help-13

[woeful isle]

And sec x * tanx goes undefined

It's alr

[vapid otter]

so then my critical number is pi/2 + pik

[woeful isle]

Yup

No but like

[vapid otter]

in this problem though

pi/2 is out of bounds

so its neglible

all i have to do is now input my bounds into the derivative to find the extremas

[woeful isle]

How to defined critical numbers?

[vapid otter]

well when it equals zero as well

umm could it be

[woeful isle]

Is it where the graph changes direction

Or where it goes undefined

Math is different everywhere

[vapid otter]

critical numbers are when the derivative equals zero or is undefined

[woeful isle]

And so are the terms

[woeful isle]

Oh ok

Then we are correct

[vapid otter]

where does secxtanx equal 0?

[woeful isle]

So sec is never zero

That only leaves tan

[vapid otter]

i think at 0

cuz tan(0) = 0

right

[woeful isle]

Yup

Also at pi

And 2pi

[vapid otter]

oh yes

[woeful isle]

So that kpi

[vapid otter]

so then its

[woeful isle]

Tan is 0

[vapid otter]

pik

right

[woeful isle]

Perfect!

[vapid otter]

ok gimme a sec

so is there an absolute minimum at pik?

oh wait

abs. max. at x = 0
abs. min. at x = pi/3

[woeful isle]

Min

Not max

Min at x =0

[vapid otter]

oh my bad

[woeful isle]

Max at x =pi/3

[vapid otter]

yeah i accidentally wrote the opposite on both

[woeful isle]

It's perfectly fine

Anything else I can help you with

?

[vapid otter]

thanks though

um maybe

give me a minute

[woeful isle]

Yeah sure

[vapid otter]

number 10

it is applicable

but i dont get the last part

"find all values c in the open interval (a, b) such that f'(c) = 0."

would it just be 0 and 2pi?

wait hold on

i think i need to do first derivative test

[woeful isle]

What is differentiation of sinx

[vapid otter]

is it also okay if i do second derivative test?

[woeful isle]

1st derivative is fine

[vapid otter]

but isnt 2nd faster?

[woeful isle]

I mean we are just trying to find where the slope of the sinx function goes to zero

We just need to figure out where cos goes to zero in 0 to 2pi

And we will be done

@vapid otter

[vapid otter]

so the critical numbers are x = pi/2 + pik

but in the interval

it would be

x = pi/2, 3pi/2

so the answer would be c = pi/2, 3pi/2

[woeful isle]

Perfect!

That's it

[vapid otter]

those are the c values where f'(c) = 0

[woeful isle]

Yup

[vapid otter]

second derivative test is only better if you are looking for the max's and min's right

[woeful isle]

Yup

[vapid otter]

its faster than first derivative test

ok

thanks

[woeful isle]

Usually

[vapid otter]

i think this will be all for now

[woeful isle]

Used to find inflection points

[vapid otter]

yes

thank you

[woeful isle]

Well had great time helping you

Have a good day

[vapid otter]

thank you

u too

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[peak wave]

hello ive got a problem understanding where i went wrong with the following question .

[peak wave]

i have the points calculated for y=24-20x as (2,4) and (1,14) but for the equation y=3-8/3x i have (2,-2.33333)(repeating) and (6,-13) i get the feeling this wrong since i cant plot negatives on the graph

and when i set the equations equal to each other i get -6.3333333335 I believe i am making a simple mistake my brain is just too fogged over to realize what exactly it is.

HALLLLLPPP!!!!!! HELP ME PLEASE!!!!!

<@&286206848099549185>

[cedar kiln]

@peak wave Has your question been resolved?

[carmine bronze]

8x+3y <= 24 should become y = 8 - (8/3)x

[peak wave]

i knew it was something simple. thank you man

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[west marlin]

for normal approx. to binomial distrubution, you have to use a continuity correction of 0.5. when do you add or subtract this correction from X value?

[worldly chasm]

@west marlin "We add 0.5 if we are looking for the probability that is less than or equal to that number. We subtract 0.5 if we are looking for the probability that is greater than or equal to that number."

From openstax

[west marlin]

ok, so if you have (P>20), you would have to find 1-(P<20) by default, would it become 1-(P<20.5) or 1-(P<19.5)?

since you are originally looking for greater than, but use algebra to get to a less than

[worldly chasm]

You're looking for X > 20, so because you are looking for "greater than or equal" you use 20-0.5 = 19.5, then you translate it into a form where you can use the cdf

So 1 - P(X < 19.5)

[west marlin]

ok cool

thanks

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[silent forum]

1st equation

1. How do I simplify this equation (im just a bit lost on this one)
2. How do I solve for y in this equation
3. How do I make desmos graph possible answers of "a" if "y" is negative

2nd Equation

1. How do I solve for y in this equation
2. How do I make desmos graph possible answers of "a" if "y" is negative

[worldly chasm]

This seems horribly complicated for it to be the answer for a standard homework question.

What's the original question just as a sanity check?

[pastel vault]

they're investigating star polygons btw

[silent forum]

finding the area of a regular polygon and finding the area of the new shape formed when 2 regular polygons of the same time are stacked on top of each other, 1 one an angle

x=length of one side
y= how many sides

[pastel vault]

yeah can you post that other image you sent me

[silent forum]

eg for the first formula solves this

the area of the ocotogon in the middle

[pastel vault]

wait, hexagon?

[worldly chasm]

Octogon

[silent forum]

mb octogon

it works no matter which polygon you use

[worldly chasm]

So your question is just how to find the area of a generalized regular polygon with shlafli symbol {m/n}?

[silent forum]

no

I already did that

right here

[worldly chasm]

And y is the area?

[silent forum]

y is how many sides

a is area

x is length

[worldly chasm]

Wouldn't {8/2} have a different area from {8/3}?

[silent forum]

the formula works for {m/2} only

[worldly chasm]

Ok

[silent forum]

i should of mentioned that...

also m can be any value apart from -2 -1 0 1 2

[worldly chasm]

Are the middle two terms being multiplied intended to be exactly the same?

[silent forum]

yes

[worldly chasm]

Ok, just checking that there wasn't a minus dropped somewhere, because I would have just expected a ² instead

[silent forum]

its unsimplifyed, im too scared to simplify it

[worldly chasm]

This is unlikely to be able to be inverted. Given that you have a function in the form tan(f(y)) g(y) = C

[silent forum]

yeah but like

[worldly chasm]

(also is your tan in degrees or radians?)

