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[alpine sable]

.clpse

.close

[lone heart]

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[dire dew]

Hi, I was trying to solve for the average distance between two randomly chosen points on a circle and came up with the following two formulas, but I don't know which one is correct or why they are different. I am assuming the second one is correct but I'm not sure. Can someone give me some insight? https://i.redd.it/avqor86msmpb1.jpg

[dire dew]

<@&286206848099549185>

[lone heart]

@dire dew Has your question been resolved?

[tacit arch]

https://mindyourdecisions.com/blog/2018/10/18/whats-the-average-distance-of-two-points-on-a-circle/

they're different because you're "randomly choosing" differently

[dire dew]

Yeah, I saw that, but why is the first one incorrect?

[dire dew]

what would I need to alter in the cartesian solution to get the second solution

[tacit arch]

I explained that here

[tacit arch]

The distribution/density

[dire dew]

I understand that, but how would I go about calculating this with a cartesian coordinate system

In other words, how would I adjust the first integral or function, to give me the right answer

[dire dew]

and, this is helpful, it's the same as my second solution, but I want to understand how I need to adjust the first one in order for it to be correct

@tacit arch

[tacit arch]

You change the distribution/density

I don't know what you're looking for

It's like asking "how do I change a fair coin to make it turn up heads twice as often"

You change the coin to turn up heads twice as likely as tails

So change the integrand in the integral

[lone heart]

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[subtle sand]

anyone knows how to do the MLE part?

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@subtle sand Has your question been resolved?

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[tough dawn]

For this problem I need to estimate the intervals on which the function is increasing or decreasing. Would this one be (-infty, infty)?

[lunar ravine]

It's not (-infinity,infinity) no

[thick lynx]

What do you mean, (-infty, infty)?

[tough dawn]

Because the arrows mean it goes on forever in both directions

I thought that was the intervals

[lunar ravine]

a function is increasing when f(x) increases as x increases

[haughty pollen]

3x+3= 3(x+3)

x^2 -1 = ???

[lunar ravine]

and is decreasing when f(x) decreases as x increases

[lunar ravine]

#❓how-to-get-help

[lunar ravine]

or you could just see it on the graph yk if the curve is going up it's increasing and if it's going down then it's decreasing

now determine these intervals where the curve is going up and the other ones where the curve is going down

[tough dawn]

You mean like where the turning points are? Like 1 and 4

[lunar ravine]

these are the points where the curve goes from increasing to decreasing and from decreasing to increasing

[tough dawn]

Right

[lunar ravine]

you need the intervals before and after these points

[tough dawn]

How do I find those?

[lunar ravine]

like (-infinity,1) is the interval before 1

basically the interval where the curve exists like its domain at that specific part of the graph yk

[tough dawn]

Oh ok so it would be (-infinity,1) and (4,infinity)?

[lunar ravine]

these and there are other two intervals between the 1 and the 4

[tough dawn]

So that would be the correct answer right?

[lunar ravine]

you have to say if it's increasing or decreasing too at each interval of these

and you have to write all intervals

[tough dawn]

So like (-infinity,1) increasing and (4,infinity) decreasing?

[lunar ravine]

yes right but there are two missing intervals still

the part between 1 and 4

[tough dawn]

Oh what would that even be? Just 3?

[lunar ravine]

Not really see that part here what is the interval of that part?

[tough dawn]

(1,infinity)?

[lunar ravine]

1 yes not -infinity tho check the x value not the y one

[tough dawn]

Oh ok

[lunar ravine]

no not infinity either it stops at that red line

[tough dawn]

Oh so is it (1,3)?

[lunar ravine]

yes right

now the interval next to it can you figure that one out?

[tough dawn]

(3,4)

[lunar ravine]

yeah

[tough dawn]

Ok so are those the intervals in between?

[lunar ravine]

they are yes

you just need to determine whether the curve is increasing or decreasing at those intervals

[tough dawn]

(1,3) decreasing and (3,4) increasing

[lunar ravine]

yes

[tough dawn]

Ok I get it now. Thank you for helping me

.close

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[serene pilot]

hannah earned 149 prize tickets playing pinball, she spends 85 tickets on a bracelet and the rest on candy that costs 10 tickets each. at most how much candy can hannah get?

[serene pilot]

so far i have calculated 140=10x+85

then i subtracted 85 from both sides

and now i am at 55=10x

[reef crown]

yeah well done

now it's simple

if you divide by 10 you get 5.5 right?

[serene pilot]

yes

[median oar]

Why did you choose 140

[serene pilot]

oh i must have pressed the wrong key sorry

[reef crown]

doesn't make sense so you have to think about how much candy you can get until you get more than 5.5, which would get out of your range

[serene pilot]

so at most she would only be able to get 5

[reef crown]

yeah correct :)

if she got 6 she would pay more than she actually has

so she can get 5 at most as you deducted

well done

[serene pilot]

i just struggle really badly with math

[reef crown]

ahaha dont worry man you solved it in the end so that's good

:)

[serene pilot]

thanks

do you mind if i do more than one?

[reef crown]

yeah ok if they're simple

[serene pilot]

they ar just like these ones

[reef crown]

ok then

[serene pilot]

trisha is saving 25 a week to go on a vacation. she determines that she will need 350 but also wants to have over 100 left in her bank account after the trip how many weeks should trisha save up?

[reef crown]

which equation did you think of?

[serene pilot]

350=25x-100

• 100 to both sides

which i ended up with 450=25x

[reef crown]

yeah well done

you have it then

[serene pilot]

so i divide 450 by 25 and ill have the answer/

?

