help-23

[oblique jay]

Or wait... Do you?

hm

So the chance of getting slot 1 is always the same: 0.25

[thorn wagon]

i think i misunderstood the question,i think it implies lesser than 1/4th chance for the chosen slot

[thorn wagon]

thats what i think i misunderstood,i should’ve considered 1/8 instead of 1/4

[oblique jay]

The chance of winning slot 1 is: 12.5% but you can only win slot 1 once

[thorn wagon]

i think the answer should be 32 i think that would be right

cause the net prob of winning 2 slots is 1/32

using this approach

so 32 times should give you one win

ie 2 slots

does that make sense?

[oblique jay]

My feeling is that it should not take more than 3 times the amount of winning one slot, because there are 4 slots...

[thorn wagon]

i mean its mainly cause its not an even distribution of chance

cause losing is 7/8 and winning is 1/8

is there no way to check the answer?

[oblique jay]

Nope, it is a real world problem 😂

[thorn wagon]

where did you get it from?

[oblique jay]

Ohhh that is really lengthy to explain. I am basically writing an algorithm that increases a trust score of a user based on chance over a large quantity of tries.

[thorn wagon]

aaah

i just took all cases to infinity it should be right

i did prob of any slot=1/4
prob of getting it=1/8
prob of not getting it is 7/8

prob of winning 2 in n turns is 1/41/8 * 1/8(7/8)raised to n-2

and just apply inf gp summation for total probability

[oblique jay]

I think I should simplify it even more first, so thinking about it gets easier.

Lets say there are only 3 slots and you have a 50% chance to win.

Winning one slot: 2 rounds

Winning two slots:

1/((3/3) * (1/2)) + 1/((2/3) * (1/2)) = 5?

sigh

Not really convincing.

[safe radish]

@oblique jay Has your question been resolved?

[thorn wagon]

then that would be 6 rounds approximately

[oblique jay]

Winning one slot:
1 / (0.5 * (3/3)) = 2

Winning two slots:
1 / (0.5 * (3/3)) +
1 / (0.5 * (2/3)) = 5

Winning three slots:
1 / (0.5 * (3/3)) +
1 / (0.5 * (2/3)) +
1 / (0.5 * (1/3)) = 11

Can someone confirm?

If I understand correctly then winning three out of three slots that have each a 50% win chance while the slots are chosen randomly requires 11 rounds on average.

[safe radish]

@oblique jay Has your question been resolved?

[oblique jay]

Still open.

[safe radish]

@oblique jay Has your question been resolved?

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[fiery wadi]

can I somehow determine an inflection point without needing to do 3 calculations
as in 1 for the 0 point when f''x is 0
1 before the 0 point and one afterwards for f''x

[safe radish]

Please don't occupy multiple help channels.

[ornate skiff]

If I understood it correctly, I think you can't further simplify the process, unless it's some specific situation, like a very simple function (a function f(x)= a polynomial of degree < 3).

[young nexus]

how did you define an inflection point?

[fiery wadi]

isnt it a switch in concavity iirc

[young nexus]

and why do you wanna calc f'' is 0?

[fiery wadi]

concave = f''x < 0

convex = f''x> 0

no?

[young nexus]

you wrote when f''x is 0 thats what i am not understanding and thats what i am asking.

[safe radish]

@fiery wadi Has your question been resolved?

[fiery wadi]

when f''x = 0

that means

it could go through 0

changing concavity

therefore inflection point

[young nexus]

oh, forget it. i misread something.

[safe radish]

@fiery wadi Has your question been resolved?

[grim flint]

I hope that explanation helps

[desert juniper]

if it's undefined at x=a, it can change concavity, but it cannot be an inflection point

[grim flint]

Ops

Yep you're right

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[graceful fable]

i dont understand limits

if i put limits 20 and 0 for my integral of my acceleration equation

what do i get?

is it the change in velocity between t=0 to t=20

or is the sum of all velocities or somethin

[safe radish]

@graceful fable Has your question been resolved?

[graceful fable]

<@&286206848099549185>

[lime sundial]

@graceful fable if you integrate with limits, you'll get the speed at t = 20

You can integrate without limits, and then substitute 0 and 20 to show that "+c" is basically 0.

You know that V_b = 0

Hope this helps

[graceful fable]

is that because integrating always gives the speed at the upper limit

or is it because my lower limit is 0

[lime sundial]

Think of it as if you integrate acceleration from period A to B, you get the change in velocity between A and B (V_B - V_A).

[graceful fable]

oh so its just change in velocitu

now that i think about it

it was pretty obvious

thanks alot for thr help

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[cobalt tapir]

how does this equal x xa =b?

[pseudo scroll]

the top was an identity that I don't remember

nvm

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[atomic veldt]

can bezeir curves be integrated?

[atomic veldt]

can bezeir surfaces be integrated to form solids?

also how

[fresh wasp]

what is that

[safe radish]

@atomic veldt Has your question been resolved?

[atomic veldt]

https://en.wikipedia.org/wiki/Bézier_surface

this

its like a curve that you can control how it goes using control points

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@atomic veldt Has your question been resolved?

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@atomic veldt Has your question been resolved?

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[cold whale]

im confused on where the 1/4 came from

[grim flint]

comes from the constant factor introduced during the integration process

[cold whale]

but thats the 4 in the 4/5

[lean otter]

partial fraction

[grim flint]

Oh true

[cold whale]

im still confused

[grim flint]

1/4 is part of breaking down 4/5 to get the final coefficient of 1/5 in the answer

When we split 4/5 I think

[cold whale]

i know but 4/5 and 1/4 are two different numbers

4/5 comes from 1/5 times 4/1

[true veldt]

no it's what happens when you break 1/(t+2)(t+6) into two different fractions

[cold whale]

i still dont get it lol

would it be 1/3

if anything

[grim flint]

from solving the coefficients A and B?

