help-38

[viscid wagon]

i thought i could do this one q but am forgetting

[carmine mauve]

i guess

so we don't need to subtract anything. because, once we start subtracting, we need to keep track of if we over-subtract, and then things get very hard to keep track of

[viscid wagon]

but the solution of counting one 7 and zero 7s then subtracting is fine right

i am putting it in the format

dy/dx + P(x)y = Q(x)

then i have to multiply everything by e^ (integral Px dx)

is that the correct first step

[carmine mauve]

and so you can simply fill out a grid to solve it

[carmine mauve]

maybe, but you'd need to keep track of if you're subtracting something you previously added back in because you previously previously subtracted to correct overcounting previous previous previously

things get hard if you try to correct by subtracting XD

[carmine mauve]

https://www.youtube.com/watch?v=DJsjZ5aYK_g

integrating factor method to solve this one

[viscid wagon]

i just did

5 digit number with no 7s : (8 * 9^4)

5 digit number with one 7:
leading 7: (9^4)
not leading 7: (4 * 8 * 9^3)

add them up subtract by total number of 5 digit numbers

[trim jolt]

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[carmine mauve]

anyway, the answer should be 7290

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[fleet notch]

why is b_n = abs(a_n) for the series (-1)^(n+1) /n

[nimble stone]

are you checking if it converges absolutely or something

[fleet notch]

im not sure

my instructors notes dont make sens

e

[nimble stone]

seems like youre doing the ratio test

[fleet notch]

he doesn't explain why b_n = |a_n| = 1/n. The 1/n would make sense in the context of a_n = (-1)^(n-1) or a_n = (-1)^n

nah this is the alternating series test

im just confused about why b_n = |a_n|. My intuition says to use pattern matching to locate the alternating term ((-1)^(n+1) in this case), and then to make b_n the rest of the expression

[ionic pendant]

if $b_n$ is always positive then [ \abs{a_n} = \abs{(-1)^{n-1} b_n} = \abs{(-1)^{n-1}} \abs{b_n} = b_n ]

[solid kiln]

cloud

[fleet notch]

why are we making the assumption that b_n is always positive via |a_n|? if b = 1/n, and n is [1, infinity), it can never be negative to begin with

[ionic pendant]

yeah that's why it works

[fleet notch]

right so then if it can never be negative, then why do we need |a_n|

because a_n contains b_n..

[ionic pendant]

because that's an easy way to denote "taking the non-alternating part", because taking the absolute value does that

[fleet notch]

okay, I guess that makes sense

why does it matter if the power of (-1) is n-1, n, n+1?

[nova spire]

it doesn't really, the sum will just change its sign

it doesn't affect convergence

[fleet notch]

weird... then why does the theorem state n-1?

if the theorem works for n-2, n, n+1, n+n, etc

[nova spire]

not n+n

[nova spire]

because

they could have chosen any of n-1, n or n+1

so they chose one

and that's it

the theorem remains valid if you change it to n

or n+1

but (-1)^(n-1) = (-1)^(n+1)

[fleet notch]

I wish he would have specified that

[nova spire]

just a quick tip

if $r\neq 0$, then $\sum a_n$ and $\sum ra_n$ have the same convergence type

[solid kiln]

rafilou is not not born in 2003

[nova spire]

so apply this to r = -1

(and if you do have convergence, then $\sum ra_n = r\sum a_n$)

[solid kiln]

rafilou is not not born in 2003

[nova spire]

so hopefully, you can see

that whether the power to which -1 is raised is n-1,n,n+1,n+2,etc...

[nova spire]

it won't matter for convergence

[fleet notch]

yeah because it's a geometric series, im sware

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[grave zephyr]

You really need some help

[grave zephyr]

Prove that limx→0 sin(x)
x = 1 with the squeeze theorem. Do not assume differentiability, sin here is defined
geometrically and not as a power series.

It didn’t paste well

So here’s an image out of the digital textbook

[nova spire]

the famous geometric proof

[grave zephyr]

I stuck

[nova spire]

have you tried doing a drawing?

[grave zephyr]

Ye

Ive been at this for an hour

[nova spire]

ok, here's a starter

for positive but small angles x, can you show sin(x) <= x?

with a simple drawing

comparing areas

[grave zephyr]

I can try

I have to go. I will answer your question when I have a chance.

[nova spire]

I have to go too, ping helpers if you're still stuck

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[trim jolt]

Please don't occupy multiple help channels.

[river cosmos]

I don't know how to do this 😭

It needs to be integrated with respect to x right?

[simple panther]

It might be better with y

If u integrate with x, u would have to take 2 parts

[river cosmos]

ohhhhh

I see

[atomic rain]

Y would be the move

[simple panther]

Wait, I have a doubt

If we integrate with y, won't the part to the left of y axis be removed from the total area?

As opposed to integrating with x

[atomic rain]

Oh

I just realized

[simple panther]

That would need 3 parts then lmao

[dapper swift]

yep, it would be easier to integrate with respect to x first

the curve with the greatest x is the purple curve, and the one with the least x is the red curve

[simple panther]

So, theoretically if the graph was shifted a bit more to the right and the area was completely in the first quadrant, y would be better?

[river cosmos]

The thing is I tried doing with respect to x in two parts (-1 to 1 and 1 to 4) but it didn't work

[dapper swift]

it's not about that

it's the fact that if you integrate with respect to y first, so your inner bounds are y = f(x) and y = g(x)

the lower bound is given by two different functions

so you would need two different integrals

[simple panther]

So u would have to split the integral into 2 parts. But if you do with y, then the upper is one function while the lower is one function. So you can just do it in 1 integral

[dapper swift]

yep!

[simple panther]

K, thanks for the clarification

[river cosmos]

Does anyone know why it didn't work though?

[dapper swift]

it definitely works though

what integral did you write out?

well, integrals

[river cosmos]

It was for a very similar question but the numbers are different

And it was wrong 😕

[dapper swift]

https://www.desmos.com/calculator/t2ny4kakth

[versed bloom]

you can’t just confuse her with double integration, nooooooo

[dapper swift]

oh I didn't know which class this was for.....

