help-39

[white estuary]

i don't know what to do next

[obtuse oar]

lol

you got the points right?

just put them on equation of line and they should satisfy the equation

[white estuary]

ohh ok

tysmmm

[obtuse oar]

:))

[white estuary]

ok

tysmm

.close

[pearl pond]

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[cedar solar]

this is my third time sending this i think

someone help

and give me a answer please

[pearl pond]

@cedar solar Has your question been resolved?

[flint basalt]

median means its in the middle. look for which one is in the middle.

[cedar solar]

im just confused on what class to take

what do i take as median class

[flint basalt]

you use the class part of the table

because its asking for the median class

also even by looking at the answer choices, you can tell which part of the table you need to use.

[cedar solar]

median class should be taken as the class having higher value than n/2 or equal and greater than n/2

[cedar solar]

how

[flint basalt]

well, look at the choices. and look at the table. which row do you think you would have to use to get the answer?

[cedar solar]

this a question

[flint basalt]

for this small of a table you dont need to use a formula you can just find the middle

especially since it has an odd number of values

[cedar solar]

30-40?

[flint basalt]

take one off from each side. see what you're left with when you have one value left

take one off from the left and one off from the right. keep crossing it off until you have one value left

[cedar solar]

one off from where? 💀

the frequencies?

[flint basalt]

the class bro, i already mentioned before that you're using the class values to get the answer

[cedar solar]

how does that work

do you know how to find the class by cumulative frequency

[flint basalt]

read the question

"The median CLASS is"

[cedar solar]

yes

but you can’t just take the middle most class

[flint basalt]

you mentioned you have to find it by frequency. so now rearrange the table so that frequency goes from least to greatest

and then find the median

[midnight arch]

@flint basalt i believe ur taking the wrong approach here

[flint basalt]

how so?

[cedar solar]

10-20
50-60
60-70
70-80
20-30
40-50
30-40

[midnight arch]

since its a cumulative frequency table we will first need to add all the frequencies and then divide it by two. which in this question gives us 45. and then we see in which class the 45th value stands at which in this question is in the 30-40th class

[cedar solar]

answer is given 40-50

it’s probably wrong

[flint basalt]

wouldnt 40-50 give you the 45th value?

please stop sending these here.

[midnight arch]

wait i might be wrong, lemme solve it again on paper once

[cedar solar]

class cf
10-20 5
20-30 20
30-40 45
40-50 65
50-40 72
60-70 80
70-80 90

[midnight arch]

im not sure why the answer shows 40-50

[cedar solar]

okayy

got it now

thank you for the help

[midnight arch]

np

[cedar solar]

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[midnight haven]

isn't this one also telescoping

[midnight haven]

oops wrong question

isn't this one also telescoping

tm gave this formula

but how would it work for this one

[toxic lichen]

this series is not telescoping in any neat fashion

it is the sum of two geometric series

[midnight haven]

how?

[graceful yacht]

It asks for the result of [n = 0, n = 1, n = 2, n = 3] and then for diverging, you'll simply reference large numbers in a series(10, 100, 1000, etc.).

Oops? I guess that's too complicated.

[midnight haven]

how would i find the sum of the series from that

[rough stream]

So hopefully you can recognize that Σ (1/7)ⁿ is geometric. That's rⁿ for some r < 1

[midnight haven]

P SERIES

[rough stream]

Not p series, because the exponent is not a constant

[midnight haven]

emm okay

[graceful yacht]

My bad. I looked it up: It actually does refer to geometric series. Can I provide an image for aid?

[midnight haven]

ok

[graceful yacht]

Although I think you could just look up an image on Google, I stole this directly from there for reference:

You have a function, right? :)

[midnight haven]

yes

i understand now thanks @graceful yacht

,close

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[near pine]

no idea wherer to start

i know rref stuf

ff

[hallow tide]

ur name means circumference?

[pearl pond]

@near pine Has your question been resolved?

[near pine]

😭 whar

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[tired vigil]

Hello, I have doubts about whether I correctly reduced my 3x3 matrix using the Gauss method.

[spare lark]

,rccw

[spare lark]

i advice you to put the line starting with a 1 on top

[tired vigil]

you mean change the second line with the first one?

[spare lark]

yes

[tired vigil]

Ohh I see

[spare lark]

its less harder to reduce after

having a pivot = 1

[tired vigil]

I'll remember it for the next reduction, thank you very much for the help!

[spare lark]

i didn't check if it was correct tho

well 2nd line is wrong

if you do -2L2

[tired vigil]

If you give me 5 minutes I can do it again, it will be less tedious.

[spare lark]

sure

retry

[tired vigil]

done

[spare lark]

what are you doing for the second line ?

