help-23

[stoic dune]

A ∩ B is the set of things that are in both A and B. Can think of ∩ as "and".

[soft fulcrum]

Yes

[dreamy marten]

oki

[stoic dune]

Do you understand what elements are in A = (-3,6]?

[dreamy marten]

https://tenor.com/view/i-just-want-answers-cardi-b-elle-i-just-want-to-know-the-answer-can-i-have-the-answer-gif-18916227

[stoic dune]

Lololololol

Funny gif. That won't happen here though

@NickNamerz#1446

[safe radish]

@dreamy marten Has your question been resolved?

[safe radish]

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[buoyant tusk]

If a line through the centroid G of triangle ABC meets AB in M and AC in N then prove that AN.MB +AM.NC = AM.AN both in magnitude and sign.

[safe radish]

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[cedar socket]

Let u be the complex number u = (1 +iV3)/2 and let F be the field F = Q(u). 1. Describe an injective group homomorphism from Z6 to the group of units in F.

[cedar socket]

Anyone have a clue?

[peak estuary]

what is u^6 ?

[cedar socket]

u^6?

its Z6 the group of integers mod 6

[peak estuary]

yes

what is u^6

the complex number u raised to the sixth power

[cedar socket]

its 1

[peak estuary]

ok. so what can you say about the subgroup of C^* generated by u

[cedar socket]

its cyclic

ahh think i got it

thanks

[peak estuary]

well tell just to make sure

[cedar socket]

you assign the integers mod 6 to u^integer power

[peak estuary]

yup

as a follow up, how does that generalise to other roots of unity?

[cedar socket]

uhh

you will have an injective homomorphism from Zn to the n roots of unity

[peak estuary]

yup

[cedar socket]

yeah

that for the help

thanks

.close

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[red yacht]

I am supposed to calculate $\sum_{k=1}^\infty \frac{1}{4k^2-1}$.
I was given the hint to solve it in a similar manor as
$\sum_{k=1}^\infty \frac{1}{k(k+1)}$, which was split up in $\sum_{k=1}^\infty (\frac{1}{k} - \frac{1}{n+1})$

This now creates a telescoping series that only allows the first and last member of it to survice.

I know the solution should be $\frac{1}{2}$ (thanks to Wolfram Alpha). I tried to split it up in two parts like this $\frac{1}{k} - \frac{4-\frac{1}{k^2}-\frac{1}{k}}{4k-\frac{1}{k}}$. But this didn't help me to create a similar telescoping series.

I would appreciate another hint a lot.

[flat frigate]

TrayKnots

[peak estuary]

do you know partial fraction decomposition?

[stray socket]

^

[red yacht]

sorry, no

[stray socket]

https://www.purplemath.com/modules/partfrac.htm

Easier to give you this than explain it lol

[peak estuary]

you can write $$\frac{1}{4k^2-1} = \frac{1}{(2k+1)(2k-1)} = \frac{A}{2k-1} + \frac{B}{2k+1}$$ for some numbers $A,B$.

[flat frigate]

Denascite

[peak estuary]

then yeah see the link on how to compute A,B

[red yacht]

oh, binomic formula?

I didn't see that

okay, I will read through the link and I think I got it. Thank you!

.close

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[snow schooner]

Hey all.

I've been working with Welch's t-test to find the probability that one population has a lower median than another.
I then realised that my distributions aren't normal (but the samples are still independent, and of different sizes).
Since they aren't normal distributions, I've been looking at non-parametric ANOVAs, such as the Kruskal-Wallis test.
As far as I know, this test only gives the probability the medians are the same.

So how can I use this test to find the probability that one population has a lower median than another?
Or is there a different, more suitable test I can use?

[safe radish]

@snow schooner Has your question been resolved?

[safe radish]

@snow schooner Has your question been resolved?

[buoyant tusk]

.reopen

.reopen

[buoyant tusk]

<@&286206848099549185> can u guys please help

[safe radish]

@snow schooner Has your question been resolved?

[safe radish]

@snow schooner Has your question been resolved?

[snow schooner]

It's probably best you ask your question in a new channel

Having two open questions at the same time can get rather confusing

I'll post it again to make it easier to read.

My question:
I've been working with Welch's t-test to find the probability that one population has a lower median than another.
I then realised that my distributions aren't normal (but the samples are still independent, and of different sizes).
Since they aren't normal distributions, I've been looking at non-parametric ANOVAs, such as the Kruskal-Wallis test.
As far as I know, this test only gives the probability the medians are the same.

So how can I use this test to find the probability that one population has a lower median than another?
Or is there a different, more suitable test I can use?

[snow schooner]

It's been 7 hours, so I think it's alright to <@&286206848099549185>. Apologies if I shouldn't have!

[plucky elk]

.close

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[safe radish]

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[vagrant sand]

Roll a seven-sided die, if I get a larger number than the opponent, I get $2.50. If I don’t get a larger number than the opponent I lose $1.25. What's the expected winning?

[vagrant sand]

i got $0.54 for this question

basically there are 49 total outcomes

since 2 seven-sided die is used

[versed gull]

math sucks

[vagrant sand]

are you gonna help????

[versed gull]

no

[vagrant sand]

than leave the server

if you hate math

[versed gull]

idk why am here

[lost patrol]

Do not message the help channels if you arent gonna help in a way

[versed gull]

math sucks

[vagrant sand]

tysm

[lost patrol]

There are discussion channels

[vagrant sand]

bro ban this person

[lost patrol]

For the same

I guess he left the server.