[versed bloom]

Sneak peak at Calc 3 then

[dapper swift]

wait you're doing single-variable calculus?? oh that makes a ton of sense

then yeah you need the two different integrals, but it'd be:

$$\int_{-1}^1 4 - (3 - x) \ dx + \int_1^4 (4 - 2 \sqrt{x}) \ dx$$

[river cosmos]

Wait dy dx?

Isn't it just dx?

[dapper swift]

corrected

[solid kiln]

south

[river cosmos]

.close

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[still star]

I have solved this question using a tree diagram and got the answer 7/27, however this is a question in a combination/probability unit, and I'm wondering if there is a alternative and more efficient solution then drawing a branching diagram. Any help is greatly appreciated!

[simple panther]

I think that's the logic

I can't think of anything else

[viscid flower]

theres 27 total possible?

yea

so 3^3

assume the cricket does return

either it visits all 4, or 3 , or only 2

oh 4 hops

so 81?

okay

[viscid flower]

enumerate these

theres 3 for 2

[simple panther]

That seems lengthier

[viscid flower]

its not really

its 3p1 + 3p2 + 3p3

over 3^4

its longer because i have to explain it

,w (3p1 + 3p2 + 3p3)/3^4

or im wrong

[solid kiln]

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Requested by janniku

[viscid flower]

thanks wolfie

[simple panther]

U'll get 5/27

[viscid flower]

no

[simple panther]

(3 + 6 + 6)/81 = 15/81 = 5/27

[viscid flower]

do get the wrong answer tho

missing

hmm

[simple panther]

I didn't get ur explanation

[viscid flower]

say we only visit 2 pad

wait impossible

so must no

[simple panther]

Why not

[viscid flower]

sorry i been up a long time

[simple panther]

1 -> 2 -> 1 -> 2 -> 1

4 hops

[viscid flower]

yea

okay

so thats 3p1

say we only visit 3

s a b a s

3p2

say we visit 4

s a b c s

3p3

right?

hmm

[simple panther]

Yes

[viscid flower]

6+6+1 isnt 15

so were even shorter

[simple panther]

But 3p1 is 3

[viscid flower]

oh lol

oh

[simple panther]

What about s a s a s

[viscid flower]

another way to visit 3

[simple panther]

s a s b s

[viscid flower]

thats right

[simple panther]

So another 3p2 here

[viscid flower]

sasas is already counted

[simple panther]

And 3p1 here

[viscid flower]

sasbs isnt

yea

there ya go

[simple panther]

Where?

[viscid flower]

in any 2

3p1

[simple panther]

No?

[viscid flower]

ann 😭

be nice im tired

lol

[simple panther]

Lol

U better sleep

[trim lichen]

wait so does this problem ask for P(cricket comes back to leaf 1 at least once) or P(cricket is on leaf 1 at the end of the 4 hops)?

[viscid flower]

its already counted sasas i mean

[simple panther]

How?

I don't think it did

[viscid flower]

thats the first 3p1

[simple panther]

Oh u mean that?

[viscid flower]

because if the guy only hits 2 pads

[simple panther]

I still think branching is better and easier

[viscid flower]

it must be sasas

noooo

[simple panther]

Than taking cases like this

[viscid flower]

its only easier if u cant count

cases are faster

[simple panther]

Like, u mean that is easier than this?

[viscid flower]

yes

you scribble

u get it

its fast

i got it and i havent slept or eaten and have no paper

[simple panther]

U got it wrong though

[viscid flower]

first answer is always wrong

you gotta check yourself

idk different strokes

this seems faster to me

[simple panther]

sasas -> 3
sabas -> 6
sasbs -> 6
sabcs -> 6
Total: 21
21/81 = 7/21

[viscid flower]

ez

[simple panther]

@still star how about this solution?

[viscid flower]

im sure there a group theory solution

but not from me

im just a counter

[zenith egret]

Ig I got it

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[gentle viper]

so i got 2≠0 which is a contradiction so the system doesn’t have a solution

how do I solve this problem correctly

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[simple panther]

@zenith egret

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[fair bison]

can you explain what the work you have so far is? it's hard for me to read and follow your logic

[gentle viper]

this is the work sorry

forgot to send the first picture @fair bison

[urban copper]

what did TA say

I think if a = 1/2 then (-1,1,-4a) is a linear combination of (1,-1,2)

[gentle viper]

im going in about 12 hours to her

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[urban copper]

what happened

[cunning cliff]

I think this problem becomes rather simple if you note the 1st, 3rd, 4th image vectors are ... related in some way

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[foggy fossil]

Pls help

Plssss

[native shuttle]

This channel is occupied right

[frail heron]

!occupied

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[fervent leaf]

I don't get what's underlined

[fervent leaf]

Why is that true

Like why is it a submatrix of the original matrix what part of that is true?

Like it is but why?

Oh god I got it

Effing hell I hate notation

Anyway

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[little compass]

help pls

[little compass]

<@&286206848099549185>

[carmine mauve]

Is that a zero or a letter O?

[little compass]

zero

its the 0 block matrix

idk where to start on this

i tried cofactor expansion but that gets very messy

[dull island]

definition of determinant with the permutations

going from bottom to top

[carmine mauve]

you can get rid of B with jordan elimination

[dull island]

only contributions from C can contribute to the determinant

[little compass]

B doesn't rly matter bc you can just make it triangular

[dull island]

then after that only contributions from A matter

[little compass]

ok ill try this

[dull island]

then fix the algebra after

that's how I'd do it, although I'm not sure how I'd formalise it

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[dim creek]

hello, i'm trying to calculate the probability of a sum of n uniform random variables x_i over some symmetric interval [-y, y] falling into that same interval. now, considering i want to use the irwin-hall distribution, i must transform the variables x_i to some other variables u_i over the interval [0, 1], and i used u_i=(x_i+y)/2y, and in the context of the sums its U_n=(X_n+ny)/2y. the problem is in the integration bounds of the pdf, im not quite sure if im supposed to set the bounds at 0 and 1 or do i calculate them by expressing P(-y≤X_n≤y) in the form of U_n which ends up being P((n-1)/2 ≤ U_n ≤ (n+1)/2). thanks in advance!!