Wait you're absolutely right

[true veldt]

if you did partial fraction decomposition you would find that $\frac1{(t+2)(t+6)}=\frac14\cdot\bigg(\frac1{t+2}-\frac1{t+6}\bigg)$

[flat frigate]

sentinel

[cold whale]

whats partial fraction decomposition

i still dont understand where the 1/4 comes from lol

[true veldt]

its exactly what it looks like, it's the method that you would use to break a complicated fraction like the one in the integral into something that's easier to integrate

[grim flint]

It's the coefficient we get when we split up the fraction into its partial fractions

[cold whale]

it follows

[true veldt]

if you use that formula then $\frac14$ comes from the $\frac1{b-a}=\frac1{6-2}$

[flat frigate]

sentinel

[true veldt]

maybe you don't already know partial fraction decomposition, but that is the technique that you would likely use to figure out the formula you posted

[cold whale]

i thought 1/b-a is 1/5 cuz of the 0,5

[true veldt]

look at the formula again

b and a are not the bounds of the integral

[cold whale]

its average value

[true veldt]

look at this formula

it's not an integral with bounds

[cold whale]

ohh

ok

its clicking

i totally get it now

thank you

[true veldt]

awesome

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[acoustic goblet]

I am in desperate need of help with this one

[acoustic goblet]

I struggle to find a fitting alpha

And the other tasks want me to use alpha.

[safe radish]

@acoustic goblet Has your question been resolved?

[acoustic goblet]

<@&286206848099549185>

[azure delta]

It's been too long and I barely remember what a form is, but I know this has to do with non differentiability in a loop around zero.

It's very classic. You should try searching for closed but not exact.

Oh, this might be a bit harder because of the dimension

[acoustic goblet]

What do you mean by dimension?

[acoustic goblet]

I mainly find it hard to find a well defined form on U1 thats closed.

I cant even do that, and then i also need to think about if its exact.

[azure delta]

Do you know the classic example of closed but not exact in R^2?

[acoustic goblet]

Yes

its this -y/r^2 dx + x/r^2 dy where r=sqrt(x²+y²).

But that does not work here, no? Since my U1 includes the z-axis

[acoustic goblet]

i was also thinking about this, but i wasnt sure

[obtuse jackal]

It seems silkster left
Feel free to try your luck in the relevant advanced channels
Should be #diff-geo-diff-top if I'm not mistaken

[safe radish]

@acoustic goblet Has your question been resolved?

[acoustic goblet]

<@&286206848099549185>

[white umbra]

You might recall that you can view this geometrically as a (co)vector field which circles around a hole

similarly in U1 we also have a hole (C1)

I would try writing a similar formula to the form that you gave that works in R^2 \ {(0,0)}, except instead of circling about the origin, it circles around some part of C1

[acoustic goblet]

Do you have a formula that you mean? I dont exactly get it.

I tried something like 1/((x²+y²-1)²+z²) but this stuff got extremely complicated

[white umbra]

(The fact that this is not exact comes from the fact that if you imagine the arrows as a current in a lake, you can sail around the circle and get faster and faster. The fact that this is closed comes from the fact that the curl of the (co)vector field (-y/r^2, x/r^2) is 0.)

[white umbra]

I would probably start by drawing a picture of what your choice of alpha might look like

[acoustic goblet]

Or does something like (x-1)/((x-1)²+z²) dz - z/((x-1)²+z²) dx ?

work

[white umbra]

Do you mean dx at the end

[acoustic goblet]

yes

[white umbra]

No that won't work

Because it's not defined on all of U1

(For example, on (1,1,0) it's not defined)

[acoustic goblet]

mhhm

[white umbra]

Maybe you could try changing variables into cylindrical coordinates? It's possible that that would help

(I don't know the solution to this off the top of my head, just shooting out ideas)

[acoustic goblet]

tried that too but i dont saw anything

The problem is:

Im super new to this and from what i understand, even x dy + y dx works?

[white umbra]

That form is exact

Do you see why?

[acoustic goblet]

But we did something in class where we proofed non exactness like this:

Okay suppose u have f s.t. df=alpha
-> consider the singular cube g : [0,1] -> R^3 s.t. g(t)=(cos(2pi*t),sin(2pi*t),0)
Because of Stokes, now I_(border of g) f = I_g df = I_g alpha
But this is I_[0,1] alpha*(y dx + x dy) = I_[0,1] 2 pi dt = 2pi, whereas I_(border of g) f = I_(0,1,0) -I_(0,1,0) = 0, therefore we have a contradiction

[white umbra]

okay

[white umbra]

so that's the formal argument of what I've written out here

if you're on a raft drifting on the lake and the arrows represent the current on the lake, then as you flow in the direction of the arrows around the circle, by the time you get back where you've started, you've increased in speed

however, if the form is exact, this shouldn't be possible—your speed should be exactly the same as your original speed

exact forms correspond to "conservative" vector fields

[acoustic goblet]

okay

[white umbra]

if you've taken any physics, this is the principle of conservation of energy

[acoustic goblet]

yes, heard of that

but does that mean that my form is exact? What is wrong in my "proof" then?

[white umbra]

That's what a proof that a form is not exact looks like; how do you prove that a form is exact?

[acoustic goblet]

Well but i want to prof that its not exact, no?

[white umbra]

I'm telling you that x dy + y dx is exact

How would you confirm that it is?

[acoustic goblet]

Waiit, but whats wrong there? Because in my opinion, this is done with x dy + y dx

[white umbra]

Oh I get what you're asking now

[acoustic goblet]

well i would sit there and try to find a function f that works, i guess

[white umbra]

The integration on the last line is wrong; you should not be getting 2 pi dt. You have a sign error.

[white umbra]

Yeah, what function might you try in order to find a form whose exterior derivative is x dy + y dx?

