help-49

[abstract cairn]

yes i get where the numbers come from now

what do i do with them

[last slate]

See v1-v4 and v3-v4 are linear mulitple of each other

If they are linear multiple they are parallel

[abstract cairn]

but can i just subtract them like that?

if they're marked as regular coordinates

ABCD

and not vectors

[last slate]

U can do that

U k complex numbers??

Do it on complex plane then procedure will be same

[last slate]

Yep vector 3d are interconvertable

Any coordinate (a,b,c) represent the position vector or free vector v=aî+bĵ+cƙ

[abstract cairn]

aaaa

so if i prove that

would that be it

or do i need to do something else to prove that its a trapezoid

[last slate]

No it's enough but if u writing a proof also prove that other two sides are non parallel

If they are parallel then it will become a parallelogram

[abstract cairn]

and how do i calculate its area?

[last slate]

Find diagonal vector

Both diagonals

Take their cross product modulus divided by two

[abstract cairn]

ohh

[last slate]

Or u can use another formula if u not comfortable with that

[abstract cairn]

is |d1| * |d2| and |d1*d2| the same

[last slate]

No

[abstract cairn]

aa

so i need to do |d1*d2|??

why not |d1| * |d2|

[last slate]

It is cross product

[last slate]

U k how to find cross product??

[abstract cairn]

OHH

yes i know that

[last slate]

Yep u need to find cross product then find its mod and divide it by two

[abstract cairn]

also how do i know here where are the arrows supposed to go

cuz its different if i do CA and not AC

and DB or BD

[last slate]

It doesnt matter take nay side with the modulus function ans will come same

[abstract cairn]

ohhh

thats great

[last slate]

It will not matter

Take any

[abstract cairn]

yayayay

[last slate]

U just need to identiy the z1 ,z2 ,z3 ,z4 and i already did that on above pic

This one

[abstract cairn]

yeah i know how to do that

great

thats it

thanks ^-^

[last slate]

Wlcm ^^ all the best 👍

[pearl hull]

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[plain kestrel]

ø: Z to S_8
ø(1) = (1426)(257)
how do i calculate the ker(ø)

[hot steppe]

ping after 15 mins

[plain kestrel]

oh my bad

[obtuse basin]

ker(phi) is exactly the set of integers such that phi(1)^n is the identity permitation, i. e., integer multiples of the order of phi(1)

Start by finding the cycle decomposition of phi(1)

[plain kestrel]

so i just do phi(1) over and over agian til i get the 12345678 back?

[obtuse basin]

You could do that

[chrome vessel]

why is there two 2 in the cycle notation?

is it product of the cycles or is it disjoint cycle notation

[obtuse basin]

It is a product of cycles

[plain kestrel]

well i get n = 12

what does that mean

because 4, and 3

you have to match to get it back

[chrome vessel]

so is it disjoint cycle notation you are using?

[plain kestrel]

product of cycles i think

[chrome vessel]

then you should take the product first and put it in cycle notation

[obtuse basin]

Show your work

[plain kestrel]

this is the question

so phi(20) would actually be phi(8)?

cuz i get ientity back after 12 turn so 20-12?

[obtuse basin]

Or at least explain why you think the order is 12

[plain kestrel]

well i jsut dedudce that since first one have 4 elements, 3 elements

1 cycle takes 4 action to be back agian

[obtuse basin]

That would work if the cycles were disjoint

[plain kestrel]

and 2nd 3 times

so inorder to syn them you woukld need 12 time?

[obtuse basin]

Does (1 2)(1 2) have order of 4?

[plain kestrel]

oh

[obtuse basin]

This is why I suggested starting with cycle decomposition

[plain kestrel]

i computeed it and i get (4532716)

like adding them together'

[obtuse basin]

Where did 3 come from

[plain kestrel]

give me 1 sec.

1 2 3 4 5 6 7 8
4 5 3 2 7 1 6 8

sorry i was just listing them

so i guess (142576)

[obtuse basin]

Right

So phi(1) is that cycle

[plain kestrel]

so 6 turn?

[obtuse basin]

What is the order of (142576)?

Yes

[plain kestrel]

what nexxt?

ker Ø = 6?

[obtuse basin]

Like I mentioned, the kernel of phi is exactly the set of integer multiples of 6

[plain kestrel]

1,2,3,6?

[obtuse basin]

Multiples, not divisors

[plain kestrel]

6,12,18,ect?

[obtuse basin]

Yes

0 and the negative multiples are in there as well

[plain kestrel]

so ...,-18,-12,-6,0,6,12,18,...

[obtuse basin]

Right

[plain kestrel]

the entire number line of a multiple of 6?

<6> i guess

to write is nicer

[obtuse basin]

All integer multiples of 6, yes

[plain kestrel]

thanks

so phi(6) is identity

so if i wanted to calculate phi(20) then i do multiple of 6 factor out

like phi(18+2)

and just calculate phi(2) = phi (20) ?

[obtuse basin]

Yes

[plain kestrel]

Thank you so much

gotta run to exam now

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[lucid delta]

how can i find the middle 50% of a population using a graphing calculator?