[dim creek]

if somethings unclearly written or explained feel free to say so, i can explain further

<@&286206848099549185>

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[brisk rain]

girlll omg I’ve been trying to get help the past 20 min too 😭

[dim creek]

they got a shortage of helpers 😭🙏

but it's okay let me close this because after staring at the problem for 30 minutes i get why it would be (n±1)/2

.close

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[twilit estuary]

How do I find u x v and how do I find the angles between u and v I thought py thag would work but I got decimals. The left one is v the right one is u

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[marsh forum]

so I was thinking

this function is the gradient of $\ln( (x^2+y^2)^{3/2})$

[solid kiln]

What a wonderful world !

[marsh forum]

so shouldn't it, being a conservative field, have the line integral along a closed loop, as 0

[simple haven]

,w ∇(ln((x^2 + y^2)^(3/2)))

[marsh forum]

hmm, the y coordinate isn't negative

ncm

[solid kiln]

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[marsh forum]

I mean the y coordinate can be solved using a 1/3 term

[simple haven]

Right scaling constants are easy to take care of

[marsh forum]

,w gradient of arctan(y/x)

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[marsh forum]

yeah, that works

[simple haven]

There we go

[marsh forum]

but how is it 2π if C^+ is any counterclockwise oreinted simple close curve around origin

[simple haven]

At (0,0) we have a pole

[marsh forum]

a what now

[simple haven]

A "team rocket is blasting off to infinity again"

Anyway, conservative fields have a number of interesting properties. One of which is that closed loops around poles pick up a constant for each circuit around them

This is similar to the residue theorem from complex analysis, and while I'm pretty sure they're the same under the hood, there might be a subtlety that I'm missing.

[marsh forum]

Hmm, okay

[simple haven]

Well, ok so "conservative"

Obviously this isn't a conservative field

[marsh forum]

This is from a section on green's theorm

[simple haven]

Because by definition

But it's close, and as long as you do not circle the origin, it works

[marsh forum]

because it isn't defined at (0,0), right

or anywhere on the y axis for that matter

[simple haven]

I'm specifically talking about the derivative.

Not arctan

[marsh forum]

Ah

wait, I thought a field was conservative, if it had a graident function

[simple haven]

We can "fix up" the arctan by adding that when x = 0 then it's pi/2 or -pi/2 depending

In this case it's only broken at (0, 0)

[marsh forum]

so how do I do part (b)

[simple haven]

Use the theorem and work the area integral?

[marsh forum]

Yeah, but doesn't green's theorm require me to know the curve

[simple haven]

$\oint_C (L , \dd{x} + M , \dd{y}) = \iint_D \qty( \pdv{M}{x} - \pdv{L}{y}) , \dd{A}$

[solid kiln]

OmnipotentEntity

[simple haven]

Pick a rectangular or circular region, and show that the result doesn't not depend on the region variables.

(except when it contains the origin)

[marsh forum]

When it doesn't contain the origin sure

[simple haven]

Like, you can do a circle centered at (x, y) which radius, r, less than √(x^2 + y^2)

And then radius greater than

I'm nearly dead tired. So who knows if I'm even coherent at this point.

There are probably smarter ways to do this?

I'm going to try to sleep a bit, all the best

[marsh forum]

thanks

.close

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[burnt sable]

permutation multiplication has me confused just need a little bit of help please

[burnt sable]

I understand the representatives from K being used to make the cosets but I am confused on the computing of the cosets themselves

[broken pilot]

you take every element in H and multiply on the right by (23) to get the coset H(23)

[marble wharf]

they changed the "order" of the elements in their solution

so first element of H times (23) isnt the first element of H(23) etc

[burnt sable]

ohh how did i not see that lol

i think it get it now

yeah that makes sense now thanks

.close

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[worldly ginkgo]

im in way over my head with piecewise scaling math. Can anybody help me?

[simple panther]

!da2a

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[worldly ginkgo]

hey bro!

can you help me maybe?

[simple panther]

just ask the question

anyone who can answer will reply

[simple panther]

I'm note sure as i didn't see the question yet

[broken pilot]

Could you send a specific question you can’t figure out? We’ll be able to help you better then

[worldly ginkgo]

yes ofc! sorry guys

so im working on some math for a game. So the characters in the game has a "stance" pie chart. There is three different stances (Fire, water and grass). The purpose of the stance chart is give a multiplier to your attack damage ranging from 0.5 (half damage) to 1.5 (50% more damage)
There is also 3 diffrent attack types (fire, water and grass)
When battle starts the stance is always 33% 33% 33% (this results in a 1.0x)
So if a fire attack is used, i want the multiplier to be 1.5 if the stance is 100% fire. If the stance is 100% water (which makes it weaker) thats where it would give a 0.5 multiplier. And grass doesnt affect it directly.

I hope im making sense. Ill gladly answer any questions ofc

[simple panther]

what help do you need in this?

and, it doesn't appear to be a math question

[broken pilot]

So what you want is a piecewise function to define the correct attack damages?

[simple panther]

oh, now it makes sense

[worldly ginkgo]

exactly piecewise yes!

im in sooo deep guys i could cry

i have been wrestling with chat gpt for hours, i dont know if this helps you guys

[broken pilot]

Can you have fire stance and water attack type?

[worldly ginkgo]

yes you can! in that case it would be a 0.5 multiplier, if its a 100% fire stance

[simple panther]

what if i'm doing to a fire attack and my stance is 80% fire and 20% water?

[broken pilot]

Chatgpt can be wrong and yet still sound right, and its outputs can mislead more than help, so I wouldn’t recommend asking it for textual answers, maybe programs would be better

[worldly ginkgo]

then you would get a multiplier that is a little under 1.5

[simple panther]

ok, makes sense

[worldly ginkgo]

water affects it negatively

[broken pilot]

so there are 3 x 3 x 3 x 3 = 81 combinations of stance and attack type, for both opponents?

oh wait it’s just an attack and defense isn’t counted?

