help-49

[twilit field]

$f(x,y) = x^2 \mathbf{i}+ y^2 \mathbf{j}$

[grand pond]

What a wonderful world !

[twilit field]

To check if f is the gradient of some potential field, let's consider the path integral along the closed path $(cos(t), sin(t)$,where t ranges from [0,2π} to R^2

[grand pond]

What a wonderful world !
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

so $\int_{0}^{2 \pi} ( cos^2(t) , sin^2(t)) \cdot ( - sin(t) , cos(t))dt$
$=\int_{0}^{2 \pi} -\sin(t) cos^2(t) + sin^2(t) cos(t)dt =0$.
Thus as the path integral along a closed path is 0, thhere exists a function $\phi$, whose gradient is $f$

[grand pond]

What a wonderful world !

[twilit field]

How does this look?

[floral junco]

$\int_0^{2\pi} (\cos^2(t),\sin^2(t)) \cdot (-\sin(t),\cos(t))\,\dd t$

$=\int_0^{2\pi} -\sin(t)\cos^2(t) + \sin^2(t)\cos(t)\,\dd t=0$

[floral junco]

(just some latex advice)

[twilit field]

Thanks

but is the general idea right

Now , I would like to check if $f(x,y) = 3x^2y \mathbf{i} +x^3y \mathbf{j}$ is the gradient of some scalar field

[grand pond]

What a wonderful world !

[twilit field]

For this , I suspect checking path independence is easier.

for this I'l consider two path from (0,) to (1,1)

(t,t)

and (t,t^2)

$$\int_{0}^{1} (3t^3,t^4) \cdot (1,1) dt = \int_{0}^{1} 3t^3 + t^4 dt = \frac{3}{4} + \frac{1}{5} = \frac{19}{20}$$
\
\
$$\int_{0}^{1} (3t^4,t^5)\cdot (1,2t)dt = \int_{0}^{1} 3t^4+2t^6dt = \frac{3}{5} + \frac{2}{7} = \frac{31}{35}$$
\
As the path integrals aren't path independent , $f$ isn't the gradient of a scalar field

Does this work

[grand pond]

What a wonderful world !

[twilit field]

?

[midnight plank]

@twilit field Has your question been resolved?

[midnight plank]

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[prime garden]

If 0<=b<=2R and R is a constant and b changes, find the value of 'b' in terms of R that maximizes the following:

b(1 + 1/x), where x = 1/2(root((2R - root(4R^2 - b^2))/R)

[prime garden]

,rotate

[prime garden]

,rotate

[prime garden]

,rotate

[lyric charm]

you can use ,rcw to rotate an image clockwise by pi/2.

or ,rccw to rotate counterclockwise by the same amount

$\max_{0 \leq b \leq 2R} b \cdot \paren{1 + \frac{2\sqrt{R}}{\sqrt{2R - \sqrt{4R^2 - b^2}}}}$

[grand pond]

Ann

[lyric charm]

is this what you're doing?

[prime garden]

Yes.

[lyric charm]

looks quite ugly

1. does calculating d(that shit)/db give anything of note?
2. where did this rather meaty looking function come from, anyway?

[prime garden]

Ans. 1: Calculus is prohibited. This is a simple algebraic question and therefore one should be able to solve it using simple algebra.
Ans 2: From my experiments with truth.

[lyric charm]

uh what.

ok i'm gonna call BS on this being a "simple algebraic question"

plenty of "purely algebraic functions" don't admit themselves to calculus-free methods of optimization

• gonna have to ask you to elaborate on this "experiments with truth" stuff. what exactly did you experiment with? what problem will this maximization solve? what, if anything, do b and R represent?

the ban on calculus appears to be entirely self-imposed...

[prime garden]

The value of b in terms of R will give the unequal side of an isosceles triangle inscribed in a circle of radius R that will such that the perimeter of this isosceles triangle will be the maximum among every other isosceles triangle inscribed in the same circle of radius R.
Ans: b = root(3) * R
So it will be an equilateral triangle.

[lyric charm]

also the other reason im asking this is that if you really don't want to use calculus, then knowing where this problem came from might actually lend some insights into a calculus-free solution...

ok so it's a geometry thing then

[prime garden]

I asked another question that was on the same question but involved trigonometry:
maximize 2 cos(x/2) + sin(x) for x in between 0 and 180.
This gives x = 60 deg which too supports that the triangle must be equilateral.

[lyric charm]

you've got a circle of fixed radius R, and you want to maximize the perimeter of an isosceles triangle inscribed into it.

what angle would be represented by x here then?

[prime garden]

angle ABC.

[lyric charm]

it might be somewhat easier or cleaner to express the perimeter in terms of angle BAC instead

[lyric charm]

i mean... in that case, problem solved, right?

or are you still on the lookout for a proof of maximality at x = 60° there?

one that, according to your taste, does not involve calculus

(the use of calculus to solve geometric optimization problems is IMO not nearly as much of a sin as you're making it out to be)

[prime garden]

Yes.

Either the trigonometry one or the one I sent today.

[lyric charm]

well i am definitely not touching that rooty monstrosity

and i would recommend you do the same

is 2cos(x/2) + sin(x) the ratio of the perimeter to R?

[prime garden]

I move AC vertically upwards such that it does not change it's x - coordinate. This way angle ABC changes linearly. This ensures every possible isosceles triangle..

In terms of angle ABC (or x) and R, perimeter = 4R cos(x/2) + 2R sin x = 2R (2 cos(x/2) + sin(x))

[lyric charm]

oh so it's the ratio of the perimeter to 2R

ok, the essence of it is the same...

[lyric charm]

so let t = angle BAC in this diagram

x = 180°-2t

so your 2cos(x/2) + sin(x) becomes... 2 cos(90°-t) + sin(180°-2t), or 2 sin(t) + sin(2t)

ok hold on. i think i have some ideas for how we could manipulate that with trig shit.

actually @prime garden are you ok with AM-GM inequality trickery

cause i think i found something that might work

[prime garden]

Yes

[lyric charm]

oh wonderful

then i have cooked something up

one moment

so first just to be on the same page here: i will be using a slightly generalized version of AM-GM

specifically weighted AM-GM

[prime garden]

okay

[lyric charm]

so like, AM-GM is tricky to apply here bc you have to tinker for a bit

with the motivation being that you want to end up with a constant on the other side, so that you may then invoke AM-GM's equality condition to find when the upper bound is achieved

(this is why i do the strange step of multiplying and dividing by 3^(3/2) in line 7)

@prime garden hows this look

[lethal path]

supremely clever, Ann

[lyric charm]

also as a bonus you get the max value of P/(2R) at the end there

as 3sqrt(3)/2

[prime garden]

All hail Ann.

