I dunno what you have done but every quadratic in 1 variable has a vertex

[past comet]

Is the definition of vertex mean that its the midpoint of the whole quadratic formula

[weary karma]

So you're trying to find the midpoint of a line which cuts a circle

[past comet]

Yes

[weary karma]

The whole function is symmetric about its axis , so in a way yes

[weary karma]

Lemme think a bit , this is the first time I've seen someone use -b/2a for midpoint of a line

[past comet]

I am processing the correlation between the vertext and the midpoint of the line

[weary karma]

Oh

[past comet]

I dont understand the correlation

[weary karma]

Hmm

I suppose a better method would be to use sum of roots

Since we can find the cordinates where it will intersect the circle through that quadratic

Wait

They have used that

Basically

x1 + x2 will be -b/a

Sum of roots of a quadratic equation

The midpoint is simply x1 + x2 / 2

Which is

-b/2a

This relates to the graph as well

If you graph the quadratic and find the midpoint of its roots , it will be the vertex

[past comet]

Then I just put the x point back to the st line equation?

Then I can find y

?

[weary karma]

Yeah

There should be a better way too

Use the equation of the line

To find a relation between x1 + x2/2 and y1 + y2/2

Should be easier

Since x1,x2 and y1,y2 satisfy the equation of the line

[past comet]

I see

I have no idea how to do that

[weary karma]

;-;

Well

y1 = -x1 +k

[past comet]

Yea

[weary karma]

Same for y2

Just add them and divide by2

Bro is NOT a chill guy

[weary karma]

y1 + y2 / 2 = -(x1 + x2)/2 + k

Use this after finding x1+x2/2 by -b/2a

[past comet]

Ahhhhh

[weary karma]

This was a pretty decent question

[past comet]

Lmao this is one of the pass papers for our entrance exams

[weary karma]

What are you studying for?

Bro is literally yapping 😭

[past comet]

Wait no its not called entrance exams

[proud arch]

Wha...

[past comet]

Is more like diploma for secondary exams

[weary karma]

Oh damn

Good luck

[past comet]

I have no idea what does any of that means

I am only finding the midpoint in terms of k

I dont need exact numbers

[weary karma]

Totally

[weary karma]

He's just saying anything bro 😭

[past comet]

Ok buddy

[weary karma]

Anyways I gotta go

Good luck

[past comet]

Yea thanks

How do I close this again?

[weary karma]

.close

[past comet]

.close

[marsh citrus]

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[peak fiber]

please post an actual question you need help with (e.g. a picture of one) in an available help channel

[marsh citrus]

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[valid cape]

any idea for this? i'm stumped

[valid cape]

initially i want to use this theorem

but again i'm not sure what function to divide it by

my thought was (sqrtx + 1)/(sqrtx + sin^2x)

[valid cape]

asks to determine convergence/divergence btw

[fervent rampart]

this is one where i think direct comparison would be simpler

[wary kite]

0<= sin^2 x <= 1

[valid cape]

yeah, but the problem asks to use comparison theorem

[fervent rampart]

does it specify which? it seems like you have comparison theorems I and II

[valid cape]

no, just comparison theorem

this is I

[valid cape]

wait quick check, does comparison theorem I work if it's an improper integral of type II?

[hushed egret]

what is type II?

[valid cape]

type I is the improper integrals where the bounds are infty

type II is the improper integrals where there's a singularity in the interval of the integrals

[hushed egret]

it does work

[valid cape]

interesting

sin^2x is only from 0 to 1

so if we pop that out, the function only gets bigger

[wary kite]

sqrt(x) <= sqrt(x) + sin^2(x)

[fervent rampart]

note that if the denominator is smaller then the overall function is larger

[valid cape]

yes i typo'd

$\int_0^1\frac1{\sqrt x+\sin^2x}\dd x \le \int_0^1\frac1{\sqrt x}\dd x$

[hushed egret]

crazy

[valid cape]

crazy i know

the rhs is divergent

[hushed egret]

is it

[fervent rampart]

double check your power rule

[wary kite]

add 1

not subtract 1

[hushed egret]

i love when my integral of x^-1/2 is x^-3/2

[valid cape]

oh wait

ok ok i see

oh it converges to 2

thus the original integral must converge

[hushed egret]

must it?