[worthy kestrel]

what information do you have

[lucid delta]

mean is 112

standard dev is 8

i need to find what range makes up the middle 50%

nvm i figured it out

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[steady trail]

[
\lim_{z \to 1+i} \left( \frac{(z^2 - 2z + 2)^3 + 3z^5 - 5z^3 + 7}{(z - (1+i))(z^4 - 4iz^3 + (1-4i)z^2 + 2iz - 3)} \right)
]

[grand pond]

Ancelotti

[steady trail]

how to solve this

[strong stump]

wtf is that

[steady trail]

A limit

[strong stump]

why are there z's?

[pearl hull]

complex numbers

[strong stump]

Where did you get this exercise from

[strong stump]

I know

@steady trail

[steady trail]

What

[strong stump]

?

[steady trail]

From the book

[strong stump]

What is the book called?

[steady trail]

"Complex Variables and Applications"

[strong stump]

I'm trying to find a PDF online

Link?

[steady trail]

Link?

Wdym

[strong stump]

Is this book online?

[steady trail]

I don't know

[strong stump]

Okay

Let me see

It seems a bit boring

Do you have the solution?

[steady trail]

Yes

[strong stump]

Can you take a photo of the solution to the book?

[steady trail]

I cant now

[strong stump]

Looks like that

[steady trail]

I only have the phone now

[grand pond]

jandro

[steady trail]

Nice

Thank you

[strong stump]

It's correct? Cuz now i have to go

[steady trail]

When I get back I'll go and see

It seems well done

[bitter tide]

should conjugate the denominator to clean it up a lil

[strong stump]

Ok let me know

[strong stump]

?

For me its clear like that

I have to go. Bye.

[bitter tide]

no u dont want complex numbers in the denominator, its not clean and hard to visualize

[strong stump]

Fix It, I have to go

[steady trail]

.close

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[steady trail]

.reopen

[midnight plank]

✅

[steady trail]

The limit should be 1

[strong stump]

@steady trail I can't help you now, I'm going to work

[steady trail]

Ok

Thanks for the help anyway

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[last slate]

2 + 5 + 12 + 31 + 86 + ................................................
find summation.

[calm wind]

2+5+12+31+86 = 136

[fossil knot]

looks like to get the n-th term, you multiply the n-1 th term by 3 and subract the n-1 th odd number

[calm wind]

oh is that supposed to be a squence

yeah you have to like, actually give us the sequence

[last slate]

[tidal token]

how the heck did you notice that ?

[last slate]

i tried that ig

[fossil knot]

idk it looked like it was sorta tripling lol

[last slate]

yeah

i did something like that which worked for few terms and then doesnt make sense

[fossil knot]

oh wait

you can just do $3^{n-1}+n$

[grand pond]

Ari

[last slate]

it works

[fossil knot]

[last slate]

but how would i add it now

[last slate]

it would have taken few hours for me to figure it out

[fossil knot]

if it continues forever it'll go to infinity

since it's increasing exponentially

[last slate]

true,
its diverging series, how the hell are we gonna do it ?

wait

[fossil knot]

if it just tells you to find a summation just write it in summation notation

bc it adds to infinity

[last slate]

lol its not till infinity( i got emotional their adding those dots ..... its upto n )

forgot n at the end

[calm wind]

are we actually ever given the series

[fossil knot]

So we want

$\sum_{i=1}^{n} (3^{i-1}+i)$

[grand pond]

Ari

[fossil knot]

$=\frac{3^{n}-1}{2} + \frac{n(n+1)}{2}$

[last slate]

2 + 5 + 12 + 31 + 86 + ...n

[last slate]

can you explain for the exponential part, ik the other ofc

[grand pond]

Ari

[fossil knot]

Let $S_n = 1 + 3 + \dots + 3^n$

[grand pond]

Ari

[fossil knot]

Then $3S_n = 3 + 9 + \dots + 3^{n+1}$

[grand pond]

Ari

[fossil knot]

Substracting the two, we have $2S_n = 3^{n+1}-1$

[grand pond]

Ari

[calm wind]

oh then the sum is 136+n

unless there are more terms you're not telling me about

[last slate]

mb

136 is the answer

[last slate]

yeah i got this,
gonna remember this thing for exponential form
its similar to what we did for sigma n/n^2

THANKS

[last slate]

the answer is correct too i just checked

[fossil knot]

[last slate]

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[floral kelp]

they flipped these two right?

[floral kelp]

like it should be -A/2 when k=1 instead

$\frac{A}{2j} \rightarrow -\frac{A}{2}j$

[grand pond]

chrelleren

[robust isle]

yeah (b) seems screwed up indeed

@floral kelp

[midnight plank]

@floral kelp Has your question been resolved?

[floral kelp]

its kinda annoying lol I have two versions of the same book available

both have this mistake..

like whatd they fix on the 2nd 💀

I just put it in paint and fixed it myself

https://tenor.com/view/gg-meme-dank-cursed-khat-gif-21280690

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[daring sluice]

How can I find sum of this series?

[bitter tide]

rewrite the interior as 1 fraction over a common denominator

[cedar vortex]

idk I'd turn it into a product

[bitter tide]

then u can use log properties

[cedar vortex]

and simplify

oh lmao

yeah it's trivial once you simplify

[daring sluice]

one sec

I rewrote it as 1 fraction over a common denominator, used log props and got this:

ln(n(n+1)-2) - ln(n(n+1))

Where should I move next?

