help-36

[severe pike]

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[daring ruin]

help me find the h[n] that would have this z-transform

[daring ruin]

whole thing factors into $\frac{1}{\left(1-\frac{1}{2}z^{-1}\right)}\cdot\frac{1}{\left(\frac{2}{3}z^{-1}+1\right)}\cdot\left(1-3z^{-1}\right)$

also i have these formulae but idk if they can fit

[soft zealot]

shiru

[daring ruin]

maybe its worth doing a PFD

-1/2 z^-1 = 1 -> z^-1 = 2
A = (1 - 6)/(4/3 + 1) = -5/(7/3) = -15/7

2/3 z^-1 = -1 -> z^-1 = -3/2
B = (1 + 9/2) / (1 + 3/4) = (11/2)/(7/4) = 44/14 = 22/7

great so its $\frac{\frac{22}{7}}{\left(\frac{2}{3x}+1\right)}-\frac{\frac{15}{7}}{\left(1-\frac{1}{2x}\right)}$

[soft zealot]

shiru

[daring ruin]

2nd term is exponential

first term is.. ??

negative exponential?

huh

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[jagged flare]

wtf

[jagged flare]

ok

hint 1:||assume a>=b>=c and prove a|bc||
hint 2:||bound lcm(a,b,c) and show that it must equal bc||
hint 3:||p|b+c implies p|bc. what does this give?||

i have no idea what hint 3 is talking about

i did 1 and 2 and simplified the problem to ab+ac=3bc

https://math.stackexchange.com/questions/4833943/textlcma-b-c-fracab-bc-ca4

also unless im dumb, all solutions here just stop at lcm(a,b,c)=bc?

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[still eagle]

idk how to do

can someone help

[severe canyon]

Start by dividing by 2 on both sides

[still eagle]

sin 2x = 1/2?

thenn what do i do

@severe canyon

[severe canyon]

Then you solve it like every other trig equation, which I supposed you already have seen

I mean, would you know how to solve sin(t) = 1/2?

[still eagle]

i forgot

[still eagle]

in calculator?

sin -1 (1/2

????

??

[tawdry kestrel]

?

[still eagle]

Hi

can u help

[tawdry kestrel]

Ye

[still eagle]

how do u know

[tawdry kestrel]

Mb

Mb

[still eagle]

sin^-1 (1/2) = 60?

[tawdry kestrel]

Sin (30°) =1/2

[still eagle]

ye

its 30

[tawdry kestrel]

Right

[still eagle]

Ritht

[tawdry kestrel]

So 2x= 30

[still eagle]

yes

[tawdry kestrel]

x=15

[still eagle]

what about ASTC rule

theres multiple answers

[tawdry kestrel]

Ye that comes

[severe canyon]

Sure

[still eagle]

thats what my teacher write

[tawdry kestrel]

Now we also know that sin 180-x = sinx

[still eagle]

How

[severe canyon]

Yeah, that's correct

[still eagle]

Is the 15 always A?

[severe canyon]

Unit circle, for example. Or sine of a sum formula

[severe canyon]

What is A?

[still eagle]

ASTC

the a

all positive

[severe canyon]

Mmh I don't know these names my bad

[still eagle]

i dont relaly understand it

[severe canyon]

Where I live we don't use them 🤷‍♂️

[still eagle]

is there an easier method

[severe canyon]

Well look at your notes. Your teacher must have explained you these things previously

[still eagle]

Ok

wait

[severe canyon]

You only need to know, in order to solve that equation, that

sin(t) = 1/2 has these solutions:

t =30° + k360° V t = 150° + k360°

[tawdry kestrel]

Supplementary and complementary trig identities

[severe canyon]

Yeah you have to learn these identities for sure, they are extremely important

[still eagle]

why isnt it -sin

[severe canyon]

Because that would be wrong

[tawdry kestrel]

Astc rule bro

Sin is positive in 2nd quadrant

[still eagle]

whats k360

[severe canyon]

The periodicity

[still eagle]

😭 sorry

[tawdry kestrel]

K= integer

[severe canyon]

Maybe you use n or other letters? @still eagle

[still eagle]

Dont u add 180 if its a

[severe canyon]

Did your teacher say this? 🤔

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[rain sentinel]

can someone explain where they got this from thanks

[muted prairie]

Every function of the form Acos(ωt)+Bsin(ωt) can be expressed that way, as a single sine function with a phase shift

If you expand out Rsin(ωt+α) using the sine of sums formula and match it to the above, you can find the values for R and α

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[minor sandal]

Thwr was a video on zero order by Stanford, which was for computer science AI students.
And I wanted to learn zero order logic for Real Analysis.

Will it be sufficient. Or should I start Hammock book

[candid pulsar]

<@&268886789983436800> korean goon

ty

[minor sandal]

Kindly help

[candid pulsar]

im not very educated in undergraduate syllabus...

hopefully someone who knows comes across this

[onyx peak]

You probably dont need a lot of logic for analysis

What helps is having some experience with proofs (and it doesnt really matter whether its logic, naive set theory, number theory or something else)

[minor sandal]

I studied 3 chapters without knowing that there is something called as"zero order or first order logic"

I knew only those quantifiers from pre university math

But when I come across , negation of a theorem , or statement. I am having some trouble accepting those

[onyx peak]

I see, in that case it probably won't hurt to learn some logic. If I were you, I'd probably just watch the vid and find some exercises focused on combining logic with natural language (so negating statements, intepreteting english sentences as logical ones..)

[minor sandal]

On YouTube , there's everything taught related to cs

And i hate those "mathematics for computer science"

And i m persuing a degree in computer science student

[onyx peak]

Then youll meet logic anyway

[minor sandal]

We have disrespected mathematics a lot in this course

[onyx peak]

So if u wanna get ahead a bit, you can start reading the book as well

[minor sandal]

I think I should study negation of a statement and do some exercise as u said

[stone flint]

Enderton has a rather decent book on mathematical logic that you can perhaps look into.

(if you want a written companion to your video resources.)

[minor sandal]

Thank u , but that's quite expensive for me .
I will refer the pdf

[minor sandal]

Thank u

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[onyx peak]

Huh

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Bot is offline it seems?

