help-38

[orchid drift]

i will make a new channel

we can continue there

.close

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[plain tusk]

help

[plain tusk]

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[arctic kite]

Can anyone give me a formula for median for frequency table?

[trim jolt]

Please don't occupy multiple help channels.

[arctic kite]

sin(53) = a/14.5 cm

[tulip violet]

Please stop opening multiple channels

[arctic kite]

Can u help?

[tulip violet]

No

[arctic kite]

Stop trolling then noob.

<@&286206848099549185>

[red mountain]

!volunteers

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Helpers are just people volunteering their time to help you. Be polite and patient.

[red mountain]

and you already have a channel open #help-28

choose one

[frosty vapor]

Bro

n + 1/2

This is basic math

[plain tusk]

yeah some1 helped me time ago lol thanks though

[thick tapir]

yes
Grouped or Ungrouped Data

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[long basin]

Aight so I got a quiz tomorrow on combinations and probability. There are some sample questions from the HW im just going through, just wanna check my work diligently.

[long basin]

The first problem i chose is:

How many 3 digit numbers xyz with all x,y,z ranging from 0 to 9
have at least of their 2 digits equal. How many have exactly 2 equal
digits?

And what I did was break it down into instructions:

1. Choose 1 digit from the 10
2. Put that digit in any 2 of the 3 spots
3. Choose 1 of the 9 remaining digits
4. It gets automatically placed in the last spot

That is,

$$\binom{10}{1}\binom{3}{2}\binom{9}{1}\binom{1}{1}$$

[solid kiln]

Umbraleviathan

[long basin]

would this be correct? (I know there are redundant factors, but its to account for the instructions I wrote)

also ping me because im bouncing between tabs

[hallow spruce]

this gives the number of numbers with exactly 2 equal digits

[long basin]

oh it didnt copt correctly

[hallow spruce]

the question specifies "at least 2 digits equal"

[long basin]

it wants only 2 digits

wait

[hallow spruce]

actually it doesn't say that either

[long basin]

yeah no i meant "how many have exactly 2 equal digits"

the textbook is weird

it doesnt copy all numbers

lol oofs

there

fixed it

[hallow spruce]

[long basin]

but then the counting is correct?

[hallow spruce]

have at least of their 2 digits equal
so then what the heck is this

if you're concerned with precisely 2 matching digits then yes

[long basin]

oh if it wants at least 2 digits equal I can juust add 10 since thats (10 1) (3 3)

but I mean

¯_(ツ)_/¯

[hallow spruce]

i don't get paid enough to make inferences

[long basin]

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[long basin]

.reopen

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✅

[long basin]

moment

Alright another counting moment

A person has 8 friends, of whom 5 will be invited to a party. How many choices are there if 2 of the friends are feuding and will not attend together?

Here are the instructions I wrote:

1. Choose 5 of the 8 friends to invite
2. If 2 of the beefing friends get put together, then 3 other people can make up thos egroups, and these arrangements need to be removed

So $\binom{8}{5}-\binom{6}{3}$

[solid kiln]

Umbraleviathan

[long basin]

Is this correct?

also ping me

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[astral bolt]

I am trying to solve a discrete mathematics problem but I am having trouble with knowing where to start

[lament reef]

have you tried just using the definition of (a choose b)?

[astral bolt]

I ran through the problem once using binomial coefficient identities but I got stuck when trying to factor and cancel the factorials

[silver glade]

we can't cancel all of the factorials

but we can reduce it down to a single binomial coefficient

[astral bolt]

ok, how can I do that?

[silver glade]

well cancel what you can

what's left is three factorials

which actually form a binomial coefficient

[astral bolt]

this is what I have so far, are those the correct three factorials im supposed to be left with?

[silver glade]

you don't need any ellipses

some of the stuff on the LHS cancels with the RHS

actually, even simpler

what identities do you have?

this makes it almost too easy

[astral bolt]

these are the identities I have

[silver glade]

ah, ok it's not there

[astral bolt]

I'm not sure if I continued it correctly

[silver glade]

oh

that's not what i meant when i said you don't need ellipses

here i'll show you

[astral bolt]

mb

[silver glade]

$\binom{a}{b}\binom{b}{c}=2\binom{a}{c}$\
$\frac{a!}{b!(a-b)!}\cdot\frac{b!}{c!(b-c)!}=2\frac{a!}{c!(a-c)!}$\
$\frac{1}{(a-b)!}\cdot\frac{1}{(b-c)!}=2\frac{1}{(a-c)!}$\
$\frac{(a-c)!}{(a-b)!(b-c)!}=2$

[solid kiln]

Axe

[astral bolt]

ahhh

that makes it so much simpler

from there, how can I finish the problem by finding all positive integers for a,b, and c which satisfy that equation?

[silver glade]

write the LHS as a binomial coefficient

then think about when a binomial coefficient equals 2

[astral bolt]

so as a binomial coefficient, it would look like a-c choose a-b

[silver glade]

yeah 👍

[astral bolt]

cool

[silver glade]

now 2 only appears once in pascal's triangle

[astral bolt]

sorry I dont really understand how to find the values of a b and c

[silver glade]

are you familiar with this?

these are all of the binomial coefficients

[astral bolt]

i dont think ive seen this in my class yet

[silver glade]

ok

[astral bolt]

how do I read it?