[bitter tide]

do u know about telescoping series?

[daring sluice]

umm yes, but idk how to remove log ;d

[bitter tide]

try to do telescoping with the log

[daring sluice]

Should I just write down a part of values and try to simplify middle ones?

[bitter tide]

expand the sum with the logs

[daring sluice]

hmmm

what do you mean, sorry

[bitter tide]

write the sum out

like at n=2

then +

at n=3

for like 5 terms

[daring sluice]

okay I will try

[bitter tide]

show me what u get

[daring sluice]

should I see some pattern or am I doing something wrong

[bitter tide]

hmm maybe split the 2nd ln into another sum, also ur n=2 is wrong for the 1st ln, it should be ln(4)

[daring sluice]

ohh sorry

one sec

now ended up with this

I have split that ln(n(n+1)) into ln(n) + ln(n+1)

[bitter tide]

hmm ngl im not sure how to do that

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[iron zephyr]

Can anyone explain this move please :)

[zinc spear]

It's basic splitting of middle term

[iron zephyr]

idk what that means 😅

[ionic orbit]

that's just expanding

isn't it

[zinc spear]

Umm see
Basically it is a method to factorise

An algebraic expression

[ionic orbit]

oh oops

[zinc spear]

Mostly used in case of 2 degree polynomial

[ionic orbit]

you can use the quadratic equation

or you can find two numbers that add to give the coefficient of x and multiply to give the constant at the end

[zinc spear]

That is what is splitting middle term method

[ionic orbit]

ok

in this case coefficient of -x is -1 and constant is -2

so two numbers that add to give -1 and multiply to give -2 would be -2 and +1

and they are the factors

so you can write it as (x - 2)(x + 1)

[iron zephyr]

Oh I think I understand

[ionic orbit]

thre are plenty of youtube videos and examples online

but sometimes this technique doesn't work

you wouldn't be able to find two numbers always

so in those cases you would use quadratic formula

[midnight plank]

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[last slate]

$$\overline{xy}$$

[grand pond]

david

[last slate]

this is what I mean

[atomic jolt]

(10x+y)/z = 20+y+6/10
(10x+y)/z = y+8/10+y/100

assuming comma is a decimal separator here

,w solve (10x+y)/z = 20+y+6/10 and (10x+y)/z = y+8/10+y/100

[grand pond]

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Requested by stipendi

[last slate]

um

[atomic jolt]

kinda sus

[last slate]

you forgot about the period

it's period 6

And period y

,w solve (10x+y)/z = 20+y+9 and (10x+y)/z = y+8/10+y/90

,w solve (10x+y)/z = 20+y+6/9 and (10x+y)/z = y+8/10+y/90

[grand pond]

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[last slate]

ugh

can someone help

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[last slate]

guys how come i haven't gotten taught any formulas for algebra 1

[midnight plank]

Please don't occupy multiple help channels.

[cedar vortex]

oh no 😦

what would you like to learn

[last slate]

I NEED TO BRO

[cedar vortex]

I think you learn some more in algebra 2

[last slate]

i feel useless not knowing them

[cedar vortex]

hmm

[last slate]

like i can only figure out a finite amount of problems in that area

with the random yt methods

[cedar vortex]

well that's all of us

[last slate]

and i don't understand how variable works

[last slate]

no no its like a very finite

with only the yt methods

[cedar vortex]

yt methods are great

it's nice that you can learn in some way

[last slate]

its good for an specific process

[cedar vortex]

I would recommend reading the first part of spivak calculus

then

[last slate]

but ion understand much of variables

[cedar vortex]

it's quite difficult

but it builds a great basis

I remember when I first read it, it seemed like I learned a lot of clever tricks

and it made me approach mathematics in a different way

esp the questions

one question I still remember is

[last slate]

ii need help with those word problems tbh

i couldn't figure out a rational problem:<

[last slate]

i've got the answer from google but that doesn't rlly help tbh

[cedar vortex]

this one of my fav parts of the book

[last slate]

like i got an answer from google

but i have not learnt anything rlly

[cedar vortex]

do you wanna share it?

[last slate]

here

[cedar vortex]

@last slate do you mind if I get a tldr

this guy is kinda a yapper

[last slate]

t d r table?

[cedar vortex]

no like a summary

[worthy kestrel]

what did i do :(

[cedar vortex]

LMAO DW

no no the problem video statement 💀

the guy went on for like 5 minutes

on absolutely nothing

like they could state the riemman hypothesis in that tine or sm

[midnight plank]

@last slate Has your question been resolved?

[last slate]

where did bro come from lmao

[midnight plank]

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[acoustic notch]

does this notation make sense?

[midnight plank]

Please don't occupy multiple help channels.

[visual tiger]

it doesn't

the argument of the sum is k

and it would take all values between 1 and... itself?

[acoustic notch]

that's what I was thinking! good to know

the number of k varies

sometimes its 2, sometimes its 4

[visual tiger]

what are all the values that k can take?

[acoustic notch]

up to infinite I guess

1, 2, 3, 4, etc

[visual tiger]

not what you wrote in " beta = An1xBn1 -..."

