help-17

[fringe scarab]

fluid flow like an actual fluid

[verbal blaze]

but also surface area in some cases

[fringe scarab]

or as an adjective

[verbal blaze]

no

bcs u can have non fluid like vector fields

[proper minnow]

Hn that's sounds interesting

[verbal blaze]

it's painful

[fringe scarab]

does "fluid" mean differentiable everywhere?

[verbal blaze]

yeah this makes sense

[verbal blaze]

fluid as in the thing ppl usually compare vector fields to

but not necessarily noncompressible

[proper minnow]

So in theory gggggf will be completely different compared to fggggg

[fringe scarab]

like i said i took some neutered ass multivariable

[verbal blaze]

oh yea tru

I had this result earlier

[proper minnow]

Without even using my initial statement

[fringe scarab]

wait

ffs

that's what they were saying earlier

[verbal blaze]

I proved that
with only one f you can always find where it is

[proper minnow]

Not just that

Ffgggggggg is always smaller than gggggggff

[verbal blaze]

yeah I see that okay

[proper minnow]

Fffgggggggg will be smaller than ggggggfff

[fringe scarab]

u got it backwards

[proper minnow]

Eh ?

No

I got it correct

[fringe scarab]

oh there're different numbers of functions

i thought the strings were the same length

[proper minnow]

If you get 1/2 in constant term first it will get get very low due to gs

[proper minnow]

They are same

I just didn't bother to count

[verbal blaze]

oh hey group theory time

bcs I see no other way to do it

[fringe scarab]

where's the group

[verbal blaze]

well

okay so it's not a group but I mean the whole like

equality

and how we don't know if it's transitive

[fringe scarab]

transitive?

[verbal blaze]

well, the inequality I mean

[proper minnow]

My proofs are just enough tbh , c can never reach negative

[verbal blaze]

my idea is that

we know that swaps are necessarily unequal

[fringe scarab]

if they're at the end

[verbal blaze]

uh

hm

can we prove if h =!= j -> kh=!=kj

[fringe scarab]

what's k

another sequence?

[verbal blaze]

arbitrary sequence

[gentle sleet]

it's anywhere
if fg ≠ gf then hfgk ≠ hgfk

[verbal blaze]

obviously where h and j are permutations otherwise it's trivial

[fringe scarab]

how so

[gentle sleet]

just left compose and right compose

I think that's legal

[fringe scarab]

composition is not commutative

[gentle sleet]

if we prove h ≠ j then we're done no?

[fringe scarab]

i mean

u dont know if h is one to one in the space of functions

[verbal blaze]

u mean k?

[fringe scarab]

yeah k mb

[verbal blaze]

I think it is

[fringe scarab]

im used to the hg != hf example ive been saying

[verbal blaze]

ah

[fringe scarab]

"i think it is"
QED

[verbal blaze]

oh yeah wait it can follow from this

[gentle sleet]

composition of one to one functions is still one to one right?

[verbal blaze]

nvm hg != hf is cleaner

wait

[fringe scarab]

idk abt in the space of functions tho

[verbal blaze]

hg != hf has to be true

[fringe scarab]

in R -> R yeah i think so

[verbal blaze]

bcs they have different numbers of functions

[fringe scarab]

oh

oh yeah

are we done then

[verbal blaze]

uh

no

[fringe scarab]

let me try

[verbal blaze]

go for it

[fringe scarab]

does fggfgfgff = gfffgfggf

[verbal blaze]

hm

[fringe scarab]

we know
hf != hg
fgh != gfh
fh != gh

[verbal blaze]

we also know
fgg
gfg
ggf
are all unequal

[fringe scarab]

ff != gf, so we have fggfgfgi and gfffgfgj

[proper minnow]

Eh I don't wanna be that guy but this won't work

Yes the condition are true but this won't help with other strings

Because these conditions start with constant as 0

But can't be said same for pattern which appear on different statement

[fringe scarab]

can we say the fgfgi != fgfgj?

[proper minnow]

Cant

I and j could be equal

[fringe scarab]

no i = ff and j = gf

[verbal blaze]

I think we have to prove hi != hj iff i != j

[verbal blaze]

yeah

[fringe scarab]

idk abt iff

[verbal blaze]

follows from function numbers

[proper minnow]

Rhey are different ofc proven by the slope

[fringe scarab]

oh shit ur right

alr

fggi and gffj; i = fgfgff, h = fgfggf

[verbal blaze]

graph theory save usss

[proper minnow]

It can't tbh

It won't help here

[proper minnow]

They could be same

You can't tell

[gentle sleet]

you can convert the functions into matricies on a <x, y, 1> vector
also not sure if it's helpful

[proper minnow]

Well ofc without my statements 😉

[verbal blaze]

wait what?