[reef crown]

yeah

[serene pilot]

i came up with 18

[reef crown]

yeah that's the result

and the equation is correct as well

[serene pilot]

ok thank you very much

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[valid void]

hey guys I need help with this for my homework.

[valid void]

I already drew the functions but im just a bit confused with the questions they are asking me

<@&286206848099549185>

[lone heart]

@valid void Has your question been resolved?

[valid void]

no

<@&286206848099549185>

<@&286206848099549185>

[valid void]

<@&286206848099549185>

<@&286206848099549185> PLEASEEEEE

[gloomy forum]

hey what is r(x) i cant see

is it 3/8 (x+5) -3 ?

[valid void]

it is

5/8

instead of 3/8

[gloomy forum]

okey i will help in 10 minutes

[valid void]

ok tha k you so much

[gloomy forum]

g is 1/3 (absolute(x-1)) right?

-5

because of 5/8 is greater than 1/3 rx will be always ahead of gx

[valid void]

yes it is

how do I write that using the interval notation line?

[gloomy forum]

i now realize that 115/23 is 5

[valid void]

ok thank you so much

[gloomy forum]

npnp

[valid void]

can u help with the second prt too?

@gloomy forum

<@&286206848099549185>

[gloomy forum]

so you need to find the values between blue lines to do that you need to equalize p(x)=r(x) which gives us x=3 and otherone is -5 as we found earlier so its (-5,3)

[valid void]

sorry im not at home right now so I cant show you how I would write my answer. How do you think I should write it?

[lone heart]

@valid void Has your question been resolved?

[lone heart]

@valid void Has your question been resolved?

[lone heart]

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[dim sundial]

im having trouble finding the vertex since
the distance between focus to directrix is only 5
units\

[dim sundial]

focus (-2,3) and directrix is x = 3

[gray isle]

why'd the distance being "5" an issue for you?

[dim sundial]

First time encountering given that is like that

uneven

[gray isle]

same principle applies

how would you do this if you were instead given
focus (-2,3) and x=4, (distance of 6)

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[cinder mist]

I am focusing on deritatives, I know that derivatives are related to the acceleration of original graph. So far I choose graph C. But I don't know which 6 answer choices it could be for the second one.

[dim sundial]

Find the midpoint?

Oops my channel timed out uhh

[gray isle]

make another one

[cinder mist]

lol

[lone heart]

@cinder mist Has your question been resolved?

[cinder mist]

There must be a lot of people who need help

[lone heart]

@cinder mist Has your question been resolved?

[lone heart]

@cinder mist Has your question been resolved?

[lone heart]

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[narrow flint]

Given the boundaries of a 2D surface area, an initial position and a brush diameter, are there algorithms that would help calculate or approximate what is the shortest trajectory that would paint all of the surface ? Is it also possible to determine the shortest distance without generating the trajectory ?
The problem can also be viewed as a Roomba robot that has to clean a room by travelling the shortest path.

[remote heron]

cool question. i hope someone else knows because im not sure lol

https://www.frontiersin.org/articles/10.3389/frobt.2021.624333/full

interesting heres discussion about the problem from a robotics pov

[fallen verge]

Are we assuming a flat area? I feel like curved surfaces might make this crazier

[lone heart]

@narrow flint Has your question been resolved?

[remote heron]

man what a bummer they ghosted

[narrow flint]

Yes a flat area for now 😆

[narrow flint]

What terms did you search to find that article ?

[fallen verge]

Ok so as i understand it for now, lets say you have a grid of pixels and you want to create a path between said pixels to minimize the distance traveled

Ive simplified the problem a lot, but its better to start somewhere

[remote heron]

coverage path planning

[fallen verge]

Unfortunately, with the way ive phrased it, this seems akin to the traveling salesman problem

[narrow flint]

Yes, but unlike a travelling salesman problem, moving to a pixel has impact on its surounding pixels.

[remote heron]

i have a friend who does robotics, this is what he asid its called

it looks complicated

[fallen verge]

This does seem complicated

[narrow flint]

I suppose you could imagine it as a travelling salesman who's marketing has influence and spreads around him, so if he were to visit New York then he doesn't need to go to Philadelphia since people have heard about him*

[fallen verge]

Give me some time to think, ill come back soon with whatever findings i have

[remote heron]

i do wonder if its solved in a really constricted case

[narrow flint]

One non-optimal solution that I thought about so far would be to move along the boundaries of the shape so that the circle always touch the edge

Then repeat using the newly formed boundary until it completely collapses

You would end up with a set of layers that you can then follow one by one.
Not optimal, but that would provide one possible path. Perhaps it's easier to optimise an existing path rather than optimising while creating. I dunno

[fallen verge]

I actually think the whole locality thing makes this harder to solve

Obviously the most optimal solution is to find a path with 0 overlap but also you want to make sure the path does not make more than 90 degree turns or the overlap of the brush is suboptimal

That means I have solutions for rectangles and circles

But not much else

Oh any polygon with interior angles more than 90 degrees it seems

[narrow flint]

What would it look like ?

[fallen verge]

Not convinced on that one

For circles you start at the center and make concentric circles

For rectangles, you start at a corner and follow a wall

Ok wait is the brush round or square?

Bc this will leave some gaps in corners

[narrow flint]

Yes it's round.
I feel like the corners of a rectangle would be painted using diagonals lines or something. I don't think I'm not sure if the answer is probably straightforward even for a rectangle since the start position could be anywhere inside, but it could be useful to study the most "basic" case first(unless it's one of these problems where it's the opposite..)

[lone heart]

@narrow flint Has your question been resolved?

[lone heart]

@narrow flint Has your question been resolved?

