help-33

[rotund blade]

I didn't mean it like that

[crystal lintel]

i will not lie courtesan is tempting

[rotund blade]

Courtesans are upper class, it's more respectable

wait I got 21496680 now

I forgot range(start, stop) doesn't include the stop value

so I just had to add 1 to it

[crystal lintel]

[rotund blade]

OverflowError: integer division result too large for a float

😭

[crystal lintel]

[rotund blade]

f(10, 10, 49, 1, 50) = 3849811561536289278422279904000

there we go

f(100, 100, 4999, 1, 5000) = 732224286710524827636120524987672874499155335247048884186065576277777758959320757415746068561542071401694225068145661829838448780394830511814910941252189678433350935628009788234884309653353129950121241674290425386757948715855022592372595068731310879211687843521103454529151297464305829009073195484506260847000290253031848172883177491052237044925323953930216878497433312353962546588541226854539428853220253906161605476233463176962229665456426327584873837386630648760145724167374369262435828144792730369028366458892150420285815135559547882075765224893735002854627191905383061943507053287441001704570766491199124392362114239758315914678046642055094876251542757987495450146321090958590494850579074961784087871451150884570956015898079969235260407151776008497795937304333172845244051144148663006146163127069101295505708914545733729008226646916861933325203785811735163690907306452418957146205052980583814732173136451298291613058684903147760465152407683145515830534744427261306338590641792708900422406668722753670265936472342701195355735739961442866096339154802156484372502876437823406253786359163542677494061606485106874216935534847200491394521901334535928755661892774276162759414499096128839301264165762970967658975405509230808801147622746893309828019308787798979640697465812271392066848969619108695798790724753541805902182299380553372631104752776024183522045273889803971373736534840007799079676397360272405065001925411693755868741741462364643613363746717345651857924471177482260985189635735721712308151944347380832899231685787281793756749614349767619866510758780553760630226058850294739224211308165530394792412792691328728006133460830021460673473730128649994200795835547908720468200833297799417937854199157770187308339291180959260043239170077221615722028090405120487234466358325811461268644006561279516482986625175303071875145149101075139582552923229535342157931405178101700180618499137092061408520534837931376035858052499698422034546466656950657705629062685810138792687992355578073568357978143620302744220768157809982691084094410656750051202058636633043845367765855890639154003079817415793687226448809170311780210217805887140101890555347051032600922502578639111097641527178285969854086174147840953308269615379198237994365573993132544896255333229000559421353974885236311281986265308103417793423008206539501569864926297621455422498712150261494448609764252781091730231223952977895994247312488948617117766828744029957745649122971647116248033468966056329723901291855272712220062570417252419204721029531231294630334341940079611020477928564209352352291412231677254908537880735672787458483995111992317457348546242013703289165471478614129188481153922072529201856648651889983714493949366462284776160606284787931579839745867192698670664552771780759336519694149310668938842411855443198165415262006254536457411182787763154426106331644531552686537488299986125355416015168861593072876609899710511203002563254382856482700824231550868192053449167029832545791801650778670101894153934606988276512389686625091887430507286579857648000000

it's pretty fast

[crystal lintel]

rip my algorithm definitely won't be able to do that

you win this time

[rotund blade]

layla: 1, Existentialistic: 1

can you give me a few more to double check?

[crystal lintel]

yea one sec

[rotund blade]

Ty

[rotund blade]

f(5, 3, 5, 1, 9) = 13560

[rotund blade]

f(20, 4, 22, 1, 57) = 942338812815947908160

f(3, 2, 2, 1, 3) = 12

[crystal lintel]

sounds legit

[rotund blade]

and just so you know I'm not copying:

[crystal lintel]

rip no follow-up comment you're copying

[rotund blade]

f(11, 7, 40, 1, 36) = 350666019090348961769952
f(11, 7, 41, 1, 35) = 307440856230705727557888
f(11, 7, 42, 1, 34) = 255659925445166718721392
f(11, 7, 43, 1, 33) = 201595909471735973143488

ValueError: Exceeds the limit (4300 digits) for integer string conversion; use sys.set_int_max_str_digits() to increase the limit

[crystal lintel]

it's so big

[rotund blade]

12044 digits

[rotund blade]

I'll show you my ||code|| if you show me your ||code||

[crystal lintel]

[rotund blade]

[rotund blade]

No pressure, whatever you’re comfortable with

[crystal lintel]

[rotund blade]

😂

@crystal lintel btw what we just did is called a zero-knowledge proof. https://youtu.be/V5uVKZn3F_4?si=Rr5-fGITf8CbUOGV

[crystal lintel]

loll

[rotund blade]

I think that was my first time using it in a practical application

[crystal lintel]

i used it yesterday!

#help-41 message

[rotund blade]

Cool!

[tight furnace]

777777 spotted early on

[crystal lintel]

69 spotted later on

[tight furnace]

Ig that's not too surprising

[crystal lintel]

for any m, n, a, b, c, f(m, n, a, b, c) = f(m, n, c, b, a) right

each valid board with parameters m, n, a, b, c, corresponds to a valid board with parameters m, n, c, b, a where the C's and T's are flipped

[rotund blade]

It should

[rotund blade]

@crystal lintel What do I need to do to see your ||code||?

Sorry I'll stop asking

[crystal lintel]

i wanted to clean it up and make it shareable but not feeling well

[rotund blade]

is it alright if I share mine?