[proper minnow]

Could be useful

Could be not tbh

Eh tbh not

[verbal blaze]

how

[gentle sleet]

that's what you did on your first try

[verbal blaze]

those weren't matrices

[proper minnow]

They won't have any help to prove further

[gentle sleet]

you can convert them into it

[proper minnow]

Using patterns aren't useful too

[verbal blaze]

or I mean ig they were but not very useful ones. you didn't get linearity or anything

[proper minnow]

Try different method if you want a good proof

[verbal blaze]

babi you haven't proved it either

[proper minnow]

Tbh I did

[cosmic cloud]

bro asked a question and left

[proper minnow]

It should cover all basis

[proper minnow]

Ikr

[proper minnow]

• you aren't even going in the right way

[fringe scarab]

all u did was show gh != fh

[proper minnow]

Using patterns just the worst method

[verbal blaze]

hey you don't know that

[proper minnow]

Hm I do

These are functions not some binary digits or pattern

You can't use pattern logic here

[fringe scarab]

ok bro

[verbal blaze]

betterert do u kno any number theory

[proper minnow]

No I proved swaps will always be different in spite of the distance of the swap

[verbal blaze]

could rlly come in clutch rn

[proper minnow]

• the distance of f

[fringe scarab]

how could number theory come in clutch

[verbal blaze]

you didn't prove it for n swaps

[verbal blaze]

constant term

deciphering the constant term

[gentle sleet]

some require more swaps and your proof isn't transitive

[proper minnow]

It will be same , you can only ever get same result if c is less than negative

[proper minnow]

Try it

[gentle sleet]

or at least you haven't proved that it is transitive

[proper minnow]

Just equate and see it

[fringe scarab]

i mean

[proper minnow]

C will always need to be less than 0 for the pattern to be same

But c won't

[fringe scarab]

with the constant term is multiplying by 3/2 and then adding 1/2 and then multiplying by 1/2
the same as multiplying by 1/2, then multiplying by 3/2 and then adding 1/2

no

[verbal blaze]

yeah no

[fringe scarab]

i mean that's what makes fg != gf

um

[verbal blaze]

g is linear f isn't

[fringe scarab]

f is linear?

[gentle sleet]

you keep repeating that but that doesn't have anything to do with my statement

[verbal blaze]

it's not a linear map I don't think

thats what I meant

[gentle sleet]

actually, with the matrix, kinda I think

[fringe scarab]

what is a linear map

[proper minnow]

Just equate the equations and see

[fringe scarab]

idk any lin alg

[gentle sleet]

here's f

[fringe scarab]

ok bro.

what is this.

[proper minnow]

You will get c to be less than 0

[verbal blaze]

thats it

[gentle sleet]

idk how to latex it ok

[verbal blaze]

ur gonna have to explain this one to the class

[fringe scarab]

is this that x y 1 u were talking about

u multiply it by that vector

[gentle sleet]

yea

[fringe scarab]

what does f being linear have to do with this

[gentle sleet]

the 1 will be the same
y will be y3/2 + 1/21
and x is just x*3/2

[fringe scarab]

owned

[gentle sleet]

isn't matricies imply a linear transformation?

[fringe scarab]

idfk

idk any lin alg !!!!

wait wtf is y

[verbal blaze]

it can't be a linear map they're tweaking

[gentle sleet]

y is the constant term

[fringe scarab]

oh ok

can we name it c then

[verbal blaze]

u can't just put a plus there

[gentle sleet]

yeye

[verbal blaze]

wait that works

ugh

[gentle sleet]

that's why we're acting on x, c, 1
so we can add stuff

[verbal blaze]

linear algebra pls save us

[verbal blaze]

I see that now

[fringe scarab]

how does adding help us

we're doing composition?

[verbal blaze]

they meant add to the constant

matrices can only scale stuff usually

also matrices r functions

[verbal blaze]

ok so we have f

can u get us g also

[gentle sleet]

now compositions are just matrix multiplication

[gentle sleet]

ye

[fringe scarab]

i multiplied F by <x,c,1> and got 3x + 1/2 c + 1

did i do something wrong

[gentle sleet]

did you make it a vertical matrix?

the x, c, 1

[fringe scarab]

yeah

row times column and then you add them all up right

[verbal blaze]

fyre

this one too

[proper minnow]

Which one showcases the constant tho?

[verbal blaze]

this gets us all the properties of matrix multiplication

[fringe scarab]

f isnt linear?

f(2) + f(1) gives 11/2
f(3) gives 10/2

[gentle sleet]

you prob did something wrong then
should be
3/2 x + 0*c + 0* 1
0*x + 3/2 c + 1/2 * 1
0*x + 0* c + 1*1

[fringe scarab]

im confused

[gentle sleet]

second line

[proper minnow]

Even if we get matrix multiplication to work what you can do is just

Show order doesn't matter for 2 of them

[verbal blaze]

yeah it's not

[gentle sleet]

order does matter
they're non commutative

[verbal blaze]

we're not talking about the inputs/outputs of the functions atp

just operations on the slope and constant

[proper minnow]

Oh yeah mb

[proper minnow]

But still same thing

[fringe scarab]

what's the +1 for

[proper minnow]

Only can prove for 2 terms

You can prove for 3 or more by equating them like I did

[gentle sleet]

so we can scale to add

[verbal blaze]

omg

[proper minnow]

This probably just a induction instead of a proof

[verbal blaze]

guys we got associativity

[gentle sleet]

we already have that, no?

[verbal blaze]

no idea

[proper minnow]

^

[fringe scarab]

yeah composition is associative

iirc

[verbal blaze]

what

since when

[proper minnow]

Since ever?

[verbal blaze]

ugh it is nvm

uhh

[proper minnow]

Hm

Just prove by induction fr

[verbal blaze]

is this invertible?

[fringe scarab]

alr ill just trust u but i dont get it

[verbal blaze]

and the other one too

[fringe scarab]

alr next we need to prove closure and we have a group

yes.