[silent forum]

if you know x and y you can find a, why cant I find y if I know a and x?

[silent forum]

degrees

[worldly chasm]

Ok, does desmos know that?

[silent forum]

yes i think

[worldly chasm]

It's a setting I think, but it's not the default

[silent forum]

its in degrees

[worldly chasm]

You can find values for y, just there isn't a way to do so systematically using only the standard functions we normally use on a day to day basis

[silent forum]

so

[worldly chasm]

There are many very simple expressions that are impossible to invert

Such as solving y = sin(x) + x for x

[silent forum]

💀 so I have been trying to solve something impossible for 3 weeks?

[worldly chasm]

Perhaps! There might be a special function that has been studied that can be used to do this.

But I'm not aware of it.

Given that tan has expressions involving exp

You might have a method of solving this using the Lambert W function

[silent forum]

whta does this mean>?

[silent forum]

whats that?

[worldly chasm]

So trigonometric functions are connected to the exponential function via complex numbers.

For instance, cos(x) = (e^(ix) + e^(-ix))/2

[silent forum]

okay

[worldly chasm]

and the Lambert W function is defined as follows:

x e^x = y implies W(y) = x

So if you can massage this expression into a form where you can apply the definition of the Lambert W function you might be able to manage this

But it would be, at the very least, extremely ugly

If not just down right impossible.

[silent forum]

damn

[worldly chasm]

tan(x) = (e^(ix) - e^(-ix))/(e^(ix) + e^(-ix))

Which is already pretty iffy

Because you need to combine these exponentials into a single one to make this work

[silent forum]

okay

do you know any like websites I can use to learn this stuff?

[worldly chasm]

Black Pen Red Pen (the YouTuber) really enjoys content involving the Lambert W function. So he has several examples of the sorts of manipulations you need to do to put things in the correct form

[silent forum]

m alr thankyou

one last question though

how do I make desmos graph answers of "a" if "y" is negative?

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[cedar kiln]

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[worldly chasm]

@silent forum how would a negative number of sides make sense?

[silent forum]

its theoretical

[silent forum]

^

.close

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[silent forum]

2 questions, why is this a thing in my graph? is it bc desmos is not willing to solve for those values?

and why wont desmos give answers of "a" when y is negative even though I have this setting on?

[silent forum]

fr last questions

[worldly chasm]

@silent forum it's because you have an implicit (x,y) input, but the complex mode gives an (x,y) output. You're asking too much. You need 4 dimensions to display the required data

(a real, imaginary pair for each (x,y) input point)

[silent forum]

so uh

is there any way to graph the equations>?

[worldly chasm]

What would you expect from such a graph?

[silent forum]

something amazing

idk I just wanna know what it looks like

im guessing this means its not possible to graph fully?

[worldly chasm]

It is possible, but it would require 4 dimensions!

[silent forum]

so not possible

[worldly chasm]

Depends on how clever you are about it

[silent forum]

so not possible

[worldly chasm]

For instance, you could get away with 3 dimensions if you plot the magnitude of the complex number instead of the full number

You throw away phase information though

[silent forum]

huh>?

im sorry idk what like half the stuff you say is

[worldly chasm]

that's ok, I'm happy to explain

So.

Currently, you are graphing curves in the x, y plane that satisfy a given value a.

But you seem to want to see what happens when y is allowed to be negative.

In other words, you want to specify (x, y) and find out a from this, yes?

If a is a real number, then we can do this in 3 dimensions.

(x, y) as input and a as the output, using desmos.com/3d or similar

But if a is a complex number you normally need two values to specify it

a = b + ci

Which means you need two dimensions for output

However, we're not really good at visualizing 4d things in general.

So we've come up with a few tricks

Instead of showing a, we can show |a| = √(b^2 + c^2), in other words the magnitude of a.

This is the "length" of the complex number. Or the absolute value.

But we throw away information about the phase or angle. We generally compensate for this using color.

[silent forum]

mm

[worldly chasm]

Like color wheel math, where red is close to the real axis and green is close to the negative real axis. Yellow is near the positive imaginary axis, and blue is near the negative imaginary axis

[silent forum]

right

[worldly chasm]

Perhaps it might be helpful to study the polar form of complex numbers.

So this seems a little bit less out of left field

[silent forum]

okay 😭

[worldly chasm]

For any complex number z, we have z = x + iy = r e^(iθ)

Where r is the magnitude or length and θ is the phase or argument or angle. Different places use different language.

x = r cos θ
y = r sin θ

[silent forum]

😭 things started to get confusing after this point

[worldly chasm]

Lol

If you have a function in two variables, then you need two dimensions for input, so to give an output you need a third dimension

So x and y are your inputs, and your output a is given by the height of the surface

f(x,y)

[silent forum]

ohhh

wait yeah i get it

[worldly chasm]

https://www.desmos.com/3d/esgt02uf6y

See also https://en.wikipedia.org/wiki/Domain_coloring

[silent forum]

so how am I suppose to apply this to the equation?

[worldly chasm]

Well, I'm honestly not sure how to make it work with desmos

I haven't played around with this sort of idea yet

And it's rather new support, not sure if it's made it to the 3d plotter

[silent forum]

okay

[worldly chasm]

You might have to use a different software.

I hope this isn't too unhelpful 😦

[silent forum]

this was very helpful

im grateful for your help

[worldly chasm]

Thanks, I appreciate it.

I know it's a lot like drinking from a firehose

[silent forum]

mhm, and now I need to learn about these new stuff so I dont have to end up asking for answers

anyway thankyou

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[olive dirge]

hi

[olive dirge]

help needed

Dont know whats wrong wit this

if anyone could help that would be great

[cedar kiln]

@olive dirge Has your question been resolved?

[olive dirge]

nvm got it

[worldly chasm]

@olive dirge check your end points

[olive dirge]

did it myself

[worldly chasm]

I think

[worldly chasm]

Nice!

[cedar kiln]

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[crimson sedge]

Hello! I was just wondering, can I somehow convert this graph into a quadratic equation?

[cedar kiln]

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.

[crimson sedge]

Oh, well I don't have an original problem (i'm only doing this out of curiosity) but the function I used to generate this is this:

[vagrant elbow]

wdym by "convert"?

you want to find a quadratic approximation?