[acoustic notch]

thats an example

[visual tiger]

but you want it to eventually stop

[acoustic notch]

here lemme post the original for you

[visual tiger]

you want a finite sum

[acoustic notch]

im using it for number of actions in a game

so each action has a value, that's (Anew x Bnew) - (Aold x Bold)

and then beta is the sum of all the values

but sometimes there are 4 actions left, sometimes there are 2

[visual tiger]

so name N the number of actions in total

$\beta = \sum_{k=1}^N (A_{new,k}\cdot B_{new,k} - A_{old,k}\cdot B_{old,k})$

[grand pond]

rafilou2003

[acoustic notch]

i like it

does that make sense for what im trying to say?

[visual tiger]

don't have the full context

[acoustic notch]

if there are 4 actions then beta = (Anew1 x Bnew1) - (Aold1 x Bold1) + (Anew2 x Bnew2) - (Aold2 x Bold2) ...

hard to describe the context but if it means that when n is 4 then ^ is true, then its good

(Anew1 x Bnew1) - (Aold1 x Bold1) +
(Anew2 x Bnew2) - (Aold2 x Bold2) +
(Anew3 x Bnew3) - (Aold3 x Bold3) +
(Anew4 x Bnew4) - (Aold4 x Bold4)

and then when N is 2:
(Anew1 x Bnew1) - (Aold1 x Bold1) +
(Anew2 x Bnew2) - (Aold2 x Bold2)

I think it makes sense

[midnight plank]

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[worthy spruce]

Hey everyone, I would love some help with this question, I'm honestly not even sure how to approach this.

[midnight plank]

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@worthy spruce Has your question been resolved?

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@worthy spruce Has your question been resolved?

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[last slate]

$\sum_{r=1}^{1}f(r)=f(1)$

[grand pond]

yajat

[last slate]

is this correct

[worthy kestrel]

yes

[last slate]

r u sure 😭

[worthy kestrel]

yes

[last slate]

okk

ok so if this is correct

Let $f(1)=1$, and $f(n)=2\sum_{r=1}^{n-1}f(r)$. Then $\sum_{r=1}^{m}f(r)$ is equal to

[grand pond]

yajat

[last slate]

how can i do this

lemme show how i did first

i got an answer but somehow it says its wrong

this is what i did

but the answer is $3^m-1$

[grand pond]

yajat

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

@last slate Has your question been resolved?

[visual tiger]

doesn't make sense at all, f(1) = 1 yet 3^1 - 1 = 2

[junior flower]

yea that answer makes no sense

[junior flower]

this looks good

[visual tiger]

oh I might be wrong wait

It's just f(n) = 3f(n-1) for n > 2

so f(1) = 1, f(n) = 3^(n-2)*2 for n>= 2

[last slate]

wdym @visual tiger i dont get it

[last slate]

i thought my answer was correct, but lets just say it is 3^{m-1} (the ans according to the answer key), and now when we take m=2, it is f(1)+f(2)=1+2=3, now from 3^m-1/2 it is 5/2 and from 3^{m-1} it is 3

so its sure that my answer is wrong

but how i dont get it :(

[junior flower]

the answer key is not right but i’m also not sure what you’re saying here

[last slate]

ok let me explain what i mean

according to what ive solved $\sum_{r=1}^{m}f(r)=\frac{{3^m}-1}{2}$ now taking m=2 $\sum_{1}^{2}=\frac{{3^2-1}}{2} \implies f(1)+(2)=\frac{6-1}{2} \implies 1+2=\frac{5}{2} \implies 3=\frac{5}{2}$

bruh fuck

all that for that

[grand pond]

yajat

[last slate]

ok we good now

but if we take $\sum_{r=1}^{m}f(r)=3^{m-1}$ we are getting 3=3

[grand pond]

yajat

[last slate]

so the answer in the answer key should be right yea?

[last slate]

according to what ive solved $\sum_{r=1}^{m}f(r)=\frac{{3^m}-1}{2}$ now taking m=2 $\sum_{1}^{2}f(r)=\frac{{3^2-1}}{2} \implies f(1)+f(2)=\frac{6-1}{2} \implies 1+2=\frac{5}{2} \implies 3=\frac{5}{2}$

[grand pond]

yajat

[last slate]

sorry the notations were kinda wrong so i did it agian

layla went off 🙏😭

[junior flower]

soz will read now

[last slate]

oh its ok

[junior flower]

was this supposed to be 3^(m-1)?

[last slate]

💀💀💀💀💀💀💀💀💀💀💀💀💀💀💀💀💀💀

yea... 💀

[junior flower]

bruh…

then yea, the answer key is right

[last slate]

ok

but how is my answer wrong

like i can clearly see theres a geometric progression building up

and it goes to m th term

first term is 1 and common ratio is 3

[junior flower]

no?

starting with n = 1, f(n) goes 1,2,6,18,…

[last slate]

but if we have to find sum_ r=1 ^ m, wont this be something like f(1)+(f2)+.....+f(m)?

so that means there are m terms in this GP

[junior flower]

yes

my point is it’s not quite geometric

[last slate]

oh wait

[junior flower]

you can just look at the 2,6,18,… part

[last slate]

YEA

just noticed that

so we can add that 1 later

[junior flower]

[last slate]

so it should be $1+\frac{{3^m}-1}{2}$ right?