1 1 2

2 1 -3

[tired vigil]

multiply the numbers by -1 and 2 to get 0 on the first line

[spare lark]

so what is 2* L1 ?

and what is -1*L2

[tired vigil]

It would be multiplying the second line by 2 and the first line by -1

I don't know if I'm making myself clear

[spare lark]

well since you swap them its 2*first line and -1*second line

so that you have 2

on first

and -2 on the second

[tired vigil]

Indeed yes, I do the calculation on the right so that there is a record of it, then I add the result of the first line and the second and they give me the value of 0 at the beginning

[spare lark]

the issue here is not the idea but the way you multiply the line

you do 2*L1 which should be 2 2 4 but you calculate it as 2 2 2

[tired vigil]

OOHHHH

You're right, JAJAJAJA I don't know how I didn't notice it

the final result is 0 0 7

[spare lark]

and now you can swap 2nd line and 3rd to end the reduction

[tired vigil]

yeaaa

It would just be a mistake then, I have to be more careful, as I rush to solve it and I make mistakes with the numbers, it is not the first time that it happens

I have to go more calmly

[spare lark]

so to keep in mind :

• try having 1 on top left if it don't imply having horrible things elsewhere
• becareful on line operations
• always recheck twice when starting the gauss method

[tired vigil]

okey, I will

[spare lark]

perfect

[tired vigil]

I appreciate you helping me with this

you're awesome

[spare lark]

you're welcome

[pearl pond]

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[frozen quiver]

hey guys quick question elementary really but when completing a square what number do you subtract form the equasion

[plush bramble]

,tex .cts

[jolly parrot]

riemann

[frozen quiver]

alright i was confused on whether I was supposed to subtract b/2 or b/2^2

its b/2^2 right?

[spare lark]

(B/2)²

Yes

[frozen quiver]

alright thanks comrade

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[oblique sphinx]

\text{I need some seroious help: I have a sequence }u \text{ defined on N with }u_n=3\text{and for all whole number n:}
\
u_{n+1}=\frac{2}{1+u_n}
\
\text{Then I have the sequence v:}
\
v_n=1-\frac{3}{u_n+2}
\

[toxic lichen]

this would be very painful latex

[oblique sphinx]

So now I need to prove that v is geometric (in the form of v(n+1)=q*v(n)

[oblique sphinx]

wdym?

[plush bramble]

sounds like induction

[toxic lichen]

the endless \text commands

that ain't it

[oblique sphinx]

I didn't know how to use the latex bot on the server

[toxic lichen]

look at the expression v(n+1)/v(n). tinker with it somehow

[oblique sphinx]

So i typed it out on a random website lol

[toxic lichen]

inshallah it reduces itself down to a constant

[oblique sphinx]

I've tried all that but just can't find a way

Let me grab a pic

[hasty edge]

<@&268886789983436800>

[oblique sphinx]

lol

[toxic lichen]

<@&268886789983436800> scam

[hasty edge]

My bad

Apparently too many people wanna troll and make their names similar to moderator

[toxic lichen]

let it also be known this guy is a huge loser for setting his name to "im scared of women"

,rccw

[oblique sphinx]

I've managed to solve the v(n+1)/v(n) down to there

But just can't find a way to make it simpler

After that I also tried to use a hypotese that q=-0.5 (gotten from calculating v(0)/v(1)), but also got stuck

I tried to get ChatGPT to solve this (I am really stuck lol) but even it won't give me a correct answer

Now I am begining to doubt the question 😭

[plush bramble]

!nogpt

[pearl pond]

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).

[oblique sphinx]

Yeah I know, but was just too stuck :(

[plush bramble]

induction from here should work

[oblique sphinx]

Currently I don't really see how I can simplify the last expression further

factorizing by -1 won't really get me anywhere

[plush bramble]

don't see where you use induction anywhere

[oblique sphinx]

Does induction mean simplify?

Or like using a hypothese to deduce something?

I've tried to continue developing the hypothese (which I suppose is also called induction?), but somehow got the result that un=1?

Which is obviously wrong, but I just can't find out where

[plush bramble]

https://brilliant.org/wiki/induction/

[oblique sphinx]

ima try now

fml doing maths at 11pm

[pearl pond]

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[supple fractal]

<@&286206848099549185>

[supple fractal]

How did it get that?

[random ermine]

expand (x+h)^3

[plush bramble]

!15m

[pearl pond]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[haughty sigil]

F(2)= 2X³-1=15

Then solve

And also expand (x+3)³

[haughty sigil]

ie at pt (2,15) f(x) is 15

[pearl pond]

@supple fractal Has your question been resolved?

[hasty edge]

$(x+h)^3=x^3+3x^2h+3xh^2+h^3$

[jolly parrot]

mathisfun

[pearl pond]

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[fallow junco]

Posted this earlier but still looking for thoughts - could not find a solution 😦

Hard Game Theory Question:

You are forced to play a game with 44 other players. Each of you are presented with 5 boxes - each labelled 1 through 5. You must place the value on the box inside one of the boxes (eg. you may choose to put 1$ into box 1, or 2$ into box 2). Players have no information on any other players' strategies, or which boxes they played (no collusion).

After everyone places their money in a box, the game master comes in, and takes the box with the most money in it, selecting uniformly at random in the case of ties.

Everyone who's money was not in that box gets their money back. Assuming all players are rational, what is the optimal strategy for this game, and what is your EV?

Is this even solvable w/o numeric methods? I tried using monte carlo as well as multinomial to define a loss function and got ~-0.597 as EV, but there r some flaws in the method i used

[pearl pond]

@fallow junco Has your question been resolved?

[fallow junco]

<@&286206848099549185>

[pearl pond]

@fallow junco Has your question been resolved?

[thorn shoal]

i don't even get the question

[fallow junco]

Wdym? sorry i can try and clarify

[pearl pond]

@fallow junco Has your question been resolved?

[half tendon]

@latent oasis hi game theory

[latent oasis]

I'm working maybe when I have a sec

[latent oasis]

if the game master selects a box 5, does everyone who put their money in box 5 loses it or is it just the player with the specific box he picked?