[crimson sedge]

yes

[coral jewel]

i present to you taylor

[vagrant elbow]

do you know about taylor polynomials?

[crimson sedge]

sounds familiar

but I haven't used it

[wraith dagger]

neon

[vagrant elbow]

Take a to be the minimum

[dusk goblet]

in other words you’re just creating a function that passes through that point while also having the same first and second derivatives at that point

[vagrant elbow]

,w minimum of gamma(x)/x + 1/x

[wraith dagger]

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[vagrant elbow]

well you're going to get a nasty equation

i hope you're fine with that

[crimson sedge]

Oh, I wasn't expecting this 🤣 Is there any like sources or tutorials that might help me understand me this even more

I'm very new to Calculus

[gentle flower]

youtube/khan academy i guess

[vagrant elbow]

https://youtu.be/3d6DsjIBzJ4?si=tX9LmD5vmSCy2x_N
If you have 20 minutes this is an insane video

[crimson sedge]

Thank you soo much!!!

[crimson sedge]

So with this, I can find the quadratic equation that can represent this graph something like that?

[vagrant elbow]

you'll find an approximation for it about a yeah

it won't be perfect

but around a it's decent

you can also make it a better approximation by increasing the degree of the polynomial

[crimson sedge]

Thank you for the info!

I initially guessed like I'm going to use the vertex form and convert it to the quadratic equation 🤣

[cedar kiln]

@crimson sedge Has your question been resolved?

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[calm hedge]

If the supply voltage is 5V, and two parallel resistors R1 and R2 have 0.7 and 0.5 ohms respectively

[calm hedge]

The current going into R1 will be?

So I need to find total current value first

I=V/R

need to find total resistance

total R = (R1*R2)/(R1+R2)

[hollow trail]

do you?

[calm hedge]

hm

i mean thats what i think, my thought process was to find total current, divide that total current by the R1 to find that current into R1

[hollow trail]

think about what quantities are common between two parallel resistors

[calm hedge]

voltage stay the same across resistors in a parallel circuit

[hollow trail]

yes

[calm hedge]

OH

god damn

I = v/r

5/0.7

7.14A

oh yeah, where was i even going with my process

ok thanks man

If three 1.2V Ni-Cad Cells are connected in parallel to supply a 10 resistive load, then the supply current will be

I was thinking maybe to get supply current

I = V/R

so 1.2/10

is that correct?

[muted crest]

multiply by 3?

[cedar kiln]

@calm hedge Has your question been resolved?

[calm hedge]

but isnt it in parallel?

should it tall be the same?

[muted crest]

i think voltage remains same through out

not the current

current should add up

[cedar kiln]

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[flat sage]

let z=1+i, then what is the main argument of Z^8? 2π?

[ancient lodge]

,w argument of (1+i)^8

[wraith dagger]

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[flat sage]

oh

[ancient lodge]

If “main” means principal, then 0 is better

[flat sage]

tnx

[ancient lodge]

But 2pi is an argument, just not the principal one

[flat sage]

okok tnx so much

[cedar kiln]

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[flat sage]

sorry i have one question

if 1+i^8, i^8 is 1 , 1+1 is 2

so the resoult shuld be 2

so why your answer is 0?

[ancient lodge]

1. (1+i)^8 isn’t 1+i^8
2. You asked for the argument, not the value itself

[flat sage]

ohhhh okok sorry

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[analog crow]

I don’t get question (ii)

What am I supposed to do after figuring out the volume of the water

[cedar kiln]

@analog crow Has your question been resolved?

[analog crow]

<@&286206848099549185>

Thanks 🙏

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[manic acorn]

how do i prove that they are equal?

[potent fractal]

Try expanding $\left(\sum_{i=0}^n a_i\right)^2$

[wraith dagger]

EQUENOS

[potent fractal]

You will get a bunch of terms like $a_i a_j$. Try grouping them together and see what happens

[wraith dagger]

EQUENOS

[manic acorn]

got it thanks

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[analog summit]

hello, how do i prove that the sequence (fn)n dont converge uniformly pls

x in ]0, 1]

[viscid ore]

Hi

[analog summit]

hi

[viscid ore]

@analog summit

where are you from?

[analog summit]

uhm why

[viscid ore]

just curiosity

[analog summit]

fr

[viscid ore]

btw, I can guide you how to do this

[analog summit]

ok thank you

so the limit function is 1

and the norm difference is (1-x)^n

now i dont know how to justify the sup

[viscid ore]

|fn(x)-f(x)|=fn(x)-1|=|1-(1-x)^n-1|=(1-x)^n

right?

[analog summit]

yes

[viscid ore]

solved?

[analog summit]

why?

[viscid ore]

to determine if fn converges uniformly to f(x)
we have to check if
sup(|fn-f|)->0(n->inf)

[analog summit]

yes

but the sup is when x approaches 0

[viscid ore]

not clear yet?

[analog summit]

i mean, if we look at x in ]0, 1], how do i give the sup then

[solid juniper]

you don’t actually need to compute the sup

you could just show that it is, e.g., greater than 1/2 for every n

which you can do by fixing n and then picking x so close to 0 that (1-x)^n is at least 1/2

[analog summit]

so i have to pick an exact x?

like 10e-3 or something?

[solid juniper]

you want to show one exists (don’t necessarily need to construct it)

you’ll need one for each n

[mental trail]

At this point showing that f_n induces a bijection (or is surjective) between ]0,1] and [0,1[ with IVT is kinda same-ish

[analog summit]

uhm

[analog summit]

dont think i get this

[analog summit]

and what is x then

[solid juniper]

something for you to think about

[analog summit]

do you mean i have to write x = n² for example?

or 1/n since n² would not be smaller than 1

[solid juniper]

n is fixed and you want an x for which (1-x)^n > 1/2

you can do some algebra to find one

[analog summit]

ok

[analog summit]

x < n-th root of 1/2 - 1?

uhm

but it's negative

[solid juniper]

you made a sign error or something

[analog summit]

oh yes

ok thank you 👍

.close

[novel ice]

well another insight could be to use the ratio of lim n=>0 abs fn+1/fn <1

then to find the radius of convergence to be null

[solid juniper]

[novel ice]

?

well enlighten my foolish mistake

[analog summit]

but does it work if i say the limit when x approaches 0 is 1 so not unif convergent?

