help-26

[lone pier]

$$60^a =(512)^a= 3$$
$$60^b =(512)^b= 5$$
$$12^\frac{1-a-b}{2-2b}$$

[thorny flame]

G. Spark

[lone pier]

$(5*12)^a = 5^a * 12^a = 3$

[thorny flame]

G. Spark

[lone pier]

Any help?

Any help?

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@dusk sigil Has your question been resolved?

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[native nova]

could i get a reminder on this integration

V infinity is that value. btw

[grave vigil]

You can rewrite the integrand as (T^1.0906)^(-3) = T^(-1.0906*3) = T^(-3.2718)

Then it's just the exponent rule for integration

int(x^a)dx = (1/(a+1))x^(a+1)

[native nova]

ah right

i was so confused

[native nova]

mind giving me numerical example

[native nova]

forgot about the +1 stuff. slowly coming to me

[grave vigil]

for example, int(x^3)dx = (1/(3+1))x^(3+1) +C = (1/4)x^4 +C

[native nova]

add 1 to power then divide by power

[grave vigil]

yup, it's the reverse of the exponent rule for derivatives

[native nova]

yh i remember that. now about 80 percent confident on this

see, ur saying initally i move all stuff bottom to top. do integral stuff yet all shit ends bottom

is that cause if it stayed top only diff would be power would be neagtvie

[grave vigil]

Yes

[native nova]

gotcha. thanks pal. tiktok fried maa brain

on the recovery. maths better 5 yr ago

[grave vigil]

No problem! good luck

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[neon iron]

https://media.discordapp.net/attachments/1182036949251412069/1182100441216061600/rn_image_picker_lib_temp_1e00a641-0891-4ba9-b1a1-45811dc3e751.jpg?ex=658377d3&is=657102d3&hm=a1d37e8199eb595cea003db34b3ee377826b9712b87e3c6c5c7995db1a24aa31&=&format=webp&width=899&height=675

[neon iron]

number 50

ill translate hold on

ABCD is a parallelogram

Points F , E are the middle of DC , AB

Prove Sdebf = Sade + Sbcf

[rigid ivy]

What does S mean?

area?

[neon iron]

yes

[rigid ivy]

Draw the line EF and show triangles AED and DEF are congruent, and BFC and FEB are congruent (in fact, all 4 triangles wouldbe congruent to each other)

[neon iron]

can you elaborate as to why i need to draw the line EF?

[rigid ivy]

for triangles DEF and EFB

[neon iron]

oh

i see

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[limpid flame]

NGL idek how to graph cos(x)

[neon iron]

Graph the original graph, find the new period, amplitude, and midline!

Midline is at -2, the amplitude is 6, and the period is 8

[limpid flame]

Idk how to graph the original

[neon iron]

y = cos(x) has a midline at 0 and a y intercept at 1

since cos(0) = 1

[limpid flame]

Wat other point is there

Idk where to put the point to get to the original

@neon iron

[neon iron]

So we have a y intercept of 1

The nearest x intercepts of the y intercepts original graph are -π/2 and π/2

[limpid flame]

How to make original

@neon iron

<@&286206848099549185>

[grave vigil]

The graph you have in the picture you sent is the original

It's the graph of cos(x)

[limpid flame]

Oh

I just made it like that cuz it looked close enough

[limpid flame]

Does this look right

@neon iron

<@&286206848099549185>

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@limpid flame Has your question been resolved?

[neon iron]

oh sorry

yes it’s

correct

[limpid flame]

Ty

[limpid flame]

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[ocean loom]

Help

[ocean loom]

I’m stupid

I suck at math

[neon iron]

Picture too blurry

@ocean loom pic too blurry

[ocean loom]

In the first minute he traveled one fourth the total distance to the North Pole. In the second minute he traveled one third the remaining distance. In the 3rd minute he covered one fourth the remainder, and in the fourth minute he covered one-half the distance left. By then he had 300 miles LEFT.

By then, how many miles had Santa TRAVELLED?

Here.

Im dumb at this type of math

Pls help

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[ocean loom]

.reopen

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✅

[ocean loom]

In the first minute he traveled one fourth the total distance to the North Pole. In the second minute he traveled one third the remaining distance. In the 3rd minute he covered one fourth the remainder, and in the fourth minute he covered one-half the distance left. By then he had 300 miles LEFT.

By then, how many miles had Santa TRAVELLED?

[exotic carbon]

What did you try?

[ocean loom]

adding, subtraction, and division

Tbh i think i think i'm doing it wrong

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[floral rampart]

Where do we get the name $\sigma$-ring from?

[thorny flame]

Zander

[topaz sinew]

@floral rampart Has your question been resolved?

[pastel salmon]

We use this term in measure theory and, consequently, in modern probability theory

[floral rampart]

Yes, but I guess why is it called the way it is

Like sure, we have a ring of sets or whatever

But why sigma?

