help-27

[visual stone]

i just guessed it was like that

[hardy trail]

Let me clarify what I'm asking for:

I want you to draw the vectors AB, BC and CD, which you have already drawn on this image.

I just don't want you to draw DA (which you have also drawn)

[visual stone]

alright

[hardy trail]

Good! So,

Based on this image, what do you think AB + BC + CD looks like?

How do you normally add up vectors?

[visual stone]

a triangle

[hardy trail]

mmm

[visual stone]

with their components

[hardy trail]

Out of curiousity, what is the vector AB + BC?

[visual stone]

AC

[hardy trail]

Good! So,

What do you think AB + BC + CD is?

[visual stone]

AD

[hardy trail]

Good!

So, based on what you said,

AB + BC + CD = AD

Right, so - what do you think we can do with this information?

[visual stone]

hm

it means that its a resultant because the arrow of AD is touching the arrow of CD

[hardy trail]

Sure (I mean resultant means total anyway, so AD is the resultant vector because it is the total of the other three vectors).

I guess what I mean to say is, how does this relate to our question?

Like, what are we trying to find?

[visual stone]

we need to find AB + BC + CD to find AD

[hardy trail]

I agree :)

So, what do we get?

[visual stone]

we get the resultant components of AD

then it says the distance

so we have to do sqrtx^2 + y^2

then u have the distance

[hardy trail]

Nice.

By the way,

[visual stone]

yeah

[hardy trail]

Is AD the vector from Diane to Adam?

[visual stone]

yeah it said to find the distance from Diana to Adam

[hardy trail]

Right. So does it matter whether we have AD or DA?

[visual stone]

yeah because if it was DA it wouldnt be a resultant vector

because its arrow touches the origin point of AB

[hardy trail]

I mean

Yeah, DA isn't a resultant vector

But the question is asking us to find the distance from Diana to Adam, so whether we have vector AD or DA, does it matter which vector we use?

[visual stone]

nah it wont cause the distance would remain the same

[hardy trail]

Perfect.

Just to let you know; AD is the vector from Adam and Diana, while DA is the vector from Diana to Adam.

[visual stone]

wait i got a question what if it didnt say one of the components but it said a norm

[visual stone]

yeah

[hardy trail]

The question does ask for the norm from Diana to Adam, but as you said, the distance doesn't depend on the direction.

[hardy trail]

If it only had a norm, then you'd need a direction too.

Since a vector has two things; distance and direction.

Using those two, it is possible to convert it back into component form (similar to what we did yesterday).

[visual stone]

oh yeah

without orientation i cant do it

alright thanks a lot dude

[hardy trail]

No problem :)

[visual stone]

.close

[devout snow]

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[pastel rose]

Hi can i have a small hint on this pls

[pastel rose]

only one i can think of is (xn) = 0

[uncut crow]

can you write out the condition with the epsilon and N in math symbols?

[pastel rose]

yea sure

[uncut crow]

not what i meant

a sequence where you can choose the N independently of epsilon

[pastel rose]

So we want a for all N?

[uncut crow]

no, you want a “there exists N”

one that works for all epsilon

[pastel rose]

Hmm alright, lemme think abt this

[uncut crow]

also just so you know there are more that meet the condition than just the 0 sequence

[pastel rose]

And they’re not constant sequences?

[delicate river]

(1, 0, 0, ...)

[uncut crow]

not quite

[pastel rose]

Oh wait yea

Hmm so it’s the tail of the sequence we look at here

But it has to be independent

[uncut crow]

yep, good

[pastel rose]

Of epsilon

[uncut crow]

ok that sounds a little off topic honestly

what is “it”?

[pastel rose]

N

[uncut crow]

what must the tail look like?

[pastel rose]

hm so my thought is, after n>=N the tail must be 0s

[uncut crow]

yea that’s right

needs proof but that is what you should try to prove

well more precisely

the sequences that meet the condition are the ones that are eventually all 0s

[pastel rose]

yep

im so confused how id even prove that? is the q asking to prove it or just state examples?

tbf proving could be an exercise in itself

[uncut crow]

i like to think about sequences in terms of the words frequently and eventually. not “eventually all 0s” means there are frequently terms that aren’t 0

where frequently means they exist no matter how far out you go

[pastel rose]

but we want eventually all 0s to satisfy our condition

[uncut crow]

yea but we want to show all the other sequences don’t satisfy

[uncut crow]

this is ez to show

[delicate river]

write out the quantifiers
hint: ||when writing out quantifiers, each variable can depend only on the variables before it; for instance, the regular definition is for all epsilon>0, there exists N; the N can depend on epsilon because it comes after for all epsilon quantifier||

[pastel rose]

Ah so we want in our case, once N is fixed, ie there exists an N in N such that for all eps >0 and all n >=N we have that |xn| < eps

[delicate river]

yeah exactly

[pastel rose]

That forces xn to be 0 then surely

[delicate river]

so what does it mean if |a|<epsilon for all epsilon>0

yeah, see if you can prove that
hint: ||proof by contradiction||

[pastel rose]

Ahhhh alright, I’ll have a go

Is that working?

I think that works intuitively 😭

[devout snow]

@pastel rose Has your question been resolved?

[delicate river]

the logic is right, it's just that the wording makes it a bit weird

like you're not using the fact that x_n converges, you're using the modified assumption you began with

[devout snow]

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[main crater]

We are learning vectors but I was pulled out of class right at the end. So all I got for the end was how example 11 started. Now I don’t know what im supposed to do next.
I just need someone to help me understand what im supposed to do on question that ask me to “write each vector in trigonometric form”

[acoustic leaf]

i'm not sure how your class expects "trigonometric form" to be written but the calculations would consist of finding the magnitude and angle as shown in example 11

[main crater]

I never got to see how example 11 ended since I left class early. And here is what she gave us for the forms I guess.

I don’t fully understand what all that mumbo jumbo means😭

[acoustic leaf]

ok so once you have $\norm{\vb p}$ and $\theta$ you can write it in the form $\norm{\vb p}\langle \cos\theta,\sin\theta\rangle$ and that is the trigonometric form

[woven radish]

cloud ☁

[main crater]

So for example 11 I’d just do this then?

[acoustic leaf]

well i would probably just leave it in the second to last form

since the last one is just back in component form again

[main crater]

Okay!👍🏽 thank you so much time to do the homework!