[lone heart]

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[tough nacelle]

translation

A.
Write an expression that calculates the area of ​​this trapezoid.

B.
Calculate the area of ​​the trapezoid,
when a = 4 cm and b = 3 cm.

[lone heart]

Channel closed

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[tough nacelle]

.close

[modern imp]

where did 'a' and 'b' go?

[tough nacelle]

.close

[tough nacelle]

well, first you have to figure the areal out first

[modern imp]

Wait 'c' here is area right

[tough nacelle]

the b?

yea i thought of it too

[modern imp]

i mean the variable c

this

[lone heart]

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[bitter cloak]

what are the 2 values of x :( i'm lost with the exponents+ factorization part

[white crater]

try letting u = x^3

[bitter cloak]

i tried looking up how to factorise it but it threw me into this

[bitter cloak]

so like

substituting the original x value?

[white crater]

nvm i dont think my idea works

[bitter cloak]

where does the 7 go? and why are there multiple x to the power of 5, 4, 3 etc

i'm soooo lost

lol

[bitter cloak]

<@&286206848099549185> sorry for ping but i desperately need help T-T i'm seeing something about a factor theorem that might be of use? i have 0 clue honestly

[stark path]

Are u familliar with reminder theorem?

[bitter cloak]

not really

mind explaining?

[stark path]

Sure

[bitter cloak]

thanks!

[timid temple]

Solve $u^2+7u-8=0$

[ocean seal]

KTMath

[bitter cloak]

i tried this just now

but it confused me later on

i tried using it in the pq formula but idk there were a lot of decimals/fractions

[stark path]

So like
In this theorem

Considering a function f(x) = something
U have to try and find for indivitual values for x that sum up to 0 in the function

Lemem give u an example
If f(x) = x^2 -2 x + 1
If u do f(1) = 0
So hence u can say x-1, is a factor for the function f

[timid temple]

Tell me two u values

[stark path]

This is actially the brief of the solution ur web has provided to u

[bitter cloak]

ohhhhhhhhhhhh

wait that makes sense

[stark path]

If u have any other ideas on factoring
U can choose that as well
It's not like u have to stick to one single formula

[bitter cloak]

thanks i'll try to do this now to see if it makes more sense

[stark path]

Rog that

[timid temple]

Sorry to interop find u first and consider cubic root of u

[bitter cloak]

yeah it's just we haven't been learning this too much using cubic roots so im not sure if that's what my teacher wants us to work with

we have mainly been using factorization

[bitter cloak]

but u is a substitution for x^3 right?

[stark path]

Yep

U can chooss that path as well

[bitter cloak]

is it more efficient?

[stark path]

That depends on whther u r assigned to factor using indivitual rules or not

[bitter cloak]

i mean both methods have been shown to us

[stark path]

Then it's ur wish to proceed

[bitter cloak]

i just havent been working with cubic roots a lot / the substitution method but my teacher has said it's fine

[stark path]

It's ok

[bitter cloak]

how would i go about if i did the substitution method? just normal pq formula?

but then once i have value of u i take the cubic root out of it?

[stark path]

Yea u can take it out once u r done factorizing the substitioned equation

[stark path]

This equation can easily factorized using middle term break rule actually

[bitter cloak]

is that like

(u-1)(u+8)=0

[stark path]

Yes

[bitter cloak]

ohhh and then the cubic root of -8 is -2?

WAIT

THAT MAKES A WHOLE LOT OF SENSE

[stark path]

Yea lol

[bitter cloak]

omg i feel so stupid doing maths sometimes 😭

[stark path]

Actually cubic root of 1 and -8 :V

[bitter cloak]

thank you both so much!!!

[bitter cloak]

OH RIGHT

but

cubic root of 1

[stark path]

No mention bud

[bitter cloak]

is still 1

no

[stark path]

Yea

[bitter cloak]

yea

LOL

okay thank you so much

srsly

saved my grade

i get it now

[stark path]

Lmao no mention

[bitter cloak]

have a nice dayyy!!

.close

[lone heart]

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[lone heart]

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[snow spade]

I have a question

[snow spade]

Whenever you're using an angle to the vector do you have to always use force or can you use like the soh cah toa as substitute?

Or that's on a different equation?

[alpine sable]

may you give me a example?

[snow spade]

Wait lemme just rephrase that actually

it might be confusing

So whenever you have force as given in your freebody diagram

to find the tension, moment or etc

Can you use soh cah toa to pair with force instead of cos and sin?

[alpine sable]

give me a couple minutes to think about this. however you have a well-grasped understanding of your subject.

[snow spade]

Like this

Can I like do 3/5.8 (200) for the x

[alpine sable]

assume the following. imagine your self on a hill, you’re going done the hill. would you feel going down or up? Within that case, in which direction you feel. you may use the trig functions. you’re correct.

[snow spade]

Thanks a lot!

[alpine sable]

sorry if it seems obvious or trivial, what other problems you might need help with

[snow spade]

That's it I just probably got confused.

Thanks a bunch!

.close

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[maiden lotus]

hi all, I tried various ways to solve this equation but I always got stuck at 4^x(2-sqrt(3))^x = 1 but I don't think it's correct.
Equation in question (2+sqrt(3))^x + (2-sqrt(3))^x=4^x

[opal jolt]

$(2+sqrt(3))^x + (2-sqrt(3))^x=4^x$

[ocean seal]

LordFelix

[opal jolt]

the solution is very obvious by trial and error, but im having trouble getting it analitically

are you expected to solve it analitically or numerically?

[maiden lotus]

analytically

I don't want a numerical solution

like I get it if it's 1

but why i

[opal jolt]

it is 1, im looking into how to properly compute it

obviously 2+sqrt3 +2-sqrt3=4

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@maiden lotus Has your question been resolved?

