help-28

[delicate torrent]

if you let H be the intersection of MP and NQ then H's the midpoint of both of them

the problem is OI

[delicate torrent]

guys

😭

[wary condor]

can you just turn it into an algebraic problem and solve for y=0

[delicate torrent]

🙏

.

i wish i could turn it into an algebraic problem

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[bright leaf]

Hi if anyone could confirm my working for part I that would be lovely and also explain how to start the second part !

I did star a line because im not sure if thats an actual thing yiu can do…

[lime ether]

looks good but the way you do this seems quite weird

you can just differentiate dt

instead of involving this dv/dh stuff

$\dv{V}{t} = 20\pi h \dv{h}{t} - \pi h^2 \dv{h}{t}$

[glossy valve]

knief

[bright leaf]

Dv on dt is given by the question

Because they say water is being poured in

[lime ether]

yes i know

thats the point

[bright leaf]

But my question asks for the depth changed so isn’t that dh on dt

[lime ether]

treat V and h as functions of t then differentiate both sides wrt t

[lime ether]

yes

which is what we want

[lime ether]

you isolate dh/dt here

[bright leaf]

Like i rearrange the equation

To find dh on dt

[lime ether]

$\dv{V}{t} = (20\pi h - \pi h^2) \dv{h}{t} \implies \dv{h}{t} = \frac{2}{100\pi - 25\pi} = \frac{2}{75\pi}$

[glossy valve]

knief

[bright leaf]

Yay okay thank you sm

[lime ether]

plug in 2 for dv/dt and 5 for h

[bright leaf]

Yes i see thank u

[lime ether]

ok so part 2 you need help with?

[bright leaf]

I’m not sure what im meant to be looking for

Is it dr on dt

[lime ether]

yes they want dr/dt

have you made any attempts so far?

[bright leaf]

um no haha

i wanted to confirm my i

[lime ether]

well we need to write V as a function of r yes?

[bright leaf]

i dont know

which equation has an r in it

[lime ether]

in order to find dr/dt we are going to have to somehow relate r to V or h

[bright leaf]

beside like A.... but idk why id use A

[lime ether]

since we know dv/dt and dh/dt when h = 5

[bright leaf]

yes but idk how to

[lime ether]

do you know the volume of a hemisphere in terms of r?

[bright leaf]

um ngl no

[lime ether]

if you know the volume of a sphere then this should be easy

a hemisphere is half a sphere

[lime ether]

do you remember this?

[bright leaf]

okay nvm its given to me

4/3 pi r^3

so for hemisphre its that divided by two right

so can i do dr/dV maybe?

ill try that

[lime ether]

no wait but V is the volume of the water

not the volume of the container

[bright leaf]

should i use

[lime ether]

so we should really try relating r to h instead

[bright leaf]

V= 1/3 Ah

and write A in terms of r^2 and pi

[lime ether]

where did you get this?

[bright leaf]

volume for sphere

so for hemisphere

1/6 Ah

[lime ether]

what is A here?

[bright leaf]

area of the water surface

[lime ether]

oh

yea i think that should work

you mean then relate it to the equation you had strictly in terms of h?

[bright leaf]

yesss

the V = 1/3 h^2....

[lime ether]

yea 1/3 pi h^3(30 - h)

[bright leaf]

okay now my answer seems ridiculous tho

[lime ether]

what did you get

[bright leaf]

its incorrect

they used similar triangles

[lime ether]

i don't think it should be pi r^2

yea

[bright leaf]

sad

i will try again

thank you for the help

the i was right tho thankfully

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[edgy ferry]

i dont understand how to solve the problem

[edgy ferry]

no info about the curve itself is given

[bright bronze]

well, do you really need the curve? All you need is O, A and B, all of which are known and do not depend on f(x) at all

[bright bronze]

draw a figure and you should be able to how you need to proceed

[edgy ferry]

so a trianlge

passing throught 3,4

so how do i use the info that the sub tan and sub norm is equal

i didnt rlly get how to put those in the figure

[bright bronze]

do you know what a subtangent and subnormal is?

[edgy ferry]

the x axis projection of the tangent and the normal

[bright bronze]

yes, and your problem tells you that those two are equal

that tells you something very specific

[edgy ferry]

so they are at an angle of 45 from the origin?

then only they will have the same projection right?

[bright bronze]

yea

[edgy ferry]

uh ok so

dang

[bright bronze]

well, at this point, you have two choices for the tangent

[edgy ferry]

an isoceles triangle

[onyx glen]

wait what are those

[edgy ferry]

with its tip as the point 3,4 is what im thinking'

[bright bronze]

lengths of projections on x-axis, between the point on curve to the point of intersection with the axis

[bright bronze]

not really, (3,4) is not A or B or O

and those three are the vertices

[edgy ferry]

at that point we know that

im thinking we take 2 of these triangles

add em to get the AOB triangle

like aob is right angled,

[bright bronze]

well, that was always gonna be the case, considering x axis and y axis form the two sides of triangle AOB

isosceles right triangle is more relevant here

[edgy ferry]

ye 😭

[edgy ferry]

why will the x and y coordinates be the same

[bright bronze]

coz the slope of line AB is gonna be 45deg

thats why I said draw a figure

[edgy ferry]

oh

so

uh ye its isocosseles\

how do we find max area now

[bright bronze]

well, you have two choices about what the tangent is gonna be, one is with positive slope, the other with negative slope

[vale sphinx]

Man i dont even know simple math im here to lock in

[bright bronze]

calculate both the areas and see which one is larger

[edgy ferry]

????

uhh

[bright bronze]

did you draw a diagram yet???

