help-26

[wet harbor]

if you but vectors C and E in a matrix and calculate determinant

the value is the area of that parallelogram

carry that up to higher dimensions

[wraith iron]

okay

its not obvious for me yet how this links to recurrence relations

[wet harbor]

a recurrence relation can be turned into a matrix

eigenvectors can help you easily calculate powers of matrices

the recurrence will turn into an arbitrary power of a matrix and you use the eigenvalue decompostion to calculate it

resulting the explicit solution to the recurrence relation

[wraith iron]

oh wow

[wet harbor]

i did that to calculate the fibonacci sequence before

[wraith iron]

how efficient is it?

[wet harbor]

i would say it's pretty good

for higher ones it would be more difficult

[wraith iron]

i see

because I think there are explicit formulas for arbitary constants coeficients

[wet harbor]

but the thing is it has a systematic theme to it

so you can see pattern and write a solution

yes there is

and these formulas can be proven with that exact method

[wraith iron]

very good

[wet harbor]

there are other methods too

you can turn the constant coeff equation into a differential equation and solve it

[wraith iron]

what?

[wet harbor]

although in many cases it can reduce to a simple algebraic equation

told you that equations can come in many shapes ?

well there is a set of equations called differential equations

those are ones which involve derivatives

these are included in calculus

[wraith iron]

yeah i wrote down a set of those for the n-pendulum

but its not solvable

[wet harbor]

there is an easy way to solve the constant coeff ones

you just substitute e^mx

and calculate m

it will result in the same characterstic equation

[wraith iron]

can we try dot x + ddot x = 0

i want to see how it works

[wet harbor]

i can't understand it

you mean time derivative ?

[wraith iron]

$\dot{x} + \ddot{x} = 0$

[thorny flame]

violet

[wraith iron]

yeah

i use that notation

[wet harbor]

substitute e^mx

[wraith iron]

because its easier

[wet harbor]

it is

derivative of e^mx is me^mx

so first one becomes me^mx and second one is m^2e^mx

the e^mx cancels out

you get m+m^2=0

[wraith iron]

oh

[wet harbor]

so either m = 0 or m = -1

[wraith iron]

could have never thought differntial equations and recurrence relations are so similar

[wet harbor]

incredibly similar

that's why one of the methods of solving differential equations is turning them into recurrence relations

[wraith iron]

and one of the methods of solving recurrence relations is turning them into differential equations?

[wet harbor]

yes xD

[wraith iron]

haha

[wet harbor]

whichever the one that is easier to solve you switch to it xD

[wraith iron]

group theory seems fun

[wet harbor]

math is fun

also group theory is related to number theory

[wraith iron]

what is the funniest math area?

[wet harbor]

idk i love all of it

i will be biased toward mathematical analysis since i spent like 4 years self studying the book of rudin

but it's all fun

[wraith iron]

isnt analysis just calculus

but more proofs

[wet harbor]

sorta like that

but it establishes the meaning of infinity

and creates a basis which you can rely on to prove anything in calculus

and more

you can prove that infinities can have different sizes xD

[wraith iron]

countably infinite

and uncountably infinite

yes?

[wet harbor]

uncountably infinite ones have infinitely many sizes

there is a theorem which says that the power set of any set can never be in one to one correspondance with that set

hence every time you calculate the power set of a set it gets bigger

[wraith iron]

what do you mean when you say a set is in one to one corresponance with another?

equal cardinality?

[wet harbor]

that is the way we define cardinality yes

if you can find a bijective map between two sets

they have the same cardinality

[wraith iron]

oh

so you can assert a set is smaller than another if the first to second is one to many

that is smart

[wet harbor]

you can find one to many sets

but the key is

you cannot find a one to one

every map you find must be many to one

if that's the case

one is bigger

[wraith iron]

why can you not find one to one sets

[wet harbor]

example, the real numbers are bigger than the integers

[wraith iron]

because they are equal by extensionality?

[wet harbor]

you mean maps

sets cannot be called one to one

[wraith iron]

yes maps

[wet harbor]

it's the map

[wraith iron]

why not one to one?

[wet harbor]

because you can prove that you can't

that's why

can you find a one to one map between reals and integers ?

[wraith iron]

no

[wet harbor]

for the purpose of this let's stick to reals between 0 and 1

[wet harbor]

did you try ?

[wraith iron]

there is a way you can make one more real number out of existing

when you have coreesponded them with the naturals

[wet harbor]

great you know the drill

[wraith iron]

taking the 1st decimal digit of the first

so on

[wraith iron]

no, because i realized

[wet harbor]

that's one example of why you can't get bijections between sets sometimes

it's called cantor diagonalization

[wraith iron]

there is something about infinity

[wet harbor]

it's crazy and I love it

[wraith iron]

is there a set bigger than R.

P(R)?

or are they equilvalent

latter i think

[wet harbor]

the power set of R is bigger

that's the point

the power set of the power set of R is bigger than the power set of R which is bigger than R

[wraith iron]

i meant R for the reals

does that change anything

[wet harbor]

nope

the set of all subsets of R is bigger than R

[wraith iron]

i thought there were only countable and uncountable infinity

[wet harbor]

and now there is infinitely many of them

[wraith iron]

so countable infinities are those that biject N

[wet harbor]

yes

[wraith iron]

and uncountable infinities are those that

um

are many to one in correspondence to N

?