[simple panther]

fire attack dmg bonus = 0.5 * (fire stance)/(fire stance + water stance) - 0.5 * (water stance)/(fire stance + water stance)

does this make sense?

[worldly ginkgo]

defence is also calculated, but its the exact same formuler i wanna use for that, just in reverse if you know what i mean

[worldly ginkgo]

not really sorry

[simple panther]

if fire = 100, water = 0; bonus = 0.5 * 1 - 0.5 * 0 = 0.5

[broken pilot]

actually, can you give some examples?

[worldly ginkgo]

yes!

[broken pilot]

we might have understood wrong, so you could explain using an example

[worldly ginkgo]

so aslong as fire and water are equal, it should always give a 1.0 multiplier

[broken pilot]

can you be even more specific

[worldly ginkgo]

For now on, lets assume we are always using a fire attack

[broken pilot]

say A is in a battle, using a fire attack, then what

[worldly ginkgo]

then we need to figure our what multiplier is added to the damage, from 0.5 to 1.5

and its all about how much of the stance is water, and how much is fire

[simple panther]

where does the above formula fall short?

[simple panther]

this formula

[worldly ginkgo]

i cant create a formular for it

let me give you more examples guys

[broken pilot]

So if, say, A is in a battle, using a half fire half water attack, on a stance which is a third of all of fire, water, and grass, and its original damage is 50, then we can calculate the actual damage?

[worldly ginkgo]

all atacks are either fire water or grass

[broken pilot]

But stances can be part of each

[worldly ginkgo]

this looks confusing, bc rage is fire, tactical is water and shadow is grass. But the end results looks kinda right

[worldly ginkgo]

ye sir

they can change doing combat

but start at 33% 33% 33%

[broken pilot]

wait, now you’ve never mentioned these terms before

[simple panther]

i think its just shadow, tactical and rage instead of water, fire and grass

[broken pilot]

are the stances called “shadow”, “tactical”, and “rage”, or “fire”, “grass”, and “water”?

[worldly ginkgo]

i changed the terms to fire water and grass to make it easier to understand sorry

[worldly ginkgo]

exactly, sorry guys im not making it easy for you

[broken pilot]

Ok, then, if B is in a battle, using a fire attack, on a stance which is a third of all of shadow, tactical, and rage, and B’s original damage is 50, then it is possible to calculate the real damage?

[worldly ginkgo]

asked chat gpt to change the terms

[mystic pilot]

can somone help me out in help-8 please 😭

[worldly ginkgo]

this would result in 50 damage

the multiplier would be 1.0

as long as fire and water are equal, it should always be 1.0

[worldly ginkgo]

this is using a grass attack btw

[broken pilot]

Now I’m confused. Are these properties of the stance, or the attack?

[worldly ginkgo]

the stance dictates the multiplier

[broken pilot]

A stance can be grass, or tactical, or rage, or water, and so on?

[worldly ginkgo]

no no, lets forget about rage, tactical and shadow. Those were just me changing th terms so it would be easier to understand for you guys (like pokemon)

[simple panther]

how did u get the number 1.375

is it random?

or is there a pattern

?

[worldly ginkgo]

let me find the calculation chat gpt gave me one sec

remember, this is always using a grass type attack

[simple panther]

Is it necessary to take a piecewise function?

won't it be better if we take a linear one?

[worldly ginkgo]

that would result in the calculations being wrong i found out after many conversations with chat gpt.

let me find a wrong one for you

[simple panther]

sure

[worldly ginkgo]

then this happens

its the same stance as before, but the multiplier gets lower then 1.0

[broken pilot]

Have you checked all of its calculations?

[simple panther]

can you see the last step

[worldly ginkgo]

im not very math heavy sorry, i dont think i would be able too

[simple panther]

its 0.5 + 0.5 - 0.125 = 1.375

i think u should be able to tell that it's wrong

[worldly ginkgo]

why is it wrong?

[simple panther]

...

[broken pilot]

,calc 0.5 + 0.5 - 0.125

[solid kiln]

Result:

0.875

[worldly ginkgo]

omg

why would it say something else=

[simple panther]

cause its ai

!nogpt

[trim jolt]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[broken pilot]

Chatgpt is built to complete sentences, not to do maths

[worldly ginkgo]

i see, didnt know that guys

[simple panther]

i think that one should work the best given your situation

[worldly ginkgo]

can you guys help me calculate some stance examples using this formular?

[broken pilot]

Bit weird that it doesn’t symmetrically include grass

[simple panther]

yeah, so for 50 fire, 25 of other each

[simple panther]

thats for fire type

,calc 0.5 * 50/75 - 0.5*25/75

[solid kiln]

Result:

0.16666666666667

[simple panther]

so its 1.16 times boost

honestly, it feels too much to have 1.3 for just 50 fire and 25 water

[worldly ginkgo]

1.16 seems a little low?

[simple panther]

idk

[worldly ginkgo]

i agree!

[simple panther]

cause fire is just 50 while water is at 25

[worldly ginkgo]

let me come with another stance example

[simple panther]

if fire would be 75 while water is 25,

[worldly ginkgo]

would that be 1.25?

[simple panther]

,calc 0.575/100-0.525/100

[solid kiln]

Result:

0.25

[worldly ginkgo]

make sense bro

[simple panther]

1.25?

it feels less

[worldly ginkgo]

we are on to something here

[broken pilot]

Your maximum amplification is 1.5, and the minimum is 0.5, right?

[worldly ginkgo]

yes sir

only way to get 1.5 is 100% grass stance

[broken pilot]

with fire attack?

oh we fix fire attack since it’s symmetrical

[worldly ginkgo]

only way to get 0.5 is 100% fire

for grass attack

and aslong as grass and fire is equal it should always give 1.0

can you try the formular on some of these stance examples?

[broken pilot]

would all water stance and fire attack also give 0.5?