[fallow junco]

Everyone who puts their money in box 5 loses it

[latent oasis]

gotcha

otherwise I think you just always put it in box 5 and it's easy

[fallow junco]

Wdym? Boxes are shared between players

[latent oasis]

oh

there's only 5 boxes

oh

I see

[fallow junco]

yeah

[latent oasis]

ok

I thought each player had 5

[fallow junco]

No no they all share it

And whichever has the most money gets taken

[latent oasis]

@fallow junco Have you determined the game master's distribution as a function of the player's distribution? Because that's what I would start with

(Assume all players are playing optimall and thus have the same distribution)

[fallow junco]

Ive tried looking at the player distribution that makes the game master indifferent to all choices - this was the solution my friend suggested

But (at least to the best of my belief) this isnt the nash equilibrium

Verified it w monte carlo too so unless i messed up my code idt its right

[latent oasis]

yeah I wouldn't think so either

we want the player to be indifferent

not the game master

[fallow junco]

What ive thought abt is find the distribution s.t. for any box, the probability the 45 (weighed) votes for each box makes it the majority * ev of box (if it is majority) is indifferent

But idk if that is computable rly

And also the probability dist s.t. if 44 other players play w/ that dist, no matter which pure strategy choice we make we are indifferent in EV

Which is what i used for monte carlo sim

to get the -.597 Ev

[latent oasis]

shouldn't the distribution for the game master be approximately 60/137 * [1, 1/2, 1/3, 1/4, 1/5] ?

as the number of players approaches infinity

[fallow junco]

idk if that holds true necessarily

Im not sure if this is true or not but that was my intuition at first too

But at least for 45 case we can see it isnt true

[latent oasis]

how would the players be indifferent then

sure it's not true for 45

it's true for infinite people though

[fallow junco]

yeah i did w/ sim

[fallow junco]

Is the players EV proportional to game masters EV? that would have to be true for this to be the case right

[latent oasis]

uhhhhh

I think so it's zero sum right?

[fallow junco]

Hmm but ur also playing against other players

[latent oasis]

it's still zero sum

[fallow junco]

Im not sure if ur wrong or not haha just thinking out loud

[latent oasis]

the sum of all money leaving is the money put in

[fallow junco]

But wouldnt the odds of who wins be more like a multinomial based on the player distribution

rather than just like adding it up and being proportional to the distribution itself

[latent oasis]

well the EV for all the players is the same

[fallow junco]

yeah

hm

[latent oasis]

like

for n players

boss's EV is -n*player's EV

[fallow junco]

yeah

Wait lets assume infinite players

then ev of playing w ur dist (assuming p1, p2, etc) ~>

0.2(60/137)(1) + 0.2(30/137)(2) + 0.2(20/137)(3) + 0.2(15/137)(4) + 0.2(12/137)(5) = 0.43

but we can deviate to only picking 1 -> 0.2(1)(1) = 0.2 exp loss

[latent oasis]

wait isn't the expected loss 60/137 ?

[fallow junco]

Wait wdym

[latent oasis]

uhh

the boss is picking box 1 60/137 times

so

we lose $1 60/137 times

[fallow junco]

Well the revenue of each box would be 60/137

so hed pick them all equally

bc if u place into box 5, ur putting 5$ into it

Wait am i missing something 😭

[latent oasis]

idk lol

[fallow junco]

yeah im confused

[latent oasis]

sorry I'm multitasking

[fallow junco]

nah u are so good

dw

thank u so much for the help ive been stuck on this all day 😭

[latent oasis]

no we don't want the boss to pick them all equally

we want the boss to pick with this distribution #help-39 message

[fallow junco]

Wait mb i misunderstood what u meant there

[latent oasis]

so that from the player's perspective the EV of each box is the same

[fallow junco]

Hmm

So find probability vector s.t. odds 1 is majority is 60/137, 2 is 30/137, etc?

[latent oasis]

I can lose $5 or I can be 5 times as likely to lose $1

etc.

[fallow junco]

Makes sense

[latent oasis]

yeah I think so

[fallow junco]

Lemme try this out rq

[fallow junco]

(used monte carlo so there may be some leniency here) but seems to not hold true necessarily - EV when everyone plays is -0.97 ish, and when i tried my old strategy against this new one, there was a pretty significant benefit from deviating to the old strategy

[latent oasis]

hmmm

[fallow junco]

Used a search algo to minimize MSE between monte-carlo generated vector representing probability boss picks the box for a given strategy vector p vs the vector you gave

[latent oasis]

wait

is the money pooled ?

like

[fallow junco]

Like wdym

[latent oasis]

we're not trying to maximize the EV per players

that's a very different problem

[fallow junco]

We're trying to maximize EV as one individual

[fallow junco]

when i did this test, i used one person playing with the "baseline" strategy i saw before, and had everyone else playing with the strategy the search algo found

[latent oasis]

oh I see

[fallow junco]

And there was a pretty large difference in EV (in favor of the one person)

[latent oasis]

how was the EV for the new strategy not -60/137 though?

shouldn't it be by construction?