[acoustic goblet]

But alpha* (y dx + x dy) = cos(2pi t)*(2pi*cos(2pi t) dt) + sin(2pi t)*(2pi*sin(2pi t) dt), or am i stupid?

and this gives 2pi

dt

[white umbra]

To be more clear, y dx + x dy = sin(2pi t) (**-**2 pi sin(2pi t) dt) + cos(2pi t) (2 pi cos(2pi t) dt)

You're not stupid but you have a sign error

[acoustic goblet]

Oh gosh

yes

okay

Now that my mind is clear, id probably use f=x*y?

[white umbra]

yup

that's correct, df = x dy + y dx

[acoustic goblet]

okay, this is simple

however

[white umbra]

in fact any form that's defined on R^2 which is closed will also be exact

[acoustic goblet]

ahh okay

[white umbra]

so it's very important that we have a "hole" in our space R^2 \ {(0,0)} which lets us have forms which are closed but not exact

[acoustic goblet]

yeah this i get, but i just didnt find anything useful for the given problem

like

nothing, i got that idea to use .../((x²+y²-1)²+z²) or something to make sure, that only points on C1 are undefined

But THEN i got stuck.

[white umbra]

Yeah that's a smart idea

[acoustic goblet]

Because everything becomes so complex with the derivatives

[white umbra]

You said you tried doing change of variables to cylindrical variables right

What did you end up doing there

[acoustic goblet]

Yeah, but i didnt got any further. I tried to use cilindrical coordinates and applied this idea to .../((r-1)²+z²) or in spherical coordinates by using 1/((r-1)²+cos²(phi)) or something

but it didnt get easier

And i am just like disgusted by this task, because its one point of 16 on this weeks sheet

[white umbra]

yeah I think ((r-1)^2 + z^2) is a smart idea

did you try using that form we talked about earlier

-y/(x^2 + y^2) dx + y/(x^2 + y^2) dy

but instead of x and y

you could use (r-1) and z

[acoustic goblet]

Ohh, can i just plug r-1 and z in?

[white umbra]

maybe!

[acoustic goblet]

Wow, let me think about this

[white umbra]

I think it would work

[acoustic goblet]

if yes, then i feel extremely dumb

[white umbra]

don't be haha

this math is pretty abstract

[acoustic goblet]

I sat on this for hours and got everything but the final step

Wait

But

well get dr and dz, dont we?

are they definded for r=0?

[white umbra]

that's a good question

I don't think it would be

hmm maybe there's some way to edit

[acoustic goblet]

i dont really see one, this was a problem i faced when i tried alpha = d_r or something

because such things are obv. closed

Or wait

can we do something like d(r-1)? haha

[white umbra]

nah that's the same thing as dr

unfortunately

[acoustic goblet]

nah, that wouldnt work either

yes

[white umbra]

yeah I'm not sure sorry :(

I'll keep thinking about it

something I'm thinking is maybe you can define the form in such a way to avoid the singularity at r=0

[acoustic goblet]

But are we sure, that it isnt something totally obv. like d_phi or something?

[white umbra]

something like this?

[acoustic goblet]

yikes

i dont even know how to put this into math language

[white umbra]

if you mean like phi being the angle in polar coordinates, then d phi is the same thing as -x/r^2 dx + y/r^2 dy

(which is the impetus for how you would come up with that form; d phi is not exact because phi is not actually a function due to being multivalued)

[white umbra]

hm I have an idea, maybe instead of graphing r, we could graph ln r

the logic is that r goes from 0 to infinity, so ln r goes from -infinity to infinity, removing the weird edge case at 0

not sure if that works or not

[acoustic goblet]

can you

draw it?

I dont really get the idea

Well

mhhm

and what now? hahaha

[white umbra]

then the singularity happens at (ln r, z) = (0, 0)

and you could do the same -x/r^2 dx + y/r^2 dy idea maybe?

plug in ln r and z

see if you get something that's defined at (x,y) = (0,0)

[acoustic goblet]

Wait

does this already include the r-1 thing?

[white umbra]

you don't have to do r-1 here

because if the singularity happens at r=1, then ln r = 0

so I guess you could say it's another way to "move" the 1 to 0

[acoustic goblet]

But

So dont we work in cylindrical coordinates anymore?

Because how would that fix dr?

Then we get smth like -ln(r)/(ln(r)²+z²)) dr

Still dr isnt defined in 0, no?

[white umbra]

hmm

-ln(r) / (ln(r)^2 + z^2)) d(ln r)

and then if we calculate it out

yeah I think you're right about it still not being defined at the origin

grr

[acoustic goblet]

it cant be that difficult

Its one point

either we miss something obvious or the question is plainly wrong

[white umbra]

well the question's definitely not wrong

I can say that much

[acoustic goblet]

Im so desperate that i even tried ChatGPT (shame on me) and this mf suggests the most weird answer all the time, he just wants this -y/r dx + x/r dy thing, even when you tell him that this is just dumb

[white umbra]

the r-word isn't allowed in this server btw

okay maybe this is dumb but what if we did (r^2 - 1) instead of (ln r)

[acoustic goblet]

oh, didnt know that! (does this word have any other meaning then "dumb"? because im not an english speaker and thus i always thought its the same)

[white umbra]

then d(r^2 - 1) = d(r^2) = 2r dr = 2r (x/r dx + y/r dy) = 2x dx + 2y dy, so that's defined at the origin

[white umbra]

a lot of ppl nowadays (over the last two decades or so) consider it an ableist term, one that makes fun of people with intellectual disabilities

[acoustic goblet]

Can you rewrite this in cylindrical coords?

[acoustic goblet]

ah oh no, then i wont use it anymor

[white umbra]

(it used to be a medical term meaning intellectual disability, but now it's often considered offensive because people just use it as an insult, so it's no longer used in medical contexts)

[acoustic goblet]

where did the z go?

[white umbra]

I'm thinking something like -x/(x^2 + y^2) dx + y/(x^2 + y^2) dy, but replace x and y with (r^2 - 1) and z, respectively

[acoustic goblet]

but switching x and y in the numerator, no?

[acoustic goblet]

mhm but

then we have dr again

or not?