[simple panther]

i think there might be a loophole if the formula i gave

[worldly ginkgo]

100% water would give 1.0

[simple panther]

if fire is 1 while water is 0, it would still mean 1.5x

ig replacing the denominator with 100 would be the best

[worldly ginkgo]

i have 3 indian programmers on my team, with pretty math heavu educations, and we cant figure this out lol

[simple panther]

is this, like, an official play store game you are making

or just a fun project?

or a web game?

[worldly ginkgo]

yes guys! you are welcome to take a look at it!

[summer tapir]

ok, i would recc doing 50% and 200% or 66.6666...% and 150%
not
50% and 150%

[worldly ginkgo]

going to be on steam is the plan

[simple panther]

i like how it defies the stereotype

that indians are good at math

[worldly ginkgo]

ill even put you guys in the credits hehe

this has been a problem for us for a long time

(well 1 month)

[simple panther]

that's waaaay too long

[summer tapir]

why are you going for 50% and 150%?

that is a bit weird as 50% * 150% = 75% < 100%

or are these additive effects?

[simple panther]

ig additive

[summer tapir]

in which case
i would reccomend
fire damage multiplier = 1.5*firestance + 1*grass stance + 0.5*water stance

[broken pilot]

Let me just restate the rules to understand them better:
If B is in a battle, using a fire attack, on a stance which is a third of all of fire, water, and grass, and B’s original damage is 50,
then since fire and water are in the same proportion,
the total multiplier is 1, so
the damage is 50

[worldly ginkgo]

ÿes sir

[summer tapir]

how is this?

[simple panther]

but ig shouldnt depend on grass

should it? @worldly ginkgo

[summer tapir]

this is grass percentage

[worldly ginkgo]

lets use grass attacks as the default now

[broken pilot]

no can we stick with fire

fire attack

[simple panther]

then why does fire have 1.5

[broken pilot]

or it’ll be confusing

[worldly ginkgo]

as long as we all agree 😄

[summer tapir]

well, because if you have full fire, then you have 1.5x damage

[simple panther]

but thats only for fire attacks right?

u said its grass

[summer tapir]

the grass stance is grass percentage

[worldly ginkgo]

okay, just so we are on the same page, We are using fire attacks as the default typ.e Fire stance invreases the multiplier, and water stance decreases it

[simple panther]

oh, thats what u meant

wolf, should fire attack depend on grass stance?

[worldly ginkgo]

not at all

[summer tapir]

grass stance + fire stance + water stance = 1
right?

[broken pilot]

And grass stance does nothing?

[simple panther]

thats what i meant here

[worldly ginkgo]

10% fire, 10% water and 80% grass should give 1.0

[summer tapir]

fire damage multiplier = 1.5*firestance + 1*grass stance + 0.5*water stance
@worldly ginkgo should do the trick

[worldly ginkgo]

15% fire, 5% water and 80% grass, should give a very minor damage boost

[simple panther]

but it is dependent on the grass stance

[summer tapir]

not exactly

[simple panther]

it shouldnt be

[summer tapir]

it doesnt exactly depends on grass stance

[broken pilot]

ok that clears it up a lot. So it seems the formula for the multiplier on a fire attack should be 1 + 1/2 x (fire stance percentage) - 1/2 x (water stance percentage), if you want the maximum to be 1.5 and the minimum to be 0.5

[simple panther]

wait, let me recheck myself

[summer tapir]

i am just using grass stance = 1- waterstance - firestance

[worldly ginkgo]

can you try giving me a few examples using this?

[summer tapir]

is water stance = fire stance
this would give exactly one multiplier

[simple panther]

we'll get 1 * (fire stance + grass stance)

and it isnt 1

[summer tapir]

in 0.15 fire, 0.05 water, and 0.8 grass, this would give 1.05 damafe

[summer tapir]

no

we would get

[simple panther]

how

[broken pilot]

Oh, your formula is now equivalent to mine using grass stance = 1 - water stance - fire stance

[summer tapir]

water stance + fire stance + grass stance = 1

[simple panther]

yea

[worldly ginkgo]

that sounds good

[simple panther]

oh

wait, i get it

nvm

[worldly ginkgo]

what about 70% fire, 20% water and 30% grass?

[summer tapir]

1.45 damage

[worldly ginkgo]

thats waay to much bro

[simple panther]

,calc 1.5 * 0.7 + 1 * 0.3 + 0.5 * 0.2

[solid kiln]

Result:

1.45

[worldly ginkgo]

remember only 100% fire gives 1.5

[broken pilot]

,calc 1 + 0.50.7 - 0.50.2

[solid kiln]

Result:

1.25

[worldly ginkgo]

looks more correct

[summer tapir]

theres a mistake in this, 70 + 20 + 30 = 120

[broken pilot]

Those don’t all add up to 100%

[worldly ginkgo]

lol i didnt notice!

sorry guys

what about 70% fire, 20% water and 10% grass?

[summer tapir]

1.25

[worldly ginkgo]

seem right!

seems*

[simple panther]

,calc 1.5 * 0.7 + 1 * 0.1 + 0.5 * 0.2

[solid kiln]

Result:

1.25

[simple panther]

yep

[worldly ginkgo]

this looks very promising!

[broken pilot]

,calc 1+0.50.7-0.50.2

[worldly ginkgo]

can i give you a few more stance examples?

[solid kiln]

Result:

1.25

[summer tapir]

sure

[broken pilot]

You can plug the numbers in yourself and see if they agree with your intuition, right?

[worldly ginkgo]

lets say 40%fire 30%water 30% grass

[broken pilot]

,calc 1+0.50.4-0.50.3

[solid kiln]

Result:

1.05

[summer tapir]

1.05

[worldly ginkgo]

seens right

80 grass, 10 and 10

[summer tapir]

1

[simple panther]

isnt that similar to the formula i gave?

[worldly ginkgo]

should be around 1.4 ish?

[summer tapir]

80 grass, 10 water, 10 fire?

that should give 1

right?