[silver glade]

um

if it's n choose r

then n is the row and r goes across

starting from 0

so for example 6 is 4 choose 2

[astral bolt]

ok

so then 2 would be 3 choose 1

[silver glade]

yeah

no

2 choose 1

n also starts from 0

[astral bolt]

mb thats what I meant

2 choose 1

and then now I just need to configure my values for a b and c to match that

[silver glade]

yeah

[astral bolt]

I got a=3, b=2, and c=1

[silver glade]

that should work but there are more solutions

[astral bolt]

how could I find all positive integers for a, b, and c?

[silver glade]

a-c=2
b-c=1

[astral bolt]

okay, so then I just solve for a, b, and c using that

[silver glade]

2 linear equations in 3 variables should give us a line

[astral bolt]

because regardless in order to equal 2 it needs to fulfill that requirement

[silver glade]

yes

you can eliminate 2 of the variables and you'll have 1 free variable

a=c+2
b=c+1
so solutions are (c+2, c+1, c) for any positive integer c

[astral bolt]

ahhhh okay that makes sense

because there are infinite solutions as long as they eventually equal 2 choose 1

[silver glade]

yes

[astral bolt]

thank you so much

[silver glade]

(3,2,1)
(4,3,2)
(5,4,3)
and so on

[astral bolt]

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[brave arch]

Hello

[brave arch]

I havea question about this rule

so this is the question

[lofty elk]

Ok wats the question

[brave arch]

This is my work for the question

My question is on STep 4, when you integrate it, in this part arent u supposed to times du/dx(shown by the rule) at the end as well(the answer key did not) or is the answer key right

[lofty elk]

Look at 1. Not 2.

With 2., you are differentiating f(u) with respect to x (hopefully my terminology is correct)

That's implicit differentiation

[brave arch]

no? both rules are in respect to x

[lofty elk]

Here, if you chuck the integral to the other side, you get d/du (f(u))

[lofty elk]

Yes

But one is d/dx (f(x))

The other is d/dx (f(u))

Do you see the difference

[brave arch]

WHy are we not looking at rule 2, we are using u

[lofty elk]

Because your integral is in du

[brave arch]

u and x are 2 differnet thigns

[lofty elk]

You are not mixing x and u

[lofty elk]

They are

I think ur a lil confused

[brave arch]

exactly

thats why we should look at rule 2

[lofty elk]

That's not how it works

Rule 2 is for handling multiple variables

Rule 1 is for handling 1 variable

[lofty elk]

Ok replace every single u with x in this equation

Then tell me which equation you think you should use

[brave arch]

first

but

we are using u sub

in the problem

[lofty elk]

That's fine

Cuz when you do the sub

You are replacing every x with some arrangement of u

You are not handling 2 variables at once

You're turning all the x variables into u variables

Does that make sense to you

[brave arch]

uh

[lofty elk]

Implicit differentiation (not chain rule) is for handling 2 variables at once

But when you've got an equation with only 1 variable, use the regular method

Substitution or whatever, if you can reduce it into 1 variable, use the regular technique

[brave arch]

What does implicit have to dow this

we arent using y

[lofty elk]

Rule 2 is called implicit differentiation

In a sense it is the chain rule, but I think it would be better for you to separately classify them

[lofty elk]

x,y and u are variables, and their names aren't of great significance

I could easily call my variables a, b and c

[brave arch]

but chain rule

[lofty elk]

Or even use Chinese characters if I want

[brave arch]

bro what

[lofty elk]

I mean, yes but no

[lofty elk]

Pretend the u is a y

Would that not be implicit differentiation

[brave arch]

OHHH

i get it

[lofty elk]

There you go

Don't let the letters confuse you

You can call them whatever you want, as long as if they serve the same purpose

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[opaque dust]

Our math final is multiple choice and we don't need to show work, this means I can use my calculator to solve most things.

[opaque dust]

Especially with the ti 84 plus CE with all of its special abilities

I need to convert standard form to vertex form

How do I do that without completing the square?

Legit any method I could use on a calc or anything

I just despise completing the sqaure

or could I just enter it into the grpahing calc and then look for the p and q

[past widget]

standard form to vertex form: parabola?
Consider 7y = 6x² - 13x + 5. If you're using a calc, you can plug 7(y - q) = 6(x - p)² and plug any two points: (0, 5/7), (1, -2/7) on the parabola

[digital bison]

you can graph it and use the max/min function to find the vertex

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[opaque dust]

Tysm

[opaque dust]

Also can someone explain each step of this

I like

80% understand

But I wouldn't remember how to do this on my final in 8 days

[regal bough]

what part r u struggling

[opaque dust]

Uhh honestly I understand besdies what the 25 means, isn't it the max width?

Can't tell what it says

I just don't complete understand every step 100%

[regal bough]

okay lets start with part a)

over there, we find two equations. Particularly the perimeter and the area.
The perimeter is bound by the fact that its max size is 100.
so we assume all fences were used and 100 = 2L + 2W
From there we get 50 = L+W, which we substitute into A=LW
this yields the final quadratic equation

For the second part,
u r right, the vertex y coordinate represents the area width possible

the x coordinate represents the width needed for the rectangle to have the max area

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[bright aurora]

the answer is in red but i have no clue how they got there

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[livid fable]

You calculate the difference between t=4 and t=5 and the difference in t=5 and t=6 and then you calculate the average of these two values.