[edgy ferry]

yes vro

mb not tryna ragebait

im js not following a bit here

[bright bronze]

well, in your hypothetical diagram, what you took to be a tangent and a normal can be swapped too

[edgy ferry]

yes thats true

[bright bronze]

so that gives you the other choice of OAB

[edgy ferry]

ull get more area only if its having -ve slope

[bright bronze]

yes

[edgy ferry]

but the point thats 3,4

[bright bronze]

and??

[edgy ferry]

is the midpoint of the isoceles right angle?

[bright bronze]

doesnt have to be

wait, what does that even mean?

[edgy ferry]

so it lies somewhere along that line

[bright bronze]

whats a midpoint of a triangle

[edgy ferry]

not the triangle, im talking about the hypotenuse

[bright bronze]

yea, well, it doesnt have to be

and it wont be

[edgy ferry]

then how do i find the area 😭

[bright bronze]

just find the intercepts of the line thats the hypotenus???

[edgy ferry]

uhh

[edgy ferry]

how do i do that

[bright bronze]

whats the equation of the line?

[edgy ferry]

its gonna be the same cuz its a isoceles triangle anyway right

[edgy ferry]

ohhhh

i see

1 sec

[bright bronze]

yea

[edgy ferry]

its x and y intercepts are both 7

so area is 49/2

[bright bronze]

yep

[edgy ferry]

i got the ans but it seems like the idea of using the line eq would not pop in my head normally

does the question hint at using the eq of line somewhere?

[bright bronze]

you dont really need to tbf

[edgy ferry]

in exam id 100% get grilled

[edgy ferry]

oh is there another approach

[bright bronze]

if you draw the diagram you can see it, you dont have to calculate

[edgy ferry]

like just pulling the point 3.4 along x and y axis?

to get both as 7

[bright bronze]

well, you see that the line is gonna be 45deg

[edgy ferry]

i understand

dang

[bright bronze]

so equation is of the form x+y or x-y

so intercepts are gonna be 3+4=7

[edgy ferry]

yeah

[bright bronze]

you dont need to rigorously find the equation and all that, it all gotta flash in your head like that, and then ofc you can write the solution on paper for your exam step-by-step

[edgy ferry]

i understood

tysm vro

[bright bronze]

np

[edgy ferry]

mb for the ragebait 📿

aight then ty

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[hoary ember]

$ABC$ is a triangle, with $M$ being the midpoint of $AB$. Point $D$ lies on the arc of $AB$ that does not contain $C$ such that $BD$ is tangent to the circumcircle of $BCM$. Point $E$ lies on $AD$ such that $BE$ is parallel to $AC$. Point $F$ lies on $CE$ such that $\angle EAC=\angle EFA$. Prove that $FM$, $CA$, $BD$ are concurrent

[glossy valve]

ihave<skissue>

[hoary ember]

i genuinely have no idea where yo start with this

[void plume]

holy shit... is this what geometry has become

uhh.... im not expert but... probably some geometry mumbo jumbo that results in the points all being on one line

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@hoary ember Has your question been resolved?

[unique perch]

I think by angle chasing you should be able to see that:
A, M, F, C are concyclic
E, D, B, M, F are concyclic

[hoary ember]

oh wait is the idea like that one thing about the 3 radical axis of 3 circles are concurrent or smth

[unique perch]

From there the concurrent point is just the radical center of the 3 circles (AMC), (BMD) and (ABC)

yea you got the idea

[hoary ember]

mm lemme think on proving this

[unique perch]

i think the latter might be a lil tricky but havent tried

[hoary ember]

wtf am i this ass at geo

i cant prove either one

[unique perch]

huh

ok might start with A, M, F, C

since we dont know the property of MF yet it's better to prove MAF = MCF

nvm AMC = AFC is much easier

[twilit leaf]

idea: it suffices to show DF||AC

not sure if thats easier

[unique perch]

umm they arent parallel

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@hoary ember Has your question been resolved?

[twilit leaf]

damn

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[turbid tide]

Can someone explain how to do these? I don't understand how to figure out the behavior of f using f'

[thorny bluff]

what do "increasing" and "decreasing" mean?

do you know?

[turbid tide]

When the slope is getting positive or negative?

[thorny bluff]

which means when the derivative is positive or negative, right?

because derivative = instantaneous slope

[turbid tide]

So for example the first one is increasing (-3,-1.5) and (1,3) ?

[thorny bluff]

no, because the problem gives you the graph of f'(x), not f(x)

you're seeing when f'(x) (the derivative of f(x)) is positive or negative

[turbid tide]

So it's increasing (-2,-1) and (2,3) because those intervals are postive?

[thorny bluff]

yep

or (2, infinity), because the graph continues

[turbid tide]

What about for maxima and minima like this problem?

[thorny bluff]

for maxima and minima, you know f'(x) = 0

the way you tell max/min apart from each other is by seeing if f'(x) goes from positive to negative, or, negative to positive

take x=-1 for example

f'(-1) = 0, so it is either a max or a min

because f'(x) goes from positive (before x=-1) to negative (after x=-1), it is a maxima

now you do the same for the other times when f'(x) = 0

[turbid tide]

Okay i got
X=-1 max
X=0 min
X=1 max
Im not sure about -2 and 2

[thorny bluff]

-2 and 2 are negative to positive

the graph kind of cuts off so its hard to tell

[turbid tide]

Hmm I see

Thank you that was very helpful

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[dim gyro]

this is from an old book

[dim gyro]

is this norm on the LHS and abs value on the RHS?