[barren hound]

oh no

[sonic crystal]

it seems you dont have the permissions. allow me to graciously assist you

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[latent dragon]

the hell

[floral zenith]

Um

Broken

[minor sandal]

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[proper dagger]

!15m please.

[final saddle]

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.

[proper dagger]

well, guess politeness is wasted on you then.
<@&268886789983436800>

[wild venture]

<@&268886789983436800> ban him (troll)

[full jungle]

what the fuck is this question

[lucid nymph]

you just posted some messy looking proof involving primes and are now asking a question like this

[tulip coyote]

How about we ban you for not following our rules instead

[warm tundra]

helpers are volunteers who are giving up their free time to help you. take an hour to remember basic courtesy and respect

[full jungle]

you're js trolling

[tulip coyote]

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[tulip coyote]

(no no, you had a point the first time!)

[wild venture]

Yes, it is what you deserve

[dense coral]

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[minor sandal]

I was learning , how can I translate my proof to logical notation.
And i got this template.

Is this correct. And objects here refers to variable and the set which to which they belong or maybe some conditions associated with these

[jagged wave]

imeanyeah a lot of statements are structured vaguely like this

[wild venture]

Looks fine enough

[minor sandal]

Is the definition of "objects" that i stated correct?

[minor sandal]

"vaguely"

[jagged wave]

well not every statement ever is gonna be covered by a universal quantifier

[minor sandal]

Okay

[minor sandal]

Objects refer to all those variable na .
Like we say "natural number >k " and such kind of statements

[onyx peak]

"All prime numbers except for two are odd" fits the template:
"Forall naturals n, n is prime and not 2 --> n is odd

"Sets are equal if they have the same elements"
Forall sets A, B, A has the same elements as B --> A = B

"Linear functions satisfy f(x+y) = f(x) + f(y)"
Forall functions f, f is linear --> f(x+y) = f(x) + f(y)

Just a couple of examples

[onyx peak]

Yeah, it's important to realize that those variables arent necessarily numbers though

they can be sets, functions, numbers, graphs, various algebraic structures, ...

[minor sandal]

Understood.

[onyx peak]

Also my examples arent perfect (e.g. the third one should explain what x and y are, using another quantifier), but they should be good enough to explain the template

[minor sandal]

But they are not conditions we can say ..
Those are descriptions

I hope u understand what I m tyeing to say

[minor sandal]

Talking about those objects

[onyx peak]

Are you trying to say that e.g. "n is prime" shouldnt be a condition but rather a description? Or something like that?

[minor sandal]

We have the statment of theorem

And from that we have to place specific parts into their places

Like objects , hypothesis and conclusions

Forget it .
I understood

Thank you

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[toxic tree]

Why is this true?

[opal plinth]

Determinant

[toxic tree]

Chartbit!!! ❤️ ❤️

[toxic tree]

Ahh ok

Thanks!

❤️

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[gloomy jasper]

how do you do 10 ^ 1.5 or 5 ^ 1.6 or something like tht

[sturdy wasp]

Put it in a calculator…?

[gloomy jasper]

but just how.....

[vital crag]

can you be more specific

are you asking how calculators do it or how to use a calculator

[sturdy wasp]

Are you asking what it means to take something to a decimal power?

[gloomy jasper]

ye

[abstract bramble]

Idk by hand but you can write it as a root raised to a power

well you can do roots by hand as well but its tedious

so you write 1.5 as 3/2

[vital crag]

you need to learn calculus

[gloomy jasper]

bruh

[vital crag]

this explains it well:
https://math.stackexchange.com/a/4803452

[abstract bramble]

lol

[gloomy jasper]

but how will 10 ^ 3/2 help more?

[abstract bramble]

you write it as the second root of 10 raised to the 3rd power

second root commonly called square root

[gloomy jasper]

okey

that explains it

[abstract bramble]

yeah you can turn 3/2 into 3 * 1/2 and according to power rules...

blah blah

1/n exponent is nth root

[next thorn]

1.5 --> 3/2
1.6 --> 8/5

with exponent rules, you can turn
10^(1.5) to 10^(3/2) to (10^1/2)^3

if.... thats what you mean

[gloomy jasper]

yeah i think i got it now

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[tranquil wharf]

.

[versed dawn]

Guys is this correct for derivatives (differentiation method)

[barren hound]

no

write √x as x^1/2 and use power rule

[final saddle]

@versed dawn Has your question been resolved?

[versed dawn]

Why is it 1/2? I applied identity rule to x so it became sqrt 1

[barren hound]

but you didn't have x

you had √x

[versed dawn]

Ohh okaay, sry it hasnt been tackled for us yet

Tysmm

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[worldly raven]

I have the question Find the Value(s) of k for which the quadratic function f(x) = 2kx^2 + kx -k +2 has real roots.
I can do all of it except work out which of the values of k have <= and which has >=. How can I work that out?

[vital crag]

Check a smaller or larger value

[worldly raven]

just substitute for k?

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@worldly raven Has your question been resolved?

[plucky rover]

Would help to show what you've done so far

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[static oak]

Feel like I am going crazy because I got the right equations but I swapped them. This is the answer key btw

[final saddle]

@static oak Has your question been resolved?

[static oak]

Also for 8 a why is the period according to the answer key 3 instead of 3.3

[tulip coyote]

Why do you think the period is 3.3?

[static oak]

Because between each of the points so far in the period it was .75. With the given points the difference between them was double that which is how I got. .75

so like .75 is 1/4 of the quarter

[tulip coyote]

0.75 is a quarter of the period?

[static oak]

I think

Because the difference between .3 and 1.8 is 1.5

Wait

ok I get it now

Sorry I’m sick I have no idea what my brain is doing lol

[tulip coyote]

It happens to us all, sometimes to me even when I'm not ill

As long as you're happy with that one

[static oak]

Yup I’m good likely will be back tho. Cus I have a test on Wednesday so today and tommrow I do all the previous homework from this unit. Well this time I am only doing it from the quiz onward since the test is mostly on th second half of the unit and it’s a longer unit

[tulip coyote]

Awwww, well you're always welcome back whenever of course best of luck for all the preparing for the test, I may see you before then

[tulip coyote]

Also were you alright with this one btw?

