help-10

[wise hedge]

It's like a digression testing my linear equation skills. Like, what? why?

[royal basin]

they arent being tested as such

its more like, are you able to recognize and handle a linear(ish) equation in an unfamiliar context

cause if you cant then maybe thats something to work on

[wise hedge]

No, I can but I think Climate hit the nail on the head. To put it in simplest terms what's needed to make the equations true is to get a -1 out of the cosinse function.

[royal basin]

climate jumped ahead a step and a half

after 30 cos(30x) = -30 you would divide both sides by 30 to get cos(30x) = -1

[wise hedge]

Any odd Pi gets you a cosine of -1 in this example.

[royal basin]

odd multiple of pi

[wise hedge]

Okay, thank you for all the help. It just struck me as out of place for this course. They didn't ask or teach much like this in the material itself.

[high lily]

algebra is pretty much present EVERYWHERE

[wise hedge]

.close

[obtuse pebble]

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[viral tree]

You can't find the root to a negative number ?

[civic bridge]

yes

[viral tree]

I'm asuming its because -*- equal +

[civic bridge]

correct

[viral tree]

And if I search for cube root?

[civic bridge]

well you will get -

because --- = -

syntax...

but -*- x - = -

[viral tree]

so its possible to search for cube root of negative number but not square root

[civic bridge]

yes

[viral tree]

Thanks

[civic bridge]

any odd numbered root too

[viral tree]

if its positive odd number?

[civic bridge]

yes

[viral tree]

you can find the root tho right

[civic bridge]

yes

[viral tree]

its just gonna be a number with decimals

[civic bridge]

yes

[rich plume]

use
\*

[civic bridge]

okay thanks

[viral tree]

.close

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[timid silo]

Hello. How can I solve this equation $\frac{x-2}{x+2}=e^{\sqrt{\lambda}}$

[warm shale]

Şêro

[timid silo]

for x

[north pebble]

multiply both sides by (x+2)

[timid silo]

And then?

[north pebble]

what did you get ?

[timid silo]

You‘ll get x-2=e^lambda • (x+2)

[north pebble]

expand the right hand side

[timid silo]

Yea?

You‘ll get x-2=e^lambda x + 2e^lambda

[north pebble]

then try to keep everything that has an x in the left hand side and take everuthing else to the right hand side

[timid silo]

It got $x=\frac{2e^{\lambda}+2}{1-e^{\lambda}}$

[warm shale]

Şêro

[timid silo]

@north pebble

[north pebble]

correct

[timid silo]

Alright

.close

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[shy vigil]

i just tried putting the value and there's an error of sqrt
so just correct wherever you went wrong

,w solve{(x-2)/(x+2)} for x=(2e^b+2)/(1-e^b)

[warm shale]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by zplus11

[obtuse pebble]

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[versed stratus]

!status

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[echo gazelle]

1

so i swap t for x right then idk what to do

[versed stratus]

yes

And then, unless I'm mistaken, differentiate both sides to find the turning points

[echo gazelle]

how do i do that?

[kind hawk]

so, what way do you know to identify whether a function is increasing or decreasing

[echo gazelle]

if f’’ > 0 increasing

or is it f’>0?

[kind hawk]

f' > 0

ok

so we want to calculate F'(x)

now imagine you had an antiderivative of the integrand

lets call it G(x). then you know that G'(x) = (x-4)/(x^2+7)

yes?

[echo gazelle]

yes

[kind hawk]

and we know that the integral is equal to G(x)-G(0)

upper bound - lower bound

[echo gazelle]

yes

[kind hawk]

so F(x)=G(x)-G(0)

from which we can then find F'(x) = G'(x) = (x-4)/(x^2+7)

yes?

[echo gazelle]

so do i find the anti derivative then?

or no it’s just x-4/…

what u typed

[kind hawk]

its just what I typed

now we want to find where F'(x) > 0

[echo gazelle]

ah okay

[kind hawk]

aka (x-4)/(x^2+7) > 0

[echo gazelle]

so how do we do that?

[kind hawk]

we can multiply both sides of the inequality by x^2+7

do we know whether x^2+7 is positive or negative?

in other words, do we have to swap the inequality sign?

[echo gazelle]

is it positive?

[kind hawk]

why?

[echo gazelle]

i’m not sure like if u graph it its increasing

[kind hawk]

x^2 is always at least 0

so x^2+7 is always at least 7

in particular positive

so we can leave the inequality sign the way it is

what do we get?

[echo gazelle]

im confused does that just get rid of the x^2 + 7 then?

if you multiply the right 0 by x^2+7?

[kind hawk]

yes

[echo gazelle]

so x-4>0

[kind hawk]

yes

[echo gazelle]

so x>4?

[kind hawk]

yes

[echo gazelle]

so now what do i do with that information

[kind hawk]

we started with F'(x) > 0

and ended that this holds for x>4

[echo gazelle]

ah okay i get it

so the largest possible open interval where F(x) is increasing is (4,infinity)?

[kind hawk]

yes

[echo gazelle]

okay got it thank you for helping me understand

.close

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[obtuse pebble]

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[wise hedge]

$30cos(30x)+14=-16$

[warm shale]

j4w4

[candid yarrow]

Can you show your work?

[wise hedge]

So, I understand this equation better now that I have had time to think about it. The left side is a trigonometric function of the form: A * cos/sin(Bx) + D = some value on the graph.

The value they chose was -16.

The only values of X that will make the equation true are -1, so 2 points on the trigonometric graph will make the equation equal -16.

Knowing that the value -1 on a cosine graph points to a minimum value then if this were a temperature graph we could say when x = -1 the temperature would be at a minimum of -16 degrees.

[candid yarrow]

Is your task to solve for x?

[wise hedge]

https://imgur.com/wXcaQyl

[errant lark]

x = -1 doesn't make the equation true.

[wise hedge]

Right, sorry, when cosine is -1

[candid yarrow]

Can you rewrite this equation to the form cos(30x) = k for some k?