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[midnight plank]

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[formal shore]

So when we have 2 vectors a and b. Given the lengths of both, saying |a| = 3sqrt(2) and |b| = 4.

AC = a + b

How can we find |AC|?

[lethal path]

that's the thing! there's no one value for |AC|

[formal shore]

That means it's a dead end? According to the book the answer is sqrt(58). I'm not even sure how to start solving the problem 😅

[lethal path]

oh is there more information

[formal shore]

I can write the complete question to give more context.

In the parallellogram ABCD, corner A = 45 degrees, vector AB = a and vector AD = b. Furthermore, the lenght of vector a = 3sqrt(2), and the length of vector b = 4.

a) Find vector AC and BD expressed with vector a and b.
b) Find a * a, b * b, a * b.
c) Find the length of AC and BD.

a)
AC = a + b
BD = b - a

b)
a * a = 18
b * b = 16
a * b = 12

[lethal path]

yeah then you need the 45 degree info really badly

[formal shore]

That what I've done so far

[lethal path]

I swear this problem is in some famous textbook cause I've seen this exact question asked before here

[formal shore]

It's a non English book, but can be well copied from another book haha

[lethal path]

yes so you need coordinates, so you have a = (3, 3) cause a 45-degree angle means x = y
and then you can let b be the horizontal side, so just b = (4, 0)

a + b = (7, 3)

[formal shore]

The main thing I struggle with understanding is that from the book so far I know that |AC| = sqrt(x^2+y^2). Yet we're not given any coordinates

[lethal path]

AB = a and AD = b as claimed

verify that |a| = 3 sqrt(2) by Pythagoras

if a = (3, 3)

[formal shore]

So how did we come to the conclusion that a = (3,3)? Did we define our own coordinate system?

[lethal path]

yep

orient the parallelogram so that it looks like the picture

let AD be the x-axis

[formal shore]

I'm just drawing it out lol

[formal shore]

This kind of how you meant?

[lethal path]

no I really did mean this

[formal shore]

alright, so if we were to say that A(0,0). What would be the next step?

If we'd take it with extra slow steps

I'm trying to comprehend what's all happening

[midnight plank]

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[cinder rain]

Hi uh could anyone solve this

2023! Is divisible by 9 so it's js 2023! = 0 mod 9

So S(S(S(2023!))) = 0 mod 9

But then after that what ._.

I got no idea what to do after finding this

[tender trench]

,rotate

[waxen willow]

S(S(S(n))) based on intuition is a very small number

In terms of n

So it is very possible that S(S(S(2023!))) is a single digit number

[cinder rain]

Yea so what is it 😭

[cinder rain]

I got no idea what to do after this

[waxen willow]

Try to prove this

[cinder rain]

I got no clue on how to do that

💀

[waxen willow]

Bound the number

S(2023!) < what?

[cinder rain]

Uhhhh

._.

S(n!)?

[waxen willow]

9* the number of digits the number has

Idk if S(S(S(2023!))) will get you to a single digit. I hope it does

[cinder rain]

😭 honestly I'm lost

[waxen willow]

You can go the other way around

Actually dont

[waxen willow]

One ok bound is

$$S(2023!) < 9(1 + \floor{\log_{10} (2023!)})$$

[grand pond]

casework

[cinder rain]

I was js thinking about like cus 2023! Is so large but then S(2023!) Reduces the digit to like few thousands S(S(2023!)) Makes it like just 2 digits or 3 digits and S(S(S(2023!))) Is like 1 digit so cus it's 0 mod 9 the anwser is 9

Is this an acceptable way or nah

[cinder rain]

Cus I got no idea what's going on here

It's 4:30am for me rn cus I been studying since 7pm so my brain is kidda broken rn

[waxen willow]

Answer is good

If its a a , b , c , d question circle 9 and you are good

If you need proof

You have to prove (in worst case)
$$S(S(2023!)) < 99$$

[grand pond]

casework

[waxen willow]

That implies S(S(2023!)) is at most a 2 digit number of form 9k. Which mean if you take another S(of that) it will be 9.

[cinder rain]

Nah I don't need proof but for me I js like having proof cus it gives me like confidence in the awnser lol

[waxen willow]

Well working with 2023! Is hard so just put $2023^{2023}$

[grand pond]

casework

[cinder rain]

Yea so uh how can I prove S(2023!) = how many digit and then how to probe S(S(2023!)) Is 2 digital

Digit

[cinder rain]

Wouldn't the value change a lot tho?

[waxen willow]

It would. But it doesnt matter

[cinder rain]

💀

[waxen willow]

2023! Is a random number. This would prob work for way larger numbers

$$\log_{10}(2023^{2023}) < 3\cdot 2023$$

[grand pond]

casework

[waxen willow]

In this form its easy to calculate the log

[cinder rain]

I mean yea

Prob 2023²⁰⁰⁰ would honestly still be higher then 2023!

[waxen willow]

So you have
$$S(2023!) < 9(1 + \log_{10}(2023!)) < 9(1 + 3\cdot 2023)$$

[grand pond]

casework

[waxen willow]

And thats like

,calc 9(1 + 3*2023)

[grand pond]

Result:

54630

[waxen willow]

So very small number

You do the same thing now

[cinder rain]

Ok ngl I'm confused on the part of where 9( 2023 × 3 + 1 ) came from

💀

[waxen willow]

$$\log_{10}(2023!) < \log_{10}(2023^{2023}) = 2023\cdot \log_{10}(2023) < 2023 \cdot 3 $$

[grand pond]

casework

[cinder rain]

Ohh

[waxen willow]

And you have
$$S(n) < 9(1 + \log_{10}(n))$$

[cinder rain]

Wait uh where did the 9 and +1 come from?

[grand pond]

casework

[waxen willow]

$1 + \floor{\log_{10}(n)}$

[grand pond]

casework

[waxen willow]

This is exactly the number of digits n has

For our sake we drop the floor because thst makes the number just so slightly bigger

And a bound on S(n) is 9*(how many digits n has)

If n has m digits. Then S(n) <= 9*m

[cinder rain]

Oh cus it's mod 9 alright

[waxen willow]

No , because 9 is the biggest digit

[cinder rain]

Kk

,calc 9(1 + 3*2023)

[grand pond]

Result:

54630

[cinder rain]

So this is js S(2023!)