[white umbra]

er yeah, sorry

[white umbra]

but the r in 2r dr cancels out the 1/r factor in dr

I'm pretty sure if you calculate this out, you'll get a form that's defined on all of U1 now, so it's the closed but not exact form we wanted

[acoustic goblet]

Is this equivalent to just using (x²+y²-1) and z instead of x and y?

[white umbra]

yea

[acoustic goblet]

ah well obviously it is

[white umbra]

so your original idea was right

[acoustic goblet]

let me just quickly calculate it

yikes it gets ugly

[white umbra]

not too ugly

[acoustic goblet]

let me check WA if its still closed, just to be sure

[white umbra]

you should get
||1/((x^2+y^2-1)^2 + z^2) (-2xz dx - 2yz dy + (x^2+y^2-1) dz)||

[acoustic goblet]

...

This looks strange

[white umbra]

haha

[acoustic goblet]

//?

[white umbra]

oh sorry

typo

[acoustic goblet]

ahhh, im stupid haha

[white umbra]

[\frac{-2xz,\dd x - 2yz,\dd y + (x^2+y^2-1),\dd z}{(x^2+y^2-1)^2 + z^2}]

[acoustic goblet]

i combined dy with dz

[flat frigate]

fly_haltERIC

[white umbra]

like this

[acoustic goblet]

tbh i dont even want to check with WA, well get 6 partial derivatives...

ill just believe it

mhm but now im curios

Do we have to integrate around the circle in the xz plane f.ex. to prove that its not exact?

[white umbra]

it's not that hard too, see here:

[acoustic goblet]

Is this equation in latex?

[white umbra]

,w curl of (-(2xz)/((x^2+y^2-1)^2 + z^2), -(2yz)/((x^2+y^2-1)^2 + z^2), (x^2+y^2-1)/((x^2+y^2-1)^2 + z^2))

[flat frigate]

Results provided by WolframAlpha

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

Requested by erictao

[white umbra]

Zero curl, so the form is closed

[white umbra]

yeah, this message was in latex #help-23 message

[acoustic goblet]

but now the integral gets ugly, no?

[white umbra]

you could consider doing an integral in (r^2-1, theta, z) space

this should somehow correspond to an integral in (x,y,z) space

[acoustic goblet]

How does this integral need to look like? like which direction do i have to integrate?

[white umbra]

what direction do you have to integrate for the classic example of -x/r^2 dx + y/r^2 dy?

[acoustic goblet]

around it in a circle

okay, so then i use g(t)=(sin(2pi t), 0, cos(2pi t) for (r, phi, z)?

[white umbra]

for (r^2 - 1, phi, z) yeah

not r

btw I think I figured out a way to visualize what this form looks like

[acoustic goblet]

Really? Need to see it

[white umbra]

okay I'm gonna show you a series of images

this first one is what the form looks like if you just plot the axes (r^2 - 1) and z (I'm omitting the phi axis)

as you can see, the form makes circles around the origin

except that r^2 - 1 must always be at least -1, so we have sort of a singularity there, which I've indicated by the dashed line

good so far?

[acoustic goblet]

Yes

Whoops

Not a reply to this one haha

[acoustic goblet]

Yes

[white umbra]

alright so then

if we change the axis from r^2 - 1 to r

it basically remains the same but it distorts the image

so now it looks like this

imagine like horizontally stretching/shrinking the graph

crucially, the graph looks horizontal once we approach r=0

good so far?

[acoustic goblet]

Yes

[white umbra]

okay and now

the last thing we need to introduce into the picture is the phi axis

so we take this whole picture and rotate it around the z-axis

and we get this

[acoustic goblet]

Oh wow

[white umbra]

so it's like

there are rings going around the red circle

which is what we wanted

[acoustic goblet]

Looks like the solution to some PDE ive seen in physics somewhere haha

[white umbra]

and the parts that hit the singularity in our previous graphs, they become a continuous loop which avoids the red circle

yeah you can imagine it sort of like if you had a loop of wire which had a changing current going through it

this would be the induced magnetic field

in fact that's basically exactly what faraday's law of induction says, I think

[acoustic goblet]

Pretty neat

[white umbra]

imma plug this into a 3d vector field plotter

okay check this out!

you can see the pattern I drew above

[acoustic goblet]

Oh wow yeeah

[white umbra]

https://www.desmos.com/3d/wuvdkmyf1f

here's the link if you want to play around with it yourself :)

(if you want to change some parameters on this Desmos link: alpha just changes how long the arrows are; Lx, Ly, and Lz just change the range that's visible on the graph; and Lengthmax truncates the arrows to not be over a certain length so the graph is less messy)

[acoustic goblet]

Ahhh okay 🙂

Still trying that integral

Its weird haha

But the pattern you found looks very nice

[white umbra]

you got this :)

as a side note, if (r^2 - 1, phi, z) = (sin(2 pi t), 0, cos(2 pi t)), then you could think about what that graph looks like in (x, y, z) coordinates

the integral is easier to do in (r^2 - 1, phi, z) space but still a nice thing to think about so you can concrete visualize what path you're taking for the integral

[acoustic goblet]

But im not sure about this. Do i write alpha in its intial form then?

[white umbra]

Yes you can write alpha as (- z / ((r^2-1)^2 + z^2)) d(r^2-1) + ((r^2-1) / ((r^2-1)^2 + z^2)) dz

it should be a fairly straightforward calculation

[acoustic goblet]

Can i sub in something for r²-1?

[white umbra]

yeah sure you can call that u

u = r^2 - 1

[acoustic goblet]

Like just rename it to u?

[white umbra]

I gtg, but hopefully this was helpful 🙂

[acoustic goblet]

Sure, you were super helpful

I need to sleep

Its 5am here

Havent closed an eye yet

Phone rings in 4 hours

Thank you

It was very fun

[safe radish]

@acoustic goblet Has your question been resolved?

[safe radish]

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[novel blaze]

I need help with a linear algebra question. I've been asking chatgpt for help with linear algebra, but I notice once in awhile it'll give an anomalous answer, so it's hard for me to know if it's giving the right answer or not sometimes.