[fallow junco]

Yeah also confused abt this - will recheck code

I did run this for only 45 players tho fyi

So maybe that makes a huge difference

[latent oasis]

I don't think it matters actually

the boss's distribution has to be the same for the nth player to be indifferent

[fallow junco]

Tbh honestly even more confused though -> ur logic seems to make sense and is the same as what I was thinking before w/ the selecting strategy s.t. we generate odds the boss picks the box are proportional to the cost of picking the box

Maybe i just fucked up the code somehow but we'll see

[latent oasis]

[latent oasis]

it's inversely proportional to the cost of the box

[fallow junco]

Yeah srry thats what i meant

S.t. the odds boss picks * cost is indifferent

[latent oasis]

yeah

[fallow junco]

Thanks for the help though ❤️ continued running tests and seemed to still hold bad performance

Hopefully will figure it out soon

my guess is that even though the boss picks each box inversely proportion to the boxes' actual value, there is a difference between that and picking it both in respect to that value asw as the expected amt of players picking that

[latent oasis]

yeah the boss should not be indifferent

or at least there's no reason to think the boss should be

[fallow junco]

by indifferent u mean picking box 1 with 60/137 etc

right

[latent oasis]

I mean like

the boss might get a higher EV by changing his strategy but he's obligated by the rules not to

if that makes sense

the boss doesn't get to choose his strategy

[fallow junco]

does he? the boss just picks greedily whichever gives highest revenue

he has full info

I dont rly see how he could improve on EV past that

[latent oasis]

oh I see

ok that's still not indifferent though

that's greedy

[fallow junco]

yeah i just assumed what u meant is as in

we'd find the strategy s.t. the vector representing odds each is the largest is [60/137, 30/137...]

[latent oasis]

yeah

[fallow junco]

w/ respect to values ofc

[latent oasis]

(ignoring ties)

[fallow junco]

yeah i just had it sample uniformly in case of tie

[latent oasis]

not sure how much those matter

[fallow junco]

Which wouldnt actually occur anyway so

[fallow junco]

Idt it does

[latent oasis]

yeah so I don't think it would matter

[pearl pond]

@fallow junco Has your question been resolved?

[pearl pond]

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[minor tide]

help

[keen quail]

do you know the cosine rule

[minor tide]

yes

r u sure i use it

[keen quail]

yes
but instead of $a^2+b^2-2abcos 30$ use $a^2+b^2+2abcos30$

[jolly parrot]

strizz

[keen quail]

as the angle given is between the vectors

for the second part

[minor tide]

i did that

i may have made a mistake tho

lemme retry

[keen quail]

yes

[minor tide]

can i replace p^2+q^2 with 28

[keen quail]

yep

[minor tide]

ok

i still got pq in the equation 😔

what do i do then

help

i did not cook

[keen quail]

wait

[minor tide]

kk

[keen quail]

you'll get pq= something write

write p in terms of q using this

then replace it back into $p^2+q^2=28$

[jolly parrot]

strizz

[keen quail]

so that it's all in one variable

[minor tide]

mhm

[keen quail]

$p^2+q^2=28 -- a\
p^2+q^2+2pqcos30 = 52 ---b\
b-a\
2pqcos30 = 24 \
pq =8\sqrt{3} \
p = 8\sqrt{3}/q \
(8\sqrt{3}/q)^2 +q^2 = 28$

WHAY

[minor tide]

LOL

[keen quail]

oh

[jolly parrot]

strizz

[keen quail]

it's solvable from here

[minor tide]

where did the 52 come from

[keen quail]

from squaring 2sqrt(13)

[minor tide]

oh ofc

wait what

when was that suppose to happen

OH SHIT

nonno

miswrite

i copied one line wrong from the one above

[keen quail]

ohhh

[minor tide]

one secc

[civic drum]

[minor tide]

wait i justed checked

i got 2pqcos30 = -24

whatdaheck

[keen quail]

no

do b-a

[pearl pond]

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[subtle fulcrum]

Is this value correct

[unkempt yacht]

,rcw

[toxic lichen]

what does this notation mean?

@subtle fulcrum

did you try to write down "the product of all natural numbers from 1 to 100 inclusive"?

@subtle fulcrum tagging you here again so you can find this more easily

[subtle fulcrum]

Sorry I mean multiplying number 1 to 100

[toxic lichen]

"the product of all natural numbers from 1 to 100 inclusive"

ok right

the correct notation for that is $\prod_{n=1}^{100} n$

[subtle fulcrum]

Thanks

[jolly parrot]

Ann

[toxic lichen]

and your value for it is wrong.

[subtle fulcrum]

Ok sorry

So what is right value

[toxic lichen]

the factorial of 100 cannot really be written in any neater way than 100! i am afraid

but if you want to know the value "in full"...

,w 100!

[jolly parrot]

Results provided by WolframAlpha

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Requested by ann_dec

[toxic lichen]

it's still a fuckoff big number, with 158 digits

[vagrant briar]

there are less atoms in the universe than this 😭

[toxic lichen]

comparing to yours...

,w (36288 * 10^12)^10 * 36288^10

[jolly parrot]

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[toxic lichen]

your number overshoots it by like 40+ orders of magnitude

[subtle fulcrum]

Thank you sir 🙏🙏🙏🙏🙏🙏🙏🙏🙏 for answer because I am only 13 year old I was just trying sir

[toxic lichen]

don't call me "sir" please.

[subtle fulcrum]

Sorry so what I call

[toxic lichen]

just my name (Ann) will be perfectly fine.

