help-39

[cursive wraith]

can you write

in one equality/inequality

what it means for x to be in V_delta(y)

[hoary nacelle]

If x in V(y) then it means x in (y-delta, y+delta)

Implying d=0

[cursive wraith]

nono

the distance between x and y doesn't have to be 0

otherwise that would mean x = y

[hoary nacelle]

X-epsilon /geq y-delta

X+epsilon >= y+delta

[cursive wraith]

mmmh no there's no need to use epsilon here

x in (y-delta, y+delta) is all you need

and maybe d = |x-y|

[hoary nacelle]

X>= y-delta

[cursive wraith]

you mean x > y-delta?

[hoary nacelle]

x<= y+delta

[hoary nacelle]

Yes

[cursive wraith]

ok

[cursive wraith]

can you rewrite both of those inequalities

into a single one

that uses d

[hoary nacelle]

X>y-epsilon +d
So x-y>d-epsilon

And reverse

x<y+delya

Which implies x-y< epsilon -d

[cursive wraith]

no need to use epsilon

where does it come from?

right you're using that delta < epsilon - d

but that's not what's we're after right now

[hoary nacelle]

X-y=d<delta

[cursive wraith]

|x-y| = d < delta

good

that's what I wanted

so if x in V_delta(y) means d < delta

x in V_delta(y) means d < delta

but we want x NOT in V_delta(y)

so how do we choose delta

[hoary nacelle]

So delta <d

[cursive wraith]

delta <= d is enough

so delta <= d, delta <= epsilon - d

[hoary nacelle]

Omg I am so happy now

[cursive wraith]

pick any value of delta that works

like the min

min(d,epsilon-d)

and we're done

[hoary nacelle]

You’re a god

Yes that’s the formulation

I must try a bit more with the method I will write a more formal version with this 🥰🥰🥰🥰 now

[hoary nacelle]

How can I learn more of these subtleties of math part

I am almost desperate

[cursive wraith]

Start studying a few math proofs here and there, keeping in mind that most proofs were built the same way:

• What do I need to prove: to get better at this part, always start your proofs by writing out exactly what you know at the beginning and what you want at the end, leaving some margin in between

So it would look something like this:
Question statement: prove that A is a subset of B.
Reformulation (to do yourself) : for all x in A, x is in B.
Proof :
Let x be in A. By the definition of A, that means...

[...]

Thus x is in B.
Thus A is a subset of B.

This is a really simple idea but it can be applied anywhere

[hoary nacelle]

Yes, I feel like my logic and formulation is not really good, and sometimes I can’t just do anything but to reapply or to redo puzzles which is simply waste time but I should revise the way I study

[cursive wraith]

Then:

• How do I prove it: this is where all the intuition part comes in, and you have to use what you already know, by following logical implications, doing some computations, applying known results/theorems, etc...

Intuition builds up over time, but in the meantime it can be helpful to ask yourself this type of questions:

• is there a theorem/property that I can use given my original data?
• I have to prove a "P implies Q" statement, do I use direct proof, contrapositive or proof be absurd?
• I have to prove "P or Q", do I prove it naturally, do I suppose not P and prove Q, etc...

[hoary nacelle]

I think I had become so desperate these days I almost never think these questions when I do puzzle I just squeeze my brain to write everything and it got messier and messier

[cursive wraith]

And as we did earlier, visualizing the problem can help, so drawings, drafts are recommended!

[hoary nacelle]

I have another irrelevant question

[hoary nacelle]

Maybe just because I felt so insecure

[cursive wraith]

Go for it

[hoary nacelle]

It is actually possible for someone transferring to mathematics be successful and is it possible to actually pass at least 4 finals in half a year plus 5 finals for my current study

4 finals from math

I felt quite addicted to this little thing and I really don’t feel like doing my current study

But I am already into second year

And I just feel super conflicting

Like I don’t know what to do, I am only good at calculus and maybe analysis considering I didn’t take course. But so many subjects I must catch up if I want some chance

[cursive wraith]

I'm not sure I can judge given the amount of finals, finals that last 4h are different than finals that last 2h for example

[hoary nacelle]

If I can actually get all the finals for math right then I can actually risk Econ finals but then I will have to risk my current study