[verbal blaze]

closure is trivial

inverses are hard

[gentle sleet]

idk I'll ask wolfram

[verbal blaze]

although

wait what

[proper minnow]

Just get determinant

[verbal blaze]

no we don't get closure

[fringe scarab]

the <x,c,1> vector right

u cant... square it

[verbal blaze]

what if we just

don't think about the vectors at all anymore

[gentle sleet]

it is for f

[fringe scarab]

are all linear maps invertible

wait

[verbal blaze]

yes

no?

[fringe scarab]

idfk

[verbal blaze]

no

[proper minnow]

And how is a group gonna help tho? Like I said we can't apply the pattern and solve

Because ffg pattern could initial have different constants to begin with

[gentle sleet]

and g too is invertable

[verbal blaze]

not all matrices are invertible

well

okay there we go, inverses

we're missing identity

[fringe scarab]

just... add it?

[verbal blaze]

uh

sure okay

[fringe scarab]

3x3 identity matrix exists

[gentle sleet]

I think you can form these from elementary matricies too

[verbal blaze]

okay we have a group

[fringe scarab]

yes ...

i dont think itll be finite though

[verbal blaze]

is that useful at all

[verbal blaze]

it can't be

[gentle sleet]

can that be formed from f and g though?

[fringe scarab]

does it matter?

[verbal blaze]

that said

[fringe scarab]

it doesnt Do anything

[verbal blaze]

prove the group is infinite and you've proved the original statement

[gentle sleet]

not sure

[fringe scarab]

oh no

i composed my function h with the identity function!

help!

[gentle sleet]

there could be 2 that's equal and it'd still be infinite

[fringe scarab]

the identity function is like uno, it comes with ur xbox

[verbal blaze]

oh yeah

uh

prove u can't divide the group

[gentle sleet]

yea ig it doesn't effect the space other than adding 1 more element

[verbal blaze]

then you've proved the original statement

[fringe scarab]

um

if u have an element in ur group a
and then an element b
and a third one, c

if ab = ac then b = c

[verbal blaze]

are we

are we done then

[fringe scarab]

well idk

are we in a group

[verbal blaze]

yeah

[proper minnow]

Are we ?

V sauce music intensifies

[verbal blaze]

the identity + the two function matrices + their inverses
generate a group

[verbal blaze]

is this statement strong enough to prove

[fringe scarab]

it's just True

in any group

[verbal blaze]

I mean strong enough to prove our original claim

[fringe scarab]

doesnt prevent like
ab = cd tho

[gentle sleet]

how's that true in every group?

[fringe scarab]

um

i proved it myself in a little book i own

i forget the proof

[gentle sleet]

won't f(x) = 2x and g(x) = x/2 also form a group?

[verbal blaze]

yea

[fringe scarab]

idk

i think so

[verbal blaze]

it's just z

it's just the whaitsitcalled

the boring one

[gentle sleet]

so how's that true

[verbal blaze]

z +

cyclic group

that's just the cyclic group of infinite order

[fringe scarab]

idk what u do with ff and gg

[verbal blaze]

I have some work I need to get done

[gentle sleet]

matricies, they're just f = [2] and g=[1/2]

[verbal blaze]

idk if we've made any progress in the last two hours except finally saying that hj != hi

[verbal blaze]

that gets you a group

just take the number 1 and multiply it by two until infinity

then divide it by half and get zero

then any of the numbers you've reached can be multiplied amongst themselves w group properties

its isomorphic to the integers + addition

[gentle sleet]

I'm just using it to disprove that that's not true for an arbitrary group

[fringe scarab]

suppose ab = ac
then, abb' = acb'
a'abb' = a'acb'
ee=ecb'
e=cb'
eb=cb'b
b = ce
b = c

[verbal blaze]

yeah

[gentle sleet]

ah flob, I read that wrong

[verbal blaze]

that makes sense

[gentle sleet]

mb

[verbal blaze]

I'm gonna go work on something I'm gonna make progress on

I've decided math isn't for me and that numbers aren't real

[proper minnow]

Lol

This is because you aren't going the right way

Groups and matrix aren't the way

Induction is

[gentle sleet]

I thought that was what Lexica was referring to when she said that we're done

[fringe scarab]

cute how any function and its inverse generate a group isomorphic to the integers over addition

[verbal blaze]

that's not true

[fringe scarab]

oh

why not

[verbal blaze]

uhhhhhhhh

seems wrong

lemme think abt it

no wait it'd have to be its own inverse

[fringe scarab]

yeah that's what i said

i mean it makes sense

[verbal blaze]

involutions

doesn't work with involutions

I don't think

https://en.wikipedia.org/wiki/Involution_(mathematics)

[fringe scarab]

ok cuz this looks like addition but yeah that would ruin it

[gentle sleet]

won't it be Z/1Z or something

Z modulus 1

or 2

[verbal blaze]

z/2z

[fringe scarab]

would that just be -1, 0, and 1

[gentle sleet]

no, just 0 and 1

[verbal blaze]

uhhh

yeah that

[fringe scarab]

oh ok cool

[gentle sleet]

-1 = 1

[fringe scarab]

yeah lol

is that the 2 element group

[verbal blaze]

yag

[gentle sleet]

or 1 if f = I
i think

[fringe scarab]

Z/2Z as the function as its own inverse
or as the function and a different inverse

[verbal blaze]

hey wait

if we assume two sequences are equal

we would be able to wrap them apart w their inverses right

[fringe scarab]

wrap them apart?

