help-28

[lapis holly]

x³+3x-5=0

Δ = (b/2)² + (a/3)³

[queen gull]

Mabye its xy...

!xy

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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.

[thick hedge]

probably is

[lapis holly]

a is 3 and b is -5

[full forum]

@lime rover Has your question been resolved?

[hidden belfry]

Isn’t there a formula used to solved cubics without a x^2 term?

[robust slate]

[https://www.wolframalpha.com/input/?i=x^3%2B3x%3D5]

[hidden belfry]

Is that what the cardano formula is?

[hidden belfry]

;-;

[lime rover]

No way.

[lime rover]

There is an easy formula but before you use it you must make the cubic equation depressed.

[hidden belfry]

How do you make a cubic equation regret its life decisions?

[lime rover]

Well the full one looks like this:
ax³+bx²+cx+d=0

And you have to use other constants like p and q

To simplify it down to x³+xy+b=0

And thats the hard part

[hidden belfry]

(It was a joke)
😊

[lapis holly]

p = 2b³-9abc+27a²d / 27a³

q = 3ac-b² / 3a²

[full forum]

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[robust gorge]

could anyone help me on this

[robust gorge]

i forgor how to do it

[quartz flare]

Do you remember what slope of the tangent line is

[robust gorge]

yo whats up cool

[quartz flare]

Or I should say how to find it in general

[robust gorge]

slope of tangent is ?

[quartz flare]

heyy

[robust gorge]

m formula?

[quartz flare]

(the ||derivative||)

[robust gorge]

oh ok

this lesson is based off of derivative

so its still tackling basics

[quartz flare]

That's what one of the big things of calculus is remembering/learning all of the alternative definitions of the derivative

Derivative = slope at a point = slope of tangent line = a few other things probably

[quartz flare]

So you'll likely need to use the limit definition to calculate this derivative if you haven't learned the rules yet

But you have done this limit before so I'm pretty sure you can do it

[robust gorge]

like this problem?

[quartz flare]

Yes a limit like what this derivative will be

Do you remember the limit definition of derivative

[robust gorge]

i dont think we studied that yet

[quartz flare]

How are you supposed to calculate the derivative then o.o

Okay have you learned something called power rule yet

[robust gorge]

we aint calculating derivative tho

we calculating slope of tangent line

[quartz flare]

.

[robust gorge]

my bad

[quartz flare]

Unless they taught you a different way to calculate the slope of a tangent line

[robust gorge]

they didnt connect the tangent line slope = derivative thing yet in lessons

wait i see a lesson video on it

imma go watch that

[quartz flare]

Okay 👍

That'll be helpful for understanding

[robust gorge]

alright imma close the channel to not keep u waiting

[quartz flare]

And I'll also give this gif

Where is it

[robust gorge]

gif?

[quartz flare]

Yeah it's one I use for derivative/tangent relationship

[robust gorge]

lets see it

[quartz flare]

https://tenor.com/view/secant-line-converging-to-tangent-line-animation-gif-18716967

There it is

It'll make more sense once why this describes it you watch the video I'm sure

[robust gorge]

the closer u look

the straighter the line

[quartz flare]

But for a short summary slope is y2-y1/x2-x1, as we move the points P and Q closer together (a limit!) we approach a single slope, that's the derivative

[quartz flare]

Yes, essentially

[quartz flare]

Sounds good

[robust gorge]

alright thank you

ill see ya later

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[onyx glen]

!status

[full forum]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[summer creek]

1

[onyx glen]

do you need explanation for any of the financial terminology or no

[summer creek]

So half yearly would be 7.5%

No

[onyx glen]

ok so let x be the initial sum borrowed on e.g. Jan 1

[summer creek]

107.5x/100 and we will pay 5547

[onyx glen]

work out the remaining balance on:

• jun 30 (5547 INR paid, interest not applied)
• jul 1 (interest applied now)
• dec 31 (second 5547 payment)

[quartz flare]

I like how this question came up relying on the formula being asked about in another help channel at the same time

What ann said

[summer creek]

Which channel?

[quartz flare]

#help-2

The guy posted a section from their textbook (I assume? but they said they were writing it) on compound interest

Featuring, naturally, the formula for compound interest that they're trying to work on

[summer creek]

If I borrow x amount on jan1 then after 6 months i have to pay

107.5x/100-5547×107.5x/100-5547

[quartz flare]

There should be some parentheses there

But yes

[summer creek]

But ann declined above

[quartz flare]

Because it wasn't right when you did it above

[summer creek]

I see

So how will i find values?

[quartz flare]

Put in the parentheses first

Then ask how much money will be left on the loan after you pay it off completely

[quartz flare]

No

[summer creek]

Then?

[quartz flare]

Start back at the steps ann told you

I just noticed that interest is applied the day after the payment not before based on what she said

[summer creek]

I didn't get it properly actually

Could you make it easy?

How may I calculate it?

7.5x/100-5547?

[quartz flare]

Do like she said, let x be the initial amount of money, then figure out the balance on jun 30, all you've done is paid 5547

[summer creek]

On june 30

Is this correct?

Is this not 107.5x/100?

Needed more help

Ann

[quartz flare]

If I started with x, and took away 5547, how much do I have

[onyx glen]

i literally said DON'T apply interest on jun 30

[onyx glen]

.

jun 30 balance is just (x-5547) only

[summer creek]

x-5547

[onyx glen]

🥀

[summer creek]

Hmm

[onyx glen]

THEN you multiply all that by 1.075

and get the jul 1 balance

[summer creek]

(x-5547)1.075 july 1

[onyx glen]

hm hold on.

i might be speaking out of my ass.

[summer creek]

(x-5547)1.075-5547 again?

[onyx glen]

yes, then that should be 0 bc you pay off by then

[summer creek]

,w (x-5547)1.075-5547=0 again?

[glossy valve]

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[summer creek]

But question is asking something different?

total interest paid?

[onyx glen]

x is the amount borrowed

amount paid is obviously 5547×2

so interest paid is?

[summer creek]

,w 5547×2-10707

[glossy valve]

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[summer creek]

What if we do the question with formula?

[onyx glen]

then you shall have to supply the formula

[onyx glen]

!noai

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[summer creek]

https://testbook.com/question-answer/surekha-borrowed-a-sum-of-money-and-returned-it-in--5fc656b6bf4f99d1a18fd7c6

We can trust this site

[onyx glen]

however it seems that this presupposes that the interest kicks in before you pay off the first installment

[summer creek]

Hmm

[onyx glen]

which is a very frustrating and notorious place to get fucked up

[summer creek]

Compound interest

I meant it will grow each day no?

[onyx glen]

because nobody fucking knows how the fuck the timing of payments vs interest applications works

[onyx glen]

not each day unless it says DAILY COMPOUNDING in the question

[summer creek]

Compounding after 6 months mean...they will charge over interest too

6 months no charge

Okay so I asked someone and told me the step where we clash were correct in terms of Ann's place... but here it is wrong

The initial amount is p

(1.075p-5547)×1.075-5547=0

,w (1.075p-5547)×1.075-5547=0

[glossy valve]

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[summer creek]

Yes that's correct answer Ann

11094-9960=

[onyx glen]

ok so in fact interest is applied first

in other words the MOMENT you walk out of the bank with the money then you already owe them 7.5% more

QUADRILLIONS IN DEBT FOR EVERY HUMAN BEING ON PLANET EARTH SINCE BIRTH

[full forum]

@summer creek Has your question been resolved?