.close

[devout snow]

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[devout snow]

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[wild phoenix]

I have a quadratics test and I dont understand how to graph from standard form
One of the main questions im struggling with is about time in the air and height
the equation is phrased like h= 2t² - 9t + 18 where h is height in the air and t is time
Im very lost and unsure how to answer it, i googled it and it left me more confused

[wild phoenix]

this is another example question

[devout snow]

@wild phoenix Has your question been resolved?

[inner summit]

-b/2a is the x coordinate of the line of symmetry. You can plug in additional points or solve for the roots and plot them

2t^2 being positive means it opens up

[wild phoenix]

so -28/-4?

[inner summit]

thats the x coordinate yeah

[devout snow]

@wild phoenix Has your question been resolved?

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[dry canyon]

why does this equal 0

[ionic harness]

$e^{-6x} = \frac{1}{e^{6x}}$

[woven radish]

Coolempire93

[dry canyon]

interesting go on

[ionic harness]

Plugging in infinity ||yields infinity in the denominator||, and since cos is ||bounded -1 to 1||, we have ||a constant over infinity|| ||-> 0||

[dry canyon]

i understood the denominator part but why does -1 to 1 and constant over infinity make it 0?

why is it 0 because cos is bounded to a range between -1 to 1

[ionic harness]

Here's a question

$\lim_{x \rightarrow \infty}\frac{c}{x} = ?$

[woven radish]

Coolempire93

[dry canyon]

i dont know how to solve that lol

[crystal dawn]

what happens to a fraction as its denom grows without bound?

that's essentially what that limit is about

[ionic harness]

Think about some real examples of fractions

[fervent sluice]

if you have a pizza and you’re sharing with people

[dry canyon]

but even if it is a 1/inf, it is still between -1 and 1 right?

[fervent sluice]

what happens to the size of pizza as the amount of people increase

[dry canyon]

how many slices do we have?

pizza stays constant?

[fervent sluice]

hopefully

[dry canyon]

wait would the pizza eventually run out?

[fervent sluice]

I like my pizza constant

nah it’ll just disappear

[crystal dawn]

I mean if you take the pizza analogy

[fervent sluice]

poof

[crystal dawn]

the more people there are, the more slices there are

[ionic harness]

(The point is ||the numerator is at minimum -1, so -1/infinity = 0, and at maximum 1/infinity = 0 so it's bounded by squeeze theorem (sandwich theorem)||)

[crystal dawn]

and the more slices there are, the less of the pizza each slice is

agreed?

[ionic harness]

Open that up once they finish explaining

[dry canyon]

got it

[dry canyon]

more slices there are ? so like pizza going from lets say like 2 slices to 8 slices for example?

[crystal dawn]

yes

[dry canyon]

okay understood so far

[crystal dawn]

so as you slice the pizza more and more times

the area becomes smaller and smaller

[dry canyon]

yes

[crystal dawn]

so what happens when eventually you slice an infinite number of slices

[dry canyon]

the slices could reach to absolutely nothing but it will never reach it 🥀

[crystal dawn]

yes, true, but as the number of slices grows without bound, the area of each slice tends to 0

and that is the key

[dry canyon]

i see, so that is why if 1/2 eventually goes to 1/inf, 1/inf "limits" to 0?

[crystal dawn]

the limit of 1/x as x tends to inf is 0, yes

but notice that

it doesn't matter what the numerator was

if you had 1000 pizzas, slicing all the way to infinity will still have the area of each slice tend to 0

[dry canyon]

so even if sin was in place of cos, it would still be 0?

[crystal dawn]

yes

because the numerator of the fraction is a finite constant

[ionic harness]

(They're both bounded by constants)

[crystal dawn]

if it was infinite then we'd have an indeterminate form

that will be a headache for another day

[ionic harness]

Hahaha

[dry canyon]

inf/inf doesnt equal 1?

[ionic harness]

💀

[crystal dawn]

oh nonononononono.

[ionic harness]

Infinities have different sizes as you'll see later

[crystal dawn]

limits is the first topic that forces you to acknowledge that our intuition regarding infinity is not as good as you think

[ionic harness]

A simple example

[dry canyon]

so thats why the cos bounding part was importatn

[crystal dawn]

and no, you cannot treat infinity like a number and cancel them off

[ionic harness]

Both $x^2$ and $x$ grow to infinity as x approaches infinity, but $$\lim_{x \rightarrow \infty}\frac{x}{x^2} = \lim_{x \rightarrow \infty}\frac{1}{x} = 0$$

[woven radish]

Coolempire93

[ionic harness]

Although sometimes it may seem like it, like here XD

[dry canyon]

im gonna throw up

[ionic harness]

||Which I mean x and x are the same infinity so same infinity squared but||

[dry canyon]

well imma close the channel now

thank you guys for your help so much!!

[crystal dawn]

[ionic harness]

No problem! I would say since you're working on all these limit problems though

It would be good for you to try and get an intuitive understanding like we worked on here

Let me get you a link

To read

[dry canyon]

got it

[crystal dawn]

also, you may want to note the seven indeterminate forms

[ionic harness]

https://tutorial.math.lamar.edu/classes/calci/limitsintro.aspx

[dry canyon]

i know a guy named paul

he was not a good guy

[crystal dawn]

if you get these, your limit is indeterminate and you need to manipulate your expression

[ionic harness]

The actual introduction to limits is quite complicated (because it's rigorous) so I would start with the Computing Limits section if what you see in the first parts aren't included in your textbook

[ionic harness]

Yep, just like that 0/0 problem you asked about before

[crystal dawn]

(even 1^inf btw!)

[dry canyon]

alright ill check out the resources

thank you guys again so much!!!

[crystal dawn]

but 1^inf is a bit of a specialty amongst the indeterminate forms because it's a bit nuanced

so another day perhaps

[dry canyon]

.close

[devout snow]

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[ionic harness]

Oh I guess as a last note if you have free time and want to do something blackpenredpen has great limit videos (on indeterminate forms as well as his 'evaluating 100 of these' in preparation for tests) and great calculus videos in general 🙂

[devout snow]

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[cerulean ruin]

hi

[cerulean ruin]

how do you approach b

i think for a you use cross ratio formula right

for b do you just plug in a point in T(z) = ...

then see if it maps to upper half?

[fast garden]

you don't need to use cross ratios

you find your T, and you should know that boundary is mapped to boundary

so check some points on your boundary to identify whether it's mapped to a line or a circle

[cerulean ruin]

oh i mean to find T, you use cross ratio right

[fast garden]

oh right

yes

[cerulean ruin]

ok suppose $T(z) = \dfrac{4z}{6-z}$

[woven radish]

toast

[cerulean ruin]

could u explain this with this T^

[fast garden]

yeah

so, hopefully you know that boundaries are mapped to boundaries?