[subtle fulcrum]

Thank you Ann

Ann do have tricks for calculate fast for multiplication

[toxic lichen]

not really

the best i can offer for multiplying tons of integers together by hand is "group factors of 2 and 5 into 10 and similar easy to calculate numbers"

[pearl pond]

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[flint fjord]

Yo

[flint fjord]

Help me dawg

I can’t do anything

[vagrant briar]

which question do you require help with

[pearl pond]

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[limpid python]

Alright so it's not as much of a problem with one question as it is in general with a variety of similar problems:
say for example, you have 52 cards and you need to pick out 3 cards one after the other from the deck without replacement, what would be the probability that you'd get 2 red cards and 1 black card?
I tried this problem in two ways:

1. I used the multiplication theorem of probabilities and did (26/52)×(25/51)×(26/50), using this I got the correct answer.
2. I did (26C2×26C1)÷52C3, from this method I got basically the above answer multiplied by 3. I don't know what the error is and any help would be greatly appreciated.

[toxic lichen]

do you want two reds and one black in that order?

or is e.g. red-black-red also OK for your purposes? @limpid python

[limpid python]

Nope

Order doesn't matter

[toxic lichen]

in that case 2) is actually the right answer

[limpid python]

yeah I was thinking that as well, but at the same time 1) was given to be the correct answer which had me confused

what's the problem with the first method in that case?

[toxic lichen]

given by whom?

can you show the problem exactly as stated? it sounds like there might be a minor detail getting lost in translation here.

[limpid python]

Alright give me a minute

[graceful yacht]

(26/52)×(26/51)×(25/50) red-black-red order

It's working on assumption that you picked them in order.

[toxic lichen]

ok so the way they're solving it

[limpid python]

Not necessarily though, whatever order you pick them you'll get the same expression

[toxic lichen]

they do in fact imply that they want red-red-black in that order

or wait no

red-black-red

but if you did not care about order you'd need to consider all 3 possible orders anyway (RRB, RBR, BRR)

and this would give you the same answer as your method 2

[limpid python]

wouldn't those orders be considered in the probabilities themselves?

like, if we take a smaller example of 6 cards where 3 are red and 3 are black, and we have to pick 2 cards where 1 has to be Black and another has to be Red

[toxic lichen]

the solution manual says unambiguously: first red, second black, third red

they consider only this draw order to be valid, and not the two others

[graceful yacht]

The probability for:
RRB RBR BRR

Versus the probabilities of:
RRR RBB BBR BRB BBB

[limpid python]

The fact that the cards aren't being replaced, could it have something to do with not counting the other draw orders?

[regal herald]

your probability in (1) has the specific of RBR

if it was purely the probability of drawing 2 reds 1 black without any order
then you have RRB, RBR and BRR, all of which have the same probability
so you get 3 times it

as found in (2)

[limpid python]

okay, I guess it's just an error on the book's part

Thank you all for helping out 😄

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[midnight haven]

I did not understand how did they solve this.

Topic: Analysis of Variance Tests ANOVA

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

<@&286206848099549185>

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

<@&286206848099549185>

[flint basalt]

dont spam helpers, just wait for someone to respond

[midnight haven]

After 15 minutes, feel free to ping @ Helpers.

[graceful yacht]

This feels like cobwebs(for me!) from a few years back but I'll ask about what you have thought about so far.

[errant cedar]

http://www.statsexamples.com/table-F-distribution-0.025.html

You want to fill this table so you can show your working to compute F and do the hypothesis test for the equality of the means, the critical value for the test can be found in a table like the one I just sent. You reject equality of the means if the F > critical value.

To fill the table you use formulas like $SS_{total} = \sum_{i=1}^k \sum_{j=1}^{n_i} (y_{ij} - \bar{y})^2$

$k$ is the number of groups, $n_i$ the number of observation per group, $y_{ij}$ the value of the observation $j$ in group $i$ and $\bar{y}$ the mean of all the observations

you should have these formula somewhere in your notes

[flint basalt]

i was told before to not ping them so thats why i said that mb

[jolly parrot]

pola_touche

pola_touche

[pearl pond]

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[proud valve]

explain which vectors are considered equivalent by this distance function

[proud valve]

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[tawdry axle]

hello

[tawdry axle]

is the answer here not wrong?

reflection over y axis means the negative in front should be inside right?

is the answer not showing a reflection in x axis?

[leaden wadi]

Yes, for a reflection across the y-axis, the negative sign should be inside the radical.

[tawdry axle]

ok so the book is wrong right?

[leaden wadi]

Yes.

[tawdry axle]

does that mean the graph is also wrong?

[leaden wadi]

Either the answer is wrong or there was a typo and the question should say x-axis rather than y -axis.

[leaden wadi]

Yes, the graph would also be wrong.

[tawdry axle]

ohh ok

[leaden wadi]

Send an email to the publisher.

[tawdry axle]

this book has so many mistakes its really annoying

thank you so much for ur help

yea

[leaden wadi]

They love getting corrections and could very likely acknowledge you in the next edition. 😉

[tawdry axle]

haha good idea

thanks

have a good one ❤️

[leaden wadi]

You as well.

[tawdry axle]

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[zenith sparrow]

I don’t understand how to do this, I don’t know what to do with the info given…can someone explain this in simple terms? I’m lowkey struggling with summation

[hasty edge]

Do you know what a Riemann sum is?

[zenith sparrow]

yeah its the area usign rectangle

[hasty edge]

State the relationship between a definite integral and a Riemann sum

[zenith sparrow]

um its the number of recntangles as the limit goes on

like the integral splits the area using recntangles

then u add the recntabgles up

[hasty edge]

Hmm.