[crystal lintel]

loll you don’t need my permission for that 😂

[rotund blade]

I wasn't sure if you wanted to try to figure it out yourself or not

I don't want to post "spoilers"

from fractions import Fraction
import sys

def count_cat_boards(m, n, c, a, t):
    count = Fraction(4*(m+n-4)*factorial(m*n -3),factorial(m*n - c - 2)*factorial(c - 1))

    sum = 0

    for i in range(1, min(c, t, 4)+1):
        sum += Fraction((-1)**(i + 1) * (2 ** i) * comb(4, i) * factorial(m*n - 1 - 2*i),factorial(m*n - c - 1 - i)*factorial(c - i))

    sum *= (m*n - 2*m - 2*n + 4)

    count += sum
    print(len(str(count)))
    print(f"f({m}, {n}, {c}, {a}, {t}) = {count}")

if __name__ == '__main__':
    sys.set_int_max_str_digits(100000)
    m = 20
    n = 4
    c = 22
    a = 1
    t = m*n - c - a
    count_cat_boards(m, n, c, a, t)```

This is the formula

[crystal lintel]

🥵

[rotund blade]

[crystal lintel]

[rotund blade]

I hope you feel better

[crystal lintel]

that would be nice

[rotund blade]

it looks like it doesn't work for n = 1 so I'll have to fix that

[crystal lintel]

i'm gonna try to make a simpler formula for that

and look at some other special cases

[cobalt bone]

.

[crystal lintel]

ok maybe i won't be able to make anything simpler

[crystal lintel]

i think i can subvert inclusion exclusion

hm no i'm basically gonna end up doing inclusion exclusion

[rotund blade]

I can explain the formula if you want

I think some of the factorials can be simplfied, they partially cancel

also

BOTTOM CHANNEL

[hasty nymph]

that grade are you guys?

[crystal lintel]

i would be 19th grade if i was still in school

[crystal lintel]

i'm gonna try make something simple but still thinking about it and got distracted making dinner

you can explain if you want though

[rotund blade]

ok I'm working currently it depends on how busy it is

[crystal lintel]

alright

[crystal lintel]

[rotund blade]

it's the same idea as my previous formula

there's two cases to consider

A is on the edge, A is in the center

if A is in the corner we can't spell cat

if A is on the edge it's relatively straight-forward since there can only be one cat spelled at a time

but if A is in the center there can be up to 4

so we have to use inclusion-exclusion

that's what the for loop from 1 to 4 is for

[crystal lintel]

i might be able to do the slayla special to subvert inclusion-exclusion

[rotund blade]

I love inclusion-exclusion

[crystal lintel]

[crystal lintel]

this will certainly make an appearance in my formula

[crystal lintel]

actually change of plans

[tight furnace]

are we doing the 1 A case?

[crystal lintel]

yea. this is a formula

[rotund blade]

sorry for the bad picture, I can take a better one or type it up in LaTeX

Also it only works when n, m >= 2

[crystal lintel]

so eepy

[tight furnace]

it shouldn't need a summation

P(A is on an edge) * P(CAT given A is on an edge) + P(A is in the middle) * P(CAT given A is in the middle)

each of those should be easily closed formable

[rotund blade]

I needed a summation to calculate P(CAT given A is in the middle)

[tight furnace]

oh right inclusion exclusion

still i feel like if you set the top bound to 4 and then added the sum to get a not-closed form

it would hold even in cases where C or T are <4

that might be hard and pointless though

ig you'd be getting 0/0 stuff

or 1/(-1)! stuff (which "should" be 0)

if you fixed all of those errors then it would work

[rotund blade]

I'm having a hard time understanding the motivation

[tight furnace]

make the formula "nicer"

[crystal lintel]

“closed form” fetish

[tight furnace]

i kinda realized it wouldn't be worth the effort as i was typing

[tight furnace]

me

[rotund blade]

what counts as a closed form? like you could use absolute values to cheat out conditional statements

[crystal lintel]

well kinda open to interpretation

[tight furnace]

probably still ok i want to say

[crystal lintel]

i fucking hate the phrase “closed form”

[tight furnace]

anything where computation time is nice and bounded (so not sums, products, and god forbid limits/integrals/etc)

[crystal lintel]

mostly because 95% of the time people say it or ask for one they have no idea what they are even asking for. but moni gets a pass today

[tight furnace]

mainly i wanted something purely with the 4 fundamental operations lol

[rotund blade]

right like my algorithm is already pretty fast

[tight furnace]

i suppose factorials are not fast to compute despite typically counting as "closed form"

though

they're as fast to compute as they are large

[rotund blade]

it seems "closed form" enough

[crystal lintel]

it’s just a min

[rotund blade]

yeah

[tight furnace]

yeah

[rotund blade]

like

[tight furnace]

since the sum is bounded

[rotund blade]

we could have three different closed formulas

depending on what the min is

four different

I mean

[tight furnace]

my point is just i feel like they all should be the "same"

[rotund blade]

I guess 5 if you want to count 0

[crystal lintel]

another reason is “closed form” often implies something that is fast (like an if statement or a min or a sort) is unacceptable which is like… why

[rotund blade]

both of your names are the same length and color I can't tell them apart

[crystal lintel]

[rotund blade]

@crystal lintel and @tight furnace

I can't tell who's saying what

[crystal lintel]

[void elm]

but how

[rotund blade]

the names blend together ok

[crystal lintel]

ah

[void elm]

oh I see

compact mode

[crystal lintel]

that makes more sense

[rotund blade]

[void elm]

[crystal lintel]

[rotund blade]

[tight furnace]

u need to hurry up and get promoted

so our names are a different color

[crystal lintel]

the only option i have is demotion or purple

[tight furnace]

yes i was also very confused until you send the compact mode screenshot lmao

gets rid of the pfps and the pronoun roles

[rotund blade]

thank god I thought I was blind for a second

[crystal lintel]

yea i was like idk what the big deal is

i’m so eepy but i need water

[tight furnace]

also helps when i'm the one that wrote half the messages

[crystal lintel]

i feel like we should try 1 x n

if we can’t express f(1,n,a,b,c) nicely then that’s just sad

and by nicely i don’t mean “closed form”

just something that can be computed for large n

[sleek lake]

i'm new to the channel, i understand the fox game but what's a,b,c

oh okay

[crystal lintel]

that’s just what i use for amounts of each of C,A,T. which is mildly confusing with the c’s but it’s easier for me

also welcome

[sleek lake]

i left, no idea 🤣

[rotund blade]

Should I edit it to c, a, t

?

[crystal lintel]

wdym?