[bright aurora]

i got it tysm

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[rigid hare]

what was the original question

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[cursive bolt]

im asked to plot some different transformations on this, and one of them involves adding x(-1)+x(1). but both have two values, so im not clear on which ones i take?

[cursive bolt]

my intuition is that since its a signals course that i would take the greater magnitude? but not sure

[supple halo]

if you add 1 it shifts to the left by 1 unit

if you subtract 1 it shifts towards the right with 1 unit

as far as i understood the query

[cursive bolt]

but like, what is the value of x(1) here

[supple halo]

oh x(1)

can you tell me which axis represents x in this

[simple haven]

x is the vertical axis and t is the horizontal, almost certainly

[supple halo]

i see

[simple haven]

@cursive bolt as long as you are consistent between the choice, the values is actually the same

[supple halo]

so @cursive bolt what is the value on the vertical axis at 1 and -1

[simple haven]

They are discontinuities

[cursive bolt]

wdym by consistent?

[supple halo]

this is just a graph

continuity is not needed here

i think

[simple haven]

Either consistently choosing the previous or next value

[supple halo]

x(1) is simply 2and x(-1) is 0

[simple haven]

At -1 we have a jump from 0 to 1, and at 1 we have a jump from 2 to 1, if we always choose the first number we get 2, and if we always choose to second we also get 2

[cursive bolt]

ahhh ok

thank you that works

[supple halo]

youre onto something

and it kinda feels wrong

[cursive bolt]

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[obsidian hornet]

Wondering if I can apply square root to c?

I think so, I dont think theres anything stopping me

basic algebra should still apply?

yes basic algebra should still apply but I cant have two different messages

so 3 is not an encryption function

[jaunty scarab]

how do you define encryption function?

[obsidian hornet]

i wrote it in purple

i think this should be it for the set of rules

but i could be wrong

[unique minnow]

You can’t really do much more than adding and subtracting without being careful with modular arithmetic, especially when the modulus isn’t prime.

Someone correct me if I’m wrong because I’m a bit rusty, but your intuition is still right. If the root exists, then should be two of them.

2. exhibits a good example of why you need to be careful. If k is chosen among Z/NZ, how can you be sure that it has an inverse? Z/NZ is not a field by itself

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[obsidian hornet]

wait, whys Z/NZ not a field by itself

[limpid dawn]

(Z/NZ, *) is not always a group

[obsidian hornet]

oh i see

(Z/NZ) is not F_p

where p is prime

cause (Z/NZ)* includes all elements from {1,…,N-1} coprime to N

[limpid dawn]

yes

and these are all units and they are invertible by definition

[obsidian hornet]

but if you have any number missing from set {1, n-1} then it’s not (z/nz)*?

[limpid dawn]

oh I think I understand

then I'd say it's a subset of (Z/NZ)* to say the least

but yea since gcd(n-1, n) = 1 then n-1 missing would imply it's not (Z/nZ)*

[obsidian hornet]

i’ll come back in a bit

got 2hr class

[unique minnow]

Yeah mb I should have been more clear. If indeed it is shorthand for Z/ZN multiplicative with all element comprime to N then you won’t have any issue

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@obsidian hornet Has your question been resolved?

[obsidian hornet]

nvm, i’m gonna close this

do more readings then come back to this question

.close

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[obsidian hornet]

.reopen

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✅

[obsidian hornet]

wait

here Z/PZ wont be a field

[obsidian hornet]

and k might not have an inverse (so k is not a unit)

but why would you use them interchangeably then

[obsidian hornet]

is this true tho\

[humble stratus]

ion kno

[unique minnow]

Z/pZ is a field for prime p

The point is that it's usual for $\mathbb{Z} / N\mathbb{Z}$ to be additive, in which case it is a group under addition, but it need not be.

If you defined $\mathbb{Z} / N\mathbb{Z}$ as the set of all numbers in ${1,2, \ldots, N-1}$ which are coprime to $N$, then it is a \emph{group} under multiplication (because every element has a multiplicative inverse), but in general it won't be a \emph{field} if $N$ is not a prime number.

[solid kiln]

Azyrashacorki

[obsidian hornet]

ok perfect

I see

so Z/NZ = F_p(where F_p is a field) IF N is prime, or I guess interchangeable, not equal to

but if N is not prime, then Z/NZ is not a field, its only the ring of integers modN, and not all elements here will have a multiplicative inverse

@unique minnow

correct?

[unique minnow]

Yeah

[obsidian hornet]

ok!

thank you!

[unique minnow]

Nw

[obsidian hornet]

wait another thing is,

I guess it doesnt even matter whether N is prime or not, because isnt Z/NZ = {0,1, N-1}?

so 0 wouldnt be a unit(have an inverse) anyways

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[marble wharf]

how many solutions are there to x^2=1 mod 8 ?