[summer creek]

Suppose ann if I borrow x money on Jan 1 with half yearly interest rate 7.5%

so if I pay you 5547 on 30 june

matter is closed?

I don't think this is a good way to lend money because anyone on the planet can borrow millions of money and pay it before one day and they will have to pay nothing

In indian terms

Simple interest term mean they will not charge any extra interest on the internet earned annually

For compound interest they will charge interest over interest after given time period what interest how applied

So if they borrow x amount then after 6 months complete i have to pay 1.075x amount to you, i cannot pay before one day,two day....

Happy to discussion

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[eager sphinx]

I want to simplify
$\frac{2}{(5-\sqrt{3})(2+\sqrt{3})}$

[glossy valve]

botou

[hoary fable]

What have you tried?

[eager sphinx]

I’ve expanded the bottom to get 7+3 root 3

$7 + 3 \sqrt{3}$

[glossy valve]

botou

[eager sphinx]

am I correct so far?

[sudden chasm]

thats good. now you can either end here or rationalize the expression

[eager sphinx]

no I need to rationalise and hence

[hoary fable]

Yes, you are

[eager sphinx]

I multiplied the top and bottom by 7 - root 3

for the numerator I got 14 -6 root 3

for the denominator I got 7 + 3 root 3

[hoary fable]

(7+3sqrt(3))(7-3sqrt(3))

Think about that

What does it look like

[sudden chasm]

it will be a difference of squares, a^2-b^2

[eager sphinx]

49 - 9(3)?

so it becomes /22

14-6root3 /22?

somethings wrong as the

[hoary fable]

Is there a common factor between the numerator and denominator?

[eager sphinx]

+++++

2?

[sudden chasm]

or, (7-3root3)/11

[hoary fable]

Yes

[hoary fable]

Factor that out of the numerator and divide, you get this

[eager sphinx]

so tje final answer is this?

[hoary fable]

Yep, you got it

[eager sphinx]

Thank you

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[eager sphinx]

.reopen

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✅ Original question: #help-28 message

[eager sphinx]

Simplify $\frac{3-2\sqrt{5}}{\sqrt{5}-1}$ in the form $p + q \sqrt{5}$

I don’t understand thenform part

[glossy valve]

botou

[sudden chasm]

again, here we have to multiply the numerator and the denominator with the conjugate of the denominator

[brittle steeple]

they want you to write it like that, but where p and q are specific numbers

[onyx glen]

your answer needs to be a fraction times sqrt(5) plus another fraction

[eager sphinx]

How can I do
That wouldn’t it be p/q

[sudden chasm]

it will be if you try this

[brittle steeple]

p and q are fractions

already

Or maybe they're integers, idk

[eager sphinx]

ikay let me try solve it as I would normally

[brittle steeple]

That should work

[eager sphinx]

I got root 5 -7 / 4

[brittle steeple]

,w (sqrt(5)-1)(sqrt(5)-7)/4 = 3-2sqrt(5)

[glossy valve]

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[brittle steeple]

Good job

Do you know what p and q would be here

[eager sphinx]

hmmm

yes

p would be 1/4 q would be -7

?

[brittle steeple]

-7/4 = q

[brittle steeple]

Also you should have used parentheses here

[eager sphinx]

I’m confused

ohhh it’s
Plus

yeah so it’s 1/4 - 7/4

root 5?

[brittle steeple]

the root 5 is on the 1/4

[eager sphinx]

huhhhh

[brittle steeple]

you got $\f{\sqrt{5}-7}4$, right

[glossy valve]

Dreyuk

[eager sphinx]

yes

[brittle steeple]

Do you know how to split this into two different fractions?

Why is texit a she

[sudden chasm]

$\frac{a \pm b\sqrt{c}}{d}=\frac{a}{d} \pm \frac{b\sqrt{c}}{d}$

[glossy valve]

T&C

[eager sphinx]

Root 5/4 and -7 / 4

[brittle steeple]

right

[sudden chasm]

now let a/d = p and b/d = q

[eager sphinx]

no let’s not do that

too many letters

can we keep it simple guys

[sudden chasm]

we're just distributing the denominator to each term of the numerator

[brittle steeple]

The form it wants is some rational + another rational times sqrt(5)

[sudden chasm]

to get a p+q root form

[brittle steeple]

Thinking of the rationals as your coefficients

[eager sphinx]

oh yes

yeah so

1/4 and -7/4

x root 5

[brittle steeple]

Right

[eager sphinx]

is that not my final answer

[brittle steeple]

uh

is that x a times

and you wrote the equation over two messages?

then no

[eager sphinx]

huh hahaha what

[brittle steeple]

the sqrt(5) was divided by 4, how did it end up being multiplied by -7/4

[eager sphinx]

can’t we get the reciprocal

or something

[sudden chasm]

what?

[brittle steeple]

I have no idea what that means

[eager sphinx]

nevermind

okay so how do we do this

[brittle steeple]

ok where are we now

[brittle steeple]

Here still?

[eager sphinx]

I’m not sure

[eager sphinx]

No we split it into two fractions

Root 5/4 and -7 / 4

[brittle steeple]

Right

[sudden chasm]

u need to find the values of p and q right?

[eager sphinx]

yeah

[brittle steeple]

ok so one of these things is our constant and the other is something times sqrt(5)

[sudden chasm]

then compare this expression with p+q root5

[eager sphinx]

yeah so we need to somehow get rid of that sqrt 5?

but why isn’t q just 1/4 then

as it’s multplilyimg with root 5

[brittle steeple]

It is

[sudden chasm]

its correct

[eager sphinx]

and p -7/4

have I not said this before???

[sudden chasm]

it wasn't making sense b4, in the way u said it unfortunately

[brittle steeple]

Where?

[eager sphinx]

okayokayoakay

thank you guys

!!

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[brittle steeple]

no problem

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[dusky hedge]

Why the order of the cycle is the length of it

[muted flare]

Context?

You mean a cycle in a graph?

[grave elm]

by definition of order

we define order to be that

[dusky hedge]

Bruh

[foggy vapor]

​If you shift elements in a circle of size m, you must shift them m times to get them back to the start

[dusky hedge]

I ask why bro

[dusky hedge]

Ik

[grave elm]

Why do we call trees "tree"? Because we gave them that name

[dusky hedge]

But is there a proof to that

[grave elm]

there is really no deep reason, you dont need a proof to define stuff

[dusky hedge]

Having cycle like that

(a1 a2 … ak)

Why the order it should be k

[maiden vapor]

Follows directly from definition of order

[grave elm]

How do you define order (of a cycle)?

[dusky hedge]

[civic bay]

https://math.stackexchange.com/questions/762322/how-do-i-prove-that-an-order-of-a-cycle-is-its-length

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[sour shore]

Let two lines be L1 and L2 . Number of lines in a family of lines (all lines passing through the intersection of L1 and L2) that are at a fixed distance from given point can be at max ?