[solid juniper]

i am not sure what that has to do with the problem

[solid juniper]

yea. that’s a non-constructive way to find an x

[solid juniper]

you can say an x exists because lim x to 0 of (1-x)^n is 1 for any n

[analog summit]

ok

[novel ice]

hmmm true my bad

[analog summit]

one last question, is it true to say that the sup is fn(0)? even when x is never equal to 0, can i still say the sup is when x approaches 0? Or does it have to be when x strictly in ]0, 1]

[solid juniper]

well i would just say for any n, the sup of f_n over x in (0,1] is 1. and then also sure f_n(0) is 1 for any n if you extend the domain a little

[analog summit]

so can i actually say sup|fn(x) - f(x)| = fn(0)?? when x in (0, 1]?

even if 0 is not even in the domain?

[solid juniper]

it depends how the f’s were introduced

if they were introduced as functions with domain (0,1] then i wouldn’t write that

[analog summit]

mmh yes makes sense

[solid juniper]

if they were declared functions with domain R then fine whatever

there’s still something about writing that that bugs me though. not because it’s wrong but because it maybe indicates misconception to me

[cedar kiln]

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[potent fractal]

Problem

Imagine a game where each of 2 players receives an infinite sequence of 1's and 0's obtained by tosses of a fair coin. Each player receives their own sequence. The players can not communicate in any way. Player 1 chooses a natural number a, and at the same moment player 2 chooses a natural number b. After that we compare a-th digit in the sequence of player 2 with the b-th digit of the sequence of player 1. Both players stay alive if the digits match, otherwise they DIE. Both players can discuss the strategy before the game starts.

If this is too verbose, read this interpretation of the problem: #help-13 message, #help-13 message

Warm-up question: is there a strategy such that the probability of staying alive is greater than 1/2?
Big question: what's the exact upper bound p of the probability of staying alive? Is it achievable with some strategy?

Progress

I reduced the problem to so-called finite strategies (more about that below) and proved that p > 0.699999 (precisely 5 nines). However, what I'd really want to achieve is at least finding an interesting upper bound for p. Currently it's just 1.

UPD: Correction of a definition from the message below (#help-13 message)
I forgot to mention that Im(f_n), Im(g_n) must be subsets of {1, ..., n}. Apologies

[wraith dagger]

EQUENOS

EQUENOS

[potent fractal]

Spent some time solving this problem during summer and decided to share it here, because it seems fun and it will be awesome to see some progress with the upper bound for p

[limber marsh]

what’s stopping both players from choosing the same number

[quiet plover]

they dont know each others sequences

[potent fractal]

They can choose the same number, but the digits might not match and they'll loose

[limber marsh]

oh they have different sequences

lol

[quiet plover]

does the process happen only once?

or several times

[rocky harness]

once by the looks of it

[potent fractal]

They play just once

[limber marsh]

sorry what happens if the players just both say beforehand “pick a number which matches up with a 1”?

[limber marsh]

i will stop asking dumb questions after this

[quiet plover]

yeah

thats confusing

[rocky harness]

the players cannot communicate in any way

[potent fractal]

Sorry, wdym?

[limber marsh]

they can strategize beforehand

[potent fractal]

Those are good questions that help to understand the objective

[rocky harness]

im not sure if that is given or follows from the statemtnt

[limber marsh]

if i can strategize beforehand, i am just going to say “if there is a 1 in my sequence, then i am picking that number”

[rocky harness]

they can think all they want independently of what strategies the other one can employ

[quiet plover]

wh?

[potent fractal]

That's a decent idea. They can, for instance, pick the number of the first 1 in the sequence and this strategy will win with a probability of 2/3

[limber marsh]

what why is it 2/3

[rocky harness]

oh they switch their numbers afterwards

if thats what youre confused about

[potent fractal]

I think an easier way to explain the game is using this definition:

[rocky harness]

so they cannot be sure that the number they pick will also be 1 in the others sequence

[potent fractal]

Player 1 has sequence X, player 2 has sequence Y. Players win if and only if X[g(Y)] = Y[f(X)]

[limber marsh]

oh i see.

great, thanks. now i think this is an interesting problem and will go back and use my reading comprehension skills

[rocky harness]

unless im missing something, isnt the set M_n finite? you have 2^n x 2^n matrices whose rows are composed of 0 and 1 blocks of length 2^k. what stops you from bruteforcing AB and finding the maximum?

[quiet plover]

ohhh

[potent fractal]

M_n is indeed finite, but bruteforcing is too brutal

[quiet plover]

so we swap

[rocky harness]

well i suppose it is much harder in practice

yeah

[potent fractal]

At some point I made an iterative process that produces sub-optimal strategies, but the complexity is something like O(n 4^n)

So I stopped at n=10 which gave me the aforementioned lower bound of 0.699999

[potent fractal]

For reference, bruteforcing is O(n^{2^{n+1}})

[quiet plover]

how can you even bruteforce

[rocky harness]

there are finitely many elements in M_n, so all possible products give you a finite set, so the maximising trace is findable by bruteforce

[rocky harness]

i see

[potent fractal]

Also I'm not sure if that process gets sufficiently close to the true optimal pair of matrices

But at least it increases tr(AB) with each iteration

Useful intuition: tr(AB) is the dot product of "flattened" A and B^T

[cedar kiln]

@potent fractal Has your question been resolved?

[haughty glen]

Probably of no use but the trace is also the sum of the eigenvalues from the matrix A*B

[potent fractal]

I tried this intuition but unfortunately I haven't obtained anything from it

[rocky harness]

Sorry , im unsure if im missing something, but arent there infinitely many strategies? How are you finding a correspondence to a finite set. I can only assume the correspondence is not injective?