[vagrant sand]

alright

i might have to repost

[thin bridge]

don't repost

[vagrant sand]

alright

[thin bridge]

jsut show your work

how did you get 54c

[vagrant sand]

there are 7 outcomes where both me and my opponent roll the same number die

so those wouldnt be counted

21 outcomes where i have a larger number than my opponent = 21/49

21 outcomes where i have a less number than my opponent = 21/49

i would than multiply 21/49(2.50) + 21/49(-1.25)

[thin bridge]

so those wouldnt be counted
not quite

[vagrant sand]

how?

[thin bridge]

consider which catergory that would fall into

[vagrant sand]

neither

[thin bridge]

if you roll the same number
are you getting a number higher than your opponent? yes/no?

[vagrant sand]

no

but its also not less as well

[thin bridge]

and what the the rules of the game state happens if you don't get a number higher than your opponent?

[vagrant sand]

you lose money

[thin bridge]

yes

[vagrant sand]

ehhh ig

ok

[thin bridge]

it felt like your originally had the correct response and then edited it

[vagrant sand]

so i was right before than

i contradicted myself

yeah

21/49(2.50) + 28/49(-1.25)

would be approx $.36

[thin bridge]

yes

[vagrant sand]

ill do more of these questions for more practice

[safe radish]

@vagrant sand Has your question been resolved?

[safe radish]

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[gritty parrot]

having a bit of issues with this

[gritty parrot]

the corerct answer is there idk how its cvorect tho

<@&286206848099549185>

pls help me :-) my exam is tmmrw

how do u do it

i laready know the dotted line thing and the straight

si t has to be > or <

[sick harness]

The shaded region is everything to the left of x=3

[tropic hound]

x > 3

[sick harness]

Wouldnt it be <

Because its everything smallet than 3

[tropic hound]

-3x > -9 -> x > -9/-3 -> x > 3

[gritty parrot]

YEA

thats what i did

but it was wrong\

[sick harness]

OH

I forgor

When you multiply / divide an inequality by a negative number

Its less/larger than gets switched

[gritty parrot]

oh

why do toyu divide it tho

[sick harness]

Youre trying to find the equation that is the same as x<3

[gritty parrot]

ok

i was sick when they taught this bruhh

im still so confused whsat

ok so

-3x>-9

x>3

so u divide everything by 3

[gritty parrot]

which number is considered the inequality>

[sick harness]

You divide by -3

-3x > -9 -> x>3

[gritty parrot]

so 3 is trhenubmer

[sick harness]

F*ck

I meant

-3x > -9 -> x<3

Yeah 3 is the number

[gritty parrot]

so u divide both sides by -3

thus changing thec sign?

[sick harness]

Ye

[gritty parrot]

oh ok

thanks

[safe radish]

@gritty parrot Has your question been resolved?

[safe radish]

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[safe radish]

Please don't occupy multiple help channels.

[steep fog]

i got 43.0543, is that the correct answer?

[grizzled shoal]

a or b?

[steep fog]

(a)

[grizzled shoal]

what are the angles BAC and ACB?

we use different terms

is for example BAC the angle at point B?

[steep fog]

no, it would be point A

as the letter A is in the middle

[grizzled shoal]

ah ok

yes i get the same

[steep fog]

what about b?

[grizzled shoal]

no idea why it is important the AB is the longest side

43 is wrong for a actaully

they want sin x

not x

43 is for b

[steep fog]

thanks a lot

[safe radish]

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[near wren]

Hi, I'm tring to map a tuple of multiple values into unique index to a combination of all posible values, but I'm not sure how to do that. Math is not my strength...
I mean, for example, given a combination of two numbers (x, y), where x has only two valid states (0 and 1) and y has three valid states (0, 1 and 2). How can I map them?
I'm trying to find a math formula which would allow me to map them.

It's for programming actually, so I could build a table with all the possible values, but a table for my real case wouldn't fit in memory (to many values hehe), so I need math for the rescue.

x y -> i
0 0 -> 0
0 1 -> 1
0 2 -> 2
1 0 -> 3
1 1 -> 4
1 2 -> 5

i -> x y
0 -> 0 0
1 -> 0 1
2 -> 0 2
3 -> 1 0
4 -> 1 1
5 -> 1 2

¿Any idea?
At least knowing the name of what I'm looking for would help me so I can research on my own. I've no clue what to google actually.

[quasi bison]

i mean this sounds like you're encoding the numbers from zero through five in base 3

at least in this toy case

how many variables do you have in your real case and how many valid states for each?

[near wren]

That would be if both x and y had 3 states, but x has 2, so each variable is a different number base.

[near wren]

a = 100, b = 100, c = 2, d = 9, e = 9, f = 9, g = 9, h = 9, i = 9, j = 9, k = 9, l = 9, m = 9, n = 9, ñ = 9, o = 9, p = 9, q = 9, r = 9, s = 9, t = 9, u = 9, v = 9

They are quite a lot 👀

[quasi bison]

...right

well ok like the fancy term for this is mixed-radix number system

[near wren]

Thanks, I'll research that

.close

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[river yoke]

In a meeting each person greets at least 2 men and at least 3 women. How many people are there at least in the meeting?

[river yoke]

I think is 7

because 3 + 4

3 men + 4 women

But im not 100% sure

[lean otter]

That's right

[river yoke]

Oh

So thanks

.close

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[south bane]

how do you find a,b and c?

[south bane]

ive got b=2 cuz vert asymptote is -2

the horizontal asymptote is -1

[nova creek]

That means the lim x->∞ should be -1, yeah?