[wet harbor]

that are not in bijection with N

xD

[wraith iron]

okay

[wet harbor]

just add the word not and it fixes everything

much like a finite set is one which is in bijection with the set {1,2,3,...,n}

and an infinite set is one which is not finite

[wraith iron]

yea

by the way, what topics do one learn in uni?

like the general

path

considering maths

[wet harbor]

i don't even know what you will learn

[wraith iron]

it will be some years before I go ( or not go)

[wet harbor]

my college is the faculty of engineering

I learnt all the math on my own

they don't do rigorous math here

[wraith iron]

how much time do you spend on maths everyday?

[wet harbor]

nowadays not a huge amount

but when I was younger I would do math whenever i'm not doing anything else

my free time was all math

[wraith iron]

thats me

[wet harbor]

yes as you get older other stuff start to kick in

you know getting job and stuff

you also often feel like you wanna turn your brain off for a while

but in your age it's normal

so make the most of it while you can

[wraith iron]

ty

[wet harbor]

^ ^

[wraith iron]

any thing you recommend me to learn in maths (self-study, no teacher)

because as I mentioned there are many areas which i dont have the knowledge

[wet harbor]

well, mathematical analysis and abstract algebra are goats

do those

[wraith iron]

how much time do they take

[wet harbor]

and anything else will be easy

time is relative

I spent 4 years on a book you can finish in less than a year

but I like to take my time because I don't read the proofs

I do them myself

and If I leave and come back I redo all of them

I have proven chapters from rudin more than 6 times

because if how many times I came back to revise

and I'm doing one more revision now as we speak

[wraith iron]

this?

[wet harbor]

no real analysis only

there is baby rudin, papa rudin and grandpa rudin

the picture has papa rudin

you need baby rudin

[wraith iron]

where does it say papa rudin?

[wet harbor]

i'm saying this for a reason, papa rudin legit builds upon the last chapter of baby rudin so you literally need it

it doesn't say that's what people call them xD

baby rudin = principles of mathematical analysis

papa rudin = real and complex analysis

grandpa rudin = functional analysis

[wraith iron]

what is functional analysis?

[wet harbor]

that's a topic for another day xD

there is something called functional

which is a map that maps entire functions

[wraith iron]

just a short introduction will do

[wet harbor]

it doesn't map value

[wraith iron]

that is crazy!

[wet harbor]

but maps the whole path

optimization over a functional is called the calculus of variations and that's where lagrangian comes in

[wraith iron]

how do you 'optimize'

or what do you 'optimize' for

[wet harbor]

you learn that in calculus

the take derivative and make it zero thing

but in a more general sense

you optimize for something you don't know

the meaning of that comes when you use it in physics

only then do you know what you are optimizing

in lagrange mechanics you are optimizing what is called the "Action"

you can look up "Principle of Least Action"

the action is a functional which relies on the function representing the path

in physics it would be the particle path or the pendulum path

it was chosen to make the action stationary

there is another principle in optics called fermat principle

says light travels in the path which takes the least time

total time spent would then be the functional and you would optimize over it

there is much more than that ofc but I'm just giving you a simple idea

[wraith iron]

i see

i was looking at the table of contents

i saw topology

dont know if i'll like it

[wet harbor]

well I'm here xD I know the whole table of contents by heart

[wraith iron]

thats one area that i no nothing about

know*

[wet harbor]

first is introduction to real numbers

then general topology which is really nice

it won't rely on anything you know

rudin defines everything

hence the name baby rudin xD

the book pretends you are a baby

in the topology chapter you will get some definitions for infinity and sequences and all that

then definition of a metric space

what neighbourhoods are

and basic definitions like limit points and interior points, open sets, closed sets etc

[wraith iron]

they dont look very friendly lol

metric space and all that

[wet harbor]

it's an old book xD

but I like that stuff

[wraith iron]

its like you build everything from ground up

[wet harbor]

yesss

in chapter 9 you jump to vector values functions

in 10 you carry integration to more than one dimensions with differential forms

and last chapter introduces a beautiful generalization of integration

the lebesgue integral

whenever i wanna prove anything I just resort back to rudin in analysis and artin in abstract algebra

a basis in algebra and one in analysis makes you able to understand anything

[wraith iron]

i see

i'll read them when im free then

i guess i should start with analysis?

[wet harbor]

start with whichever you like first

I did analysis because it was more useful to me

it shows up more in engineering

also I liked the crazy concepts of infinity

but the two fields will cross from time to time

chapter 9 in rudin requires matrices

and in artin an introduction to solving differential equations with matrices required some analysis

[wraith iron]

just to make sure

artin's abstract algebra

[wet harbor]

yes

vector spaces then linear transformations

then chapter 5 is symmetries

[wraith iron]

linear operators

not transformations

they may well be the same

though

[wet harbor]

they aren't

but it speaks about both

linear transformations are from one vector space to another

linear operators are from one vector space to itself

so you say the map operates on a vector space

hence the name operator

[unreal quest]

So this one's a mapping and the other one a transformation?