[cerulean ruin]

ye

[fast garden]

right

but your mobius transformation maps circles and lines to circles and lines

[cerulean ruin]

yes

[fast garden]

so, you are looking at where your boundary is mapped to (that is, your im(z) = 0 line)

it can only be a circle or a line

would you know how to do that

[cerulean ruin]

hmm

[cerulean ruin]

so like

in the question

it maps points from the real axis to the real axis

[fast garden]

right, so your new boundary is also the real axis

how do you check which region it actually is?

[cerulean ruin]

plug a point then?

[fast garden]

yes exactly

[cerulean ruin]

so this is where you plug in i for exampel

[fast garden]

yeah

your favourite number from the original region that is not on the boundary

[cerulean ruin]

OHHH wait

its asking where does T map the upper half plane

so this is our input

[fast garden]

:)

[cerulean ruin]

so we just use a random point from the upper half plane and plug into T

[fast garden]

as long as it is not on the boundary, yeah

do you know why that is

[cerulean ruin]

not sure tbh

[fast garden]

well, the boundary points are not in the region anyway, but if it was >= 0 instead, it is important to know

[cerulean ruin]

i learned this mobius transformation stuff approxmaitely 3 hours ago and my final is at 10 am

😭

[fast garden]

right, well, we are trying to find where the region is

[cerulean ruin]

ye

[fast garden]

we have alreayd identified it maps the real line to the real line, but if we plugged in a boundary point, what would we get back

[cerulean ruin]

a point on the boundary?

[fast garden]

exactly

[cerulean ruin]

ahh

[fast garden]

but we already know where the boundaries are, so it's superfluous

[cerulean ruin]

ic

and the boundary here is just the real axis

[fast garden]

if your transformation is correct, yeah

[cerulean ruin]

got it

[steel sage]

His transformation is correct

[cerulean ruin]

so in this question

this is mapping a circle -> line?

and i guess to see what line specically we use boundary points of |z| = 1

[fast garden]

yes

[steel sage]

Though when it’s a region not a circle you still need to use an extra point. Like f(z)=4z/(6-z) you found that f maps x-axis to x-axis, but the image might be either upper or lower plane. Do decide which you still need to try one point, z=i for example

[cerulean ruin]

so here we can plug in like -i, i, -1

[cerulean ruin]

oh rlly

[fast garden]

oh mb i thought i explained that

[steel sage]

Yeah. Möbius map maps a circle to what circle you need three points, maps what region to what region you need four

[fast garden]

you could yeah

[cerulean ruin]

so we get -0.5, -0,5 +- i0.5 so x = -0.5?

[steel sage]

Oh he said that

[fast garden]

yh seems right

although

this makes me want to ask, how did you deduce it was a line?

[cerulean ruin]

the pole is real

probably not good enough tho

😭

[fast garden]

right, but that is not enough to deduce, since if i shrunk the circle down to say |z| = 1/2, it would no longer be a line

[cerulean ruin]

ah

[fast garden]

the point is that this pole is on your circle so there is a point at infinity

[steel sage]

To see what f(z) maps unit circle to, it’s the same as finding {z: |f^-1(z)|=1}, which is solving |z+1|/|z|=|1+1/z|=1

[cerulean ruin]

oh

so that makes a bit of sense

if a pole exists on the boundary

then it must map to a line

[fast garden]

in general, you need 3 points to determine
2 points is not enough because it could just be a circle passing through those 2 points
the 3rd point determines if they are all colinear or whether it really is a circle

[cerulean ruin]

cuz there would be a point at inifity?

[fast garden]

yes

[steel sage]

A line can be viewed as a circle whose one point is infinity

[cerulean ruin]

ooo

that makes sense

so just out of curiosity

[cerulean ruin]

here it maps to a line and we plug in different points on the circle to find that line

could there be an example where it just maps to differrent real values?

would the line be a slope in that case

[steel sage]

I think given three distinct points as input and three distinct points as their image it can always determine a möbius transformation

[cerulean ruin]

ah

[steel sage]

Three distinct points, needless to say.

[cerulean ruin]

ah

and thats how you determine a line

OHHH

this makes so much sense i think..

[cerulean ruin]

and

my understadning is that

these 3 points should always form something linear

(if we are mapping to a line)

so probably something went wrong

with the mapping

[steel sage]

It’s simply that three distinct points determine a circle (line is viewed as a circle too)

[cerulean ruin]

ah that makes so much sense now

i was wondering why we needed 3 instead of 2, but here we just treat a. line as a special circle with points agt inifnity

so if we are suppsoe to map to a line (i.e a pole on the boundary), ,and we dont get collinear points then somethign went wrong

but if we do get collinear points, then we can determine what line the transformation maps to

[steel sage]

Yes. And infinity is very real to us actually, the complex plane adding this infinity is just a sphere S^2

[cerulean ruin]

got it :) thank you so much this makes alot more sense now haha

i guess this is why these thigns are called generalized circles

[steel sage]

Yeah

[cerulean ruin]

alright 😭 thats prbs all i need to know for mobius transformations on the exam

[fast garden]

good luck!

[steel sage]

Rock the exam

[cerulean ruin]

thank you!

i probably have some toher questions on the review

.solved

[devout snow]

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[cerulean ruin]

lowkey forgot how to do this problem

😭

is it realted to ML inequality/?

[steel sage]

Use Cauchy’s estimate

I mean f(z) has Taylor series on the whole plane, and use Cauchy’s estimate to estimate f^(n)(0)

[cerulean ruin]

is it this?

[steel sage]

Yes

M_R in the picture is the maximal |f(w)| for w in the closed ball, radius R, center z_0

closed disk I meant

[cerulean ruin]

so we let |z| = R

oh waittt

so they gave us the poklynomial of degree one

that means f^(n)(z_0) where n >= 2 = 0

or i guess

[steel sage]

Yeah you need to show that

[cerulean ruin]

that is what we are trying to show

[steel sage]

z_0=0

[cerulean ruin]

why do we use z_0 = 0 versus any other point?

[steel sage]

f(z)<=4+5|z|^(3/2)

If the condition is f(z)<=4+5|z-z_0|^(3/2) certainly we would consider z_0

[cerulean ruin]

ohh

bc its centered at 0

$|f^{n}(0)| \leq \dfrac{n!}{R^n}\bigg(R + 5R^{3/2}\bigg)$

[woven radish]

toast

[steel sage]

Yeah

[cerulean ruin]

oops 4 + 5...