[sour shore]

2?

[brittle sun]

What does it even mean for a line to be at a fixed distance from a point

[onyx glen]

!xy

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[brittle sun]

Beat me to it lol

[sour shore]

Perpendicular distance between a point and a line?

[brittle sun]

There's only one of that

Show the original question please

[sour shore]

There's no ques. This is js a conceptual doubt.

[twilit leaf]

well the intersection of L1 and L2 is a point

[sour shore]

Let two lines $L_1$ and $L_2$ intersect at a point $O$. Consider the family of all lines passing through $O$. What is the maximum number of lines in this family that are at a fixed perpendicular distance $D$ from a given point $P$?

I suppose this is better worded

[twilit leaf]

so what youre really asking is if you have a point P1 and a second point P2 which is the center of a circle of radius r, how many tangents to that circle go through P1

[brittle sun]

Oh you're fixing the distance too?

In that case the first part doesn't matter

It's just asking, we have a point. How many lines exist such that the perpendicular distance from the point is x

[sour shore]

I get it....a lil bit

[twilit leaf]

tangents are perpendicular to the radius at that point

so perpendicular distance being the same is like have a tangent to a circle

[sour shore]

Thtd be infinite...

[brittle sun]

And that's correct

[sour shore]

@twilit leaf can u maybe explain with these variables?

[sour shore]

Idts

[twilit leaf]

what do you want to call the fixed perpendicular distance

[sour shore]

Let's say D

[twilit leaf]

ok

[glossy valve]

ch3rry

[twilit leaf]

so you have a line going through O that needs to have a perpendicular distance D away from P

[onyx glen]

don't these lines form a cone

[brittle sun]

Oh wait I'm stupid sorry

[sour shore]

Yep

[onyx glen]

what are the roles of L1 and L2 here

[brittle sun]

All yours garlic

[twilit leaf]

are we in 3d or 2d

[sour shore]

2D

Wouldn't 3D js be infinite?

[twilit leaf]

yes, and that would form a cone

but anyway

what we can do is draw a circle with radius D at P

[sour shore]

Okok i got it i think

Only 0,1,2 no. Of lines r possible ryt?

2 for external point, 1 for point on circle, and 0 for internal point?

[twilit leaf]

nice

exactly that

the point referring to O because P is always the center

[sour shore]

Yes yes

Thanksies

.solved

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[indigo frost]

I don't know how to find the expressions for p and q

[zenith kernel]

For example

(2p+3q)^e_1

Other than (2p)^e_1 and (3q)^e_1, all other terms are divided by pq

Two terms left, same for the second equation

[indigo frost]

I don't quite get what ur saying

[zenith kernel]

Sorry made a mistake

I meant

[indigo frost]

What , how

I mean it should be newton binomial

[zenith kernel]

(2p+3q)^e binomial expansion, any term other than (2p)^e, (3q)^e has a pq in it

[indigo frost]

I see

[indigo frost]

So this should be the end result ?

I.e there's no way to improve it ig

[zenith kernel]

I guess

Further step might be this

But it might not further improve the calculation

[indigo frost]

I see thx

[indigo frost]

Btw this have nothing to do with p,q being primes

[zenith kernel]

Yeah

[indigo frost]

I mean it works for any integers

[zenith kernel]

Yeah

[indigo frost]

Ok

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[feral forum]

What’s wrong in my approach for part c

[feral forum]

…

[lone berry]

I believe your mistake is at the end, you multiply a 3x1 matrix with another 3x1. If my knowledge of matrices is correct, that's not allowed

[onyx glen]

that is intended as a dot product not a matrix product

[feral forum]

What Ann said

[lone berry]

yus, my matrices knowledge has been severly depleted, my bad haha

[bitter star]

looks conceptually correct and i dont really see any mistakes in your calculations, 14/9 seems to be the right answer

[onyx glen]

i think i see it

sign error?

BA = [-2, -1, **+**2] by the looks of it (from part a)

[bitter star]

but arent we using BC in part c?

[onyx glen]

oh

wait nevermind

i was looking in the wrong place sorry

[bitter star]

the only thing i can think of is you didnt calculate the length but ig you just didnt include that part in the picture cause this was your mistake

[lone berry]

I mean apart from using C as their point in the L_BC, that gets you the 5. I don't see anything wrong.
Also, is the red notation the correction?

[bitter star]

i am guessing your teacher or whoever checked it might have used line CB and point C when finding OX which would lead them to get -5/9 not realising that the point X would remain the same

[full forum]

@feral forum Has your question been resolved?

[feral forum]

The value of lambda is wrong

-5/9 is the correct value

[feral forum]

But I believe the value of lambda should turn out to be the same no matter if i do that?

[feral forum]

Idts, if you flip the signs from part a

[bitter star]

no lambda will be different but the point x will remain same

try solving the question with line CB or point C just a different configuration from the one you did

you will get different lambda

but the point X would remain the same

try with line CB and point C and find OX, you will see that it will remain same

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@feral forum Has your question been resolved?

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[night musk]

how could we ever choose an N_2(P) that would work then

[night musk]

also for the geometric aspect why is he not including that red line I added for p since that is where the functions y value is at the bounds of A

[uneven birch]

There may never be a N2 that works for all N1

as the sentence just before suggests, N2 depends on N1 (usually)

[lime ether]

not sure i understand your question? you first choose N_1 then are able to find an N_2 based on that N_1, of course if N_1 was chosen differently the N_2 might not still satisfy it is what they are saying

[lime ether]

because it’s outside of the neighborhood of p so we don’t care about it

we care about the images of points within neighborhoods of p

[slender onyx]

as a concrete example, if N2 works for N1 then N2 also works for 2N1 but might not work for 0.5N1

[night musk]

so he says that if N_2(P) works for an N_1(A) then it will work for a greater N_1(A) which makes sense since our original N_2(P) will be part of the new N_2(P). But then he says if we choose a smaller N_1(A) it may not work. Shouldn't the answer be that it never works. Because by saying it may not work that implies that there could be an example where it does but I can't think of any

[night musk]

why might not shouldnl't always be not

[lime ether]

yes because N_2 might not be contained in a smaller choice of N_1

[slender onyx]

take f to be a constant function

[lime ether]

that doesn’t mean that if we chose N_1 smaller we still couldn’t find a corresponding N_2

[slender onyx]

then for all N1 any N2 works

[night musk]

ok I see because any neighberhood of x values will always have the same output

[lime ether]

for the constant function yea

[night musk]

but isn't the P part of the neighberhood determined by all the y values in N_1(A)?

shouldn't this y value be part of our neighborhood?

[slender onyx]

the definition of limit only cares about points inside N1

[night musk]

shoudn't it be similar to this?

[lime ether]

you don’t care about finding the largest such neighborhood around p, just find a neighborhood around p

[slender onyx]

not all functions look like sqrt

[night musk]

I understand that the p neigberhood here is decided by all x that give the y values in the a neighberhood

[lime ether]

i wouldn’t say the p neighborhood

[night musk]

isn't any point why the corresponding y value in that neigberhood?