[waxen willow]

Its a number bigger than S(2023!)

[cinder rain]

I mean yea

I think 2023^2000 would still be bigger

So

,calc 9(1 + 3*2020)

[grand pond]

Result:

54549

[cinder rain]

Eh meh

,calc 9(1 + 3*2000)

[grand pond]

Result:

54009

[cinder rain]

Lol

So js 54,000

[waxen willow]

Probably. But its easier to argue $2023^{2023} > 2023!$ than $2023^{2000} > 2023!$

[grand pond]

casework

[cinder rain]

I mean yea

[waxen willow]

Well basically do this again

[cinder rain]

Rn we can js assume prob S(2023!) Got like 54,000 digits

[waxen willow]

Not that it has 54000 digits. But that its about 54000

2023! has about 54000/9 digits

Which is around 6000

[cinder rain]

So uh 233²

= 54000

[waxen willow]

Now get thet S(n) is a 2 digit number if n <= 54630

That would prove that S(S(2023!)) is a 2 digit number

And it has to be less than 99

Because if its 99 then S(S(S(2023!))) Could potentionally (but it isnt) be 18

[cinder rain]

Chances aren't that good for it to be 99

[waxen willow]

Well yes they arent

[cinder rain]

So rn we got uh S(S(44630)) right?

Or uh am I js wrong

[waxen willow]

I mean you dont work with a discrete number now but you just assume that S(2023!) < 54630

[cinder rain]

If yes then it's easy to prove cus it's js 4+4+6+3 which is S(17) which is 8 so it's a 1 digit number so js 9 ig?

[cinder rain]

Ye so it's js S(S(S(2023!))) < S(S(54630))

[waxen willow]

You dont work with 44630 but rather a number less than 54630. So an arbitary 5 or less digit number

[cinder rain]

Oh us this was a typo

[waxen willow]

And if you take that S(n) <= 9*(number of digits)

[cinder rain]

I meant to write 54630 lol

[waxen willow]

You can get upper bound of S(S(2023!)) Pretty easily

[cinder rain]

It could also be like 59999 lol (it can't cus 54530 it already above S(2023!) Which makes it S(41) which is 5 also again a 1 digit

So ig doesn't matter what it's a 1 digit number

Anyways tq for ur help

Imma sleep now it's 5 am

Been studying for 10 hrs straight

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[waxen willow]

Yes 5 + 4*9 is also ok bound

But when you realise that you can do just way bigger numbers without changing a thing you can do much worse bounds like 5*9

[midnight plank]

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[verbal kindle]

I've gotten to (iii)

and I want to make sure the only condition is a^2 = 3b? is that it?

<@&286206848099549185>

[fallow scarab]

show your proof

[midnight plank]

@verbal kindle Has your question been resolved?

[verbal kindle]

@fallow scarab

[verbal kindle]

Man I gotta sleep

:/

.close

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[potent forum]

Could someone tell me if this would be an acceptable formula in L?

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[potent forum]

<@&286206848099549185>

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[dusk pier]

Hello, I remember someone needing help with this problem earlier. I was stumped on it until today, so please do tell me if my method is incorrect. I will send the working after I put down the question.\
If $f\qty(\frac{x}{x+1})=\frac{f(x)}{2}$, and $f(1-x)=1-f(x)$, then calculate $$\sum_{n=1}^{\infty}f\qty(\frac{2}{2n+1}).$$

[grand pond]

mathisfun

[dusk pier]

From the first equation, we have $f\qty(\frac{2}{2n+3})=\frac12f\qty(\frac{2}{2n+1})$. Thus we can transform the sum into $$\sum_{n=1}^{\infty}f\qty(\frac23)\qty(\frac12)^n.$$

[grand pond]

mathisfun

[dusk pier]

Is this valid?

Because finding f(2/3) is really easy.

[lament knoll]

you took x=2 ?

[dusk pier]

What do you mean?

[lament knoll]

how you used the first equation

[dusk pier]

I just plugged in x=2/(2n+1). Lol

Pretty neat.

[lament knoll]

k i’m dumb lol

[dusk pier]

But does it look good so far?

All I have to do is use the reflexive properties.

$f\qty(\frac12)=1-f\qty(\frac12)$, $f\qty(\frac12)=\frac12$ from the second equation

$f\qty(\frac13)=\frac12f\qty(\frac12)=\frac14$, from the first equation

[grand pond]

mathisfun

mathisfun

[dusk pier]

$f\qty(\frac23)=1-\frac14=\frac34$, from the second equation.

[grand pond]

mathisfun

[dusk pier]

Then it's just a simple geom series.

$\frac34\cdot\frac{\frac12}{1-\frac12}=\frac34$

[grand pond]

mathisfun

[lament knoll]

you proved $f(\frac{2}{2n+1})=\frac{1}{2}^n f(2/3)$ using induction?

[grand pond]

tm

[dusk pier]

Oh.

Well, I was just doing a calculation.

[dusk pier]

But doesn't this relation directly imply a geometric series?

$u_{n+1}=au_n$, where $a\in\bR$, is a geometric series.

[grand pond]

mathisfun

[lament knoll]

oh i didn’t see it 😭😭

you right

its not (1/2)^n-1 ?

[lament knoll]

here

for the sum

[dusk pier]

We are starting at n=1, no?

Hmm.

[lament knoll]

its 3am for me i’m tired sorry 😭😭

it seems good sir

[midnight plank]

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[undone marlin]

did i reach the answer the right way? trying to understand it conceptually but im kinda confused

[lethal path]

inverse function theorem, so suppose the slope of f(x) at some point is $\frac{\Delta y}{\Delta x}$

[grand pond]

south

[lethal path]

for the corresponding point on the inverse of f, we just swap x and y, so the slope there is $\frac{\Delta x}{\Delta y}$

[grand pond]

south

[lethal path]

and $\frac{\Delta y}{\Delta x} \frac{\Delta x}{\Delta y} = 1$

[grand pond]

south

[undone marlin]

uhh ngl all i understand is f^-1 (x) = 1/f`(y) and vice versa but i dont know why

[lethal path]

read the explanation I gave above one more time

[lethal path]

so yes, I am claiming that $f'(x) f^{-1} ' (y) = 1$

[grand pond]

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

[undone marlin]

ohhh okay

my brain didnt process that it was the derivative of the inverse

sorry about that 😅 ty for the explanation!

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[lethal path]

ahhhhhhhh no wonder

no worries!