[grand pond]

yajat

[last slate]

wait

but now the total terms are m-1

and the first term is different too

[junior flower]

[last slate]

so it should be $1+\frac{\cancel{2}(3^{m-1}-1)}{\cancel{2}}$

[grand pond]

yajat

[last slate]

and now i get 3^(m-1)

ok finally 😭

[junior flower]

[last slate]

yes

thanks a lot

[junior flower]

[last slate]

nvm i cant use it

[junior flower]

lol sucks

[last slate]

lmao fr 😭

[pearl hull]

Has your question been resolved?

[last slate]

yes

mb

.close

[midnight plank]

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[graceful stag]

hello I just want to understand how the given answer apparently to this question was 2/3

at first I tried to check that it might be about conditional probability

laying event X to either have A<B or B<A, then Y having A<C B<C C<A C<B

I thought I arrived at the answer of 1/2 bc X had 100% chance of having one card smaller than the other

and 2/4 for Y, althought I'm having a hard time viewing this the right way, or atleast arriving towards the right assumption for an answer of 2/3

[viral dagger]

not sure if this is correct but just writing the possibilities we get 4/6=2/3

[torpid warren]

You can think of it like this:
Since we're only taking the smaller card after the first comparison, then asking if it is still the smallest after the second comparison, our probability is just what are the chances we picked the smallest number out of all 3 in the first two cards

If we did pick the smallest number in the first comparison, the probability is 100%, it will be smaller than both others (as they are distinct)

If we did not, then probability is 0%, since the last card is the smallest value, and the "smaller" card from the first comparison is actually the middle number

The probability we pick the smallest card in the first two tries is 1/3 + 2/3 * 1/2 = 2/3

[graceful stag]

I like your explanation about it, but I'm having a hard time grasping it

[torpid warren]

That's fair, it's a bit confusing sorry

I'll try and walk you through it

Let's say we named the three distinct integers x,y,z, in order without loss of generality

That is to say, x < y < z

[graceful stag]

alright

[torpid warren]

If we picked x in the first two cards, it doesn't matter if the other card is y or z. x will be the smallest

Furthermore, in the second comparison, x will still be the smallest

Is that understandable so far?

[graceful stag]

yepp

[torpid warren]

What happens if we do not pick x in the first two cards?

Well, that means we picked y and z. y is the smallest, so we go on to compare y and x. Remember that y=B and x=C in this case, so our result is a fail, since C is the smaller one

[graceful stag]

we could still have y as smaller than z

[torpid warren]

That is correct, but the question only asks about "the probability that it (smaller of A and B) is also smaller than the value on card C"

If we had A and B be y and z, the probability is that the smaller of A and B is smaller than C (x) is 0, since x is the smallest

[viral dagger]

i think you can just do this but pick one randomly and the random one you picked is C

[torpid warren]

That might work, this is just the way I thought of it

[cobalt sigil]

Let the 3 cards be already named A,B,C
we assign values values to them consider 1,2,3
Now this can be assigned in 3*2 = 6 ways
Now it says we have already chosen the one card and another card, this can be in 3 ways like (1,3) or (1,2) or (2,3), now it says to find the probability that we have chosen the least value in the picking

[viral dagger]

so for C=x, the smaller of the rest is y and x is smaller so no
for C=y, the smaller of the rest is x and y is bigger so yea
for C=z, the smaller of the rest is x and z is bigger so yea

[graceful stag]

pretty much, just kind of wanted to understand how he thought of it, I think(?)

[torpid warren]

Both of the examples above are good, and probably easier to understand

[graceful stag]

this makes more sense

[cobalt sigil]

So its 2/3 [conditional probability]

[graceful stag]

alright, thanks for the input guyss!

.close

[midnight plank]

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[midnight plank]

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[raven sparrow]

The matrix question.
I found out the eigen values ans vectors.
How do I proceed

[tribal temple]

What are the eigenvalues/vectors you have?

[raven sparrow]

Eigen Values are : 4,9

Ohh I just realized I did the vector part wrong

Brb

[tribal temple]

No worries, take your time

But anyways, if those are distinct, enough to deduce that $A$ is diagonalisable, so there's a diagonal $D$ and invertible $P$ (made up how you expect) such that $A = PDP^{-1}$, from which, hopefully, finding $A^k$ is easy for any nonnegative integer $k$?

[grand pond]

@tribal temple

[raven sparrow]

Tysm I understood how to do it!!!
:))))

[tribal temple]

Perfect

[raven sparrow]

have a great day!

[tribal temple]

(@flat spire lurking to learn linalg too, I see )

[tribal temple]

You too

[raven sparrow]

.close

[midnight plank]

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[flat spire]

lmao never

[tribal temple]

We shall see about that

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[scenic urchin]

Anyone knows stuff about eliptical functions and like crypto currencies

[midnight plank]

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[last slate]

4th

[midnight plank]

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

[last slate]

idk how to start

[small jasper]

Integration by parts

[last slate]

hmm

so we have to solve it indefinately

anyway ?

[small jasper]

Yeah

[midnight plank]

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[midnight plank]

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[shrewd arrow]

is the ▽C supposed to be ▽C₀ ? (along with all the C inside the array)

[shrewd arrow]

from the video https://youtu.be/tIeHLnjs5U8?si=HoEWwP5BjVU3Y7ii

5:33

they are talking about this the entire video

and then show this without explaining anything else

they must talk about the same expression, right?