[slender onyx]

the def of limit only requires ‘if x in N2 (x!=p) then f(x) in N1’

[lime ether]

sure you can find larger open set of points around p that map into N_1 but you don’t care about doing that

[slender onyx]

there is nothing required to be satisfied if x is not in N2

[lime ether]

the point of the limit is that you can always find some neighborhood around p that maps into it

[night musk]

ok so any of these neigberhoods are good if they provided y values in N_1(A)

[night musk]

why not?

[lime ether]

because there isn’t just one neighborhood

[night musk]

but the x for N_2 are determied by what we choose N1 to be

[lime ether]

if you take any open interval with radius delta > 0 around p that works then all positive numbers less than delta also work and provide neighborhoods of p

[slender onyx]

this doesnt contradict anything in the definition

[night musk]

ok

[slender onyx]

again for a concrete example take f to be constant

for all N1 pick N2 to be radius 1

then f(x) is in N1 for all x in N2

sure it holds for all x in R but we only care about N2 to satisfy the def of limit

[night musk]

ok I see

thx

[slender onyx]

np

[night musk]

.solved

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[torn jolt]

maybe im too tired to thik properley

[torn jolt]

but can anyone tell me what is 10c) actually saying

[rapid rain]

if m = 4n-1 is composite, then there exists p = 4k-1 prime that divides m

[somber violet]

with p prime, yes

[torn jolt]

ah ok

[slate violet]

You can deduce c) entirely from a) and b)

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@torn jolt Has your question been resolved?

[torn jolt]

ah ok thanks i was to lazy to do b) that's uh

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[torn jolt]

my faulty

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[plucky pagoda]

guys idk if this is more of a math question or coding question but im just wondering if this python syntax is correct? its suppost to be a calculus AI

[plucky pagoda]

oh, the other file i sent was swedish. heres the english version

is it correct?

[quartz flare]

From a syntax standpoint just looking it looks like it should run

No guarantees about if it'll actually do what you want and without running it I can't actually say that it'll run
But at first glance it looks fine

[uneven birch]

You can always try to make it run and see if it works

[plucky pagoda]

ok, because idk if the AI works and if it can do all types or just answare special prompts

[uneven birch]

Ai can be useful in computer science, but always double check what it does and see if it works/does what you want it to

[plucky pagoda]

ok

[uneven birch]

By looking at it there are no syntax errors i can spot, however whether it gives a coherent answer or not, you can try it and see if it works

Or I can look at it closer but it'll take more time

[plucky pagoda]

im kinda new to python, but i have experience in other coding languages like C++, html, assembly etc

[plucky pagoda]

sure

[uneven birch]

what im saying is, can you at least try to see if it works first 😅 and if it doesnt i can help you

[uneven birch]

well as i said, the syntax looks good so there shouldnt be any problems with syntax per se

[plucky pagoda]

i runs but idk if it can do the math correctly

thats kinda the main problem

aight

.close

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[waxen raven]

why can you multiply by dx in u sub?

[modest obsidian]

what do you mean by why

[warped frost]

you mean why we can do stuff like
u = 2x
=> du/dx = 2
=> du = 2 dx
?

[modest obsidian]

its because that's how we get the differential

[waxen raven]

say int (3x+1)² dx

with u = 3x+1, you get du/dx = 3

so du = 3 dx

why can I multiply by dx like that?

[waxen raven]

yes but du/dx isn't a fraction

[modest obsidian]

df/dx=f' so $\Delta f = f' \mathrm{d}x+\xi$

[waxen raven]

also what is the mathematical definition of dx as a differential?

[warped frost]

so if you have never heard of it being stated this way

[glossy valve]

parabolicinsanity

[warped frost]

d{whatever variable} refers to an infinitestimally small change in that variable

[modest obsidian]

from this you get the expression df=f' dx

which is called its differential

[warped frost]

so when you do things like dy/dx, what you're actually saying is that it is the ratio of an infinitesimally small change in y to an infinitestimally small change in x

[modest obsidian]

i hate epsilon so i used xi but this is the real reason for "multiplying by dx"

[waxen raven]

hang on sorry can we start with $\lim_{\delta x \rightarrow 0}\frac{\delta y}{\delta x} = f'$

[glossy valve]

Moonful

[waxen raven]

I know how that works but I'm looking for uhh

[modest obsidian]

uhh okay, this means that in a vaccum, you can say what i wrote $\delta y = f' \delta x +\xi$

[glossy valve]

parabolicinsanity

[waxen raven]

I suppose I'm looking for a rigorous proof

[modest obsidian]

from the definition

[waxen raven]

but where did the limit disappear?

[modest obsidian]

because its not the limit perse

the epsilon term accounts for that

[waxen raven]

ah

right

[modest obsidian]

or ξ in my case

[waxen raven]

so then how do you go from delta y to dy and delta x to dx?

[modest obsidian]

then the ξ will disappear once you approach the point

[modest obsidian]

its whats called "a linear approximation"

thats how they formalise df=f'dx

as coming from the linear approximation or first order taylor expansion

[waxen raven]

wait sorry

is a linear approximation not just using the increments formula to approximate the height of f?

[modest obsidian]

just know that $f'(x) \mathrm{d}x + \xi = f'(x)\Delta x$

[glossy valve]

parabolicinsanity

[modest obsidian]

it will make sense from that onwards

i think

iirc

[modest obsidian]

i think you're referring to something else

[waxen raven]

the increments formula is $\delta y \approx \frac{dy}{dx} \delta x$ when x is small

say you have y = f(x)
then $\delta y \approx f'(x) \delta x$

now if you want to approximate the value of f at x=a+k given you have f(a), then $f(a+k) \approx f(a) + \delta y \approx f(a) + f'(a)*e$

say you want it at some x given f(a)

then $f(x) \approx f(a) + \delta y = f(a) + f'(a)(x-a)$

[glossy valve]

Moonful

[waxen raven]

okay so this makes sense

[modest obsidian]

i'd rather use $\xi$ than $\approx$, its more formal

[glossy valve]

parabolicinsanity

[waxen raven]

I see

ok so then

[waxen raven]

from here how do you use the linear approximation to go from delta y to dy and delta x to dx

[waxen raven]

sorry idk why I wrote *e there i meant to write *k

[modest obsidian]

yes

then the error term disappears

and you get your differential

[waxen raven]

sorry how do you do that

[modest obsidian]

??? just say $\lim_{x \to x_0}\xi =0$

[waxen raven]

as x goes to 0?

[modest obsidian]

no, as x approaches a point say x=x_0

[glossy valve]

parabolicinsanity

[modest obsidian]

hope that made sense why we multiply by dx

[waxen raven]

so you're also saying that $\lim_{x \to x_0}\delta y =dy$?

[glossy valve]

Moonful

[modest obsidian]

erm in a vaccum yes but that in particular isn't rigourous

[waxen raven]

what is the rigour behind it?

[modest obsidian]

dy in a vaccum is not rigourously defined

because like without the quotient its meaningless

that is why we habe delta y = f' delta x + epsilon first

[waxen raven]

this is the problem im having

how can we use it if its not rigorously defined

[onyx glen]

it's a symbolic device to keep things straight when doing the integration equivalent of the chain rule

have you ever had it explained to you in the general case

[waxen raven]

the int f'(g(x)) g'(x) dx = int f'(u) du?