The question is

[novel blaze]

it's answer was

but I got

[light shoal]

"once in awhile it'll give an anomalous answer" - you're lucky if it's only once in a while

[novel blaze]

right

[burnt notch]

By the way your answer is correct

[novel blaze]

it gave an upside down version of my answer

[burnt notch]

It's just a linear combination of the vectors given from chatGPT

[light shoal]

there are infinitely many valid answers

[burnt notch]

Which is fine

[novel blaze]

I mean, I get it, I've just been using it as a last resort, comparing it to the powerpoints and textbook answers to see which times it's right and slowly understanding

but yeah, it's definitely a sloppy way to go about learning, I know

[burnt notch]

Actually, the fact that you got a correct answer by yourself is very good 👍

[novel blaze]

thank you

I'll ask more questions in this server when something comes up, but thank you for confirming the answer for me!

.close.

.close

[safe radish]

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[void blade]

Ok my biggest thing is that idk what do with actually solving each equation

[vapid glen]

one way, but might be confusing, is multiply the first equation by 2 to make a 2x^2 term appear, then subtract the second equation from it, and that gets rid of the x completely

try that out and see how far you can get, if you get stuck let me know

[void blade]

Alright

It’s just perfectly fine to multiply it by 2?

[vapid glen]

yeah as long as you do the same thing to both sides you're ok, since 2 has an inverse

you could always undo it by dividing both sides by 2 is what I mean

if you tried multiplying both sides by 0, then you can't undo that, so it's like "throwing away" info

[void blade]

Ah

So what happens with y?

[vapid glen]

it just adds on

[void blade]

oh?

wait

so what it becomes 4y^2

[void blade]

LOL

[vapid glen]

ah it would if it was 2y^2 but it's 2y

[void blade]

HUH

wait

im confused how the heck

Btw this was the actual answer idk how

[vapid glen]

$2x^2+2y^2=2$ when you subtract that second equation from it you get
$$2y^2-2y=0$$

[flat frigate]

Merosity

[vapid glen]

can you solve that 2y^2 - 2y = 0 now

[void blade]

doesnt it just leave it as is

so it just stays as 2y^2 - 2y = 0

cause they arent like terms

[vapid glen]

yup exactly

[void blade]

thats something

uh what do i even do now

cause of the other equation

and like i cant even find y

[vapid glen]

oh we can solve for y in this equation now

you could do it the "hard way" and use the quadratic formula

but you can also get rid of the 2 and factor it

[void blade]

so you divide the entire thing by 2?

to get rid of it

but then it just leaves y^2 - y = 0

could completing the square work?

[vapid glen]

you could but I wouldn't for this

you can factor y out of both terms

[void blade]

as like a gcf?

[vapid glen]

I mean like y(y-1) = 0

does that make sense or is that too much of a leap

[void blade]

nah i think thats basically what i meant

[vapid glen]

oh ok cool

[void blade]

y is the gcf between the entire thing

[vapid glen]

ah yeah I see what you're saying now

[void blade]

now my biggest thing

how do you even get plot points from it

[vapid glen]

ah well we're almost done finding the y values, y(y-1)=0 what does this mean y can be

as a hint, if a*b=0 then a=0 or b=0

[void blade]

uhhhhhhh

im guessing 1 or -1

just cause its the only number to really pull from

[vapid glen]

close so we don't have to guess I'll show you how we can figure it out

y(y-1)=0 means y=0 or y-1=0 since one of those things has to be 0 to multiply the other and make the whole thing 0

[void blade]

ohh i see

so it should be 1 then

i think

[vapid glen]

could be y=0 or y=1 both work

so now for each y, you can plug them in to either equation, and solve for the x values that go for that y value

[void blade]

i see so basically im plugging in 1 and 0 into both equations?

[vapid glen]

yeah, just don't do both at the same time

first equation is probably easiest one to work with since it doesn't have all those 2s

[void blade]

Right

Alright let’s see

So I guess something like this

And then plug each x value into the other question

Question mark

[safe radish]

@void blade Has your question been resolved?

[vapid glen]

yeah that's good, just one slight thing when you square root x^2=1 you could also get +1 and -1

since (-1)^2=1 too

[void blade]

Oh

I suppose so

I’m cooked man I’m so tired

Tank you for helping me

And sorry if I stressed ya out

[vapid glen]

yeah all good that was a lot, you're welcome

[vapid glen]

haha no I'm good 😛 any time ping me later if you need help again

[void blade]

Will do

https://tenor.com/view/bocchitherock-bocchi-hitori-gotou-ぼっちざろっく-anime-gif-26998601

[safe radish]

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[trail otter]

Let $A=\begin{pmatrix}
2&2\3&3
\end{pmatrix}$ and $B=\begin{pmatrix}
4&3\6&1
\end{pmatrix}$ be matrices over $\mathbb{Z}_7$. how to solve $AX^T=B$. i know that $A,B$ are not regular matrices so i cant just use inverse. how should i do it then?

If $X=\begin{pmatrix}
a&b\c&d
\end{pmatrix}$ ill get \begin{align*}
2a+2b&=4\
2c+2d&=3\
3a+3b&=6\
3c+3d&=1
\end{align*}
represented by matrix
[\begin{pmatrix}
2&2&0&0&4\
0&0&2&2&3\
3&3&0&0&6\
0&0&3&3&1\
\end{pmatrix}\sim\begin{pmatrix}
1&1&0&0&0\
0&0&1&1&0\
0&0&0&0&1\
0&0&0&0&0\
\end{pmatrix}]
but expanded matix has greater rank thatn original thus there should be any solution. what am i doing wrong?

[flat frigate]

Slowaq

[safe radish]

@trail otter Has your question been resolved?