[marble wharf]

see above

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[feral gale]

i was wondering how my teacher got these coordinates, specifically the x-values

[past widget]

Looks like equation for y = 18 + 8sin(100pix - pi/2). I believe your teacher looked at the peak and trough points and figured they're at phase shift + period

it's not hard figuring soln for 26 = 18 + 8sin(100pix - pi/2)

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[feral gale]

Does this require a calculator

cause im pretty sure the question is no calculator

And why did she randomly get 100 i just dont understand if they werent values mentioned in the problem

[past widget]

no it does not

[past widget]

if you're confused how she did it so fast, that's coz of familiarity with this question, or similar types.
ow, if you're asking how to figure when the sine curve approaches peak, center, trough, you figure phase shift, then you add 2pi/b is period and so on

[feral gale]

Is there anyway you could write the work out so I can better understand

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.reopen

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[past widget]

You have y = 18 + 8sin(100pix + pi/2)
Now, what is sin maximum? 1 right? So y max is 26. when is sine maximum, let's say you start plotting with x ≥ 0, you immediately have sine max at x = 0, so (0, 18) is a point on your wave
When is the next time the wave hits 18? that's 0 + period of sine.. Now for a wave function y = L + Asin(wx +c), the period is 2pi/w, so your wave hits the y = 18 line every integer multiple of 2pi/w

also, now you have an understanding that the period T = 2pi/w

you can say that since the wave started at peak position, it reaches center at t = T/4, and trough at t = T/2 for the first time

Add an integer multiple of T to those values and you have the points your teacher wrote

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[silver wolf]

anyone can explain this ?

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@silver wolf Has your question been resolved?

[visual wigeon]

44*2+1+1=90

[silver wolf]

hmmmm

[visual wigeon]

You have the numbers from -44 to 44 whose sum is 0

and then 45 itself

the numbers from -44 to 44 are 44*2+1(not forgetting the zero) so 89 and then 45 itself makes it 90

[silver wolf]

hmm i still don't get it

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[brazen rock]

my answer is incorrect haha

[brazen rock]

how do I do this

[visual wigeon]

formula

[brazen rock]

sample or population

[silver wolf]

how can there is -44 ?

owh it is integers

so it can be negative

momok kuda ngomongo jancok

🤔

[silver wolf]

popolotion maybe

.close

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[oak shadow]

Need help chat

[pure fiber]

have you already tried doing something? if yes could I see your work?

[oak shadow]

Alr wait

Found it but it isnt neat or nun

And the answer wrong

[past widget]

Recall: sin(60 + x) sin(60 - x) sin x = sin 3x

Now, rewrite given equation: sin(60 + x) sin(60 - x) = cos 2x => sin 3x = sin x cos 2x => 2sin 3x = sin 3x - sin x => sin 3x + sin x = 0

@ me if you don't get

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[gaunt temple]

my prof wrote this but i dont understand the notation...<x,y> is the smallest subgroup of G containing x and y, right? so it must contain all the powers x^0, x^1, x^-1, x^2, x^-2,... all the powers y^0, y^1, y^-1, y^2, y^-2,... their products and the inverses of the products

[gaunt temple]

but if we have something like x^3*y^4, then that means n_0 = 3, m_1 = 4 and all the other exponents have to be 0

but m1...mk and n1...nk are in Z star, so they cannot be 0

[marble wharf]

k=0

only n_0 and m_1 appear

[gaunt temple]

oh yea right

im so stupid

thank you

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[marble wharf]

easy to miss

[dull temple]

.(deleted modping)

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[hallow hedge]

I want to find the rank of this 3x3 matrix?
The rank that I find is 3 while most of the bots say its 2. But I feel the answers by bots were not correct because upon studying the solutions provided. I noticed, they had basic + - issues. So they are not reliable.

[ 4, 5, 6 ]
[ 1, 2, 3]
[ 7, 8, 9]

[uncut aspen]

row reduce it and find the number of pivot columns

[wraith hinge]

it's 2 yeah

1st row = 2nd row + 3 * all ones
3rd row = 2nd row + 6 * all ones

[hallow hedge]

This is my process:

1/4(R1)
R2 - R1
R3 - 7(R1)
4/3(R2)
R3 + 4/3(R2)
6/7(R3)

[wraith hinge]

if you do R3 - R1 you get all 3's in row 3
then if you do R1 - R 2 you get all 3s's in row 1

[hallow hedge]

okay. and what do I do next?

[wraith hinge]

well R1 and R3 are same

so rank = 2

you can reduce it further but dont need to

dont wanna sound rude but you should watch this
https://www.youtube.com/playlist?list=PL221E2BBF13BECF6C

[hallow hedge]

Thanks. I'll have a look

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[cedar adder]

@past widget

Any idea about 3-b?

Something like this?

[past widget]

honestly, I have no clue what you've done

[cedar adder]

This is from another exercise.

Is the solution something similar?

[cedar adder]

.

[past widget]

well, if you've done 3(a), you can use that

z + z{bar} = 0 or z - z{bar} = i

[cedar adder]

Well, actually it is i × Im(z)

[past widget]

which one?