[valid cape]

i mean, yeah?

no?

you did say comparison theorem I also works for type II imp. integrals

[hushed egret]

it was a trick question you were supposed to say yes

[valid cape]

okay thanks, glad to hear

.close

[marsh citrus]

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[valid cape]

.reopen

[marsh citrus]

✅

[valid cape]

quick question about this theorem

it said that "Suppose that f and g are functions with f(x), g(x) >= 0"

should they be nonnegative on only the integral's interval, or on its domain?

[fervent rampart]

it does specify that on the slide

[fervent rampart]

wait i misread. but it does for this slide, and it's the same idea

the behavior of the function is only relevant on the integration interval

[valid cape]

i see. thanks

.close

[marsh citrus]

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[median stag]

if 4l^2-5m^2+6l+1=0 then show that the line lx+my+1=0 touches a fixed circle. Find the centre and radius of the circle

[marsh citrus]

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[median stag]

<@&286206848099549185>

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@median stag Has your question been resolved?

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[still temple]

I don't understand this reasoning

[desert dirge]

which part

your argument of a is (-3pi/4)

for a purely imaginary number we want the arg of a^m to be +-pi/2 or +-3pi/2, the +-pi/2 is not really doable eg if -3mpi/4=pi/2 then -3m=2 so m isnt an integer

so we go for 3pi/2 or -3pi/2 (or equivalent angle)

arg a^m = -3pi/4 * m, lets say we go for -3pi/2 then we can choose m=2, -3pi/4 * 2=-3pi/2 so a^2=i|a|^2

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

I understand your working, but why must a^m be +-pi/2 or +-3pi/2 ?

[desert dirge]

because thats the argument of i and -i

for it to lie on the imaginary axis

[marsh citrus]

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[heavy blade]

how do I expand cubics/quartics fast?

For example (x+1)(×+2)(×+3)
Usually I would just expand them per 2 groups but for a topic in fm (roots of polynomials) I cant be spending ages expanding out brackets where I have alpha and beta and gamma and delta, especially when the risk of making mistakes is high when doing this

[cloud iron]

Have you heard of the binomial theorem?

Guess it works better for polynomials raised to a power but it has implications here

[heavy blade]

yes but that would only be for when it's like (2x+3)^3 etc

If that's what ur thinking?

[cloud iron]

It is

And while it isn't directly applicable here, it has implications about how you can approach this

[heavy blade]

how so?

hmm perhaps is there another way to approach this whilst there being low implications?

[cloud iron]

I guess it's not as similar as I initially thought but there's a similar process

The first and last coefficients are easier to get as they involve the first and last terms of each binom directly

We know the first will be x^3 and the fourth will be 3*2*1

[heavy blade]

Yupp

[cloud iron]

There is a pattern that the middle two follow that can be used for problems of this nature

Try expanding (x+a)(x+b)(x+c) and see what you get

[cloud iron]

Yup

That applies to any problem like this, I imagine knowing that would be the fastest way of doing these

"Any problem like this" being three binomials multiplied together

[heavy blade]

basically I'd have to memorise this form 🥲

[cloud iron]

If your goal is to expand those problems fast, yeah

Unless you get really quick at multiplying them out

[heavy blade]

damnn okok ill see if I can get my head around this, tysmm!

.close

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[marsh citrus]

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[digital gate]

I already did part a) and b)

I am not sure how to proceed for task c)

• the eigenvalues are v1=(1,-1) and v2=(0,1)

• for task b) i have the following solution: first case: stable node; second case: marginally stable (bifurcation point); third case: saddle

---

according to gpt it should be stable but i don't know how to show it mathematically

[digital gate]

#dynamical-systems

[marsh citrus]

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[marsh citrus]

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[digital gate]

same question as before

[digital gate]

<@&286206848099549185>

[marsh citrus]

@digital gate Has your question been resolved?

[marsh citrus]

@digital gate Has your question been resolved?

[marsh citrus]

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[limpid pond]

I solved for sin(2x)=cos(x) where
0 =< x <= 360
and got 30°,90°,150°,270° are my answers correct ?