[zenith sparrow]

thas what my teacher say

[hasty edge]

$\int_a^b f(x)dx=\lim_{n\to\infty}\sum_{k=1}^n\Delta xf(a+\Delta x\cdot k)$, where $\Delta x=\frac{b-a}{n}$

[jolly parrot]

mathisfun

[zenith sparrow]

he told me to use this one example from earlier, but idk how to use it

can i show ?

[zenith sparrow]

yeah its this. formul but a little different

it is f(ci)delta x

[hasty edge]

Yes

[hasty edge]

So you can sum here

[zenith sparrow]

but i dont understand how to use this for that proble m

[hasty edge]

$\int_0^2 f(x)dx=\lim_{n\to\infty}\frac{2(n+1)(3n+2)}{n^2}$

[jolly parrot]

mathisfun

[zenith sparrow]

oh

so

It’s sum like this?

[hasty edge]

Yes

[zenith sparrow]

okay wait can i try to do it and then send it here to see if im doing it right

wait

[zenith sparrow]

the numerato would be 6n^2+10n+4/n^2

cant i just use that limit rule

since 6n^2/n^2 = 6/1 so its 6?

[hasty edge]

Yes

[zenith sparrow]

oh

omg

[zenith sparrow]

so whenever i get something like this

i only have to multyply the top out

look for big exponent and use that one limit rule ?

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[zenith sparrow]

um

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[pearl pond]

@ivory wasp Has your question been resolved?

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[full prism]

Exercise 21. (0–3 points)

Paul cut out a right-angled triangle ABC from cardboard, with legs measuring 12 cm and 16 cm (drawing I). Then, he connected the midpoints of the longer leg and the hypotenuse with a dashed line parallel to the shorter leg and cut triangle ABC along this line into two figures. He then rearranged these figures to form trapezoid PRST (Drawing II).

Calculate the difference between the perimeters of triangle ABC and trapezoid PRST. Show your calculations.

[full prism]

T is 16 cm

[hasty edge]

A point is 16cm?

Be specific.

[full prism]

Yes.

Also

16 cm*

[unborn hearth]

!status

[pearl pond]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[unborn hearth]

Hint: || Review Mid Point theory ||

[pearl pond]

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[full prism]

Hello

[robust kraken]

hello

do you need help?

[unborn hearth]

Is it this one?

[full prism]

Y

[unborn hearth]

What does it mean? A yes?

[pearl pond]

@full prism Has your question been resolved?

[unborn hearth]

@full prism Please share the problem you need help with

[pearl pond]

@full prism Has your question been resolved?

[latent quail]

Exercise 21. (0–3 points)

Paul cut out a right-angled triangle ABC from cardboard, with legs measuring 12 cm and 16 cm (drawing I). Then, he connected the midpoints of the longer leg and the hypotenuse with a dashed line parallel to the shorter leg and cut triangle ABC along this line into two figures. He then rearranged these figures to form trapezoid PRST (Drawing II).

Calculate the difference between the perimeters of triangle ABC and trapezoid PRST. Show your calculations.

[latent quail]

Original question 👆

[pearl pond]

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[river cliff]

im stuck at log(a^5 b)^1/2

because where i would make it 5 log a normally now theres also that 1/2

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[whole lynx]

I dont understand how to do the second part of this question (Find x and p so that Ax = p is the best approximation to b in C(A).

[whole lynx]

This is my work for the first part of the question

I'm unsure where to go from there

[sharp vigil]

you can use the normal equation to find the least-squares solution

[whole lynx]

sorry, im not sure what that is

would you mind showing me what you mean?

[sharp vigil]

https://textbooks.math.gatech.edu/ila/least-squares.html

[whole lynx]

alright i'll take a look!

[whole lynx]

am i doing this correctly?

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[scarlet glade]

Does this look right?

[scarlet glade]

Oh uh... I cropped the original problem

[flint zenith]

what you have looks correct

but I think sin is bounded by -1 and 1 if the original problem doesn't specify anything

[scarlet glade]

Ok, good the sin in the numerator made me cautious

Oh I guess that makes sense

[flint zenith]

because the range of sin(x) is -1 <= y <= 1

[scarlet glade]

Yeah, idk why I choose 2

[flint zenith]

alg but I would use -1 and 1 if you have to submit this for an assignment

[scarlet glade]

will do, thanks!

[flint zenith]

anytime 🙂

[scarlet glade]

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[jaunty jacinth]

how do you do this problem

like theres only y we can plug in no x

and theres multiple like "slopes" for every y

[heavy snow]

if you look closely, at every fixed y the slopes appear to be the same for all x

[jaunty jacinth]

u mean like

but

[heavy snow]

for example these are all the same slope

in the circle

and the same goes for every "row"

[jaunty jacinth]

oh

so if we plug in 5?

[heavy snow]

yeah youd plug in 5 for y and the x coordinate wouldnt matter

[jaunty jacinth]

i see

i have 1 more question if u dont mind

[heavy snow]

what is it

[azure notch]

the slopes all seem pretty steep. in (b) and (c) at y=1 for example, the run is greater than the rise. but in the graph at y=1 the rise seems much greater than the run

[jaunty jacinth]

for this question

[azure notch]

x is relevant if it weren't a multiple choice question lol

the derivatives are all functions of y

[jaunty jacinth]

where is the 1/10 coming from

[heavy snow]

average

you divide by the length of the time interval, which is 10 seconds

[azure notch]

do you know how to find the avg value of a function over an interval?