[wise hedge]

Yes I guess so but that would take it ouf of the trigonometric standard form which I would rather not do.

You would be moving the midline to the other side, taking it away in essence.

It's already an oddly non-trigonmetric question in my opinion.

[candid yarrow]

In order to solve this equation, this consideration is unnecessary

[wise hedge]

Yes, I understand but it's a bizarre question so making it less trigonometric in nature baffles me even more.

I can solve it by keeping it in the trigonometric standard form.

[candid yarrow]

So can you show your work based on this method?

[wise hedge]

There is not much work to be done other than first realizing that -1 is the required factor to solve the equation which is pretty easy to see.

-1 for cosine that is

[candid yarrow]

What do you mean by the required factor?

[wise hedge]

And so knowing that angle 180 degrees cosine is = -1, it's pretty easy.

The factor to multiply against 30 to get -30

[candid yarrow]

Do you mean that you found that cos(30x) = -1?

Given that 30cos(30x) + 14 = -16

[wise hedge]

Well, there's a variable x in there still so I'm not sure that makes sense, you have to choose an x first.

Don't you?

[high lily]

what would be the general solution to
$$\cos(\theta) = -1$$

[warm shale]

ℝam()n()v

[wise hedge]

I'm saying that when x = -1 then the equation becomes true.

Sorry, when cosine = -1.

[candid yarrow]

cosine of what?

[wise hedge]

angle 180

Pi

[candid yarrow]

cos pi = -1 is an identity

[high lily]

refer to my question in the above image

[wise hedge]

I won't argue that but what am I to learn here? Was I missing that all along?

[polar fossil]

cosine is a function, it can't equal a number
you mean "when cos(30x) = -1 then the equation becomes true"

[wise hedge]

when cos(whatever) is = -1 then the equation is true, so what value gives -1 as a cosine? Pi

[candid yarrow]

have you gotten similar questions right before?

[high lily]

pi is one of the values such that applying cos to it gives -1

what would be the general solution?

[wise hedge]

No, I've been doing this Khan course for like 6 months and I thought I was a master almost and then they throw this wacky question in on me and it has me baffled.

I'm getting it now more and more it's just the strangest of questions in this trigonometry course though I must say.

[polar fossil]

i must inform you that this is a very standard question, in fact the weirdest part about it is that there's only one solution in [0, 2pi)

but, there are still more solutions when you go outside that interval

[wise hedge]

Well, that might be so, but not in the Khan trigonometry course. They never exposed me to one like this in the material. This is popping up on a course review.

[polar fossil]

yes, this is asking you to synthesize the concepts from algebra with the concepts from trigonometry

[high lily]

if its in the review, similar quesitons would've come up before

[wise hedge]

Nupe, I've been doing this my friend, like I said for 6 months. The course is very legit, very thorough, but they did not expose these questions before.

They are humans who make mistakes.

No big deal to me.

[high lily]

have you done stuff with general solutions before?

[wise hedge]

Outside the course?

[high lily]

anywhere

[wise hedge]

Of course, I recognize your interval notation that you used. But they don't teach that in this course.

[high lily]

what would be the general solution to
$$\cos(\theta) = -1$$

[warm shale]

ℝam()n()v

[wise hedge]

Let me look at this a second.

Well, that would 180 degrees or Pi

[high lily]

as mentioned above

that's only one solution

the general solution describes all the posible solutions

[wise hedge]

Well, as far as I know, there is only 1 way to get -1 from a cosine function and that is when the x-coordinate is 180 degrees or Pi. So, I'm lost.

[high lily]

so you haven't done stuff with general solutions?

[wise hedge]

Maybe not

[hidden compass]

What is cos(-180°)? And cos(540°)?

[wise hedge]

This question has thrown me for a big loop

-180 is the same as positive 180

[high lily]

i suggest first looking up a vid on general solutions trig

[wise hedge]

I may need to do that then, a short coming of the Khan material

[high lily]

pretty sure its in the khan material

[wise hedge]

Why, when did you last take the course?

I've been doing it the most recent 6 months off and on. When time permits.

I can tell you there is no topic called "General Solutions" in the trigonometry course.

[polar fossil]

unit 4 lesson 2: sinusoidal equations

[wise hedge]

Oh, so under a different name then.

[hidden compass]

Yes, there's no need to make a separate chapter

[wise hedge]

Well, I'll have to give that another look I guess. It never hurts to refresh the memory. Thank You all for the help.

.close

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[timid silo]

Anyone know how to go about answering this? I’m not getting the correct solution.

[timid silo]

@polar fossil

[polar fossil]

um hi what makes you think i know anything about Fourier transforms

[tacit scarab]

what's the correct solution?

[timid silo]

Lol

[timid silo]

That in red

I partially have the correct solution after some chcange of signs and complex number

But no idea where sin(k)/pi(k) has come from

[timid silo]

What do you know?

[polar fossil]

nothing

head empty

[timid silo]

Case of ‘lights are on but nobody’s home??’

[timid silo]

Any idea

[tacit scarab]

nope

[tardy epoch]

Don't ping individual helpers

[tranquil geyser]

$\theta = 180, 540 etc$

[warm shale]

David

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[timid silo]

how to solve this using u sub

[gilded needle]

do you know how to integrate 1/(1+x^2)?

(without the 9)

[timid silo]

oh

arctan

[gilded needle]

yep

[timid silo]

I see it now

thanks

[gilded needle]

sure

[timid silo]

.close

[obtuse pebble]

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[robust sleet]

someone nice?

[wooden cipher]

You get what you get i guess

[royal basin]

probably code for not me

last time i tried to guide you through the problem you just went and ghosted me

[robust sleet]

solve for x?

ohh

[wooden cipher]

So are you familiar with clearing denominators

Sure

[royal basin]

without warning

[wooden cipher]

Anyway

Anyway

[frank monolith]

personal grudges in dms please

[wooden cipher]

.

I dont help in dms

[frank monolith]

for what?