[wraith iron]

I think those books will take a while, they dont look easy

i'll have a read anyways

hopefully i can get through them

after all i have got 3 years before i go to uni

i have to go now, see you later mohsen

many thanks for your teaching

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[wet harbor]

.reopen

[wet harbor]

i mapping is a transformation

[wet harbor]

@wraith iron happy to help ^ ^

.close

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[unreal quest]

From what I've heard, a transformation maps an element of a set onto an element of that same set while mapping is the more general term

Could be wrong though

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[neon iron]

Hi, how do I prove that:
8cot(pi/4) + 4tan(pi/8) + 2tan(pi/16) + tan(pi/32) = cot(pi/32)
I've attempted to rewrite it as 8 + 4tan(4pi/32) + 2tan(2pi/32) + tan(pi/32) and use the double-angle formula tan(2x) and the quadruple-angle formula tan(4x) but so far I haven't yielded any useful results
Can you suggest some approaches to this problem

[topaz sinew]

@neon iron Has your question been resolved?

[neon iron]

<@&286206848099549185>

I wonder if there is a nice combinatorial proof with generating function. I will keep this question on the back burner 🙂

[neon iron]

I attempted using algebra and trig identities but it didn't seem like it would lead to a useful result

[topaz sinew]

@neon iron Has your question been resolved?

[topaz sinew]

@neon iron Has your question been resolved?

[neon iron]

$8cot(\frac{\pi}{4}) + 4tan(\frac{\pi}{8}) + 2tan(\frac{\pi}{16}) + tan(\frac{\pi}{32}) = cot(pi/32)$

[thorny flame]

Sukiyaki

[neon iron]

For anyone who wants to help him

[neon iron]

@neon iron

I dont know if this is the correct answer.

but $8cot \frac{\pi}{4}$ should be $\frac{8}{tan\frac{\pi}{4}}$ as the Reciprocal rule for COT should be $\frac{1}{tan \theta}$

[thorny flame]

Sukiyaki

[neon iron]

So if you reciprocate the cot pi/4, sum it all up

you can prove that it is the same as cot pi/32

[neon iron]

I am pretty sure double angle formula for tangent will work. Try it with induction, and build up the identity.

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[slate shoal]

hi

[slate shoal]

can anyone help me

hi

?

[verbal hound]

Hi

[slate shoal]

well

dont get mad ok ?

[verbal hound]

Ok?

[slate shoal]

prof that

1+1=2

[verbal hound]

Hold one of your fingers

[slate shoal]

my brain is not braning

HOW

U JENUIS

TYS?

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[vital dune]

Is the answer 3^11/11? We have 11 spaces where we can choose 3 colours in each, hence 3^11. We divide by 11 because of the rotations

[whole geode]

@vital dune let's do a smaller example with 2 colors and 4 placements

let's say we have red and blue, and so our combinations are RRRR, RRRB, RRBB, RBRB, RBBB, BBBB

[vital dune]

Ok

[whole geode]

we would expect 2^4/4 from your formula above, but that is 4. And we have 6 here. What went wrong?

[vital dune]

I think the division by 4 is fine, the problem might be in 2^4

[whole geode]

well, not quite

it's actually the other way around

There's only one way to make RRRR, but we divided this possibility by 4

similarly with BBBB

[vital dune]

Ohh yeaah

[whole geode]

There are only 2 ways to make RBRB, similarly

thankfully, 11 is prime, so we only have to worry about edge cases, seemingly

(also it should immediately bother you that 3^11 / 11 is not an integer)

[vital dune]

Yes good point

[vital dune]

Why is that?

[whole geode]

if we consider why there were only two ways to make RBRB, it's because it was a pattern with period 2. Which divides 4

so only patterns with periods that are divisible by the total number of beads we need to handle carefully

but for 11 that's only 1 (all one color) or 11 (any other kind of pattern)

[vital dune]

Ahh

[whole geode]

So in total you'll have REDACTED

if I didn't make a mistake

[vital dune]

So the edge cases are all Red, all Yellow, all Blue?

[whole geode]

yup

[vital dune]

Oh right, I am still processing what you said about the periods lol

[whole geode]

I shouldn't post the answer though, tbh

[vital dune]

Oh no problem

So basically if it was 12 then we would need to work more to get the answer

[whole geode]

yes, because you'd have to handle the following cases:

period 1, period 2, period 3, period 4 but not period 2, period 6 but not period 2 OR period 3, and period 12

[vital dune]

So period n is for those when rotating n times in the same direction gets you the same as the original one?

[whole geode]

exactly

for instance for a period 6 you could have RRBBYYRRBBYY

[vital dune]

But then why not period 7 for example?