[jaunty jacinth]

oh wait i remember now

ty

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[lucid oasis]

.reopen

[azure notch]

did they ever resolve the first question?

idts

[lucid oasis]

nah they didnt

how old are ygs

[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

[fervent tartan]

This approach could work, but proving that U is finite-dimensional actually turns out to be rather hard here

A better approach could be to essentially 'build' a basis for U, starting with an empty set of vectors and adding independent vectors to the set until it spans U; This process must eventually stop, since the size of the set can never grow larger than dim(V)

As far as I can tell, the approach I gave is the best, at least assuming we're working from the ground up

If you mean the "set can never grow larger than dim(V)", that part is trivial since a set of size, for example, dim(V)+1 could not possibly be independent in V

[lilac jackal]

the reason the process stops relies on the fact that an independent set is no longer than a spanning set, and this itself takes some time to prove

sure, i usually phrase it as if v1..vk are independent in V and u1..uj span V then k<=j

the strategy here is already given by 823

you make a recursive process for building an independent set in U

yeah quite a few preliminary facts about independence and span (in finite dim spaces) can be proven by recursive processes, its very neat

you should justify with one line why the new set is independent

also its not clear why the process stops

maybe youre trying to invoke this but not clearly

if v1..vk are independent in V and u1..uj span V then k<=j

ill rephrase the fact as "an independent set is no longer than a spanning set"

so i'd say the process ends as n=dimV (any spanning set of V has size n) and by the above fact we cannot build an independent set longer than n

no prob

823 gave you the strat

more recusion

but if you mean easily

then the book already gave a very short proof

you can prove 2.46 directly using recursion (and my fact)

or prove 2.45 using my fact then use 2.45 to prove 2.46

yeah recursion is probably the ONLY way for 2.40

induction might work

i never bothered bc the recusive algo appealed so much to me

yes

fortunately my prof didnt use LADR and fully developed determinants in lecture

its one of the few crap that properly develops it

his lecture notes

he started at postulating properties of an "oriented volume" function

and showed a construction of the function and how it led to determinants

i attribute my fortune to my prof's physics oriented mind

i had the same prof for my dynamical systems course

books have wide variety of quality

some do that and some just do definition lemma theorem repeat

bourbaki algebra

https://www.cmat.edu.uy/~marclan/TM/Algebra i - Bourbaki.pdf

this book will kill u with dryness

its simply that bourbaki books arent intended for undergrads

(btw bourbaki refers to a group of authors, its a fun rabbit hole if u have time)

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[lilac jackal]

np

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[sour sluice]

sorry locke

^^ am I supposed to find the equations for f(x) and g(x) based on the graphs before starting the problem?

[fossil jewel]

They already give you f and g

You don't need to find an explicit equation, no

[sour sluice]

so for a.

P’(2) would be product rule, so g(2)f’(2) + f(2)g’(2) ?

[limpid lily]

Right.

[sour sluice]

but won’t that just be zero?

[limpid lily]

It might be.

Why do you think it's zero?

[sour sluice]

derivative of any number is zero/ I see that g(x) is zero at x = 2

I am walking through the dark here

[limpid lily]

No, that's not the derivative of 2.

That's the derivative of f(x) at x = 2 and the derivative of g(x) at x = 2.

f(x) is not 2. g(x) is not 2.

[sour sluice]

oh wait

[limpid lily]

It's like if f(x) = x^2.

f'(x) = 2x.

f'(2) = 2(2) = 4.

You only get zero with something like this:

q(x) = 10, so q'(x) = 0, so q'(2) = 0.

You're differentiating f(x), and then filling in 2 for x.

Or you're differentiating q(x), and then filling in 2 for x.

With f, the derivative becomes 2x and then you fill in 2 for x.
With q, the derivative becomes 0 and then you fill in 2 for x.

So, it's the differentiation step that can give you zero, not the argument to f' or q'.

[sour sluice]

i’ll be back with an attempt at the problem, thanks

.close

[pearl pond]

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[limpid lily]

No problem.

[pearl pond]

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[sour sluice]

we’re back

does this look good for a.

[limpid lily]

Yes, that looks good.

[sour sluice]

heckayes

appreciate it

[limpid lily]

No problem.

[sour sluice]

.close

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[jolly kernel]

None of these match

I am confused

I tried solving it by hand and none made sense either

[sharp vigil]

you may need to change the zoom to match the zoom shown in the charts

[jolly kernel]

is it V?

I got it now thank you

I forgot about zoom/windows

.close

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[pearl pond]

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[raven geode]

did they do the reduction formula twice

?

[summer imp]

Yes

[raven geode]

is there a better way to solve this crap other than using the reduction formula

[summer imp]

Probably is with trig identities and stuff, but I doubt any of it will be less annoying than reduction

[proven sable]

im not sure about sec and its powers

but i know for cos and sin powers

[proven sable]

you can use demoivre for them to get some binomial expansion

which you can then get cos^n in the form of multiple angles of cos

then integral each term like that

so z = cos theta + i sin theta
(by demoive), z^-1 = cos theta - i sin theta
then 2 cos theta = (z^1 + z^-1)

then you can raise that up to the power of n and its all good

im not sure if it could work with negative powers

RHS expansion wont work though for negative powers to get it cleanly

[pearl pond]

@raven geode Has your question been resolved?

[pearl pond]

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[sour sluice]

scared to continue this problem. I need a helping hand.

[woven garnet]

Positive slope means positive y values

[sour sluice]

true.