[verbal blaze]

yeah

like

in our group

fgfg g'f'g'f' or smth

however it is

[fringe scarab]

oh

[verbal blaze]

ykwim

[fringe scarab]

u mean invert the sequence?

[verbal blaze]

could we prove that the unwrappings are unique

you could mayb prove that f g' or g f' can never occur

[fringe scarab]

well do all the elements we generate have inverses

like does fg as a matrix have an inverse

if u take the product of 2 invertible matrices do u get an invertible matrix

[verbal blaze]

yeah

[verbal blaze]

yeah

[gentle sleet]

so is f'g' = (fg)'?

[fringe scarab]

no, (fg)' is g'f'

u reverse the order

[gentle sleet]

oh yea that makes sense

[fringe scarab]

oh

then ofc the product of two invertible matrices is invertible

it says it right there

what are we trying to say in the context of groups then

that an element only shows up in 3 places? both products with itself and the identity and then one time in the table?

i dont have much experience with infinite groups

[verbal blaze]

inverses are unique

so

qed gg

[fringe scarab]

im going to bed

[verbal blaze]

I think we did it

more or less

[gentle sleet]

nah, we've been running around in circles

[verbal blaze]

I think this proves it

well

no

[gentle sleet]

but can it be formed by only one sequence of inverses

[verbal blaze]

ugh

[gentle sleet]

yea

[verbal blaze]

wonderful use of3 hours

[vast shale]

that was entertaining so yes

good job u lot

[vocal sleet]

@prisma rain Has your question been resolved?

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[molten kettle]

Hello, sorry if my english is bad, i'm not a english native so my spelling in some words could be wrong.
I'm looking for an explanation on how to use the property that something hasn't been completely discovered on a field of mathematics to make use of it and create a new, using bases of maths that we already knows

[molten kettle]

The thing is i don't want to show the work behind to not having someone taking the credits for it, so i'm just looking forward for an explanation on general, because i'm kinda "new" on the fields of maths and all since i'm on physics normally

So just looking for an general explanation

[trail mesa]

"i don't want to show the work behind to not having someone taking the credits for it"

uhh what

if you use progress made by other people, you should credit them

[molten kettle]

Yes i do know that, but that isn't all to proof something, is it?

I mean, putting references of other people, can this be enough?

[trail mesa]

yeah, ofc

[molten kettle]

So with this i can make a new thing just by citing references of formulas that's already exist? Or there more steps?

@trail mesa

Which tools do i need to proof to say that something works, even though it hasn't been proved

Alright got my answer by someone else, have a good day!

.close

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[tired dock]

Can anyone explain the " Multiply to get common denominators " step please?

[brittle umbra]

when ur adding two fractions the denominator has to be the same

for ex the first term u multiply by x+2/x+2 which is essentially 1 (multiplicative identity) so it’s not changing the value of the term but we are manipulating its form

such that you can now add them tgt

[graceful wave]

uh help

[vocal sleet]

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).

[graceful wave]

uh

mb

idk how this works 😭

[tired dock]

so we multiply each number by the denominator of the other number (divided by itself to equal 1 and the result doesn't change) to get the same denominator on both sides?

[tired dock]

ask there

[brittle umbra]

Yea exactly

[tired dock]

Thank you for helping me, your explanation style was smooth ❤️

.close

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[vast shale]

I am stuck on the first question. The only thing I’ve thought to do is get A+B+C=158

I know we need two equations

But I am stuck I was thinking of volume but then when taking the partial derivatives I get everything =0

[astral pilot]

what do you think is the formula for volume?

[vast shale]

V=ABC

[astral pilot]

yeah

substitute C = 158-A-B

[vast shale]

oh maaa god. Silly me. Thank you.

[astral pilot]

glad, I could be of help

[vast shale]

Ok still stuck sorry

This is the working I did. I took partial derivatives and then set =0. Then got two equations

Made A the. Subject of one and B the subject of other

Then I tried plugging that into v but failed…

[astral pilot]

You get linear equations in two variables

158 - 2B - A = 0
158 - 2A - B = 0

solve them simultaneously

Or, a simpler way, 2B + A =158; 2A + B = 158
Therefore, 2B + A = 2A + B, i.e. A = B

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[haughty nexus]

I have a question about designing LTI-systems. The question is about is making a stabile low pass filter that has a cutoff frequency of 30 Hz by putting 2 poles and 2 zeros on a graph (reel, imaginary as axis). I dont understand the relationship between where the poles and zeros are placed on the graph and the output of the system. How do I know where to place these poles and zeros so that they give me the desired cutoff frequency while having the characteristics of a filter i want?

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@haughty nexus Has your question been resolved?

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@haughty nexus Has your question been resolved?

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@haughty nexus Has your question been resolved?

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@haughty nexus Has your question been resolved?

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[bitter holly]

why f(x) = 2x is injective and not surjective, when defined from N to N

[bitter holly]

oh maybe because on the codomain there are gaps from one step to the other

[versed bane]

we can use the definition of injective and surjective.

[versed bane]

yes essentially

do you know the definitions

[bitter holly]

like f(1)=2 f(2)=4

then there is the element 3 of the codomain

ok perfect thanks

[versed bane]

for injectivity 2a=2b means a=b, because you can divide by 2

[bitter holly]

but I am thinking that if we had a function from N to a set defined as {x|x in N, k in N and x=2k} then it would be surjective too

[versed bane]

yes, though i would describe the set differently (i am assuming you mean even whole numbers). that’s called corestriction of the function. we took the codomain and made it smaller.