[onyx glen]

yes

thats kinda just about it

du = g'(x) dx is bookkeeping, and little more

[waxen raven]

that makes sense but now suddenly what happens when you get int x(x+1)⁹ dx

with u=x+1 it becomes trivial but it's not in the form f'(g) g'

[onyx glen]

but it is

just not obviously so

u-sub lets you not worry about form any more than is needed

you get $\int (u-1)u^9 \dd{u}$

[glossy valve]

Ann

[waxen raven]

so its straight up a substitution and nothing else

[onyx glen]

substitution with "extra steps" vis-à-vis a purely algebraic one, but yeah

[waxen raven]

then why does du = g'(x) dx even work if it means nothing

[onyx glen]

int f'(g(x)) g'(x) dx = int f'(u) du

[waxen raven]

oh wait I see

hang on im still a little bit confused tho how do I answer someone when they ask "why can you go from du/dx = 3x to du = 3 dx"?

[onyx glen]

3x dx, not 3 dx

[waxen raven]

sorry yes

[onyx glen]

anyway like... du = (du/dx) dx

[tired kite]

You're just solving for du; so you can multiply dx onto the other side to get 3x dx.

[onyx glen]

where du/dx is to be understood as the derivative in the usual sense, written leibniz style

[waxen raven]

I understand that but du/dx isn't a fraction so why can you treat it like one

[onyx glen]

the problem is "multiplying by dx" is not rigorous

[tired kite]

True.

[waxen raven]

what does this mean? like I get it's the chain rule but what does du and dx mean by themselves (besides small change in)

[onyx glen]

nothing

du = g'(x) dx
and
du = (du/dx) dx
are the same shit

[waxen raven]

ah

[onyx glen]

like really there is nothing to explain here rigor-wise\

thats just how the notation works

[waxen raven]

it feels so circular reasoned though idk why

[tired kite]

It makes more sense with practice in my opinion.

I feel like most teachers don't go super technical with it either; hence the confusion can exist.

[waxen raven]

of course I have lots of practice

im way past integration by substitution, and i understood easily why you solve for dx, but I never understood why that's even correct

[timber veldt]

do you know implicit differentiation?

[waxen raven]

yes

use of chain rule

[onyx glen]

its a historical holdover of notation thats what it is

[waxen raven]

I don't like it

😭

[onyx glen]

¯_(ツ)_/¯

[timber veldt]

i misread the question my bad

every answer is right funnily enough. but it's true du is simply defined as (du/dx) dx where dx is a one form

[waxen raven]

so its just a definition

what is a one form sorry😭

[timber veldt]

a linear functional, or something that maps a vector (not simply the geometric vectors you might be used to, but any element of a vector space) to a scalar

[waxen raven]

hmmm I see

thank you very much

oh also I had another question

1 sec

[timber veldt]

basically, for $u(x)$ being a differentiable function: fix a point x. take a small change $h$ at the input and consider the resulting change in output. as you should be aware, the derivative can create a line that linearly approximates that function around the neighborhood in x. that infinitesimal change $h$ can be mapped to the actual change in output, aka $h \mapsto u'(x) h$. this mapping is the differential du. $du(h) = u'(x) h$

[glossy valve]

Misanthrope

[waxen raven]

we define $\int_a^b f(x) dx = \sum_{n=1}^{\infty} f(x_i) \delta x_i$ right?

[glossy valve]

Moonful

[timber veldt]

dx is simply h in this case.

[waxen raven]

I seee

[waxen raven]

yes I'm also aware of the riemann integral

its definition with the partitions and the darboux sums

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@waxen raven Has your question been resolved?

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[hoary ember]

im stuck on b

so i have 3 ideas:
a. have a list of 1225 steps of like either its the +1, or /2, and have the end equal tk 1
b. for a number, write in base 2, then find something that gets you the jolliness of the number
c. from 1, try to reverse the process of +1 and /2, so that you do -1 and ×2 instead, and repeat 1225 times and find how many diffrent numbers it goes to

for idea a, you cant have two of +1 back to back, so you can think of selecting objects (which of the steps are +1) from the 1225 objects (steps), and that the objects arent adjacent. if im not mistaken the formula is the nth fibbonanci number. theres a few tricky things though, like i think (?) the last two must be a /2 and /2, but that just becomes the 1223th fibbonanci number and i doubt thats the answer

for idea b, ive been trying things, and im guessing it has something to do with adding the length in base 2 to some variable? idk i havent been able to find something that can find the jolliness from it

for c, honestly im abit iffy on what to do here, like from 1, it has to be a ×2 and ×2 im p sure, ok actually while writing this, its just idea a but flipped so nvm

any pointers?

[zenith kernel]

I think you directly calculating {(f^n)(1225): n} till it reaches 1 isn’t hard
And comments on your ideas:
a) didn’t get it
b) might work
c) doesn’t work I think since f is not injective, f(2k-1)=2k=f(4k)

[hoary ember]

wdym {f^n(1225): n}

you need to do this by hand btw

[zenith kernel]

Yeah and it won’t be long

[hoary ember]

whats {f^n(1225): n}

[zenith kernel]

1225, f(1225), f(f(1225)),…

Till it reaches 1

[hoary ember]

whats f?

[civic bay]

why would you apply the iteration to the number 1225? you want to count the number of integers that reach 1 after 1225 applications of the iteration, not what happens to 1225 itself

[zenith kernel]

f: N->N
f(2k-1)=2k
f(2k)=k

Notice that (f^2)(m)<=max {(m+1)/2, (m/2)+1}=(m+2)/2. thus you twice operation roughly make the value half

So it really isn’t a long list

[civic bay]

how would that help solve the problem? the goal isn't to trace the number of iterations it takes for 1225 to reach 1

[zenith kernel]

1225 is close to 1024 , so you do roughly 20 times you can obtain 1

[zenith kernel]

The number of steps I know

And it won’t be long.

[bright bronze]

no its not

[civic bay]

that's not what we're trying to find

[hoary ember]

for clarificarion, like let a town that has jolliness 1225 be represented as ADDADADADDDADADDDD... (theres 1225), with A standing for a +1 operation, and D standing for a /2 operation. for example 5 is represented as ADADD. i think (?) the last two of the sequence must be a DD, and there cant be adjacent A's, so i think its the 1223 fibbonanci number (tho im kinda doubting this works)

[bright bronze]

the question is how many numbers x exist such that f^1225(x) = 1

[zenith kernel]

Shit I am dumb

Misunderstood it

[blissful sail]

Happens

The problem is really interesting tho

[civic bay]

if you let A_n be the set of numbers that first reach 1 after n steps, can you find a way to write this in terms of A_{n-1}?

[zenith kernel]

Yeah A_n={2a: a in A_n-1} union {b-1: even b in A_n-1}

[civic bay]

right, based on the way the function is applied, undoing it is a matter of either doubling or subtracting 1

this is kind of the approach you were suggesting in c, but instead of doing it 1225 times which isn't feasible, you can find a recurrence

[hoary ember]

uhhh

[civic bay]

does this statement make sense?