[quiet anvil]

sin(2x) = 2sin(x)cos(x) = cos(x)

So cos(x) = 0 or 2sin(x) = 1

Did you get the above?

@limpid pond

[limpid pond]

Yea

[limpid pond]

I did

[quiet anvil]

Cool, your solution looks correct to me, sorry I had to double check.

[limpid pond]

Ok thanks

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[marsh citrus]

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[still temple]

i tried solving it by using a function but i didn't found a way to prove that the derivative is positive for every number in this interval. any help ?

[wary tulip]

salut

[still temple]

hi

[wary tulip]

tu es francais?

[still temple]

nah actually im morroco and we study sci things in french

[wary tulip]

ok sorry i have very bad in maths

[mellow crag]

does taylor expanding around x=pi/2 work?

[still temple]

im not allowed to use it

[still temple]

it's okay 🙂

[mellow crag]

wait so what's ur defn of sin then?

[wary tulip]

but I think you should pass the sin(x) to the other side and expand then factored

and after doing the variation table and after using Maths gpt to check

[still temple]

sin is sin nothing defferent im just not allowed to use the taylor expanding cuz it is not a part of the chapts i studied this semester. i think the arcsin will work because of the interval and stuffs but i dont know how to use it in this case

[mellow crag]

as in how do you define sine?

normally sine is defined in terms of its taylor expansion

[mellow crag]

just like sin = y-coordinate on whatever?

[still temple]

ooh talking abt it we dont have a def for the sine not yet we just work with it as a periodic function and to find a special point we just use the formulase to add or multiply sine ans cosine (as i know i will have the taylor series next semester)

[still temple]

i tried it but it doesn't work (i dont know how to prove it )

[mellow crag]

if you shift everything to one side to get f(x) = sin(x) - 3x/pi + 4x^3/pi^3

you can show that f is concave

f(0) = f(pi/2) = 0

so it must be >= 0

although i'd imagine there's probably some other way of doing the problem tho

[mellow crag]

cus the RHS feels motivated

but i don't see the motivation

[still temple]

i dont really know how to show that it is concave (i know how to prove it in general but i dont know how to prove it in this cas becuase of the sine thing)

[mellow crag]

differentiate twice to get 24x/pi - sin(x)

remember ur small angle approximations: sin(x) ~ x

in fact, recall that sin(x) <= x

so that gets you f is concave

[still temple]

great if so it is solved then

[mellow crag]

this feels unintended tho so there'll probs be some alternate soln

[still temple]

yeah this is what i believe too, but i didnt found any (i have been thinking abt it for the last three days)

[marsh citrus]

@still temple Has your question been resolved?

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[marsh citrus]

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[wise oxide]

How I do this

[meager escarp]

either product rule or quotient rule

and chain rule

$u=1+x$

[elfin berry]

Tenant

[novel juniper]

Pretty sure that's not needed here

[meager escarp]

oh sorry

[novel juniper]

IBP is needed IMO

[meager escarp]

I thought it was a derivative

my bad

[wise oxide]

So I was suppose to move e^2x up

?

[novel juniper]

Well, yeah, as $e^{-2x}$

[elfin berry]

ƒ( wai ina teacup)= I don't know

[novel juniper]

Also, this isn't a test, is it?

[meager escarp]

yes

[marsh citrus]

@wise oxide Has your question been resolved?

[meager escarp]

$\int (1+x)/e^2x dx = \int dy$

[elfin berry]

Tenant

[meager escarp]

$\int \frac1/2 dx = \int dy$

[elfin berry]

Tenant

[meager escarp]

sorry i can't use latex

$\int \frac1+x2 dx = \int dy$

[elfin berry]

Tenant

[meager escarp]

frick

[novel juniper]

$\int e^{-2x} (1+x)$

[elfin berry]

ƒ( wai ina teacup)= I don't know

[novel juniper]

right, so how would you integrate this

[meager escarp]

good format

oh its {} not ()

ohh

anyway

I'd say LIATE

[novel juniper]

yup

[meager escarp]

A = (1+x)

err

A stands for alg

u = 1+x

dv = e^-2x

$u = 1+x$

[elfin berry]

Tenant

[meager escarp]

$dv = e^{-2x}dx$

[elfin berry]

Tenant

[meager escarp]

$\int dv = \int e^{-2x}dx$

[elfin berry]

Tenant

[meager escarp]

$v = -\frac12 e^{-2x} + c$

[elfin berry]

Tenant

[keen mist]

Who’s even the one asking the question anymore?