[midnight plank]

@shrewd arrow Has your question been resolved?

[shrewd arrow]

no one's doing ML math 😔

[wispy compass]

This is just a definition of that reversed triangle

Just a vector of partial derivatives.

[shrewd arrow]

is it valid for ▽C₀ as well ?

[wispy compass]

Yeah then it would be dC_0/dw's

See 'gradient'

[shrewd arrow]

all the dC_0/dw and all the dC_0/db right?

I think I get it
Also a few more questions abou this.. is the variable C₀ named that way throughout the video because it's for the first training example?

Assuming the answer to my previous question is correct:
It seems C₀ equals to a cost function without the part that sums all the training examples and divides by them(i think because the video doesn't showcase multiple training examples)
if we'd want to do the same process with multiple training examples, would we do the same thing but with a derivative of C instead of C₀?

@wispy compass

[wispy compass]

I haven't seen the video

[shrewd arrow]

Okay thanks. Hopefully someone else will be able to help with this

[midnight plank]

@shrewd arrow Has your question been resolved?

[last slate]

I haven't seen the video sorry

[midnight plank]

@shrewd arrow Has your question been resolved?

[timber tundra]

they did explain everything else

scroll back 10 seconds; C is an average of the cost function over all training examples C_k

which includes C_0

so what the gradient ▽C contains is the average values of the partial derivatives of the cost function w.r.t every weight and bias

ie these

theyre all averages over every training example

[shrewd arrow]

omg i don't know how i missed that frame.

thank you

[timber tundra]

yw

im on a quest to watch this series and read http://neuralnetworksanddeeplearning.com this summer so fun to see others doing it too lol

[shrewd arrow]

yeah it's a cool topic
i made this year a model that classifies 18 emojis from hand drawn drawings with transfer learning and fine tuning
can't share the model rn but i made the dataset available if you want to see: https://github.com/FlafyDev/emoji-drawings

i learned from coursera's courses

[timber tundra]

oh this is cool

is the model itself unfinished

[shrewd arrow]

it's finished

i can send a link to it tmr tho

[timber tundra]

👍

[midnight plank]

@shrewd arrow Has your question been resolved?

[midnight plank]

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[tidal turret]

How many units apart is any pair of parallel sides of a regular hexagon with side of $6$ units? Express your answer in simplest radical form.

[grand pond]

studying_calc_real_analysis

[tidal turret]

may I ask a geometry question?

[last slate]

6 root 3

[tidal turret]

can you elaborate

@last slate

bro you cannot dissapear after spoiling the ans

you gotta explain the tricks you used, theory etc with examples and kind words

[last slate]

Ohhh ok ok

I wi explain

[tidal turret]

<@&286206848099549185>

[last slate]

See the diagram

[last slate]

Hexagon will have 6 equailateral triangle within it

[tidal turret]

65°?

[last slate]

60 degree

[tidal turret]

60 60 60

[last slate]

Equilatera triangle

[tidal turret]

yes

[last slate]

Yep

[tidal turret]

how is 6 one of the sides tho

like of the equilateral triangle

is that due to similarity

[last slate]

Cuz in equilateral triangle all sides are equal

And side of hexagon is given as 6

[tidal turret]

but

OKAY OKAY

how did we got this is equilateral?

why 2 of the inner angles are 60 btw?

[last slate]

2pi/n

It is the angle formed by n side regular polygon side on center

And hexagon lie on circle

With center same as center of hexagon

[tidal turret]

pi/3 = 60

[tidal turret]

according to who?

[last slate]

Symmetry

[last slate]

N sides polygon will divide polygon in n equal parts and and complete angle is 2pi so the angle formed at center by each side will be 2pi/ 6

[last slate]

Bro i joined it with center too

[tidal turret]

okay so I understood this is equilateral, now what?

[last slate]

Now find the altitude

And twice of it will be the answer

Basically diameter of inner circle

[tidal turret]

wdym altitude, perimeter?

oh you mean height

a moment

we can use pythagorean theorem

,, 36 = 9 + b^2 \

[grand pond]

studying_calc_real_analysis

[tidal turret]

,calc 36-9

[grand pond]

Result:

27

[tidal turret]

b = sqret(27)

this is the altitude

,, 2\sqrt{27}

[grand pond]

studying_calc_real_analysis

[tidal turret]

is the answer?

@last slate }

but its saying that I need to explain in simplest radical form

that is impossible here this is not a fraction

[lime plover]

to get simplest radical form notice 27 = 3*9

$2\sqrt{27} = 2\sqrt{3*9}$

[grand pond]

Jaxx

[lime plover]

$=6\sqrt{3}$

[grand pond]

Jaxx

[tidal turret]

okay thanks

.close

[midnight plank]

Channel closed

Closed by @tidal turret

Use .reopen if this was a mistake.

[tidal turret]

.reopen

[midnight plank]

✅

[tidal turret]

can I ask other geo question?

Let $O$ be the circumcenter of triangle $ABC$. What is the measure of $\angle BAC$, in degrees?