[hoary ember]

gimme a sec

oo yea

[civic bay]

in words, the numbers that reach 1 in n steps are either twice the numbers that reach 1 in n-1 steps (in which case the n/2 rule was applied) or one less than the even numbers that reach 1 in n-1 steps (in which case the n+1 rule was applied)

[civic bay]

so consider cardinalities, we ultimately want to find |A_1225|

we know that {2a: a in A_n-1} and {b-1: even b in A_n-1} are disjoint, do you see why?

[zenith kernel]

Oh , find |A_1225|. I thought find A_1225. This becomes very clear then

[hoary ember]

yeah, for an odd element in A_(n-1), it must come from a /2 in A_n (the first bit), and for an even element in A_(n-1), it can either come from a /2, or a +1 (repreaented by both the first and second)

[zenith kernel]

You will have a 2 by 2 matrix, you can let s(n)=number of odd elements of A_n, t(n)=number of even elements of A_n. x(n)=(s(n), t(n))^t, you will end up with x(n+1)=Mx(n) for some matrix M. Power of M use canonical form of M

[civic bay]

it's actually even easier than that lol, the matrix won't be necessary cause you'll recognize the recurrence

[hoary ember]

we basically need to find the number of even things in A_i right?

[civic bay]

since A_n={2a: a in A_n-1} union {b-1: even b in A_n-1}, and those two sets are disjoint (the first contains even elements and the second contains odd ones), we know that |A_n| = |{2a: a in A_n-1}| + |{b-1: even b in A_n-1}|

[zenith kernel]

Oh that matrix itself is almost a canonical form

Binomial expansion is enough

[civic bay]

what's the cardinality of that first set?

[hoary ember]

just A_(n-1)

ohh wait

[civic bay]

right, so |A_n| = |A_n-1| + |{b-1: even b in A_n-1}|

[hoary ember]

the second one is A_(n-2)?

[civic bay]

mmhm, the only even elements in A_n-1 come from some unique element from A_n-2

so |A_n| = |A_n-1| + |A_n-2|

could you find some initial values for |A_n|, like |A_1| and |A_2|?

[hoary ember]

|A_1| is 1, which is 2

|A_2| is also 1, which is 4

[civic bay]

right

[hoary ember]

wait, so A_n is just the nth fibbonanci number right?

[civic bay]

mmhm

[zenith kernel]

No

You need to find the coefficients

[civic bay]

coefficients?

[hoary ember]

of?

[zenith kernel]

|A_n|=u x^n+v y^n where x,y= (1+/-sqrt(5))/2 plug in two n to find u,v

[civic bay]

we wanted to find |A_1225|, we know |A_n| = |A_n-1| + |A_n-2|, and we know |A_1| = |A_2| = 1. so |A_n| = F_n where n is the nth fibonacci

[zenith kernel]

Oh the initial terms match the Fibonacci, sorry

[hoary ember]

ohh hold on bruh what

the number of ways to select n objects so that no two are adjacent is F_(n+2)

[zenith kernel]

If you want a specific expression you can plug in finding out u,v are 1/sqrt(5), -1/sqrt(5). but sure if the question takes F_n as answer won’t need to

[zenith kernel]

What do you mean

[civic bay]

the number of length n binary strings without consecutive 1s is F_{n+2}, yes

[civic bay]

idk about the "no two are adjacent" claim

[hoary ember]

arent they the same

but yea thats what i meant

[zenith kernel]

No two 1 are adjacent you meant?

[civic bay]

well 0s are allowed to be adjacent in the example i gave, for example 0000 is a valid string of length 4 but 0011 isn't

[zenith kernel]

Yeah 0x
1x
1x cases further divided to 10y and 11y sub cases

[hoary ember]

but fuck so theres 1223 left, so it should be the (1223+2)th fibbonanci number of 1225 fobbpnanci number

ok a shouldve worked D:

[civic bay]

yes there is a way to frame this in terms of binary representations and use the F_{n+2} thing

[hoary ember]

oh yeah

[civic bay]

i think the set approach is nicer imo so you don't have to worry about mapping back and forth between binary strings and town routes

[hoary ember]

yea i agree

okay guys, thank you!

.solved ❤️

[full forum]

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[zenith kernel]

I don’t think they are the same though, different recursion
0 x : a(n-1)
1 0 y: a(n-2)
1 1 y: 2^(n-2)
it will be of the form ux^n+vy^n+2^n
Nothing, having no adjacent 1… I thought having adjacent 1…

[hoary ember]

huh?

[full forum]

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[night musk]

Can someone explain what is this N(p) he is reffering to when he says that N_2(p) subset N(p) and how that rellates to G. Also for the inequality it says that any g(x) is between 0 and h(x) so any input is either the same as h is smallerl meaning as close as h is to 0 or less but how does relate to values of g(x) getting closer to 0 as x get closer to p?

[zenith kernel]

N(p) is given in the statement

Context of theorem 3.3

[night musk]

Why can we assume that it is a subset of N(p)?

[zenith kernel]

You mean why N2 is a subset of N?

You can replace N2 with N2 intersection N

[night musk]

i don't see it

[zenith kernel]

lim G(x)=0 is proven by definition
lim t(x)->0 means any neighborhood N1(0) of 0, there exists neighborhood N2(p) of p such that t(N2(p)) is contained in N1(0) (then obviously any open subset of N2, like N2 intersection N, image of under t is still contained in N1(0))
Now G(N2(p)) is contained in H(N2(p)) is contained in N1(0). Done

[night musk]

G?

[zenith kernel]

If you prefer G
Any N1 there exists open N2 such that G(N2) is contained in N1
is equivalent to
Any N1 there exists an open N3 being contained in N such that G(N3) is contained in N1

[night musk]

I htought it was H

[zenith kernel]

Any function, whatever letter you prefer

[zenith kernel]

Former->latter: the N2 in the former, let N3=N2 intersection N
Latter-> former: the N3 in the latter, simply let N2=N3

[night musk]

how did you get this step

[zenith kernel]

Your picture

0<=G(x)<=H(x) implies that G(N2(p)) being contained in H(N2(p))

[night musk]

ok maybe this is a bad question but how do we know that it is the N2(p) points. is it because x is just any general point

if that is the case I can see if we say that the only points we are looking at are p in N2(p) then all of g(N2(p)) < H(N2(p)) and am I right to say that since g's outputs are less then they can be just as close to 0 as h. But then where does N2(p) subset of N(p) come in?

[zenith kernel]

^

[night musk]

I still don't see how that connects to what I said above

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@night musk Has your question been resolved?

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[rotund pebble]

can someone please explain to me contour integrals and Cauchy's residue theorem

[full forum]

@rotund pebble Has your question been resolved?

[rotund pebble]

😢

[drowsy finch]

do you mean conceptually (what they are) orr how to compute with them?

[full forum]

@rotund pebble Has your question been resolved?