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[empty ivy]

How do you do part b

[lethal path]

you can just expand

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[twilit field]

@wary thorn graph theory

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[molten bay]

How can I start this problem?

[violet dune]

since its absolute value, start by figuring out where its negative and where its positive and separate the integral in two parts based off of that

[midnight plank]

@molten bay Has your question been resolved?

[molten bay]

It is negative less than 1

[1/2,1][1,2]

[violet dune]

yes

now set up two integrals with those bounds

[midnight plank]

@molten bay Has your question been resolved?

[molten bay]

@violet dune

[tribal temple]

(If you know how to integrate ln(x), and you hopefully do, then you can convert the integrals you’re dealing with in terms of ln(x) instead )

[midnight plank]

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[midnight plank]

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[dusty portal]

[dawn dagger]

maybe there is a way completing the square inside the square roots

[dusty portal]

Nope

They sent this yesterday, I tried doing it for an hour, couldn't do anything

[dawn dagger]

Oh good thing you said that

[midnight plank]

@silk forge Has your question been resolved?

[simple field]

yeah me too 😭

there has to be some trick, the answer is so nice

[simple field]

whats the source of the problem ?

[humble wolf]

!showwork

[midnight plank]

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

[humble wolf]

if you don't want to do so, but just want a solution, you may consider
https://www.integral-calculator.com/

[simple field]

tbf in this instance i dont think that'll be of help to anyone

[humble wolf]

???
to ask a question one has to show work
that's to help helpers to know what OP has already known

so as to avoid duplicate efforts for both sides, and helpers don't have to repeat what OP has already known

that can save time for both sides, so that OP can get help quicker

in addition, helpers would know what OP has already known
so that the guide can adapt better to OP's level

[simple field]

yeah no i understand why generally speaking

just for this particular question, 2 of us already tried at this integral for a decent amount of time and got nowhere (and im sure its much a similar story for the OP)

[humble wolf]

then OP has to show what they've already tried

so that future helpers would know that the way that OP has tried wouldn't help

[simple field]

u = 1/x didnt help, u = 1-x didnt help, no trig sub seemed promising, trying to rewrite the integrand by squaring it (or a portion of it) got nowhere

[humble wolf]

that's to save helper's time, so that OP can get help quicker, so it'll eventually save OP's time

[simple field]

@dusty portal anything else you tried ?

[humble wolf]

i mean OP should post their calculations, so that helpers can verifiy whether they've made a (careless) mistake, which might lead to a blockage

[simple field]

this is a hard integral, i dont think op has made any progress on it

[dusty portal]

Nah

I have school and I’m prepping for a comp tomorrow so I probably won’t be doing this integral

[simple field]

fair dos

[midnight plank]

@silk forge Has your question been resolved?

[simple field]

sqrt(1/(4x^2) + 1/x - x) - 1/(2x) has a nice (ish) series expansion but i cant imagine its the way forward

[midnight plank]

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[hasty holly]

they got one expression each with substituting y then z, how did they got the final expression?

[midnight plank]

@hasty holly Has your question been resolved?

[silk forge]

its just some small youtube channel he posts maths questions

[silk forge]

couldnt rly do much tbh

ntg worth showing tbh

[midnight plank]

@hasty holly Has your question been resolved?

[simple field]

Does he not go through the solution? Anyhow drop the link pls !

[silk forge]

i just wanted to to see if there were any other sols

its in hindi

idt i can send it here ill dm

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[timber tartan]

Q23

[timber tartan]

I only know that one angle is the same

[dusty portal]

Think in terms of units.

If you have a ratio of 9cm:4cm, then what will the ratio of perimeters be?

[visual tiger]

wrong question

[visual tiger]

Think about the angle DBC and the angles inside that add up to it

then remember the sum of angles of a triangle (for example in triangle DBM)

[timber tartan]

I think i got it

i didnt read ir msgs yet i was looking at the q lemme send u my ans first

First we prove that DCB~DMB

D is common
Angle DBC= Angle DMB so theyre similar

which means d=mbc

And d=mbc and dmb=bmc

[visual tiger]

sure this works

[midnight plank]

@timber tartan Has your question been resolved?

[timber tartan]

yay thanks

[midnight plank]

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[radiant roost]

i don't understand what this is asking me to do

[solid iris]

prove that $\bigcap_{\text{subring } S\ni a}S=\lbr{\brc{a^n:n\in\bZ^+}}$

if youve ever seen alternate characterizations of span in linear algebra, this is very similar

[radiant roost]

wouldn't ∩S also include elements like a^2+a^3?

[fleet moss]

the RHS should be the subgroup generated by that set

and the LHS should also be the subgroup generated by the intersection i think

which is the same set but without the multiplication structure

[solid iris]

oh typo

[grand pond]

𝓡𝓞𝓚𝓔𝓣𝓣𝓞─୨ৎ─❥ ♡ <𝟹❤

[radiant roost]

i wonder if i should interpret "the subgroup generated by {a^n | n in Z}" as the intersection of all subgroups containing that set

[fleet moss]

this is immediate if you have the fact that the intersection of subgroups is a subgroup

[solid iris]

yeah theres like two defs of generated subgroup

not sure which one ur allowed to use

one uses intersections and the other uses finite products

[radiant roost]

i think we proved it earlier

[radiant roost]

hmm

[solid iris]

ig i should ask which one youve seen

[radiant roost]

i posted the definition

from the book

[solid iris]

intersection?

[radiant roost]

immediately before that definition they stated as a theorem that the intersection of subgroups is a subgroup

after that definition they talk about finite products as a theorem

[solid iris]

ok then you have access to both defs

[fleet moss]

yeah so "subgroup generated by S" is axiomatically the smallest subgroup containing S. so if you have the fact that subgroup cap subgroup equals subgroup, its straightforward to show that your construction works.

[radiant roost]

i'm kinda lost though

[fleet moss]

try to imagine what $\langle{a^n\mid n\in\mathbb{Z}^+}\rangle$ actually looks like

[grand pond]

esca

[radiant roost]

it's finite products of powers of a

well

finite sums

[fleet moss]

why finite

[radiant roost]

that's what the theorem says

i'll post it

[fleet moss]

okay yeah thats right mb

[radiant roost]

oh actually the elements could have coefficients like 2a^2 + 5a^7

[fleet moss]

define $A^-\coloneq\langle {a^n\mid n\in\mathbb{Z}^+}\rangle$ and $A^+\coloneq\langle R_a,+\rangle$. we need to show $A^-\subseteq A^+$ and $A^+\subseteq A^-$. since $A^+$ is, by definition, the smallest subgroup containing $a$ and $A^-$ is a subgroup containing $a$, im pretty sure that gives us the second inclusion.