[grand pond]

studying_calc_real_analysis

[slender walrus]

apply inscribed angle theorem

[tidal turret]

can you guys explain what it means?

why or?

oh, two ways of saying the same thing

how do I find coa

also, how do you know this angle is inscribed

,calc 360-210

[grand pond]

Result:

150

[tidal turret]

,calc 150/2

[grand pond]

Result:

75

[tidal turret]

answer is not 75

I need more hints

circumcenter?

<@&286206848099549185>

[cinder dome]

Isoceles triangles

As O is circumcentre

@tidal turret

[tidal turret]

ohh

,calc 360-210

[grand pond]

Result:

150

[tidal turret]

oca is 150

,calc 180 -150

[grand pond]

Result:

30

[tidal turret]

coa is 15 and oac is 15

,calc 180 -110

[grand pond]

Result:

70

[tidal turret]

boa is 35 aob is 35

,calc 15 + 35

[grand pond]

Result:

50

[cinder dome]

Did u get it?

O

Ye

It's right

👍

[tidal turret]

Yeah I got it

the part of the isoceles was key

but yeah all lines that start from circle and end in circumcenter are radii

really nice geo exercise

.close

[midnight plank]

Channel closed

Closed by @tidal turret

Use .reopen if this was a mistake.

[tidal turret]

.reopen

[midnight plank]

✅

[tidal turret]

The first square below is in position ABCD. After rotating the square $90$ degrees clockwise about its center point, the second square is in position DABC, as shown. Next, square DABC is reflected over its vertical line of symmetry, resulting in the third square in position CBAD. If the pattern of alternately rotating $90$ degrees clockwise and reflecting over the vertical line of symmetry continues, in what position will the $2007$th square be? Write your answer starting with the lower left vertex and continuing clockwise with the other three vertices. Do not use spaces or commas when entering your answer.

[grand pond]

studying_calc_real_analysis

[tidal turret]

can I ask about this one?

<@&286206848099549185>

[last slate]

!15mins

[midnight plank]

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.

[cinder dome]

Need to find after how many times it repeats

5th square is the same

As first square

[tidal turret]

2007 times?

[cinder dome]

So 2005th sauare

Will be the same

As first sauare

Square*

[tidal turret]

[cinder dome]

So it will look like the third one

[tidal turret]

ok let me sketch it out

the fifth square

,w 2007 mod 5

[grand pond]

Results provided by WolframAlpha

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

??????????

will be 2nd square

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

,w 2006 mod 5

[grand pond]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

,calc tan0

[grand pond]

The following error occured while calculating:
Error: Undefined symbol tan0

[tidal turret]

,w tan0

[grand pond]

Results provided by WolframAlpha

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

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

,w tan255

[grand pond]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

.close

[midnight plank]

Channel closed

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[midnight plank]

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[twin ridge]

Anyone feel like doing this with me? Would be nice to have someone to bounce ideas off of.

[kind zealot]

It might be a good idea to assume, without loss of generality, that a < b. This could help you pick a suitable ε

[twin ridge]

Without loss of generality meaning this would work the exact same way for b<a?

[kind zealot]

Yes

[twin ridge]

Then I could say that a=b+x, for some number x that fills the void, then prove that epsilon is smaller than x?

Smaller than half of x, I guess.

[sage olive]

yes

not even prove

just choose epsilon to be slightly less than half of x

[twin ridge]

Oh ok. That sounds pretty simple.

But it also looks like it should work so I'm gonna go with it because this is hour 8 on this assignment lol

4 part problems should be illegal.

[sage olive]

you can just say let epsilon = 0.49 * |a - b|

ez

done

[twin ridge]

I don't think my teacher would appreciate a 1 sentence proof lol.

I would though.

[sage olive]

there might be some more rigor in showing that they don't intersect

[twin ridge]

I'm gonna do the first one cuz I'm lazy.

[kind zealot]

https://tenor.com/view/todd-howard-it-just-works-bethesda-this-all-just-works-gif-20598651

[twin ridge]

You know how many times I said that when I was doing set proofs and just drew a Venn diagram?

[sage olive]

proof by trust me bro

[twin ridge]

My teacher doesn't trust anyone lmao

Although I'm gonna have to pick up the pace because my other math class starts next week.

And my 3rd one starts in two weeks.

[kind zealot]

Choosing open ε-intervals around a and b that share 1 end-point would quickly show this

[twin ridge]

Might be what I'm supposed to do tbh.

But I'm lazy.

[sage olive]

something like let epsilon = |a-b| / 2, then let the epsilon neighborhood around a be all points x such that |x-a| < epsilon?

then the point a + |a-b| / 2 is shared but not included

[twin ridge]

Pray for me neil.

I'm about to have 3 math classes and one other in 2 weeks time all at once.

I'm not even spiritual, but I need every force on this planet to get me through this semester.

[twin ridge]

I lowkey forgot this was a "there exists" proof.

Did I notate this correctly?

[torpid warren]

Just want to add my interpretation here:
(Informal) Basically, we just want to show that we can make two small bubbles around two distinct points a and b such that the bubbles are smaller than some size, and they don't touch each other

[twin ridge]

Right. It's intuitive, but I gotta actually show it.

Since you can get infinitesimally small.

[torpid warren]

Well, like the other helpers have mentioned prior, consider making the bubbles less than half of epsilon

[twin ridge]

Just did. I'm just a slow typer when it comes to these because I'm not perfect at LaTex.