[split fox]

she sleeps early i think

she’ll be back in the morning for sure haha

[tulip coyote]

I figured, I've worked with her before

[static oak]

Sorry just saw this I’m still confused on that one

And already have another question for this one. I just don’t know how they got the angle measurements this is from the first half of the unit so it’s been a while

[shell condor]

Are you asking how they calculated arccos(0.7), well for that you generally need a calculator

[static oak]

No how they got 45.6% as a result maybe it is just putting in a calculator but I did not get that when I put it I

[shell condor]

It's in radian

[static oak]

Ah

well glad I wasent doing anything wrong besides the calculator issue

[shell condor]

You gotta make it in degrees, which you can do either through a setting on your calculator or by multiplying the radian answer with 180/pi

[tulip coyote]

(also, when you're ready, maybe you could explain how you went about deciding the equations for these? )

[static oak]

Yea I’m ready

[tulip coyote]

Cool cool I'm wondering if maybe you have sec and csc mixed up for that one, maybe?

[static oak]

Sec is the reciprocal of sin right?

You know what

That might be it

Csc is sin right…

[tulip coyote]

Yep, that's it

[static oak]

Ok I’m glad that was nothing super complicated

Most of my errors on my tests are little stuff like that which is good because it’s not tons of points but also very irritating because I know I know it

[tulip coyote]

Yea that's what makes them even more frustrating, it's not like it's completely impossible so you know you can't, it's like it's so close and yet, just, arghhhh

Mostly a matter of time, practice and all

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[raven loom]

Can I get help on solving this? I know it's the 3rd time, but I still can't get it

[raven loom]

Hello back, day later

[barren hound]

Here are three things you can do to an equation like this:

• append " º g^-1" to the end of both sides
• prepend "g^-1 º " to the beginning of both sides
• cancel "g º g^-1" into nothing

where g is whatever function symbol you want (so you could do it with f or h as well)

so for (a) if you wanted to isolate f, you could try appending º g^-1 to both sides and see where that gets you

[raven loom]

And how can I do that?

[barren hound]

you can just kinda

$f \circ g = h$

$f \circ g \orange{\circ g\inv} = h \orange{\circ g\inv}$

[soft zealot]

schrödinger's kitten

[raven loom]

What is the rule between ordering the functions?

[barren hound]

it's like in algebra if you start with 2x + 3 = 7
you can do 2x + 3 - 3 = 7 - 3

[raven loom]

Like what order should I apply

This is what triggers me the most

[barren hound]

ummm the main rule is that you can't replace f º g with g º f

[raven loom]

Yes because it changes the whole composition

[barren hound]

we're not really evaluating any of these functions here; we're working with them as objects

just like if we had 2x + 3 = y and we wanted to solve for x

[raven loom]

Yes but I don't know why is

[barren hound]

we'd get x = (y - 3) / 2 but at no point did we ever have a real value for x or y

[raven loom]

$h \circ g^{-1}$ g after h, not before

[soft zealot]

JPuXIUim6x

[barren hound]

ahhh okay yea

the idea here is that you need to be consistent with both sides

you can do whatever you want tho like

i could do uh

[raven loom]

YOu can't

Because if you change the order, then you change the composition

Therefore changing final output

[barren hound]

$f \circ g = h$

$f \circ g \blue{\circ f} = h \blue{\circ f}$

that second equation is a true equation. It's just not very useful.

[soft zealot]

schrödinger's kitten

[raven loom]

Ok so I cancel out the g by applying inverse and therefore doing identity

[barren hound]

yes, exactly; you can apply the inverse (on the right) and then replace g º g^-1 with the identity

[raven loom]

Is identity here just 1?

Or is it nothing

[barren hound]

it's often written as i or sometimes e? or you might not have a symbol for it

like you could go from $f \circ g \circ g\inv$ to something like $f \circ \iota$ or you might just go directly to $f$

[soft zealot]

schrödinger's kitten

[raven loom]

Will it make a difference if I place it like $f \circ g^{-1} \circ g$?

[soft zealot]

JPuXIUim6x

[barren hound]

going directly from f º g to f º g^-1 º g isn't an operation you're allowed to perform

[raven loom]

Why?

[barren hound]

things you are allowed to do:

• compose a function on the left or on the right
• collapse g º g^-1 to the identity

well it's just like regular algebra, you can't go from

3x + 4z = 7

to

3xy + 4z = 7y

[raven loom]

I don't even know what happend in this equation tbh

[barren hound]

the one i just wrote? it doesn't really make sense; how did i go from 3x to 3xy? i multiplied by y. But then i should have done that to 4z as well

[raven loom]

Still not clear, but going forward I think

What about solving the B part

[barren hound]

in your functional equation f º g = h

you can compose both sides with g^-1, but trying to insert it in the middle somewhere doesn't really make sense -- how would you apply the same operation to the other side?

you can't reach into the guts of h and shove a g^-1 in there

[raven loom]

So If I want to cancel out f, in the second question

I cannot put it between normal f and g

[barren hound]

that's correct

[raven loom]

I must put it at the end?

[barren hound]

which end?

[raven loom]

$f \circ g \circ f^{-1} = h \circ f^{-1}$

[barren hound]

that is a valid operation; let's see where it gets us

[soft zealot]

JPuXIUim6x

[raven loom]

Like this?

[barren hound]

we have $f \circ g = h$

$f \circ g \orange{\circ f\inv} = h \orange{\circ f\inv}$

[soft zealot]

schrödinger's kitten

[barren hound]

right, so now do you see in here anything you can cancel? any instance of f º f^-1, or f^-1 º f?

[raven loom]

We can cancel f

[barren hound]

how?

[raven loom]

Because with f and f^-1 we will just revert the operation after completing g

So might as well just not do it

[barren hound]

do you happen to have a rubiks cube?

[raven loom]

Nope

[barren hound]

how about this, do you have a piece of paper handy?