[floral rampart]

And even this, why did we agree on the word "ring"

It seems like some kind of distributive lattice

[pastel salmon]

there is also sigma - field

and

sigma - algebra

those names inherit asbtract algebra terms

[floral rampart]

Yes, sure, but even these do not live up to the algebraic terms they are given: an algebra over a set is not an algebra over a ring

My issue is why do these names still get used?

So I guess, my issue may be with universal algebra, not say measure theory

But at the end of the day, measure theory predates universal algebra

[floral rampart]

Not well, though. This is my issue

[pastel salmon]

you know, i have accepted those various definitions, such an automatic approach an di know those definitions etc, but i am not interested to much in
the etymology of those words 🙂 maybe we wud need to dig in math literatus so much to find , assume first article contains such terms. do you bother their etymology? or you search explanation why those terms are different in other branches?

[floral rampart]

well, when I hear the word ring of set, I want to think about rings, not lattices perse

i used that at the begining since it is the biggest offender of the issue

[pastel salmon]

in the history of mathematics, we have a lot of terms that originated, it seems in the wrong way, but over the years, they have become accepted and no one wants to change it now 🙂

[floral rampart]

my initial thought was like: oh we like to do measure theory at first over a sigma algebra and then go to a sigma ring which complicates a bunch of theory, maybe some sort of parallel exists. however, I can't really find one

[pastel salmon]

it does not interrupt me,

but I do understand you, maybe we should change names of them but who will decide about it ?

[floral rampart]

I have found that this is really an issue in analysis, frankly. When I do work in algebra, everything seems quite consistent, maybe I am just used to the convention though

[floral rampart]

I can buy sigma algebra, I guess

Since there are a few reasons why that name is fine

But I couldn't find one about sigma ring

[floral rampart]

I can dig my heels in representation theory here

[pastel salmon]

you know, maybe also, someone might have taken it incorrectly, started using it, publishing it, etc., etc., and now no one knows why or why.

[floral rampart]

i sure hope that isnt it haha

[pastel salmon]

you wud be very suprised

[floral rampart]

maybe ill go ask an analyst tomorrow! because it has been bugging me for a while

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[thorny stag]

How do I determine this?

[thorny stag]

,rotate

[thorny stag]

<@&286206848099549185>

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[whole prawn]

is this not true?

[flat kindle]

it's true though

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@whole prawn Has your question been resolved?

[whole prawn]

but the e is raising just the d right?

[flat kindle]

yes

[whole prawn]

👍 thanks

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[waxen iris]

I am on ii)
i have done the working out and there is no remainder, where have i gone wrong

[gilded robin]

you can find the remainder by plugging in $P\left(\frac{-b}{a}\right)$ where $P(x) = x^3 - 2x^2 - 4x + 8$

correct me if im misinterpreting your question

[waxen iris]

https://cdn.discordapp.com/attachments/885151153258635354/1182367842679656558/rn_image_picker_lib_temp_487c0f1f-d8bb-464c-83a2-75aef1f2e35d.jpg?ex=658470dd&is=6571fbdd&hm=f4bb4bcf6939dd5848aaf36c43e93cec8af7a890b2cf1f30d0c2a7f42a1a2485&

[thorny flame]

Xerunox

[waxen iris]

we are being asking to use long division to solve it

[gilded robin]

that is

interesting

what i would do in my case is work backwards, i would do $x^3 - 2x^2 - 4x + 8 = (x+2)(ax^2+bx+c)$ and then multiply those 2 brackets together to get a series of multiple unknowns in a polynomial, which you can then use to create several equations

it's my closest idea to how you would do long division, to be honest

[thorny flame]

Xerunox

[waxen iris]

ive got it thanks

.close

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[waxen iris]

where do i start

[vale furnace]

Induction

[drifting swift]

nah, induction will be hard here

[vale furnace]

Hm yeah

[drifting swift]

@waxen iris do you know anything about modular arithmetic

[vale furnace]

Just do mod

[waxen iris]

nah...

i think they want us to use deduction

[drifting swift]

... deduction?

[waxen iris]

yea

proof by deduction

[drifting swift]

wonder what they mean by that

[waxen iris]

thats just the only proof method weve been taught i think

[drifting swift]

is it mentioned by that name in your textbook

if any

anyway i got the idea that you could look at n even vs. n odd separately

[waxen iris]

yes thats what it would likely be

so how would i got about that

would i use 2n as n

[drifting swift]

no

[waxen iris]

and then 2n + 1

[drifting swift]

not 2n

any other letter except n

you would look at n = 2k and then separately at n = 2k+1

[waxen iris]

ok so 2y
then 2y + 1

ok...

once i get answers what would that prove

would i be that the divisible one would be the even one

would i do
2k^3 +2 = 0?