[meager escarp]

idk

lol

I'm just tryna figure out latex

[marsh citrus]

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[keen mist]

#latex-testing

[marsh citrus]

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[honest creek]

gcd(x,y) and gcd(201x + 2y, 100x + y), they’re not always equal right?

[honest creek]

since the determinant isn’t 0

wait

they’re always equal cuz the determinant is 1

?

[still temple]

yes

[proud basin]

they are probably equal

[honest creek]

how do you show this

[honest creek]

without this

just a moderately rigorous “proof”

[proud basin]

euclids division algorithm?

[honest creek]

it’s a multiple choice question

[novel juniper]

bezout's lemma, probably may work

[honest creek]

uhhh looks like effort though

what’s the fastest handwavy way to make an assertion

[still temple]

gcd is preserved under transformations with determinant 1 matrix for 201x + 2y and 100x + y determinant is 201 × 1 - 2 × 100 equals 1 gcd property holds since any linear combination of x and y has the same gcd check using matrix transformations and gcd definitions

[proud basin]

gcd(201x+2y,100x+y)=gcd(201x+2y-2(100x+y),100x+y)=gcd(x,100x+y)=gcd(x,y)

[honest creek]

it’s not a proofs class so i don’t need to “prove it”, just need to show they’re always equal (to an extent)

[honest creek]

hmm okay yeah this works

okay then thanks guys

.close

[marsh citrus]

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[buoyant jetty]

George is about to get a certain amount of change less than one dollar from the cash register. If he gets the most quarters possible and the rest in pennies, he would need to receive $3$ pennies to meet the amount. If he gets the most dimes possible and the rest in pennies, he would need to receive $8$ pennies to meet the amount. What is the sum, in cents, of the possible amounts of change that he is trying to get?

[elfin berry]

938c2cc0dcc05f2b68c4287040cfcf71

[buoyant jetty]

like

25Q + 3 = 10D + 8

Q is number of quarters and D is number of dimes

25Q + 3 < 100

because 100 cents is a dollar

[still temple]

the amount x satisfies two conditions:

1. x ≡ 3 (mod 25)
2. x ≡ 8 (mod 10)

solve this system of congruences to find the possible values of x

[buoyant jetty]

how

[still temple]

substitute x = 25k + 3 into x ≡ 8 (mod 10) then solve for k

[buoyant jetty]

10D + 8 < 100

10D < 92

25Q + 3 < 100

25Q < 97

[buoyant jetty]

(25k+3) = 8 mod 10

25Q + 3 = 10D + 8
25Q - 10D = 5
5(5Q - 2D) = 5
5Q - 2D = 1

(5Q - 1)/2 = D

Case Q = 1

Case Q = 1, D = 2
Case Q = 2, D = Impossible
Case Q = 3, D = 7

28 + 78

,calc 28 + 78

[elfin berry]

Result:

106

[buoyant jetty]

.solved

[marsh citrus]

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[kindred kernel]

This is 1

[kindred kernel]

This is 2

Can someone check this for me?

[kindred kernel]

I forgot to add the floor in last two lines, assume it is there🙏🏼

[kindred kernel]

I am evaluating the integral at the top of this page

[marsh citrus]

@kindred kernel Has your question been resolved?

[kindred kernel]

<@&286206848099549185>

[light island]

Riemann zeta function, ζ(i) for 𝑖 > 1 i>1.

Right?

[kindred kernel]

Yeah but a finite zeta

Yeha ig i could have zeta approximated too

But euler-maclaurian seemed straightfoward so i did it

[light island]

One second. I dont wanna take a photo so imma whip my notepad out

The expansion is consistent with the formula, but the higher-order corrections could use clearer notation to show B little 2(x-⌊x⌋)

think Bernoulli polynomials

B subscript 2(x−⌊x⌋)=(x−⌊x⌋) exponent 2 −(x−⌊x⌋)+ 1 over 6

rewriting the integral term would look like

While 𝐵 subscript 2 is well-defined, higher-order corrections (like 𝐵 subscript 4) are computationally intensive, so explicitly truncating the series or providing a bound on the neglected terms is helpful

fhis approximation works

but it disregards the slower-decaying terms, which could be significant for small 𝑖

As for the second one, it looks consistent, yet I am not 100% sure.