[shy beacon]

Actually nothing, because A doesn't have inverse
You dont have enough equations to solve for a b c d

[trail otter]

solution should be $\begin{pmatrix}
2&0\5&0
\end{pmatrix}+span(\begin{pmatrix}
6&1\0&0
\end{pmatrix},\begin{pmatrix}
0&0\6&1
\end{pmatrix})$

[flat frigate]

Slowaq

[trail otter]

according to my teacher

[frozen shoal]

Yes, you are almost done when you computed the four linear equations.

So basically, from the four linear equations we have that a+b=2 and c+d=5, right?

[safe radish]

@trail otter Has your question been resolved?

[trail otter]

well we have a+b=2, c+d=3/2, c+d=1/3

and this doesnt have solution

[frozen shoal]

What is 3/2 and 1/3 mod 7?

@trail otter

[trail otter]

ah true shm stupid modulo

3/2=3*2^-1 = 3*4 and 3^-1=5

so yes ill get a+b=2 and c+d=5

[frozen shoal]

great, do you want to give it a try with that info?

This is basically now a standard linear algebra question.

[trail otter]

yes now i should be able to do it thanks

.close

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[burnt sequoia]

I need help

[burnt sequoia]

w a work sheet

i got exam soon

i need someone to explain to me how to solve it

[fathom adder]

Isolate x in one of the equation and replace x by the value in the other equation

Then solve for y

[vagrant ice]

bonjour

[fathom adder]

And once you have y value

[fathom adder]

Yo le s

[fathom adder]

Replace it in the x equation to have x value

[safe radish]

@burnt sequoia Has your question been resolved?

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[zinc pier]

I have isolices triangle AB=BC = 10 cm and AC = 12 cm

Find height a and area

[zinc pier]

.close

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[placid garnet]

does this check out?

[vagrant ice]

correct method but

$1 - (r^2 + 1) = 1 - r^2 - 1 = -r^2$

[flat frigate]

south

[vagrant ice]

so yeah you need to write your working properly

[placid garnet]

ill keep it in mind, thanks

.close

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[burnt grail]

i’m confused with 2.7

[burnt grail]

would parallel and up be on the left side of the block

[manic radish]

yes

[burnt grail]

then where would friction be?

[burnt grail]

and does the arrow point left or right

[burnt grail]

oh alr

[manic radish]

The weight of the box is a force downwards. You first need to express that in components parallel and perpendicular to the inclined plane

[burnt grail]

oh

[burnt grail]

so is friction parallel and down the inclined plane?

[manic radish]

friction acts parallel to the plane, "normal force" acts perpendicular

[burnt grail]

uhhh is that a yes then

[manic radish]

i.e. the plane pushes on the block (counteracting its perpendicular weight component), and friction acts parallel to the plane (counteracting it sliding due to it's parallel weight component)

[burnt grail]

so 30N is its parallel weight component?

[manic radish]

no, that's another entirely different force

[burnt grail]

oh

is that called an applied force

[manic radish]

yep, someone is applying a 30N force

[burnt grail]

and friction counteracts that?

[manic radish]

the block's weight also applies a force

[burnt grail]

oh yes

[manic radish]

friction counteracts all forces acting parallel to the plane

[burnt grail]

then for 2.7(a), where it asks for the force of friction, wouldn't it be 30N as well?

[manic radish]

no

[burnt grail]

or -30N?

but wouldn't newtons i think third law apply here?

[manic radish]

because part of the blocks weight acts parallel to the plane

look at the force diagram

[burnt grail]

this?

is the one going straight down its weight?

wait no it shuld be the one next to it

given my diagram at least

[manic radish]

you need to work out

1. the perpendicular force from the weight of the block (which equals the normal force provided by the plane)
2. the sum of parallel forces (which equals the friction provided by the plane)

red - forces from the weight of the block
blue - normal force from the plane
green - an external force acting on the block

[burnt grail]

how am i meant to solve for the perp. force

if we don't know the normal force

[manic radish]

you need to work that out by decomposing the weight of the block into two components - one acting parallel, the other acting perpendicular.

you know the angle of the plane, and you know the mass of the block. this is enough to label the triangle.

(this triangle with bold lines)

[burnt grail]

is the perpendicular line 1.25N?

so the largest red line

[manic radish]

nope. what's your working?

[burnt grail]

5xsin30

[manic radish]

5kg is the mass, not the force

[burnt grail]

oh

then 30 x sin30?

[manic radish]

why sin?

[burnt grail]

bc opp/hyp

and opp is the largest red line

[manic radish]

where's the angle?

[burnt grail]

there

[manic radish]

which is where on the triangle?

[burnt grail]

oh

crap

so is 30N the shortest red line?

[manic radish]

I think you're doing the right calculation now, but not quite the right answer.
What's the calculation you're using?

[burnt grail]

if we're solving for that big red line

let that be x

so sin30 = x/30N

x = 30 x sin30

x = 15

[manic radish]

draw the angle on the triangle

[burnt grail]

nvm im wrong

it cant be 15N

cus im thinking of 30 degrees from the plane

like 30 degrees inclined plane

but idk any angle values in the triangle

[manic radish]

Which is the 30 degree angle, green or blue?

[burnt grail]

green?

[manic radish]

yes

So, what's the force of the long red line?

[burnt grail]

30cos30

wait

shit no

30/tan30

[manic radish]

why tan?

[burnt grail]

isn't black the hypotenuse?

[manic radish]

yes

[burnt grail]

so opp/adj?

[manic radish]

and the long red line is?

[burnt grail]

like are u asking for the actual value of it

[manic radish]

no, is it hyp, opp, or adj?

[burnt grail]

oh adj

to the green angle

[manic radish]

which trig formula has hyp and adj?

[burnt grail]

cos

[manic radish]

yep

what's the length of hyp?

[burnt grail]

30N

ah

i got mixed up

[manic radish]

how did you get that?

[burnt grail]

cus there's a force of 30N acting on the 5kg

[manic radish]

no, that's the green line

#help-23 message

the black line is from the weight of the block

[burnt grail]

then it'd be 5?