[raven loom]

I have A4 paper

Right now in front of me

[barren hound]

go ahead and draw a heart on it, nice and dark so we can see through it because we're gonna flip it over in a sec

what would happen if we:

• rotate it 90º clockwise
• then, flip it horizontally
• then, rotate it 90º counterclockwise

[raven loom]

I have a qustion

I have it flat laying on m table

If I rotate it 90 degrees it fill just flip orientation, is that it/

[barren hound]

it should turn it like sideways so the heart has its point facing to the left

[raven loom]

So the heart is on the other side of the paper

And it's upside down

[barren hound]

yep good

okay now put it back to how it was when we started, with it upright facing you

and now i don't want you to rotate at all, just flip it horizontally

[raven loom]

Ok

[barren hound]

now what does it look like? which way is it facing?

[raven loom]

It is upside down again

And not facing me

[barren hound]

it should be facing you but on the other side of the paper

[raven loom]

I mean yes

On the other side

[barren hound]

yep so the result is different, right?

[raven loom]

No

wAIT

wHAT IS horizontal flip

No I was doing it correctly

[barren hound]

um like put your arms out, grab the sides of the paper, then cross them over each other

[raven loom]

OK what is the point here

[raven loom]

THis is not correct/

[barren hound]

these are the results i would expect you to see

[raven loom]

YOu are saying flip vertically

Not horizontally

[barren hound]

oh sorry :3 yea i figured that might cause some communication issues, my bad

[raven loom]

What is G doing?

[barren hound]

g is flipping

[raven loom]

Horizontal flip?

Ok so the sides aren't equall

So I messed up

[barren hound]

i guess i could colour it darker to denote that it's on the backside of the paper

the point is that if you do f, then g, then f^-1 you end up in a different spot than if you just do g

[raven loom]

Shit

Ok what did I do wrong

[barren hound]

i'm... not sure? maybe you're flipping a different way to try to "fix" it

if you follow the way i've drawn out the steps, do they make sense to you?

[raven loom]

Yes

Function is just giving an output

So the operation with the paper

And I messed up the operation order, that's why it's not equal to h

[barren hound]

yes, the operations with the paper are examples of functions

in this case, f is "rotate by 90º clockwise" and g is "flip vertically"

[raven loom]

I don't understand wht it's not working

[raven loom]

So this is not correct

I don't know why

[severe canyon]

It is correct, but not useful at all

[barren hound]

that is a totally valid thing that you can do; you can compose f^-1 on the right all you want. it just won't be very useful

instead, let's compose it on the left

[raven loom]

Why would I compose it on the left

And what it changes if it's composed on the left

[barren hound]

starting with $f \circ g = h$, we can compose on the left to get:

$\blue{f\inv \circ} f \circ g = \blue{f\inv \circ} h$

[soft zealot]

schrödinger's kitten

[raven loom]

But the order doesn't change nothing

[barren hound]

and now look what we have!

[severe canyon]

To cancel your undesired function, since you need to isolate g

[raven loom]

Ok why I can't put it infront of f?

[severe canyon]

It does!!!

[severe canyon]

??

[barren hound]

like why can't you go from f º g to f º f^-1 º g?

[raven loom]

Why I can't put f inverse infront of f

[severe canyon]

...

[barren hound]

well... because what are you going to do on the other side?

[raven loom]

So $f \circ g$ is one composite

[barren hound]

if you're trying to shove f^-1 into the middle of the left side, you'd have to shove it into the middle of the right as well

[severe canyon]

That's what has been done here: #help-36 message

[barren hound]

and again you can't do surgery on h

[soft zealot]

JPuXIUim6x

[raven loom]

And I am theoretically creating new one with $f \circ f^{-1} \circ g$?

[soft zealot]

JPuXIUim6x

[severe canyon]

No

[barren hound]

there's just kinda not a consistent way to go from

$\rsq \circ \bsq$ to

$\rsq \circ \psq \circ \bsq$

[soft zealot]

schrödinger's kitten

[barren hound]

like it's hard to say what you did to get there

[raven loom]

Ok makes sense in some way

[barren hound]

and especially it's impossible to do the same thing to the other side of the equation

[severe canyon]

These are the ONLY allowed operations @raven loom
Do not forget that!

[raven loom]

But that means, with g I have to do the same thing?

I can't shove it in the middle?

[barren hound]

yep, you can only append or prepend

[severe canyon]

You can never put things in the middle exactly

[raven loom]

Oh that makes sense

But why in algebra I can?

Oh no I can't, if there are brackets

[barren hound]

generally in algebra we're working with commutative operators -- stuff like how a + b is the same as b + a

[raven loom]

So will it be the same framework in question C?

[barren hound]

so if we have 3x + 4 + 7y = 12, we have no problem going straight from there to 3x + 7y = 8, because we subtracted 4 from both sides

[proper dagger]

one way I can see to show why it's impossible is to think of f and g as putting on two different pairs of socks and the inverse being to take it off. if you left-compose (append to left) the inverse of f to f \circ g, then you are putting on sock g, then putting on sock f, then taking off sock f. this works. but then h would be putting on both socks as a unit at once, so appending f^-1 (taking off sock f) works there too. what doesn't work is asking h to take off sock f in the middle - there is no middle in the instructions for h because the two socks are put on together!

(if it helps. sorry if it doesn't, I'll nuke the message)

[barren hound]

heck i should have gone for the shoe analogy why didn't i think of that haha

and yep question C is the same, it just requires multiple operations

you'll want to make sure to do one at a time

[raven loom]

$f^{-1} \circ h^{-1} \circ f \circ g \circ h = f^{-1} \circ h^{-1} \circ k$

[soft zealot]

JPuXIUim6x

[raven loom]

Like this?

[severe canyon]

No....

How are you gonna cancel the function composition on the left??

[raven loom]

By removing them?

[severe canyon]

Do you see any $f\circ f^{-1}$ ?

[raven loom]

No?

[soft zealot]

Alberto Z.

[barren hound]

this is a valid operation you can perform, but again you'll find it's not very useful. Try dealing with one thing at a time (f or h)

[severe canyon]

And do you see any $f^{-1} \circ f$ ?

[soft zealot]

Alberto Z.

[raven loom]

No!

[proper dagger]

the problem with that, OP, is that you are asking a reader to do this

[severe canyon]

So how are you gonna cancel function f ???

[raven loom]

With erasing them?

[severe canyon]

BUT HOW???

[raven loom]

Well isn't this being told to me since 50 minutes?