[whole geode]

because 7 doesn't divide 12

if you tried you'd get 7, then 5 left over

but you can't fit a copy of the 7 in those 5 spots

[vital dune]

Ahh

So it would be (3^12-3-6-12-18-36)/12+…

[whole geode]

so for 12 it would be 3 (period 1) + (3^2 - 3)/2 (period 2) + (3^3 - 3)/3 (period 3) + (3^4 - 3^2)/4 (period 4) + (3^6 - 3^3 - 3^2 + 3)/6 (period 6) + (3^12 - (3^6 - 3^3 - 3^2 + 3) - (3^4 - 3^2) - (3^3 - 3) - (3^2 - 3) - 3) / 12 (period 12)

[vital dune]

Huh

That’s tough, didn’t even get near the answer

[whole geode]

and the above doesn't actually seem to be correct

I must have a typo

[vital dune]

How did you get that?

[whole geode]

oh, I see my error.

ok so, let's work this through one period at a time

period 1, there are three possibilities RRRRRRRRRRRR, BBBBBBBBBBBB, and YYYYYYYYYYYY

so that's just 3

period 2, we have all of the following possibilities RBx6, RYx6, BYx6, but how do we count them?

well, first recognize that we can choose any two colors for the two positions. That's 3^2

but if we leave it here, we'll also be overcounting, because we would have counted RRx6 and so on

So we need to exclude RR, BB, and YY because we've already counted them, that's 3^2 - 3

[vital dune]

Oh I think I get it. Can we see it for period n, dividing the circle into n parts and subs tracting the periods that are divisible by n?

[whole geode]

and finally, we divide by the period because RBx6 is equivalent to BRx6

similarly for period 3

but there's a complication for period 4

for period 4 we need combinations that don't contain period 2 or period 1

so we have 3^4 - (3^2 + 3) - 3

which is 3^4 - 3^2, then divide by 4, so (3^4 - 3^2)/4

next for period 6.

that's (3^6 - 3^3 + 3 - 3^2 + 3 - 3)/6 = (3^6 - 3^3 - 3^2 + 3)/6

where we subtracted the 3s, and the 2s, and the 1s, but we're left with an overall add 3.

are you following so far?

[vital dune]

Why we need to add 3?

[whole geode]

have you ever heard of the inclusion exclusion principle? I think something similar is happening here.

[vital dune]

Yes

I thought of that when you started explaining it to me

Ohh, I got it

Okok, thanks @whole geode for your help!!!

[whole geode]

no worries, glad to help

[vital dune]

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[jolly oar]

Could someone explain me how to multiply huge decimals like 0,2981/0,6729 * 1,7480/0,1872? Numbers are purely arbitrary, so might not be correct - I just can't grasp the concept. Am I supposed to do the division first or multiply numerators and denominators seperately and then divide? I can multiply them successfully but dividing such huge decimals is a big hassle for me

[keen venture]

Doesn't matter

You can multiply the numerators and denominators, THEN do the division if you wanted

Otherwise you can simplify both fractions and THEN multiply them together. Same answer.

[jolly oar]

I mean I do multiply them successfully but when both numerator and denominators are huge decimals with many numbers, I just fail to divide

[jolly oar]

Does simplfying first work? Sounds cool. I'll give it a chance

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[clever citrus]

How would I find the bounds for theta

[finite moth]

0 to pi, then multiply the integral by 2

[topaz sinew]

@clever citrus Has your question been resolved?

[clever citrus]

wait what

why multiply the integral

why even bring an integral when finding the bounds for theta

[fringe grove]

I think the limits would be from 0 to pi and if you need an integral for something it would not be necessary to multiply by two since it is not like a circle with a center (0.0), correct me if I'm wrong

[clever citrus]

bro why r u guys talking aboit integrals when finding the bound for theta

Im acc so lost rn

this was the original question. Im getting a triple integral for this

int(a to b) int(0 to -5sintheta) int(0 to 6(1-r^2/25)) r dz dr dtheta

Im just not sure what the theta bounds would be so I put in a and b for now

but are you guys trying to say that

[rigid cloak]

Guyz

[clever citrus]

it would be something like:

2(int(0 to 1) int(0 to -5sintheta) int(0 to 6(1-r^2/25)) r dz dr dtheta)

[rigid cloak]

Bro u got that integral?

Imma gonna try it now

[clever citrus]

it looks wrong?

ngl I just freestyled after watching a few vids on the topic

[rigid cloak]

You do this one

Imma gonna solve that one

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[rigid cloak]

.reopen

[clever citrus]

Given D, we can get

x^2 + (y + 5/2)^2 <= 25/4

(rcostheta)^2 + (rsintheta + 5/2)^2 <= 25/4

r(r + 5sintheta) <= 0

solving for r we get 0 and -5sintheta

r bounds:
0 <= r <= -5sintheta

and then I gotta solve for theta

(? not sure how 😭 )

Now for the upper half of the ellipsoid

we can solve it for z to get

z = pm 6(1 - x^2/25 - y^2/25)

Since we want the upper half we can ignore negative:

z = 6(1 - r^2/25)

0 <= z <= 6(1-r^2/25)

.reopen

[topaz sinew]

✅

[clever citrus]

this is my draft so far

[clever citrus]

wait

all of this is one question btw

[rigid cloak]