[drifting swift]

no

on two counts

1. there is no reason to equate n^3 + 2 to zero. who said we wanted to know when it's zero?

2. n^3 is (2k)^3 not 2k^3

[waxen iris]

but if n = 2k

then n^3 is 2k^3

[drifting swift]

sorry, typo fixed

[waxen iris]

thats what my thought process it

[drifting swift]

2k^3 means 2(k^3)

order of operations

but you are cubing 2k not just k

so n^3 is (2k)^3

you understand (2k)^3 ≠ 2(k^3) right

[waxen iris]

yes thats what i meant

just forgot to bracket it sorry

[drifting swift]

these brackets are of vital importance, unfortunately for you

anyway

what does (2k)^3 expand to?

[waxen iris]

8k^3

[drifting swift]

right

so then, for n = 2k we have n^3 + 2 = 8k^3 + 2

can you reason why 8k^3 + 2 is not a multiple of 8?

[waxen iris]

yea

[drifting swift]

ok

[waxen iris]

so now i do it with 2k+1

[drifting swift]

the second case is actually like three times easier if you know anything about even and odd numbers

[waxen iris]

well since ur adding 1, then it would be adding 3 to 8k^3, which would always be odd and never be a multiple of 8

[drifting swift]

right(ish) answer bs reasoning

well since ur adding 1, then it would be adding 3 to 8k^3
what

(2k+1)^3 is not equal to 8k^3 + 3

[waxen iris]

thats not what i meant

you have to add 2 after (2k+1)^3 since the question is it n^3+2 is divisible by 8

so it would be 8k^3 + 3

[drifting swift]

(2k+1)^3 is not equal to 8k^3 + 1 either.

[waxen iris]

is
(2k+1)^3 + 2

which is (8k^3 +1 )+2

[drifting swift]

no

(2k+1)^3 is not equal to 8k^3 + 1

(a+b)^3 ≠ a^3 + b^3

you fell victim to the freshman's dream

[waxen iris]

why do u keep saying squared?

[drifting swift]

typo, sorry.

[waxen iris]

that was confusing me

[drifting swift]

two counts of it.

(2k+1)^3 ≠ (2k)^3 + 1^3 still

[waxen iris]

yes its
(2k +1)(2k+1)(2k+1)

or use binomal expansion

right

[drifting swift]

in principle yes

however this is not needed here

do you know that the product of two odd numbers is also odd?

[waxen iris]

yea

ive got the answer all good

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[waxen iris]

thanks

.close

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[neon iron]

hello

[neon iron]

let me type out the equation

why does $3a^2(b+c)+3b^2(a+c)+3c^2(a_b)+6abc$ simplify to $3(a+b)(b+c)(a+c)$

[drifting swift]

that looks mistyped

3a^2(a+b) would give you an a^3 term which (a+b)(b+c)(a+c) does not have

[jagged hatch]

yep a^3 is dead

[neon iron]

WAIT

i did mistype it thanks let me fix it

[thorny flame]

schopenhoe

[neon iron]

this is from yesterday sorry

i feel like there's a factoring rule that i missed out on

so i see they factor our the 3

but where does

(a+b)(b+c)(a+c) come from

[mellow arrow]

factor theorem:
f(a,-a,c)=0
f(a,b,-b)=0
f(-c,b,c)=0

[jagged hatch]

wth

[neon iron]

so a function with those inputs equal 0?

i'm misunderstanding something

so they all equal 0

i understand how to distribute (a+b)(b+c)(a+c)

im still stuck sorry

i don't know how to apply that

so do you use division?

huh maybe i see where this is going

<@&286206848099549185>

[snow prawn]

uhm, im sorry, im not covering this topic rn myself (im 10th grade)

[neon iron]

im gonna mess around with it and probs ask math stack exchange

[snow prawn]

alr sure

[neon iron]

thanks who ever sent me this but i dont understand how to apply it to multivariate polynomials but i can sort of see where this is going

llike

does f(-a, a,c)= 0 relate (idk if this is the right word) to the factor of (a+c) and so on

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[jaunty citrus]

I need to calculate the electric field strength at all three points, but i'm not sure how to start this.

I know intuitively 3 is zero because there's nothing going through it, but i thought 1 is zero as well but its not. can someone guide me through this?

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[jaunty citrus]

@knotty finch

[jaunty citrus]

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[drifting swift]

can you remind us what ω(G) means

[silk kite]

clique size

max clique size 4

and complement of g with clique size 3

[drifting swift]

ah

right yeah

that looks like it checks out then

[silk kite]

ive been staring at it for so long now i cant tell 😭

just want to get one more person's opinion 😄

<@&286206848099549185>

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[silk kite]

<@&286206848099549185> 🙏

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[random shell]

For part d, how is catching a deer at random times during the day an improvement

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[polar cove]

How do I prove that (a, b) ∩ Q is countable

[polar cove]

I have already shown that there are infinite many fractions in the interval

[snow nimbus]

Do you already know that Q is countable?