[bitter holly]

yeah

[versed bane]

it would be surjective onto that set yeah

[bitter holly]

would the description of the set {2k | k in Z} be better?

[versed bane]

yeah that’s better, k would be in N in this case because we only need positive whole numbers. but yes i would describe even integers that way

[bitter holly]

ok, thanks

[haughty nexus]

no but it is what it is 😞

[bitter holly]

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[raw coral]

Help

[raw coral]

Could anyone help me out solving this

[wild ivy]

all ? one?

[raw coral]

ii) and iv

I tried internet but I can't get it

[wild ivy]

$tan(alpha) = sin(alpha)/cos(alpha)

[raw coral]

bruh you know what this stffs is deleted for the latest publishers of the same book but our school still uses the ol

...

What is alpha here ? Is that theta*

i just got the trigonometry table and 3 identities how do solve it using those

@wild ivy

[wild ivy]

ya. theta

just minute.

[raw coral]

Ok

[wild ivy]

are u clear?

[raw coral]

Yup thanks a lot

i got it all

can you solve no iv too

[wild ivy]

ok.

[raw coral]

you did it so smoothly the one which my teacher did was kinda long and boring

[wild ivy]

what are u doing?

[raw coral]

i genuinely doing the trigonometry last exercises for revison

i haven't touched it for like weeks

[wild ivy]

talk about that alot.

talk in DM.

[raw coral]

k

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[vestal fractal]

is there a way to make desmos show work in square-wave "quantized" steps

[vestal fractal]

it should be stepping, non-continuous

[flat whale]

google is your friend

non-continuous is "discrete"

[vestal fractal]

would appreciate you not assume i haven't googled before i type here :)

[flat whale]

nah

[flat whale]

i dont' see this word in your search

[vestal fractal]

I am one of those "practical men" as described by Mao. Very little formal education in mathmatics

less than highschool lol

[flat whale]

¯_(ツ)_/¯

googling isn't a skill taught specific to math

[vestal fractal]

to know the term "discrete" to refer to a "square wave step" is

[flat whale]

great i told you

[vestal fractal]

yeah so just using the floor function does what i wanted

thank u

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[craggy imp]

I need help with this problem. I have no idea where to start.

[junior notch]

!15m

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[junior notch]

Sorry

[craggy imp]

!15m

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[craggy imp]

<@&286206848099549185>

[tawny sorrel]

is this for calc?

[vocal sleet]

@craggy imp Has your question been resolved?

[craggy imp]

yes, optimization

[tawny sorrel]

ok start with the equation for the length of the ladder

[craggy imp]

i dont know how to do that

i think i need to use pythagoras

[tawny sorrel]

you basically have two triangles, the sum of their hyptonuse is the length of the ladder

you see what two triangles im talking about?

your writing the equation for the length of the ladder in terms of the angle the ladder makes with the ground

[craggy imp]

i see the 2 triangles but I dont know how to make the equations in term of the angles

[tawny sorrel]

using trig functions

if x is the angle the ladder makes with the ground

you have opposite side is 8ft for the the first triangle

so what is the hypo for the first triangle

[craggy imp]

how would i find that if i dont have the other side

the angle isnt given either

[tawny sorrel]

you aren't finding the length right now

you are just setting up the general equation

[craggy imp]

just in terms of x?

[tawny sorrel]

yes

[craggy imp]

sin(x) = 8/y

[tawny sorrel]

yeah but you want it in terms of y, length of ladder

so rearrange it

[craggy imp]

y = 8/sin(x)?

[tawny sorrel]

yes

now do the same thing for the second triangle

[craggy imp]

cos(x) = 4/y

y = 4/cos(x)

[tawny sorrel]

yeah so the sum of the two equations is the length of the ladder

write the full equation

[craggy imp]

8/sin(x) + 4/cos(x) = hyp

[tawny sorrel]

ok good

now use the first and second derivative to find the minimum

[craggy imp]

so find f'(x) of f(x) = 8/sin(x) + 4/cos(x)

[tawny sorrel]

yeah

[craggy imp]

f'(x) = -8cos(x)/sin^2(x) + 4sin(x)/cos^2(x)

[tawny sorrel]

yea set that equal to 0 and solve for x

[craggy imp]

from what domain would i do it from

[tawny sorrel]

what do you mean

[craggy imp]

theres a lot of values that would fit

i tried it on desmos and theres a lot of x intercepts

[tawny sorrel]

just find one for now

simplify the equation first and you'll see it doesn't matter

[craggy imp]

i have no idea how to simply from there

[tawny sorrel]

$\frac{8\cos x}{\sin^2x}=\frac{4\sin x}{\cos^2x}$

[twin meteor]

parm

[tawny sorrel]

get rid of the denominators

[craggy imp]

cos^2(x)(8cos(x)) = sin^2(x)(4sin(x))

[tawny sorrel]

yea which is just 8cos^3(x) = ...

[craggy imp]

4sin^3(x)

2cos^3(x) = sin^3(x)

[tawny sorrel]

yeah one more step

[craggy imp]

2cos(x) = sin(x)

?

[tawny sorrel]

no

[craggy imp]

i cant take 3rd root?