[kindred kernel]

Sure but R_p in Euler-Maclaurin’s can be adjusted no?

And here since we just need a closed form

Shouldn’t this be sufficient(further terms decay tremendously fast)

[kindred kernel]

Also btw pre-uni doing higher order bernoulli🙏🏼

So op

🛐

[light island]

Im in geometry

my dads a math professor at the university in my town

Okay wait back to locking in

[light island]

If precision is required for small 𝑖 the higher-order corrections or numerical summation may be necessary. For large 𝑖 the existing approximation is indeed sufficient.

I can elaborate on practical ways to adjust 𝑅 subscript 𝑝 if needed maybe

@kindred kernel

[kindred kernel]

Hmm yeah that would help, since the form im getting now( that infinite sum of g(i)) doesn’t seem to converge

[kindred kernel]

“In geometry” as in?

[light island]

are we talking first or second

[light island]

As in geometry one 😭 I just needed an extra math credit

[kindred kernel]

Op

[kindred kernel]

First pic

I mean thats actually the second page

But yeah

[light island]

agh I need to re go through it one second

Im thinking it has to do with the periodic nature or slower-decaying terms that weren't sufficiently like addressed in the approximation

Higher-order terms involving said B subscript 4 (x−⌊x⌋), 𝐵 subscript 6 (𝑥−⌊𝑥⌋) etc., would add similarly structured corrections. These are crucial when g(𝑖) is being ssummed over a like range where the periodic terms accumulate without decaying quickly

If 𝑔(𝑖) represent a contribution from the periodic correction terms, and the resulting sum doesn't converge, then I think the issue if there is any is that of insufficien\t Damping

Periodic terms like ⌊𝑥⌋−𝑥 decay slower than the main terms in the expansion, especially for small 𝑖 Including higher 𝑅 subscript 𝑝-terms will provide additional damping.

or Improper Approximation of Remainder

If 𝑔(𝑖) is derived from an incomplete 𝑅 subscript 𝑝-term (e.g., only using 𝐵 subscript 2 without adjusting higher-order corrections) could lead to divergence I think

[marsh citrus]

@kindred kernel Has your question been resolved?

[light island]

These terms rapidly damp out for large 𝑖 but also ensure convergence for smaller 𝑖

If the same divergence continues, maybe try techniques like zeta-function regularization for the periodic components to handle the slow decay

Excuse me if im wrong, im not too good at this

@kindred kernel What are we thinking

[kindred kernel]

Ahh ok so we move back to zeta

Nice

[kindred kernel]

Oh yeah i tried adding the B_3 term and doing again, still getting divergence

So i’ll try zeta later

Thanks a lot

[light island]

Im all over the place w ts

[light island]

Sorry 😭

[kindred kernel]

Nah its fine i’ll try zeta in a bit

Thanks

.close

[marsh citrus]

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[wispy ledge]

How do I do iii

[slate yarrow]

use that fact that $1+\cos {2\theta}=2\cos^{2}{\theta}$

[elfin berry]

convergence

[marsh citrus]

@wispy ledge Has your question been resolved?

[wispy ledge]

Ok one sec

[marsh citrus]

@wispy ledge Has your question been resolved?

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[unkempt violet]

Can someone please explain why we aren't considering the D in the middle of PDP^-1

OOOH NVM I GET IT

Lol

LMFAO 🤦 hahaha

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[spark hazel]

What differentiation rule is used to get from the 1st image to the 2nd?