[gusty flicker]

what even is $x$, i need more context

[glossy valve]

Sunny — ping me plz

[full birch]

seems like it is 1-1

but maybe x is something else

[gusty flicker]

like the natural assumption is to assume $x$ is real, in which case i dont know what you're talking about with norm

[dim gyro]

its from the spivaks calculus on manifolds book

[full birch]

you might wanna show us the whole picture of the question

[dim gyro]

that is the question

[full birch]

does the question directly start with 1-1 and a fullstop?

[dim gyro]

thats the context. this is a book from the 70s

because is this is an abs value on the RHS this makes sense

[gusty flicker]

is $x^i$ notation for like the $i$-th term?

[glossy valve]

Sunny — ping me plz

[full birch]

it seems more of a condition for every term from i = 1 summation till n.

that x to the power of i is the condition for every term that we get from the summation function

[full birch]

and we prove that x is less than Or equal to it

[gusty flicker]

this would make sense, and in which case, yeah, it would work

[leaden ermine]

so the length of a vector is always less or equal to the sum of its absolute components, interesting

[full birch]

It would

[dim gyro]

it makes sense, its how i read it but it took me 20 minutes to come to that conflusion

[gusty flicker]

since this is manifolds, i dont think this is necessarily "length of a vector"

[gusty flicker]

love the accidental typo

[leaden ermine]

pardon me, i should have used norm

[dim gyro]

my kid dropped a soda thats jammed up my keyboard :P'

[gusty flicker]

i'll now officially use "conflusion" as a portmanteau of confusion and conclusion

[full birch]

haha.

[dim gyro]

okay yeah this is easy to prove then.... take the square of each and move down

[full birch]

Mhm

[dim gyro]

Thanks all!

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[full forum]

@lean rock Has your question been resolved?

[lean rock]

or for codiagram?
A pushout of a copair of C-coarrows with a cocommon domain is a colimit in co-C co-for co-the codiagram!

[lean rock]

this seems. a bit tricky:

for pushout which arrows pushout out along which arrow?

this is the original:

c--g-->b--f'-->d
c--f-->a--g'-->d

f' is pushed out along g? so f arises?
g' is pushed out along f so g arises?

or
f' is obtained by pushing out f along g
g' is obtained by pushing out g along f

        g             
   c ───────▶ b 
   │          │
 f │          │ g′
   ▼          ▼
   a ───────▶ d
       f'

g' is the pushout of g along f
f' is the pushout of f along g

⧈

Oh I didn't reverse the arrows,, the diagram seems the same diagram as that of pullback

[lean rock]

I think it's co-qed

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[glass gull]

Can someone help me with qn 5 (IV) and (v)

[frigid carbon]

What have you tried?

[errant tiger]

use d(u/v) = (vu' - uv')/ v²

[glass gull]

Hmm alright thankss

[wide sundial]

think of f/g as fg^-1

and then use product rule

[void nova]

Assuming they know how to diff g^-1 😅

I mean, I believe they've learnt chain rule so maybe yeah

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[dry eagle]

Can someone help me understand maths from absolute scratch based on complete logic.

[lime trellis]

no

[gusty flicker]

please be more specific — what do you really want

[dry eagle]

Allow me to rephrase

I'm looking for guidance on how to learn mathematics from the absolute basics with a logic-first approach. I need recommendations for books, the sequence to follow, and how to study, obviously I don't want to be spoon-fed, just direction.

[gusty flicker]

right, that's much better than "Can someone help me understand maths"

uh

I'm not sure actually — and I would ask the same question

[dry eagle]

Yeah, I agree. It was sloppy english lacking intent. One could clearly tell I myself had no idea what I was talking about.

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@dry eagle Has your question been resolved?

[solar bison]

Go on youtube as for us we went though primary to high school and this varies from country to country so i think going to youtube or even on reddit or stack exchange past posts u will get your answer

[dry eagle]

I'm in high school. But math sucks here. Half-ass explanations and stuff.
I wanna learn it properly.

[solar bison]

It is high school if u go to univ u will understand in more-detail assuming u will be doing maths at uni.

[dry eagle]

I want to learn it from basics.
Like High-school doesn't teach maths correctly. At least my school doesn't for sure.

[solar bison]

Did u refer to youtube beucase they some famous mathematician there

[dry eagle]

Could you tell me the name of that mathematician?

[solar bison]

https://youtu.be/pTnEG_WGd2Q?si=-8AZgfSJwRLEz2CV

[solar bison]

The Maths Sorcerer

[dry eagle]

Thanks a lot Beluga.
Or Mr.Beluga.

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[blissful fox]

the cost of two pies is the same as the cost of 3 patties. one pie and one patty tigether cost 3.50. what do they each cost

[frigid carbon]

Cost of pie = x
Cost of patty = y

Form equations

[blissful fox]

i dont understand it

[frigid carbon]

Assume cost of one pie is x

Cost of 2 pies = 2x

You follow?

[blissful fox]

yes

[frigid carbon]

K

Assume cost of one patty is y

Cost of 3 patties = 3y

[blissful fox]

ok

cost of 2 pie = 2x

[frigid carbon]

Now we are said, cost of 2pies = cost of 3 patties

=> 2x = 3y

This is first equation

You follow?

[blissful fox]

yes

[frigid carbon]

K, now cost one pie + one patty = 3.5

What does it imply?

[blissful fox]

1x + 1y = 3.5

right

[frigid carbon]

From 2x = 3y
=>can write x =(3/2)y

Substitute this in other equation

[blissful fox]

ok so

[frigid carbon]

Go on

[blissful fox]

u have to subtract 2x = 3y from another equation

[frigid carbon]

You prefer subtracting?