[polar cove]

No

[snow nimbus]

Have you heard of Cantor's diagonal argument?

[polar cove]

I got it

But what about (a, b)\Q ?

[snow nimbus]

That has to be uncountable as (a,b) for a<b is isomorphic to R and R is uncountable

[polar cove]

yes

but it isn't a proof

or is it

[snow nimbus]

It skips out a bit of details

[snow nimbus]

What parts are you bothered with? So that I can elaborate on further?

[polar cove]

I can show that there are infinite many s in (a,b)\Q
I just need a strict proof that (a,b)\Q is uncountable

[snow nimbus]

So you already familiarized with (a,b) n Q having countably many elements?

[polar cove]

yes

Ok, I found this proof
https://proofwiki.org/wiki/Uncountable_Set_less_Countable_Set_is_Uncountable

[snow nimbus]

Yes, that's what I wanted to show you

[polar cove]

Right

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[vapid tangle]

Good evening

[loud ingot]

Goodnight.

[vapid tangle]

6. Use chain rule to derivate

I'm doing the first one and wanted to confirm that it is right

Let me take a picture of it

Re sent picture

Please let me know if I'm skipping a procedure or doing smth I shouldnt

[loud ingot]

Looks cool

Although you could simply (10x-2)/2 to 5x-1

[vapid tangle]

Oh I see what youre saying

Thanks!

I'll close this channel and come back later lemme do the other ones

[loud ingot]

,w differentiate (5x^2-2x)^(3/2)

[thorny flame]

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Requested by diesel5243

[vapid tangle]

There it is

What you said hahaha

Something I did not understood

At a certain point you will have (3/2)(5x^2-2x)^(3/2)(10x-2)

You multiply the 3 for the whole thing and divide only the 10x-2

[loud ingot]

What do you mean by whole thing?

[vapid tangle]

You have 2 dividing the 2 functions

And 3 multiplying the 2 functions aswell

Ok eureka

Theres no parenthesis

So you can multiply the 3/2 with any of the 2

(30x-6)/2

15x-3

3 common factor

(5x-1)3

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[gritty matrix]

.close

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[neon iron]

I'm so confused by these questions

for example, the first one has an answer of 2pi/3

it's negative, correct?

[somber socket]

Theta?

[neon iron]

so wouldn't it make sense for the radian value to be from a negative quadrant for cosine?

[somber socket]

No

[neon iron]

no just simply the answer is negative

[somber socket]

That is in quadrant 3

-1/2?

Are you familiar with your unit circle values

[neon iron]

I just did cos-1(-1/2)

got 2pi/3

drew it

[somber socket]

Yes

Ur calc

Has domain restrictions

Recall

[neon iron]

ok

so it must be in either q3 or q4

why not q3

[somber socket]

Cos inverse is restricted to quadrants 1 and 2

[neon iron]

why is it in q4

[somber socket]

Draw it out

[neon iron]

thats how you'd get the answer

I have

it's in the s quadrant

q2

[somber socket]

It’s not in q4?

[neon iron]

2pi/3

[somber socket]

Cos is negative in quadrants 2 and 3

You can’t plugged these into your calc

[neon iron]

yep, so you would choose a quadrant where it's negative aswell correct?

[somber socket]

I don’t understand your question

[neon iron]

I apologize

hard to explain

[somber socket]

The answer is negative right

[neon iron]

with text

yes it is

[somber socket]

Costheta= -1/2

[neon iron]

mhm

[somber socket]

At what two values is cosine -1/2

2pi/3 is one

[neon iron]

first is 2pi/3

[somber socket]

And what is the other

[neon iron]

2nd would be 2pi/3+pi

[somber socket]

Are you familiar with your unit circle?

[neon iron]

somewhat

[somber socket]

It makes these easier

[neon iron]

I hate using it though

purely memorization

[somber socket]

It’s not.

[neon iron]

which should never be something you do in math

[somber socket]

If you just memorize one quadrant

[somber socket]

You don’t have to memorize it if you don’t want to

[neon iron]

yeah I generally know what a unit circle looks like

[somber socket]

You can create your triangle 30 60 90 triangles

And 45 45 90 triangles EVERYTIME if you like

Or you can just memorize 3 diffirent cord points.

[neon iron]

being 0, pi/2, pi, 3pi/2, 2pi?

or are you talking about in the

quadrant

I don't do triangles lol

[somber socket]

Just memorize the unit circle.

It saves you time

[neon iron]

,unitcircle

[somber socket]

Here is one

[neon iron]

anyways what are you trying to get to?

[somber socket]

If you know your unit circle values

It makes this literally a 2 second question

[neon iron]

ok

so it's in q2 rn

[somber socket]

No….