[odd orchid]

chain rule

and derivative of ln

[spark hazel]

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[stoic slate]

I think you have a typo in your second integral

[narrow bay]

what do you think it should be

let me check

[main idol]

Show the original question

[stoic slate]

It should be squared

[narrow bay]

yes you're right

yes

my bad

[buoyant pond]

vsr

[glossy ibex]

vsr

[brave spire]

(vsr = volume of solid revolution)

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[void pebble]

there is a function in x,y say $f(x,y)$ we would like to decrease or minimize so I will take the partial derivative of $f$ wrt $x$ and $y$
$$
\begin{align}
\text{let } x' & =\dv{f(x,y)}{x} \
y' & =\dv{f(x,y)}{y} \
\end{align}
$$
here $x'$ and $y$' are in the direction in which the $f(x,y)$ increases so we will do $f(x-\Delta x',y-\Delta y')$ then is it guaranteed that $f(x-\Delta x',y-\Delta y')<f(x,y)$ for $\Delta$ small enough is not can you give me an example

[elfin berry]

greenflame41
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

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[quick sandal]

Asta, Bronya, and Clara all have different numbers of cats. They had the tollowing conversation.

i. Bronya to Clara: You have the most cats.

ii. Asta to Bronya: I have exactly 30% more cats than you.

iii. Asta to Clara: The number of cats you have is the average between the number of cats I have and the number of cats Bronya has.

iv. Clara to Asta: You have at least 4 more cats than me.

However, not all of these statements are true. When speaking to someone with fewer cats, the speaker always tells the truth, and when speaking to someone with more cats they always lie.

[quick sandal]

Wanted to check that I did this right

[silk cave]

looks fun

[quick sandal]

Statements 1 and 4 are contradictory

So they’re false

If clara had more cats than bronya then she would be lying meaning it’s false

So bronya has less cats than clara

[silk cave]

what if the two has the same amount of cats

[quick sandal]

It says they have different numbers of cats

[silk cave]

the negation of > is < ^ =

[still temple]

That would be a catastrophe wouldn't it?

[silk cave]

ah

[still temple]

ba dum tiss 🥁

[quick sandal]

So we can use same logic for statement 4 right

[silk cave]

maybe I'm stupid but I think you can only conclude that they can't be both true

or equivalently at least one is false

due to contradiction

why are they both false?

[quick sandal]

For statement 4

[silk cave]

oh wait the contradiction is from the rules

not between the statements

[quick sandal]

If clara was telling the truth, then asta has more cats than clara which would mean she is lying.

[silk cave]

okay mb

yep 1 and 4 is false

[quick sandal]

So the order is BCA

And then we just use equations

A = B * 1.3

And C = (A+B)/2

This is where I got a bit confused

[silk cave]

yea?

[quick sandal]

I think Bronya would have to be multiples of 10

B = 10 doesn’t work

[silk cave]

just to make A an integer?

[quick sandal]

B = 20 ends up working

[silk cave]

sure

[quick sandal]

Yeah

So idk if that’s the way to do it

Seems a bit weird wondering if there’s a nicer way

It was asked in a quant trading interview

So I’d expect it’d have a nice solution

[silk cave]

c = 1.15B anyway this requirws B to me a multiple of 20

wait rhis means c>b a conteadiction

[quick sandal]

No c>b is correct if 1 is false then then clara has more then bronya has less than clara

[quick sandal]

Yh true ig

[silk cave]

c > b implies the first is true

but you concluded its false

anyway I might gtg sry feeling sleepy now

[quick sandal]

If Clara has less than bronya then then bronya would be telling the truth to Clara

In statement one

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[prisma dock]

So I have the following problem. I have troubles writing a formal, rigorous proof that what I've came up with so far is indeed correct - I don't really know how to start going about it.

[prisma dock]

I'm sorry, there should be a n^2th root, not just square

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[proud basin]

i got the idea to change it to 16((cos50-cos10)^2+cos10cos50) but im not sure how to continue

[main idol]

Is this the original expression

[proud basin]

yes

[pliant siren]

that’s in degrees right?

so cos^2(10deg) etc.

[pliant siren]

you could expand the square

[remote barn]

I think you can do cos10^2+cos(60-10)^2-sin(60-20)sin(60+20) and then like relate the 20 with the cos10 thing maybe?

cos10^2+(cos60cos10+sin60sin10)^2-(sin60cos20) ^2-(cos60sin20)^2 could lead to smth ig

[marsh citrus]

@proud basin Has your question been resolved?