[south bane]

would you be able to use another term other than limit? havent really learnt much abt it

[nova creek]

I can try a whole nother concept. What's the horizontal asymptote for 1/x?

[south bane]

it would be 0

[nova creek]

So we need to shift the graph vertically down by 1, right?

[south bane]

ye

[nova creek]

How would we do that?

[south bane]

the c would be -1?

[nova creek]

Yep

As for a, you can just pick a point and plug in

[south bane]

ah I see

so the horizontal asymptote is always the c variable? and if there is no c variable then it goes back to the powers of the x variables in the equation?

[nova creek]

The horizontal is always c. I dunno what you mean by "if there is no c"

[south bane]

like if it were just y=a/2+x

the asymptote would just be 0 for that right?

[nova creek]

If you mean a/(2 + x), then yeah

No constant a will change the horizontal asymotote

[south bane]

alright

would u be able to explain limit quickly?

[nova creek]

Well, when you take the limit x->∞ of a/(x + b) + c, it means x gets really really big. Since x is in the denominator, it'll make the whole fraction really really small, until it's 0. Then you're just left with c, which ought to be -1, since the limit should equal -1

[south bane]

ah i see

[neat kiln]

I have a question Jojo boss

[nova creek]

If it's a math question, ask in a different channel. If it's not a math question, ask in a different channel

[neat kiln]

Is limit x->infinity the same as x-> negative infinity?

[south bane]

is b 6? as thats the amplitude, or would it be 3

[nova creek]

Not in general

[lean otter]

this is help chat not general

[nova creek]

The amplitude is the distance from the center of the wave to the peak

[neat kiln]

But in this case it is right?

[nova creek]

Think of it as the average between the lowest point and the highest

[south bane]

ah alright the average

so amplitude is 3

[nova creek]

This isn't a general chat, but in this case, yes

[neat kiln]

Ok

[south bane]

how would you find a and c?

[nova creek]

Sorry, not the average, but half the distance between the peak and troph

[south bane]

ah alright

[nova creek]

So in this case, 3

The average will be the y value of the midpoint of the wave

a is the vertical shift. Since the midpoint of the wave is normally y = 0, the shift in a will make the midpoint y = a

[neat kiln]

Is c the horizontal shift?

[nova creek]

Taking that in reverse, a is the midpoint of the wave

[nova creek]

Yes

Also called the phase shift, though it's the same thing

Kinda

[south bane]

is the circled part the a value?

[nova creek]

No

That's the troph

The midpoint is halfway between the peak and troph

[south bane]

alr so midpoint would be 3

and then half of that?

[nova creek]

The midpoint isn't 3

Thats the amplitude

That's the distance from the midpoint to the peak/troph

[south bane]

halfway between the peak and troph

is this in distance or like coordinates or something else?

[nova creek]

Like coordinates. This is where you take the average

[south bane]

im kinda confused, so the max in this graph is y=4.5 and the min is -1.5

oh and then you add them 2 and divide 2?

[nova creek]

Yep

That'll give you the y value of the center of the graph

[south bane]

ah alright

[nova creek]

Thats also how much the graph is shifted vertically

[south bane]

alright

how would you find c?

[nova creek]

This one's a bit trickier

This is a good example for plugging in points

(0, 0) looks like an easy one

[south bane]

yep

[nova creek]

I was worried there wouldn't be any easy points, but seems like its fine

[south bane]

so just put in (0,0) as the x and y values?

[nova creek]

Yep

Solve for c

Sin is periodic, so there will actually be an infinite number of possible values. Either pick the simplest one, or whichever one your calculator spits out

[south bane]

or sorry theres a restriction in the question mbad

c < pi

[nova creek]

Ah. Then pick the c that fits that requirement

[south bane]

so the equation should be 0 = 1.5 + 3Sin(-c) right?

[nova creek]

Yeah

[south bane]

hmm my calc isnt giving anything

[nova creek]

Wdym

[south bane]

its giving: x=6.283... * constn(1) + 0.52459...

[nova creek]

If the calc ain't working, this is a nice one to do mentally

[south bane]

so ive gotten it to -1/2 = sin(-c)

is that right?

[nova creek]

Looks like that's 2 π * n + 0.52459 rad, where n is an integer. That second bit is what we need, but we want the exact value

[nova creek]

Yes

You can simplify further by saying sin(-x) = -sin(x)

[south bane]

can you cancel out the negatives so then it'll just be 1/2 = sin(c)?

[nova creek]

Yep

[south bane]

by canceling i mean like multiplying both sides by -1

aight so c=30

which is pi/6 in radians

[nova creek]

Yep

[south bane]

alright ty, can i leave this open for a bit so I can ask you some questions if I have

[nova creek]

Dunno the rules on that. Might be better to close it and just open a new channel if another problem arises. I would try to do the questions on your own a bit before resorting to this channel, though

[south bane]

alright, i'll give u a ping or dm if I got any other q, is that alright?

[nova creek]

Eh, it'd probably be easier to just make another channel. If I see it, I'll try and help. If I don't, someone else will.

[south bane]

alr easy

ty again

.close

[safe radish]

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[muted jungle]

hello I was wondering how to get the answer on the first part and also what the second part is asking?

[muted jungle]

the answer for b is 3 (i just guessed) but if anybody is able to explain that would help a lot

<@&286206848099549185>

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@muted jungle Has your question been resolved?