No

[clever citrus]

wait wym

[rigid cloak]

Remember yesterday

[clever citrus]

which

the

sin(x + arctan(x) - 1)

[rigid cloak]

sin(x-1+arctanx)

Yes

[clever citrus]

oh bro

I dont think I gave the qhole question

the context

for that one

[rigid cloak]

You do this imma get that done

What

[clever citrus]

maybe I was approaching it in a wrong way

[clever citrus]

wait nah can u help me with this for now

that one imma focus on it at the end

[rigid cloak]

Ok let's see

[clever citrus]

I think I got everything covered beside the bounds for theta

[rigid cloak]

Okok good

Let's see

The original question

[rigid cloak]

Expand this

r^2+25/4+5r/2 sintheta

It should be + right?

Okok it's corrext

Correct*

Let's check forward

@clever citrus

Did u try this

,w range -5sin theta

[thorny flame]

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[rigid cloak]

-5 to 5

,w graph x^2/25+y^2/25+z^2/36=1

[thorny flame]

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[rigid cloak]

Ok now we can get a clear idea for theta

[exotic grove]

What coordinate system are you trying to write your integral in??

[rigid cloak]

Polar coordinates

[exotic grove]

Ok, get rid of the z btw

[clever citrus]

my bad had to find my charger

im back

going to read chat one sec

[rigid cloak]

Okok

u want to get rid of z

[exotic grove]

You shouldn’t be solving for z

[rigid cloak]

So he should be solving for x or y?

[exotic grove]

No he needs to solve for r

[rigid cloak]

What

[clever citrus]

that cant be the bound for theta right?

[rigid cloak]

Yes

See

[exotic grove]

Polar is rho, phi, theta

[rigid cloak]

-5<=-5sintheta<=5

This is the range of -5sintheta in common

[clever citrus]

yes but im talking about theta specifically

like

my integral at the end

[rigid cloak]

Bro see this has some relation

[clever citrus]

would require the bounds for theta aswell no?

[rigid cloak]

With u calculating the upper half

[clever citrus]

oh

[rigid cloak]

Like how theta varies

[exotic grove]

If I asked you to write an iterated integral finding just the volume under the ellipsis what would you say?

[rigid cloak]

And like how bro is saying phi rho and theta

[exotic grove]

Or even just, what if it’s a sphere

[clever citrus]

in this question they had me deal with the upper half of ellipsoid so what I did was I solved for z and discarded the negative solution

[rigid cloak]

,w draw x^2+y^2+z^2=1

[exotic grove]

Ok but the upper half part is covered by phi

[thorny flame]

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[clever citrus]

$\phi$

[thorny flame]

Calc III Victim

[rigid cloak]

A sphere we have

[exotic grove]

Do yk what rho, theta, and phi are?

[clever citrus]

wait tf is phi suppose to mean here

[rigid cloak]

,w define phi

[clever citrus]

$\rho$

[thorny flame]

Calc III Victim

[rigid cloak]

No it did not give the polar definition

[clever citrus]

we dont rlly use these in class

[exotic grove]

Oh wait you said Polar not spherical

[clever citrus]

so im not sure

[clever citrus]

ye

[exotic grove]

Am stupid

Ok

[rigid cloak]

Yes

Polar coordinates

Theta

Fr

,w define polar coordinates

[clever citrus]

so how would u do it

[thorny flame]

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[rigid cloak]

a clear definition

[exotic grove]

Nvm lmao I changed my mind

[clever citrus]

,w -5sin(theta) <= 0

[thorny flame]

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[exotic grove]

3d polar is spherical

💀💀

[rigid cloak]

Why he is using r and theta

[rigid cloak]

Wait

Wait

[exotic grove]

https://byjus.com/maths/polar-coordinates/#:~:text=3d polar coordinates or spherical,r%2C Φ%2C θ).

[rigid cloak]

💀☠️

[exotic grove]

Go read this first

[rigid cloak]

,w define spherical coordinates

[thorny flame]

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[rigid cloak]

A clear definition

[clever citrus]

bro

[rigid cloak]

For ellipsoid

I knew it had some relation with u calculating the upper half

[clever citrus]

man what r u guys talking about

Im saying

[rigid ivy]

@clever citrus what is your question?

[clever citrus]

alr one sec lemme type the whole thing out

[rigid cloak]

This

[exotic grove]

Surely you should start with something easier

Not in offensive way, but your head will explode if you do spherical too early imo

[clever citrus]

For this question what I have done so far is

I found the bounds for r and z which are

x^2 + (y + 5/2)^2 <= 25/4

(rcostheta)^2 + (rsintheta + 5/2)^2 <= 25/4

r(r + 5sintheta) <= 0

r bounds:
0 <= r <= -5sintheta

Now for the upper half of the ellipsoid

we can solve it for z to get

z = pm 6(1 - x^2/25 - y^2/25)

Since we want the upper half we can ignore negative:

z = 6(1 - r^2/25)