4/R^n + 5/R^{3/2 - n}

[steel sage]

n-3/2, in your notation

And n!

[cerulean ruin]

yepp and when n >= 2 and we take the limit as R -> infinity

we get that the derivative <= 0

[steel sage]

=0

[cerulean ruin]

so thats why its a polynomial t most degree 1

oh ye

[steel sage]

Yeah

[cerulean ruin]

kk that makes alot of sense tysm

[steel sage]

Np

[cerulean ruin]

im guessing 1. is an applicaiton of louisvilles theorem?

wait

[steel sage]

Yeah

[steel sage]

Correct

[cerulean ruin]

kk

i see

louisvilles theorem is like if a function is entire and bounded then it is constant

[steel sage]

Yeah

[cerulean ruin]

so i guess we define g(z) = |f(z) - i|

or hm

[steel sage]

Your g is not bounded

1/it will be bounded though

[cerulean ruin]

ahh

[steel sage]

I mean if the statement is false, 1/g will be bounded

[cerulean ruin]

so this is kind of liek a proof by contradiction

[steel sage]

Yeah

[cerulean ruin]

so assume that for all z |f(z) - i| >= 1

then the function g(z) = 1/|f(z) - i| is bounded by 1

so g(z) <= 1

then

[steel sage]

Though I would drop | | in g

[cerulean ruin]

note that g(z) is entire

oh ok

bc g(z) is entire and bounded, then we can represent g by some constant alpha by louivilles

[steel sage]

h(z) entire doesn’t imply z|-> |h(z)| entire

[cerulean ruin]

oh ok

ur right

but then this implies f(z) = 1/alpha + i

but we assume f(z) is nonconstant

so a contradiction

[steel sage]

Exactly

[cerulean ruin]

thanks!

and b is also a little weird

[steel sage]

It’s just that cos(Z) has no singularity, and 1/z does have, 0

So when integral around 0, one will be 0, another one will be 2πi

[cerulean ruin]

ah okay

why is the domain relevant

[steel sage]

I don’t think it does. It looks like that form only to make sure it contains a closed curve around 0

[cerulean ruin]

got it

tysm

.sovled

.solved

[steel sage]

No problem

[devout snow]

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[cerulean ruin]

oh

ty!

[devout snow]

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[soft umbra]

Hiii

[soft umbra]

Hi

Do you have any question to ask?

bcoz you just open a channel

[devout snow]

@wheat moss Has your question been resolved?

[soft umbra]

.solved inactive channel

[devout snow]

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[wicked thorn]

Some angle's sides are going through the circumference. We need to find and prove which arc is bigger

[tender wharf]

!xy

[devout snow]

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[wicked thorn]

as i assume there are two cases. We need to prove that arc CD is bigger than arc AB

[tender wharf]

arc closer to the vertex should be bigger

[wicked thorn]

we need to prove it

[tender wharf]

so none of sides is tangential?

hmm..

angles with vertex outside the circle measure = 1/2[ bigger arc - smaller arc]

[devout snow]

@wicked thorn Has your question been resolved?

[devout snow]

@wicked thorn Has your question been resolved?

[devout snow]

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[wicked thorn]

Some angle's sides are going through the circumference. We need to find and prove which arc is bigger

[sand pumice]

!xy

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[sand pumice]

wait shit use the arc length formula

[wicked thorn]

why?

[tender wharf]

to prove

[wicked thorn]

that one arc is bigger than another?

[tender wharf]

ye

[wicked thorn]

wait no

there are no measures given, we need to prove geometrically

[sand pumice]

arc length = $\theta \pi r$

[woven radish]

Green

[pseudo basin]

this sounds weird and understated

[pseudo basin]

no pi there

[sand pumice]

by 180*

[sand pumice]

in degrees

[pseudo basin]

ok now you have to specify theta is in degrees

yes ok

[sand pumice]

mb

[wicked thorn]

we must prove that arc AB is bigger than arc CD geometrically

[tender wharf]

this would apply if MA = MC(vertex is M)?

[pseudo basin]

are you sure you have that the right way around

cause i think you may have meant CD>AB

[gloomy aurora]

bigger arcs subtend larger angles at centre, if that's what you want?

[wicked thorn]

what can be wrong?

[pseudo basin]

.

[wicked thorn]

ok

[sand girder]

what does the problem give?

or is it just that?

[wicked thorn]

it's just that, some angle is near one circumference

forming two secants

and four arcs

AB and CD are one of them

[pseudo basin]

<@&268886789983436800> troll

[modest dagger]

Don't troll in the help channels.

[pseudo basin]

maybe draw lines AO and BO or something

and see where they intersect the circle

rn that's the only thing that is coming to mind

[tardy edge]

i feel like you can do that with central and peripheral angle

draw lines CO and OD, thats the central angle at O, and the angle above points A and B is the peripheral angle that also includes the CD arc

but then with that proof you can only prove CD arc > AB arc, not the other way around, since the central angle is two times bigger than the peripheral

so it's arc length would also be 2 times bigger

[devout snow]

@wicked thorn Has your question been resolved?

[ionic harness]

The way my book does it is

Let's name the intersection point outside the circle K just so I have something to call it

1. Construct AD. measure of CAD is half of measure of arc CD
2. Since K-A-C is a straight line, angle KAD = 180 - measure of angle CAD
3. Measure of ADB is half of measure of arc AB
4. Therefore since mADB + mKAD + mAKD = 180 because AKD is a triangle, 180 = arc AB/2 + 180 - arc CD/2 + mAKD
   and
   mAKD = arc CD/2 - arc AB/2
5. Because the measure of an angle cannot be negative, arc CD has a larger measure than arc AB 👏

[devout snow]

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[desert grove]

Hello, I am trying to solve this equation:

[ionic harness]

Solve for s?

[desert grove]

Given also, a(s)=k/s²

[desert grove]

Solving for t where s(t)=0

[wind mason]

physics problem i assume?

[ionic harness]

Ah

That makes more sense

[desert grove]

I tried and i got only as far as a differential that was y²*y"=k

In this case, k=gravitational constant*mass of the body, but that's irrelevant

[ionic harness]

Are there supposed to be two integrals but only one constant of integration?

[wind mason]

k=mg…

I feel like something is tickling me here

[desert grove]

Yes, youre integrating acceleration first to find velocity then space

[wind mason]

This might be known

[wind mason]

but why a(s(x))?