[lone heart]

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[minor nebula]

Here:2x=18+2

[winter light]

I don't get your point, only 2 is subtracted

[gray isle]

(also though gpt is correct in this specific question, it is mostly unreliable and shouldn't be used for math)

[winter light]

If you have a + before a parentheses you can delete the parentheses and add what you had inside. On the contrary, if you have a - before parentheses, you delete the brackets and change all signs of what you had inside brackets

[gray isle]

the presence of () here don't really do anything here

if you want to bring stuff like associative property into this

[winter light]

Also, remember that in math:
$$\text{stuff} = +\text{stuff}$$

[ocean seal]

Alberto Z.

[gray isle]

$\begin{aligned} x + (x - 2) &= x + (x + (-2)) \
&= (x+x) + (-2) \
&= (2x) + (-2) \
&= 2x - 2 \end{aligned}$

[ocean seal]

ℝam()n()v

[gray isle]

or just drop the () since they were unnecessary and
x + x - 2
naturally simplifies to 2x - 2

[minor nebula]

My point is why x is outside of parenthesis when its supposed to be substracting with 2

[minor nebula]

When i wrote that equation i had meaning in my mind

[winter light]

Subtraction occurs only between 2 things

In this case the x (the one in the brackets) and the 2

[minor nebula]

Question: sum of two consecutive values which are even is 18 then find find those numbers

[gray isle]

its unclear what you issue still is

[minor nebula]

Why is x outside of parenthesis

[gray isle]

that's how you wrote it

[minor nebula]

In 2nd step

[winter light]

Why should it be a problem?

[gray isle]

because the () pretty much do nothing

[minor nebula]

Why and how

[gray isle]

if you want to bring stuff like associative property into this
look at the above image

x + (x - 2)
is essentially the sum of three values:
x,x and -2
and from the associative and commutative properties of addition you can add those in any order you want

[minor nebula]

Oh ok

Wait what if we there is division in parenthesis

@gray isle

[gray isle]

write what you mean

[minor nebula]

2+(14÷7)

16÷7 is not equal to 2+2

[gray isle]

well that would violate the order of operations

as you have division between 14 and 7
that division takes priority over addition

again those () were uselss here, division takes priority regardless

[minor nebula]

How do u solve x+(x÷2)

[gray isle]

because there associative property of addition is applicable

here it isn't

[minor nebula]

It isnt applicable when there is division but it is with multiplication and substraction?

[gray isle]

$2 + (14 \underbrace{\divisionsymbol}_{\textbf{NOT} +} 7)$

[ocean seal]

ℝam()n()v

[gray isle]

do you know what the associative property for addition states?

[winter light]

Read about PEMDAS or BODMAS (for example in Wikipedia, if you wish)

[minor nebula]

Ok

[gray isle]

$$a + (b+c) = (a+b) + c$$
that doesn't apply here

[ocean seal]

ℝam()n()v

[minor nebula]

How do u solve x+(x÷2)

[stark cosmos]

x/2

first

[gray isle]

you could combine into a single fraction consider common denominators

[minor nebula]

x+(x/2)=15

[winter light]

This @minor nebula

[minor nebula]

Oh ok

Where could i get all important and even micro rules all together

Anyway thx everyone

[lone heart]

@minor nebula Has your question been resolved?

[lone heart]

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[chilly idol]

Need some help on 40 and 41

For 40 I'm assuming that since their is a absolute value applied for x that all points below the x axis, will have their x coordinate taken in the absolute value

I'm guessing for 41. I do the same thing I did with 40 but I take anypoint that is on the left side of the y axis and do the absolute value of its y coordinate.

I just want to know if im on the right path

The answer are kind of weird though

[stark cosmos]

f(abs(x)) = {f(x) if x=>0, f(-x) if x<0

what happens if you plug in x = -2 in this function?

[chilly idol]

You would get f(2)

[stark cosmos]

right, so now it means that for all negative x:s then the function should look something like the positive side

right?

[chilly idol]

Yeah

[stark cosmos]

okay now just do some values for example

x = -4, x = -6, x = -8

what would you get?

[chilly idol]

Well I would get x=4, x=6, x=8

[stark cosmos]

plug those values in your f(x)

and draw dots on your graph

[chilly idol]

When we plug it into f(abs(x))

[stark cosmos]

does it look familiar ?

[chilly idol]

hold on

let me just make a rough sketch

@stark cosmos So everything is getting smushed into the right side of the x axis

Because we are doing absolute value to all of the negative x values

[stark cosmos]

i am not sure what you mean

but if you make dots

and draw them together

you will see it is a reflection of the positive x axis

[chilly idol]

Like if we took (-9,-3) and did abs value of the x value, we would get (9,-3)

[stark cosmos]

almost my friend

so

let us look at the negative x - values

[chilly idol]

ok

[stark cosmos]

f(abs(-4)) = f(4) = -2

so a point would be

(-4, -2)

[chilly idol]

(4,-2)

[stark cosmos]

x will still be the same

[chilly idol]

oh ok

[stark cosmos]

only the function will do something

it will do... f(abs(x)) = y

now do this for the points that you see in the positive x axis

(x =4, 6, 8)

but do them for the negative values

[chilly idol]

ok

[stark cosmos]

x = (-4, -6, -8)

then you make dots on the graph

and you should see a pattern

[chilly idol]

Ok, I'll try right now

I also do this for x=-9, -3, -6

[stark cosmos]

you don't need to

[chilly idol]

ok

[stark cosmos]

you will see the pattern quite fast

have you done it?