[torpid warren]

Ah, so have you solved it?

[twin ridge]

Looking at this makes my eyes a tad blurry.

I know how to solve it, I just gotta type it.

[torpid warren]

No worries, good luck!

[twin ridge]

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[manic bison]

dunno where to start

[slender walrus]

negative exponent law

[manic bison]

i got it now lmao

[spark cedar]

you transfer x to the numerator and change the exponent to a negative one

[manic bison]

yea

[spark cedar]

and shorten x

[manic bison]

the main step was taking log 3 on both sides i was stuck on that

got it now tho

ty

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[manic bison]

didnt understand the second step

understood the alegbra

just not how logn a . logn b is logn c

[cyan sequoia]

The second line is subtraction of fractions

[manic bison]

i understood the alegbra

not the multiplication of logs

[cyan sequoia]

What do you mean

[manic bison]

if we simplify the thing above we will get 2 logs being multiplied for the denominators of the top and bottom fractions

somehow the turn into lognc and lognb

idk how

[cyan sequoia]

Log N b is erased

Hmm

Just substitute log N a = A like that and calculate it

[manic bison]

Like I'm getting this

The bottom equation

How do I proceed further now

WAIT

MB

GOT IT

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[young jolt]

I don’t know how to approach it

If I directly apply the formula which you can see I applied it by blue colour

The whole result will be 0 cuz sin0 is 0

What should I do

Yes

[hexed elbow]

how

[marsh galleon]

Have you tried using trigonometry?

(Like forming a right-angled triangle with the appropriate lengths from cot(x) = -5/11)

[young jolt]

I can’t find an answer by that

I used the 3rd formula

[marsh galleon]

Try expanding maybe?

[young jolt]

I got 0

[marsh galleon]

It looks like a cos(A+B) kinda thing will pop out?

[young jolt]

Ohhhhhhh

Shit

I’m blind let me try again

I see my mistake

[young jolt]

I wrote the cos^2x-sin^2x identity wrong

[marsh galleon]

I think every identity on the page was wrong 😭

[young jolt]

This page yeah

[marsh galleon]

Even the triangle btw

(theta is larger than pi/2)

[hexed elbow]

youve written sqrt 11?

[marsh galleon]

cot is just 1/tan

I'd say just start again

fresh mind :)

[hexed elbow]

perhaps this simplification might help?

[young jolt]

Sorry the question has cot-5/root(11)in it I wrote it wrong

[marsh galleon]

yea I was thinking something like this would pop up

[young jolt]

Actual question

[hexed elbow]

you can find sinx and cosx values from cotx

use half angle identities

to get it for x/2

[young jolt]

Ohh yeah

Let me try

[hexed elbow]

sure

[hexed elbow]

it should be B

try and see if you can get it

[young jolt]

Okay

[marsh galleon]

I also got B

fun question

[hexed elbow]

thats why its a hw question

[marsh galleon]

it really makes you use a lot of the stuff you know :)

[midnight plank]

@young jolt Has your question been resolved?

[young jolt]

I did it

Thanks

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[radiant vector]

I put d cause if the score is less than 9 (-1.5 z score) Id think that the z scpres would range from 0 to -1.5 but apparnetly not

[marsh galleon]

I'm not very knowledgeable on statistics/prob but it looks like Z is simply a transformation of X?

Some affine transformation (scale and then translate)

[radiant vector]

in a way

z score is basically a\how many standard deviations away from the mean

in his case the mean is 12 so lets say we have a score of 14

[marsh galleon]

that's 1 standard deviation above the mean

[radiant vector]

the z score is then 1 since 1 SD away from mean (sd is 2)

[marsh galleon]

so a z-score of 1?

nice!

[radiant vector]

yup

[marsh galleon]

so the probability that X is less than 9 is 3/sigma standard deviations below the mean

I see how you got to -1.5 as a z-score yea

I guess you can exploit the symmetry of the normal distribution?

the probability of having a z-score < -1.5 is the same prob as having one > 1.5

maybe?

[radiant vector]

I think ur right

cause since symmtry the values on each side r the same

[marsh galleon]

I'm a bit of a noob so I'm only thinking of these things as areas under the graphs

[radiant vector]

I think u got it thou

thanks lmao

[marsh galleon]

let's goooo

🙏

[radiant vector]

W

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[waxen silo]

what am I doing wrong?

[waxen silo]

doesn't cosh(u)/sqrt(1+sinh^2(u) make arcsinh(sinh(u))??

[jagged saffron]

That can be simplified

Alternatively you use their identity to simplify sqrt(1+sinh²(u))

[waxen silo]

I didn't use their identity bc I used a maplesoft if you know what that is

[jagged saffron]

Well you should be able to simplify arcsinh(sinh(u))

What is arcsinh?

[waxen silo]

1/sinh ?

[jagged saffron]

No, it's the inverse function of sinh

[waxen silo]

oh sinh^-1?

[jagged saffron]

Yes

Not (sinh(u))^(-1)

[waxen silo]

oh so does that mean sinh^-1(sinh(u) = u?

[jagged saffron]

Yes

[waxen silo]

since I would assume they cancel out

ah

[jagged saffron]

But really you should be using the identity to simplify the denominator as that's kinda the point of the sub

[waxen silo]

now I look at it, it's much easier