[ionic harness]

Right

Because that should be a(t)

[desert grove]

Because acceleration is a function of space

[ionic harness]

Ah

They gave it to you as space

Got it

[desert grove]

The closer you are to the body the more you accelerate

[wind mason]

o

[ionic harness]

That solves half the problems but there should be a dxdx after the integral regardless

[desert grove]

I see,thanks

[ionic harness]

Well dxdt

[wind mason]

my smol brain is too smok to comprehend this

[desert grove]

Is it? Shouldn't it be dxdx

[ionic harness]

But yeah it should come out to something of this sort

[ionic harness]

Yeah but the bound is in terms of t

[wind mason]

lessee here though

[ionic harness]

So your result will be in t

[desert grove]

Then how do I solve that given the initial value of s(0), s'(0) and s"(0)

[wind mason]

its definite, so it should be dxdy, no? (Anyways i feel its a bit pointless to argue about this anywyas)

[ionic harness]

Right but now I'm noticing we have s(t) = ... s(x)

I'm wondering what's a placeholder and what's not

XD

[desert grove]

This is essentially "given non uniform acceleration, how long will it take for a body to a accelerate from a given distance to an object, to reach such object"

[ionic harness]

Yep

In which case easiest way is to differentiate using FToC and get your differential equatio

[wind mason]

why do I feel like you can make a differential equation from this?

[desert grove]

As you can see above it is a differential

[ionic harness]

Because that's the standard way to do it 👍

[young spade]

Might i ask, has this problem given to you / read it as is from a book / etc... or its something of your own?

[ionic harness]

Although with acceleration as a function fo s is different

[ionic harness]

Good question

!xy

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[desert grove]

I couldn't find any info on this problem so I tried seeing if a solution could be found

[wind mason]

where is it from though

[young spade]

mostly because DEs for gravity are remarkably well known and documented

[wind mason]

or did you ask yourself this question?

[ionic harness]

Very true

[desert grove]

I tried looking and someone even recommended me something but it was less than understandable

[young spade]

And really easy compared to this integral

[desert grove]

I mean, I still wanna see how the differential can be solved, on its own it seems intriguing enough

[young spade]

I suppose you were looking at Newtons equation, correct?

[desert grove]

An earlier incorrect version of this problem was built upon the laws of motion

And of course gravity isn't constant so it was wrong

[ionic harness]

s'(t) = -int[0,t]a(s(x))dx
s''(t) = -a(s(x)) = k/s^2(x)
Yeah that reaches me to where you were

y^2 * y'' = k and it's a 2nd order nonlinear

Which we can just backsolve by integration

[young spade]

If we are talking about 2 bodies

Yrying to calculate the position of celestial bodies newton found that there is technically two ways to treat the problem.
You could think about it in terms of cartesian (XY) or polar coordinates

iirc cartesian is practically unsolvable as is, polar coordinates work under the assumption that bodies without extreme mass follow elliptic orbits, which have the nice property of keeping Angular Momentum as a constant

[desert grove]

Calculating a free fall as a very eccentric orbit feels rather wrong

Conceptually

But I guess it is technically correct

[ionic harness]

Isn't that the definition of free fall 🤔

[desert grove]

Well, it does not seem to be an orbit of any sort to me

[young spade]

If youre trying to calculate free fall for objects not** at extreme altitude, just assume g is a constant

[ionic harness]

I mean I guess if you define orbit as stable free fall

And not just directly equate them

[desert grove]

What if the fall goes from extreme altitudes to smaller ones

So you have both cases

Then it would only make sense to indeed have a formula which works universally

[young spade]

Just for clarity, g drops to 90% at about 400km up.

Anyways, yeah, the DE is somewhat "simple" even then.

Lets say that our model is attempting to show the time it takes for an object to reach ground, we have to know a few things from there

y as the altitude
t as time
We need y(0), aka the initial value of altitude, and y'(0), the initial velocity

From here it depends on how many assumptions you wanna make
Can we treat gravity scaling linearly with respect to distance to ground?
Is air drag a thing?
Etc...

[desert grove]

for the simplest case, y'(0) can be assumed to be 0, or at least I intended it to be

[ionic harness]

That makes things nice

[desert grove]

Air drag, of course, is a terrible thing that should never be considered

Joking of course but yeah, simplest case still applies

[young spade]

Knowing that we have 2 variables here, in reality we have 2 differential equations here

Which we usually call a system

[ionic harness]

\begin{align*}
y^2 \cdot y'' &= k \
\int y^2 \cdot y' dy &= \int kdt \
y^2y' - \int 2y^2 dy &= kt + y'(0) \
y^2 y' - \frac{2}{3}y^3 &= kt + y'(0)
\end{align*}

[woven radish]

Coolempire93

[ionic harness]

oh you seem to know more than me

I'll leave it to you then

[young spade]

dy/dt is just = v aka velocity.

Since position changes purely with velocity

Now the relatively hard part, dv/dt = f(y,v)

Aka, velocity is setup as a differential equation based on height (y) and current velocity (v)

[desert grove]

Ohh that makes sense

[young spade]

If we assume no air drag, velocity basically doesnt play a part

We just end up with dv/dt = f(y)

Heres the relatively hard part, what is f

For most practical purposes, we just assume that f(y) = g
Aka, acceleration (rate of change of velocity) is equal to some constant, but you change it to be whatever

On the most technical sense, yes, k/y^2 is the most accuratr

[desert grove]

Yeah

[young spade]

You end up with two equations you have solve, and each depend on the other

$$\frac{dy}{dt} = v$$
$$\frac{dv}{dt} = ky^{-2}$$

[desert grove]

I guess not considering the first formula might have hindered any attempts to solve it

[young spade]

This is also why both y and y' are needed as initial conditions

Also, consider, yes, this system can be simplified into a single equation

Since dy/dt = v

dv/dt = d²y/dt²

[desert grove]

Yeah, the one i had originally proposed

Though I used a roundabout way to find it

[young spade]

In another words:
$$\ddot y = ky^{-2}$$

[young spade]

Its a relatively simple DE, you just have to know how to solve em.

[desert grove]

I for the life of me could not find a solution, i even asked a whole mathematician but i think it was not their specific area of study

[inland carbon]

Multiply both sides by y dot i think

[devout snow]

@desert grove Has your question been resolved?

[desert grove]

Do you have any link or guide on the how?

[young spade]

If you hardly get DE's getting it on your own would be basically impossible, and it requires some tricky substitutions

https://www.quora.com/How-can-I-solve-the-differential-equation-yy-2-k

Heres an implicit

[desert grove]

Thanks!