[chilly idol]

yeah Ill send a picture

,rotate

[chilly idol]

Blue is orginial graph

[stark cosmos]

yeah i see

now can you see that the purple looks awful a lot like the right side?

[chilly idol]

I see a reflection of the y axis

[stark cosmos]

kind of like it is mirrored or something

[chilly idol]

yeah

[stark cosmos]

there is your answer

now for abs(f(x))

[chilly idol]

yeah

[stark cosmos]

abs(f(x)) = {f(x) if f(x) >= 0, -f(x) x<0

so all values of f(x) must be positive

[chilly idol]

ok

[stark cosmos]

just do all the f(x) positive then

and you are done

so for example:

[chilly idol]

So all y values must be positive

[stark cosmos]

abs(f(-3)) ) = abs(-9) = 9

yes

[chilly idol]

ok

[maiden lotus]

I proved it

[chilly idol]

,rotate

[chilly idol]

@stark cosmos I got it

Is it correct?

[stark cosmos]

nope

[chilly idol]

hmm ok

[stark cosmos]

abs(f(-3)) = abs(4) = 4

you don't need to do what you did before

just plug in all the x values

and look what you get

[chilly idol]

oh ok

[stark cosmos]

the trick is that when ever f(x) < 0

meaning when f(x) is negative

it has to be positive

[chilly idol]

ok

[chilly idol]

i think i get it from this example

[stark cosmos]

and ofc abs(f(-9)) = abs(-3) = 3

[chilly idol]

I did the graph, sending picture right now

,rotate

[stark cosmos]

you are right

[chilly idol]

Blue line is the new graph

[stark cosmos]

but there is one place

where it is wrong

[chilly idol]

hmm ok

[stark cosmos]

this place is already positive

the y axis is positive ther

e

so you do not have to change anything there

[chilly idol]

ok

wait so I leave it as (4,-2)

[stark cosmos]

oh i thought that it was the old one

the red line

[chilly idol]

oh ok

yeah the red line is the old one

blue is the new one

[stark cosmos]

alright then it is okay there

[chilly idol]

ok

Thank you for your help! @stark cosmos

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[clear comet]

I got -2*sqrt(a) and wolframalpha too but answer key says:

is the answer key wrong or wolframalpha ?

[nimble fern]

it seems like -2√a to me

[sharp thorn]

answer key is incorrect

[clear comet]

okay, Thanks :)

!close

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[opal jolt]

i'm... not seeing it proven there

[maiden lotus]

Well no mb, not proven solve for x I mean

Check bottom left

[opal jolt]

yeah, but i'm not seeing that either

[lone heart]

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[deep hornet]

Question about the fixed point theorem in numerics, p. 796 i Kreyszig 10th edition.

Am tasked in an exercise sheet to prove that a given g(x) for a given equation has a unique solution for x >= 0.
I've found a fixed point by doing the x_n+1 = g(x_n) iteration steps. additional iterations with this x_n i found gives the same value.
Analyzing the derivative of g i've found that it is strictly < 1 (it converges towards 0) for any x >= 0.

Am i missing anything at this stage? Do i just insert the x_n i found into the original equation and see that it holds?

[deep hornet]

I guess my real question is just at what point have i "shown that is has a unique solution for x>=0", what criteria are there. I've read the relevant pages in the textbook and looked at the examples. Our lectures aren't recorded and the notes aren't very helpful

[proper leaf]

Youve covered the nessacery steps to show g(x) has a unique solution for x greater than or equal to 0 by finding a fixed point through iteration and confirming that the derivative of g(x) is strictly less than 1 for all x greater than or equal to 0 you have addressed important aspects of proving uniqueness

[deep hornet]

hm okay, idk it just feels like there is supposed to be more but I couldn't see what else there was to do. If those are the only requirements I'll move on to the next one, Thanks!

[proper leaf]

Welcome!

[deep hornet]

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[surreal meteor]

Since f''(x) is not > 0 for all x in the interval I

does this mean we have that f(x) does have an inflection point at the point A on the graph

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@surreal meteor Has your question been resolved?

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[atomic bobcat]

I feel like Im missing something with the acceleration. In order to do part b I need to find the tension first.

I get ma = ft, but I have a = 0 because the velocity of the cart is constant so I get Ft = 0 which doesnt seem right

[wary stream]

The velocity of the cart is not constant, it's linear

Constant would be a horizontal line

[atomic bobcat]

True, you're right

Its early, I missed something stupid like that lol

I appreciate you pointing that out for me

[wary stream]

Happens to the best of us

[atomic bobcat]

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[fickle mesa]

how do I calculate this bearing?

[wanton umbra]

define bearing

[fickle mesa]

already figured it out, ty tho :D

.close

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[fickle mesa]

These 3 are all very confusing to me

[gray isle]

what about bearings is confusing you?

[fickle mesa]

I'm just not sure how to calculate them

[gray isle]

are you able to identify the angle they want you to find?

[fickle mesa]

U from T?

[gray isle]

or the definition of bearing

[gray isle]

like visually, on the diagram mark the angle they want

[fickle mesa]

48

[gray isle]

no

that 48° is given,

not the angle they ultimately want

the (true) bearing of A to B, or B from A
is the angle measured clockwise from the North axis from A to B

can you apply this and mark like the red arrow i used
(only mark, don't calculate yet)
the required angles for your questions

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@fickle mesa Has your question been resolved?

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[lone heart]

Please don't occupy multiple help channels.