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[muted jungle]

.reopen

[safe radish]

✅

[cinder viper]

@muted jungle It would be better if you sketch the graph

[muted jungle]

okay

I don’t know what part 2 is asking

[cinder viper]

If you draw a horizontal line through the parabola anywhere, it intersects twice

We need it to intersect one for the function to be invertible

so you divide the parabola into two parts

One for x>=4, one for x<=4

and then the resulting functions are invertible

x>=4 and x<=4 are both correct options here, but they chose the right part of the parabola, so x>=4

[muted jungle]

thank you

.close

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[hasty roost]

Hi

[hasty roost]

I need to solve these 3 qs

[safe radish]

@hasty roost Has your question been resolved?

[safe radish]

@hasty roost Has your question been resolved?

[hasty roost]

@lean otter

[lean otter]

Sorry but I'm not really qualified to answer these 😅

[hasty roost]

😭😭😭 I understand lool pls recommend someone who can

[lean otter]

You may ping helpers role now (since it's more than 15min), someone may help you

[hasty roost]

<@&286206848099549185>

[karmic ravine]

Bruh these are those type of questions

I don't know how to do these

[hasty roost]

Pls who can help me?

I really need assistance

Is there like a super smart guy here who could help?

[brazen jungle]

just put values of x and you'll get corresponding values of y use them as abscissa and ordinate to plot, f(x)=0 to find x intercept

for cosx and sinx you should have the graphs memorized

and graph of sinx remains same when squared

[safe radish]

@hasty roost Has your question been resolved?

[sour musk]

do you know the trapezium rule ?

idk how the rules work here but its super easy, draw the cosine of x between -90 and 90 on the graph (it should look like a bell), divide the diagram into 4, the best places to do this would be at +-pi/4 or +-45 as this just makes it easier. then using the trapezium rule:

sub in the values. Consider whether or not the trapeziums would go under or over the graph - this will help you to understand if its going to be an over/underestimation

[safe radish]

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[hasty roost]

No

[hasty roost]

So what are the values tho?

Like what is h

[sour musk]

ill send you a video cuz you need to learn what the trapezium rule is to do this type of integration

https://youtu.be/4lSuPwccd8Y

[hasty roost]

Im watching it now

But for now here is my cos graph

It’s loading @sour musk

[sour musk]

have you sent it here ?

[hasty roost]

Sending now

[sour musk]

k

[hasty roost]

Almost done

[sour musk]

thats a standard cosine graph

[hasty roost]

Ya

[sour musk]

you need to restrict it between -pi/2 and pi/2

[hasty roost]

Ohhhh

Lemme Google that

[sour musk]

draw out for me what you think it may look like

[hasty roost]

Np

It looks like an upside down quadratic

From -1 to 1

It’s sending so slowww lool

@sour musk

[sour musk]

yes thats right

[hasty roost]

Ok imma continue watching vixen maths

[sour musk]

good plan !

[hasty roost]

Bicen*

So now I understand

Convex is when sec derivative is greater than or equal to 0

And concave is the other

@sour musk

What do I do from here to find the y values or create the pillars for the graph

[sour musk]

you've got to create 4 equal sections,

so the best thing to do is to spilt the graph every 45 degrees or pi/4

[hasty roost]

? On what way

Just create them

Randomly

Like 4 lines

[sour musk]

they're all vertical strips on the graph

[hasty roost]

Can you show me what that means?

[sour musk]

mhm gimme a sec

[hasty roost]

Ty

90 percent of the other questions represent the graph as a function

Not the graph as an image

Like sqrt(x-1)dx from limits 3-7

The other tutorials I mean

oOH I NEED TO KNOW THE AVLYES OF COS

but I don’t know where to put them on. The graph

@sour musk

[sour musk]

mhm

[hasty roost]

I only have +-90•

Im low-key lost lemmme write the values of cos

[sour musk]

you need to find the values of y at each value of x

hence at -90, -45, 45 and 90

[hasty roost]

Sooo

Sqrt of 2/2

And sqrt if 0

@sour musk

Is x

No is y?

[sour musk]

is y

[hasty roost]

0.7

0.71

[sour musk]

leave it in the surd form

[hasty roost]

Ok

Therefore it looks like this

@sour musk

[sour musk]

yes but the values are still positive

[hasty roost]

Ooohhh ok

[sour musk]

and yes that looks great !

[hasty roost]

Thanks imma use desmo or my calculator to do this

But my family got me running errands

I’ll get back to you soon

Thanks bro

[sour musk]

nws

[hasty roost]

@sour musk

1.110720735

[sour musk]

yup!

well done !

epic maths !

[hasty roost]

Ok @sour musk

Therefore it’s convex?

Since it’s greater than 0

0

[sour musk]

think so

wait

lemme refresh myself on that real quick

yes it is

[hasty roost]

@sour musk thanks so much maybe I’ll do 7 and 8 tomorrow

Im mentally finished

[sour musk]

nws !!

i gotta get back to my own maths revision

got me big exams next week

[hasty roost]

.close

[safe radish]

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[craggy stirrup]

help

[craggy stirrup]

so this is what i have dont so far

and then i am stuck @.@

[safe radish]

@craggy stirrup Has your question been resolved?

[craggy stirrup]

<@&286206848099549185>

[safe radish]

@craggy stirrup Has your question been resolved?

[plucky elk]

it's just the binomial expansion of
$\sqrt{1+x} = (1+x)^{1/2}$

[flat frigate]

riemann

[safe radish]

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[bright jungle]

Hi there! I have a kinematics derivation problem I have been stuck on for too long.

In essence, the standard kinematic model is (dv/dt) = a. In my case, I am simulating drag as well which is a product of the velocity. So my equation looks like: (dv/dt) = a - rv where r is drag and v is velocity.