[grand pond]

esca

[radiant roost]

hmm Ra is the smallest ring containing a not the smallest group

[fleet moss]

ah yeah youre right mbmb

[radiant roost]

hmm

any element of A- looks like some finite sum like 2a^2 + 5a^7 + ... by theorem 7.6, and this element must be in A+ by closure in Ra

so A- is a subset of A+

i guess

[fleet moss]

i think so too

[radiant roost]

cool

idk how to do the other direction though

well what if we prove that A- forms a ring?

then A+ is a subset of A-

[fleet moss]

does Z+ include 0 here? or does the book define rings as not necessarily unital? otherwise theres the question of whether A- contains 1$

[radiant roost]

oh good point

it doesn't require unity for a ring

[fleet moss]

okay then yeah i think i agree with you, nice

[radiant roost]

thanks, i appreciate it 🙏

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[dark wharf]

How do I prove x=10 is the only solution here?

[midnight plank]

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[safe onyx]

How can I put this in desmos?

[safe onyx]

I was trying to use sigma notation, but it isn't working well

I got it now chatgpt

.close

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[inland patio]

Exercise: Suppose $\mathsf T:\mathsf R^2\to\mathsf R^2$ is linear, $\mathsf T(1,0)=(1,4)$ and $\mathsf T(1,1)=(2,5)$. What is $\mathsf T(2,3)$? Is $\mathsf T$ one-to-one? \

I wonder what I'm doing wrong in my solution. I have this theorem to my disposal that a linear transformation is determined by its action on basis vectors. Since $(1,0)$ and $(1,1)$ form a basis of $\mathsf R^2$, I'd simply like to conclude that $\mathsf T(a,b)=a(1,4)+b(2,5)$. I'm afraid this doesn't give the right linear transformation. What am I doing wrong?

[grand pond]

psie

[fallow scarab]

bT(1, 1) = T(b, b)

[inland patio]

true by linearity indeed

[fallow scarab]

you did T(a, b) = aT(1, 0) + bT(1, 1) = T(a, 0) + T(b, b) = T(a + b, b)

[inland patio]

ah ok, it's the second equality that is just incorrect, right? So we want T(a,b)=aT(1,0)+bT(0,1) instead 🙂

thanks

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[empty ivy]

I don't get part b, like i know it's a semicircle but how do i figure out the area between x=1 and 2?

[vast ginkgo]

have you learned trig sub

[empty ivy]

it's part b i don't get

not part a

[vast ginkgo]

oh

[grim vector]

Whats the area of a disk?

[empty ivy]

I don't know what you mean by a 'disk'. Like a circle?

oh, pi r^2

For the second bit though, how do I know what the angle is for the sector? Or is there another way to do this?

[vast ginkgo]

plug in x = 1 into the semi circle

(then you have opposite and adjacent)

oh sine works too

forgot had radius

yeah

easiest wouldve been cosine probably

1/2

well for cosine you pretty much have 1 already

[empty ivy]

Ohh i got it

[vast ginkgo]

sine and tan you need the y value

[empty ivy]

Thanks for your help!

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[molten bay]

For collinear i check that vector a=lambda vector b

a-b+c=x(4a-7b-c)

there is no such x so it is not collinear

For coplaner i checked cross product of two vector are not paralell with third vector

So they are on the same co plan

But the determinant of three vectors is non zero

[midnight plank]

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[misty wagon]

how would i do this in an exam condition?

i did try to just plug every inputfrom S into f(x), and i got {0,2,7,8} as solutions, but apparently thats wrong, so perhaps i miscalculated
but is there a better way than that?

[fleet moss]

set $f(x) = 7$ and $f(x) = 1$. the solutions here will form $f^{-1}(T)$

[grand pond]

esca

[carmine sigil]

This quadratic is irreducible over mod 11.

So there really is no better way than just plugging in all values and seeing which ones give you 1 or 7

[fleet moss]

oh i missed the mod 11 part

yeah

[misty wagon]

then i mustve miscalculated ig

cuz i did plug all the x in S to find when f(x) = 1,7

i did f(0) = 0 + 0 + 7 mod 11, so 0 is a solution

f(1) = 14 mod 11 = 3 mod 11, so 1 is not a solution

i did f(2) = 16 + 6 + 7 mod 11 = 7 mod all, so 2 is a solution

ig ill check again

[carmine sigil]

It might be useful to construct a table for the operation of x^2

Then for each member of the image of f(x) = x^2 construct a table for 4*x

Construct a table for 3x

And then use the tables to perform the computations

[misty wagon]

so i did miscalculate for f(5), whoops

[misty wagon]

do u think u could draw an example?

[carmine sigil]

Sure.

x | x^2
------
0 | 0
1 | 1
2 | 4
3 | 9
4 | 5
5 | 3
6 | 3
7 | 5
8 | 9
9 | 4
10 | 1

Notice that you only get 0, 1, 3, 4, 5, and 9

So when we make a table for 4x we only need entries for these values

x | 4x
-------
0 | 0
1 | 4
3 | 1
4 | 5
5 | 9
9 | 3

And so on

The idea is that you get all of the calculations out of the way while the patterns are fresh in your brain

Then we can chain together the tables. f(x) = 4x^2 if we want f(3) we look at the 3 row in the x^2 table to get 9, and the look at the 9 row in the 4x table to get 3. So 4(3)^2 = 3

@misty wagon ^ does the above help?

[misty wagon]

thats so interesting!

alr ty for that, i was jus trying it myself- take some getting used to but it seems fast

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[subtle peak]

Could somebody help me find my error?

[subtle peak]

I hate linear algebra

[drifting dust]

on the second line, that last term should be minus 8 * matrix

[subtle peak]

Thank you! I still get a wrong answer but that's closer

[feral sedge]

if i was asked to compute a 4x4 determinant on an assignment, i would hate linear algebra too

[subtle peak]

This teacher sucks

[drifting dust]

I'm getting the same number between that last line and the true determinant

[subtle peak]

Do you get 504?

[drifting dust]

yea

[subtle peak]

Bruh am I crazy?

[drifting dust]

wait what? 😭

[subtle peak]

Yeah, right?

I'm confudled

[drifting dust]

what program is that?