[potent fractal]

The set of finite strategies is finite. Sorry, I've just realised that I didn't specify that $\Im(f_n), \Im(g_n) \subset {1, \dots, n}$ in the definition

[wraith dagger]

EQUENOS

[potent fractal]

So basically we play a game where the sequences are truncated down to n digits

[rocky harness]

Ah so you are limiting the strategies to be such that the swapped numbers obtained from the strategy are in {1,..,n}

[potent fractal]

yes

[fair geyser]

someone asked this question fairly recently i just told them what smay said

[potent fractal]

That's kind of mind-blowing because I came up with this question myself based on a similar question I saw in a russian youtube video

[fair geyser]

didn't notice they swap

huh

must have been you then

[potent fractal]

Nope, it's my first time occupying a help channel

Well, I posted this question in #1021175428326633542 over 2 months ago but I doubt that's what you meant

[fair geyser]

same youtube video maybe

[potent fractal]

oh wow

Probably

Although it's a nerfed version

The answer to their question is "let both players pick the index of the first 1 in their sequence"

This will produce a probability fairly close to 2/3

[fair geyser]

i don't get it

it works unless one of them has all 0 and the other one doesn't

can't be 2/3

[potent fractal]

If all digits are 0 then we just say 1

[fair geyser]

cuz they don't swap

that's the nerf, not the 100

[potent fractal]

Oh, they don't swap in that problem?

[fair geyser]

no it sounds like they don't

[potent fractal]

I meant that the nerf is that we're interested in some strategy, not the optimal one

[fair geyser]

i see

how on earth is it 2/3

hmm

cuz they often pick the same position got it

[potent fractal]

The most accurate explanation is just a straightforward computation

[fair geyser]

yeah but it's also like that, it's 50% if they don't pick the same position and 100 if they do and that happens often

so it's significantly above half

unless it's not 50%

[fair geyser]

yeah it is ok

[cedar kiln]

@potent fractal Has your question been resolved?

[cedar kiln]

@potent fractal Has your question been resolved?

[cedar kiln]

@potent fractal Has your question been resolved?

[cedar kiln]

@potent fractal Has your question been resolved?

[cedar kiln]

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[tepid delta]

is there any way to do this other than the long way

[cedar kiln]

@tepid delta Has your question been resolved?

[cedar kiln]

@tepid delta Has your question been resolved?

[tepid delta]

<@&286206848099549185>

[crimson sedge]

count how many 7,70,700 there are

wait nvm

[cedar kiln]

@tepid delta Has your question been resolved?

[cedar kiln]

@tepid delta Has your question been resolved?

[crimson sedge]

lol

[queen stirrup]

try to instead find the sum of all numbers that contain seven in their base 10 representation

and find a pattern

[vague sparrow]

work it out in a smaller case than 2024 so you can check it by hand, I'd suggest using inclusion-exclusion

[cedar kiln]

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[cedar kiln]

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[drowsy mountain]

what does it mean to differentiate with respect with x?

[drowsy mountain]

and if you can find an example too :]

.close

[cedar kiln]

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[drowsy mountain]

i think i got it

[azure stirrup]

differentiate in terms of x only so eg u have to differentaite ax^2 with respect to x, x is the only variable affected by differentation, not a, which will give u the answer of 2ax

if it was differentiate ax^2 with respecr to a, itll be just x^2

[drowsy mountain]

thank you!

.reopen

[azure stirrup]

or even if thers no second variable it essentially means whatever ur differentating with respect to (variable) only that variable will be affected by the differentiation

[cedar kiln]

✅

[drowsy mountain]

i didnt get it haha

[azure stirrup]

nwwws

[azure stirrup]

its okay, does it make sense now?

[drowsy mountain]

yes!

how do i apply that to uhhh

implicit differentiation

im just learning it, so i dont really know anything about it

there’s like an example question

which is

x^2+y^2=r^2

and its ike

like* differentiate that with respect to x

idrk how to do that 😅

[azure stirrup]

um

we could start off with

ur essentially differentiating both sides with respect to x yeah?

we could start off from the right side first since its easier

so u have d/dx r^2

do yk what that would be?

[drowsy mountain]

0

cause r is a constant

[azure stirrup]

yes

u got one part done

now for the lhs

we wanrt to d/dx (x^2 + y^2), we could expand it into d/dx x^2 + d/dx y^2

lets do the first part first, d/dx x^2

[drowsy mountain]

2x

right?

[azure stirrup]

mhmm

now

so far we ahve 2x + d/dx y^2 =0

forthe d/dx y^2 part

we need to apply chain rule

do yk what chain rule is?

[drowsy mountain]

yep

du/dx * dy/du= dy/dx

[azure stirrup]

yes

as of now we cant rlly do anyhitng with d/dx y^2 as thers no x terms

in order to differentiate y^2 we want it to be d/dy (ie differneated with rrespect to y)

so we could do d/dx times dy/d times d/ dy y^2

the reason why we can multiple an additional dy/d times d/dy is bc it will equal 1

[drowsy mountain]

uhhh

okay yeah im following

[azure stirrup]

from here when we multiple d/dx with dy/d first, itll give dy/dx, so now we have dy/dx times d/dy y^2

[azure stirrup]

sorry idrk how to explain it that well 😭

lmk if any parts doenst make sense

[drowsy mountain]

will do!

dont worry!

[azure stirrup]

for the d/dy y^2 part, itll just be 2y yeah?

[drowsy mountain]

yep

[azure stirrup]

so now we have dy/dx times 2y

back to the original, we went from d/dx (x^2 + y^2) = d/dx r^2

to

2x + dy/dx 2y = 0

[drowsy mountain]

how do we get dy/d

is it just 1/2y?

[azure stirrup]

so yk how the only way to differentiate y^2 is if its with d/dy?

[drowsy mountain]

yep

[azure stirrup]

wait let me draw smt

[drowsy mountain]

thank you!

that would help more :]

[drowsy mountain]

im lost from this point on

if that helps!

[azure stirrup]

still drawing

[drowsy mountain]

take your time!

im still here :]

so

its gonna be

lemme draw too

[azure stirrup]

have a read of this and lmk if it clears it up

[drowsy mountain]

oh it’s plus not times

[azure stirrup]

remember the original equ was x^2 + y^2

[azure stirrup]

yesss

[azure stirrup]

the times only applies to the derivative with the term

[drowsy mountain]

do i then re arrange for dy/dx

[azure stirrup]

it’ll be 2y times day/dx

[drowsy mountain]

oh right

[azure stirrup]

remember how we wanna like d/dx (x^2 +y^2)

u can think of it kinda like

expanding ur brackets

[drowsy mountain]

is 2x+2y in brackets?

[azure stirrup]

d/dx times x^2 plus d/dx times y^2

[azure stirrup]

since we have expanded the brackets already there is no longer any brackets

[azure stirrup]

does this kinda make more sense?