[tawny sorrel]

well you gotta root the 2 as well

[craggy imp]

oh yeah

[tawny sorrel]

you can

[craggy imp]

3rdroot(2)cos(x) = sin(x)

sin^-1(3rdroot(2)cos(x)) = x

[tawny sorrel]

no youre trying to get it to one trig function

you have sinx one side and cosx the other

how you get it to be 1 trig func

[craggy imp]

is it the trig identity for sin?

[tawny sorrel]

remember tan = sin/cos

[craggy imp]

tan(x) = 3rdroot(2)

[tawny sorrel]

yeah last step

[craggy imp]

now do i inverse tan?

[tawny sorrel]

yes

[craggy imp]

x = tan^-1(2^1/3)

[tawny sorrel]

that should be the answer but you have to verify if its a minimum

[craggy imp]

i have to take second derivative?

[tawny sorrel]

yeah

or idk if youre teacher is fine with it just pick a value just to the right of x and see if its larger

[craggy imp]

in the first derivative?

[tawny sorrel]

and to the left of it see its larger as well

in the original function

you should probably just do the second derivative

[craggy imp]

ok

im just getting 0 because its a constant

[tawny sorrel]

whats 0

[craggy imp]

a min or max?

critical number

[tawny sorrel]

no what do you mean you are just getting 0

[craggy imp]

the tan^-1(2^1/3) derivative

im getting 0

should i do (tan^-1(2^1/3))/x

derivative of that?

[tawny sorrel]

you shouldn't be getting 0

im getting 51.56

[craggy imp]

can you show me how

[tawny sorrel]

like just put it in your calculator what is tan^-1(2^1/3)

oh hold on you mean second derivative

[craggy imp]

yeah

[tawny sorrel]

you wanna find the derivative for the general formula

[craggy imp]

oh ok

[tawny sorrel]

tan^-1(2^1/3) is not the function that is the answer for f'(0) =

[craggy imp]

im getting 51.52 degrees when i plug it in calculator

[tawny sorrel]

yea that should be the answer for where the minimum is but you got to verify that it is a minimum with the second derivative test

[craggy imp]

what equation should i take the second derivative from

x = tan^-1(2^1/3)

this?

[tawny sorrel]

no the original equation for the derivative

x = tan^-1(2^1/3)
this equation came from when you were solving f'(x) = 0

[craggy imp]

cos^2(x)(8cos(x)) = sin^2(x)(4sin(x))
this?

[tawny sorrel]

yeah that

no

not that

that is still f'(x) = 0

[craggy imp]

8/sin(x) + 4/cos(x) = hyp

this one?

[tawny sorrel]

this one

[craggy imp]

ok

[tawny sorrel]

its gonna get really ugly

[craggy imp]

(8sin^3(x)+16sin(x)cos^2(x))/sin^4(x) + (4sin^3(x) + 8sin^2(x)cos(x))/cos^4(x)

[tawny sorrel]

i'd just put it into wolfram alpha to check i didn't calculate it

[craggy imp]

eyah i did

some solver

[tawny sorrel]

k now you gotta check if it is positive 51.56

[craggy imp]

how

[tawny sorrel]

you just plug in that value for x in the second derivative

[craggy imp]

ok

i got 49.80 using 51.52 degrees

[tawny sorrel]

so its a min

[craggy imp]

how did you know that

[tawny sorrel]

first derivative is 0

first derivative 0 means it is a turning point happens at the original f(x)

second derivative is positive which verifies its a min

if it was negative its a max

[craggy imp]

ok so now would i plug this into the original equation

the x value that i have

[tawny sorrel]

yes

[craggy imp]

8/sin(x) + 4/cos(x) = hyp
in this one right

[tawny sorrel]

yes

[craggy imp]

THANK YOU SO MUCH

ive been stuck forver

[tawny sorrel]

youre welcome

[craggy imp]

if you have time can you help me with one more?

[tawny sorrel]

i dont really have time but jost post the question ill tell you what to look for

[craggy imp]

im pretty sure i use similar triangles

[tawny sorrel]

this is gonna take me too long i cant help close this channel and open snd ask in a new one

[craggy imp]

howdo i close it

[tawny sorrel]

.close

[craggy imp]

.close

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[mighty jackal]

I would like to know what part I was wrong about. In the module it says the answer should be (e-1)

[mighty jackal]

The topic is Areas in the Plane, Calculus

[marsh charm]

elne - e = -e

not 0

prob one reason why u got it wrong

[outer warren]

draw a graph and you should see what you missed

[outer warren]

it is 0

[mighty jackal]

lne=1
e(1)-e

[marsh charm]

r u sure

i could be wrong

cause i actually dont know

but

this shit never lies

[mighty jackal]

yea but e(1)-e is e-e which is 0

[marsh charm]

i see

i didnt see

[mighty jackal]

but i think i didnt miss anything

[marsh charm]

e1 anywehre

[mighty jackal]

Maybe the module is weong

[outer warren]

the module is fine

[muted plaza]

yeah i think it is, factor out an e

[mighty jackal]

Ok

[muted plaza]

e(lne-1)

e(0)

=0

[outer warren]

the integration work you've done is fine
you just missed a component for the total area

[mighty jackal]

Oh

[outer warren]

draw a graph and you should see what you missed

[mighty jackal]

idk what i missed

Ok

I still dont get it

[outer warren]

can you show what you drew

[mighty jackal]

i didn't draw anything

hmm

Ok wait

[outer warren]

note that the integral you have doesn't give the area of the whole desired region

only part of it

specifically the part from x=1 to e

[mighty jackal]

Yea the area doesnt make sense

Hmm

So the 1x1 area was not included in the e-2?