[rotund pebble]

i know a bit of both but i'd like a proper explanation if possible

[quaint prawn]

If you ask a more specific question about them I can help

[rotund pebble]

well maybe about the use cases of the residue theorem and how it can help making integration easier, and also if you could give me an integral to try and use it on

[drowsy finch]

contour integrals don’t measure area or distance they measure how singularities twist the function

the residue theorem says: add up those twists multiply by 2πi and you’re done

if that makes sense

they're mainly useful for hard real integrals, especially $\int_{-\infty}^{\infty} f(x),dx$
rational functions like $\frac{1}{x^2+a^2}$ or $\frac{x^2}{x^4+1}$
integrals with $e^{ix}$, $\sin x$, or $\cos x$ (rewrite using complex exponentials)

[glossy valve]

anflo

[rotund pebble]

but then we have integral of - inf to inf of 1/1+x^2 is equal to pi but multiplying the sum of singularities (which are in this case i and -i) by 2ipi gives us 0 which is not true since the integral is equal to pi

i don't undertand how

[drowsy finch]

you don’t sum all singularities you only sum the singularities inside the contour you chose

for $\int_{-\infty}^{\infty} \frac{dx}{1+x^2}$ you don’t integrate around the whole plane you close the contour with a semicircle in the upper half plane

[glossy valve]

anflo

[drowsy finch]

the poles are at $z=i$ and $z=-i$, but $z=i$ is inside the contour and$z=-i$ is outside the contour so only one residue contributes

[glossy valve]

anflo

[rotund pebble]

so multiplying 2ipi by i which is inside the contour still gives us -2pi which is different from the wanted result

why so

[drowsy finch]

you don’t multiply by the location of the pole ($i$) you multiply by the residue at the pole

[glossy valve]

anflo

[rotund pebble]

how do i get the residue at the pole

[drowsy finch]

for a simple pole at $z=z_0$, the residue is $\text{Res}(f,z_0)=\lim_{z\to z_0}(z-z_0)f(z)$

[glossy valve]

anflo

[drowsy finch]

so at $z=i$: $\text{Res}(f,i)=\lim_{z\to i}\frac{z-i}{(z-i)(z+i)}=\frac{1}{2i}$

[glossy valve]

anflo

[rotund pebble]

ohhhh so it basically goes back to evaluating a limit

oh so now i understand

thanks so much for your help @drowsy finch !

.close

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[drowsy finch]

yep exactly for simple poles computing residues is literally just evaluating a limit

[rotund pebble]

what does simple pole mean

there's different types of poles ?

.reopen

[full forum]

✅ Original question: #help-28 message

[drowsy finch]

oh no 😭

[drowsy finch]

yes

a simple pole is a singularity where the function blows up like $\frac{1}{z-z_0}$ (first order)

[glossy valve]

anflo

[drowsy finch]

more generally if $f(z)\sim \frac{1}{(z-z_0)^n}$ near $z_0$ then it’s a pole of order $n$

[glossy valve]

anflo

[rotund pebble]

but the residue theorem still applies the same way no matter the degree of the pole right ?

[drowsy finch]

yes

the only thing that changes is how you compute the residue

[steel solar]

I would suggest a problem set giving the contour to you

[full forum]

@rotund pebble Has your question been resolved?

[glass niche]

we good?

[rotund pebble]

oh i actually went to do something urgent and forgot to reply and close the channel

thanks guys for helping me especially you anflo

.close

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[neon yoke]

"The system is in multiuser state if and only if it is operating normally."$$m \leftrightarrow n$$"If the system is operating normally, the kernel is functioning."$$n \rightarrow k$$"The kernel is not functioning or the system is in interrupt mode."$$\neg k \lor i$$"If the system is not in multiuser state, then it is in interrupt mode."$$\neg m \rightarrow i$$"The system is not in interrupt mode."$$\neg i$$

[glossy valve]

Sean [Ping On Reply Please!]

[neon yoke]

Can I use the rules of inference to do this one

I got m, not k, and not n by using the last one in the middle three propositions

[upbeat knot]

yes

[upbeat knot]

you got m and not n

not n <-> not m from (1)

m and not m, thus inconsistent

[neon yoke]

alright

[neon yoke]

btw in this one, does it make sense to conclude that the first two professors do not want coffee and the third one does

is it the opposite

[upbeat knot]

if the first professor doesnt want coffee, what would he have replied instead

[onyx glen]

it is in fact the opposite

[neon yoke]

cool

alright that makes sense

what about this one

[onyx glen]

do you know the "one guard tells the truth the other only lies" story

[neon yoke]

haha no

but im thinking its gonna be like 'suppose this path leads to the jungle, and if i ask you if it is a jungle, would you say yes'

[zenith kernel]

I think yours won’t work. Either case they answer yes

[upbeat knot]

hmm ideally you want to force the same answer out of the lier and truth teller, which can be done with a double truth / double lie, i.e. the back 2/3s of this statement

but the first 1/3 of the statement needs to be tweaked to relate directly to the current fork and not a imaginary fork

[rapid rain]

and more damning, this doesn't answer whether the path you pointed indeed goes to the jungle

[zenith kernel]

Yeah

[rapid rain]

which is what you wanted to know

[zenith kernel]

Can you explain more about this double true double lie idea? Or do you have other examples

[rapid rain]

I think that in order to grasp the answer to that puzzle

you can start looking at the guards puzzle, which might somehow give you more incentive as to how to solve it

You are a prisoner in a room with 2 doors and 2 guards. One of the doors will guide you to freedom and behind the other is a hangman–you don't know which is which, but the guards do know.

One of the guards always tells the truth and the other always lies. You don't know which one is the truth-teller or the liar either. However both guards know each other.

You have to choose and open one of these doors, but you can only ask a single question to one of the guards.

What do you ask to find the door leading to freedom?

The key thing that isn't explicitely said in the original puzzle is that ||you have access to both a truth teller and a liar, even if you don't know which is which||

[upbeat knot]

wait i might have a solution:

if i ask you if i am walking towards the ruins, would you say yes?

[rapid rain]

Unless you already know where the ruins are

that doesn't help you

The truth teller will say the correct path to the ruins

and the liar will lie to you

[upbeat knot]

Case 1: walking towards ruins

Truth-teller says yes
Lier would lie to first part of question, "no", then respond "yes" to lie for second part of qn

no matter what, resonse is yes

[upbeat knot]

the trick is that the lier must lie to both parts of the qn

[rapid rain]

?

[upbeat knot]

Part 1: I ask you if I am walking towards ruins

Part 2: Would you say yes

lier would have to lie to both parts

[rapid rain]

You're asking them whether A => B

then

you don't get them to answer both A and B

[upbeat knot]

but the A itself is a question

so the lier has to lie to both A and the implication

[neon yoke]

If I ask you to tell me if this is the path to the jungle, are you gonna say yes

wat abt this one?

[upbeat knot]

thats basically the same as my qn, WLOG on jungle and ruins

[neon yoke]

whats wlog

[rapid rain]

ok yes that works

[upbeat knot]

wlog means without loss of generality

never mind, for simplicity, your qn is identical to my qn

[rapid rain]

someone saying the truth about something they said that was true is the same value as someone lying about something they said that was false

[upbeat knot]

yeah, that's a double truth vs double lie

in logic terms

A == not(not(A))

[zenith kernel]

Oh oh || I ask them “is the statement “if I ask you whether this path leads to the jungle, you would say yes” true?”||

[upbeat knot]

unfortunately adding the "true?" at the last part results in different answers for truth teller vs lier

[rapid rain]

nono that's still a hidden double truth vs double lie

[zenith kernel]

I think same answer

[rapid rain]

you would think at first hand with the way it's worded there are "3 statements" in a chain

but you're still inquiring about whether they would have said yes to "does this path lead to the jungle"

[upbeat knot]

oh right

this replaces "would you say yes?" with "you would say yes. True?"