[cedar adder]

z - zBar = x + iy - x + iy

=2iy

"Y = Im(z)"

[past widget]

Sure, however you wanna simplify :|

[past widget]

solving these give the set of points :|

[cedar adder]

i × 2iy + 1
-2y +1
y = 1/2

[past widget]

the first part gives z = ib forall b \in R , the second gives z = a + i/2, forall a \in R

[cedar adder]

@past widget

[past widget]

Yes

The equations are: Im(z) = 1/2 or Re(z) = 0, if that sates you

[cedar adder]

I just started the lesson, I can't understand that Z - Zbar = "i" only

And ty

[past widget]

well z = a + ib, z{bar} = a - ib, so what I was writing was 2ib essentially

alr tho, I'll remember :p

[cedar adder]

Maybe I will reach ur level after entering university

Ty again

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[ocean quail]

how do i do this?

[ocean quail]

substituting sin and cos doesnt work

because we would get that n=0

which it says its not

[digital bolt]

what do you get when you substitute sin and cos?

[ocean quail]

cos pi/4, sin pi/4; -sin pi/4, cos pi/4

[digital bolt]

ok

[ocean quail]

which when raisen to the nth power

we get cos npi/4, sin npi/4, -sin npi/4, cos npi/4

[digital bolt]

yes

[ocean quail]

its just the standard rotation matrice

[digital bolt]

yes

[ocean quail]

but it doesnt help u

[digital bolt]

so for what n is cos(npi/4) = 1

[ocean quail]

0

[digital bolt]

what other n

[ocean quail]

ooooh, can u write it as 2pi + pi/4?

or wait

no

[digital bolt]

close

the trick is to use 2pi

[ocean quail]

hmmm

[digital bolt]

ok

so cos(2pi) = 1

[ocean quail]

tuff

[digital bolt]

so for what n is cos(npi/4) = 1

[ocean quail]

yah

ooog

8

right?

[digital bolt]

yeah exactly

[ocean quail]

thats amazing

thx dude

[digital bolt]

so all this is saying is that if you rotate 45 degrees 8 times, you get 360 degrees

which is 0 degrees

classic

[ocean quail]

i got more matrices i have trouble with

[digital bolt]

i gtg, someone else may or may not show up

[ocean quail]

yup yup

no prob

cya man

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[analog pier]

To prove that a function is not injective is it enough to show that the intersection between the images of each piece is non-zero?
Here for example the first image is [0,4] and for the second function is (-inf,4). Since the intersection is [0,4) the function is not injective

[limpid dawn]

You can do this simpler if you take a look at x^2

[safe loom]

Show that there exists distinct x1, x2 such that f(x1)=f(x2)

[analog pier]

how can i do it simpler

[safe loom]

x² = c (0 < c ≤ 2) => x=±√c, that already offers 2 distinct value

[analog pier]

this was a bad example mb

i should've mentioned that both function should be injective individually

[analog pier]

if i do that then i would have x1^2=8-2*x2

and that is a lot harder to determine

[safe loom]

No, they dont necessary have to be at both functions

As @limpid dawn and I mention, look at x² and consider its properties

[safe loom]

Also if I do that then i can't prove the injectivity

(Actually its still possible through graphing)

[analog pier]

ok lets take another example. first function is for x<=0 and the second one is for x>0

now both are injective but so now I just have to check if there is an x <=0 and one >0 that have the same f(x)

how would I go about that

besides graph

[limpid dawn]

x for x >= 0 and -x for x < 0 both are injective but together they are not

[safe loom]

You can also prove that the function is not always increasing/decreasing on the entire domain, it takes at least one extremum to break the injectivity

You can also examine the point where the function is not continuous

[analog pier]

what would you do here?

if we replace the 2 with 0

[safe loom]

For me, I examine the discontinuity

Specifically x=0

lim f(x) as x -> 0+ is 8
lim f(x) as x -> 0- is 0

Along with the fact that each function is injective, there can exist a line y=c (0 < c < 8) where it intersects at both points, with each lying on different branch

[analog pier]

that is a good idea

thanks

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[ocean quail]

i need help with this exercise, i know that d) is wrong, but how do i prove that the rest are right

[trim musk]

hi gustavo

[ocean quail]

yooo

[hidden dew]

There’s two of them now

[ocean quail]

someone already tried to solve it and now i know why a) and b) are correct

[ocean quail]

huh?

[hidden dew]

Meant the 2 Gustavo’s

[ocean quail]

oh, thats because i closed the last channel because XOR couldnt help me

or nevermind, theres apparently 3 gustavos in this server

who knew

@trim musk @hidden dew do you guys have any ideas on how to solve this?

or anyone?

[wooden plover]

Right so you know what the sizes of AB and BA are now I assume ? @ocean quail

[ocean quail]

yea

i figured out a) b) c) and d) so far

[wooden plover]

Ah right ok

[ocean quail]

i only need to understand e) and f)

@wooden plover are you still here?

[wooden plover]

I gtg get my pizza i'll brb

[ocean quail]

oh right sure

ill wait no prob

[ocean quail]

did you get it?

[trim jolt]

@ocean quail Has your question been resolved?

[wooden plover]

yea now I'm back

[ocean quail]

nice

[wooden plover]

I don't think there's a super nice way for d and e

[ocean quail]

e and f you mean

[wooden plover]

yea

compute the two sides and see if they match

[ocean quail]

they dont

how can you even multiply a matrice sized n x n and 1 x 1

i thought it was impossible

[wooden plover]

1x1 is essentially a scalar

[rustic flicker]

Haha I fucking hate math

[zinc ginkgo]

#chill

or just leave the server

[rustic flicker]

math pfp

[ocean quail]

so?