[young spade]

If youre interested in an answer and not an equation

Learn runge kutta

[ionic harness]

lord

[desert grove]

I guess i shall

Thanks

[desert grove]

Is it hard?

[young spade]

Rk4 aint that bad tbh

[ionic harness]

Not necessarily

Keeping in mind I learned it in a class, so I had to do it by hand

And build a calculator to do it myself in desmos

[young spade]

Runge Kutta is a family of numerical methods that allow you to find approximations to DEs of all kinds

And they are arguably the most stable

[desert grove]

Oh, so it's an approximation

[young spade]

Yep, really precise and elaborated one, but yeaj

If youre somewhat worried about the precision, the error on RK4 is 4th order, aka, if you double your numerical precision, you get an answer "16 times closer"

[desert grove]

That is quite amazing

[young spade]

As to my knowledge, is the standard for most physics models nowadays, some people prefer higher order Runge Kutta (RK8 or some other types of RK4) for models that are supposed to stretch a lot of time with a lot of elements.

But in your case, even RK1 (Also known as Eulers method) could be enough with a time step small enough

[ionic harness]

haaaa

I can't believe I had to do euler's method, euler's improved method (RK2) and RK4 all by hand

And then code calculators in desmos for all of them

[young spade]

Its a pretty useful technique

[devout snow]

@desert grove Has your question been resolved?

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[arctic delta]

how do u do this

[dire iris]

do what? the derivative?

[arctic delta]

yeah

[dire iris]

do you know the chain rule?

[arctic delta]

yes

[dire iris]

so, how does that work here?

you take the derivative of the outer function, then multiple by the derivative of the inner function

what would the outer/inner functions be?

[arctic delta]

is outer e^sin and inner (cos(x^2) ?

[dire iris]

yeah, but it would also be easier to have outer as e^x and inner as sin(cos(x^2))

here, you will have to apply the chain rule 3 times in a row

when you take the derivative of the inner expression

[arctic delta]

how would that work? I only ever dealt with 2

[dire iris]

well when you apply the chain rule once, what do you get?

g(x) = e^x, h(x) = sin(cos(x^2)), f(x) = g(h(x))

so f'(x) = ???

g'(h(x)) * h'(x) right @arctic delta

so what would that be?

[arctic delta]

one moment

this is confusing

[dire iris]

you can keep h'(x) unsimplified

that is where applying the chain rule more times comes in

what would g'(h(x)) be?

[arctic delta]

is g'(h(x) the same as f'(g(x)) ?

[dire iris]

no, g is e^x, h is sin(cos(x^2))

what is the derivative of e^x?

[arctic delta]

I think just e^x

[dire iris]

yeah, so what is g'(h(x))?

just the same thing as the original function right

[arctic delta]

isn't that just e^sin(cos(x^2)) ?

[dire iris]

okay, then you need to figure out what the derivative of sin(cos(x^2)) is right

and multiply by that

how would you do that?

[arctic delta]

g'(h(x)) * h'(x)

cos(cos(x^2) (cos(x^2) ?

[dire iris]

derivative of cos(x^2) yes

what would that be?

you apply chain rule yet again

[arctic delta]

-sin(x^2)(2x) so -2xSin(x^2)

[dire iris]

yeah, so all together, what is f'(x)?

you just gotta multiply all of this and simplify

[arctic delta]

was this correct for the sin(cos(x^2)) part?

[dire iris]

well that last part should be d/dx(cos(x^2)) but yeah the first part is correct

[arctic delta]

so it was supposed to be cos(cos(x^2)) ?

[dire iris]

yes

[arctic delta]

wait do u substitute the cos(cos(x^2) and -2xsin(x^2) into the x in e^x ?

[dire iris]

its easier to see if you work it out on paper

[arctic delta]

yeah im doing that

[dire iris]

you originally had this times h'(x)

so you substitute h'(x) in to what you found

[dire iris]

and h'(x) is this

times

[dire iris]

this

you're essentially just successively multiplying by the derivatives of the inner functions

[arctic delta]

so this?

[uneven coral]

That's correct

[arctic delta]

ok thanks

.close

[devout snow]

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[devout snow]

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[white vine]

prove if 1/1+x^2 is injective, surjective orbijective over (0, 1].

[dire forge]

what have you tried so far?

[white vine]

I have been proving it with f(a)=f(b) from my understanding

[dire forge]

which one do you want to focus on first, injective?

[white vine]

yes.

[dire forge]

okay, so you start from f(a) = f(b) and then what did you do after that

[white vine]

set it up as aformal proof

f(x_1) = 1/(1+x^2_1) = f(x_2)=....

but other thoughts were to use monoticity to prove it?

[dire forge]

okay let's go ahead with your first thought

so we have

[\frac1{1+a^2} = f(a) = f(b) = \frac1{1+b^2}]

[woven radish]

fishERIC_information

[white vine]

uh basically yes

[dire forge]

that's our assumption, what's the thing we have to prove at the end, if this function is injective?

[white vine]

yes.

This is where I got stuck so a, b > 0

[dire forge]

sorry that wasn't a yes or no question, I mean what's our goal at the end?

the function is injective if we can demonstrate what?

[white vine]

Prove whether or not is injective in interval [0, 1)

(0, 1]*

[dire forge]

yes

but what is the definition of injective?

you only said half of it

[white vine]

maps one to one

[dire forge]

f(a) = f(b) implies what?

[dire forge]

yeah exactly

we want to show that a=b

if we know that 1/(1+a^2) = 1/(1+b^2), then what's something we can do to both sides of the equation that will get us closer to that?

[white vine]

set them equal?

the denominator

(1+a^2)=(1+b^2)

1+a^2> 0 | 1+b^2 > 0

a^2 = b^2?

a = b, a = -b

[dire forge]

yeah exactly

and so how do we know that a=b and not that a = -b

[white vine]

injectivity requires a=b?

it does not specify in the theorem a = -b?

[dire forge]

that's what we want to prove though, we don't know yet that it's injective

so we can't say it's because it's injective

we have to give another reason why a = -b is not possible

[white vine]

this is where i got stuck. Im not sure if I need to go and find a value that disproves this or if i can do it generally

[dire forge]

well let's look at the original question

there's still a piece of information that we haven't used yet

we said "over (0, 1]"

so that means a and b are both in (0, 1], right?

[white vine]

yes

[dire forge]

can we use that somehow to help us

[white vine]

domain restriction then monoticity?

[dire forge]

think simpler

why is a = -b not possible?

if a and b are both in (0, 1]

[white vine]

all positive numbers

[dire forge]

yeah exactly

so then we know that a=b

so, since a=b, what can we conclude?