[wooden cipher]

Yes

[royal basin]

did you really need to point this out

it's not the flex you think it is

[wooden cipher]

This is so cursed, goodbye

[royal basin]

ok now you did it

[gilded needle]

you don't seem like someone who actually wants nice people to help you

[royal basin]

i did see it

[tardy epoch]

I think he meant nice as in just give solution

[royal basin]

@junior sphinx don't troll.

[gilded needle]

that's actually mean, because he won't learn that way

[royal basin]

@timid silo don't feed trolls

[tardy epoch]

Yea that's true from the helper's perspective

[gilded needle]

(and his perspective, in the long run)

your wish is not my command

[royal basin]

if you recall yesterday that's exactly what i was gonna try to do. guide you toward the steps.

we stopped on the following:

you declared that you wanted to divide both sides by 21. i told you to do it. you complained that you could not. i asked why, and didn't receive any answer.

[wooden cipher]

Well you have the chance to answer now

[royal basin]

note i didn't say ghost now

i only said i didn't receive an answer

my way?

the division was YOUR idea, ftr.

i did not even give you any approval or disapproval, only a "do it and see what happens"

but in case you are still overthinking the ``do it'' part: dividing both sides by 21 in $$2 - \frac{2x-1}{3} + \frac{1-3x}{7} = 7-2x$$ would give you $$\frac{1}{21}\paren{2 - \frac{2x-1}{3} + \frac{1-3x}{7}} = \frac{1}{21}(7-2x)$$

[warm shale]

Ann

[royal basin]

well

you see how dividing by 21 was unhelpful right

multiplying both sides by 21 instead should do you more good.

i recommend that you first write it down unsimplified like i did, then only start simplifying on the next line.

break this recommendation at your own risk!

the original equation, but with both sides multiplied by 21 with no other black magic done to them.

❌❌❌

do you see how i wrote the division by 21?

i wrote 1/21 (LHS, AS-IS) = 1/21 (RHS, AS-IS)

i am suggesting you do the same, but instead of multiplying by 1/21, multiply by 21

write it on PAPER and send it here

or in microsoft paint or other image editing software

generally you should accustom yourself to not trying to do several things at once, simplification wise

do as i say then ill explain what i meant concretely

have you done what i told you to do? @timid silo

...

ok, bad mousewriting aside,

the top line is good

the bottom line is WTF

don't exaggerate its impact

i have seen far worse

big ask

go watch khanacademy or ochem tutor or both

i was gonna show you how i would write down and proceed with the next step, then sign off and go to sleep

i would probably be able to give you some kind of detailed intro if i had more time and more patience

both of which are in low supply

don't have one.

i gave you two yt channels that came to mind immediately, but i do not have any relevant books on the radar.

anyway

then go learn about the distributive property (singular, there is only one)

(two if you wanna be pedantic about left vs. right distributivity, which i know for a fact you don't)

the distributive property, basic algebra, and how to work with & simplify fractions

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[wanton dagger]

how would I do part a) ?

[grizzled shore]

Do you know what topic this is?

[wanton dagger]

yes, exponential growth and decay

[obsidian isle]

Can you write the DE?

[wanton dagger]

so we start from M=Ae^kt?

the bacteria doubles every 3 hours

so would that be M=Ae^3t?

[obsidian isle]

Should your base even be e?

Think about it. We are doubling

Lets consider a simple doubling pattern

We start with 1

Then 2

Then 4, 8, 16...

@wanton dagger

What comes next?

[unreal musk]

I mean if you choose k appropriately...
[assuming that t is supposed to represent time, grumble grumble denote your variables and something]

[obsidian isle]

Yeah but this is much simpler

You don't want to mess with logs yet

Everything should be expressed in terms of rational parameters

[obtuse pebble]

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[sharp ivy]

I was recently taught Sine, Cosine and Tangent, got this question and keep getting it wrong.
Given that sine is Opposite divided by hypotonus wouldn't the answer be 0.32?

[obtuse pebble]

Show your work, and if possible, explain where you are stuck.

[polar fossil]

,calc 7.78/23.9

[warm shale]

Result:

0.3255230125523

[shadow tangle]

well since the digit to the right of 0.32 is 5, round up. i think that is where your mistake was, assuming rounding is necessary in your case.

[obtuse pebble]

@sharp ivy Has your question been resolved?

[obtuse pebble]

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[green sphinx]

Can someone give me an image on when I would use "adding both arcs and dividing them by 2" ?

[green sphinx]

@zenith spade (sorry for ping, you had pre existing knowledge about what im talking about)

i cant seem to find any images on problems for when to use adding both arcs and dividing both of them

[zenith spade]

Pardon the drawing but something like this

[green sphinx]

which angle would that solve for

the blue?

[zenith spade]

Yes

[green sphinx]

oh ok

thanks again lol

.close

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[humble umbra]

I don't think I'm understanding the proof here. This section is talking about how the expected value can be used to find a mean.

[tardy epoch]

What's the question

[humble umbra]

I don't think I understand what the theorem is saying and the proof didn't help me.

So, what is theorem 3.2 saying

The expected value of g(Y) (a real-valued function) = the summation of g(y) p(y).

also I thought this was weird. p * (g_i)

[forest sinew]

they explain it

g(y) might not be one to one

so you accumulate them into a single object

[forest sinew]

does that help explain that part at least?

[humble umbra]

What does that do exactly

[humble umbra]

Not yet

[forest sinew]

like say g1, g2, ... are 1, 2, 1, 2, ....

then P(g(Y) = 2) makes sense to be the sum of all y values which could cause g to take on the value 2, right

I guess im maybe not being precise enough

theyre really just grouping them up, though

[humble umbra]

na it's me

My math maturity isn't there

[forest sinew]

its not totally you these prob ones always suck

the prob that g(Y) = g_i isn't a problem of a single y value is the point

[humble umbra]

they're grouping them up because g(y) could map onto multiple values right

[forest sinew]

well the issue is that g_i might be a result of g on multiple y values

like in the case of g_i = 1, 2, 1, 2, 1, 2 ....

so you have y_2, y_4, y_6, ...

all of these cause g(Y) to take on g(Y) = 2

[humble umbra]

yeah

I follow

that

[forest sinew]

so you really need to total them up

P( g(Y) = 2 ) = the prob that we get a y_i that makes g(y_i) = 2

because any of them will make the inner condition satisfied

[humble umbra]

you want to total them up to get around something like this^ ?