[proper dagger]

put on sock h, put on sock g, put on sock f, take off sock h, take off sock f. well, taking off sock h, with two other pairs of socks on top of it? that ain't gonna work, is it?

[raven loom]

Ok so what If I want to cancel out operation in the middle?

[barren hound]

you can't :P

[severe canyon]

@raven loom

DO NOT FORGET THAT

[barren hound]

you can only cancel stuff on the edges

[raven loom]

Ok now it makes sense

Let me try again

[severe canyon]

It's been told you more than once!

[barren hound]

alberto please

[raven loom]

I am not the smartest person here

[severe canyon]

Sorry

[raven loom]

And I do not have the highest intelligence here

[barren hound]

to be tbh i didn't really understand this stuff until i was working with matrices in college

[raven loom]

But I greatly appreciate the motivation to become a listener

lul

I had matrices already im in 11 grade

[severe canyon]

It was not about that, I just wanted to remind you why what you're doing is somehow non sense

[barren hound]

awesome :) matrices and matrix multiplication are an example of an operation that (often) has an inverse and (usually) is not commutative

[proper dagger]

I will be here as a standby if necessary

[barren hound]

in fact matrix multiplication and function composition are almost exactly the same thing

[raven loom]

$f^{-1} \circ f \circ g \circ h \circ h^{-1} = f^{-1} \circ k \circ h^{-1}$

[soft zealot]

JPuXIUim6x

[barren hound]

yea that seems good to me :) so now you can spot the places where you can cancel stuff down to identity

[raven loom]

BUt wait

Why $f^{-1} \circ f = f \circ f^{-1}$?

[soft zealot]

JPuXIUim6x

[barren hound]

ah yes, that one is i guess a kind of special case, because both of those are the identity function

[raven loom]

So those follow the definition?

[barren hound]

the "inverse" of a function f is defined to be the function that undoes its operation, both on the left and on the right

[raven loom]

$f(x) = y <-> f^{-1}(y)=x$?

[soft zealot]

JPuXIUim6x

[barren hound]

so if the inverse of f exists, then we know that $f \circ f\inv = i = f\inv \circ f$

[soft zealot]

schrödinger's kitten

[barren hound]

yea this is correct as well

[proper dagger]

for now you can assume that composing a function with its inverse anyhow will give you the identity

[raven loom]

It's not true in reality?

[barren hound]

there is a concept of so-called "left inverses" and "right inverses" but those come up in different structures, not so much in functions like these

[proper dagger]

not all functions satisfy this relationship

[barren hound]

ow

[plucky rover]

That gif hurts my head wtf

[proper dagger]

but to bother you with them requires knowing left vs right inverses and so this is off-limits

[barren hound]

im deleting that for accessibility reasons

[plucky rover]

Thank you

[raven loom]

Wait

Does it reall do something

[barren hound]

and because it does my head in

[raven loom]

What did I just do

[proper dagger]

same I think bina is watching

[raven loom]

What did I just do

With this gif

[barren hound]

some people have a lot of issues with flashing lights and stuttering gifs like that

[raven loom]

Oh shit I didn't know

Im sorry

[barren hound]

np, but yea if it just says "f inverse" then that means it cancels both on the right and on the left

[raven loom]

So it works on function have inverses?

[barren hound]

that's correct, if a function isn't invertible then you won't really be able to do algebra with it like we've been doing

[raven loom]

How can a function be non-invertible

[barren hound]

well, an example of a non-invertible function is f(x) = x^2

[proper dagger]

try inverting x^2... damn it hayley

that snipe

[raven loom]

I inverted a quadratic equation yesterday

It had x^2

[proper dagger]

oh??

[barren hound]

and did you find multiple solutions?

[raven loom]

No It just asked me to do so

[plucky rover]

Wut

[plucky rover]

Was it like "find an inverse that works in this interval"

[barren hound]

ahhh, they restricted you to x ≥ 3

[raven loom]

Yes domain limit only

[plucky rover]

Oh they did nvm I'm blind

[raven loom]

But I did

[barren hound]

yep, and that function restricted to that domain is invertible

but if we considered f(x) = x^2 + 4, where the domain is the real numbers, then that would not be invertible

[raven loom]

Ok makes sense

Thank you for 60 minutes of pure confusion

And absolute zero clarity

[barren hound]

haha no problem. if you do ever come back to matrices and matrix multiplication, you'll find that they feel very very similar to this

[raven loom]

I wish I was so nice to help people so much

[raven loom]

So this is not associative?

[proper dagger]

[barren hound]

they are associative but not commutative

[raven loom]

I don't remember term meanings

Yes ok

[barren hound]

associative is like (ab)c = a(bc)

commutative is ab = ba

[stone flint]

associativity has to do with bracketing. commutativity has to do with order.

[raven loom]

there was a third one

[stone flint]

transitivity?

[raven loom]

nu

[barren hound]

there's the distributive property, are you thinking of that? that's mostly specific to addition and multiplication

where a(b + c) = ab + ac

[raven loom]

Yes

[stone flint]

distribution only makes sense if you have two separate operations to talk about. then you would say one operation distributes over another if they follow something like a(b + c) = (b + c)a = ab + ac.

[barren hound]

you have implicitly assumed that operation is commutative as well :P

"most" operations that you work with will be associative. One example that isn't associative is exponentiation.

[raven loom]

OH yes blast university stuff in my channel

Where algebra is even confusing something

I love it

[calm ermine]

Abstract algebra is wild

[stone flint]

exponentiation over multiplication is also one of those weird examples where distribution works only one way.
(ab)^c = (a^c)(b^c), but a^(bc) \neq (a^b)(a^c).

just a piece of trivia, no need to cook yourself over this.

[raven loom]

IM talking about high school algebra

[calm ermine]

👍

[barren hound]

abstract algebra is algebra but with different rules :P

[raven loom]

[stone flint]

I'd say abstract algebra is elementary algebra but you are more interested in the structure and ruleset over concrete applications, which I suppose is why it is never really taught in school.

[calm ermine]

Sometimes its taught in hs but thats usually only an optional module but yea anyways we should prolly close if this is finished

[stone flint]

sorry for derailing the conversation OP. do you have anything left unanswered? let's get back to those.