[white vine]

it is injectivity?

injective*

[dire forge]

yeah

excellent

[white vine]

is there a proper way to formulate that explanation mathematically?

[dire forge]

let's summarize what we did

[white vine]

f(a) = 1/1+a^2 = f(b) = 1/1+b^2

1 + a^2 > 0 | 1 + b^2 > 0 -> Take reciprocals

1+a^2=1+b^2

a^2 = b^2

[proud perch]

.close

[dire forge]

please don't comment in other people's channels

[white vine]

square root both sides a = b

or a = -b but how do you explain domain restriction mathematically

I can explain it using words

[dire forge]

yea, and that works because a > 0 and b > 0

so, since a^2 = b^2, we know that a^2 - b^2 = 0, or (a+b)(a-b) = 0

[white vine]

because the domain restriction 0 to 1 only includes positive numbers, it is not possible to have a=-b

[dire forge]

so from that we know that a+b = 0 or a-b = 0

we can rule out a+b = 0 because a, b > 0, so a+b > 0

therefore, the only possibility is that a-b = 0

which means that a = b

does that make sense?

this way avoids talking about square roots or domain restrictions and keeps everything in terms of simple operations

[white vine]

yes

[dire forge]

awesome

so now let's talk about surjectivity

do you think this function is surjective?

[white vine]

probably?

[dire forge]

why?

[white vine]

Wait. Also the formulation I got not injective for the first one? On a earlier attempt.

Oh my correction.

Looking at another problem.

Find the range and then compare with codomain?

or no?

[white vine]

Plug in 0 and 1 into the function?

[white vine]

Ive gotten

[dire forge]

yeah, so it's not going to be surjective

you can prove that by picking a value that's not in the range and then proving that it's not in the range

[white vine]

then lastly bijection?

because its not surjective but is injecetive its not bijection?

[celest socket]

Well ​a function is bijective if and only if it is both injective and surjective.

So yeah, its not bijective

[devout snow]

@white vine Has your question been resolved?

[devout snow]

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[hasty zodiac]

is there any difference between x^1/2 and sqrt(x)? (over complex numbers)

[dire iris]

that is the definition of sqrt(x)

[hasty zodiac]

i heard z^1/2 is multivalued, while sqrt(z) implies the principal root

so it's single valued

[dire iris]

well that is a notational difference

not a mathematical one

it depends who you ask basically

[hasty zodiac]

so it's subjective?

i see

[acoustic leaf]

different people and books may adopt different conventions on that

[hasty zodiac]

thanks

.close

[devout snow]

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[devout snow]

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[dry canyon]

why is this equal to 0

[pseudo basin]

do you mean that like "what the hell is going on" or "i think it should be something else and not 0"

[dry canyon]

i just entered 0 into the answer and it was correct (i remembered from lesson), but i forgot why the answer was 0

[tender wharf]

0 * anything becomes 0

[dry canyon]

so any limit going towards 0 is 0?

[pseudo basin]

poor explanation

[tender wharf]

Mb lol

[pseudo basin]

not even close

anyway, have you heard of the squeeze theorem?

this is a tool that can help you here

[dry canyon]

heard of it but dont really understand concept

[mild comet]

it’s because
[ -|x| \leq x \sin \left(\frac 1x\right) \leq |x|]

[woven radish]

soup_norm

[pseudo basin]

in some other languages it is called the two policemen theorem

with the suggested imagery being that if two cops are dragging a drunkard between themselves and they both are going to the police station

then the drunkard will also end up there

here the drunkard is the "middle" function and the cops are the "squeezing" functions

not a terribly good bit of imagery but at least it is iconic

[mild comet]

you can intuitively see why from this graph

[dry canyon]

are the squeezers able to touch the drunkard?

[mild comet]

Yes

[dry canyon]

i see

[mild comet]

So the limit at 0 is forced to be 0

[dry canyon]

ill go watch a video on the squeeze theorem and come back later, do u guys have videos?

[mild comet]

looks like the organic chemistry tutor has one on that

[dry canyon]

got it, ill check that out

[mild comet]

the squeeze theorem is a consequence of the fact that limits preserve weak inequalities

[dry canyon]

well, thank you for helping me out

ill close the channel now

.close

[devout snow]

Channel closed

Closed by @dry canyon

Use .reopen if this was a mistake.

[pseudo basin]

as many times as they like

[ionic harness]

This is another version of that bounding idea we discussed last time 👌

This one's at the end of that computing limits article, but yeah a video is best

[ionic harness]

Not another version, it's actually what we applied I should say

In that case it was -1/inf <= limit <= 1/inf, since both sides converge to 0 our limit was 0

Hopefully now that you've learned it fully what we did then makes more sense

[devout snow]

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[robust bobcat]

in 18 b we showed that p(x) = 1/2(x^2-x). Then I want to use the sub x- floor(x) = u. I was thinking that then you can isolate for x and plug that in. In this solution what they did is they used the fact that it is has a period of 1 and then they used r as the upper bound. Which method is correct?

[robust bobcat]

This is for part c

[steel sage]

P is periodic not because f itself is, but integral of f on [k, k+1] is 0, cancelled.

I think you need to expand the actual expression of f on [k, k+1] and integrate

Without the [] I meant

On [k, k+1) you can remove it

[robust bobcat]

im not following. How can we expand f? f is jsut x - floor(x) - 1/2

[faint hearth]

Floor can be expanded further in [k,k+1)

[robust bobcat]

?. Isn't floor just n <=x < n+1. and then floor(x) = n

How can there be an expansion then?

[faint hearth]

What do you mean by which method is correct? You showed only one method

[robust bobcat]

the way I see it is that we found that P(x) = 1/2(x^2-x). For any x we have x- floor(x) = u. Then we have x = floor(x) +u and sub that in for x and we have 1/2((floor(x)-u)^2 - (floor(x)-u))

Then I see that we have shown that P(x) has a period 1. Then x-floor(x) = u is in one of these periods. Meaning if we use r as the upper bound we will have 1/2(r^2-r)

and then plug in for r

[faint hearth]

You have to use the fact that P(x-floor(x))=P(x) because P has a period of 1.

x-floor(x) is in [0,1], and for any r in [0,1], you have proved that P(r)=1/2(r^2-r).
So P(x-floor(x))=1/2((x-floor(x))^2-(x-floor(x)))
So
P(x)=1/2((x-floor(x))^2-(x-floor(x)))

[robust bobcat]

ok so then my issue was that I did not treat x as a general x but as a specific one

thx for the help

[faint hearth]

It doesn't really matter how you "treat x". anyways does it make sense why

P(x)=1/2((x-floor(x))^2-(x-floor(x)))?