[forest sinew]

oh geez i havent read that far ahead yet

[humble umbra]

wait

I'm just trying to follow haha

[forest sinew]

im just explaining the equals

[humble umbra]

oh boy

[forest sinew]

lets see what they do here

E[g(Y)] = sum g_i p*(g_i) just makes sense

[humble umbra]

what does p*(g_i) mean though

that notation looks weird to me

cause it's a function

[forest sinew]

the nice thing i think is that like

[humble umbra]

I don't think I've seen f*(y) before

[forest sinew]

hmm

are they implying that g_i is the set of distinct values g takes on

i guess that makes sense

okay that makes more sense

so here we actually just have g_1 and g_2

in the example i was giving

while you have as many y's as you want

does this make sense

y_i = 1, 2, 3, 4, 5, 6, 7, ....

they are saying let g_i be the distinct set of outputs of g on Y

so g_1 = 1 and g_2 = 2

then $P(g(Y) = g_1 = 1) = p(y_1) + p(y_3) + \dots = p^*(g_1)$

@humble umbra

[warm shale]

jan Niku

[humble umbra]

wait let me break that down

[forest sinew]

note the piece " m <= n "

[grizzled shore]

Honestly I don’t fully get the last 2 lines too so jan I’m listening

[forest sinew]

frosst is typing

[grizzled shore]

[forest sinew]

oh no

i thought you were gonna take over

well the way i see it

[grizzled shore]

No I’m here to learn

[humble umbra]

So, P(g(Y) ) is the Prb of the prob distribution Y. This equals g_1 which equals 1

[forest sinew]

what

[grizzled shore]

No no

[forest sinew]

P( g(Y) ) means nothing

[grizzled shore]

g_1 = 1

g_2 = 2

[forest sinew]

lets make a smaller set of y's

[grizzled shore]

g_3 = 3

Etc

[humble umbra]

oh

[forest sinew]

Y = {1,2,3,4,5}

[forest sinew]

no

[grizzled shore]

What

[forest sinew]

we are using a specific example

[humble umbra]

you're just mapping the values ?

[forest sinew]

heres what were doing

they are using this stupid notation

Y = {1,2,3,4,5} okay

y_1 = 1, etc

[humble umbra]

yes

[grizzled shore]

Oh it doesn’t have to be 1 2 3

[forest sinew]

lets say that g(y_1) = 1

g(y_2) = 2

g(y_3) = 1

[grizzled shore]

It’s just whatever distinct values of g(Y) are

[humble umbra]

following

[forest sinew]

g(y_4) = 2

g(y_5) = 1

[forest sinew]

yes

im just creating an example so we have something to point at

[grizzled shore]

Yes go for it

[forest sinew]

the stupid notation the problem is proposing

note that n is 5 here, right?

|Y| = 5

y_1, ..., y_5

but, g(Y) for any y_i can only take on two different values

so create a set g_1, ...., g_m, with m<=n

that are all the values g(Y) can take on

here, g_1 = 1 and g_2 = 2, say

m = 2, n=5

are we on the same page there?

[grizzled shore]

Obviously m <= n since you have at most n outputs of the g function

[forest sinew]

sure

but the bold makes me feel like im making sense

[forest sinew]

what u think

[humble umbra]

reading through rn

why is there a restriction of only two values that y_i can take on?

[forest sinew]

theres not

neither in our example or in general

y_1 are the values that the discrete random variable Y can take on

[humble umbra]

replying to this

[forest sinew]

in our case its 1, 2, 3, 4, 5

[humble umbra]

right right

I follow that

[forest sinew]

im just making up an example

[humble umbra]

oh okay

[forest sinew]

were saying g(Y) is either 1 or 2

[humble umbra]

OKAY

[forest sinew]

but not that doesnt have to be the case

[forest sinew]

https://tenor.com/view/cat-cry-gif-crying-cat-gif-cat-gif-meme-cry-cat-cat-cry-gif-25865339

u dont have to yell

[humble umbra]

I totally follow 😄

❤️

[forest sinew]

okay so p*(g_i) lets us group stuff together

does that part still make sense

this guy

i cant remember where we were stuck

[humble umbra]

I was not sure what p * (g_i) means

[forest sinew]

its sort of like

[humble umbra]

I understand the lhs and middle

[forest sinew]

for some g_i

you might have multiple y's that make g(Y) = g_i

so the chance that g(Y) = g_i

is the chance that Y is one of those y's that makes g(Y) = g_i

[grizzled shore]

Let $G$ be the set containing ${g_1, g_2, …, g_n}\$
$\sum_{i=1}^{m}\sum_{\text{all } y_j\text{ such that }\newline g(y_j)=g_i}g_ip(y_j) = \sum_{i=1}^m \sum_{g(y_j)\in G} g(y_j)p(y_j)$

Is this what happened in the last 2 lines

[warm shale]

Frosst

[forest sinew]

idk to me it looks like they just moved a term under the sum then used the

used the

[grizzled shore]

Idk how to new line on the bottom of a summation 😦

[forest sinew]

oh youre right that is weird

they did just move it under the sum right

hold on i gotta pee

[grizzled shore]

Idk how they did the summation manipulation so if anyone knows I’d be interested

Anyway keep going with whatever part your on atm

[humble umbra]

yes plz

[forest sinew]

so how do you feel about p*

[humble umbra]

I hate it!@

jk

I'm getting there

let's continue

and I'll come back to it

[forest sinew]

well i mean they just use that defn

that gets you to the place me and frost are confused about

they apply a defn to get E(g(Y))

then substitute the definition for p*

[humble umbra]

this def

[forest sinew]

right

[humble umbra]