[raven loom]

Hopefully we will never meet eachother, if I am going for applied mathematics

I am done for today with questions probably

[raven loom]

Talkibing about pure mathematics

[raven loom]

Oh yes I need this in my life

Does everyone go through this?

[stone flint]

I can only see the last panel. I think I am very much screwed.

[barren hound]

if you're doing math in college then abstract algebra will be a required class, even for applied math. but college math in general is dramatically different than high school math

[raven loom]

How can I survive it?

[barren hound]

you'll be prepared by the time you get there

[calm ermine]

Its not that bad tbh

[barren hound]

it was one of the easier classes, you have fun constructing these silly little worlds

[raven loom]

I hope

Thank you everyone

[barren hound]

also i'm not saying you have to get a rubiks cube but like

[raven loom]

I don't know the count of people in this help channel

[calm ermine]

Not the rubiks cube 🥀

[raven loom]

Probably at least 10

.close

[final saddle]

Channel closed

Closed by @raven loom

Use .reopen if this was a mistake.

[raven loom]

.reopen

[final saddle]

✅ Original question: #help-36 message

[raven loom]

You know I might have more questions

Even though I know how to solve this compositions, but It's like doing this because Ive memorised it

The only rules that I can recall is 1. it works like algebra, 2. You can't put stuff in the middle, 3. Cancelling out doesn't matter the order

Is this enough or I lost something?

[barren hound]

yea i mean that's the basic idea. again it's very similar to matrix algebra

[raven loom]

Ok

Another question

Is anything regardign a variable a function?

Even a standalone equation 2x+1?

[stone flint]

this, without an equal sign, is not an equation any more. it's just an expression.

[raven loom]

Ok

[stone flint]

it can be made a function of x by explicitly stating that it is, but as it stands it is just an expression.

[raven loom]

OK thanks again

.close

[final saddle]

Channel closed

Closed by @raven loom

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[minor sandal]

I cannot understand (iii) part.
How to follow this

[tired walrus]

what do you mean by (iii)

[minor sandal]

Sorry (c)

[tired walrus]

do you mean that you can't even parse statement c? or that you don't understand how it connects to one or more of the others?

[strange sparrow]

it is just the statement that

,tex $limsup_{n to infty} x_n = inf_{n geq 0} sup_{m geq n} x_m $

[minor sandal]

I understood it's first part .
u_m .
But I cannot picture or imagine how x* is equal to infimum and limit

[tired walrus]

in your photo there is a part that says "b implies c". you should look at that and also at whatever comes next that says "c implies d"

[strange sparrow]

You know that lim sup x_n = lim u_n right?

but since u_n is decreasing, lim u_n is the same thing as inf u_n

[minor sandal]

u_m is the set of all Supremum , of mth tail of the sequence

And we choose lower bound of this set (infimum) .

How does it equals x*

[strange sparrow]

did you read what I wrote at all

[minor sandal]

?

[minor sandal]

This words i understood.
But how it equals x*

I cannot relate inf(u_m) and x*

[strange sparrow]

Can you state your definition of limit superior

Also it is not clear what you're confused about as Ann said

Are you having trouble following the proof? Or are you not sure about why intuitively it's true?

[minor sandal]

Intuitively

[strange sparrow]

Can you follow the proof given that this is equivalent to the inf(sup x_m) definition?

[minor sandal]

The first line is causing some confusion

"Sufficiently large m".
Why do we require such m

[strange sparrow]

Why do we require?

Do you mean "how do we obtain"?

[minor sandal]

Leave it

Sorry

.close

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Channel closed

Closed by @minor sandal

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Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[soft heart]

where am i wrong?

[proper dagger]

,rccw

[ebon agate]

What does at mean here?

[barren hound]

i can't even tell what your answer is

[soft heart]

i didn't solve it completely

acc. to the solution..
the 4 and 5 i wrote in second line..
has to be reciprocal

so 30:5

6

then 6 will be multiplied by 4

8x = 24

x = 3

does that make sense?

[barren hound]

i also got 3

[soft heart]

ok but i didn't get the logic

[barren hound]

i went with uhh
we bought them at 15 for $8
so to make a 25% profit we need to sell them at 15 for $10
dividing by 5, we get 3 for $2

[final saddle]

@soft heart Has your question been resolved?

[soft heart]

.close

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• Show your work, and if possible, explain where you are stuck.
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• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[ornate kestrel]

can some1 help

[ornate kestrel]

Let

[
S_n:=\sum_{m=1}^n m^2\ln m.
]

By manipulations involving the Barnes (G)-function we obtains the exact identity
\begin{equation}\label{eq:Sn-G}
S_n = n^2\ln n! - (2n-1)\ln G(n+1) + 2\sum_{j=1}^n \ln G(j),
\end{equation}

We already know
\begin{equation}\label{eq:P2}
P_2(n)=\left(\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}\right)\ln n-\frac{n^3}{9}+\frac{n}{12}.
\end{equation}
And

[
\ln A_2 = \lim_{n\to\infty}\Big[-P_2(n)+S_n\Big].
]

The standard result is

[
\ln A_2 = -\zeta'(-2),
]

and numerically (-\zeta'(-2)\approx 0.03044845706).

Standard known expansions:
\begin{itemize}
\item Stirling (for large (n)):

[
\ln n! = n\ln n - n + \tfrac12\ln(2\pi n) + \frac{1}{12n} - \frac{1}{360n^3} + O(n^{-5}).
]

\item Barnes (G) asymptotics (we write only the pieces used):

[
\ln G(n+1)=\frac{n^2}{2}\ln n - \frac{3}{4}n^2 + \frac{n}{2}\ln(2\pi)-\frac{1}{12}n + \zeta'(-1) - \frac{1}{240n^2} + O(n^{-4}).
]

Replacing these in $S_n = n^2\ln n! - (2n-1)\ln G(n+1) + 2\sum_{j=1}^n \ln G(j)$ identity we had and simplifying we get:

\begin{equation*}
\boxed{\sum_{k=1}^{n-1} k^2\ln k = \left(\frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n\right) \ln n - \frac{1}{9}n^3 + \frac{1}{12}n^2 - \frac{1}{12}n + \zeta'(-1) + \frac{11}{360} + \frac{1}{360 n} - \frac{1}{240 n^2} + O(n^{-3})}
\end{equation*}

(That constant (11/360) is the Euler--Maclaurin contribution coming from Bernoulli numbers and the specific integrand.)
\end{itemize}

Putting this result in:

[
\ln A_2 = \lim_{n\to\infty}\Big[-P_2(n)+S_n\Big].
]

The $P_2(n)$ term completely cancels out and we are left with:

[
\boxed{\ln A_{2}=\zeta'\left(-1\right)+\frac{11}{360}}
]

And as we already know $\ln A_{2}=\zeta'\left(-2\right)$

So we get:
[
\ln A_{2}=\zeta'\left(-2\right)
]

Whats wrong in this?