[manic radish]

almost. 5kg is the mass of the block, not its weight

[burnt grail]

so i have to convert that into its weight?

[manic radish]

yes

[burnt grail]

do i multiply it by 9.8?

[manic radish]

yes

[burnt grail]

so 49

kg

[manic radish]

49 N

[burnt grail]

or is it N

ight

[manic radish]

Kg is mass, N is force (or weight)

[burnt grail]

ah

[manic radish]

remember, weight is a force.

[burnt grail]

alr

[manic radish]

so, 49N at 30 degrees is (for the long red line)

[burnt grail]

42

to 2sf

42 N

[manic radish]

yes

and the short red line?

[burnt grail]

24.5 N

or 25 N to 2sf

[manic radish]

yep

So if the block is pushing inwards with 42N, what force is the place pushing back out with?

[burnt grail]

42N

by newtons third law?

every action has an equal and opposite reaction

[manic radish]

yep

Let's add everything onto our diagram

[burnt grail]

so if 42N is getting pushed inwards, then 42N must be getting pushed outwards

ohhh

[manic radish]

So, what's the force of friction provided by the plane?

[burnt grail]

25N?

[manic radish]

no

[burnt grail]

wait

55N?

[manic radish]

no

[burnt grail]

cus sum of parallel forces

oh

[manic radish]

red points down the plane, green points up the plane

[burnt grail]

but if it points up the plane, that means it's going parallel to the plane isn't it?

so 30 - 25?

[manic radish]

yep

[burnt grail]

to 5N is the friction

[manic radish]

yep

[burnt grail]

hurray

man thats so confusing

[manic radish]

[burnt grail]

am i expected to draw allat

like in an exam format

[manic radish]

Keep drawing diagrams and labeling everything

that's the trick to these questions

[burnt grail]

also for this

yk how u drew the 30N on the right hand side

should i do that or is it fine to keep the 30N on the left hand side too

[manic radish]

it doesn't really matter

I'm used to drawing all forces acting on an object from the center of that object

[burnt grail]

ah alr

so (b) would be 42N

[safe radish]

@burnt grail Has your question been resolved?

[burnt grail]

.close

[safe radish]

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[safe radish]

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[lavish lichen]

a is a fixed non zero constant

[lavish lichen]

How do you find the derivative
I tried using the quotient rule but I don't seem to be arriving at the answer

[rugged patio]

its occupied

[lavish lichen]

the answer given in the key is
cos a /cos^2 x

[grave mantle]

help me

[lavish lichen]

what's the question say

also the channel is already occupied

[sleek wolf]

!occupied

[safe radish]

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).

[lavish lichen]

when can i @ the helpers?

[grave mantle]

owh where i can find the non occupied?

[grave mantle]

The equation of the tangent line to the curve, which is perpendicular to the line

[lavish lichen]

Math help (available)

[lavish lichen]

<@&286206848099549185>

[safe radish]

@lavish lichen Has your question been resolved?

[lavish lichen]

.close

[safe radish]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
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[umbral tulip]

i forgot how to do this and google is useless

[thin bridge]

start with a general quadratic equation

sub in the points and solve a system of equations

[umbral tulip]

i did it for the first one and it comes out to c=25

[thin bridge]

how are you getting c=25

[umbral tulip]

well it’s 0

[thin bridge]

what does your c represent,
can you take a pic of your work

[umbral tulip]

it js cancels out no?

cs anything multiplied by 0 would js be 0

[thin bridge]

why are you squaring the 5

[umbral tulip]

sorry

i meant c=5

[thin bridge]

yes

[umbral tulip]

alright what do i do w that

do i js already have the value of c

And i do the system of equations for the other 2 equations

[thin bridge]

yes

[umbral tulip]

what would i do here

since i alr have the value of c

[thin bridge]

use it

[umbral tulip]

i js plugged it in then subtracted it on botch sides

[thin bridge]

now solve the system

[umbral tulip]

would i multiply the top by 3 and then the bottom by 2 so i get 6b and then subtract them to solve for a?

[thin bridge]

that works

[umbral tulip]

alr thanks

it’s so early in the morning idk y im trying to do this

.close

[safe radish]

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[jaunty finch]

this is equal to

[jaunty finch]

1/n+1/2n+...+1/n^2

i want to find a sequence bigger than this that has the limit 0 but i cant find it

any ideas?

[austere cypress]

wym bigger

[jaunty finch]

i mean

i have to find the limit of that number

using the criterion of

i dont know the name

claw criterion or sum like that

i found the inferior sequence that has te limit 0

and i want to find a bigger sequence that still has teh limit 0

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[viscid void]

is there a formula to get the probability density function of a given x^n function? i'm trying to do Inverse transform sampling

[plucky elk]

!original

[safe radish]

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.

[viscid void]

"given a random n-sized sample of numbers, from 0 to 1, arrange them such that their distribution fits the area under the curve x^-2.5"

i was trying to use inverse transform sampling

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@viscid void Has your question been resolved?

[viscid void]

<@&286206848099549185> ?

[plucky elk]

did you learn this formula

https://brilliant.org/wiki/inverse-transform-sampling/

follow the example with exponential function

once you find the cdf of x^n, you find its inverse function

[viscid void]

Thanks

[viscid void]

.close

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[lapis breach]

wtf am i supposed to say for e

[lapis breach]

is it just f’(x)?

ifk how to put in context of geaph

Its just limit definition of derivative..

[median vigil]

it's the derivative at which value of x?

[lapis breach]

oh f(1)?

[median vigil]

it's f'(1), yes

[lapis breach]

thats just zero no?

F(1)=-1 d/dx constant= 0

[median vigil]

how do you get that it's a constant at x = 1?

[lapis breach]

or well -4

f(1) =-4

on the graph?

No?

[median vigil]

yes. but we are looking for the derivative (slope) at x = 1

[lapis breach]

yeah thanks

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[finite dawn]

Hey, can someone help me solve sin^2 2x + sin^2 3x = 1?
In my book, it says the answers are: +-18 +72k, +-90 +360k.
Thank you!