I do see that as well

[forest sinew]

alright youre up to where were confused

gimme a sec

[grizzled shore]

Please unconfuse me jan

[forest sinew]

idk it makes sense in the case of our example

the switch from m to n is just it right

those two sums are gonna make sure you leaf out to n

[grizzled shore]

No

I turns into j

[forest sinew]

since you have m-1 divisions of Y

or partitions or whatever

[grizzled shore]

And one of the summations disappeared

[forest sinew]

right

but thats fine

[grizzled shore]

I’m guessing it disappears due to nothing being dependent on the iterator after the substitution

[forest sinew]

youre really iterating over n with the two sums

yea thinking of the iterator on the second sum is screwy

thinking of it instead as something thats pulling out a subset of Y makes more sense

because you just have m-many of those subsets

and the sum of the cardinality of those m-many subsets is just n

since theyre all for sure gonna be distinct

[grizzled shore]

Shouldn’t there be n subsets?

[forest sinew]

nah

youre grouping y_i's into subsets based on what they produce in the image of g

[grizzled shore]

Oh

I mis wrote

Let $G$ be the set containing ${g_1, g_2, …, g_m}\$
$\sum_{i=1}^{n}\sum_{\text{all } y_j\text{ such that }\newline g(y_j)=g_i}g_ip(y_j) = \sum_{i=1}ⁿ \sum_{g(y_j)\in G} g(y_j)p(y_j)$

Should say m

Not n

For the set G

[forest sinew]

i feel satisfied

you know

[warm shale]

Frosst

[forest sinew]

i feel like they could have named stuff a lot better here

like using sets lol

[grizzled shore]

Lol

[forest sinew]

or idk

i hate that second iterator or whatever that is

@humble umbra so idk if i can explain but

[grizzled shore]

All y_j such that g(y_j) = g_i?

[forest sinew]

you okay up to here?

[forest sinew]

adopt their fucked up notation and let Y^*_j be all y_j such that g(y_j) =g_j

idk

thats a bad idea

[grizzled shore]

No stars for me thank you

[forest sinew]

stars are alright sometimes

sometimes u gotta do something

[grizzled shore]

I think it’s just the g for the function and g_i for the range

[forest sinew]

and when u gotta do something u might as well use a star

[humble umbra]

they just rewrote p*(g_i) using the def 3.4

I follow that

[forest sinew]

right

okay, next line is also nothing

[humble umbra]

[forest sinew]

this is just distribution

[humble umbra]

yes

[forest sinew]

okay

[humble umbra]

for summation right

this is a summation rule

[forest sinew]

sure you can think of it that way

its just like

well idk we dont need to contrive another example

its really just distribution

but if it helps to think of it as a rule then yes

[humble umbra]

ok

[forest sinew]

the last line is the hard part

but it helps a lot of you understand what each sum is doing

the first sum is picking out the distinct values that g can take on, given inputs from Y

and the second sum is saying, okay, so given some distinct g_i value, pick out all the pieces in Y that make g take on this value

does that make sense?

the second sum will never hit the same collection of y_i's twice

hold on, im gonna draw a picture

[humble umbra]

Just letting you know I read this and I'm waiting.

I haven't stepped away

[forest sinew]

trying to think of how to draw this

okay so

the first sum picks from the right circle

and the second sum picks from the left

so say first, the outer sum says, look at g_1, the yellow

then the second sum, the inner one, is used to go through all the y's inside the yellow circle

make sense?

like, you need two sums, because you might have multiple y's for each g

first you pick a g, then you go through all the y's

[humble umbra]

Yes actually

[forest sinew]

[humble umbra]

That actually makes a lot of sense

[forest sinew]

okay there is only one more jump to make

[humble umbra]

Cause before I was thinking something else when we started hahah

[forest sinew]

you need to believe that all of the y's inside Y are gonna be inside some circle

[grizzled shore]

Ohhhh

Ahh

Nice picture jan

[forest sinew]

[humble umbra]

Seriously

[forest sinew]

what you need to believe is that no y in Y is gonna be left out of a circle, and each y in Y is only gonna be in one circle

which should hopefully make sense, each y_i can only make g(y_i) take on one value, because thats part of the definition of a function

and g is given to be defined for all y in Y

[humble umbra]

Super clear

[forest sinew]

so each y in Y goes in one colored circle, and when youre done putting them in circles, youve used them all up

and since theres n total y's,

youre gonna end up doing n total multiplications in these two sums

once for each y, whichever circle its in

you hit each y in Y one time, find the appropriate g_i its matched with, multiply them, and move on

so really, its a single sum, over 1 to n, the number of y's you have in Y

[humble umbra]

Okay

[forest sinew]

thats this switch here

seeing that the double sum is really just a sum over the y's, each time picking the appropriate g

[humble umbra]

Right

[forest sinew]

and thats it

[humble umbra]

Okie

[forest sinew]

[humble umbra]

I probably would never have thought that far

That was super helpful

Ty

I'll read this again later.

.close

[obtuse pebble]

Channel closed

Closed by @humble umbra

Use .reopen if this was a mistake.

[obtuse pebble]

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[oak aurora]

https://math.stackexchange.com/questions/4743513/the-is-an-element-of-relation-and-axiom-of-substitution for full context.

Was looking through stackexchange and saw the post linked above, followed by the comment in the picture, and im kind of confused how logic works and the difference between an axiom and a definition.

Whats wrong exactly with defining X and Y to be equal iff every element of X is an element of Y and vice versa. Why is it necessary for us to make this statement an axiom instead?

[oak aurora]

The comment here talks about how you "need an axiom to insure that the first example holds" but doesn't it just still hold just by definition?

[dapper bloom]

I think the issue is in describing different types of equality.

In first order logic you might have a notion of equality called syntactic equality

Where writing a=b means a and b are the same strings of symbols.