[neon iron]

no?

[somber socket]

We are quadrants 3 and 4

For this problem

And cosine is negative in quadrant 3

And positive in quadrant 4

[neon iron]

2pi/3 is in quadrant 2

which is the value you get

from calculating cos-1 -1/2

[somber socket]

…… I don’t think you’re hearing me

That has no relevance

[neon iron]

yeah so we'd shift

[somber socket]

To this problem

[neon iron]

it in q3

and q4

because of

[somber socket]

Just q3

[neon iron]

it's restriction

ok q3

why not q4?

[somber socket]

This should be a no calc problem

Because cosine is positive in q4

[neon iron]

perfect

so it would be 2pi/3+pi in that instance

[somber socket]

That would put you in q4

…

And not q3

[neon iron]

hmm

wait I want it in q3

[somber socket]

Put down your calc for this problem

This is a simple problem with all unit circle values

[neon iron]

3pi/3

adding 2pi/3

[somber socket]

That’s pi

[neon iron]

would make it 5pi/3

which is q4

[somber socket]

No we aren’t looking for an answer in q4

[neon iron]

wait what would we do then

we certainly cant subtract

[somber socket]

We would refer to our unit circle

At what theta value is cosine -1/2 in quadrant 3

[neon iron]

7pi/6.. but I'd rather do it through calculations

is that possible?

[somber socket]

???

No

That’s wrong

Cosine is 1/2 when it pi/3

[somber socket]

Cosine is x

….

[neon iron]

oh sorry

4pi/3

but still, what would you do to calculate that

[somber socket]

Yes

I wouldn’t run through the calculations

[neon iron]

is there no way to do it?

because that's jut what I'm used to

[somber socket]

There is bjt quite frankly there is no reason…

You will waste time

[neon iron]

surely can't be that bad

[somber socket]

Just memorize the unit circle pattern

And you will save yourself time

[neon iron]

if I do 6pi/3 -2pi/3

I get 4pi/3

but that's subtracting from the x-axis in q4

[somber socket]

I don’t quite understand why you say you want to run through it

And then use another unit circle value

It doesn’t matter sense.

Put down your calc for unit circle values ….

[neon iron]

I'm not using it

[somber socket]

Then how did you find 2pi:3

2pi/3

[neon iron]

I used it at first for that

but then for the rest

I math it out

by getting a common denominator

and adding

[somber socket]

If you really want

Just subtract 2pi - 2pi/3

It’s just going the opposite direction

[neon iron]

hm

why would that work though

would that not put you in q4?

[somber socket]

No

2pi/3

Is in q2 right

[neon iron]

yes

[somber socket]

The first square in q2

So if we go -2pi/3

That would move us over to q3

And the first square in the opposite direction from there

Instead of going counterclockwise like we normally do

We went clockwise 2pi/3

From the positive x-axis

[neon iron]

hm

so 2pi/3 - 2pi/3

puts you in q3?

or -2pi/3

puts you in q3

not sure what you meant

yeah I guess it would

if it's -ve

-2pi/3 is q2

q3*

counter clockwise

or the opposite way it usually goes

[somber socket]

Just memorize the unit circle…..

This problem should take two seconds

No calc needed at all

[neon iron]

ok say I did

what would your thought process be

for this question

[somber socket]

I would draw a little sketch

[neon iron]

starting from the very beginning

[somber socket]

Like this

[neon iron]

cos -1/2

[somber socket]

Ok I’m in q3 and q4 only

Then put this little x

Where cos is -1/2

[neon iron]

ok

[somber socket]

And that’s where 4pi/3 is

Or you could mark both places

[neon iron]

but how would you know that

[somber socket]

Where cosine is -1/2

[neon iron]

if you only know the top quarter

[somber socket]

Wydm

[neon iron]

like sure it's easy but only if you have the entire thing in your head

[somber socket]

You just memorize the patterns of theta on the unit circle

And if you memorize the cord sets

And apply what you remeber about when x and y are positive

In what quadrants

[neon iron]

idk, I just feel it's not for me

[somber socket]

Before you attempt these problems just memorize the unit circle…

[neon iron]

like sure it's a pattern

but I don't memorize too well

[somber socket]

It takes 10 mins

And you use it for the rest of math

[neon iron]

yeah idk

just the math part

seems logical

like i can justify it

why you'd add 2pi

[somber socket]

But you waste so much time….

[neon iron]

to each their own I guess

[somber socket]

You are justifying your answer using unit circle values it doesn’t make sense

[neon iron]

hm, so the reason you'd do 2pi-2pi/3

is because if you draw it out

[somber socket]

Honestly

If you want

[neon iron]

you'd have to go -2pi/3 from 2pi?