0 <= z <= 6(1-r^2/25)

but this should be a triple integral right

int_(bound for theta) int(0 to -5sintheta) int(0 to 6(1-r^2/25)) r dz dr dtheta

the only thing Im having trouble is coming up with the bounds for theta

[rigid cloak]

I feel

Funny man is correct

[clever citrus]

bro

[rigid ivy]

[rigid cloak]

But then I feel it's given

above the region in x-y plane

So it changes my mind

[clever citrus]

ive done polar coordinates alrdy its just that I forgot most of the shit I learnt in calc 1/2 so im having trouble solving

-5sin(theta) >= 0

[rigid cloak]

-5sintheta>=0

Is

sintheta<=0

[clever citrus]

right

[rigid cloak]

,w solve sintheta<=0

[thorny flame]

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[clever citrus]

how would I write the bounds for theta given that

[rigid cloak]

Wait let's get a quick review of

[exotic grove]

Brother this is not polar coordinates in 2d though

[rigid cloak]

,w graph x^2+y^2+5y<=0

[thorny flame]

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[rigid cloak]

This means that

[exotic grove]

Here try this, find the volume of the top half of the sphere x^2 + y^2 + z^2 = 1

[rigid cloak]

All the region inside the circle

[exotic grove]

Would that be like very easy for you?

[rigid cloak]

x^2+(y+5/2)^2=25/4

[topaz sinew]

@clever citrus Has your question been resolved?

[rigid cloak]

,w graph x^2/25+y^2/25+z^2/36=1 , x^2+y^2+5y<=0

[thorny flame]

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[rigid cloak]

,w draw x^2+y^2+5y<=0 inside x^2/25+y^2/25+z^2/36=1

[thorny flame]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
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Requested by oppailol.

[rigid cloak]

Aah

[clever citrus]

something like this right

[rigid cloak]

For real

Yus

[clever citrus]

?

[topaz sinew]

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[exotic grove]

.reopen

Ok

Well

[agile harness]

.reopen

[topaz sinew]

✅

[exotic grove]

This is called cylindrical coordinates

Polar coordinates implies spherical which is completely different

I’m pretty sure your theta bounds are wrong

You need to find the theta values where the two graphs intersect

Those will be your bounds

You just found bounds of D, remember it also needs to be inside the ellipsoid

[topaz sinew]

@clever citrus Has your question been resolved?

[exotic grove]

Ok jk I’m crazy my bad, I think you are right, the cylinder is contained inside the ellipsoid so it’s good

Thought the cylinder was bigger

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[strong sable]

Let the function ( u(x,y) ) be continuous on the closed region ( D ). Inside ( D ), the function ( u(x,y) ) has continuous second partial derivatives, and satisfies the conditions ( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 ) and ( \frac{\partial^2 u}{\partial x \partial y} \neq 0 ). Then is the following statements correct?
The maximum and minimum values of ( u(x,y) ) are both obtained on the boundary of ( D ).

[thorny flame]

riyobi

[sour bloom]

looks like a harmonic function

[strong sable]

Hmm any ideas?

[sour bloom]

seems to be on wikipedia https://en.wikipedia.org/wiki/Harmonic_function#Maximum_principle

but it requires the closed region to be compact

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@strong sable Has your question been resolved?

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[low monolith]

8xy+10xz-14xw

[low monolith]

how do I factor this

"factoring the sum or difference of two cubes"

[cerulean lily]

this is not a cube

[pearl peak]

do you have a screenshot of the q

[cinder sequoia]

the best thing you can do here is pull out a gcf

[low monolith]

yes

no.17

[pearl peak]

ok cool yeah, x is common in all the terms, and also take out the gcf of 8,10,14

[low monolith]

the gcf of it is 2x

the answer would be 2x(4y+5z-7w)

but how do I split it

or not

[topaz sinew]

@low monolith Has your question been resolved?

[cursive patrol]

yes this looks good

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[foggy igloo]

i dont know where to start

[cursive patrol]

first rewrite that logarithm

[foggy igloo]

2 ln y

what do i do after

[neon iron]

implicit differentiation

[foggy igloo]

2/y dy/dx?

[neon iron]

yes

well do that for the other terms too

4yx and 2

[foggy igloo]

-4y-4x dy/dx

2 is 0

[neon iron]

that looks correct yeah

now algebraically solve for dy/dx

,, \frac 2y y' - 4(y + xy') = 0

[thorny flame]

!Kiz__

[foggy igloo]

can i do
dy/dx(2/y-4x)=4y

4y/2/y-4x

[neon iron]

write this out properly but yes

[foggy igloo]

idk how to use that

[neon iron]

well u don't have to

[neon iron]

i just meant you should write it out unambiguously on paper or on ^

[foggy igloo]

ok

y'=2y^2/1-2xy

[neon iron]

,w simplify (4y)/(2/y - 4x)

[thorny flame]

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[foggy igloo]

my answer is undefined wtf

[neon iron]

💀

[foggy igloo]

i think dr frost just shat itself

[neon iron]

well all you need to know is you're correct

;p

[foggy igloo]

ty

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[royal imp]

Is this a mistake? How did 2x/sqrtx become 3x^1/2?