[soft zealot]

ts pmo

[ornate kestrel]

by whats wrong with this i mean why dosent it numerically give the correct answer

$\begin{equation*}
\boxed{\sum_{k=1}^{n-1} k^2\ln k = \left(\frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n\right) \ln n - \frac{1}{9}n^3 + \frac{1}{12}n^2 - \frac{1}{12}n + \zeta'(-1) + \frac{11}{360} + \frac{1}{360 n} - \frac{1}{240 n^2} + O(n^{-3})}
\end{equation*}$

[soft zealot]

ts pmo
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[ornate kestrel]

full latex for the equation incase u want the full picture

[final saddle]

@ornate kestrel Has your question been resolved?

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[ornate kestrel]

can some1 help

nOOOO

I DINT MEAN TO CLICK YES

NOOO

brh

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[plain rain]

Hi. Suppose we have a vector field, V, and a field, F. A homomorphism from V to F usually takes every unique equivalence class of (applied) vectors in V and maps each one as a single point in F, or am I far off from that definition?

[plain rain]

the equivalence relation being ambiguous, perchance

ill go ask in the linalg channel

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[clear vessel]

ignore the question. just looking at the diagram

[clear vessel]

to find the red angle why is it

and not over -0.05?

[vital crag]

Depends what tan(theta) is used for

looks like the red angle is reference angle:
https://www.cuemath.com/geometry/reference-angle/

[clear vessel]

alr

i figured out why just now

but thanks

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[fallen ocean]

How do I go about getting that answer?

[fallen ocean]

Oh that looks rly weird whoops

[muted prairie]

The limit shouldn't depend on x and y

this can be solved by factoring the numerator

[fallen ocean]

Oh it's sort super simple isn't it

Ok I'm just slow

😑

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[worldly spruce]

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[eternal pelican]

chat can i get a tad bit of help over here

[eternal pelican]

so i kinda get the way of doing 1p

but 2ps is actually kinda cooked

[tired walrus]

what do you mean by "1p"

do you mean like when the letter p appears in only one coefficient?

[eternal pelican]

yes

[tired walrus]

so you know how to work out the discriminant in that case, yes?

[eternal pelican]

yes

[bronze patrol]

Yo, I’m shit at maths

Help please

[eternal pelican]

the discriminant is just -b^2 -4ac no?

[bronze patrol]

I’m failing

[eternal pelican]

create a new ticket pelase ^w^

[tired walrus]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[tired walrus]

b^2 - 4ac, no minus on the first term.

but anyway,

[eternal pelican]

oh right mb

[tired walrus]

what if i told you the method was exactly the same

[bronze patrol]

Shit mb I just joined

[tired walrus]

no matter how many times or where p appears

[eternal pelican]

oh?

hm

mind blowing

[tired walrus]

you still can and should work out b^2 - 4ac (in terms of p obviously)

[eternal pelican]

i lowk cant imagine it

okay let me try

uhh

where does the p go

oh wait

wait wait

would it be

(-6 * p)^2 -4(9)(-8)?

or how

😓

🥀

<@&286206848099549185>

it feels so wrong pinging helpers

[icy tulip]

nah its fine

wait a few mins i am waiting for the question to load

[eternal pelican]

would it be
(-6 * p)^2 -4(9)(-8)?

this is how far i got

[icy tulip]

this the one right

[eternal pelican]

yes

[icy tulip]

oh ok thanks

so you should solve for D = 0

im sure you know what D is right

[eternal pelican]

mhm

[icy tulip]

and how to find

[eternal pelican]

is this right?

does this =D?

[icy tulip]

one minute, im stilll thinking

intel pentium core processor is processing

[icy tulip]

uhh no

you forgot something

first term is alright

something in your second term needs to be checked

the -4ac part

[eternal pelican]

(-6 * p)^2 -4(9 * p)(-8)

[icy tulip]

yessir

[eternal pelican]

okay

yay

[icy tulip]

now its a simple quadratic equation

im pretty sure you can do that youselves

[eternal pelican]

-36p^2 - 128p

[eternal pelican]

hold on i kinda forgot

[icy tulip]

buh

whats p-6 whole square

basic algebraic identities here

[eternal pelican]

oh

right

i actually forgot what to do from here

🥀

[icy tulip]

nah its fine i can help

[eternal pelican]

yes pls

[icy tulip]

so you can simplify (p-6)^2 right

[eternal pelican]

oh by rooting everything?

so we are left with

-6p - 8 sqrt(2)

[icy tulip]

p^2-12p+36 + 288 + 32p?

i shouldn really be feeding you the simplification tho

you dont need to root it just multiply add and factorise

[eternal pelican]

OH

I GET IT

wait

let me cook

[icy tulip]

i may have messed up so do it yourself again

[eternal pelican]

oh okay

[icy tulip]

kbai

[eternal pelican]

okay thanm u

baibai

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[wraith glen]

hi

[wraith glen]

i just don't know the first step to solving the question

if this was my exam i woulda guessed like -1 and 1

[ebon agate]

Notice the stuff in summation is same just that there's a slight change in limits

[wraith glen]

i see

i noticed that ye

it's like Z has two extras X summations

[ebon agate]

Z kind of "contains" X

[ebon agate]

Yeah

If Z=X

[wraith glen]

ooh shoot

[ebon agate]

Those two extra summations must add to?

[wraith glen]

0?

[ebon agate]

Yes

[wraith glen]

but then that would just mean that X and Z are equal for all N right

[ebon agate]

No

Why?