[drowsy mountain]

yeah, i just dont know where you got dy/dx(2y)

[azure stirrup]

does the second line of working out makes sense

[drowsy mountain]

its a bit confusing

[remote trail]

I prefer to think about it as d/dx[f(y)] = d/dy[f(y)]*dy/dx if that helps at all

^using chain rule to change d/dx to be derivable in terms of y

[drowsy mountain]

it makes sense when i look at the bottom

so after i get uhhh

2x+dy/dx(2y)=0

do i re arrange for dy/dx

[azure stirrup]

umm its like the only way u can derive y^2 is with d/dy, but there is no d/dy so we can use chain rule where we multiple the already d/dx y^2 with d/dy times dy/d on top, so theres a d/dy we can use to deririve the y^2

[drowsy mountain]

oh right

[azure stirrup]

so u have d/dx times dy/d times d/dy times y^2

u can kinda think of it like fractions

[drowsy mountain]

mhm :]

[azure stirrup]

the d/dx times dy/d will give dy/dx

and the d/dy times y^2 = 2y

so were left with dy/dy times 2y

does this clear it up better?

[drowsy mountain]

yes!

wait

[drowsy mountain]

d/dy * y^2?

or y^ differentiated with respect to y?

or is that the same thing

[azure stirrup]

so i introuced the purple (chain rule), the bliue highlgiher from the second to third line is the same thing,

[azure stirrup]

same thing

[drowsy mountain]

oh okay, that makes sense

[drowsy mountain]

thank you!

[azure stirrup]

and i left the d/dx times dy/d untouched from the second to third line

does the working out from the second to third line make sense now?

[drowsy mountain]

brb i have break

[azure stirrup]

um

[drowsy mountain]

yes!

[azure stirrup]

i need to go soon

[azure stirrup]

yay

okay

were nearly done

[drowsy mountain]

after the third step of your drawing, do i uhhh

rearrange?

[azure stirrup]

the third line the d/dx times dy/d gives dy/dx (works like fractions)

[drowsy mountain]

7

[azure stirrup]

left with 2x + dy/dx times 2y = 0

arrange so dy/dx is on one side only

dy/dx times 2y = -2x

dy/dx = -2x/2y

[drowsy mountain]

-x/y

[azure stirrup]

dy/dx = -x/y

[drowsy mountain]

yep!

thank you! i think i get it now

[azure stirrup]

and thats ur asnwer

[azure stirrup]

hurray nwws

[drowsy mountain]

.close

[cedar kiln]

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[cedar kiln]

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[thorny ferry]

Hello!! I need help with an exercise:

write down all the divisors
a²-36b²

[thorny ferry]

I don't understand what i have to do basically

[hot crag]

factorize, i believe

[ocean grove]

a²-b²= (a+b)(a-b)

[thorny ferry]

Oh thank youuu

[ocean grove]

np!!

[cedar kiln]

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[potent fractal]

I fell asleep and this channel closed. 1 sec, I will post a link to the original message

[potent fractal]

#help-13 message

I will probably keep it open for several days and then let it close again (unless I miss bot's notification again)

If someone will be interested we will move to a forum thread

[cedar kiln]

@potent fractal Has your question been resolved?

[proper mortar]

any intuition why its not always just 50%?

or, what would be the strategy for n=2? There are 4*4 possibilities. If there is a 1 and they both choose the first one, and with no 1 they just choose index 1, then they win when both have any combination of (1, 0), (1, 1), (0, 0) (9 ways), or both have (0, 1) (1 more way), which gives a total of 10/16 I see

[fair geyser]

they point at the same spot alot of the time

1/3 precisely if infinite length

rest of the time it's 50%

[proper mortar]

so this strategy reduces it to the probability of both having the 1 at the same index, and if they don't, they still have a chance to win when their numbers match by coincidence. With 1 based indexing, for an infinite sequence we have P(X(k) = 1) = (1/2)^k. Then we need to sum up P(X(k) = 1 and Y(k) = 1) for any k, they are independent events, so it's just the sum of (1/2)^(2k) = (1/4)^k, geometric series and it yields 1/3, and then we add 50% times the probability of indices not matching (2/3), and it does indeed yield 2/3, interesting

how did current brute forcing attempts go?

[proper mortar]

(if the indices don't match, then one player with the lower index will guaranteed pick a 0, the other player will then indeed pick randomly 50/50, so the case that indices don't match does not influence this probability)

I don't really get how T_k are composed and what these blocks exactly need to look like

[cedar kiln]

@potent fractal Has your question been resolved?

[cedar kiln]

@potent fractal Has your question been resolved?

[cedar kiln]

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[ornate rune]

what's the difference in using parenthesis in matrix and brackets

[crimson delta]

you mean () vs [] ? nothing

[ornate rune]

wait so its just a matter of choice?

[crimson delta]

pretty much

[ornate rune]

is n k a matrix here then

[crimson delta]

no

thats a binomial coefficient

$\binom n k = \frac{n!}{k! (n-k)!}$

[wraith dagger]

Denascite

[crimson delta]

a number

[ornate rune]

why does it look like a matrix

[crimson delta]

ask whoever came up with the notation ages ago

[ornate rune]

well how do i know if the notation represents a matrix or not

[modern compass]

context

(x+y)^n is a number.
if that was a matrix or vector then the sums would be a vector. A number equaling a vector doesn't make sense. So it's not a matrix.

[cedar kiln]

@ornate rune Has your question been resolved?

[ornate rune]

i see thanks

[cedar kiln]

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[pulsar blade]

This is my problem

[pulsar blade]

So I understand that bounded functions f,g have some bounds M and N say, and U(f,P)<=M(b-a) and U(g,P)<=N(b-a) just not sure how to relate them

Everytime I try to connect them, the inequality signs are wrong

<@&286206848099549185>

[cedar kiln]

@pulsar blade Has your question been resolved?

[pulsar blade]

[potent fractal]

Full bruteforcing takes too much time. However, I made an iterative process that converges to suboptimal strategies (complexity O(n 4^n), as opposed to O(n^(2^n)) of true bruteforcing) and I stopped at n=10, getting 0.699999

[pulsar blade]

You gonms help me?