2pi/3*

[somber socket]

Cos =1/2

It gives you pi/3 in ur calc

And add pi

To that answer

Because that transfer you from q1 to q3

[neon iron]

okay

so the -ve simply moves you

from q1 to q2

so the way to do these questions is

take the positive, inverse cos

[somber socket]

Is to use the unit circle.

Not waste your take using a calc

[neon iron]

I'm not giving in 😭

[somber socket]

Because no teacher is giving you a calc on this section

You will fail.

If you rely on a calc

You will not succeed, im being honest

If your test is no calc, and you don’t have your unit circle memorized

You will struggle

Unless you want to draw your triangles every single time .

[neon iron]

I already had an assessment on this

with no calc

had the unit circle memorized and all

but I forgot it

[somber socket]

How do you forget the unit circle

What

[neon iron]

it was like a month ago lol

the assessment I'm having on this is generally calc allowed

it's just a broader test

so my teacher will likely use

questions that calcs dont instantly solve

but I just wanted to grasp this minute concept

so answer me this, you plug in cos inverse of the positive value, then add pi or subtract from 2pi depending on where the value is positive/negative?

is that all there is?

[somber socket]

I’m not going to entertain this way of solving it

It’s inefficient

And you will waste time

By the time you finish the first one

I will be done with them all

[neon iron]

I can't just get a tattoo of it on my arm lol

[somber socket]

You can take 10 minutes to learn it

And understand where it comes from

[neon iron]

maybe one day

but for now

[somber socket]

You’re memorizing a process right now

It doesn’t make sense.

[neon iron]

it does because it follows a logical system

[somber socket]

And the unit circle is more logical

And just accounts for what you’re doing.

[neon iron]

alr man 😭

I can't win this debate

thanks for helping though

[somber socket]

I beg for you just to memorize the unit circle

It will make your life easier

And if you are ever given a no calc test, which are common

[neon iron]

if the test is no calc then I'll memorize it

otherwise I'll disregardi t

because I don't memorize anything

[somber socket]

Whatever suits you i guess

[neon iron]

i can't

[somber socket]

You just did

??

[neon iron]

lmao ok

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[lucid junco]

I don’t understand the steps in red. Where did the 2x come from and why do you multiply it to the numerator and denominator

[sweet shard]

It's to make du more obvious

2x/(2x) = 1

[lucid junco]

Could you also do dx = du/2x and then get the same 2x/2x multiplied to the integrand

Or is that different

[sweet shard]

dx = du/(2x) makes no mathematical sense

[lucid junco]

Well how I normally integrate functions is find dx = du…. And replace dx with that like I haven’t done du = dx….before so I don’t really know how that works

Also if u = x^2

Then du/dx = 2x

then dx = du/2x

No?

[sweet shard]

Yea you can keep doing that

Just know it doesn't make mathematical sense

[sweet shard]

Which is why they multiply top and bottom by 2x instead

[lucid junco]

So I’m confused it’s illegal but I get the right answer with it?

So it’s wrong??

or

[sweet shard]

Yes

It's mathematically wrong, but you still get the right answer

[lucid junco]

I’ve been doing that method for the like past 2 months since I’ve started integration is that something I should change

[sweet shard]

Do whatever you want

[lucid junco]

Because i don’t want it to cause like serious problems later on

where I can’t integrate or whatever because of it

[sweet shard]

idk your future

[lucid junco]

💀

But you know maths

will that cause knowledge gaps

[sweet shard]

Why would I know what math classes you're gonna take

[lucid junco]

I don’t think that matters but

Whatever

Doesn’t matter

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[terse trellis]

can someone explain to me how my teacher did the blue marked step

[terse trellis]

how did she bring the sqrt to the denominator

[ruby tree]

sqrt(x)/x = 1/sqrt(x)
(just dividing both parts by sqrt(x))

[neon iron]

[
\f{a^n}{a^m} = a^{n-m}]

[neon iron]

its this pretty much

[pastel oracle]

factor a sqrt(1 - x^2) from the top, then simplify everything

[ruby tree]

Note that the integral got split as well

[terse trellis]

yeah

[ruby tree]

Which btw makes the LHS incorrect, there shouldn't be 'dx' twice in there

Or there should be an integral sign before the first blue part

[terse trellis]

wait

still dont get it

ill write it down maybe ill understand then

holdo

[ruby tree]

$\frac{\sqrt{1-x^2}}{1-x^2} = \frac{\sqrt{1-x^2}}{\sqrt{1-x^2}^2} = \frac{1}{\sqrt{1-x^2}}$

[thorny flame]

Nel

[terse trellis]

did this

replace + with * btw

before the integral

Oh ok think i get it

[terse trellis]

gonna have to remember this one 👍🏽

[ruby tree]

It's pretty basic fraction manipulation

[ruby tree]

If it makes it easier do it like that

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

[thorny flame]

Nel

[terse trellis]

Just dividing by sqrt(x) way easier imo

Thxxx

[ruby tree]

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[neon iron]

is there a theorem where you can easily calculate $\sum_{n=1}^{k} \frac{1}{n}$

[thorny flame]

Retribution | ᵇᵃˢᵉᵈ

[neon iron]

🙏

is it this btw:$\ln(k) + \gamma + \frac{2k}{12k^2} + \mathcal{O}\left(\frac{1}{k^4}\right)$

[thorny flame]

Retribution | ᵇᵃˢᵉᵈ

[vale furnace]

There’s no closed form at least not simple ones.