[dapper bloom]

But that's not the same thing as saying A and B are the same things as sets

Or matrix equality or function equality and so on (ignoring that matrices and functions can be defined from sets).

You can introduce new objects into your formal logical system

But in the case of sets we want to be able to describe when two sets are equal in a sense that means more than just "are the same sequence of symbols"

Or another example is logical equivalence itself

Logical equivalence describes when two logical formulas are "equal" in a certain kind of sense that isn't the same as syntactic equality or set equality.

[oblique yacht]

?

[oak aurora]

i have almost no background in logic so its probably a me problem but even after reading this im still having alot of difficulty explaining why "defining X and Y to be equal iff every element of X is an element of Y and vice versa" is not enough, would you be able to give a more direct explanation in the context of this?

[dapper bloom]

So, in plain first order logic at most you have a notion of equality called syntactic equality.

[oak aurora]

yes so n=n+0, we say this two are syntactically equal right?

[dapper bloom]

What this means is you can see a=b for some strings of symbols a and b if and only if a and b are the exact same sequence of symbols more or less.

[dapper bloom]

No these are unequal in that sense

Because n and n+0 are different strings of symbols.

Just like AU{} and A are syntactically unequal.

[oak aurora]

oh so they are equal in the sense of them being equal numbers instead

[dapper bloom]

Yeah exactly. The notion of equality between numbers allows us to say two numbers are equal in some cases where they may not be the exact same strings of symbols.

[oak aurora]

ok i think im following thus far

[dapper bloom]

Equality between sets, strings of symbols, numbers and so on may be different relations depending on how you formalize things is all.

Without saying what makes two sets equal when you start formalizing set theory in ur zfc axioms, all you'e have left at best is syntactic equality.

(I mean like if you leave out extensionality)

[oak aurora]

How would the axioms of equality, transitivity/substituion/reflexive come in from this

[dapper bloom]

Those are just equivalence relation properties. Ideally we'd want different notions of equality we introduce to obey those properties.

For example, it would be really weird for a=b to be true but b=a to be false.

[oak aurora]

ah okay so that is sort of a separate thing entirely

we just happen to want equality to always be equivalence relation, something like that

[dapper bloom]

Well it's kind of like when we define equality between objects we make sure to only do it in ways that give us reflexivity and so on.

Mmm, modular arithmetic is another good example of this stuff.

When we say a=b in the context of the integers mod 7 we mean something different from a=b in the context of real numbers. The two usages of = are different.

But we don't use = when we mean a<b in the real numbers

There's weird exceptions to this like with big oh notation but I've never seen it in normal undergrad math.

[oak aurora]

but we don't have to say that a=b in the context of the integers mod 7 is an axiom. This one is allowed to be a definition since we can probably somehow use the definiton of a=b in the context of real numbers to construct a=b in the context of intergers mod 7 correct?

[dapper bloom]

Depends on what set of axioms you're starting from

[timid silo]

this axiom also insures that {1, 2} = {2, 1} and {1} = {1, 1}, so a set is only characterized by which elements it does and which elements it does not contain. There is no such thing as order of elements or a count for each element

[dapper bloom]

That's a good point.

[oak aurora]

your trying to give an example of how syntatic equality is not enough to define equality of sets correct?

[dapper bloom]

Syntactic like syntax

[oak aurora]

and since i only have syntatic equality, i ahve to make equality of sets an axiom

since I dont have a way to construct this type of equality

[dapper bloom]

What you've said is basically how I look at it at least lol

[timid silo]

yeah, like, im not sure but i think syntactic equality wouldn't ensure that {} = {} U {} for example

[dapper bloom]

Yeah it def wouldn't

[oak aurora]

I think my confusion now is the difference between an axiom and a definition, im just understanding it right now as an axiom lets us do new stuff while a definition lets us do stuff with stuff we already have, would that make sense?

or would this be too informal of an explaination

[dapper bloom]

I'm not entirely sure but with axioms you're assuming a particular thing is true. When you define something you're usually saying what certain things that (hopefully) already exist are, what you're going to call them, what notation you're gonna use for them and so on.

Like, the empty set axiom is an axiom.

It's a logical statement you're assuming when you do set theory.

The notation for the empty set is a definition.

You already know the empty set exists (because of the axiom), so you've said, "I'm gonna use this notation to refer to that thing".

Sometimes you have to be careful with definitions tho.

Like, you can't say I'm going to use the notation % to refer to THE even integer, because there's more than one even integer.

With the empty set you can show it's unique

So there's no issue with saying THE (singular) empty set and using a single symbol to refer to it.

[oak aurora]

using equality of sets right?

[dapper bloom]

Yep, and the empty set axiom and some logic.

I guess you can use definitions for things that don't exist too probably they'll just never apply.

Like, if you say "define B to be the set containing all nonempty empty sets", B will just be the empty set.

Or if you try and define a set to be a nonempty empty set I suppose you'd be assuming a contradiction.

So, you can't do any useful non-vacuous reasoning from it.

[oak aurora]

So going back to my very original question if we were only to define A=B as true if they share the same elements, this would sort of be valid but im using "=" sign abit weirdly

its the other direction A=B implies they share the same elements that needs the axiom part

[dapper bloom]

I'm not sure I understand your question entirely.

So, like the A=B iff A and B have the same elements, I'd say the important thing here is = is equality of sets.

It's not equality of some other weirder structure.

For example we could have a really boring set of axioms like that we only have two distinct objects a,b and that both of them satisfy some predicate R(x) but only B satisfies some other predicate G(x).

[oak aurora]

ok not too sure what i was asking either oops.

[dapper bloom]

We don't have some new definition of equality here.

Our logical system may or may not include a built in syntactic equality that allows us to say things like a=b is false.

[oak aurora]

I think i will just leave it at this, i still have some confusion but i think I have gained quite abit of insight

thanks alot for the past hour, its been very helpful

[dapper bloom]

No problem. It's an interesting question.