[potent fractal]

Sorry, not now

I'm currently going home

[pulsar blade]

😥

[potent fractal]

For any $t_i$ from $P$ you have $\max_{t\in [t_i,t_{i+1}]} f(t) \le \max_{t\in [t_i,t_{i+1}]} g(t)$. The rest follows immediately

[wraith dagger]

EQUENOS

[potent fractal]

The inequality is due to the fact that if $\tau = \text{argmax}{t\in [t_i,t{i+1}]} f(t)$, then $g(\tau) \ge f(\tau)=\max_{...} f(t)$, and of course $g(\tau) \le \max_{...} g(t)$

[wraith dagger]

EQUENOS

[pulsar blade]

But how do you show that these maxes relate to their upper sums?

[potent fractal]

I thought that upper sums are sums of rectangles such that the height of each rectangle is just max of f on the corresponding segment

[pulsar blade]

Sounds right

[potent fractal]

So i-th rectangle's area is just $(t_{i+1}-t_i)\max_{t\in [t_i,t_{i+1}]} f(t)$

[wraith dagger]

EQUENOS

[pulsar blade]

So it just can follow directly from the definiton of their upper sums?

[potent fractal]

(a)? Yes

[pulsar blade]

You can say this?

M is max on that interval

[potent fractal]

Yes

[potent fractal]

Where inequalities for max follow from this^

[pulsar blade]

Oh should I just mention something like since its true for each interval?

What about the other parts?

Also can you try to explain infinum and supremum intuitively

[potent fractal]

A set of real numbers, for example, can have a lower bound, so some number x such that all elements of that set are >= x.
You can try increasing x, as long as it keeps being a lower bound. At some point you won't be able to increase it anymore. That's when you reached the exact lower bound. That's called infimum

Supremum is the same thing but with upper bounds

[pulsar blade]

So the infinum of sinx is -1?

[potent fractal]

By definition of inf, inf_P U(f, P) <= U(f, Q) for any Q

[potent fractal]

Yes

If you mean the set of all values of sin(x) with real x

[pulsar blade]

Did you mean P for the 2nd one and inf_P would just be U(f)?

[potent fractal]

In the second one I used Q because P is overloaded in inf

The notation inf_P U(f, P) means "the exact lower bound of {U(f, P) : P is a partition}

[pulsar blade]

Is this necessarily true?

[potent fractal]

Yes

Wait the second inequality isn't true

But try avoiding the 3rd number

[pulsar blade]

Im not sure how else to show it

[potent fractal]

After the first inequality you can directly use (a)

[pulsar blade]

But I need to show U(g)>=U(f)

[pulsar blade]

But since f(x)<=g(x), wouldnt it have a smaller upper integral?

[potent fractal]

It directly follows from (b)

[pulsar blade]

How

[potent fractal]

Essentially (b) is saying "U(f) is a lower bound for the set of all U(g, P)"

And therefore it's <= exact lower bound, which is U(g)

[cedar kiln]

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[solar venture]

Hi

[solar venture]

.close

[cedar kiln]

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[cedar kiln]

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[fading garden]

I'm not really sure how to set this up

[fading garden]

I'm pretty sure I need to parametrize C. So I would probably do r(t)=<t, 2t^2> as my parametrization, where 0<t<1. The unit tangent vector would be r'(t)=<1, 4t> divided by sqrt(16t^2+1) I think. But I'm not sure what to do about the rotation by pi/2

[cedar kiln]

@fading garden Has your question been resolved?

[dreamy void]

I think what is meant by "rotated clockwise by pi/2" is if you took the tangent vector at say (0,0).
It would be like this ➡️.
So I think by rotating they want you to rotate it now looking like ⬇️ 90 degrees clockwise which would resemble a vector that is normal to T.

So instead of <1,4t> we would do <4t,-1>

[fading garden]

Oh

Ok

[dreamy void]

This also would explain the notation for n since it's normal to T

[fading garden]

That’s true. Although admittedly I didn’t look that far into the meaning of n in this problem

I’d just need to parametrize F now. So I think that would give F=<t^3, 2t^2+t>

[dreamy void]

It is rather doing F o n ds = F(r(t)) o n(t) dt

[fading garden]

Is this not F(r(t))?

[dreamy void]

it is, just you saying parametrize felt wrong lol

[fading garden]

Oh lol

[fading garden]

I mean fair enough though. I see why you say that

Anyway I should be able to solve the rest of this later since I have a class that starts in 5 minutes. I’ll come back to it after and see what I get

[dreamy void]

https://www.desmos.com/3d/5kcryef0rw

I am wondering though if one could also say r(t) = (t, 4t², 0) and then simply take as unit normal vector (0,0,1)

Ok I don't think you can, because we have F(x,y) instead of F(x,y,0)

Ok I noticed you would just end up with 0 because the dot product would yield 0 lol

[cedar kiln]

@fading garden Has your question been resolved?

[fading garden]

how would it be 0?

we have <t^3, 2t^2+t> o <4t,-1>/sqrt(16t^2+1) right?

[dreamy void]

I was talking about something else forget it

[fading garden]

oh ok

[dreamy void]

Just don't forget ds = ||r'(t)|| dt

[fading garden]

thats just sqrt(16t^2+1) right?

[dreamy void]

yes, you can basically cancel them out

[fading garden]

cool

[dreamy void]

||r'(t)|| = ||n(t)||

[fading garden]

Im glad you said someting because id actually forgotten lol

[dreamy void]

This is basically Green's theorem here

[fading garden]

i havent learned that yet. we learn it tomorrow

[dreamy void]

You will see

[fading garden]

For my final answer, I got 4/5-2/3-1/2. Which is -11/30

[dreamy void]

seems right

also

,texsp [ \int_C \mathbf{F} \circ \hat{\mathbf{n}} : \dd s = \int_C M \dd y - N \dd x ]

[wraith dagger]

𝔸dωn𝓲²s

[fading garden]

I'm assuming M=1/2xy and N=y+x?

[dreamy void]

F = <M,N>

yes

one could easily see why if you start on the right side with dt/dt

Then you basically end up seeing the dot product between F and n

[cedar kiln]

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[tame berry]

This is a physics question, but I am really second-guessing myself here. In the y direction component should ma stayed instead of inputing zero bc gravity is technically an accelerator? And did I set up the components correctly?

[pulsar scroll]

Yes the net force should be the sum of her force, the friction and gravity, and they should total up to 0

[tame berry]

Because of the constant velocity right?

[pulsar scroll]

yes basically

[tame berry]

One last question

So the question gives the value for friction force here

But then asked me to find the coefficient of kinetic friction

And got the same thing lol

What was the point of asking? Is it a trick question?