[subtle peak]

MATLAB

[drifting dust]

eek 😭

yea I dunno

but i'm getting 504

[subtle peak]

yeah, but it looks right. So I'm like so confused

Yeah, I get that when I ask for the determinant too

[drifting dust]

I mean with the last line

[drifting dust]

I think MATLAB is trolling you tbh

XD

try it in another calculator ig bc idk what's going on 😭

[subtle peak]

I can't find my TI

Sobbing

[drifting dust]

desmos or wolfram it is XD

[subtle peak]

I'm getting a friggin souped up Casio calculator one of these days

[drifting dust]

lolllll

[subtle peak]

I need it. My soul needs it xD

[drifting dust]

I still wonder what's going on 😭

[subtle peak]

Yep. Desmos is saying it's 504. I'm so confused

[drifting dust]

who knows

[subtle peak]

I'm asking my classmates. I need to be able to use MATLAB

It's the bread and butter of engineering at universities

Also wrong somehow

I thought you subtracted the ones on the right diagonal?

nvm my work is right, I put it into MATLAB wrong

The usual suspect

[drifting dust]

yea I don't think I learned Cramer's rule last semester so I can't help there 😭

[subtle peak]

I gotchu ❤️

Exhibit A

[drifting dust]

yea i dunno if I screwed myself since I took a proof-based lin. alg. class instead of escencially "lin. alg. for engineers"

who knows XD

[subtle peak]

My linear algebra is regular linear algebra with a teacher who thinks that her way of doing things is the diety of math itself

She docked me 70% of the credit for a question where I got the answer right but didn't put the math notation she said we should use once in class

[drifting dust]

umm...

WHAT

[subtle peak]

And i was like it is there, it's there IMPLICITLY and I'm going to literally try to get her fired

I KNOW

I got all the answers right on the Diff Eq exam, and got a 78.5% in the class

The highest friggin grade of the whole class

She's a pile of garbitch

[drifting dust]

😭

[subtle peak]

I'm literally traumatized

Plus I have circuits and systems 1, intro to physical electronics, engineering prob and stats (the writing intensive version), microcomputer labs, circuits ans systems lab, linear algebra

Like bro

I don't have time to be dealing with her crap

[drifting dust]

classic EE schedule

or CE, I can't tell

[subtle peak]

EE cause of the semiconductor/quantum mechanics crap they have in intro to physical electronics

Semiconductor/quantum mechanics/chemistry

Like... I'm doing it all

[drifting dust]

ahh

[subtle peak]

Yeah. It's waaayyy too much. And most of the time the thing holding me back is no more will to live and fight the good fight, you know what I mean

[drifting dust]

yeah 😭

[subtle peak]

yep. hbu? what's your major?

[drifting dust]

CS engineering

so it's like basic ||bitch|| engineering XD

[subtle peak]

oh, yeah. I started off in CS, and then I felt like no one was interested in giving me internships

So then I switched to CE, and was like, this is a subset of EE with 10k less a year and all that

[drifting dust]

yea

[lavish venture]

too many people watched instagram reels telling them theyd be the next mark zuckerberg/bill gates if they studied cs

[subtle peak]

I knew a CS guy that built an entire electric car for FSAE by himself and was project manager at his internship that he made, which altogether got him a job faster than a MechE with similar qualifications.

[subtle peak]

Naw I just saw that EVERYBDOY was going into CS

[lavish venture]

yea hella oversaturated because of this though

[drifting dust]

so tip #1 on how to be a good CS major, copy the ME majors XD

[subtle peak]

And everybody I know that's like a higher executive in CS as work aren't actual CS majors, so it felt like the degree itself was not helpful

[subtle peak]

I love MechE's

[subtle peak]

I got the wrong answer for this... I got 2852 for x_1

[lavish venture]

there’s been plenty of talk about how CS is the highest paying job or whatever and everyone with no passion for it whatsoever flocks to it

[drifting dust]

I just like to code 😭

[subtle peak]

Meh. Well like for EE, we start of at ~95k at a low-paying government job in my city, whereas this CS guy started off at 80k

And we max out at like 250k with Kratos and companies like that

Lockheed Martin is funny cause people think they pay well, but really they cap off at like 150-180k

My dad was trying to hire an EE from nowhere Tenessee for his consultant company, and he was demanding like 120k a year and work from home and a lot of benefits, and he was freshly graduated too

Like EE's are in a great spot rn with the 50 billion chip bill or whatever it was

I can't remember the amount

[subtle peak]

Nevermind, I'm a literal idiot

But it's cause I've been working on hw since 7 am this morning

[drifting dust]

omg 😭

[subtle peak]

Sorry, orange

[drifting dust]

why are you doing this to yourself

[subtle peak]

I wasn't trying to diss you

You'll do great

[drifting dust]

wait what? I don't see how XD

[subtle peak]

Because ... honestly no EE's know why we're doing this to ourselves

[drifting dust]

that's fair

[subtle peak]

Oh good. I love CS majors... except the way y'all think... I just can't communicate with y'all

I feel like I can understand MechE's ish, and CE's obviously is basically EE's with some sauce

I don't know...

I always have issues with talking to my CS friends at work

[drifting dust]

tbh I feel like I think more like a math major than a CS major lol

[drifting dust]

or it's the autism speaking over the CS XD

[subtle peak]

You mean my autism? I'm pretty sure EE's have the worst autism of any of the majors

but we're like... self-aware. Your brain has to change so much to be able to think in a way that will get you through school in EE

That's what i've noticed. i went into this degree being able to communicate D= :

[drifting dust]

I think y'all are just scared of us bc we understand recursion XD

idk I didn't have a response lol

[subtle peak]

That's funny. That's adorable

I understand recursion

[drifting dust]

XD

[subtle peak]

Want me to explain it to you?

You can test me

[drifting dust]

no lol

[subtle peak]

Sobbing. i wanted extra credit points

CS credit points xD

[drifting dust]

ok fine i'll test you

[subtle peak]

I was a CS major first

Back when I was like 16, but let's ignore that part

[drifting dust]

solve this

XD

[subtle peak]

HAHAHA that's not faaaiiirrrr

That's definitely an environmental issue

No I could solve that

It would be like a bunch of testing

[drifting dust]

hehe

[subtle peak]

I already have enough testing with my embedded systems... micro computer is death itself. My lab partner and I are the only ones who haven't used our extension

[drifting dust]

schrodinger's equation, it's both 504 and -12176 at the same time XD

[drifting dust]

🫡

[subtle peak]

Don't get me started with Schrodinger

[subtle peak]

Thank you for your support. It means so much to me

[drifting dust]

ofc 😊

[subtle peak]

Ugh you're adorable. can I adopt you as a bestie?