[rotund blade]

Edit the original question

Change abc to cat

[crystal lintel]

it is cat in the original question isn't it

[rotund blade]

I mean the variable names lol

[tight furnace]

closed form it must be

[crystal lintel]

[rotund blade]

I just realized I don’t have abc in the original question lol woops

[rotund blade]

Clopen form

[void elm]

oposed form

[tight furnace]

f is number of possibilities, right

not probability

[fickle field]

<@&286206848099549185> when I start measuring a circle where its radius is 10 and its base is 50.
What is the measure
considering Pi=3.14

[rotund blade]

originally it was probability but possibilities is probably more meaningful

[coral path]

wdym base?

[rotund blade]

please open a new channel lol

[crystal lintel]

yea that's what i've been treating it as

number of valid boards

my f just prints all the relevant information out anyway

[tight furnace]

if the number of Cs, As, and Ts weren't fixed this would be easy 😭

[crystal lintel]

yea 😭

if it was just equiprobable draw one of the 3 letters for each tile

[tight furnace]

or even according to an uneven distribution as long as the draws are independent

[marsh citrus]

@rotund blade Has your question been resolved?

[tall pivot]

.close

hi can anyone tell me the basic rules for graphing tangent and primitive function like if there is a turning point at f(x) that means ... on f'(x)

<@&286206848099549185>

[unborn pelican]

.close

[marsh citrus]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[rotund blade]

[snow palm]

real

[main idol]

.close

[marsh citrus]

Channel closed

Closed by @main idol

Use .reopen if this was a mistake.

[void elm]

.reopen

[marsh citrus]

✅

[void elm]

oi, why’re we closing this

I’m pretty sure Existentialistic, Layla, and Dreyuk aren’t done?

[main idol]

it's been 3 days

[void elm]

@rotund blade @crystal lintel @tight furnace can you three confirm that you’re done here?

[main idol]

they can post the relevant parts and reopen themselves

[void elm]

they can also decide that they’re done here and close on their own

[knotty trellis]

if they havent forgotten about this channel already

[void elm]

I’m sure they haven’t

[main idol]

.

[knotty trellis]

This is open for 18 days already

[sharp vessel]

@tight furnace

[knotty trellis]

crazy

let's ust keep it open to set the record

[main idol]

this is just being further channel bloat for helpers who are actually trying to look to help people and helpees looking for their channel

[void elm]

I mean, it was left at the bottom

surely it can’t be bloat if it’s always there

[main idol]

wrong

[void elm]

I do not agree, but idk

[knotty trellis]

it's true that there are too many open channels right now

some of them had been open for more than a week

[main idol]

this is not the place to have the conversation

[void elm]

right, I apologize

[main idol]

free up the channel for people who are active

.close

[marsh citrus]

Channel closed

Closed by @main idol

Use .reopen if this was a mistake.

[rotund blade]

😭

[void elm]

oh hey

[main idol]

post your question in a new channel

and progress

and don't leave it alone for days on end

[rotund blade]

Sorry I’ve been busy

[rotund blade]

We lose the history lol

[void elm]

just repost the question here, no!

[marsh citrus]

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>.
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• 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.

[ripe glen]

having trouble figring out where t o from here

[marsh citrus]

@ripe glen Has your question been resolved?

[marsh citrus]

@ripe glen Has your question been resolved?

[marsh citrus]

@ripe glen Has your question been resolved?

[marsh citrus]

@ripe glen Has your question been resolved?

[marsh citrus]

@ripe glen Has your question been resolved?

[ripe glen]

<@&286206848099549185>

[ripe glen]

<@&286206848099549185>

.close

[marsh citrus]

Channel closed

Closed by @ripe glen

Use .reopen if this was a mistake.

[marsh citrus]

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.

[heavy blade]

im not sure how to convert surd solutions into the form (x+a)(x-b) for part b
my solutions for part a were -2+ root15 and -2-root 15

[raw hawk]

(x-a)(x-b)
Where a and b are roots?

[heavy blade]

yes but im not sure how to get my solutions into that form

[raw hawk]

,w solve x²+4x-11

[elfin berry]

Results provided by WolframAlpha

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

Requested by ishigamisenku_

[heavy blade]

yep

[raw hawk]

Well distribute the minus sign.
Like for the first one
(x-(-2-√15))

(x+2+√15)

@heavy blade

[heavy blade]

yea i understand but would it just be like that? bc thats what im thinking is one of them but it doesnt look right

[dreamy pollen]

I think this one will be
(x-(√15-2))

[raw hawk]

Yeah

[raw hawk]

Yes

[heavy blade]

rlly? oh okok i get it now

[raw hawk]

But are you told whether a>b

[heavy blade]

nope, thats literally all there is to the question

[raw hawk]

Okay

That should be all then.

[heavy blade]

so the final answer would be (x+2+root15)(x+2-root15)?

[raw hawk]

Yes

[heavy blade]

alrighty, tysm for ur help!

.close

[marsh citrus]

Channel closed

Closed by @heavy blade

Use .reopen if this was a mistake.

[marsh citrus]

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>.
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• 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.

[vague ferry]

How do I get all answers for cos4x = cos(x+45) without using graphs?

[vague ferry]

I know that one answer is 15 degrees since cos4x = cos(x+3x) which means 3x =45 but there are more answers, how do I find them?

[marsh citrus]

@vague ferry Has your question been resolved?

[marsh citrus]

@vague ferry Has your question been resolved?

[sour anchor]

cos(A+B) = cos(A)cos(B) - sin(A)sin(B) might help

[vague ferry]

not really, i get that cos4x = (cosx-sinx)/sqrt2 and then what?

[keen mist]

You could use the rule that cos(x) = cos(360 - x) along with cos(360+x) to get different solutions

[vague ferry]

yeah, but that is just adding a period onto 15 degrees

as you see if you graph it you get 7 diffrent solutions within a period

I'm wondering how to get the rest of the solutions without using a graph

[keen mist]

I mean you get x = 15 from 4x = x + 45

You get x = 63 from 4x = 360 - (x + 45)

You can get the other ones similarly

[vague ferry]

you're right, thanks

.close

[marsh citrus]

Channel closed

Closed by @vague ferry

Use .reopen if this was a mistake.