[pulsar scroll]

The friction coefficent is the force of friction divided by the weight, it should not have a unit. You seems to have subtracted her pulling force from the weight so that seems to be incorrect.

[pulsar scroll]

It should be 20.0N/F_g,

[tame berry]

Ohhh lol Dangg

[pulsar scroll]

That should solve it right?

[cedar kiln]

@tame berry Has your question been resolved?

[tame berry]

Yeah, it's all solved. Gracias bro

Appreciate it

[cedar kiln]

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[carmine terrace]

Is the cardinality of irrational numbers (Q^c) equal to ℵ0? If not, is it larger or smaller?

[carmine terrace]

ℵ is aleph

[crimson sedge]

what have you tried/what are your thoughts on this?

[carmine terrace]

I think its bigger

since there are Aleph null numbers from 1 - 2

and you can make a new number by taking the first digit of the first number, second digit of second number, and so on forever to make a new number

[crimson sedge]

if i recall correctly, you should be able to find the cardinality of the reals (R) and the rationals (Q)

[carmine terrace]

I think the cardinality of real numbers is Aleph Null

and rational are Aleph 1?

[crimson sedge]

other way?

[carmine terrace]

?

[crimson sedge]

reals are aleph1 and rationals are aleph0

[carmine terrace]

oh yeah

[cedar kiln]

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[crimson sedge]

Need help. This is just a practice quiz to train for ASVAB. I already took the quiz and didnt do well. Just going back over questions

what have you tried?

I dont get what the whole 4 dogs are sold and now its 1:1

[crimson sedge]

All i got is 4:3 and dont know how to continue

[crimson sedge]

so we can suppose that there were 4x dogs and 3x cats originally

[crimson sedge]

Yes

how many cats and dogs are there after 4 dogs have been sold?

(in terms of x)

[crimson sedge]

I have no idea

there are originally 4x dogs, and 4 have been sold

so there are now 4 less than 4x dogs

[crimson sedge]

0?

not quite

X

[crimson sedge]

Is the answer 28

did you guess and check that?

you’re correct, yes

[crimson sedge]

I dont know how to set it up as a fraction. I just seen 16 - 4 = 12. So thats 12 to 12 or 1:1. and 16 plus 12 is 28.

[crimson sedge]

Is this meant to be set up as a fraction?

if there were 4x dogs and 3x cats initially, there will be 4x-4 dogs and 3x cats after four dogs were sold

since you know that this is a 1:1 ratio, what can you say about 4x-4 and 3x?

[crimson sedge]

What im confused by is you have this equation right?

4x-4 = 3x

How did you know to multiply 4 on both sides? If thats what you did?

Im asking for reference to some more difficult problems thats arent as obvious as this one.

[crimson sedge]

i did not multiply by 4 on both sides

i moved all instances of x over to the left and everything else to the right

[crimson sedge]

You set up an equation for that?

4x-4 = 3x is already an equation

[crimson sedge]

So whats your next step

4x - 3x = 4

[crimson sedge]

Now how did you find out what x was

[crimson sedge]

simplify 4x - 3x

[crimson sedge]

Okay

what did you get?

[crimson sedge]

I dont know how to simplify

do you know the distributive law?

[crimson sedge]

No, ill look that up real quick

[crimson sedge]

Im lost

hi lost, i’m micabo

[crimson sedge]

Is it somethinh like x(4+3)

very close

[pulsar blade]

I'm sorry but what is this question

[crimson sedge]

there is a - in front of the 3

[crimson sedge]

part of a “practice quiz to train for ASVAB”, apparently

[pulsar blade]

it's a bad question imo, it's confusing

[crimson sedge]

What should it have been

[crimson sedge]

Im just confused cause they threw me a curveball. This one isnt like the examples

[crimson sedge]

X(4-3)

[crimson sedge]

it is perfectly clear to me

[crimson sedge]

yes, and what’s 4-3?

[pulsar blade]

I get it now, but I don't think this is clear at all

[crimson sedge]

1

[crimson sedge]

i am interested in where you think the ambiguity arises

[crimson sedge]

so it’s equal to 1x, or just x

so 4x-3x = 4 means that x=4

[pulsar blade]

esp when you look at the next question is literally a simple ratio problem 😭

[crimson sedge]

are you complaining that it’s too hard?

[pulsar blade]

I just don't like its wording

[crimson sedge]

because that does not make a question confusing

[crimson sedge]

Im just bad at math. And never learned fractions and rations until this week

[crimson sedge]

i have not found that the wording is unclear, but perhaps that’s just me

[crimson sedge]

that’s ok

[pulsar blade]

it's easy to think math is easy when you understand it

[crimson sedge]

So now you just have x and 1 right

[crimson sedge]

i never said that it was easy; i said that the wording was not confusing to me

perhaps it is easy to think that questions are confusing when you are confused

[crimson sedge]

yes, and x*1 is equal to x

[pulsar blade]

but we're literally helping a person that doesn't understand the question

[crimson sedge]

that doesn’t inherently make the question confusing

[crimson sedge]

So now how do you get to 28

sorry, i should specify: when i see a question being described as “bad because it is confusing”, i assume that there is some ambiguity in the question that could cause it to be answered in multiple different and valid ways, or even that there is not enough information to answer it at all

[pulsar blade]

I think a questions wording should never make a question harder than the content inside it

[crimson sedge]

if a question is more difficult than it should be, or has a trick or trap built into it (e.g. something holding for all x=0), then it is not “bad because confusing” but simply “difficult or has a pitfall”

[crimson sedge]

i don’t think that it did in this scenario

solving worded problems has always been a part of math

and learning to convert between the information provided by the question and mathematical equations is equally a skill that many tests will seek to assess

for me, i do not think there was an easier way to state the problem that would have conveyed the information in the same way

[crimson sedge]

initially i set there to be 4x dogs and 3x cats

if x=4, how many dogs and cats is that?

[crimson sedge]

Where did you find out x was 4?

[crimson sedge]

I know x is 4. But in your equation how did you get that

ok

do you first understand how i found the equation 4x-4 = 3x?

[crimson sedge]

Because 1x=4

[crimson sedge]

yes, so x=4

[crimson sedge]

Okay. I forgot about the = and the 4 on the other side. This is called the distributive law right?