[neon iron]

i though euler had one I just googled

google might be bs tho

[vale furnace]

Eulers constant?

[craggy haven]

that gamma is a number that doesn't really have a closed form

it's like 0.537...

[thick arrow]

now. idk the formula or anything. but, isnt this sum upto k terms for a Harmonic Progression?

[neon iron]

😦

On wiki it says harmonic series has no theorem

[thick arrow]

☹️

[vale furnace]

[ \rainbow{\ga + \varphi(k+1) } ]

[thorny flame]

Pure

[vale furnace]

Or smth like that

Where phi is the digamma function

[neon iron]

oh maybe that'll do

[vale furnace]

Basically, the closed form is horrendous but sometimes it’s denoted as H_n

[thick arrow]

?

[vale furnace]

?

[neon iron]

that might work

i'll test it, thanks

.close

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[cursive thorn]

.this is an approximation apparently

[vale furnace]

Yes it is.

[vale furnace]

Oh

.close

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[stray dew]

Why csnr I send my picture bruh

[stray dew]

What's going un

[bitter pollen]

is it .png?

[stray dew]

ok here

i just dont know how to go on about solving this double integral

i was ihnking of splitting it apart cuz its being added

the 24r^2 i alreaady did and got 48pie/3 (after double integral and bounds)

but idk how to do the left side of the equation

<@&286206848099549185>

[sweet shard]

,rotate

[sweet shard]

Show how you got this

[stray dew]

24r^3/3

[sweet shard]

You treat r and theta as independent variables. So one is constant with respect to the other

[stray dew]

plug in r values

which gives me 24/3 but with theta integral its 24/3(theta)

and bounds are 2pie and 0

[sweet shard]

pi, not pie

[stray dew]

so its just 24/3 times 2 pie

[stray dew]

mb

which gives me 48pi/3

[stray dew]

yea but idk how to do it

[sweet shard]

Why not

When you integrate r, you just factor out the theta terms

[stray dew]

which one is the constant

[sweet shard]

,tex .int rules

[thorny flame]

riemann

[sweet shard]

The one that's not being integrated

[stray dew]

oh so wait

for the first integral is it just 18r^4/4(cos62thetasintheta)?

only thing we do is the integral of r^3 but everything else is the same?

@sweet shard

[desert zealot]

hi please i want help to find the area of the purple surface. using integrales

[sweet shard]

!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).

[sweet shard]

I don't know what "it" is

[stray dew]

okok i get it

kinda

for this

do i need to memorize the double angle formula thingy

@sweet shard

[sweet shard]

You don't need to for this problem

Try a u sub

[stray dew]

With u as sin?

Dude ur so smart

[sweet shard]

No

[stray dew]

Wait what

Oh u as cos

Cuz its cos^2

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[neon iron]

How do make it so a reimann sum doesnt count area behind y axis negative

[neon iron]

Bc im trying to find $\int_{-4}^{4} (4 - \sqrt{x^2}) , dx$ but im supposed to use the reimann sum method instead 😑

[thorny flame]

Retribution | ᵇᵃˢᵉᵈ

[neon iron]

and whenever I use this method the area is always 0 bc the negative area is the same as the positive area and the stuff cancels out

when the answer should really be 16

[craggy haven]

,w graph 4 - |x| from -4 to 4

[thorny flame]

Results provided by WolframAlpha

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Requested by kitten.in.a.teacup

[craggy haven]

it should really be 16 yes

[neon iron]

but idk how to get the reimann sum not to make it 0

it basically counts the area behind y axis as negative area

[craggy haven]

ok, it shouldn't do that

what are you doing?

[neon iron]

I did $\sum_{i=1}^{n} \left(4 \Delta x - |i \Delta x|\Delta x\right)$

[thorny flame]

Retribution | ᵇᵃˢᵉᵈ

[neon iron]

thats not right

hold on

wait nvm it is

I did this bc $\sum_{i=1}^{n} (4 - |i\Delta x|)\Delta x=\sum_{i=1}^{n} \left(4 \Delta x - |i \Delta x|\Delta x\right)$

[thorny flame]

Retribution | ᵇᵃˢᵉᵈ

[neon iron]

I made it 4 -|i deltaX| bc im using circumscribed rectangles in the sum on the right and inscribed rectangles on the left

I wish I had an image so I could explain this easier

[fossil zephyr]

still need help? because i managed to create a sum that just gave me 16, using the normal definition.