[charred summit]

Um what is the question

[alpine sable]

I dont know what the photo means 🫨

[charred summit]

What do you mean by don't know what the photo mean

[alpine sable]

It's just pythagorus thereum

$$a^2+b^2 = c^2$$

[ocean seal]

What should I do

[alpine sable]

in this case

each side is either $a$, $b$, or $c$

[ocean seal]

What should I do

[charred summit]

Is it used to get the value of something like
$$\sin^{-1}(tan(\frac1{\sqrt{n}}))$$

[ocean seal]

Sherif Player

[charred summit]

@alpine sable
Where did you find the photo?

[lone heart]

@alpine sable Has your question been resolved?

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[barren portal]

If Sup A $<$ Sup B, Prove that, $\exists$ b $\in$ B that is an upper bound for A

[ocean seal]

.doc

[wanton umbra]

supB is one

[marsh rapids]

Not in B

But the definition of sup suffices to show existence

For a good choice of epsilon

[wanton umbra]

oh yes, sorry

supB \neq maxB

[barren portal]

I feel like b= SupA+ (k), where k<(supB-SupA) will work

that’s just the geometry i saw

[barren portal]

Does this make any sense?

[wanton umbra]

depends what A and B are, ig

[barren portal]

b must be such that supA<=b<=supB right?

[wanton umbra]

like if A {1,2,3} B {2,4} then there is no such b

[barren portal]

max B is an upperbound for A right?

[marsh rapids]

Max B may not always exist
But when it does, it equals sup B

[marsh rapids]

Then b = sup B works

[barren portal]

so I can guarantee ‘b’in B such that b is upper bound for A?

[wanton umbra]

lol me messing up upper bound and least upper bound ie sup

[marsh rapids]

That's literally the question

[barren portal]

I mean, i’m looking for a hint actually

[marsh rapids]

Yes
Just rewrite this in such a way that it clearly follows from the definition of sup B

[marsh rapids]

(Also the inequality must be large)

[barren portal]

@marsh rapids

Whcih definition of sup should I use?

[marsh rapids]

Whichever one you were given

[barren portal]

I have two, first one having 2 conditions, 1) upper bound 2) uniqueness

[marsh rapids]

Or an equivalent characterization, depending on how much theory you did

[marsh rapids]

Well the "least upper bound" being the worst one for this

[barren portal]

second def, for any e>0, s is sup of A iff there exist m in A such that s-e < m

[tired bone]

yeah i need help with 374 x 85

[wanton umbra]

completely roasted

[marsh rapids]

#❓how-to-get-help

[marsh rapids]

any e>0 is misplaced but that's the one I had in mind, so go ahead

Also that's not 2 definitions
That's 1 definition and 1 property

[barren portal]

This is not correct right?

[marsh rapids]

No

[barren portal]

I don’t think b<= sup B can imply b in B

[marsh rapids]

You're implying [sup A, sup B] is in B

[barren portal]

The desired b will be such that supA <=b <= SupB right?

[marsh rapids]

Yes

[barren portal]

Can I say choose b such that this?

and show b holds the conditions ie it’s upper bound for A and in B

[marsh rapids]

You don't choose b
You show such a b exists

[barren portal]

I think i might need a little more hint with this

sorry

[wanton umbra]

just look at def of supB and that supA<supB

[barren portal]

Definition of Sup B tells, $b \leq$ supB $\forall$ b $\in$ B

[ocean seal]

.doc

[barren portal]

also, if b’ is any other upper bound, then SupB $\leq$ b’

[ocean seal]

.doc

[barren portal]

Sup A less than sup B will mean SupB is an upper bound for A right?

[wanton umbra]

but supB is not always in B

[barren portal]

yes

[wanton umbra]

and I meant the property of sup

[barren portal]

which the episilion one?

[wanton umbra]

yea

[barren portal]

for any e>0, SupB -e cannot be an ub for B

ie there is some b in B such that supB -e $\leq$ b

[wanton umbra]

\geq

[ocean seal]

.doc

[barren portal]

my bad, b has to be right of supB -e

[barren portal]

But this wouldn’t restrict going below supA

or what epislion should I take ?

[wanton umbra]

small 🙂

[barren portal]

🙂

[barren portal]

I mean, I need this to be between Sup A and Sup B right?

[wanton umbra]

something like e<supB-supA

[barren portal]

yes

so how do i proceed?

[barren portal]

I’ll just say choose epsilon such that this?

[marsh rapids]

You can set it too

Rather than argue it must exist
Always better when the definition is simple

[barren portal]

now that i’m guranteed for any epislion

[barren portal]

I’m trying

I’m very noobie in analysis

[marsh rapids]

That's expected of people learning about sup I think

[marsh rapids]

(sup B - sup A) / 2 is very simple

[barren portal]

makes sense, is that how we actually prove?

or you are mentioning alternate proof

?

I mean, is it always about choosing epislion?

[marsh rapids]

Still doing the same proof

[marsh rapids]

Now we have an epsilon, which lets us use this

[barren portal]

sure

[wanton umbra]

also property is strictly less

[marsh rapids]

You'll see that sometimes, finding the right epsilon (or rather, the right delta or the right N) can indeed be difficult. Especially when it's not the only challenge

[marsh rapids]

Doesn't matter

[barren portal]

$supB - \frac{supB- SupA}{2} \leq b$

[ocean seal]

.doc

[barren portal]

does this makes sense?

[marsh rapids]

Well rewrite it in a more useful way for our intent

[barren portal]

Sure

hold on $\sup{A}$

[ocean seal]

.doc

[barren portal]

Great

[marsh rapids]

Didn't need to latex that one

[barren portal]

$\frac{Sup B - Sup A}{2} \leq b$

So I need to show this quantity is greater than sup A right?