[wooden plover]

scalar x matrix makes sense

[ocean quail]

whats the answer to scalar x matrix?

is that equal to matrix^2 no matter what?

[wooden plover]

multiply all entries of the matrix by the scalar

[wooden plover]

no

[ocean quail]

so the matrix will be sized n x n no matter if its AB or BA?

[wooden plover]

BA is 1x1, it's a scalar

AB is nxn

[ocean quail]

what if it waas 1 x 2

would AB be n x n still?

[wooden plover]

what was 1x2 ?

[ocean quail]

like, whats the rule to seeing what size they are

[ocean quail]

what if you multiplied a matrix n x n with 1 x 2

what would you get

or some random stuff like n x n times 3 x 8?

[wooden plover]

(n x m) multiplied by (m x p) gives a (n x p) result in general

[ocean quail]

yah

i know that

[wooden plover]

why are you asking the question then ?

[ocean quail]

because nxn multiplied by 1x1 does not have the same property

because n should be equal to 1

[wooden plover]

yeah scalars do be that way

scalar x any matrix works

[ocean quail]

goddamnit

so then how would you even check for (AB)^2

?

[wooden plover]

well (AB)^2 is still nxn, a bit painful to compute

just compute one or two entries to get the idea

what would be at top left position (1, 1) for example ?

you know what the entries AB of look like right ?

[ocean quail]

x1x2y1y2

or wait

no

it would just be the sum of the square of each element

or no

fuck

this is painful

[wooden plover]

just write the matrices out really

1st row of AB x 1st col of AB

[ocean quail]

damn, i really gotta do all that?

[wooden plover]

I'd do it

[ocean quail]

sure one sec

so for the first element i got x1y1*(x1y1+x2y2+...+xnyn)

[wooden plover]

yeah

and x1y1 + x2y2 + ... + xnyn is ?

[ocean quail]

its me?

who is it?

oh its BA

[wooden plover]

yea

[ocean quail]

i cant believe this

this is kinda tuff ngl

[ocean quail]

i got another question

if BA is a scalar

than BA * AB = AB* BA?

[wooden plover]

yeah

[ocean quail]

niceeeeeeeee

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[rich stag]

Idk these terms very well

C. Is variable right

Rn I have
A. Pop parameter
B. Observation
C variable
D. Sample stat

Do u guys think this is right

OK WAT ABT
A. Pop parameter
B. Sample stat
C. Variable
D observation

[restive harbor]

A is correct, B is a Sample stat, C is an Observation and D a Variable

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[rich stag]

OK TY

I see it

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[wraith hinge]

How to find this determinant

<@&286206848099549185>

[past widget]

!original

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Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[zinc ginkgo]

without more information, induction

[wraith hinge]

That is a determinat

[wraith hinge]

Huh?

Ok

[zinc ginkgo]

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

[wraith hinge]

Ok i will try

!close

Close!

.close

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[half scarab]

I need help with 13 and 15

[split chasm]

apply inscribed angle theorem
start with 15, its easier

[half scarab]

I don’t know what that it

This text book just has its thereoms as numbers

Like

12-3

Or 5-8

[empty orchid]

an inscribed angle A is half of the measure of the arc it subtends

let me draw it

the red is the arc that it subtends

[half scarab]

I know that CB for 13 is 96

Wait

I get it now

[empty orchid]

👍

[empty orchid]

yep!

[half scarab]

360 - (96-110)

Then divide it by 2

AB=77 right?

[empty orchid]

what

[empty orchid]

nah don't divied it by 2

[half scarab]

Oh I thought it meant the line

Thank you though

[empty orchid]

👍

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[stoic basin]

can someone please check if my ans is correct?

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[stoic basin]

.reopen

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✅

[silver glade]

i got a different answer

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[jovial thorn]

is there a way to evaluate Ei(1) without using the power series

[jovial thorn]

$\int_{-\infty}^1 \frac{e^x}{x}$ $dx$

[solid kiln]

kronium_

[jovial thorn]

<@&286206848099549185>

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[jovial thorn]

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[low cloud]

im having a hard time visualizing this which is making it hard to start

[solid jewel]

make a sphere and then make another one of the same size with its centre being on the surface of the first sphere and the overlapping volume is being asked

[low cloud]

like this?

[solid jewel]

yes

[low cloud]

okay so for something like this i would integrate with disks right because there isnt a different center radius that would be needed for washers

[solid jewel]

yes you could integrate with discs from the height of the point of intersection uptil the end of the sphere and then double that volume cause it is symmetrical

[low cloud]

so this formula?

[solid jewel]

idk i am not familiar with that formula

[low cloud]

this is from my notes if it helps

[solid jewel]

ye thats correct

[low cloud]

okay so in this instance r(x) is the sphere

[solid jewel]

no

thats the radius at that point

of the disc element

[low cloud]

but we just have radius r

[solid jewel]

the radius of the disc is varying as we integrate

make a disc at the lower end of the common volume and notice how if you keep stacking tiny discs on top of it, their radius keeps on changing

[solid jewel]

should be thin discs

[low cloud]

well usually in other problems the radius depends on the x value

[solid jewel]

it does in this one too

[low cloud]

umm how do i have have an x when i dont have a function with x

[solid jewel]

you have to make one yourself

assume x distance above the centre and determine the radius of the disc at that point as a function of x

[low cloud]

would it look like this?