[wraith glen]

you said the two extra summations will equal 0

[ebon agate]

They should provided x and z are equal

[wraith glen]

hold on let me try this question real quick

[ebon agate]

Not necessarily

[ebon agate]

Sure

[wraith glen]

i just got N = 28

i don't get what to do now tho

i can N = 28 and N = 48?

bruh the answer key sayd 4

but is 48 right?

[ebon agate]

I mean can you show your working

[wraith glen]

hold on

i skipped steps so lmk

[ebon agate]

It would be (2n+1)i*pi

[wraith glen]

thats crazy

not intuitive for me bro

[ebon agate]

n in 2n+1 is not the N in question

Basically 2n+1 means an odd number

So pi,3pi,5pi,7pi,...

[ebon agate]

Why?

[wraith glen]

i get it tho thank u bro

but why can't the answer be N = 56* also

they said it's N = 4 and N = 28

[ebon agate]

Why would it be 56?

[wraith glen]

it's periodic

do I always need to divde?

[ebon agate]

It's not

1/N is periodic

It doesn't mean N is periodic

[ebon agate]

?

[wraith glen]

naw thank you big bro I got it

.closed

.solved

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[vital crag]

<@&268886789983436800>

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[night sierra]

how come all differential functions are continous but not all continous functions are differential?

[vital crag]

proof: https://proofwiki.org/wiki/Differentiable_Function_is_Continuous

the simplest way for the converse is to come up with a counterexample

[modest sequoia]

easy example is y = |x| where it is continuous through out (meaning there are no breaks, jumps, or holes). however, it is not differentiable at x = 0 since you cannot draw a tangent line.

[final saddle]

@night sierra Has your question been resolved?

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[ornate thistle]

Need a help understanding how to solve radical/quadradic equations.

[ornate thistle]

I have pulled up a website where i can practice them but I had a hard time understanding the lecture and graspign the concept.

[mint orbit]

say we started with the first one, do you have an idea how to proceed?

$2x = \sqrt{x+3}$

[soft zealot]

jan Niku

[ornate thistle]

Sorry

was waiting so i was doing some off screen prep

I know the first step is we need to isolate the radical(s) on either side of the = sign. If there were more then one but since there isnt more then one and its already on one side and the only term all thats left to do is square both side of the equation to remove the radical.

[mint orbit]

okay, what do you get

[ornate thistle]

$4x^2 = x+3$

[soft zealot]

Zolo

[mint orbit]

can you turn it into a root finding problem

[ornate thistle]

How do you mean?

[mint orbit]

I just mean, how do you want to proceed

usually if we have something that looks like some polynomial terms, we collect them all on one side

leave = 0 on the other

[ornate thistle]

well I would assume now we make it a 0 = ax + bx + c equation

[mint orbit]

yea

[ornate thistle]

Which should give us

$0 = -4x^2 + x + 3$

[soft zealot]

Zolo

[mint orbit]

lets flip it, $0 = 4x^2-x-3$

[soft zealot]

jan Niku

[ornate thistle]

You would factor it by -1 right?

[mint orbit]

for no particular reason

yea, just divide by -1

[hexed bison]

Yeah, it's usually easiest to work with when the first term is positive

[mint orbit]

flip all the signs

you dont have to, just for my sanity

You could solve this by inspection, if you wanted

maybe its easier to use the formula?

[ornate thistle]

So now its a quadradic

[mint orbit]

yea

[craggy wing]

its probably because u multiply it by -1

[hexed bison]

you can think of it like that, or just moving the terms to the other side of the equation

[ornate thistle]

How would you kow whats best to do because I knwo you can either factor it or use the formula

[mint orbit]

i would try inspection for 30 or 45 seconds

then use the formula

personally

we'd have to factor out the 4, right

$0 = 4\qty(x^2-\frac x4 - \frac 34)$

[soft zealot]

jan Niku

[mint orbit]

we need two numbers that add to -1/4 and multiply to -3/4

[ornate thistle]

Which is tricky

so formula would be eaiser

[mint orbit]

theres some obvious things to try

but yea, if it doesnt come to you pretty quickly

id just use the formula

always good practice though

[ornate thistle]

what syntax does the bot use

im unsure what qty is

[mint orbit]

its (simplified) latex

qty is part of the physics package, which is default here

it makes brackets and parens auto scale

(the fractions require bigger parens)

[ornate thistle]

Anyway.

How would you write the formula for quadratics

i know its -b +- sqrt(b^2 - 4ac)/2a

[mint orbit]

$ax^2 + bx + c = 0$ implies that $x= \frac{ -b \pm \sqrt{b^2 - 4ac}}{2a}$

[soft zealot]

jan Niku

[ornate thistle]

in this case
a = 4
b = 1
c = 3

so

[mint orbit]

not quite

b = -1 and c = -3

[ornate thistle]

I thought the signs were irrelevent within the formula

[mint orbit]

no, signs are actually really really important

[ornate thistle]

ok so

$x= \frac{ 1 \pm \sqrt{-1^2 - 4(4)(-3)}}{2(4)}$

[soft zealot]

Zolo

[mint orbit]

thats impressive you got that on the first try

the only thing i'd change is the first piece under the root

(-1)^2 instead of -1^2

[ornate thistle]

$x= \frac{ 1 \pm \sqrt{(-1)^2 - 4(4)(-3)}}{2(4)}$

[soft zealot]

Zolo

[mint orbit]

nice

[ornate thistle]

Now we reduce

$x= \frac{ 1 \pm \sqrt{1 + 48)}}{8}$

[soft zealot]

Zolo

[ornate thistle]

$x= \frac{ 1 \pm \sqrt{49)}}{8}$

[soft zealot]

Zolo

[ornate thistle]

$x= \frac{ 1 \pm \sqrt{49}}{8}$

[soft zealot]

Zolo

[mint orbit]

you know sqrt(49) off the top of your head?

[ornate thistle]

$x= \frac{ 1 \pm 7}{8}$

[soft zealot]

Zolo

[mint orbit]

nice

[ornate thistle]

So

$x = \frac -68$

[soft zealot]

Zolo

[ornate thistle]

whoops

[mint orbit]

you can wrap it in {}

frac{}{}

[ornate thistle]

$x = \frac {-6}{8}$

[soft zealot]

Zolo