[blissful fox]

well when the 2 signs are the same u subtract right

and when the 2 signs are different u add'

[frigid carbon]

Yes but first need to make coefficient equal

Write it as
2x - 3y = 0
x + y = 3.5

Can you make coefficient of x in second equation sane as 1st?

[blissful fox]

yes

you multiply by 2

[frigid carbon]

Multiply then subtract two equation

[blissful fox]

so 2x + 2y = 7.00

[frigid carbon]

Now subtract 2x - 3y= 0

[blissful fox]

0x + -1y =7.00

[frigid carbon]

-1y ?
When you subtract 2y - (-3y)

It becomes 2y + 3y

[blissful fox]

oh

[frigid carbon]

What does it become

5y =?

[blissful fox]

yes

5y = 7.00

[frigid carbon]

Now y = 7/5

= 1.4

Substitute y in any equation

x + 1.4 = 3.5
x = 3.5 - 1.4

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@blissful fox Has your question been resolved?

[blissful fox]

ok i got it

ty

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[azure junco]

am i right?..

[azure junco]

btw forgot the parenthesis

[merry seal]

Is it 37*(2/5):3 in the first steps denominator?

[azure junco]

yes

[merry seal]

(2/5):3 can be written as 2/5/3 or 2/3*5 or 2/15

[azure junco]

. is multiply

[merry seal]

: is ratio right

And yes . Is multiply so you multiply 37 by 2/15

Results 74/15 in the denominator

[azure junco]

yes

5 . (x+0,28) /// 74/15

right?

??

[dense lantern]

why 5 * (x + 0.28)

that 5 was from ur previous fraction which was incorrect

it should just be the
(original numerator) / (74/15)

[azure junco]

x+0,14 . 2 / 74/15

?

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@azure junco Has your question been resolved?

[merry seal]

wait

[merry seal]

= (x + 0.14 x 2) / (74/15)

= (x + 0.28) / (74/15)

> 15 moves up to the denominator, same logic as the one you used in your original solution
= 15*(x + 0.28) / 74

= (15x + 4.2) / 74
I'm guessing this is sufficient as the final answer

[full forum]

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[hallow walrus]

Is there a methodical way of finding the expected number of coin tosses before seeing some generic pattern of heads and tails?

[hallow walrus]

The methods for a specific pattern depend on some casework and that doesn't really seem to generalise well to the generic case

[lament mango]

well u can use bernoulli trials

[hallow walrus]

well sure, each coin toss is a bernoulli trial

[fast peak]

set up the markov chain

I mean sure its a bit of casework but its methodic

[hallow walrus]

I don't really know what markov chains are

[fast peak]

unlucky

[hallow walrus]

what

[lament mango]

but here it is an independent event, isnt it?

[fast peak]

#help-6 message

see for example this discussion

to maybe get an idea

[hallow walrus]

ohh I see

that's actually pretty intuitive

yo markov chains are sick

[lament mango]

ikrrrr

i just posted a question on it

#help-12

[hallow walrus]

.close

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[clear wyvern]

nthg is working

im using the central difference

formula

and getting -1.8

[rough chasm]

thats correct no?

only the sign is flipped

should be 1.8

can you show me what you did maybe i can spot where you messed up

[clear wyvern]

oops i did put 1.8

and its still wrong

[full forum]

@clear wyvern Has your question been resolved?

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[robust slate]

discord.gg/chemistry

[shrewd holly]

.close

[full forum]

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[shrewd holly]

.reopen

That chemistry server is dead

I’m just trying to clarify something with chemistry although this is a math server, should it be Na3PO3 + H20 for the products

Pls help

[brittle sun]

Yeah

[shrewd holly]

When I ask ChatGPT it said NaHPO3 that’s wrong?

[torn sand]

!nogpt

[full forum]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[shrewd holly]

Ok

.close

.close

[torn sand]

this isn't balanced though

i dont think

[shrewd holly]

Ik

Tahts what im gonna do after

[torn sand]

okay

[shrewd holly]

But I’m just not sure if it’ Na3PO3

Wait I got it nvm

.close

.reopen

.close

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[spring vigil]

howd they make

1/ 6 - x to

1/ 5 - (x- 1)

so i see the (x- 1) part because they want the C = 1

but what happened to the 6 howd it become 5

[slate violet]

5 - x + 1

[spring vigil]

?

[unreal wren]

The - sign distributes

Into -x + 1

[spring vigil]

so it means if we try to simplify we go back to 6 - x

.close

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[torn jolt]

can anyone help

[gritty rose]

There are n choose r ways of selecting r things from n objects

1. is by definition a simple n choose r with n and r given

[torn jolt]

i don’t understand at all

[gritty rose]

Which word confuses you

[torn jolt]

im trying to relate permutations and stuff

[torn jolt]

its that i try to build on previous knowledge

[gritty rose]

Then don't?

[gritty rose]

This is all you need

0 <= r <= n of course

[torn jolt]

but arent you supposed to connect this with things about permutation counting

[gritty rose]

Not if you keep getting stuck

[torn jolt]

can you explain it though

im stuck cause i dont know how to make the connection

[gritty rose]

.

[gritty rose]

.

[torn jolt]

but i want to

because thats how you understand how it works

[gritty rose]

Then do problems that relate them. Your problem doesn't

[torn jolt]

okay can i ask what the intuition is behind r choose n

The reason im attempting to relate it is cause i don’t understand the intuition im trying to figure it out

[lime ether]

do you understand permutations?