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[toxic stirrup]

i have a vague idea of how to do this, but im not sure what exactly

[toxic stirrup]

basically each rook has to occupy its own row and column, that no other rook occupies

so just looking at the columns, the first rook has 8 choices

the next rook has 7 choices, etc

so 8!

hm are the rooks unique

if not, then i have to divide by 8!

which gives me 1: 8!/8!

[wary tulip]

they do not look unique

but

[wary tulip]

are you sure about this?

[toxic stirrup]

yes?

i might be wrong though

[wary tulip]

what if they are e.g. all in the same column?

say all the in the A file

[toxic stirrup]

oh crap that also works

do i have to use stars and bars

8 rooks, 7 bars

[wary tulip]

probably not have to

here’s how i would think about it

you have the case where all the rooks are in different rows and columns already

what if a column has exactly 2 rooks in it?

actually

nah forget about that

[toxic stirrup]

oh

so should i just use stars and bars?

[wary tulip]

you can try it, i will try to think of a simple way to think about it

ah - if you are missing a rook in any column, you need to have a rook in every row

[toxic stirrup]

oh yeah

[craggy haven]

that's what i was trying to prove

[toxic stirrup]

and we can just count it for columns and multiply by 2

so what should i do

[wary tulip]

we want configurations where not all columns are used now? X out one column (say the A file first). how many ways can you set the rooks?
edit: and ugh we’d need to require all the other columns be used or do some PIE bs

all rows must be used

hmm this will get a little ugly with all the ways we could not use columns

i have a stupid thing i want to try

[toxic stirrup]

this might be PIE lol

[wary tulip]

[rustic cedar]

!show

[topaz sinew]

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

[toxic stirrup]

maybe we should do some sort of complementary counting?

what cases do not work

[wary tulip]

cases that don’t work are when not all columns are used and not every row is used?

[toxic stirrup]

wait we have to use pie. 8^8 ways to put one rook per row and 8^8 to put one rook per column. this double counts the ways that have one rook in each row and column. so for that there are 8 ways to choose the column for first row, 7 for second, etc. thus it is 2*8^8-8! ?

[junior pumice]

wouldn't this work?

[wary tulip]

every row is used there

[junior pumice]

oh i see you meant both not all columns and not all rows

my b

[toxic stirrup]

does this work?

[wary tulip]

what are you computing there?

[toxic stirrup]

total number of ways to place the rooks

[wary tulip]

ic

that might be right

[topaz sinew]

@toxic stirrup Has your question been resolved?

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Remember:
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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.

[neon iron]

how would a phase shift affect a transformation of a trig identity?

[neon iron]

does the graph physically cut off at any point before the phase shift

or does it simply start there

?

[austere sable]

sin(x+a)

a moves the function

https://www.desmos.com/calculator

[austere sable]

this question doesn't make sense

trig identities are just equalities

[neon iron]

trig equation sorry

mixing two concepts I learnt lol

[austere sable]

$a sin(bx+c)$

[thorny flame]

THAT'S MY QUANT, MY QUANTITATIVE

[neon iron]

with the cut off approach i mentioned earlier, it would be possible to affect the shape of the sin or cosine graph

which is why I doubt that is the answer

i.e instead of starting from min, you'd start from max

you could*

[austere sable]

a makes the wave up and down bigger, b increases the fequency, c shifts the phase

[neon iron]

just depending on where you cut off

okay

I get that

just how does it shift

[austere sable]

think of it like this

y=x

now the RHS can be a function

so lets say y = f(x)

f(x) = x

so now we have y = f(x)

(remember, f(x) is still just x)

then we "shift" the function

y=f(x+a)

which is just y=x+a

there is no cutting off

you're just moving the function

it's like y=x^2

and moving it up 1

to be y=x^2+1

[neon iron]

so it simply just moves

and starts at that point

[austere sable]

yes

there is no start

[neon iron]

so nothing happens before it?

[austere sable]

did youclick on the link I sen t

[neon iron]

so what difference does the phase shift make

[austere sable]

to desmos

[neon iron]

yes

[austere sable]

write sin(x)

there is no beginning and end

it goes from x=-inf to +inf

[neon iron]

so what's the purpose of a phase shift

that's my question

if it has no overall affect

on the graph

[austere sable]

it does

so for y=sin(x)

if I shift by pi/2

to $y=sin(x + \frac{\pi}{2}$

[thorny flame]

THAT'S MY QUANT, MY QUANTITATIVE

[austere sable]

then what is y at x=pi/2

compared to y=sin(x)

[neon iron]

hm it does look different

so when graphing

what would I do

when I see a phase shift