[robust bobcat]

yes we know that P(x) is periodic and r will be in one of these periods and then we sub in for r

[faint hearth]

Ok, so do you have any questions?

[robust bobcat]

so your saying that you could still treat x as x= floor(x) +r

is my reasoning right then that we cannot sub in for x because it is a general x?

[faint hearth]

You could, but that won't get you anywhere without using the fact that P(r)=P(x)

You can sub in but it's unnecessarily complicated

[robust bobcat]

isn't the whole objective to have floor(x) in the expression and by using that subsition we would have floor (x)

[faint hearth]

The objective is to find P(x) when x is not in [0,1]

You know for x in [0,1], P(x)=1/2(x^2-x)

And you also know P is 1-periodic

[robust bobcat]

but that doesn't change the fact. Say x not in [0,1] x = floor (x) + r. sub in for x

[faint hearth]

What does that do?

You can't say P(x)=1/2((floor(x)+r)^2-(floor(x)+r))

But you can say P(r)=1/2(r^2-r)

[robust bobcat]

express in terms of floor(x)

[robust bobcat]

why?

[faint hearth]

Because floor(x)+r isn't in [0,1]

You need a quantity in [0,1] to substitute

[robust bobcat]

oh

everythign is clear that would only work if x was an integer

I see now. R is in these intervals. floor(x)+r is not. Meaning we cannot use it to express every P(x)

[faint hearth]

Yes

[robust bobcat]

ok thats clears everything up thx

.close

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[rain coral]

this

[rain coral]

doesnt make sense

ok

if u differentiate

ull get

3p^2 -4p+3

ok

[rain coral]

thats for gradient of tangent

if u flip it over

ull get -1/ (3p^2-4p+3)

so if u get any value of p

u ccld make it

so that the gradient of the normal/M is basically

infintely small no?

[crystal agate]

how are there 4 pi electrons guys

[knotty sage]

No, the function of M(p) (the gradient of the normal line) has a minimum value, a p where M is minimized occurs if you take the derivative of M and equate it to 0.

[knotty sage]

!occupied

[devout snow]

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).

[crystal agate]

oma gawd

[devout snow]

@rain coral Has your question been resolved?

[devout snow]

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[hard sable]

hey

[hard sable]

why do we need to consider the force at point B

if i imagine applying that force to the ship (pulling it) at point B, there is no rotation effect about P

[glossy dew]

very small torque but it exists

[hard sable]

so that force would

pull it anti clockwise?

[tender wharf]

you can say pull but clockwise

[glossy dew]

mostly pulling it away from the hinge and slightly dragging it clockwise

[hard sable]

and you know this from the diagram?

[glossy dew]

yeah
just resolve the force along and perpendicular to the line of action (black line marked in the diagram)

[hard sable]

.close

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[vivid zephyr]

Hello, I need help figuring out if what I am trying to do is possible to do or not:
I have a rotation matrix

[0: cos(z)cos(y)-sin(z)sin(x)sin(y),   1: -sin(z)cos(x),  2: cos(z)sin(y)+sin(z)sin(x)cos(y) ]
[3: sin(z)cos(y)+cos(z)sin(x)sin(y),   4: cos(z)cos(x),   5: sin(z)sin(y)-cos(z)sin(x)cos(y) ]
[6: -cos(x)sin(y),                     7: sin(x),         8: cos(x)cos(y)                    ]

which I need to extract the values x, y and z from.
Right now I am doing this as
x = atan2(m[7], sqrt(m[6] * m[6] + m[8] * m[8]))
y = atan2(m[6], m[8])
z = atan2(m[1], m[4])
, which works correctly when x is not approaching π/2 or -π/2.
When this happens, the floating point values of m[6], m[8], m[1], m[4] become too small and big errors start to occur (such as sign changes etc).

In this condition, the rotation matrix contains

[0: cos(z)cos(y)-sin(z)sin(y),   1: 0,  2: cos(z)sin(y)+sin(z)cos(y) ]
[3: sin(z)cos(y)+cos(z)sin(y),   4: 0,  5: sin(z)sin(y)-cos(z)cos(y) ]
[6: 0,                           7: 1,  8: 0                         ]

which is what my question is about: *is there a way to figure out y and z in this situation?

[outer quarry]

holy this channel makes me feel dumb

[vital edge]

Would be helpful to send images of the matrices cuz it doesn't render very well on mobile

[vivid zephyr]

ok, give me a min

[vivid zephyr]

first one

second one

[faint gorge]

Why not use quaternions

[stone stump]

why so difficult for x

you have sinx as an individual entry

and then cosxcosy

and coszcosx

[vivid zephyr]

you mean using quaterions instead of rotation matrices? I need to use the rotaiton matrix since it's the input I need to process,

if you mean that I can compute the quaternion representation, I'd have to look into it

[vivid zephyr]

I am not a mathematician, I am a computer scientist and I am trying to solve things the best I can 😅
(I tried it like this after reading this https://msl.cs.uiuc.edu/planning/node103.html)

[faint gorge]

Yeah it's worth looking into it because you don't encounter gimbal lock issues iirc

they are also numerically more stable, faster and compact

[vivid zephyr]

so what you are saying is that I should go from rotation matrix (my input) to quaternion to euler angles (what I need to compute)?

[stone stump]

all of those entries are some trig identity

eg top left is cos(y+z)

so you can convert to those

[faint gorge]

yes

[stone stump]

uhh wait all of those only depend on y+z

hmm

[vivid zephyr]

I did not know trigonometric identities were a thing , I will read more about them from https://en.wikipedia.org/wiki/List_of_trigonometric_identities thanks

[tardy edge]

I'm surprised people learn rotation matrices before trigonometric identities

[vivid zephyr]

well, I think I have solved it thanks, now I am computing atan2(sin(z)cos(y)+cos(z)sin(y)),(cos(z)cos(y)-sin(z)sin(y)), which is equal to
atan2(sin(z + y), cos(z + y))

this is sufficient given that in this situation there is already a gimbal lock situation, in which rotations by y and z are "summed" so to speak

[mystic scarab]

Really? That's completely nonsense imo

[vivid zephyr]

well idk if this is normal but in my CS courses I only encountered rotation matrices in relatively niche courses

anyhow, thanks everyone for the help, I learnt something new with this one 😄

.close

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