I have successfully derived a working equation for predicting velocity at time t (annotated as v2). Where I'm failing is further integrating this to solve for position at time t. Can anyone with strong integration/kinematics help me out? 🙂

[lean otter]

let me get this right

you have the dif equation (dv/dt) = a - rv

which descrbes the motion you observe

where a is acceleration (dv / dt)

r is a constant

and v is, well v for the sake of the equation

so you have (dv/dt) = (dv/dt) - rv

or did I missinterpret this?

[bright jungle]

That is correct! I started with vf = (vi + at)(1 - rt). I had substituted "v + dv" for vf, and "v" for vi. I factored it out and came to the above differential equation you just mentioned (dv/dt) = a - rv

[lean otter]

what is vf?

I assume vi is v initial

[bright jungle]

Sorry - my annotations for vf is "final velocity" where vi is "initial velocity"

[lean otter]

I see

[bright jungle]

In my picture I have just interchanged v2 == vf and v1 == vi for clarity. My handwriting isn't the cleanest

[lean otter]

ok, so far so good, and you solved that to get the above equation

[bright jungle]

Correct .. I was able to integrate it this first time to get that. I have verified it solving for both v(t) and t given velocity parameters and it checks out. I can't for the life of me take it a step further to be able to solve for position

[lean otter]

which is v = (1/r) * ( dv/dt - Ce^-rt )

[bright jungle]

Yes that is correct

[lean otter]

and now you need to solve it for position

I assume your position is one coordinate, like x?

[bright jungle]

Exactly, and this is where I'm very stuck

[lean otter]

well we know that v = dx / dt right

[bright jungle]

agreed

[lean otter]

so we plug that in and get

dx/dt= (1/r) * ( d^2x/dt^2 - Ce^-rt )

right?

[bright jungle]

I am tracking you writing this down, let me upload real quick

[bright jungle]

Okay perfect, I'm with ya

[lean otter]

now we need to solve this

that's the tough part

obviously

let's rearange a bit first

would this be a correct rearangement?

[bright jungle]

It looks right to me, much cleaner

[lean otter]

now id multiply all by r

since it's a constant

even cleaner still

and now we can recognize that this is a diff equation of the type

[bright jungle]

Okay I'm tracking!

[lean otter]

where a, b and c are constant

a = -r

b = C

and c = -r

so now we just have to solve this diff equation and we should get our function x(t)

are you familiar with solving diff equations of this type?

[bright jungle]

I wish I could say I was good at differential equations, but the second it hits d^2x/dt^2 I slow down hard and need to use online resources.. This is already in a dramatically different form that what I've been trying to solve for (embarassingly, months). The fact it's even here makes sense that it would be this difficult. All of my other attempts were coming out too simple and were (obviously) way off

[lean otter]

I've done nothing special

I just plugged in v = dx/dt

[bright jungle]

My google warrioring tells me this is a second-order linear ordinary differential equation?

[lean otter]

and cleaned it up a bit

I don't know how to solve it by heart either

I use a lookup site

nothing wrong with using a lookup site

[bright jungle]

I will see if wolfram likes this ... In all honesty the derivation is beautiful but I'm so set on trying to get a working solution. In this result, pardon my ignorance but where is k1 and k2 coming from?

[lean otter]

I assume those are the integral constants

k2 should be initial position of x

so if x starts at 0 k2 = 0

as for k1

well I wouldn't be able to tell right away which initial condition it refers to

[bright jungle]

okay right, so it's in the form of p2 = ____ + p1 akin to the v2 = something something v1

[lean otter]

yeah

[bright jungle]

Okay I'm going to screenshot dump real quick. I bought wolfram like 10 years ago before it was cool so let me spell out the solution and see if I can walk through it w/ you to make sense of it. This is incredible.

or maybe it was cool then, probably

[lean otter]

just a second

in the first equation

which you used for v

did you plug in a as a constant?

[bright jungle]

a is a constant in this simulation. The constant I ended up substituting was a - rv

so I said C = a - r*v. I couldn't include the first a because I couldn't seperate it from e^(-rt)

[lean otter]

does that mean a is not dv/dt?

[bright jungle]

Would I able to screen share here or is that too much? I can definitely show how I have the simulation setup

[lean otter]

oh so C has a in it

then what we just did isn't good

since we didn't account for that variable change

[bright jungle]

The initial simulation parameters are like this:
vo = 15
po = 0
r = 1
a = -6
t = 2

The object starts at position x=0 and has a velocity of 15m/s. A constant acceleration of -6m/s^2 is applied to the object for a duration of 2 seconds. I'm then predicting position at time t = 2.... so I am not 100% but for a not being dv/dt would seem true since it's constant acceleration and shouldn't be changing... maybe this is where I'm confused because mathematically it's dv/dt but I'm keeping it constant

[lean otter]

well in this case acceleration is constant

[bright jungle]

shoot

[lean otter]

for the purposes of this simulation

but that is even easier then

if we know a is constant

we just integrate this

easy integral

[bright jungle]

x = 1/r^2 (a * r * t + Ce^-rt) + x0?

[lean otter]

why is r squared?