[solid jewel]

sorry the diagram is a bit scuffed

you can find R(x) using pythagoras

and then the only thing left to determine are the intervals of x

[low cloud]

oh no🥲

okay so the intervals of x for my other problems were found by setting the two functions equal to each other

[solid jewel]

in this you can denote the volume from the plane formed by the points of intersection and above as V/2 since the volume below it is the same and evaluating the top one and doubling it will give us the total volume

then the upper limit of x = r

[low cloud]

so would the bounds go from -r to r

[solid jewel]

no

r/2 to r

[low cloud]

im sorry im just so confused

[solid jewel]

ye the figure aint all that great

you are integrating discs made on the volume of the lower sphere from r/2 to r

and the doubling the volume cause it is symmetric

[solid jewel]

you can derive the lower limit of r/2 using geometry

draw the cross section view and try

[low cloud]

i think im going to save this question for later and ask my TA for help im so sorry thank you for trying to help me but i just cant do this rn

[solid jewel]

ye its alright

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[lilac cloud]

Hi. Trying to prove that this statement is true, but struggling to show that the subset is closed under vector addition.
I assume that f,g are periodic functions where f is periodic in p and g is periodic in q. Im trying to show now that f+g is periodic in something. I feel like it is LCM(p, q) or just pq, but not sure how to show algebraically that (f+g)(x+pq)=(f+g)(x)

[lilac cloud]

I already can show that the additive identity exists and that its closed under scalar multiplication

[grim sparrow]

are you 100% sure that the statement is true?

[clear cloud]

But its not in math

Take sin(x) + sin(x*sqrt(2))

[lilac cloud]

Oh i see

That makes sense

[clear cloud]

Also

Something rare

But it can happen

Two periodic function such that f(x) + g(x) = x

[lilac cloud]

How would I state formally why $\sin(x)+\sin(\sqrt{2}x)$ is not periodic

[solid kiln]

Luca M

[grim sparrow]

one way is to derive a contradiction

[trim jolt]

@lilac cloud Has your question been resolved?

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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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[rapid meadow]

for the integration of this question, why is only the cos answer accepted? i thought there was sine as well

[empty orchid]

it...is.

[rapid meadow]

huh

[empty orchid]

like, there are two answers to this question, using sine and cosine

[rapid meadow]

and i put into like an integral calculator thing and it only gave me the sine answer?? but the answer in my homework only says the cos one is correct

[rapid meadow]

yes

ik

but is it supposed to be like

only cos accepted

cuz my answer sheet says that

[empty orchid]

no.

[rapid meadow]

so both sine and cos are accepted... right

[empty orchid]

cosine would be the more standard form here, but it should not be the only answer.

[rapid meadow]

ohh

ok

wait why though

why is cos the more standard form

i thought every question had two answers

[empty orchid]

$\dv{x}\cos^{-1}(x)=\frac{-1}{\sqrt{1-x^2}}$

[solid kiln]

;(

[empty orchid]

yeah, but im saying it might be the only accepted answer for them because it might be the more standard form for them.

imho, there is no standard form, and it is always better to use arcsin.

[rapid meadow]

oh

errr

wait then is it ok if i write both😭

okay thanks

.close

[trim jolt]

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[trim jolt]

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• Show your work, and if possible, explain where you are stuck.
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[viscid flower]

so you did c earlier

[viscid flower]

which one are you looking at?

[vast viper]

Yes

Now i am looking for X

Phi is complete

So all the elements in any set will be open and close both?

[wooden plover]

X is open and closed yes

[viscid flower]

Well, we were talking about this before

[wooden plover]

ah

[viscid flower]

Did you feel alright about the open part?

[viscid flower]

ah i mean, me and other ppl were

[wooden plover]

okok

[viscid flower]

Here it doesn't really matter, that X is closed. Are you confused about one of the exercises you posted? or that X is open and closed?

[vast viper]

I am confused about X is open and closed

I want to understand how it satisfied both properties

[viscid flower]

Before we were talking about X is open, you felt okay about that?

[vast viper]

Actually No

I want to feel it again

It was too late night for me actually yesterday it is morning right now

So i just woke up

[viscid flower]

Okay. Do you have a definition you like for open set? Or should I post one.

[viscid flower]

how about this

[viscid flower]

this is close but not really what we need, an open ball is just one kind of open set

[vast viper]

So it should contain some elemets

[viscid flower]

if a set is open, then we can create a neighborhood around any element of the set, and have it only contain elements of the set

[viscid flower]

this site uses this image

[vast viper]

[0,1]

[viscid flower]

not quite

can you see why [0,1] is not open?

[vast viper]

Around 0 we can not make

Any small ball

[viscid flower]

right

[vast viper]

(0,1) is open

[viscid flower]

because any small ball will contain a negative number

and any negative number is not in [0,1]

[viscid flower]

it is, yea

[vast viper]

So we want every small ball inside it?