[worldly ginkgo]

oh i was thinking grass attack. lets go with fire

80 fire and 10 10

[simple panther]

,calc 1 + 0.8 * 0.5 - 0.5 * 0.1

[solid kiln]

Result:

1.35

[worldly ginkgo]

seems right

[broken pilot]

what was yours?

[worldly ginkgo]

50% fire and 25 25

[simple panther]

that thing,
0.5 * fire/100 - 0.5 * water/100

for bonus dmg

[broken pilot]

oh, yes, it’s the same

[simple panther]

oh, ok

[broken pilot]

If you know algebra, you could substitute them yourself, right?

[simple panther]

1.25 ig

iirc

[worldly ginkgo]

im so bad at math bro. skipped all classes in highschool lol

[simple panther]

we did it earlier too

[worldly ginkgo]

bites my ass now

[simple panther]

lol

[worldly ginkgo]

okay we got it guys, thank you sooooo much

the indians are going to loose their mind when i come with the solution

[simple panther]

haha lol

[worldly ginkgo]

if you guys want to be in the credits of the game, send me a dm with what name you want. could either be online name or irl name

[summer tapir]

also, rememebr, this only works if your multipliers are additive

[broken pilot]

I suggest that you learn how to substitute values for variables in equations or functions, it’s not that hard to learn

[simple panther]

it wouldnt make sense to make them multiplicative

everything would be less than 1 (as its a multiplier) and it would end up decreasing

[broken pilot]

if you’ll put anything, just put soup_norm, thanks

[worldly ginkgo]

noted

[summer tapir]

it does

[simple panther]

unless you do 1.2 * 1.5 directly

[summer tapir]

yes

thats what im saying

[simple panther]

ok

[worldly ginkgo]

is there a problem with the formular?

[simple panther]

And for me, Suika is fine

[summer tapir]

not rlly, just you might want to change it, if you are doing multiplicative later

[broken pilot]

No, just discussing some other related things

[worldly ginkgo]

noted

[broken pilot]

wait, we haven’t even done the grass attack case, right??

or the water attack

[simple panther]

its similar, isnt it

just swapping out the variables here and there

[summer tapir]

it is

[worldly ginkgo]

exactly!

[summer tapir]

ok, but one thing, you do realise 0.5x is much lower then 1.5x is high

[worldly ginkgo]

okay, now i have to give the indians a formular

what do i tell them exactly?

[broken pilot]

the formula we put above

either of the three work

[simple panther]

but thats just how it is in games

[broken pilot]

they’re all equivalent

[summer tapir]

not in all games

some games do 0.5x and 2x

[simple panther]

ah, you mean like that

[summer tapir]

or 2/3x and 1.5x

[simple panther]

yeah

they usually go from 1.5 to 3

or some exotic weapons stuff go to 5

[summer tapir]

no that

[simple panther]

somehow

[summer tapir]

i am talking relative to base damage

[simple panther]

okay

[worldly ginkgo]

im debating with myself if i should change the range to 0.5 to 2.0

[broken pilot]

either of:

bonus damage = 0.5 * fire/100 - 0.5 * water/100
multiplier = 1 + 1/2*fire - 1/2*water
multiplier = 1.5*fire + 1*grass + 0.5 * water

[simple panther]

i meant accessories

[worldly ginkgo]

but the thing is, that the one recieving the damage also uses the stance system but in reverce

[simple panther]

some are just cheatcodes

[broken pilot]

If you learned a little algebra, it would be much easier to tinker with these things yourself

[summer tapir]

then do multiplicative

[worldly ginkgo]

its a dead game bro. cant learn it lol

[broken pilot]

That’s no obstacle to defense, unless the systems for defense and attack must interact

[summer tapir]

if you do that
try
(2)^fire*(1/2)^water

[broken pilot]

What do you mean

[worldly ginkgo]

the first calculation is how much damage is recived (thats what we have been discussing here)

the next calculation is what stance is the recieving part, but thats just this formular in reverse

[worldly ginkgo]

im so bad at math, ive given up man lol

sorry guys, but what is the formular exactly?

bonus damage = 0.5 * fire/100 - 0.5 * water/100
multiplier = 1 + 1/2fire - 1/2water
multiplier = 1.5fire + 1grass + 0.5 * water

is it this?

[broken pilot]

I’d really recommend trying to learn, because at this point it’s a little like you’re in a wheelchair and we’re wheeling you around

[broken pilot]

it is this for fire attack

[worldly ginkgo]

thats my level you know, cant change that

[broken pilot]

It is always possible to improve

anyway

[broken pilot]

this doesn’t factor in the type of attack; it assumes that the type is fire attack

[worldly ginkgo]

yearh, so changing fire/grass/water to the specific attack will fix it, right?

[summer tapir]

yup

[broken pilot]

I’m not clear on how the interactions will change

[summer tapir]

also, pls keep this noted down as an alternative

in case you want to do
0.5x to 2x

[broken pilot]

fire stance boosts fire attack and water stance does the reverse for fire attack, but what happens for water attack and grass attack?

[worldly ginkgo]

depends on how the stance % is allocated

[summer tapir]

i assume its
grass->water->fire->grass

[worldly ginkgo]

yes sir!

just like pokemon

can we do an example where fire is 20%, water is 50% and grass is 30% ?

[broken pilot]

What is the attack type?

[worldly ginkgo]

fire!

[simple panther]

,calc 1 + 0.5 * 0.2 - 0.5 * 0.5

[solid kiln]

Result:

0.85

[worldly ginkgo]

hmmm

[simple panther]

So 0.15 less

[worldly ginkgo]

seems like too much no?

i would expect it to be lower

[simple panther]

But it's only 50 water while still having 20 fire

It's linear so it's fair and square, it's just human mind messing with us

[worldly ginkgo]

is we reverse fire and water, it should give 1.15 right?

if*

[simple panther]

Yes, I think so

,calc 0.5 * 0.5 - 0.5 * 0.2

[solid kiln]

Result:

0.15

[simple panther]

1.15

Yes

[worldly ginkgo]

why do you put exactly these numbers in?

im really trying to understand, sorry if i seem stupid bro

and does this look correct?

[trim jolt]

@worldly ginkgo Has your question been resolved?