[marsh citrus]

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>.
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• 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.

[fluid fulcrum]

Can you help me solve this?

[broken zealot]

do the parentheses denote the gcd?

[fluid fulcrum]

Yeah

[broken zealot]

well for the first one, you know that p might divide b twice, but a only once so it could be p^2

or p

so i guess, try to think write the prime factorization of a,b

or do the p-adic valuation of them

[fluid fulcrum]

How do I know that p might divide b twice and a once?

[broken zealot]

well ok, let c and d be the power of p in the prime factorization of a and b respectively

then you know that min(c,d) = 1, right?

because taking the gcd is taking the minimum of the exponents

[fluid fulcrum]

Can you show me the full solution?

can you say something to let me know if you can show me the full solution?

.close

[marsh citrus]

Channel closed

Closed by @fluid fulcrum

Use .reopen if this was a mistake.

[marsh citrus]

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.
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• 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.

[vernal herald]

where did the 1/2 coefficient come from?

[hushed egret]

thin air

random constant less than 1 that might work

[next raft]

snow!

[hushed egret]

you could probably choose any positive constant less than 1

[next raft]

yea

[hushed egret]

obviously 1/2 is the rational choice with least denominator

[next raft]

as x becomes large, x^5 is the largest term so we just look at the coefficient of that term

[next raft]

so what snow said

[wary kite]

what snow said

[vernal herald]

ok thank u

.close

[marsh citrus]

Channel closed

Closed by @vernal herald

Use .reopen if this was a mistake.

[next raft]

yay

[night mica]

hi snow!

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sharp lion]

coukd i get clarification on these four problems ?

[marsh citrus]

@sharp lion Has your question been resolved?

[sharp lion]

<@&286206848099549185>

[sharp lion]

<@&286206848099549185>

[sharp lion]

<@&286206848099549185>

[silk sable]

Do you understand how a Taylor Polynomial is generated?

[sharp lion]

for the most part i believe

i wont fall asleep this time lol

mb for yesterday

[silk sable]

No worries. Do you recall the formula I posted?

[sharp lion]

i didn't see it

[silk sable]

f^(i) is notation for an ith derivative.

[sharp lion]

mhm

[silk sable]

An important thing to note here is that a second-degree Taylor Polynomial will almost always result in a second-degree polynomial.

[sharp lion]

can you elaborate on how we get +, -, or 0? im confused on that part

[silk sable]

Let's say that T_2(x) = ax^2 + bx + c. What would its graph look like?

[sharp lion]

quadratic

parabola

[silk sable]

And would it face upwards or downwards?

[sharp lion]

up

[silk sable]

Why? Wouldn't it depend on the value of a?

[sharp lion]

a is positive since it doesn't have a negative a

[silk sable]

a is just a variable. It could be negative.

[sharp lion]

what's next

[silk sable]

Which variable; a, b, or c; describes the direction the parabola is facing?

[sharp lion]

a

[silk sable]

Is that the term with the largest degree in that equation?

[sharp lion]

oh mb. c

[silk sable]

And should c be positive or negative?

[sharp lion]

positive

[silk sable]

The parabola is facing downwards.

[sharp lion]

so c is negative?

i see

[silk sable]

Yes.

[sharp lion]

what's our next steps

[silk sable]

Determine if b is positive or negative

[sharp lion]

0

[silk sable]

Why do you think it is zero?

[sharp lion]

the graph is in the center of the x axis

[silk sable]

The horizontal shift is typically governed by when x - c = 0.

[sharp lion]

so positive

wait

negative

[silk sable]

In this case, you have x - 1 so the graph is shifted to the right one unit.

[sharp lion]

b negative

[silk sable]

What makes you say that?

[sharp lion]

x could be negative

[silk sable]

Looking at the provided graph, where is the vertex?

[sharp lion]

(0, 0)

[silk sable]

And what is the slope of f(x) at (0,0)?

[sharp lion]

negative

[silk sable]

You are thinking of the second derivative.

[sharp lion]

?

[silk sable]

The first derivative would be zero because the tangent line at (0,0) would just be y = 0.

[sharp lion]

ohh ok

[silk sable]

And because you know that the slope at x = 0 is zero, you can take the derivative of f(x) and solve for some values.

[sharp lion]

gotcha

[silk sable]

Are you finding the derivative of f(x)?

[sharp lion]

that i have yet to do

ok

-2a(x-1)

[silk sable]

That does not look correct.

[sharp lion]

b+c(x-1)

[silk sable]

Correct.

Now you know the derivative equals zero at x=0, so you can substitute x with 0. This allows you to solve for b in terms of c.

[sharp lion]

b-c

[silk sable]

b - c = 0

And therefore,

b = c

What was the sign of c?

[sharp lion]

positive, when b = c
negative when both equal to 0

[silk sable]

The parabola is facing downwards.

[sharp lion]

negative

[silk sable]

Which means that b is also negative.

[sharp lion]

so a is 0?

[silk sable]

No.

[sharp lion]

thanks for the clarification

this one must be c then right?

[silk sable]

Yes.

[sharp lion]

i was able to answer the other similar questions. can we focus on this last one?

[silk sable]

Sure.

The equation for the Taylor Polynomial given to you is for the fourth-degree Taylor Series.

[sharp lion]

yup

[silk sable]

You only want the second-degree Taylor Series. What would that equation be?

[sharp lion]

x^2 / 2!

[silk sable]

1 - x^2/2!

You cannot forget the terms that come before it. In thise case, the second term zeroed out.

[sharp lion]

ohh gotcha

[silk sable]

Plot some points for different values of x to find its corresponding y values.

[sharp lion]

the parabola's downward

[silk sable]

What does T_2(0) equal?

[sharp lion]

1

[silk sable]

Which graphs contain the point (0,1) on the parabola?