[drifting dust]

ofc, I can friend if u want XD

[subtle peak]

They pay us for our pain. Suffering. Misery. It's horrific

It's the worst major you could choose but has the best pay in exchange

[subtle peak]

Thank you, bestie!

❤️

[drifting dust]

ofc ❤️

fr sent :)

[subtle peak]

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[tacit kelp]

Can i get help with question 7

I know that c = 3

And i know a = second difference/2

And to find b you sub in information

[lyric charm]

mmm not quite.

that's true if the x values in your table have a step of 1

but here the step is 2

[tacit kelp]

What do you mean by that??>

[lyric charm]

look at the x values in your table

they are 0, 2, 4, ...

[tacit kelp]

Mhm

[lyric charm]

and not 0, 1, 2, ...

they go up by 2 and not by 1

[tacit kelp]

So if they go up by 1 then a = 2nd difference/2

But if they go up by 2+ then no

[lyric charm]

if they go up by anything other than 1, then no

[tacit kelp]

Alright

So how do i find a then

Would i sub in c and a random value??

Then solve simultaneously??

[lyric charm]

well let's try to think about how the step would affect things

[tacit kelp]

Hmm how??

[lyric charm]

best to make a table, hold on...

[tacit kelp]

Is this wrong??

[lyric charm]

patience lmao

let me draw up a thing here

[tacit kelp]

All good

[lyric charm]

this is how you find a when the table has a uniform x-step different from 1

[tacit kelp]

Can you explain please

[tacit kelp]

How would i solve this then question 7 then

[lyric charm]

time to number each and every line...

hold on

ok, here's my thing again

tell me the earliest line number where you get confused

well, that or we could do it the painful way

[tacit kelp]

All of it

[lyric charm]

even line 1?

[tacit kelp]

How does this relate to the table questions??

[tacit kelp]

Yes I haven’t learnt that

[lyric charm]

you haven't learned the equation y = ax^2 + bx + c for a parabola?

[tacit kelp]

Yes I’ve learnt line 1

Everything else i dont know

[lyric charm]

... then the correct answer to "give me the earliest line number where you got confused" would be 2

right

[tacit kelp]

Yes

[lyric charm]

so i calculate the first difference of y

for a step of 1

which means i take the value of y at x+1 minus the value of y at x

and simplify that

that's what lines 2-4 are

do you understand this bit

[tacit kelp]

I guess a little bit

But how does this correlate to the questions im doing??

[lyric charm]

patience!!! dear lord

if there was a direct correlation at this stage, do you think i would have written that entire sheet?

what i'm doing in lines 1-5 is rederiving the thing you already know

that when the $x$-step is 1, you can find $a$ via $a = \frac{\Delta^2 y}{2}$

[grand pond]

Ann

[lyric charm]

right here, you said it yourself

the first half is rederiving the "a = second diff / 2" thing you already know @tacit kelp

do you understand this yes/no

[tacit kelp]

I know, I appreciate you helping me and all, but i really want to get started on the question as i have lots of things to do, and i dont need to know about redriving the question

If that makes sense

[lyric charm]

what, do you want me to just hand me the general formula to get a as a blackbox?

if you're going to deliberately reject explanation like this, then ignore everything except the very last line

[tacit kelp]

Yes please

I really just want the question explained

That’s how I learn

[lyric charm]

in fact i will make it easier for you and just crop all the way down to only the final formula

[tacit kelp]

If that makes sense

[lyric charm]

well you won't learn where the formula i just sent comes from

and if you forget it, then you are cooked

[tacit kelp]

I dont need to know that

[lyric charm]

arguably, yes you do.

[tacit kelp]

I dont though

Can i just get help with question 7??

[lyric charm]

well, i've just given you the formula for how to get a for question 7. Δx is the x-step, which for question 7 will be 2.

you can also kind of use it in question 8 but you have to know where to look

[tacit kelp]

Thank you

[lyric charm]

im out otherwise tho

[tacit kelp]

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[twilit field]

Trying to find the tangent plane to this surface at (-1,2,4)

[twilit field]

The plane would be given by

$z= f(-1,2)+ \pdv{z}{x} (x+1)+ \pdv{z}{y} (y-2)$

[grand pond]

What a wonderful world !

[twilit field]

$z = 3 + 8(-1) (x+1) + (-2(2)+2)(y-2) = 3 -8(x+1)-2(y-2)$

[grand pond]

What a wonderful world !

[twilit field]

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[drifting root]

im confused on how to set up an equation for 3a

[midnight plank]

@drifting root Has your question been resolved?

[drifting root]

<@&286206848099549185>

[midnight plank]

@drifting root Has your question been resolved?

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[regal barn]

In triangle ( ABC ), ( a = 2b ) and ( |a - b| = \frac{\pi}{3} ), then angle ( c ) will be?

[grand pond]

Tom

[lyric charm]

not enough info

at best you can figure out that the lengths of sides a and b are 2pi/3 units and pi/3 units respectively, but we know nothing else

[last slate]

Did you try the cosine rule?

[regal barn]

yup

[last slate]

what did you get?

[lyric charm]

so it's the ANGLES at A and B that differ by pi/3.

your use of lowercase instead of uppercase was confusing and wrong!

[regal barn]

ohh my mistake sorry

|A-B|=pi/3 what does it help?

i meant how can I use it?

[lyric charm]

make a diagram first and foremost

at least mark the sides a and b and the angles A and B

[regal barn]

angle c

$$
\cos C = \frac{a^2 + b^2 - c^2}{2ab}
$$

$$
= \frac{5b^2 - c^2}{4b^2}
$$

[grand pond]

Tom

[lyric charm]

the 2a label should be 2b

[last slate]

consider two cases, case one where A- B = Pi/3

[lyric charm]

you know A is the greater angle actually

no need to consider two cases

[last slate]

Fair enough

[lyric charm]

A lies opposite the larger side

[last slate]

Since it’s modulus we must also conside B-A = pi/3

[regal barn]

in a and b angle a is big

so how does it help?

[last slate]

Is the answer pi/3?

[regal barn]

yeah

[last slate]

cool, then do this, make two cases where A-B = pi/3 and where B-A = pi/3

[regal barn]

sure and how will it solve?