Channel closed

Closed by @worldly ginkgo

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[broken pilot]

Because 20% = 0.2 = fire stance percentage, and 0.5 = 50% = water stance percentage

[simple panther]

U said "if we exchange the 50 and 20 in fire and water" right

[broken pilot]

You're never stupid if you're genuinely trying to learn

[simple panther]

Yes

[trim jolt]

@broken pilot Has your question been resolved?

[worldly ginkgo]

im trying to write down what to tell the guy that has to implement this, and what ive written down is this

but im not sure how to explain it exactly

can it be simplified somehow?

[simple panther]

Pick the one that feels the most intuitive for you

[worldly ginkgo]

so all three lines basicly does the same?

[simple panther]

Yes

I would say the one in the middle is easier

[worldly ginkgo]

you guys have been super helpfull. thank you all of you!

[trim jolt]

@broken pilot Has your question been resolved?

[broken pilot]

.close

[trim jolt]

Channel closed

Closed by @broken pilot

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[ember mortar]

can anyone point me in the right direction?

currently i worked out that a.b=0
so i have 5x+3y+z = 0

[ember mortar]

I tried turning them into parametric form but it doesnt seem solvable like that

I also have that
L2 is
$r = i + 5k + u(xi+yj+zk)$

[solid kiln]

outletproblems

[wraith hinge]

we know two things:

1 .the dot product of the direction vectors of L2 and and L1 is clearly going to be 0

2. Line L2 passes through <1, 0, 5>

[ember mortar]

agreed

[wraith hinge]

can you try to use this information to find L2?

[ember mortar]

I tried, but the direction vector i cannot seem to find

The position vector is OK, that is simply A.
but the direction vector results in

$5x+3y+z = 0$

[solid kiln]

outletproblems

[ember mortar]

$a.b=0\n$
$(5.x) + (3.y) + (1.z) = 0$

[solid kiln]

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

[ember mortar]

how do i new line

[wraith hinge]

get your point mate

dw

[ember mortar]

ok my bad sorry

but im slightly lost since there is not much else to work with

you can try put them in parametric form, but that results in 5 unknowns and 4 equations so its unsolvable

[wraith hinge]

You have an infinite number of solutions mate

Like there are an infinite number of responses you can givve

Have you worked with infinite solutions before @ember mortar

Let me give you an illustration of an answer

[wraith hinge]

@thorn crown please post this in a different channel

this one is already occuipied

[thorn crown]

oh sorry

[wraith hinge]

refer to #❓how-to-get-help

Let us say the direction vector of $L2$ is $d2$ and $L1$ is $d1$

[solid kiln]

Edmund Cloudsley

[wraith hinge]

then we know that
$$d1 \cdot d2 = 0$$

[solid kiln]

Edmund Cloudsley

[wraith hinge]

$$<5, 3, 1> \cdot <x, y, z> = 0$$

[solid kiln]

Edmund Cloudsley

[wraith hinge]

let us say for convenicence, I take the values of x to be 0 and y to be 1

then

$$5(0) + 3(1) + z(1) = 0 \implies z = -3$$

Therefore I have

[solid kiln]

Edmund Cloudsley

[wraith hinge]

$$d2 = <0, 1, -3>$$

[solid kiln]

Edmund Cloudsley

[wraith hinge]

and then I combine this with the fixed point

$$r_2 = <1, 0, 5> + \mu <0, 1, -3>, \text{ where } \mu \in \mathbb{R}$$

[solid kiln]

Edmund Cloudsley

[wraith hinge]

you can verify for yourself that:

1. the direction vector and its dot product with the cirection vector of line 1 yields 0
2. this line passes through (1, 0, 5)

[trim jolt]

@ember mortar Has your question been resolved?

[ember mortar]

ok i think i understand

we are inputting values for the variables and comparingthem, right?

[wraith hinge]

what grade is this btw?

[ember mortar]

advanced level

undergraduate

i think

[wraith hinge]

We are just

--> picking random values for x and y --> finding a value for z

If you want a GENERAL solution:

[ember mortar]

everything in terms of z and then replace with another variable

right

but yeah that makes sense

[wraith hinge]

you know that:
$$5x + 3y + z = 0$$

[solid kiln]

Edmund Cloudsley

[wraith hinge]

and we also know that

there is a fixed point

we can keep either x, y and z as the free variable

and then get a general solution for the direction vector

[ember mortar]

one other thing

how do you know that you can use 0, and 1 for x and y?

Or can you use that since it is a direction vector it isnt bound as a position on the line?

does that make sense

x y z can be any value and be valid in other words?

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[marsh forum]

Trying to find the Global extrema of this function.
\
This is my working:
\
\
We start by computing the gradient
\
$\grad{f} = (2x+2y, 2y+x^2)$
\
\
Equating the gradient to 0, we find that the only critical point is on the origin.
\
\
nvm

[solid kiln]

What a wonderful world !

[marsh forum]

oops

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[main sigil]

The questiion seems to be incomplete

[marsh jacinth]

we have to show that after applying this operations a finite number of times, both urns can be made empty

[main sigil]

The only way you can empty them is if they both have equal number of balls

[marsh jacinth]

no

[main sigil]

i mean in one step

[marsh jacinth]

this is what I have done till now

[main sigil]

if it can be equal, what if you got stuck in an infinite loop situation?

[marsh jacinth]

So now I have to prove that in the second bowl I can get the number of bowls to be in the power of 2.
The number will lie between 2^n-1 and 2^n obviously, so lets say that the number is 2^n-1 + a then we have to bring one of the urns to 2^n-1 and subtract 1. After repeating this process 'a' times the number will be in power of 2. QED?????

[marsh jacinth]

I meant it always decreases or remains the same

ignore that though, that is not relevant

[main sigil]

Alright, i thought that it's how you're arguing that the algorithm will end in finite time

[marsh jacinth]

if I just add this to what I wrote is the proof complete?