[flint stump]

yea looks like a typo

[dense lily]

Yes

[royal imp]

the final answer should be x^-1/2 right?

or -sqrt(x)

[dense lily]

Yes

[royal imp]

ok tysm for the help 🙏

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[rustic wren]

can someone tell me why this point would not be on the circle

bc

i calculated and got

(x-2)^2 + (y-5)^2 = 5^2

and then i plugged in the point

so (3-2)^2 + (2-5)^2 = 10

so 1+9

WHICH EQUALS 10

but the thing said the point does not lie on the line

so yeah

oh nvm

five squared is 25

rookie mistakr

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[hushed geyser]

need help on i and ii

i have no clue where to start either

[flint stump]

for 1, extend the line AB

[hushed geyser]

?

[flint stump]

since they want you to prove the angle is theta + 60, find a line that divides that angle in two parts such that you can prove one part is theta and the other is 60deg

[hushed geyser]

idk how u do that tho

[flint stump]

like i said, extend the line and use the two divisions of the angles that are formed due to this

[hushed geyser]

bro

ur not making sense

extend the line??

js make it longer??

idk what u mean bro

idk what u mean by the second part of ur sentance either

[flint stump]

extend like this

and so you have divided the angle DBC into two parts as a consequence

[hushed geyser]

i still dk what u do after 😭

[flint stump]

find which part is 60 deg angle

which part measures to theta

[hushed geyser]

ive done i

how do i do ii

[flint stump]

use sin(a+b) formula

and the definition of sine

[hushed geyser]

.close

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[hard kelp]

Shouldn't Lucy's acceleration be negative I think?

[ruby tree]

Is eastward positive?

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[near root]

hi

[near root]

Im getting 40 since for each value of B from 0 to 10, we have 4 possible values of (A,C)

I wanted to double check with someone if this is the right approach

[prisma mesa]

be careful about double counting

252

also I don't think you have 4 possible values

[near root]

thats b= 5

so 252 652 256 and 656

?

[prisma mesa]

wait, your approach is somewhat messed in general

let's start from scratch

[near root]

ok

[prisma mesa]

B can range from 0 to 9

so there are 10 possible values

now how many possible values are there for A

[near root]

2 for each value of

b

[prisma mesa]

Really?

what if b was 4

[near root]

uh

but a cant be 0

[prisma mesa]

oh, right, sorry

didnt read whole problem

[near root]

or c

[prisma mesa]

okay so 2

[near root]

yeaj

[prisma mesa]

alr, so you are right with 40 except for the double counting

[near root]

where is the double counting?

can u give me an exampe

[prisma mesa]

wait, you are probably right

sorry

[near root]

all goofd

[astral blaze]

I think it is indeed 40

[near root]

are u able to think of a scenario where double counting is happening

[prisma mesa]

yeah, i was confused because of the a,c!=0 condition

[near root]

oh okay

thanks

[astral blaze]

you can try coding it to check your result

[prisma mesa]

I think it's not necessary, but i will actually try it

huh, what did I do incorrectly?

oh i see

!= 0 instead of != "0"

[near root]

oh

waht do u get after fixing it

[prisma mesa]

40

it works

[teal patio]

it is 40

I got it as well

also say for question 6 I am struggling on the part where multiplying one of them cancels out d

I still dont see how its canceling d out

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[burnt swift]

does anyone know what this means?

like we need to find a subspace S in R4 such that S is a proper subset of H and S direct sum span(-1,1,1,a) = T orthogonal

[topaz sinew]

@burnt swift Has your question been resolved?

[topaz sinew]

@burnt swift Has your question been resolved?

[opal vault]

didn't we find S earlier

and you know (-1,1,1,a) is in T orthogonal

so (-1,1,1,a) . (1,1,-1,1) = 0

[burnt swift]

can we go slower

and after we are done can we make a summary of the steps we made

or is it too much to aak

[opal vault]

ok

but at least recap what we did last time

that you understood

and the parts you didn't understand

[burnt swift]

I felt like i still can't defend myself with this exercises

S is orthogonal?

[opal vault]

to what?

S is a subspace of H

[burnt swift]

I'm confused

what does orthogonal mean

[opal vault]

A and B are orthogonal

if dot product of a and b is 0

for any a, for any b

THE orthogonal of a set A, is the subspace that contains ALL vectors orthogonal to A

[burnt swift]

.

no but

what does this mean btw

T is a span

like the direct sum of S and this shit <(-1,1,1,a)> is equal to T orthogonal

whatever that means

[opal vault]

yes

meaning

each vector in T orthogonal

is the sum of a vector of S and a scalar multiple of (-1,1,1,a)

span = smallest subspace that contains ...

[burnt swift]

span vs basis?

in my country span is called differently

[opal vault]

basis is not the same thing exactly

anyways if you have vectors v1,v2,...