[torn jolt]

yes

[lime ether]

ok so perhaps you should think of it in terms of that

[gritty rose]

Read https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Applied_Discrete_Structures_(Doerr_and_Levasseur)/02%3A_Combinatorics/2.01%3A_Basic_Counting_Techniques_-_The_Rule_of_Products

And https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Applied_Discrete_Structures_(Doerr_and_Levasseur)/02%3A_Combinatorics/2.02%3A_Permutations

[torn jolt]

🤔

[gritty rose]

And https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Applied_Discrete_Structures_(Doerr_and_Levasseur)/02%3A_Combinatorics/2.04%3A_Combinations_and_the_Binomial_Theorem

[torn jolt]

I understand product rule

and permutations

[lime ether]

ok so divide the number of permutations by r! because there are r! different orderings of the objects you select

[torn jolt]

how would that look like here

We are looking for how many 4 subsets there is right

[lime ether]

so let’s label the people first

[torn jolt]

Okay

[lime ether]

say $a_1, a_2, \dots, a_{12}$

[glossy valve]

knief

[lime ether]

we have 12 choices for the first person to pick, then 11 for the second, then 10 for the third, and 9 choices for the fourth. however, within this group of 4 people we just chose we can shuffle them 4! different ways. so we could’ve for example chose the first guy second and the second guy third and the third guy first and the fourth guy stays the same etc. hence we will overcount and get a count of 4! ways corresponding to this particular group of 4 people. this is true for every group of 4 people we choose as well hence why we divide the total count 12 * 11 * 10 * 9 by 4!

$a_1a_2a_2a_4, a_1a_4a_2a_3, a_2a_3a_4a_1, \dots$ all correspond to the same group

[glossy valve]

knief

[lime ether]

this is true for every group

so we have an extra factor of 4! in our count for the number of groups

note that this matters because their positions in the group are indistinguishable, it’s not like one person is the president and another is the vice president and another is a secretary and the last is a treasurer or something

[lime ether]

these are all considered the same

whereas if we had different "positions" like president, vice president, etc. then the orderings within the group would matter

does this make sense?

maybe you should consider a smaller example where it’s easier to list out the permutations

[torn jolt]

then youd look at the number of 4 permutations of the set right

[lime ether]

yea that would be a permutation problem

[torn jolt]

But you’re saying that because order doesnt matter you divide by 4!

[lime ether]

yes

[torn jolt]

but why does that work like that

[lime ether]

because each group we chose actually gives us 24 different ways when we count like 12 * 11 * 10 * 9

[lime ether]

i just explained it

how about we use a smaller number

[torn jolt]

Okay

[lime ether]

let’s say i have 4 students a, b, c, and d and i want to know the number of ways i can make a group of 2 students. so i have 4 choices for the first and 3 choices for the second giving me 12 total permutations
ab, ba, ac, ca, ad, da, bc, cb, bd, db, cd, dc. now we treat the groups ab and ba as the same since they have the same students and likewise we treat any permutations with the same two letters/students as identical. now for each group of two distinct students there is 2! =2 ways to order them so each individual group gets counted 2! ways in my original count of 12 ways to form the group of 2 students from 4. hence i have 2! times the amount of "groups" i’m looking for in my count so to correct for this i divide my original count by 2!

[torn jolt]

and what does all this have to do with mapping permutations to subsets?

because thats what theyve been talkin about too

[lime ether]

if i select r objects then there are r! permutations of those objects that map to a particular subset consisting of distinct elements

so my count has r! times the amount of ways i want

so we divide by r! to get the subsets that don’t care about the ways i can order those r folks

[torn jolt]

wdym by r objects

[lime ether]

how is that possibly confusing

r is just a variable

n choose r

i was relating it back to that

in the general case

[torn jolt]

i think the frustration is making this confusing so ill give it some time

[lime ether]

are you in uni?

[full forum]

@torn jolt Has your question been resolved?

[torn jolt]

you could say that

[lime ether]

whats your major?

[torn jolt]

Undecided yet

[lime ether]

any plans?

[torn jolt]

thinking of STEM

wbu @lime ether

[lime ether]

im a math major

[torn jolt]

niceee

can i go back to our question real quic

i though about it for a little

[lime ether]

sure

[torn jolt]

why do we divide by r! I dont get that part

[lime ether]

because counting the permutations gives r! times the amount of subsets of distinct elements where order doesn't matter

[lime ether]

see here

[sacred yarrow]

cs’s a good choice

a bit off topic

[torn jolt]

oh nah too competitive and market is going down

also they require like a 3.9 GPA or something

[full forum]

@torn jolt Has your question been resolved?

[full forum]

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[near storm]

hi

[near storm]

im trying to solve t^2y'' + 2ty' - 6y = 0

using this sub: t = e^x

so y(t) = y(e^x)

y' = dy/dt = dy/dx . dx/dt = dy/dx . (1/t)

y'' = dy'/dt = d( dy/dx . (1/t) )/dt

so y'' = -1/t^2 . (dy/dx)

1/t . d(dy/dx)/dt

so what is the last term?

d(dy/dx)/dt

is it y'.(dy/dx)?

.close

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[signal solar]

not sure how to get the derivative of 10e^5t

[spice knot]

you here?

[signal solar]

yeah

[spice knot]

can you find the derivative of 10e^t?

[signal solar]

well i guess thats just a constant times e^t

so 10e^t

[spice knot]

alr

and whats the derivative of 5t

[signal solar]

5

[spice knot]

now have you learned chain rule?

[signal solar]

yep

[spice knot]

can you try it?