[sharp lion]

last three graphs

[silk sable]

Ok, so you need to narrow it down more.

Given the interval for x, pick a value for x and see which ones have a point on the parabola that matches.

[sharp lion]

this graph yeah?

yup it is

thanks for your help and patience!

[silk sable]

yw

[sharp lion]

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[marsh citrus]

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[sinful pier]

Idk how to do this

[halcyon fulcrum]

What’s the actual question

That looks like a MS

You are being asked to find a list of inputs right

[sinful pier]

No…?

I moved onto my other assignment

<@&286206848099549185>

[humble sundial]

Substitute x+3 into the expression for f(x) aka… f(x)=-2x^2+3x

So it would turn into

f(x+3)=-2(x+3)^2 +3(x+3)

Thats what it would turn into ^^

Then you want to expand the expression

(That would take a long ass time to type out)

But your answer would be the image I provided

[sinful pier]

That’s not what the answer key says…

@humble sundial

Because it’s not f multiplied by g

It’s f of g written in some fuck ass way that confuses anyone or thing trying to help me >:(

It’s f(g(x/4))

[marsh citrus]

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[astral ether]

Can someone help me complete the square of this problem:

[still temple]

Any tries ?

[astral ether]

I did but I just stopped cause I wasnt sure

If I was doing it properly

[still temple]

Just send it

[astral ether]

1/2 ( x^2 + 3/2x) - 2

[still temple]

You are on the right path

Go on

[tight kite]

you forgot to distribute the 1/2 to the 4 when taking it out of the parenthesis

[astral ether]

Ah

I see

What would I do now though

[tight kite]

x² + (3/2)x + (9/16) can be factored

(x + a)² = x² + 2ax + a² so try rewriting to fit this form

[astral ether]

whatcha think about this

[marsh citrus]

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[marsh citrus]

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[midnight pasture]

Hi I need some help calculating the surface area and volume for this please

[proud basin]

find these 2 lengths, and after that its just combinations of shapes

[glad frost]

like he is saying find the shape area of the side star the multiply by the height for volume

[midnight pasture]

hmm okay so I got
yellow line = 3
red line = 3

volume = ((6x5) - (3x3))x 8 = 168cm^3
surface area = (3x8) + (3x8) + (6x8) + (8x2) + (7x8) + (5x8) + ((6x5) - (7x3))x2 = 226cm^2

is that correct

[marsh citrus]

@midnight pasture Has your question been resolved?

[midnight pasture]

whoops it's 218

for surface area

[marsh citrus]

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[dreamy spoke]

what is the area of element strip cut on hollow sphere at an angle theta (i am getting 2pi r^2 cosx dx )

[fervent rampart]

can you show a picture?

[marsh citrus]

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[night lion]

how is c correct if

[night lion]

as using cramers rule i found that

[marsh citrus]

@night lion Has your question been resolved?

[glass silo]

If lambda is not 1/2, your system has a [unique] solution

[night lion]

yeah but i ticked on infinite one

[marsh citrus]

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[fallow furnace]

how can i find function f(x) that has limit x to 0 equal to infinity, function g(x) doesnt have limit as x approaches 0 but f(x)g(x) has some limit for x to 0?

[fallow furnace]

does it even exist?

[tepid temple]

That’s not possible as far as I know

[marsh citrus]

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[fallow furnace]

Why?

[marsh citrus]

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[wet oriole]

What do the fractions we make with similar triangles represent?

[wet oriole]

Like ik we use it to find da missing variable

But what is the geometry behind the ratio of height over base

.solvedf

.solved

[marsh citrus]

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[night lion]

when u take cos x to sin half angle this becomes determinate form right

[night lion]

0/1? = 0

then i get a =0 but ans just rationalises it

[eternal fog]

you get 0/0

so you should use lhopital rule

[marsh citrus]

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[marsh citrus]

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[fallen osprey]

idk how to do c help pls

[fallen osprey]

like i understand how to do the rest of them its just the wording of c is confusing

[winter anchor]

z - score ?

Exactly what are you studying

[fallen osprey]

statistics

[winter anchor]

aw :(
I wish I could help but I haven't studied stats

[fallen osprey]

n yes its z score

[winter anchor]

sorry

[fallen osprey]

oh ok

[north bison]

i'll have a look then

let me just read the question

[fallen osprey]

im just solving for z scor

[north bison]

yep okay

what have you done so far

it's just part c right?

[fallen osprey]

ya i only need help w z

c

the rest i did

[north bison]

are you allowed to use the tables

at least in the uk we're allowed the standard normal distribution table

have you got the cumulative probability ||(you should get ~0.937?)||

[fallen osprey]

no idk where that comes from

my class allows me to ues an app to calculate i dont have to look at a table

[main nest]

Hey I need help with these tasks please, I am usually ok with maths but this has been difficult for me:

1. The number 15.27^10 has the following floating point representation (representation: sign bit exponent mantissa): 0 100010 111010001 (IEEE 754 standard). Assume 31 as bias. Convert this floating point representation to a decimal representation. Calculate the rounding error R. This follows the formula: A = B + R, where A is the value of the number in the decimal representation and B corresponds to your determined, calculated value. Enter the rounding error R! Separate the decimal part (if present) with a period and omit unnecessary zeros. If there is no rounding error, please enter exactly "0.00".

2. The number 20.36^10 has the following floating point representation (representation: sign bit exponent mantissa): 0 100011 010001011 (IEEE 754 standard). Assume 31 as the bias. Convert this floating point representation to a decimal representation. Calculate the rounding error R. This follows the formula: A = B + R, where A is the value of the number in the decimal representation and B corresponds to your determined, calculated value. Enter the rounding error R! Separate the decimal part (if present) with a period and omit unnecessary zeros. If there is no rounding error, please enter exactly "0.00".