[oak aurora]

.close

[obtuse pebble]

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Closed by @oak aurora

Use .reopen if this was a mistake.

[obtuse pebble]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[grizzled shore]

Does the B here have to be symmetric?

And is that the only requirement?

[kind hawk]

B is not even square

[grizzled shore]

Oh true

But A is symmetric

So then B must be in a way symmetric too no?

[kind hawk]

no

[grizzled shore]

Or is that what the transpose is already doing

[kind hawk]

yeah

probably helps to just do an explicit example

see that basically any B works

[grizzled shore]

So B is just anything?

Well, with 0 or 1 entries

[kind hawk]

yes

B just says which vertices you should connect

for any B you can draw a graph

[grizzled shore]

There must be some sort of theorem about randomly naming your vertices then if there exists a row swapping esque matrix that puts you in this form then it’s bipartite

But row swapping will mess up the symmetry

So I guess only row swapping half the matrix (split down the main diagonal)

[kind hawk]

you would have to do row swapping and column swapping at the same time

which corresponds to switching names of vertices

[grizzled shore]

That’s true

I guess all that’s saying is there is a way to rename the vertices

Which is just not an issue if we don’t start with randomly naming them

Sounds like these bipartite matrices have got to be similar in some sense

[kind hawk]

you mean two isomorphic graphs have similar matrices?

[grizzled shore]

Yes!

[kind hawk]

something like that is probably true, yeah

[grizzled shore]

That is what I mean

[kind hawk]

although similarity is stronger

[grizzled shore]

But maybe not similar in the sense of matrix similarity

[kind hawk]

maybe there is something like permutation similarity

[grizzled shore]

That sounds like it should exist and very elementary so someone should’ve named it

[kind hawk]

P A P^-1

or P is even orthogonal, so P A P^T

[grizzled shore]

What’s what?

[kind hawk]

A is the adjacency matrix and P a permutation matrix

https://en.wikipedia.org/wiki/Matrix_similarity

actually mentioned at the bottom

https://i.imgur.com/SftVrlp.png

[grizzled shore]

P is orthogonal because it represents a change of basis?

[kind hawk]

no

[grizzled shore]

(I think I should go read about orthogonal matrices, last time I saw them was in QR decomp)

[kind hawk]

yeah useful to know in linear algebra

they are stuff like rotations in R^n

[grizzled shore]

I just took intro to discrete math and I’m already knee deep back in LA

[kind hawk]

😄

[grizzled shore]

Hmm, so if B happens to be a permutation matrix then we can define the graph as a bijective function?

Or is that only if B is also square

Wait permutation matrices are always square

[kind hawk]

B being square means that the two parts of the graphs are the same size

[grizzled shore]

And being a permutation matrix means it’s “1 to 1”?

And also onto I suppose

[kind hawk]

yes

[grizzled shore]

Which makes it bijective

I feel like I’ve seen a permutation matrix before

It’s almost like a random vector

But 2d

[kind hawk]

well they come up from time to time

[grizzled shore]

Ooh bistochastic matrix

Some probability thing

[kind hawk]

yeah

all of those matrices can be written as a convex combination of permutation matrices

iirc

[grizzled shore]

(No idea what convex combination is)

But that’s ok

[kind hawk]

linear combination but the coefficients are in [0,1] and sum to 1

[grizzled shore]

Anyhow, back to the original topic

If the adjacency matrix can be written in that form on Wikipedia then the graph must be bipartite

[kind hawk]

yes

[grizzled shore]

And what about the permutation thing

Isomorphic graphs have some sort of similarity in their adjacency matrix?

[kind hawk]

yes

by just reordering columns/rows

aka renaming vertices

maybe put into other words, for isomorphic graphs there exist two orderings of the vertices such that the respective adjacency matrices are equal

[grizzled shore]

Hmmge

Wadafak is the phi

[kind hawk]

the graph isomorphism

given by which vertices it maps to which

easier to see with a drawing

[grizzled shore]

Ok so it first looks at the line 1 and says this vertex has degree 2

If they are isomorphic it would’ve been sent to another row also with degree 2

Therefore it is now either at row 3 or row 4

So now we’ve reduced the problem to swapping rows with equal “sums”

So if we sum each row

[kind hawk]

(thats a necessary condition but not sufficient)

[grizzled shore]

Not sufficient?

Take 2 matrices (hopefully it’s the same size!)

Sum along the rows and create a new vector

If there exists a permutation between these 2 vectors then the graphs are isomorphic?

[grizzled shore]

Hopefully they are both symmetric too

[kind hawk]

just having the same degree does not mean that you can just send the vertices to each other and everything works out

[grizzled shore]

Hmm

It works for this case

[kind hawk]

well its a small graph

as an example, take two graphs. one is a big ring, the other is two smaller rings

both have the same number of vertices and the same degrees (everything has degree 2)

[grizzled shore]

Yes

[kind hawk]

but they clearly are not isomorphic

[grizzled shore]

Yes

Fuck

Why aren’t they

;-;

[kind hawk]

cause that would be too easy

[grizzled shore]

But this doesn’t work

FUCK

The bottom one isn’t symmetric

[grizzled shore]

Does this not work then

[kind hawk]

what exactly is the question right now?

[kind hawk]

to check whether the graphs of those two matrices are isomorphic?

[grizzled shore]

Yeah

Given 2 matrices, determine if they represent isomorphic graphs

Obviously the matrices need to be the same size, be square, contain only 1 and 0’s and be symmetrical

[kind hawk]

no

[grizzled shore]

Well, they don’t need to be 1’s and 0’s

[kind hawk]

they dont need to be symmetrical

[grizzled shore]

They just need to be natural numbers

Don’t they need to always be symmetric for undirected graphs

I thought adjacency matrices of undirected graphs are always symmetrical

[kind hawk]

oh you just want undirected

then yes

[grizzled shore]

Well let’s first do that case it sounds easier

Ok let’s not worry about non-simple graphs either

Let’s suppose we have 2 matrices representing 2 simple undirected graphs with the same amount of vertices

How can we know from the matrices if the graphs are isomorphic

[kind hawk]

check all options

graph isomorphism is a hard problem

[grizzled shore]

Ok my graph has 50trillion nodes

[kind hawk]

np complete probably

or np hard or whatever

[grizzled shore]

(I’m not a computer scientist so those words mean nothing to me)

[kind hawk]

hard problem

[grizzled shore]

Ok I’ve googled and it gives me a lot of words about it being hard

Which is unfortunate

[obtuse pebble]

@grizzled shore Has your question been resolved?