[bright jungle]

Oh oops, okay so

x = (a * r * t + Ce^(-rt))/r + x0

[lean otter]

oh wait

was doing it in my head so I think I mistoook

ill write it down

[bright jungle]

Please do, don't trust my integration it hasn't worked for months lol

[lean otter]

now I'm really curious if this'll work

can you run your simulation to check?

wait

C = a - rv

is this v

initial velocity

or current velocity

[bright jungle]

Yeah it is, that v is the starting velocity. In the case of simulation parameters above it would be 15

[lean otter]

good

almost had a heart attack

[bright jungle]

Given v0 = 75, p0 = 0, r=1, a=-6, and t=2 the actual position at time t is 56.99. Let me plug in the above and see if it predicts

This is what I have been running into ... I can't quite grasp what is going wrong unless it's the code I actually wrote to implement this

[lean otter]

hmmm....

how did you do it for the velocity

it's clearly wrong

[lean otter]

so something isn't good

with position

[bright jungle]

which that one was v(t) .. and I also did t as a function of velocity which has been working too

I'm stumped because surely this is possible to express as a differential equation ... I have currently implemented these solutions for the standard continuous kinematic equations like x(t) = xo + vt + 1/2at^2 just fine. I'm running the simulations in Unity and it has a time step of 0.02 seconds, which has been small enough to still result in around 99.95% accuracy even though it isn't technically continuous

[lean otter]

the equation we got is wrong

no way can a body decelrate from 75 to negative in 2 seconds

with decelaration of 6

and yet the integration is ok

can you show me the initial equation again

the one you used to get

everything

what equation did you model your system on?

[bright jungle]

Sure -

vf = (vi + at)(1 - rt)

I substituted "v + dv" for vf, and "v" for vi. It factored it out to

(dv/dt) = a - rv

This is what then integrated to

vf = 1/r * (a - (a - r * vi) e ^ (-r t))

[lean otter]

hold on

[lean otter]

you have (dv/dt) = a - rv

how did you solve this?

[bright jungle]

Okay here is my mathematics exchange question covering this. Short answer, I didn't - I am trusting another derivation and it has checked out when I actually run all of the simulations

This is just a giant review of what we already have https://math.stackexchange.com/questions/4119031/help-deriving-kinematic-equations-with-drag-element/4119053?noredirect=1&newreg=e7f7e37444ef40da9e4fc6886d1deb32

The one answer walks through initial derivation which is 1:1 with the continuous kinematic formulas. I have tried running the integration through wolfram and it comes out slightly different, and is very incorrect when it comes to the simulation. The dv/dt = a - rv integrating to v = 1/r(a - Ce^(-r*t)) works like a charm for predictions

[lean otter]

ok, just a sec

I'll try to run my own simulation

in excel

[bright jungle]

Also this was the older version where I'm chasing time ... and it works. I just can't get position

okay wonderful!!

[lean otter]

what velocity do you get in your simulation

[lean otter]

for these numbers

[bright jungle]

I'm getting 4.962158

[lean otter]

does thatmake sense to you though

that a body goes from 75 m/s

to around 5 m/s

with a deceleration of -6

does it sound intuitive

it sounds to me like it slows down waaaay too much

[bright jungle]

It wouldn't sound intuitive if it weren't for the drag factor of 1 ... If I run it with a drag of 0.05 (it successfully predicts) the velocity to be 56.44 which is closer to intuitive

Though if drag goes to 0 the equation breaks and I use continuous motion which continues to be correct

[lean otter]

so let me just get this straight

the simulation is basically just iterating the formula

v i+1 = (vi + at)(1 - rt)

over and over

or to be more precise v i+1 = (vi + a dt )(1 - r dt )

is this correct?

[bright jungle]

That is a true take - it is iterating in 0.02 second intervals

[lean otter]

I just did that in excel

and it comes up as 60 for the final velocity

ill check to make sure

but I think I did it ok

oh nvm

mb

i messed up with one variable

ok so

Here is what I get

[bright jungle]

I'm with you, I'm getting the 4.96 still on t=2 for velocity

[lean otter]

They behave the same for the most part

so it's safe to say the formula is accurate

[bright jungle]

You're hitting my assumption which was that I could substitute continuous time in and ignore the margin of error which is like 0.05%

I've based everything off of that and it's seemed ok. It's just the integration of this I'm wondering why it goes so wonky

[lean otter]

what equation did you use to model position?

[bright jungle]

position with drag? Or without it?

[lean otter]

the position of the body being simulated

[bright jungle]

Can you rephrase? I am trying to derive that given the working velocity one ... but the model I use for when drag is 0 is p(t) = p0 + vt + 1/2at^2

[lean otter]

is this your position

at the end?

i just did current velocity * time span (0,02) + previous position

to model it

[bright jungle]

Is that given the same variables for velocity check? v0=75, r=1, a=-6, p0=0?

[lean otter]

yes

I just told excel to take the simulated velocities and multiply them with time

then add em up step by step

so we have a movement each step

just being added up

[bright jungle]

Okay yes that's what it is coming out to

Does this have to be solved through a series like that then and it cannot be expressed as a differential equation?"

[lean otter]

I just decided to model it likethat

since I already have the velocities at every point

I can then just calculate the movements between them

and add em up

so there's that

[bright jungle]

What do you think would be wrong with the differential equation we made for predicting position being so wrong? I have tried to wrap my head around which assumptions are wrong since it is just not accurate like the velocity prediction equation is

[lean otter]

now

this is where it gets fun

reaaaaaly fun

[stray socket]

Phootmath time!