[fallen osprey]

girl

u need to post it in an avail channel im pretty sure

[main nest]

Oh, I thought it would be in the help

[fallen osprey]

ya but my channel is occupied

think its fine tho

[winter anchor]

This one's occupied by @fallen osprey

[main nest]

Oh sorry

[fallen osprey]

its ok dw

[winter anchor]

light mode

smh

[fallen osprey]

its pink

my whole laptop is pink lmao

only reason i have nitro

[winter anchor]

damn

[fallen osprey]

^^^^^ someone pls help 😓

[north bison]

yes

i'm here sorry

[fallen osprey]

hi

i'm allowed this app calculator so i don't really have to do much actual calculating work i just need to understand where to put each value

[north bison]

yep but if you have that cumulative prob is 0.937 you could work out z value?

[north bison]

well the area on the left is going to be what?

[fallen osprey]

yeah but where did u get the value of 0.937

0.5

[north bison]

yeah

[fallen osprey]

but it says the area between z=0 so i'm confused

[north bison]

well 0.5 + 0.437 is 0.937

the area between z= 0 and the unknown z (let's call it z') is 0.437

and you know that the left side is 0.5

[marsh citrus]

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[frank falcon]

The question is to solve: lg100√10

On top of the page I got 2√10 which is incorrect but I know that the correct answer is 5/2, my question is how we got there?

[frank falcon]

And if there's anyone who could explain log and lg in general pls hmu)

[split comet]

Are you aware of the property of log that is

log(a × b) = log a + log b

[frank falcon]

yes

so i take 100 sqrt10 and solve it with that formula?

[split comet]

Let's break log 100√10 that way

[brave spire]

you evaluated the log as if the $\sqrt{10}$ was outside the logarithm

[elfin berry]

rak³en

[tight kite]

the square root is inside the log

[brave spire]

but clearly you have been asked to evaluate $\log \left(100 \sqrt{10} \right)$

[elfin berry]

rak³en

[brave spire]

from where after u use this property

[split comet]

?

log 100 + log √10

[brave spire]

yeah thats what I am saying

another more direct way would be to recognise that $100 \sqrt{10} = 10^2 \cdot \sqrt{10} = 10^2 \cdot 10^{\frac{1}{2}} = 10^{2 + \frac{1}{2}} = 10^{\frac{5}{2}}$

[tight kite]

not necessary tho, you can just write 100√10 as 10^(2 + 1/2)

[elfin berry]

rak³en

[split comet]

Certainly

[brave spire]

after which you can directly apply log on both sides to conclude $\log 100 \sqrt{10} = \frac{5}{2}$

[elfin berry]

rak³en

[frank falcon]

honestly very confused rn

im up to this point i think?

[split comet]

now what's log 100?

[frank falcon]

we put it to 10^2 and the sqrt10 10^1/2
i think?

[split comet]

Yes certainly

After that it becomes 2+(1/2)

[frank falcon]

oh holy fuck

ty

[split comet]

No problem

[frank falcon]

.close

[marsh citrus]

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[velvet yarrow]

,root

Last part

[marsh citrus]

@velvet yarrow Has your question been resolved?

[blissful zenith]

you need help chef?

[velvet yarrow]

yh

[velvet yarrow]

.close

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[pine condor]

Hello! For C, how do i compute +- infinity?

[gilded cape]

you can pass the limit to the exponent (since the exponential function is continuous)

[pine condor]

Yeah

but I dont understnad this bit tho

How does 30000x become that?

(1-30000/x^2 ?

[gilded cape]

they factored out -x^3 from both terms

[pine condor]

I dont have -x^3 in both terms ?

I have a x

That just blows my mind

[gilded cape]

they divided each term by -x^3

hence the -x^3 becomes 1, and the 30000x becomes -30000/x^2

and you get a factor of -x^3 on the outside

[pine condor]

ohh okey

but how can one then factor it out again if -x^3 / -x^3 becomes 1?

[gilded cape]

what do you mean factor it out again?

factoring is only happening once here

[pine condor]

That process

[gilded cape]

$a+b = c\left(\frac{a}{c} + \frac{b}{c}\right)$

[elfin berry]

Tushar

[gilded cape]

this is the property being used here

it follows from the distributive property of multiplication over addition

[pine condor]

ohh

and how come it then will be 0 ?

-(+infinity^3)*(1**0)

[gilded cape]

they are saying that precisely the -30000/x^2 part will become 0 since the denominator becomes arbitrarily large as x tends to infinity

[pine condor]

yeahhh now i get it so the x^2 "wins" over the -30000

[gilded cape]

correct

[pine condor]

So as x^2 gets large as fuck the function tends to 0

it just get smaller and smaller

[gilded cape]

not the way I would phrase it, but yes I agree

[pine condor]

hahahha but the other one then

gimme a sec i guess it will be the same process ?

Ill try it

-infinity

[gilded cape]

by "the function" you mean just the -30000/x^2 part

since the rest of the limit remains to be computed

[pine condor]

Ye but why dont we consider the -x^3?

[gilded cape]

you do

[pine condor]

Is it becuase -(+inf)*(1-somethingthatgetscloseto0)

Like can we put in the infinity ? will it make it easier for us ?

Instead of X ?

[gilded cape]

since the -x^3 goes to -inf, it becomes e^(-inf), which only then gives you 0

[pine condor]

Oh alr so the 30000/x^2 gets close to zero so we have -x^3*(1-0) = -x^3 and this will just become -infinity and e^-infinity is 0

[gilded cape]

precisely!

[pine condor]

can you stay for 1 min

Ill try the other one

alr

so i get that

same procecdure i guess for the 30000/x^2

it goes to 0 ?

but the exp(-x^3)

-(-infinity^3)

• infinity?

[marsh citrus]

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[marsh citrus]

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[silk temple]

The question with my attempt
Any help would be appreciated

[silk temple]

Also please ping if you reply

[still temple]

you got the general term wrong over here

@silk temple

it should be x^2n+1

[silk temple]

Oo yup I see that now

In the next line I have integrated it as if it was x^2n+1 so it shouldn't matter

I made a mistake copying my solution from my rough notebook

And I was not able to get very far from this approach
And could not think about anything for the second bracket

[marsh citrus]

@silk temple Has your question been resolved?