[marsh rapids]

May I point out the suspicious panda face

[wanton umbra]

oh yeah, you can make e smaller

[barren portal]

I see

[marsh rapids]

This remark basically holds for any epsilon inequality in analysis

[barren portal]

$\frac{Sup B + Sup A}{2} \leq b$

[ocean seal]

.doc

[marsh rapids]

Better

If you struggle to finish
||Remember what you need to prove, visualize what you have, deduce how to get there||

[wanton umbra]

so close

[barren portal]

Sup A < Sup B

ie supA + Sup A < Sup B + Sup A

2SupA $\leq$ SupA + Sup B

[ocean seal]

.doc

[barren portal]

SupA $\leq$$\frac{ SupA + Sup B}{2}$

[ocean seal]

.doc

[barren portal]

by definition of sup, quantity in right side is a upper bound for A

ie SupA $\leq$$\frac{ SupA + Sup B}{2}$ $\leq$ SupB

[ocean seal]

.doc

[barren portal]

@marsh rapids is it correct?

[marsh rapids]

Nothing wrong here

[barren portal]

@wanton umbra

[wanton umbra]

/spoiler but b...

[marsh rapids]

Even closer now

Just took a step in the wrong direction

[barren portal]

oops

[marsh rapids]

When your conclusion is the hypothesis, you know that's not where you should have gone

[barren portal]

wait

ie SupA $\leq$$\frac{ SupA + Sup B}{2}$ $\leq$ b

[ocean seal]

.doc

[marsh rapids]

Cmon now
Everytime I think you're 1 step away, you manage to do half of the step
Xeno issues

This should end nonetheless

[wanton umbra]

we are approaching!

[marsh rapids]

There's only so many times he can halve it before reaching the elementary step

[marsh rapids]

No offense btw

[barren portal]

my bad

[marsh rapids]

I'm just doing some geometric series jokes

Showcasing my incredible sense of humor

[marsh rapids]

Stuck ?

[wanton umbra]

be more strict

[barren portal]

I just want to make sure I know where I should get this to

[marsh rapids]

Trans

[barren portal]

I think i don’t know how far i’m or close

[marsh rapids]

It's the third time we say this

[barren portal]

I didn’t see it myself unluckily

I see that the there some b in B, such that b is less than the choice of epsiliom we had

[marsh rapids]

itivity

[barren portal]

and that choice is bigger than Sup A

and less than or equals b

[barren portal]

This would mean

That choice is a upper bound for A

and is in B

[marsh rapids]

Yes

[barren portal]

by def existence in B was already guaranteed

now being bigger than supA means it’s upper bound for A.

[wanton umbra]

/spoiler elementary step shall be repeated up to inf

[marsh rapids]

No

[barren portal]

This is correct right?

[marsh rapids]

I'm waiting for you to realize you're done

[barren portal]

transitivity holds?

ie sup A less than choice and choice less than b

this means sup A less than b

[barren portal]

This means choice less than b

[marsh rapids]

Q.E.D

[barren portal]

This sup A less than choice

[wanton umbra]

man, you are DoNE

[barren portal]

Thus transitivity of order

[marsh rapids]

He's gone

[barren portal]

Dude, I didn’t realised

[wanton umbra]

inf loop

[marsh rapids]

Sorry, you were correct

[barren portal]

I don’t how to thank 😭

[marsh rapids]

For taking away 90 minutes of your life you'll never get back ? Don't
Happy existential crisis

[wanton umbra]

there is enough humour in here to prolong my life by more than 90 mins

[barren portal]

sorry for being a clown

[marsh rapids]

Alright let's recap (I'll introduce a slight optimization)

Let e = sup B - sup A > 0
By characterization of sup B, there exists b in B such that sup B - e <= b
hence b >= sup b - e = sup A
By definition of sup A, that means b is an upper bound of A
Q.E.D

[marsh rapids]

And I could have read some topology instead, but decided RA is good

And had a good time

[barren portal]

Nice work

Thank you

I don't know what to say. May both of your goals in life come true.

.close

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[marsh rapids]

.bold of you to think I know what that is

[wanton umbra]

oh no, 0 goals for me

[wanton umbra]

.close

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[marsh rapids]

A message starting with . doesn't open the channel

[mystic vortex]

.oh that's cool

[marsh rapids]

Doesn't matter when it's closed though

[mystic vortex]

.yeah but what if it went into the open channels right as I was typing

[marsh rapids]

Gotta be on the lookout

[mystic vortex]

.yeah fair enough

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[polar plaza]

I got part a but I need help with parts b and c

[unkempt vapor]

i guess you know what is mean?

you just have to explain what mean gives here

for c, ig you have to plot a graph

[polar plaza]

Hmmmm but it says to explain though

[unkempt vapor]

explain how the graph goes

[polar plaza]

Hmmm okay

Thanks

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[long panther]

Why does the solution use pi/3

and not the root of unity, which is 2pi/n and n is 3 so 2pi/3

In every other example I've used e^2pi/n and the solutions have also followed this

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[rustic coral]

Send the original question

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[long panther]

Sorry yeah I should've

[near apex]

!status

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[sacred tundra]

Hi

Could anyone help me out on some probability trees

[near apex]

!occupied

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[alpine sable]

hello i need some help with math for my game.
lets say i have 2 cameras of which the views are split evenly on a plane vertically.
i also have one fake camera, which defines the position, orientation and fov.
i need to change the position, orientation and fov of the 2 real cameras to make it look as if it was one singular view.