[main sigil]

After repeating this process 'a' times the number will be in power of 2.

this part is not entirely clear to me

[marsh jacinth]

if we repeat this process 'a' times then from 2^n-1 + a we will get 2^n-1

[main sigil]

Are you sure about that? Why is it?

it will probabably not even be 2^(n-1) anymore

it would be some lower power of 2 i think

[marsh jacinth]

you understood that in one urn we get 1 and in the other urn we have someother number?

[main sigil]

Yeah

[marsh jacinth]

lets call them urn 1 and urn 2

urn 1 has 1

urn 2 has some unknown number

if the number of balls in urn 2 is in power of 2, then we can just double the balls in urn 1 until they both have an equal number of balls

and then we can just subtract

we have to prove that somehow we can get the number of balls in urn 2 can be brought to a number in the power of 2

that number will lie between 2^n-1 and 2^n

lets say that number is 2^n-1 + a

then we can double urn 1 until it reaches 2^n-1 and subtract 1 from both urns

we just have to do this 'a' times

[main sigil]

i think this is false

[marsh jacinth]

mb

we double urn 1 so urn 1 will have 2 right?

[main sigil]

yeah, sure

[marsh jacinth]

and urn 2 has 2^n-1 + a right?

[main sigil]

2^(n-1) + a, no?

yeah

1 --- 2^(n-1) + a
2^(n-1) --- 2^(n-1) + a
1 --- a + 1

i think that you'll reach this

[marsh jacinth]

so now subtract 1 from both sides

[main sigil]

oh, right, i see now

yeah, thats better

[marsh jacinth]

thats what I was trying to tell

[main sigil]

well here it seemed different

[marsh jacinth]

so if I add this to what I had written is the proof complete?

[marsh jacinth]

that was wrong

look at all the msgs after I said 'mb'

[main sigil]

Yeah, if you explain the method properly then it will work

[marsh jacinth]

okay

thank you

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[solid kiln]

waffle

[turbid brook]

Is this correct?

[dapper swift]

a, b, d are correct

[turbid brook]

Oh sorry

[dapper swift]

for e you're so close: it's $4! \times (2!)^4$

[solid kiln]

south

[turbid brook]

Yes

[trim lichen]

c skipped?

[turbid brook]

I need help with (c)

[trim lichen]

ah.

[turbid brook]

I think it's similar to the one we discussed before

[trim lichen]

this is the same as when you were dealing with the 3 boys and girls

[dapper swift]

I mean I could have not spoilt it, but it's that answer for the reason you have described (each of the two couples)

[trim lichen]

FMFMFMFM or MFMFMFMF are the only possible gender arrangements

[turbid brook]

Is it 4! times 4! times 2!?

[trim lichen]

yes

[turbid brook]

Okay

[dapper swift]

nice strategy

much better than subtracting every arrangement that breaks

[trim lichen]

bit late to the punch there

[dapper swift]

haha it happens

[trim lichen]

was talking to shreyjanha

[turbid brook]

.close

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[dapper sparrow]

how can i get the intersection between the first two functions, like i get its always f(x) = g(x) , but honeslty my brain has been kinda fried recently, so if could someone help me find the intervals for the first two id rlly appreciate that..

[dull pilot]

um

desmos and wa are your best friends

[dapper sparrow]

yeah but my teacher wants me to calculate it by hand

bc im doing it for a project and he said this will boost my mathematical skills or smth

[split chasm]

logs and polynomials don't mix well

[dull pilot]

is he russian. /hj

[dapper sparrow]

yeah ik itll be difficult but that what i could get to model the guitar as u can tell.. thats why im asking… bc idfk how to do log and poly

so if anyone knows lmaoao

[dapper sparrow]

nah but slavic yeah

[dull pilot]

that explains it

[split chasm]

no practical way to solve
you can use stuff like newtons-raphson method
or midpoint approximation
to get decent approximations of the intersections

[dapper sparrow]

hmm..

[trim jolt]

@dapper sparrow Has your question been resolved?

[trim jolt]

@dapper sparrow Has your question been resolved?

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[marsh forum]

Trying to prove this

nvm, just realized the problem said assuming the divergence therom .. , which is in $\R^3$

[solid kiln]

What a wonderful world !

[marsh forum]

.close

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[marsh forum]

.reopen

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✅

[marsh forum]

yea jan?

[viscid flower]

nothin

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[marsh forum]

We start by re-iterating that $\int_{C} (f_y,f_x) \cdot (dx,dy) = \iint \pdv{f_x}{x} - \pdv{f_y}{y} dA$
\
As the laplacian of $F$ is $0$,it follows that $\pdv{f_x}{x} - \pdv{f_y}{y} =0$
\
Thus we have $\iint_{S} 0dA=0$
Thus proving the path independence

[solid kiln]

What a wonderful world !

[jolly saffron]

yeah

[marsh forum]

Cool

Thanks

one more question

I'm almost sure this is 0

using green's theorm is probably a horrrible idea here

,w gradient of 1/\sqrt{x^2+y^2}

[jolly saffron]

god what is that lol

i mean you could clean it up

[marsh forum]

I could yeah

but Just trying to figure out what teh function looks like

the denominator is causing troubles

[jolly saffron]

$\mathbf F(x,y) = 2\left (x - \frac{y}{x^2 + y^2} \right ) \mathbf{i} + 2\left (y - \frac{x}{x^2 + y^2} \right ) \mathbf{j}$

[solid kiln]

dyxn

[marsh forum]

oh

wow

that makes it simpler, I suppose?

Thanks

[jolly saffron]

I guess but the partials are still painful to compute

,w differentiate y/(x^2 + y^2) wrt y

[solid kiln]

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Requested by dyxn06

[jolly saffron]

hmm

[marsh forum]

,w differentaite x^2/2 - yln(x^2+y^2) wrt x

[solid kiln]

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Requested by math_rocks

[jolly saffron]

$-2\iint_A \frac{\dd{A}}{x^2 + y^2}$

[solid kiln]

dyxn

[jolly saffron]

ah

@marsh forum remember the first problem I helped you with

(I think)

the 2pi if it encloses the origin thing

[marsh forum]

yea

oh right