[signal solar]

oh sorry i know how to do that

sorry i went afk

i meant to say idk how to get the antiderivative of that

[spice knot]

can you find the integral of 10e^t?

[signal solar]

im not sure

[spice knot]

ok think about this

the antiderivative should be the opposite of the derivative, right

[signal solar]

yeah

[spice knot]

and you said the derivative of 10e^t is 10e^t again

so why cant the antiderivative of 10e^t just be 10e^t again?

[signal solar]

i guess it is right

[spice knot]

well it isnt

you forgot about the +C

but other than that its right

the antiderivative of 10e^t is 10e^t + C

in general the antiderivative of f'(x) is f(x) + C

[signal solar]

yeah

[spice knot]

do you know where the +C comes from?

[signal solar]

yeah

[spice knot]

alr thats good

now we need to think about finding the antiderivative of 10e^5t

to do this, we need a reverse version of the chain rule

a common example is called u-substitution

have you learned u-sub yet?

[signal solar]

yeah

is trig sub another as well and the others

[spice knot]

so with 10e^5t

what do you think should be the u?

[signal solar]

5t

[spice knot]

alr,

so can you do the antiderivative now

[signal solar]

so it would just be (10e^u)*1/5

so basically 2*e^u

im kinda stuck on how you would rearrange this

because each term is cubed

[spice knot]

back
you need to write down the full integral as 2e^5t + C

[spice knot]

its inside of a ln, you dont see any ideas with that?

[signal solar]

log properties but since its an infinite sum im not sure

like could u do 3^n times ln (1+1/n) +etc?

idk

[spice knot]

thats not how you pull 3 out of this

what does it look like instead after you pull the 3 out?

[signal solar]

how else can u pull 3 out?

[spice knot]

this is not at all correct

think about ln((xy)^3) = ln(x^3 y^3)

thesea re the same thing because (xy)^3 = x^3 y^3

but the rule you stated thinks 3 ln(xy) = 9 ln(xy) instead

review the logarithm rules more closely to confirm this

[signal solar]

yeah

[spice knot]

ln(x^3 y^3) = ln(x^3) + ln(y^3) = 3(ln(x) + ln(y)) = 3 ln(xy)

[spice knot]

so here, what would the result be instead

[signal solar]

why does the ln y term not have a 3

[spice knot]

nino

look at it again

I inserted some ()s around the ln(x) + ln(y) while you werent looking

without even editing the post (wow!)

[signal solar]

XD

[spice knot]

now go back to this problem and factor the 3 out, what do you get

[signal solar]

i still dont feel like i fully understand how to apply it

[spice knot]

ok lets practice

first, you need to remember two rules

,,(abcde)^3=a^3b^3c^3d^3e^3

[glossy valve]

mtt

[spice knot]

,,\ln(x^3)=3\ln x

[glossy valve]

mtt

[spice knot]

at a minimum you only need these two rules to factor the 3 out of the expression

do you know both of these rules?

[signal solar]

yes

[spice knot]

second,

[spice knot]

do you know what a limit actually does?

[signal solar]

yes

[spice knot]

then is the sum practically infinite?

or actually infinite?

[signal solar]

im not sure what that like means but the 1/n terms will approach 0

[spice knot]

ok no thats not what Im asking about

you didnt want to pull out the 3 because the product or sum "looked infinite"

the purpose of a limit is to rewrite down an otherwise infinite thing as a pattern of finite things

so any time you do any algebra on this

n is finite

if you attach this on the left

n is still finite

n is not equal to infinity

so the sum is not infinite

the sum is always finite

[spice knot]

the only difference that a limit makes is that we are asking for what pattern this goes to

as n goes to infinity

[spice knot]

the way you asked this made it sound like you couldnt pull the 3 out, because then you would need to be considering an infinite sum, which is undefined

because we are using a limit, the sum is not undefined, it is finite

[signal solar]

but without the limit it'd be undefined?

[spice knot]

you know I have to wait for your response right?

you can afford to type faster

its been 40 minutes already

[signal solar]

i know i just try to solve it but i dont rly know

[spice knot]

pump out some fast responses, I dont need thought-out theses

first

[signal solar]

ok

[spice knot]

finite or infinite n?

[signal solar]

finite

[spice knot]

good

second,

[spice knot]

this works for finite products, right

[signal solar]

yes

[spice knot]

now

[spice knot]

look at this

finite or infinite terms?

the limit is cut off

quick yes or no for me

[signal solar]

infinite?

[spice knot]

not correct

n is finite

there are exactly n terms

the first term is (1 + 1/n)^3, the last term is (1 + n/n)^3

[signal solar]

but if n is infinite?

[spice knot]

n is never infinite

the purpose of a limit is to avoid infinity by describing it in a very particular way

at all times, n is finite

[signal solar]

ok

[spice knot]

we merely have n approach infinity

we do not have n be infinity

[spice knot]

n is not infinity

yes?

[signal solar]

ok yes

[spice knot]

there are n terms, a finite number of terms

[signal solar]

but how would you use this for n terms

[spice knot]

now go use this on the finite product

[signal solar]

(abcde..n)^3=a^3 etc ..n^3?

[spice knot]

yea nino

these are basic exponent laws

are you expecting something strange?

we're still working with finite terms

[signal solar]

no its just like a lot of terms

[spice knot]

would having a lot of terms change the rule?