[lean otter]

look at this

the final results are completely off

you'd say the solution is wrong

but

look at the graphs

for the simulation

and the formula we got

the formula is correct

we're just at the wrong initial position

if we adjust for it

and find what it needs to be

we'll have it

we dropped the integration constant

I assumed it would just be '

0

I was wrong

we need to calculate the constant

and we'll have the result

we weren't wrong

we were just not looking at the whole picture

just the result isn't enough

you need to model the whole process to see what's going on

every step of the way

we didn't finish solving the integral

[bright jungle]

Could you share the excel doc you made with the model so I can track with you? I'm lost when it comes to solving that last constant

[lean otter]

I named it k

for the sake of not messing up with C

so we know that at time t = 0

x =0

well then

let's plug that in

and solve for k

[bright jungle]

Okay tracking!!! 🤯

[lean otter]

ladies and gentlemen

we got em

kinda

my simulation is a little wonky maybe

so it's alittle off

[bright jungle]

Let me plug this in and check it

NO WAY

I have been tweaking value and it is tracking 🤯 ‼️

[lean otter]

we solved it

and it was all

because I

idiot

forgot

that C

doesn't have to be 0

for initial value of 0

[bright jungle]

Definitely all you. I am so extremely thankful. Maybe it's against the rules but could I buy you dinner or something for your time?

[lean otter]

lol

[bright jungle]

This has been in my head for a year. So elegant when it comes out like this

[lean otter]

they're not fully overlaping

maybe a smaller time step will help idk

well

[bright jungle]

So the real final equation is p(t) = a * t / r + C * e^(-r*t) / r^2 - C/r^2 + pi ... or something

[lean otter]

this was emberassing

how long it took

and how much I stumbled

[bright jungle]

My eyes are burning with how beautiful that is. Shade I cannot thank you enough. What a rollercoaster

[lean otter]

it was a shitshow

obviously

ill send you the excel in a bit

[bright jungle]

I think the timestep makes sense .. it seems to converge closer in some conditions. Considering
v0 = 75
P0 = 0
r=0.5
a=-6
t=2

the prediction is nearly spot on

[lean otter]

ye

[bright jungle]

https://tenor.com/view/baby-yoda-babyyoda-gif-20491479

[lean otter]

to all you lurkers who watched this shit go down

(yes I've seen you type)

don't judge me for those mistakes

it's 3 am

i need sleep

@bright jungle

Note that I may have made some errors in this excel file

but it looks good enough

[bright jungle]

Words can't thank you enough. I never thought a spreadsheet named "Eh" would be so close to my heart lol. T h a n k y o u

[lean otter]

np

good luck

[safe radish]

@bright jungle Has your question been resolved?

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[pastel arch]

i was wondering if someone can help me understand this, thanks!

[final halo]

,rccw

[pastel arch]

?

im stuck w question number 3

[karmic ravine]

@pastel arch which one do u want

Oh

Do u know what a period is and what a periodic function is?

[pastel arch]

the period a complete pattern?

is*

[karmic ravine]

Yeah no

The period of a function is the shortest value of differences in input whenf of x repeata

So the period in sin of x is 2pie

And in tan of x is pie

Because it repeats after every pie radians

[karmic ravine]

And the period for this function is 15 and ur given two values

@pastel arch do u get the idea?

[pastel arch]

yes that makes more sense than what i was looking at before

i’m stuck on how i’m supposed to find out what f(130)=

[unique bison]

f(x) = f(x+15) for all x

[karmic ravine]

That's the first one tho lol

Try , make that brain work my man if it don't ping me again

Did u do the rest?

[pastel arch]

i have jus this page left, ima shower n finish it. ty for the help

[karmic ravine]

@pastel arch if ur problem is solved the close , if it's not then ask for help

Also the answer to f(130) is -5 , because it is f(40+15(6))

[safe radish]

@pastel arch Has your question been resolved?

[safe radish]

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[manic elbow]

In an older physics textbook, what does the apostrophe next to the summation symbol mean?

[vocal portal]

found some info indicating that sum is finite

[manic elbow]

Seems weird considering the sum doesn't have to be finite

[vocal portal]

force should be

[manic elbow]

Yeah it should but there could be infinite mass points contributing to the force

Actually yeah

Thanks

.close

[safe radish]

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[low nest]

https://cdn.discordapp.com/attachments/543991146062544897/982059363084697681/unknown.png
Did I do this right? Point out what im doing wrong so I can fix

[safe radish]

@low nest Has your question been resolved?

[safe radish]

@low nest Has your question been resolved?

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[lean otter]

For a) find the turning point, completing the square or differentiating... whatever

io can find b

but ion know for a

So B?

how do i find b a with 100

ik the x val is 100

,w turning point of -0.002x^2 + 0.4x

[flat frigate]

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Requested by Yustina#9692

[lean otter]

Remember, there is an angle

yeah what do I do with it

cos 10 = x/100 so x = cos 10 * 100

,w cos 10 degree * 100

[flat frigate]

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Requested by Yustina#9692

[lean otter]

wait what

is there an easyier way

[hasty wagon]

This seems to be a completing square problem

[lean otter]

its on a grade 10 test

but yeah

[hasty wagon]

Hmmm test? As in you've finished the test and you wanna know how to do it? Or is it just past paper?

[lean otter]

no its a practice test

real test tom

[hasty wagon]

Okay so

I will interpret a and b ad finding the vertex of the parabola

[lean otter]

ok

[hasty wagon]

Lemme quickly draw it out

[lean otter]

kk

[hasty wagon]

As you can see, it's in an upsidedown U shape

So, the vertex is a maximum point of this parabola

And referring back to the question:"maximum height"

[lean otter]

yes so its a negitive a

I just need to know how to find the height when i have the distances of the vertex

so like finding the y when i have x

[hasty wagon]

Let's say, the vertex is (h,k)
h will be the maximum height and k will be the horizontal distance when it reaches maximum height

[lean otter]

yes

[hasty wagon]

Ring any bells on "Completing Square Method"?