[alpine sable]

im going mentally insane

!status

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[alpine sable]

1

[wanton umbra]

draw a diagram

[alpine sable]

i haven't a clue where to start

im trying to, but i just cant visualize it

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[lyric light]

hi plz i neeed help

[lyric light]

.close

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[native bramble]

When writing a cosine function, is the periodicity found with 2 x pi / 2 or 3 x pi / 2

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[jagged cobalt]

what do you mean by that? The standard period of a cosine function is 2pi
if it were something like cos(1/5 x) then the period becomes 10pi
if it were cos(5x) it becomes 2pi/5

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[terse lichen]

for (ii), what if a | y and not c? is that solution right

[raw fractal]

a | (bx + cy) for every value of x and y. so you can choose any value you want for x and y to make it so that you get a | c

[terse lichen]

what i mean is: how does a | (bx + cy) mean that a | c. I understand how it does the other way around, but how do we know a | c instead of something like a| y

[raw fractal]

we are given that a | (bx + cy) no matter which values of x and y we have and we want to show that a | c. so we are allowed to choose x and y. we want it so that bx + cy = c. then you can immediately conclude that a | c. you should be able to see what to choose for x and y here.

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[valid halo]

"Can you prove the 4th Pythagorean identity shown from 4:35-5:05 in the video?"

[valid halo]

is this just a^2 + b^2 = c^2

and then sub in the identity

[blissful whale]

Simplify both right hand side and left hand side by using sin and cos only.

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[granite rapids]

i need my answers verified

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Please don't occupy multiple help channels.

[granite rapids]

Hi

Just want someone to verify if these answers are correct.

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[twilit gull]

Is the pulley 2 moving? I cant imagine the motion here.

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[night mountain]

Not sure where I'm going wrong here. I wind up with the equation m1gsin26+0.446m1gcos26+m2gsin26+0.446m2gcos26-m3g

which I think accurately depicts what is going on in the image.

assuming left, down the ramp as being positive.

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.close

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[stable warren]

f

[stable warren]

Just a minor quetsion

it can also be simplified as..

$sqrt32$

whaaa

i am saying

[vale wigeon]

can you state your question again? maybe write it out on paper...?

[stable warren]

sqrt of 32 can be simplified as 2sqrt8?

[stable warren]

ok

[vale wigeon]

sure, these two are equivalent.

[stable warren]

rly?

if i solve like any of them both are correc?

[upbeat gorge]

“Simplify” usually means the number remaining inside the square root has no perfect square factors other than 1

[vale wigeon]

i would probably not consider 2 sqrt(8) fully simplified

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[stable warren]

how?

[vale wigeon]

because it can be simplified further into 4 sqrt(2) ...?

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[subtle sun]

Hello, I don't understand how I should solve (a). Does it mean coordinates by domain? I'm just confused about the question in general, and don't know how I should get started on this problem.

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[unkempt robin]

Hm, did they forget - at infinity in D1?

So that it's D1 = (-∞, a]

Let me ask you this

Does the function as a whole have an inverse?

[subtle sun]

I'm calculating the inverse function but it's proving hard for me to do so. I replaced x with y (switched them) and I am stuck at y^2(y-2) = x + 2

[unkempt robin]

You didn't answer my question

[subtle sun]

I'm guessing yes, the next question is telling me to answer it so..

[unkempt robin]

Hm.

Do you know when a function is invertible?

[subtle sun]

No, honestly no

[unkempt robin]

Have you heard of bijection?

[subtle sun]

I don't think we've done that yet so no

[unkempt robin]

Hm, if you haven't been taught what makes and doesn't make a function invertible you shouldn't have been given this assignment

[subtle sun]

well the teacher told me to "think" lol but I guess I'll try n look up bijection

[unkempt robin]

Hm

A function is bijective (and consequently has an inverse) when it is both injective and surjective

It is surjective if every element in the set it is mapping to is represented by at least one x

[subtle sun]

Ah, I see I was able to understand.

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[unkempt robin]

And it is injective if no two different x's map into the same y

Can you reopen?

[subtle sun]

kk

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[subtle sun]

I was able to understand the concept, ty. But I am still confused about the question.

[unkempt robin]

Yeah I am getting there

[subtle sun]

Sorry sorry, but yes thank you for the clarification

[unkempt robin]

Graphically, injection means, if you can draw a parallel line to the x -axis and it intersects the graph more than once, it is not injective (0 and 1 is okay, >1 is not okay) and consequently not bijective and doesn't have an inverse

Do any parallel lines to the x -axis intersect the graph more than once here?

[subtle sun]

Yes it does. I'm guessing it does not have an inverse function in this case.

[unkempt robin]

That's correct. Because in this case, if you draw a line y=1, all x=-2/3, x=1 and x=1/2 (approximate) go into y=1

This is okay in this case, because different x's can go into the same y in functions

But in its inverse you'd have the same x going into different y's

Which is impossible

[subtle sun]

Ah I see, that explains a lot ty

[unkempt robin]

So, let's get back to the assignment

You have figured out the function as a whole does not have an inverse

However, a) is asking you to split the graph into 3 "sub-graphs" that will allow each to be invertible

Do you know which a and b to pick now?

[subtle sun]

hm..

[unkempt robin]

Okay, let me ask you this

As x approaches 0 from -∞, is there a problem with injection?

[subtle sun]

I don't think so. The line does not intersect the graph twice

[unkempt robin]

Correct

Another way to say this is the function is strictly growing

If a function is either strictly increasing or strictly decreasing it is always injective

But then what happens when x crosses 0 with the graph

Is it still increasing?

[subtle sun]

no it isn't

But yes? after a certain point?

[unkempt robin]

And this let's you make a parallel line to the x - axis which let's you know it isn't injective

[unkempt robin]

Yeah we'll get there. That's why you were given 3 intervals

[subtle sun]

Ah, I see it now