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@finite dawn Has your question been resolved?

[finite dawn]

<@&286206848099549185>

Please if anyone can help (:

[light shoal]

maybe move one of the sin^2 terms to the other side

and use a trig identity

[finite dawn]

I tried using the identity of
sin^2x = (1-cos2x)/2, but it gives me the answers of
90 + 180k, and 18+36k

[finite dawn]

but the book says its :
+-18 +72k, +-90 +360k

[light shoal]

show your work, what did you do?

[finite dawn]

Using the identity it becomes:
(1-cos4x)/2 + (1-cos6x)/2 = 1
(1-cos4x+1-cos6x) = 2
2 - cos4x - cos6x = 2
-(cos4x + cos6x) = 0
cos4x + cos6x = 0
2cos(4x+6x)/2cos(4x-6x)/2 = 0
2cos5xcosx = 0

cos5x = 0
5x = 90+180k
x = 18+36k

cosx = 0
x = 90 +180k

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[wind wind]

what exactly does 'fully factorise' mean, and how does it work with fractional and negative indices?

[wind wind]

i have dy/dx = -2x^(-3/2) + 2/3 x^(-3) but not sure if i need to do anything else

[lime dust]

Fully depends on the context

Can be fully over integers, fully over reals, over complex…

If nothing is being said we usually consider over integers

[wind wind]

hmm okay

so leaving it like that should be fine?

[timid ridge]

well my teacher always said never leave negative exponents

[wind wind]

.close

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[limber dove]

can someone help me

[heavy frost]

what part

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[peak shore]

Hello, I need help with a question about vectors:
-20/3=-|w|cos135+2| r|cos297
-30=|w|sin135-2|r|sin297
How do I find the norm of the vectors w and r?
(-20/3,-30)

[peak shore]

the single bar is the norm

someone help?

[median vigil]

you have a linear system of equations with two unknowns and two variables

[peak shore]

ok and how do i solve it?

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[mental kindle]

Could someone help explain this? I’ve tried a couple different answers which were wrong and I’d like to know how to solve these types of problems?

[versed wave]

thats an ugly ass sine wave

[clever gale]

Points of inflection for f occur when f'' equals zero

[mental kindle]

So the only point of inflection would be where conceits changes on the x-axis?

concavity**

[clever gale]

Ohw ait

I misread

Mixed it up with stationary points

[clever gale]

You have the function of the derivative, f'

[mental kindle]

On a graph how do I tell/create the graph of f’ and f” without the equation?

[clever gale]

But you're given the derivative, so check when the derivative changes sign and equals zero

[mental kindle]

so in this case f”(c)=0 would be at the point (8,0)

[clever gale]

f''(8) = 0

Yeah

[mental kindle]

and would the graph be CCD on [8, inf]

wait sry

i meant

[clever gale]

I can't see the right side of the graph

[mental kindle]

CCD on from [2,8]

[clever gale]

It might cross the x-axis again

[clever gale]

Yeah

[mental kindle]

ohh ok thank you! I think i understand now 😄

[clever gale]

But that's a poorly drawn graph

doesn't even start at 2 if you look closely

[clever gale]

nw

[mental kindle]

The website we use for my calculus hm is awful 😭

and that graph looks horrible

apparently these are both still wrong 😭

[brave wave]

it's correct. is there a certain way you're supposed to format the answer?

try (8,0)

[mental kindle]

i’ve written it as (8,0) [8,0] and (2,8) and [2,8] and the website tells me “your answer should not be in the form of a list”

[safe radish]

@mental kindle Has your question been resolved?

[brave wave]

it looks like you have a limited amount of attempts, so if you have the ability to, just wait and ask your teacher for how to format the answer. Otherwise, I would try 0 in the first answer box. I think it's a problem with the engine, and not you.

[mental kindle]

Ok thank you bc i’ve tried multiple times for different problems and it likes to gaslight me like this 😅

[mental kindle]

Would someone explain how to do part b?

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[haughty sorrel]

Im unsure how i would go about solving this question.
Suppose R is a relation on set A where |A| = n
If R is an equivalence relation with n equivalence classes what can you say about R? what is |R|? explain

[haughty sorrel]

So an equivalence relation is reflexive, symmetric and transitive. An equivalence class is an element x in A where xRa. Im not sure where to go from here

the cardinality or size of set A is n. So wouldnt that mean if there are n equivalence classes it would be a partition and each equivalence class is size 1?

so |R| would also be n?

.close

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[hoary lava]

this is the question and the work, is it correct?

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@hoary lava Has your question been resolved?

[hoary lava]

<@&286206848099549185>

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[hoary lava]

<@&286206848099549185>

chat pls help it’s mathematical induction

[devout scroll]

what is the difficulty?

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[desert juniper]

you forgot the denominator in the 2nd to last step

other than that, seems fine

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[stable kestrel]

how to solve?

[versed wave]

usub

[stable kestrel]

what

[versed wave]

what

[stable kestrel]

?

[versed wave]

do you know what usub is?

[stable kestrel]

no

[versed wave]

https://www.youtube.com/watch?v=sdYdnpYn-1o

[stable kestrel]

thank you

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[lean otter]

Can someone please help with matrix algebra?

[lean otter]

can R2x2 > R be a linear transformation? for matrix algebra

[stoic dune]

There exists linear transformations between R2×2 -> R

[lean otter]

how would you be able to multiply the 2x2 matrix with a 1x4 in the end?

[stoic dune]

One example, let the input be:
[a b]
[c d]

And then the output would be c

[lean otter]

would u be able to help w this 1

mb for bothering :v

[stoic dune]

What's the question? Just whether or not it is linear?

[lean otter]

yea

transforming the standard gives 1, 0, 0, 1

[stoic dune]

Make sure you have the definition of a linear function in front of you

[lean otter]

but
(a b) (1)
(c d) x (0)
(0)
(1)
is not possible

like satisfying scalar multiplication and addition?