[still temple]

check this step again

@silk temple

[silk temple]

Yup its incorrect
It should be x^2n+3?

And the denominator should end at (2n+2)

[still temple]

yes

[still temple]

i dont understand if its 1 or 3

[silk temple]

I have written it as 2n+3 but I now think that is a mistake and it should be 2n+2

[still temple]

no

why would i denoninator change ?

[silk temple]

Over here I I put 2n+3 in the second line while it was 2n+2 in the first line

Made a lot of mistakes writing this 😅

[marsh citrus]

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[still temple]

@silk temple whats the question again?

[marsh citrus]

@silk temple Has your question been resolved?

[marsh citrus]

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[frank eagle]

In \textit{Introduction to Topology} by Gamelin and Greene, I'm working an exercise to show the equivalence of norms in $\mathbb R^n$. This exercise succeeds another exercise where various equivalent formulations of equivalent norms have been given, e.g. that two norms $|\cdot|_a,|\cdot|_b$ are equivalent iff the identity map from $(\mathbb R^n,|\cdot|_a)$ to $(\mathbb R^n,|\cdot|_b)$ is bicontinuous. \

Now, in showing that all norms in $\mathbb R^n$ are equivalent, the authors show a given norm $|\cdot|$ is equivalent to the $1$-norm. I have showed already that the identity is continuous from $(\mathbb R^n,|\cdot|_1)$ to $(\mathbb R^n,|\cdot|)$. To show that the inverse of the identity map is continuous, the authors claim that the unit sphere in the $1$-norm is compact. I'm getting hung up on this statement, since I don't know how to go about this without using that the norms are equivalent already. How would one show the unit sphere in the $1$-norm is compact?

[elfin berry]

psie

[marsh citrus]

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[dusty kelp]

is this right for determining the area of a koch snowflake after i iterations?

[still temple]

its like a composant question , try first +00 its gonna be exp(+00)=+00

and second for +00 its gonna exp(-00)=0

[dusty kelp]

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[obsidian bobcat]

Did I get this right? Finding f prime

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[wild perch]

Line L1 has equation r1 = (3,-1,2) + s(2,1,-1)
Line L2 has equation r2=(2,4,0) + t(3,-1,4)
Line L3 passes through the point with position vector a=8i-2j+4k and intersects with lines L1 and L2.
Find the equation of L3 in the form r3=a+ub where b is a vector to be determined

I have figured out that I can construct some equations for the intersections by equating L1 and L3 or L2 and L3: (the first 3 are for L1 and L3 and the other 3 are for L2 and L3)
3+2s=8+ux
-1+s=-2+uy
2-s=4+uz

2+3t=8+ux
4-t=-2+uy
4t=4+uz
However I'm stuck here because I don't know how to keep going. I'm pretty sure you're not allowed to use the same u between the first set of equations and the second because the intersection points are different (just to be sure, I checked and L1 and L2 do not intersect)

[marsh citrus]

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<@&286206848099549185>

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[lethal marlin]

Hey, I got some help on this problem earlier this morning and thought it made sense at the time, but now I'm still struggling with it and hoping someone might be able to help clarify it again. Basically, I'm trying to figure out how to convert a point on a convex quadrilateral to a normalized coordinate within the quad. Similar to the above image.

I got some help and was given this useful desmo which shows how to interpolate a value given the normalized input into the actual coordinates on the quad. https://www.desmos.com/calculator/jjrfsbr1ga but I still can't figure out how to get the inverse of this function and pass in the coordinates, and get the normalized input.

I was told something to do with mapping the unit square onto the quad, and solving a system of equations... but it's not really coming together for me.

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[main idol]

Well did you try looking up the radius of earth

[fervent rampart]

add 6 meters (actually 2pi meters) to the circumference an you must add 1 meter to the radius

no matter what the size of the circle

2pi(r + 1) = 2pi r + 2pi

[proud basin]

let r be the radius of the earth, the circumference would be $2\pi r$, if you add $2\pi$ total you can get $2\pi r+2\pi=2\pi(r+1)$

[elfin berry]

Skill_Issue

[fervent rampart]

really we are adding 2pi, but 2pi is about 6

,calc 2*pi

[elfin berry]

Result:

6.2831853071796

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[copper basin]

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[marsh peak]

!show

[marsh citrus]

Show your work, and if possible, explain where you are stuck.

[copper basin]

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[marsh peak]

In general, the tangent line to the curve y = f(x) at x = a is given by y = f'(a)(x - a) + f(a). So start by differentiating f to see find f'(1)

[copper basin]

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[marsh peak]

You are forgetting chain rule

[copper basin]

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[marsh peak]

Just multiply the right hand side by pi/4

[copper basin]

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[marsh peak]

That's y', yes

[copper basin]

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[marsh peak]

Now evaluate y(1) and y'(1)

[copper basin]

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[marsh peak]

That's correct

[copper basin]

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[marsh peak]

Yeah

[copper basin]

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[marsh citrus]

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[marsh citrus]

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[distant yoke]

Why is it looking like that ?

[dire spire]

view the error

[crystal lintel]

too many &’s?

[distant yoke]

Do you think that's the reason?

Imma try

I'm typing 2 slashes at the end, but when I send the message it get deleted

[crystal lintel]

hm?

[distant yoke]

It's like the * and the _ in Discord, when you put 2 of them in your sentence they'll get deleted and your text will be in italic

I think it's the same problem with the \

\\\\\\\\

[crystal lintel]

should eqnarray not have dollar signs around it

[distant yoke]

Everything mathematical need to be written with dollar signs around it

[crystal lintel]

try it without

[distant yoke]

Look, in reality I typed like 20 \ and it shows me only 8

[distant yoke]

Ok ok

[crystal lintel]

also looks like this wasn’t the issue i just didn’t know eqnarray syntax

[distant yoke]

I'm gonna keep the &, and what's happening with the (1) and (2) ?

Like is it always like this ?