[outer warren]

i.e. you've only calculated the area of this part

[mighty jackal]

Yea

Okay i got it

how do i write it in my solution

[outer warren]

at the start, A = 1 * 1 + (your integral)

leading to
A = 1 + e - 2
= e - 1

[mighty jackal]

Okay thank you so much

[outer warren]

and provide a rough sketch as well

[mighty jackal]

ok ok

oh and can i ask a question

Check the final part of the solution of my other problem

Was what i did correct?

I changed it to positive 8/3 cause.. the area cant be negative?

can anyone answer me? just a quick simple question

<@&286206848099549185>

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@mighty jackal Has your question been resolved?

[mighty jackal]

.close

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[errant sigil]

Not sure where to go from here

Solution

Not sure how this arises

Partial derivatives btw

[mighty nacelle]

for dA/dr, you forgot to use the chain rule for v

for partial derivatives, you have to treat the other variable like a constant

imagine its pi or something

so you dont need product rule for dA/dtheta

[errant sigil]

Oh, was I meant to do that before the product rule?

I thought we did that at the end

Hmm

Lemme try that again

[errant sigil]

@mighty nacelle when u derive A with respect to r, you also take account of what’s inside the bracket since its chain rule

But

Fourth line

It’s the same thing, minus the R ofc

So does that just mean u can’t derive tan theta or..

I hope I’m making sense

[mighty nacelle]

im not really understanding

[errant sigil]

Ignore all of that

[errant sigil]

Why multiply tan theta (4th line)?

Basically is what I’m asking @mighty nacelle

[mighty nacelle]

you have r*arctan(r tan(theta)) right

left term is the first part of the product rule

right term is r*derivative of arctan(r tan(theta))

outer function is arctan

inner function is r*tan(theta)

do the chain rule

[errant sigil]

Essentially when u derive tan theta, is it just tan theta then?

[mighty nacelle]

differentiate r*tan(theta)

what type of function is that?

remember that theta is a constant

[errant sigil]

Can u write it down 😭

Sorry

[mighty nacelle]

take a second and think about it

[errant sigil]

I’m js confused cus ur saying u don’t need the product rule,, what I’m also confused about is why it’s multiplying tan theta at the end

[errant sigil]

I’m trying😭

[mighty nacelle]

ok lets try this

differentiate arctan(2r)

[errant sigil]

1/1+(2r)^2 ?

[mighty nacelle]

f(x) = arctan(x) and
g(x) = 2x
what is f(g(x))?

[errant sigil]

arctan(2x)

[mighty nacelle]

ok, whats the derivative of f(g(x))

[errant sigil]

^

[mighty nacelle]

no, use the function notation first

[errant sigil]

Okay one sec

f’(g(x))= 2/1+4x^2

…

Right?

[mighty nacelle]

derivative of f(g(x)) is f'(g(x))g'(x)

this is important

[errant sigil]

Oo

So

[mighty nacelle]

if i replace the g(x) with x*tan(y), what is g'?

[errant sigil]

x/1+tany^2

sorry i’m so exhausted but if that’s wrong ill pick it up tomorrow morning

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@errant sigil Has your question been resolved?

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[junior brook]

you can do isolation of variables technique since the left and right sides are both products of x and y

the right side is correct, 2xdx, but the left side is going to be dy divided by: (e^-tan(y) cos^2(y) )

so that will turn into e^tan(y) *sec^2(y) dy

since 1/e^(-x) is e^x

and 1/cos is sec

ye

then afterwards its an easy u sub to complete the integral on the left side

you got this!

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[torn prawn]

how do i find the nth degree of this? i've tried like 20 different things and they're all wrong, including stuff like x^4 or 4. i'm not really sure what the second part is asking.

[meager shoal]

what is the mclaurin series in e^x

[junior brook]

find a pattern that changes for each recurring term of the sequence

[meager shoal]

as summation symbol

[torn prawn]

what does that mean

[junior brook]

the question is asking for a general term of the series, not something specific like 4

[torn prawn]

ohh okay

[meager shoal]

note that series is just a sum of sequence 👀

[torn prawn]

.close

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[acoustic turtle]

Hey there I'd like to understand how we get the fraction -2/7 I simply know the steps but I don't understand why it works.

[junior brook]

Hi! So first you want to substitute the value 1/2 in for x. the denominator becomes 1/2 - 4, which can be rewritten to 1/2 - 8/2. doing this subtraction you will get -7/2 in the denominator.

1/(-7/2) just flip the bottom fraction and you will get -2/7

[acoustic turtle]

thank you owoLight for responding so quickly

I apologize I'm not seeing the connection "rewritten to 1/2 - 8/2."

[junior brook]

4 can be rewritten as 8/2

so you can change it to 8/2 in order to make the math easier

[acoustic turtle]

yes that's true

[junior brook]

1/2 - 8/2 they have similar denominators so you can just subtract the numerators

[acoustic turtle]

gonna write this down real quick if that's fine I think I'm understanding you.

[junior brook]

of course. take your time

[acoustic turtle]

you got finals week as well lol

[junior brook]

yeah bro 😓 not looking forward to it

[acoustic turtle]

I think you'll be fine!

[junior brook]

thank you for the support! im sure your finals will be awesome as well (if you have them)

[acoustic turtle]

so I've got 1/2 - 8/2 that'll be -7/2 right?