<v1,v2...> is the smallest subspace

that contains all of those vectors

here

v = (1,1,-1,1)

T = <(1,1,-1,1)>

T is the smallest subspace

that contains that vector

so T contains every vector x*(1,1,-1,1)

[opal vault]

or (x,x,-x,x)

[burnt swift]

every basis is a span

but not every span is a basis

[opal vault]

span is not the same as generating family

span refers to subspaces

generating family refers to the "system" of vectors

[burnt swift]

strange

[opal vault]

wait

ok that's weird

ok maybe "sistema generador" is the same thing as a span

[burnt swift]

how would you define span

[burnt swift]

.

[opal vault]

A = {v_i}

we write S = span(A) or <A>

the smallest subspace that contains all vectors of A

so

<(1,1,-1,1)>

[burnt swift]

what is $\mathbb{K}$ btw

[thorny flame]

938c2cc0dcc05f2b68c4287040cfcf71

[opal vault]

it's R here

the field you're working on

[burnt swift]

okay

can you quiz me

I dont think i get it

[opal vault]

ok

write down the expression of every vector in T

[opal vault]

.

[burnt swift]

a(1,1 -1,1)

[opal vault]

yes

what does a vector in <(1,0,1,0),(0,2,3,0)> look like?

[burnt swift]

a(1,0,1,0)+b(0,2,3,0)

[opal vault]

yes

so last thing

just notice that (1,0,1,0) and (0,2,3,0) GENERATE <(1,0,1,0),(0,2,3,0)>

but they're only a basis BECAUSE they are linearly independent

[burnt swift]

yeah its a generador

[opal vault]

yes, for example, <(1,0,0,0), (2,0,0,0)>

though (1,0,0,0), (2,0,0,0) generate it

it's not a basis

[burnt swift]

yeah

not linearly independent

my confusion was regarding the direct sum

like what does this mean

[opal vault]

direct sum

[burnt swift]

S direct sum a generator = t perp

[opal vault]

means that intersection is {0}

S and <(-1,1,1,a)> have no elements in common, except 0 vector

[opal vault]

so

S + <(-1,1,1,a)> is the sum of those subspaces

KNOWING

that they have only 0 vector in common

[burnt swift]

yeah direct sum

what does T perp mean though?

THE orthogonal of a set A, is the subspace that contains ALL vectors orthogonal to A

[burnt swift]

complemento ortogonal de S en M

means orthogonal complement of S in M

[opal vault]

yes

denoted $S^\perp$

[thorny flame]

rafilou2003

[burnt swift]

where is M mentioned

[opal vault]

M is the vector space

R^4

[opal vault]

$T^\perp$ is the orthogonal complement of T

[thorny flame]

rafilou2003

[burnt swift]

how do you know

[opal vault]

it's

the DEFINITION

of the notation

T^perp literally means "orthogonal complement of T"

[burnt swift]

how do u know <(-1,1,1,a)> is orthogonal to T

[opal vault]

it's a subset of T^perp

[burnt swift]

why you say its subsst why

subset how

I dont see it subset

how

[opal vault]

if A + B = C

then A and B are subsets of...

[burnt swift]

C

[opal vault]

then <(-1,1,1,a)> is subset of...

[burnt swift]

okay how do I find S and a tho

[burnt swift]

T perp

[opal vault]

ok so let's start with a

<(-1,1,1,a)> subset of T perp

so (-1,1,1,a) is in T perp

T perp is the orthogonal complement of T

so (-1,1,1,a) is perpendicular to...

[burnt swift]

S

[opal vault]

it's not S here

[burnt swift]

im not sure

[opal vault]

S is a random subspace

[burnt swift]

im confused and afraid

I dont really know

[opal vault]

let me rewrite that with notations that are completely related

[burnt swift]

yeah can u do a drowing

[opal vault]

Si $T$ es un subespacio vectorial de $\bR^4$, el complemento orthogonal de $T$, $T^\perp$, está formado por los vectores que son perpendiculares a todos los vectores de $T$.

[thorny flame]

rafilou2003

[burnt swift]

fair enough

I mean I understood that bit

but the direct sum is complicating me

and also how do I find a

[opal vault]

then replace it with a normal sum

[burnt swift]

is just normal sum where the intersection is zero vector

idk why they overcomplicate

[opal vault]

"+ where the intersection is zero vector"

versus

"$\oplus$"

[thorny flame]

rafilou2003

[opal vault]

guess which one is easier to write

[burnt swift]

s is orthogonal to T

[opal vault]

yes

[burnt swift]

a + b = c

a subset c

b subset c

[opal vault]

Btw drawing of orthogonal if you didn't understand

[burnt swift]

okay

what about a?

[opal vault]

(-1,1,1,a) orthogonal to T no?

[burnt swift]

yeah

[burnt swift]

inner prod =0

?

dot prod

[opal vault]

with any vector of T

[burnt swift]

what about it

[opal vault]

scalar product of (-1,1,1,a) with ANY vector of T is 0

have you thought about writing what scalar product of (-1,1,1,a) and (1,1,-1,1) is?

[burnt swift]

,w (-1,1,1,a)*(1,1,-1,1)

[opal vault]

using x is bad idea

can't you compute it yourself?

[burnt swift]

let me try

[topaz sinew]

@burnt swift Has your question been resolved?

[topaz sinew]

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[sonic rune]

i need help with how to start this