[signal solar]

nope

[spice knot]

so having a lot of terms doesnt change the rule

look how quickly we're finding and fixing the sources of confusion here

we're on good pace, now go try the n-term version of this on the finite terms

theres a quick way you can type the math
go to https://desmos.com/calculator and type in what you want
then copy the line of math and paste it here

[signal solar]

a^{3}b^{3}c^{3}d^{3}e^{3}...n^{3}

does that work

i guess

[spice knot]

thats latex, thats the language desmos and the discord bot texit uses

[signal solar]

dont u use some text thing to do it faster

[spice knot]

to get texit to "render" or "display" this latex code, you type ,, before it

,,a^{3}b^{3}c^{3}d^{3}e^{3}...n^{3}

[signal solar]

ah

[glossy valve]

mtt

[spice knot]

see here it rendered the stuff right, it should look similar to desmos's version

desmos displays things slightly differently but the font should otherwise be the same

[spice knot]

wdym by this

[signal solar]

nothing i just meant that

u need the bot or other things

typst

anyways

[signal solar]

this has a lot of terms ,,\left(a+b\right)^{3}\left(c+d\right)^{3}

,,\left(a+b\right)^{3}\left(c+d\right)^{3}

[glossy valve]

nino

[signal solar]

how can a bot be female

lmao

[spice knot]

strange how youre seeing that, I dont see the red diamond thing anymore

oh wait they hid it lol

maybe its like referring to it like a ship or a car

[signal solar]

i guess

[spice knot]

keep in mind you can place anything here in place of a

that includes something like (1 + 1/n)

a = (1 + 1/n)
b = (1 + 2/n)
etc.

[signal solar]

OHHH

ok if you do that then you can pull it out

[spice knot]

yea I didnt want to spoil it but there it is

yes you do it like this

then theres only one ^3 that you use ln(x^3) = 3 ln(x) on

[signal solar]

right

[spice knot]

its the same thing I showed earlier with ln(x^3 y^3) = ln((xy)^3) = 3 ln(xy)

[signal solar]

yes

,,\frac{3}{n}\ln\left(\left(1+\frac{1}{n}\right)...\left(1+\frac{n}{n}\right)\right)

[glossy valve]

nino

[signal solar]

that would pretty much just be 1 inside right

after applying the limit

or 2 actually

[spice knot]

try not to take a guess on what it is for now

its going to be hard to predict

alr keep in mind they want you to write this as a riemann sum

now are you familiar with those

[signal solar]

mildly

[spice knot]

in short, you have that:

[signal solar]

will the inside of ln be the riemman sum

[spice knot]

not exactly

you need a sum

right now we have a product

[signal solar]

can i turn it into a sum

[spice knot]

what idea do you have

[signal solar]

nothing

none

[spice knot]

as before,

you have ln(abcde...n)

now this time theres a different logarithm law you can use

what logarithm law turns this into a sum?

[signal solar]

oh right

yes

but what im confused about is how to write that as a riemman sum

once you have it as a sum of logs

[spice knot]

write down the sum first

[signal solar]

ln(1+1/n)+..ln(1+n/n)

[spice knot]

alr

so now we have:

now a riemann sum tells you that an integral can be seen as the limit of a sum

this would be called a "right riemann sum"

the idea is that if you want to integrate something from 0 to 1,

you can draw n rectangles that mostly cover the area you want

these are each thin rectangles with width 1/n, so that putting n of them together looks something like this (this picture has n=4, and youll need to ignore the x-axis for this one)

you can see here that the height of the rectangles has to use the f(x) from y = f(x) to go up to the correct height

so for n rectangles,

you can imagine them going f(1/n), f(2/n), ..., f(n/n)

the heights would then line up with how high the function is at

[signal solar]

are integrals always right riemman sums?

[spice knot]

they can be left riemann sums too

or middle riemann sums

right riemann sums are just one way of distributing and setting up the rectangles

the integral also doesnt have to always be from 0 to 1

[spice knot]

you can see in the image I chose that this one is a right riemann sum with n=4
but its going from 0 to 2 instead

still the same idea

in some cases, theyll tell you the rectangles dont even need to be of equal widths

the general idea is still that, the more rectangles you use, the thinner and better they can be at covering or representing the area under the curve

and so as n -> ∞, the area of the rectangles -> the area under the curve

for a right riemann sum from 0 to 1,

each rectangle has width 1/n and height f(k/n)

so adding them up gets you:

the term inside the limit is a right riemann sum with n rectangles as said earlier

as n -> ∞, this area will approach the integral, or the area under the curve

in fact an integral can be defined this way, that is why it is called a "riemann integral"

its just that riemann sums are very hard to work with, so youre better off doing antiderivatives to find their value

what do you think about this so far

[signal solar]

i understand the general sense

[signal solar]

i dont fully understand how these are equal

[spice knot]

you can view the left side as being more like this

each term is the area of a rectangle

n terms for n rectangles, width 1/n, height f(k/n)

[signal solar]

oh ok

1/n could stand for dx here right

ok that makes sense

[spice knot]

heres a picture

this is an example with n = 10

you can see here the areas would be calculated as:
0.1 * f(0.1)

• 0.1 * f(0.2)
• ...
• 0.1 * f(1)

does that calculation make sense

[signal solar]

yes

[spice knot]

so in general for n rectangles, this is what the area calculation would be

[spice knot]

factor out the n and you get this

[signal solar]

yup

[spice knot]

alr so in short, you have an intuitive reason to believe this now, right

[signal solar]

yes

[spice knot]

now lets look at this

now to make this slightly easier, Im going to move the 3 out

alr now look at the kind of sum we're adding together

do you think you can guess at what f(x) is here?

[signal solar]

ln(1+k/n)

[spice knot]

youre close, but remember f(x) is a function of x instead of k/n

ln(1 + what instead)?

[signal solar]

x_k?