[acoustic turtle]

Yep!

[junior brook]

yep, exactly

[acoustic turtle]

and then to get rid of the bottom

I simply multiply the reciprocal with 1 so that'll be -2/7

[junior brook]

yep

[acoustic turtle]

damn lol

I missed a lot of this stuff when I was younger.

thank you for the help man.

!close

[junior brook]

any time

[acoustic turtle]

.close

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[white tusk]

$89.95 cm^{2} = \frac {23.5*h}{2}$

[twin meteor]

Beata Szydło

[white tusk]

How do I find h?

Do I have to divide 23.5 by 2?

And then subtract from 89.95 cm^2?

@loud walrus

@mellow void

@mental falcon

[mental falcon]

don't ping random people

for no reason

that'll just make people less inclined to help

[white tusk]

Ok, can you help me quickly?

Im actually in class.

@steep crater

[steep crater]

Wat

Y ping

[white tusk]

How to find h?

[white tusk]

To help me.

[steep crater]

Dude stop ping random ppl

Anyway is the unit cm^2 on RHS

[white tusk]

Whats rhs

[steep crater]

Right hand side

[white tusk]

Idk what that even is

The area of a trapeze is 89.95 cm^2

a = 13 cm
b = 10.5 cm

and so I used the formula for trapeze

[steep crater]

Then solve ig

Multiply 2 divide 23.5

Theres h

What is this

[white tusk]

$\frac{(a+b)*h}{2}$

[twin meteor]

Beata Szydło

[white tusk]

I move 23,5 cm to the other side?

[steep crater]

What

Just divide

[white tusk]

Show me @steep crater

[steep crater]

Times 2 both sides

[steep crater]

Latez on phone sucks

[white tusk]

89,95 cm^2? : 23,5?

@steep crater

[steep crater]

What

Nono

[steep crater]

Times 2 both sides dude

89.95 x 2 = 23.5 x h

[white tusk]

Ohhh and the. Divide

Ooooohhhhh

Ty

[steep crater]

Yess

[white tusk]

It makes sense

Ooohhh

I get it

[steep crater]

Np

[white tusk]

To get rid of 2 you need to divide by 2

[steep crater]

Ya

[white tusk]

.close

[vocal sleet]

Channel closed

Closed by @white tusk

Use .reopen if this was a mistake.

[vocal sleet]

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[elder fable]

Can anyone help me with eliptic curve crypthography ?

[elder fable]

and ElGamal protocol

import random

Curve parameters

p = 257
a = 0
b = 7

Operations

Inverse

def inverso(a, p):
return pow(a, -1, p)

Sum

def soma(P, Q):

if P == "O":
    return Q
if Q == "O":
    return P

x1, y1 = P
x2, y2 = Q

if (x1, y1) == (x2, -y2 % p):
    return "O"

if (x1, y1) == (x2, y2):
    u = ((3 * x1 * x1 + a) * inverso(2 * y1, p)) % p
else:
    u = ((y1 - y2) * inverso(x1 - x2, p)) % p

v = (y1 - u * x1) % p
x3 = (u**2 - x1 - x2) % p
y3 = (-u * x3 - v) % p

return [x3, y3]

Multiplication by a scalar

def multiplicacao(k,P):
R = "O"
for i in range(k):
R = soma(R, P)
return R

ElGamal Protocol ECC

Alice public and private key

a = random.randint(2, 255)
A = multiplicacao(a, [109,229])

Bob encript "m"

m = 3
M = multiplicacao(m,[109,229])

r = random.randint(2,255)
c1 = multiplicacao(r, [109,229])
c2 = multiplicacao(r, A)
c2 = soma(c2, M)

Alice desencript

if c1 == "O":
c1_sim = "O"
else:
c1_sim = [c1[0], (-1 * c1[1]) % p]

M_2 = multiplicacao(a, c1_sim)
M_2 = soma(c2,M_2)

print(M,M_2)

i have this python code

My english is also not the best i apologize for that

when Alice desencripts the whole thing she gets the same encripted point that Bob used to encript "m" , so they share a secret which is that point M

and if she or an "attacker" wants to find out "m" they need to solve the discrete logarithm problem which is "impossible"

but isnt the whole point of this to make Alice find out "m" ?

In this specific Protocol ElGamal ? Cause like in Diffie-Hellman if my understanding is correct the goal is to give both parties a "secret" which only they share and with that they can encript messages with other systems using that secret as a key , but here they specifically encript "m" so i thought that the goal should be, Alice getting "m" without other people knowing.

[vocal sleet]

@elder fable Has your question been resolved?

[vocal sleet]

@elder fable Has your question been resolved?

[vocal sleet]

@elder fable Has your question been resolved?

[flat whale]

try #computing-software

[elder fable]

ok ty

[vocal sleet]

@elder fable Has your question been resolved?

[blissful sentinel]

you might also want to try a cs server instead, see #old-network

[vocal sleet]

@elder fable Has your question been resolved?

[vocal sleet]

Channel closed

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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.

[rare solar]

i cant figure out the domain on this bad boy

[still mica]

Is h the length of the base?

[rare solar]

nno

h is the height

b is the base

[still mica]

So, $V > 0$ => $\frac{4800b - b^3}{4} > 0}$?

[rare solar]

Yes

[still mica]

Did you solve that inequality?

[rare solar]

wait no

i think your function is wrong

[still mica]

4800b mb

[twin meteor]

Adarsh
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