I will adjust that

actually

i’ll fill it out best I can

and return shortly

@woven garnet

please tell me

it was a noble effort

[fair creek]

worry not

if it looks linear

the derivative is a straight horizontal line

if it looks parabolic, the derivative is a straight line with slope depending on concavity

[sour sluice]

which is what I did :3

[fair creek]

well you've done that much

whats the problem then

[woven garnet]

The slope is decreasing on the curve

Should look like a \

[sour sluice]

the sketch of dy/dx h(x)

[fair creek]

btw for the parabola

the entire inverval has a straight line derivstive

[fair creek]

your solution is wrong due to the parabolas since you made the parabolic derivative discontinuous

they are continous and a singular straight line

[sour sluice]

my solution?

and are you saying that if

f(x) is continuous then f’(x) must be as well?

[fair creek]

not necessarily

but the parabolic derivative is

[sour sluice]

the parabola on the left is 1/2x^2 ain’t it?

[fair creek]

uh?

[sour sluice]

negative*

[fair creek]

i'm not sure what it's x^2 coefficient is but

yes it is a negative value

[sour sluice]

1/-2 is -1/2

approx

[fair creek]

okay then that's the value

why did you split the derivative into two then?

[sour sluice]

because there’s a zero at (-4,4)

approx

horizontal tangent line

[fair creek]

that just means that the slope is zero at that point

not a total zero 💔

or a discontinuity

[sour sluice]

hmm

chewing my pencil so hard rn

we need an expert

<@&286206848099549185>

[fair creek]

look here's a similar graph with the same zeroes

this expands to some quadratic polynomial

oop

if you take it's derivative

you'll notice that is a continuous straight line

because the derivative is continuous!!!

for a parabola

@sour sluice

[sour sluice]

ok ok I got schooled

[fair creek]

so just redo the quadratic intervals

[sour sluice]

working on it

[fair creek]

and btw the last interval is

sinusoidal

so it's derivative is also sinusoidal

[sour sluice]

that really sucks ngl.

this means that

the base equation is like

degree three polynomial?

[fair creek]

dw the derivative is just that function but shifted by half cycle

[fair creek]

no it's a sinusoidal equation

[sour sluice]

what makes it sinusoidal?

[fair creek]

y=Asin(wt+phi)

[fair creek]

same amplitude of turning points

[sour sluice]

woahhhh that’s perceptive

having trig flashbacks rn

[fair creek]

if you count the distance between zero and turning points

they're equal too

[fair creek]

so uhm try again for the parabolic and the sinusoidal interval

happy mathing

[sour sluice]

i’m gonna print some more copies of my sheet and then use pen

i’ll be back soon

[sour sluice]

for the sinusoidal…

will it be inverted in the graph of f’(x)

[fair creek]

inverted?

[sour sluice]

yeah

[fair creek]

don't think too hard, just shift it by quarter cycle

[sour sluice]

like what would the derivative of an unknown sinusoidal function look like?

[fair creek]

quarter cycle difference

in periodicity

then just look at the tangent at that point

and find out the initial slope at the start of the interval

[sour sluice]

how’s this?

original

[fair creek]

kinda alrighht but

the function is at the half ofthe cycle at the start of the interval

so it should not be taking a maxima at that point

[pearl pond]

@sour sluice Has your question been resolved?

[sour sluice]

yeah I think I see what you mean. as far as everything else, look ok?

and should I put open circles anywhere else, such as at x=-1
and x=-7

[fair creek]

myabe yeah

[sour sluice]

maybe?

<@&286206848099549185>

tldr- close to finishing this problem

verification on last two images (f(x) and f’(x)) needed

[fair creek]

yes, since the derivatives don't equal at those points

[snow arch]

guys help me with my MIT integral math homework question(idk how)

[pearl pond]

@sour sluice Has your question been resolved?

[pearl pond]

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[lunar bone]

given that A is a point on fx and B is a point on the line y, find the minimal value of AB if AB is vertical to the line y

[lunar bone]

ive tried using the formula of the distance between a line and a dot, and that the product of their slopes should be equal to -1 and that didn't help

[snow sail]

i dont think youd need this

youre given the slope of the minimal line

so you should also know the slope of both functions an the intersection of them and AB

is that enough of a hint?

[lunar bone]

so i need to do
f'(x) = -1?

[snow sail]

no

[lunar bone]

and work my way through there

oh

[snow sail]

wait am i right

[lunar bone]

then its not

[snow sail]

even easier

you have the formula for the minimal distance line

just find th.

the intersection with f directly

then you can use the distance formula

@lunar bone hopefully that seems a little easier

[lunar bone]

I still didnt understand how?

With the derivative?

[snow sail]

no

you know the slope of AB

and you know a point

so you can use point slope to get the formula of AB

at which point its just algebra to find the intersection between f and AB (there might be more than one)

[lunar bone]

What point do i know???

[snow sail]

oh my bad, i thought you were given B

[lunar bone]

Lol

Ill try using the -1 in the derivative

To get a point

[pearl pond]

@lunar bone Has your question been resolved?

[tardy reef]

To expand on what jan niku said, since you know that the distance AB is minimum, AB must be normal to the curve, or a line perpendicular to AB is tangent at A (which makes it parallel to y = x-3).

[pearl pond]

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[odd mauve]

can anyone help me

[odd mauve]

take the derevative of V = (pir^2((2025-r^2))^0.5)/3

[random horizon]

i can help u

please explain about problem

[odd mauve]

hi

i will send the problem

wait

here

find the maximum volume

with 3 significant digits