[void nova]

There can't be any k or n in the expression of f(x)

It's more straightforward

[spice knot]

for example f(x) = x_k^2 doesnt make any sense

we need f(x) = x^2 or something like that instead

Im asking for a function of x here

the reason for that is because choosing the letter x doesnt really matter

f(x) = x^2 and f(y) = y^2 both represent the same function

they represent the act of squaring something

so what I want you to write down is whats being done to the k/n

[void nova]

f(1/n) = ln(1 + 1/n)

f(2/n) = ln(1 + 2/n)

f(3/n) = ln(1 + 3/n)

f(4/n) = ln(1 + 4/n)

f(5/n) = ln(1 + 5/n)

f(x) = ??

@signal solar

[spice knot]

each time the same action is being done

[signal solar]

yeah im choosing a different letter i meant for representing it as a sum so you'd have the sum of k=1 to n of that times 1/n

[spice knot]

youre not supposed to use sum notation without defining the sum first

you need to say that x_k = k/n first

otherwise youre not saying it right

you also need to say what f() is

[void nova]

Hope this helps @signal solar

[signal solar]

im not sure what you mean alberto

f(x)= ln(1+1/n)+...ln(1+n/n)

[void nova]

I don't know how to dumb it down more than that without giving the solution

[void nova]

No no. ...

[spice knot]

it sounds like you want to try helping nino from here on out

I need to go do things, cya

[signal solar]

cya

[void nova]

Can't you see a pattern here?

I even used bold font

[signal solar]

f(x) = (1+x/n)

[void nova]

Let me add more terms

[signal solar]

f(x) = ln(1+x)

[void nova]

#help-28 message

[void nova]

Yess!! There you go

[void nova]

So now this is equal to which integral?

[void nova]

Recalling this

[signal solar]

integral of ln(1+x)*1/x

[void nova]

Why the 1/x??

There's only written the integral from 0 to 1 of f(x)

[void nova]

And f(x) was this, without a 1/x

[signal solar]

1/x is dx

[void nova]

Who said that?

Wait do you know how to calculate integrals??

[signal solar]

yeah i do im just funky on that

ok lets say i have integral of ln(1+x)dx

what do i make the bounds

[void nova]

Yeah you indeed have this (don't forget the 3 though)

[signal solar]

the 3 is outside

[void nova]

Read #help-28 message again...

[void nova]

Sure

[signal solar]

why does it have to be 0 to 1

[void nova]

Because that's how right Riemann sums work, with this setup
I believe Mtt has already explained it

[signal solar]

with riemman sum you can have a to b

[void nova]

I give up, sorry

[signal solar]

i mean sure u basically already made the other guy give up too

@void nova

,,3\sum_{k=1}^{n}\left(1+\frac{k}{n}\right)\cdot\left(\frac{1}{n}\right)

[glossy valve]

nino

[signal solar]

this is what i have

forgot the ln

,,3\sum_{k=1}^{n}\ln\left(1+\frac{k}{n}\right)\cdot\left(\frac{1}{n}\right)

[glossy valve]

nino

[signal solar]

the interval should be from 1 to 2

not 0 to 1

it literally goes from (1+ approx 0) to (1+ approx 1)

i dont understand why you would just pent out ur frustration at whatever ur method is and then take charge over the other helper to just also leave urself

I'm tired of this like I actually had a really cool person

.close

[full forum]

Channel closed

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Use .reopen if this was a mistake.

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[narrow mica]

idk where to start here especially since none of these are necessarily integers either

[narrow mica]

the observation i have is that x, y, z can all be swapped with each other

idk if that makes sense but

[onyx glen]

add them all together

[narrow mica]

oh yeah

and then you get (x+y+z)^2 right

well not exactly that

you'd have a constant

yeah ok

[narrow mica]

that wasnt obvious to me

for some reason

[onyx glen]

you get that=9

[narrow mica]

yeah ok

thanks

im not used to (a+b+...)^2 im usually used to seeing (a+b)^2 still

idk why

ok let me do that now

so basically all the solutions are x+y+z = 3 or -3

right

[delicate torrent]

Yep

Now you gotta think what to do with that 💀

[delicate torrent]

One choice is that from this equation, you substitute

[narrow mica]

yeah

i was thinking this too

you could substitute x as (3-y-z) or (-3-y-z)

[delicate torrent]

Well, try it to see if it works

[narrow mica]

ok

[delicate torrent]

You need to substitute in 2 of the equations and check if the 3rd one satisfies

The check is important because your values you get might not satisfy the 3rd eq @narrow mica

[narrow mica]

ok i got this

doesn't look super nice to try and simplify

[narrow mica]

yeah

[narrow mica]

this is lit just the first equation 💀

[narrow mica]

this q is gonna take me a while but can i simplify this

beyond just doing something like -5 = (y+z)^2 + 2yz - 6(y+z)

so -5 = (y+z-6)(y+z) + 2yz

idk what to do with the 2yz

maybe i made a mistake expanding (3-y-z)^2

idk

[delicate torrent]

Yeah

[delicate torrent]

Not looking so nice

I was going to do a quad on y but i don't think that's so nice

[delicate torrent]

Try making a quadratic of y

[keen vector]

yo what u typing in #help-8 ?

[delicate torrent]

Your name suits you so well, freak

[keen vector]

what

it's freaky, not freak.

[delicate torrent]

That's what you are

[delicate torrent]

Now if you can't help then get out of the channel

[keen vector]

i don't think you know what that word means

kid

[narrow mica]

wdym

[delicate torrent]

Like, Ay^2 + By + C = 0