[last slate]

since Ann already said, A is clearly greater, so we’ll consider A-B = pi/3

[regal barn]

my teacher said napier

[regal barn]

yes

[last slate]

A comes out pi/3+B

Now use the angle sum property

[regal barn]

ohh and b will be half of it

[last slate]

no unfortunately it doesn’t work like that

Using angle sum you get 2pi/3 - B = C

[regal barn]

2B+C=2pi/3

[last slate]

Yeah

Now use the sine law

2b/SinA = b/SinB

[regal barn]

sinA/a=sinB/b

sin(pi/3+B)/a=sinB/b

[last slate]

Yes, now convert a into b as given

[regal barn]

sin(B-pi/6+pi/2)/a=sinB/b

just like that?

[last slate]

no no

[regal barn]

then?

[last slate]

get it?

from that you can figure out B and then C

[regal barn]

nice

[last slate]

If your doubt is resolved, you may close the channel

[regal barn]

right

.cloose

.close

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[regal barn]

.reopen

[midnight plank]

✅

[regal barn]

could you help how can i solve for angle B

2sinB=sin(pi/3+B

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[lyric charm]

,rccw

[cinder rain]

For number 15 I found the total possible ways to get the sum of 12 is 165 but this includes the fact that the dice can become over 6

So it's 165 - (total outcomes where dice is consider a 7 and above)

And idk how to do that part

[pale tiger]

Since the sum is 12, there cannot be more than 2 dice with a value of 7 or above

therefore, the total number of invalid cases is 4 * (number of ways to assign 7 or above to 1 dice and 1 or above to the other 3)

Can you tell what the next step is?

[cinder rain]

Wait wdym 1 or above to the other 3?

[pale tiger]

well you can't roll a 0 with 6-sided dice

[cinder rain]

Oh that's what u mean 💀 I got confused as if there was like another meaning to jt

I mean honestly I was thinking about using the star and bars method for this cus a+z+b+y+c+x+d+q = 12 abcd being the minimum dice roll which is 1 and zyxq being extra so the max zyxq is 5

And so z+y+x+q = 8 after taking out abcd so I thought of like since max here is 5 or the dice roll would be 7 and above

Z = 5

So y+x+q = 3 and using the star and bar it becomes 5C2 which is uh 30

But then that's all I can think about

[cinder rain]

So uh is this ok

Or is this logic completely wrong

[pale tiger]

wouldn't it be z = 6 here, since you're looking for the cases where one dice is 7 or above?

the logic itself is fine, but you'd also have to consider z=7 and z=8(in other words, the cases where one dice 'rolls' a 8 and 9 respectively)

[cinder rain]

Oh then it would be like 4C3 = 4 then

Then uh if z = 7 then it's 3C3 which is 0

Then Z = 8 also 0

[pale tiger]

no, it would be 4C2 = 6 for z=6 (2 stars and 2 bars)

[cinder rain]

But then that means we only have to minus 4 tho but I asked my teach eariler about the awnser and i5 would be 125 so it would have to be minus 40 not 4 💀 so I'm js lost

Oh wait yea

4C2 but that's still js 12

[pale tiger]

in fact, if you go all the way, it's 6 for z=6, 3 for z=7 and 1 for z=8, which sums up to 10

so the total number of invalid cases is 4 times that, which is 40

[cinder rain]

Wait huh

Isn't it if Z = 6 it's 4C2

So uh 4x3x2 / 2x1 which is 12 so how did u get 6?

[pale tiger]

4C2 = 4! / (2! * (4-2)! ) = 24/(2*2) = 6

you might be confusing it with 4P2, which is indeed equal to 12

[cinder rain]

Wait huh wat

💀 wait I'm so confused now cus of this

[cinder rain]

So like if it's 2023C3 it's not 2023x2022x2021/3x2x1?

Like smthing like that

[pale tiger]

it is, that's the same as 2023! / (3! * (2023-3)!)

and 4C2 would just be 4*3/2*1, there's no reason to have an extra 2 in the numerator

[cinder rain]

💀 I'm so confused now lemme js ask chatgpt

Oh wait yea

I just did it wrong 💀

4C2 should be 4x3/2x1

Which is js 6

[pale tiger]

yep

[cinder rain]

Alright tysm for helping me

Bye

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[last slate]

what happened here

how did we go from that to that

[idle glen]

this is not even abuse of square root, first line is simply breaking rules of equality

[last slate]

No reason you would

maybe she meant square root of -9?

[molten pivot]

I think Yes.

[last slate]

alright thanks

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[molten bay]

Cosx cos6x=-1

[molten bay]

solution will be (2n+1)π and (2n-1)π are both same?

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[molten bay]

Cos7x+cos5x=-2

[molten bay]

so here when cosx belongs to the interval [-1,1]

Cos7x=-1

Cos5x=-1

(2nπ+-π)/7 and (2nπ+-π)/5

But given answer is (2n+1)π

[main current]

Tricky part is this is an AND condition, not the usual OR

x has to be of the form (2nπ + π)/7 AND (2mπ + π)/5, for some n and m.

This is hard to reason with, so I don't think it works as a final answer

[midnight plank]

@molten bay Has your question been resolved?

[solar moon]

your question have choices or not?

[molten bay]

No other choices

[pale tiger]

let x = y*pi/35.
if x is of the form (2npi+pi)/7 for some n, (y/35)*pi = ((2n+1)/7)*pi for some n, so y = 5*(2n+1) for some n; in particular, y is an integer and divisible by 5.
similarly, if x is of the form (2mpi+pi)/5 for some n, (y/35)*pi = ((2m+1)/5)*pi for some m, so y = 7*(2m+1) and y is divisible by 7.
therefore, y is divisible by 35, and we can now put y = 35z for another integer z.
finally, y = 5*(2n+1) implies that 35z=5*(2n+1), so 7z = 2n+1 for some n and therefore z is odd.
substituting y = 35z into x = ypi/35, we obtain x = zpi for odd z, which is of course equivalent to x = (2n+1)pi for integer n.

[pale tiger]

basically, this is the same condition as x being of the form (2npi+pi)/5 and (2npi+pi)/7, it's just hard to show

[midnight plank]

@molten bay Has your question been resolved?

[charred belfry]

Maybe?

For cos function to be -1 the value inside of cos has to be pi, 3pi, 5pi…., (2n-1)pi

Any negative number multiplied with negative number is still negative

I think

Yh

Idk bro

I think 5 and 7 are just there as indicator that as long as x is negative then the value infront of pi is also negative maybe

So maybe 5 and 7 could be disregarded when finding the general formula of x

Or whatever its called

Where n= 1,2,3,…,inf

[molten bay]

I see

Tq very much