[crystal lintel]

you can change eqnarray to eqnarray* to remove those

[distant yoke]

Oh yeah you're right, thank's !!

[fathom ridge]

you should use cases anyways

[elfin berry]

krypton

\[ \begin{cases} y = x^2 - 1 \\ 2y = - x^2 + x + 1 \end{cases} \]

[distant yoke]

I just typed the same exact thing and it doesn't work ☠️

[fathom ridge]

there

[marsh citrus]

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[pine condor]

Alr I think the teacher have done wrong here correct me if im wrong

I just did it and at the end i got 1/9 * (u^3)

That -integral of 1/3*u^2

part

We can pull out a 1/3 and just do the integral of u^2 right

U^3 / 3

so its 1/3 * 1/3 * u^3

1/9 * u^3

[red nimbus]

Yea

[pine condor]

So this is incorrect then

Since she integrates the u^2 but doesnt add a 1/3, she adds a 1/2

Tjat is how i got it

[red nimbus]

Yea

[pine condor]

XD

Thank you

.close

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[untold wyvern]

Hi im having trouble following why the $p-r_i$ are distinct

[elfin berry]

somethingwrong

[devout mauve]

the r_i are all distinct and if you subtract distinct numbers from a fixed number, you get distinct numbers

[untold wyvern]

hmm why are the r_i distinct

the r_i are the remainder of a,2a,3a,...,((p-1)/2)a that exceed p/2 right

[devout mauve]

yes

those multiples of a are all distinct

[nova wigeon]

.

[untold wyvern]

ohh

[nova wigeon]

Want me To explain It in a Happy Way?

[untold wyvern]

if for example $2a \equiv 3a \pmod p$ , we would have that $2 \equiv 3 \pmod p$ since $(a,p)=1$ which is a contradiction is the idea right?

[elfin berry]

somethingwrong

[devout mauve]

yes

[nova wigeon]

We’re looking at residues modulo an odd prime ( p ), specifically those greater than ( \frac{p}{2} ).

1. Residues: We have residues ( r_i ) that are greater than ( \frac{p}{2} ) and others ( s_j ) that are less than ( \frac{p}{2} ).

2. Unique Values: When you calculate ( p - r_i ), it gives a value that falls below ( \frac{p}{2} ). If two residues were equal, it would create a contradiction.

3. Conclusion: Therefore, all the values ( p - r_i ) are distinct and unique. No repeats!

Mathematics drama at its finest, right? If you want more detail on any part, just ask!

But I don't I might me Wrong:(

[elfin berry]

...a

[untold wyvern]

okay thanks both

.close

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[near elbow]

Guys pls help me check my work. The question ask me to solve that equation for 0<=x<=180 degrees, i think i made a mistake bcs the answer key says the answers are 38.4 and 111.6 degrees

[keen hinge]

nevermind

[vast magnet]

i think the error is here

You used $\tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$ but put $\tan(-x)$ there instead of $\tan x$.

[elfin berry]

@vast magnet

[vast magnet]

nice handwriting, btw

lmao

[near elbow]

OHH

soo the x shouldn't be negative?

[near elbow]

thankss hehe

[near elbow]

thank youuuu

[vast magnet]

nope, either you use tan(a + b) and take x as negative

or use tan(a - b) like you did and take x as positive

[near elbow]

ohhh i see i see

[near elbow]

thank you so much

ohno guys

i still get it wrong

[vast magnet]

show

[near elbow]

im kinda confused here

[vast magnet]

$\tan x(1+\tan60^\circ\tan x)=\tan x+\tan 60^\circ\tan^2x$

[elfin berry]

@vast magnet

[vast magnet]

tan60 is root3

[near elbow]

why do we add them not multiply?

[vast magnet]

uh

dont u have two terms

in the brackets

so you distribute tan x to both of them

[near elbow]

ohh

[near elbow]

where is the tan^2 x from?

[vast magnet]

oh wait

rip

[near elbow]

😭

[vast magnet]

$\tan x(1+\tan60^\circ\tan x)=\tan x+\tan 60^\circ\tan^2x$

[elfin berry]

@vast magnet

[vast magnet]

this is what i meant

[near elbow]

ohhh i seee

okayy thank you so much

i'll try againn

ohno guys i still get it wrong

[vast magnet]

nvm

sorry

ignore

,w sqrt(3)x^2+3x-2sqrt(3)=0

damn

[vast magnet]

b^2 isn't sqrt(3)^2, its 3^2

[marsh citrus]

@near elbow Has your question been resolved?

[near elbow]

oh ure right😭

[near elbow]

thank you so much

sorry for causing so much trouble

[vast magnet]

that's not so much trouble

you managed to write the entire thing twice

soo :)

[near elbow]

hehehehe thank youu thoo

i got the answer alreadyy!!

[vast magnet]

🔥

[near elbow]

i'll close the channel

thanksss

.close

[marsh citrus]

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[jolly burrow]

Yo can someone quickly tell me the answer.

Is it 110° or 70°>

.

[vast magnet]

construct another line through that intersection point parallel to the other 2 lines

[jolly burrow]

Okay:) , then what answer do I get? 🥰

[vast magnet]

!noans

[marsh citrus]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[jolly burrow]

Ok

Correct?

[cold pecan]

yes

[vast magnet]

yes

[jolly burrow]

Oh , thanks dear.:) , I'm never trusting ai again since all ai's are telling that the answer is 70°>

Thank u so much

🙂

Have a great day

[vast magnet]

never do math with ai

they suck

[jolly burrow]

Fr. I failed my exam bc of their wrong concepts.

[vast magnet]

lmao

[cold pecan]

it's over for ai

[jolly burrow]

I wish I had a good teacher.

[cold pecan]

[vast magnet]

youtube 🔥

[jolly burrow]

Ai's never replacing human

[vast magnet]

especially ||c.ai||

[jolly burrow]

My teacher doesn't even explain us

[cold pecan]

not until they can do math

[vast magnet]

=never

trol

[jolly burrow]

😀thanks

.close

[marsh citrus]

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[cold pecan]

!done