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[obtuse pebble]

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[timid silo]

can someone help explain to me why this is cos(110) and not cos(20)

cos θ = x cordinate

sin θ = y cordinate

[twilit cloud]

Yes because

[versed stratus]

what angle does the line make with the positive x-axis

[twilit cloud]

Angles start from the first quadrant

Imagine they start from point (1,0)

And they keep going counter clockwise

[timid silo]

oh

so that’s not the total angle is what i’m getting?

alright thanks y’all

.close

[obtuse pebble]

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[trim trail]

How do you solve part b (ii) onwards?

[tardy epoch]

What's theorem 3.28

[trim trail]

"Let $p$ be a prime and let $d$ divide $(p-1)$. Then, there are $\phi(d)$ distinct integers that have an order (mod p) equal to d."

[warm shale]

StatisticalCat

[kind hawk]

and is there a proof of that?

[trim trail]

in the textbook, yes

do you want to see the proof?

[kind hawk]

no

but you should check the proof and try to apply it to this specific case

[trim trail]

hmmm

ok lemme try

thanks

[obtuse pebble]

@trim trail Has your question been resolved?

[trim trail]

<@&286206848099549185> nope, not sure how to go about this

I understand that there is another theorem that states that if $d$ divides $p-1$ then there are exactly $d$ solutions to the equation $x^{d} \equiv 1~(mod~p)$

[warm shale]

StatisticalCat

[trim trail]

sorta used that to narrow down the values of k in b (ii)

[kind hawk]

do you not have a theorem which gives you the order of a^k based on gcd(k,19) ?

[trim trail]

nope

like there is one that could be interpreted like that ig

$if d = (b, c)$ and $a^b \equiv 1(mod~d)$ and $a^c\equiv 1(mod~d)$, then $a^{d} \equiv 1(mod~d)$

[warm shale]

StatisticalCat

[kind hawk]

send the proof of 3.28

[trim trail]

ok

I just wrote this for ii

purely of intuition

From the laws of modular arithmetic, the elements $a^{k}, a^{2k}, ..., a^{lk}$ will be primitive roots of $19$, where $l$ is an integer and $lk \leq 18$.

[warm shale]

StatisticalCat

[dusk relic]

Where can I assist you?

[trim trail]

not sure how to prove it... I mean I'm not sure my proof is rigourous enough

[trim trail]

here's what i wrote for b ii

[dusk relic]

So the proof you provided is an excerpt from a number theory theorem about the existence of distinct integers modulo a prime p with a given order (mod p) ?

[dusk relic]

Is that right?

[trim trail]

nope that's the theorem I quoted for denascite

I didn't use that ultimately

just something i tried out

[trim trail]

I just used modular arithmetic (multiplication)

[kind hawk]

damn I really dislike that proof. it doesnt tell you anything about whats going on

[trim trail]

EXACTLY

[kind hawk]

even though you could do it perfectly well constructively

[trim trail]

That proof took me so long to go through

[dusk relic]

Ok

[kind hawk]

thats false

[trim trail]

ohhh

[kind hawk]

so you want to know when (a^k)^d = 1

[trim trail]

yes

[kind hawk]

(a^k)^d = a^(kd)

what do you then know about kd

[trim trail]

kd congruent to 1mod p?

I mean that's kinda what I used in part i

[kind hawk]

what did you do in part i

[kind hawk]

kinda

but not exactly

what is b here

that is something you know explicitly

and then think about the order of a^4

[trim trail]

$b = ord_{n}(a)$

[warm shale]

StatisticalCat

[kind hawk]

I know that

but n=19

we know that

we know exactly what number b is

[trim trail]

wouldn't it just be b/4 if b is divisible by 4?

[kind hawk]

but what is b

[trim trail]

ok wtf im not getting what you're trying to say

[kind hawk]

b is an explicit number. you can say what value it is

[trim trail]

but the value of b depends on the value of a

[kind hawk]

it doesnt

[trim trail]

OH wait

no I still dont get u nvm

[trim trail]

how?

[kind hawk]

a is a primitive root mod 19

what do you know about primitive roots

[trim trail]

that their order is just $\phi(n)$

[warm shale]

StatisticalCat

[trim trail]

oh yeahh ok

so we know the value of b

FUCK

ok yeah got it

let me try solving this whole sum on my own now

thanks a ton for your help

i'll get back to u tomorrow if I can't solve it : )

tysm

[obtuse pebble]

@trim trail Has your question been resolved?

[obtuse pebble]

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• Be polite and have a nice day!

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[halcyon hull]

how do i find out the type of graph just by looking at the table?

[halcyon hull]

if the y values values go up then come back down, as the x axis increases, can i just assume its quadratic?

since theres be a curve, i guess

with this one though, the y values dont decrease as the x values increase

it keeps going up

so i guess that explains why its linear

[oblique nebula]

In quadratics y = x^2 and 1^2 is not equal to -1

[wooden cipher]

Are you familiar with nth order differences?

[halcyon hull]

but in the first one, the y values werent equal to the x values

and why is this considered as neither a linear nor quadratic

[oblique nebula]

In linears if you measure the y position every fixed x distance, it will increase(or decrease)by a fixed number

And you can deduce this is not quadratic by checking that y is not equal to x^2 in all cases listed in the table

[wooden cipher]

Thats not the only quadratic

[oblique nebula]

Maybe I missed something learning english 🙂 , I think I will learn something too

[halcyon hull]

wait a minute