[proud basin]

hm ok

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[proud basin]

.close oh i got it tyy

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[nova dagger]

I need help with these two questions:

[proud ice]

You can depress a general cubic $Ax^2+Bx^2+Cx+D$ with the substitution $x=y-\frac{B}{3A}$

[elfin berry]

SWR

[nova dagger]

interesting... what does this formula mean

[proud ice]

If you replace x with y-B/3A, the x² term will disappear

[nova dagger]

alr

i managed to get y^3 - 18y +12 = 0

help

[proud ice]

That solves b for you, does it not?

[marsh citrus]

@nova dagger Has your question been resolved?

[nova dagger]

oh alr

[nova dagger]

ok but for 8, idk how ur meant to prove the following equation by completing the square. the only way i could prove it is by substituing those values and eventually making both sides equal

[proud ice]

you have an $a^2$ term, a $b^2$, and an $ab$ term. By completing the square correctly, you can get an $(a+b)^2$ term. You also have an $c^2$ term and an $ac$ term. So you turn that into an $(a+c)^2$ term. How you do that exactly is where you need to use the completing-the-square technique.

[elfin berry]

SWR

[nova dagger]

hmm

[proud ice]

where did I lose you?

[nova dagger]

wait hold up

ok so i got:

[solemn island]

is this some kind of formula?

or did you come up w it

[nova dagger]

(sqrt3a+2b)^2 + (a+3sqrt2c)^2 = 4ab+12ac+a^2+4sqrt3ab+6sqrt3ac

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[tropic bolt]

implicit differentiation idk where to start

[novel juniper]

what's the differential of $x^2$ wrt x

[elfin berry]

ƒ( wai ina teacup)= I don't know

[tropic bolt]

2x

[novel juniper]

now differentiate $-5xy$ wrt x

[elfin berry]

ƒ( wai ina teacup)= I don't know

[novel juniper]

remember to use the product rule

[tropic bolt]

5xy'+5y

[novel juniper]

cool

now 3y^2 wrt x

[tropic bolt]

6y*y'

[novel juniper]

cool

so put them all together to get?

[tropic bolt]

2x-5xy'+5y+6y*y'=0

[novel juniper]

cool

Can you find $y'$ now

[elfin berry]

ƒ( wai ina teacup)= I don't know

[tropic bolt]

y'=(-2x-5y)/(-5x+6y)

is this correct

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[cursive flower]

can someone help me ex 109

[wraith cobalt]

For part a), I think it would suffice to write all of those squared norms as inner products and expand both sides

and part b) says to me triangle inequality but I haven't tested that theory out (as does every problem in anything adjacent to real analysis, linear algebra, or any field really)

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@cursive flower Has your question been resolved?

[cursive flower]

i cant prove

part b

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[kind iris]

Is there any formula for this?

[marsh citrus]

@kind iris Has your question been resolved?

[modern sorrel]

Just some things that come to mind that may or may not be useful

If P=S then trivially the answer is 1.

If S = P - 1 then the answer is 3.

[kind iris]

if S <= 1 then No solution, and T<1 then also no solution

[modern sorrel]

A complete graph has $\frac{n(n-1)}{2}$ edges, so the lower bound of subgraphs you need for a full cover is $\frac{P(P-1)}{S(S-1)}$

[elfin berry]

LooseEthics

[kind iris]

$\frac{P(P-1)}{T \cdot S(S-1)}$?

[elfin berry]

S1p3k

[kind iris]

well, the thing is that this is not the soltuion every time and doesn't have to satisfy the result all the time

[modern sorrel]

Not necessarily, since some subgraphs can't coexist

[kind iris]

So is there a solution?

[modern sorrel]

I mean obviously there is, worst case scenario you can just brute force it

[kind iris]

how?

[modern sorrel]

Found the name for this, it's called the clique problem. The minimum number of cliques that form a cover of the complete graph is what I posted before. Accounting for multiple tables then the solution should be $\lceil\frac{P(P-1)}{min(T, \lfloor\frac{P}{S}\rfloor)S(S-1)}\rceil$

[elfin berry]

LooseEthics

[kind iris]

if T = 1, S = 500, P = 2, then it is min(1, floor(2/500)) that is min(1, 0) that is 0 so the denominator is 0?