[lean otter]

tbh i never rly learned that one

I use axis of symmerty method

[hasty wagon]

Hmmm, them this site might help

https://www.mathsisfun.com/algebra/completing-square.html

[lean otter]

is there a way I can do it with aos method

[hasty wagon]

Well, there is

But then you'll have to find (b) first and then find (a), which is not exactly what the question is teaching I guess.

[lean otter]

like ik what to do when i have an equation

[hasty wagon]

By anyway if you insist

[lean otter]

but ion got one here so I need to find it

[hasty wagon]

No problem

Remember what's b/(2a)?

[lean otter]

yeah

[hasty wagon]

That will be the x-coordinate of the vertex

[lean otter]

yeah its 100

i need to find the yu

y

[hasty wagon]

Note that with x-coordinate of any point on the parabola, you can use the parabola itself to find the y-coordinates.

[lean otter]

yeah i need the equation thi

[hasty wagon]

We have
h=-0.002d²+0.4d

[lean otter]

how do i make that

[hasty wagon]

This is the equation

[lean otter]

so just fill it in

[hasty wagon]

Yep

h = -0.002(100)²+0.4(100)

[lean otter]

okay bless thats all i needed

that h=-0.002d²+0.4d\

Thats what I was trying to figure

[hasty wagon]

Last but not least, I highly recommended to take a look at completing square method

[lean otter]

okay i will

[hasty wagon]

Cheers!

[lean otter]

(b/2)2 right

thx

[hasty wagon]

(b/2)2?

[lean otter]

for the compleung the squrae

[hasty wagon]

Oh

(b/2)²

Yea, that's a part of it

[lean otter]

ye okok

thx for da help

A Frisbee is thrown straight from a position 2 m above ground level. Because of the wind pattern, the height h in metres, after time t, in seconds is given by the formula h = 2+6t-2t2. What is the maximum height the Frisbee will reach?

[plucky elk]

.close

[safe radish]

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[lean otter]

what does the expression in the second image mean?

[lean otter]

i think it means the set of all elements of Q' that are not equivalent to each other

but im not sure

[quasi bison]

no, it means the set of equivalence classes of Q' under tilde

[safe radish]

@lean otter Has your question been resolved?

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[analog frigate]

i know that i have to find the lCM but after i find that idk the next step

[neat kiln]

What is icm?

[analog frigate]

oh sry LCM

for question a

[spice grove]

LCM of what?

[ornate dome]

Oh so I’m assuming you made them all whole numbers?

[analog frigate]

yea

[ornate dome]

Cause then you can isolate one variable

[spice grove]

You could make coefficient of one of the variables same in both the equations then subtract the equations simply.

Less work when dealing with fractions.

[analog frigate]

how would you do that?

[spice grove]

https://youtu.be/oKqtgz2eo-Y

[analog frigate]

.close

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[pseudo zenith]

so

[pseudo zenith]

i havea rather simple maths question im struggeling with

the task is; show that the total length of the fence an be expressed with

G=(2pi+3)r (idk how to get the pi sign)

where G is the length of the fence and r is the radius in the circle

<@&286206848099549185>

[white swallow]

2pi*r is the total length of circumstance, 3r is 3 of those fences extended from the centre of the circle

[dull sequoia]

ping helpers after 15mins...

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@pseudo zenith Has your question been resolved?

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[forest fossil]

hello

[forest fossil]

Can some1 help me to understand this please

why did we did absolute value x-u on the left hand
and on the right hand, why did we substituted k as a standard deviation

[quasi bison]

chibishibs

on the left the actual probability is calculated, using the gaussian CDF

$|X - \mu| \leq 2\sigma$ is the inequality that says ``$X$ falls within 2 SDs of the mean''. the mean is $\mu$ and the SD is $\sigma$

on the right they appear to simply find what lower bound chebyshev's theorem gives for the same probability, but they do not actually find it

@forest fossil does this answer your question

[flat frigate]

Ann

[forest fossil]

oh yes I see

[forest fossil]

just trying to understand this

[forest fossil]

I mean

the question gave us standard deviation is 2

[quasi bison]

no

[forest fossil]

and we don't have K value

[quasi bison]

we were not told "the standard deviation is 2"

[forest fossil]

what's within 2 std of the mean referring to

[quasi bison]

"X falls within 2 SDs of the mean" means "the distance from X to the mean is less than or equal to 2 times the stdev"

whatever the stdev may be

[forest fossil]

oh so 2 k

because 2 bounds

right ?

[forest fossil]

I just noted it's written like that in the notes LMFAO

[quasi bison]

what do you mean "because 2 bounds"???

this feels like a language barrier or maybe you just got yourself deeply confused

[forest fossil]

I mean like how i draw it in the cyan

2 bounds either to the right or left

[quasi bison]

2 standard deviations.

you're misusing the word "bound".

[forest fossil]

oh ok mb

so this means 2k, right ?

[quasi bison]

no

[forest fossil]

so how did they say k=2 in the right side

[quasi bison]

$P\paren{|X - \mu| \leq k \sigma} \geq 1 - \frac{1}{k^2}$ is true for any positive $k$

[flat frigate]

Ann

[quasi bison]

setting $k$ to 2 gives us something about the event $|X - \mu| \leq 2 \sigma$, which is what we want

[flat frigate]

Ann

[forest fossil]

didn't we replace k with 2 here?

[quasi bison]

we did