[zenith kernel]

|| if this leads to jungle they both say “yes”. If this doesn’t lead jungle they both say “no”||

[rapid rain]

I had a different answer, more inspired from the 2 guards, but it's a similar logic approach

you used the fact that even if you don't know whether you have a "truth" or "lie" gate

applying the same gate twice gives you a "truth" gate always

I tried to use the fact that if you have both a truth and a lie gate, but you don't know which is which

then combining them give you a lie gate

[zenith kernel]

I somehow completed a always true gate

[upbeat knot]

same

[rapid rain]

So "Had I asked someone that replies the opposite truth value from you "....", would they have said" ..."?"

but it's too inspired from 2 guards

and instead if you just added a lie gate to that, you would get back to truth gate

[zenith kernel]

Is this well defined though, which step switch answer. They might never halt, keep switching answers in their heads

[neon yoke]

what's going on 💔

[zenith kernel]

We have answers we are exploring other ideas

[neon yoke]

I dont want the answer anymore, this question too hard

[upbeat knot]

ooo that is neat

by inquiring them about their opposite truth identifier

[rapid rain]

Well define it by "lies if you always tell the truth, says the truth if you always lie"

[neon yoke]

1. Break down the statements into logicLet's assign variables to each person to represent them telling the truth:$B$ = Butler$C$ = Cook$G$ = Gardener$H$ = HandymanNow, translate the clues:"If the butler is telling the truth then so is the cook"$B \rightarrow C$"The cook and the gardener cannot both be telling the truth"If $G$ is true, $C$ must be false (and vice versa).$G \rightarrow$ not $C$"The gardener and the handyman are not both lying"At least one of them is telling the truth.$G$ OR $H$"If the handyman is telling the truth then the cook is lying"$H \rightarrow$ not $C$

[glossy valve]

Sean [Ping On Reply Please!]

[neon yoke]

can we do this one

[zenith kernel]

But you need to specific layer, like switch answer of the statement starting from the third/second if kind of things

[neon yoke]

can i do . close if yall dont mind

[rapid rain]

We can talk about this don't worry

We'll banter in another help channel

[neon yoke]

"If the butler is telling the truth then so is the cook"$B \rightarrow C$"The cook and the gardener cannot both be telling the truth"If $G$ is true, $C$ must be false (and vice versa).$G \rightarrow$ not $C$"The gardener and the handyman are not both lying"At least one of them is telling the truth.$G$ OR $H$"If the handyman is telling the truth then the cook is lying"$H \rightarrow$ not $C$

[glossy valve]

Sean [Ping On Reply Please!]

[upbeat knot]

B -> C
not(C and G)
not(not G and not H)
H -> not C

[neon yoke]

Why is not the both lying one an exclusive OR?

[rapid rain]

because they can both be telling the truth

that would also mean they're not both lying

because neither is lying

[neon yoke]

it says cannot both be telling the truth

[zenith kernel]

Specify which layer might work. If I ask you to switch your answer to the opposite in my following statement starting from the second “if” I say, if … if … would you …. Kind of thing

[neon yoke]

the cook and gardener

[rapid rain]

but it's not the cook and gardener which are said "cannot both be lying"

it's the handyman and gardener

[rapid rain]

so "G or H", with the or being inclusive

[neon yoke]

Can you tell me about the cook and the gardener

I think it is an XOR

[rapid rain]

it's not a xor either

they're said that they can't both tell the truth

nothing is said about them not both being able to lie

so it's not(G) OR not(H)

with the or being inclusive too

[neon yoke]

they CAN lie at the same time?

[rapid rain]

if you only take this statement into account, yes

the statement "they can't both tell the truth" is exactly "at least one of them lies"

[rapid rain]

||I mean to explain it logically: if M is your villager (a logic gate that either always tells the truth or always lies), you can define M composited with a lie gate (a "not" gate). If there's no problem asking questions like "If I asked M whether ....", there should be no problem asking questions like "If I asked not(M) whether ..."||

||But again it joins back to my analysis of overcomplicating the ruins problem. While in the 2 guards problem you're inclined to ask questions like that, in the villager problem you had to resort to creating that "anti villager" which is just more complicated than the other answer||

Careful Tom tom, even if spoilered it's not a good idea to show the entire solution like that

@neon yoke where are you at in the problem now?

Did you manage to convert every statement correctly into logic?

[rapid rain]

Using the insight previously given

[neon yoke]

Can you give me a case, for which exclusive or becomes inaccurate

like maybe some truth values or smth

I understand that cannot both be truth means at least one is lying

but why is exclusive or unsuitable here

[rapid rain]

So if the statement is "at least one is lying"

then when both are lying

[zenith kernel]

I think I found three solutions though

[rapid rain]

then one or more people are lying then

so the statement is true when both are lying

[neon yoke]

yeah

in case of XOR isnt it also true

0 XOR 0 is 0

[rapid rain]

xor means "exactly one is lying"

if both are lying

[neon yoke]

0 XOR 0 is 1?

[rapid rain]

the xor makes it false

[zenith kernel]

Got it

[rapid rain]

we're talking about them lying, not about them telling the truth

[neon yoke]

both are lying then both are false, F XOR F is false

[rapid rain]

so not(0) xor not(0)

[neon yoke]

but it should have been true

right

[rapid rain]

when both are lying

[zenith kernel]

||B, C: F, F but I found out G, H: TT, TF, FT all three cases work||

[rapid rain]

1 xor 1 = 0

so using xor makes it false

but not(0) or not(0) = 1 or 1 = 1

which is the correct truth value

[neon yoke]

I am doing the 'cook and gardener cant both be telling the truth'
I did:
c XOR g

this means that only one of them is telling the truth

[rapid rain]

but why couldn't both of them lie

none are telling the truth in that case

so they're also not both telling the truth

[neon yoke]

so you are saying that the proposition should evaluate to true when both are false?

[rapid rain]

once again, "C and G aren't both telling the truth" is the same as "at least one of C and G lies"

[neon yoke]

so not C OR not G?

[rapid rain]

exactly

[neon yoke]

alright

[rapid rain]

you can also interpret original statement as not(C and G)

as in 'it's not the case that both C and G tell the truth"

so when you remember your De Morgan laws

[neon yoke]

okay now we have
b implies c
not c or not g
g or h
h implies not c

[rapid rain]

not(C and G) = not(C) or not(G)

[neon yoke]

nice

this one is more intuitive

[neon yoke]

where to go after this

[rapid rain]

there are many ways to do this, maybe for now we could do some case by case to see what could work and what doesn't

[neon yoke]

i assumed that b is true

that means c is also true

[zenith kernel]

Exact opposite I think

[neon yoke]

but the last proposition says that c is false

after following through the chain

what does that mean