Or just a single ball?

[viscid flower]

we arent even really worried about balls specifically

they are just a visualization technique for neighborhoods

what we want is to be able to move a little bit in some direction from any point

we want to be sure if we start at SOME place inside the set, we can wiggle a little bit around that place, and not leave the set

[vast viper]

I see

[viscid flower]

if you start at 0

and youre in [0,1]

then you cannot really wiggle in any direction

[vast viper]

Left part of 0

[viscid flower]

you have to be careful

its enough that it doesnt work for 0

we can start at 0.5 in [0,1] and wiggle just fine

[vast viper]

Other thing is that 0,1 are limit points

[viscid flower]

they are

so, if you feel okay on open set

[vast viper]

Interior point are???

(0,1)

Opps i am confusing myself

[viscid flower]

sorry I was looking for a definition thats similar in format to the other ones were using

[viscid flower]

an interior point is any point that we can wiggle around and not leave the set

is that definition okay?

[vast viper]

Yes

[vast viper]

👌

[viscid flower]

here's the picture we keep in mind

but the balls can be as small as we want

or not even balls, just some neighborhood

okay so, if you feel good on open sets in general

we can look at X

We want to convince ourselves that X is open

what can we try? what should we check?

how do we start

[vast viper]

So here we will check each element of it

[viscid flower]

sure

[vast viper]

And we will see if we can wiggle around it

[viscid flower]

lets take some general element x in X

and, lets try to take a neighborhood around it

[vast viper]

Okay i make

Singletons are open

Right?

Opps no

They are closed in metric space

D(x,r)<=1

The distance function is very confusing

[viscid flower]

why's that?

lets just try an example

lets use R a number line

[vast viper]

Definition are clashing in my mind. Maybe i will need some time

[viscid flower]

so (R, d) where d is the normal metric

[vast viper]

Hmm sure

[viscid flower]

d(x,y) = |x-y|

[vast viper]

So each number will be okay

And we can move wiggle around each element

[viscid flower]

yea but lets be more careful

lets take some number on the number line, r in R

we wanna make sure we can wiggle, so lets try to construct an neighborhood around it

[vast viper]

(x-epsilon,x+epsilon)

[viscid flower]

lets try, radius of 1

so lets build the neighborhood

we want all p in R such that |p-r| < 1

[vast viper]

In R each number will be okay

[viscid flower]

yea but its important to see, and im not doing a good job of it

[viscid flower]

when we make this neighborhood, we are drawing from R, right?

[viscid flower]

its just a common thing to be confused about

so say we look at metric space (R,d), the number line, with normal metric

and now [0,1]

man

@wooden plover you wanna try it was so much better in my head

[viscid flower]

@vast viper maybe its hard to see but here is the trick

when we take some r in R, and we construct a neighborhood around that r, we take points from R, again

do you see what i mean?

[viscid flower]

This thing you are confused about now I think, it's a common thing for people to become very hung up on. It's worth the effort to struggle through, and you are not alone in being confused about it. I was too

[vast viper]

Sorry i went outside in the middle of learning needed to throw something

[vast viper]

So now |p-r|<1

[viscid flower]

we are building a neighborhood of our point in R

[vast viper]

[0,1] if I include 0 then i can surely do this

|0-r|<r

It will fail

Because |0-1|<1

[viscid flower]

you need to think about the neighborhood more carefully

[vast viper]

Can we not make nbd around 0 and 1?

[viscid flower]

we can

[vast viper]

Yes

[viscid flower]

lets be careful

Let's say were looking at the entire number line, normal metric, so (R,d)

lets look at [0,1] in this space

[vast viper]

Yes

[viscid flower]

okay, lets try to create a nbd around 0

lets use epsilon

[vast viper]

(0-epsilon,0+epsilon)

(0+epsilon) is okay

[viscid flower]

so our neighborhood is all $r \in R$ such that $d(0,r)<\varepsilon$

[solid kiln]

jan Niku

[viscid flower]

so we can construct a neighborhood around 0

but it includes points that are not in [0,1]

but, if we do the same thing with R

[vast viper]

Yes

We will get R

[viscid flower]

we are looking at the number line, normal metric, (R,d)

[solid kiln]

jan Niku

jan Niku

[vast viper]

It will be in R

[viscid flower]

Can you see why this is more simple

do you see why this is more obvious that it will be open

not just visually

[viscid flower]

do you see that all p are in R, already

[vast viper]

Is it not like close?

[viscid flower]

the distance function almost doesnt even matter

because all p are coming from R anyways

they will all be in R

it cannot fail

[vast viper]

Ohh let me say

X is close and open both because it satisfy both definition

D(x,r)<epsilon--->open

D(x,r)<=epsilon-->close

[viscid flower]

yea, but I guess you should have expected that

I mean, it satisfies both definitions

[viscid flower]

so you see why X is open?

[vast viper]

Yes

[viscid flower]

its important to say

X is open in X

[viscid flower]

so, yes, lets say, X is open in X we are okay with that

Now we want to convince ourselves X is closed

lets use the number line

(R,d)

[vast viper]

Suppose we take R and [0,1]

[viscid flower]

for which?\

[vast viper]

Ohhh

R is closed in R