I naively picked a course using Folland’s book and I just can’t drop the face to let my friend know I am not doing it anymore so I can do more math subjects finishing advanced LA or even some number theory

Despite I feel maybe I have long time to prepare for Folland’s but even I am gonna be luck that I passed it it’s just one course

And I feel like, with my current motivation

Going Econ is almost unrealistic

I almost only do different math stuff

And because of these, I almost have no patience for whatsoever

.close

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[toxic lichen]

<@&268886789983436800> scam

[pearl pond]

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[midnight haven]

hi

[midnight haven]

$$\abs{x} + \abs{y} < r$$

how can i figure out that this is a graph of a tilted square?

i dont know where to start

[jolly parrot]

CUPS

[vestal tapir]

idk you can't

it's manhattan metric circle

so it would be symmetrical around origin

but i don't see how you can just know it's a square

[midnight haven]

oh

[rustic gate]

??? what do you mean you can't

just check each of the quadrants

[midnight haven]

what

[rustic gate]

split into cases based on whether x and y are positive or negative

[midnight haven]

ohhh

thats the step i was looking for

thanks

[vestal tapir]

you can't tell in a roundabout way

[rustic gate]

i don't know what you're on because you can quite easily figure out the shape with some basic manipulation

[vestal tapir]

but not without

[rustic gate]

yeah you can't do anything without doing anything

very helpful

[midnight haven]

lol

anyway, thank you both 😎

!close

.close

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[royal mantle]

evaluate ∫2 to infinity dx/(x^2-1)^(3/2)

[flint zenith]

What have u tried

[royal mantle]

I'm unsure where to start but I was going to use trig sub and then I got lost

x = sec theta and dx = sec theta tan theta

[pearl pond]

@royal mantle Has your question been resolved?

[royal mantle]

<@&286206848099549185>

where should I go from here

[proud apex]

You can group the cos/sin as cot and the 1/sin as csc so it'll be integral of cscx cotx

[pearl pond]

@royal mantle Has your question been resolved?

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[hybrid trail]

,calc 2 - (log(65.1))

[jolly parrot]

Result:

-2.1759245492145

[hybrid trail]

why is this not working?

[toxic lichen]

what are you trying to calculate?

[toxic lichen]

for this bot, you need to use log10() for decimal log.

,calc 2 - log10(65.1)

[jolly parrot]

Result:

0.18641901143181

[hybrid trail]

oh, I see

thanks so much!

.close

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[lunar schooner]

okay here is where i am and im wondering if i can take the 7x outside of the limit

[limber pivot]

you can take |7x| outside the limit

[lunar schooner]

sick then the limit is 1 right

okay last question now i need to plug in -1/7 and 1/7 to see if they converge or not

[pearl pond]

@lunar schooner Has your question been resolved?

[lunar schooner]

why is this not divergent? the answer was [-1/7,1/7] but i got {-1/7,1/7)

[humble lintel]

You always have to check the end points separately

That is where the test is inconclusive

[merry carbon]

(and how did you check the endpoints? Assunedly you did, re here)

[merry carbon]

Is this one yours?

[lunar schooner]

i plugged in -1/7 and 1/7 for x in the original series then solved

[lunar schooner]

ya this was for when x=1/7

[merry carbon]

What does your version of the "p" test state?

[lunar schooner]

shit i just read the inequalities wrong again

[merry carbon]

It happens

[lunar schooner]

ya thanks!

[lunar schooner]

can i ask for help on another question?

[merry carbon]

Of course

[lunar schooner]

am i doing this right?

i think that the denominator goes away but also we haven had a question like that before so its making me feel like i did something wrong

wait i copied it wrong

its 9^n not 9n

[merry carbon]

happens sometimes, do you wanna try it again and send it?

[lunar schooner]

it would be like this right

but i dont know what to do next actually

ususally i would take the x terms outside of the limit

but how would i do that without needeing to multiply the numerator and getting a 8nx term?

becasue i have no idea what to do with that

[merry carbon]

What do you mean "multiplying the numerator"?

[merry carbon]

(in this case, you actually can take terms out of the limit!)

[humble lintel]

They want to FOIL the numerator and factor out only x instead of the entire term (8x-9).

[lunar schooner]

oh i can take out the whole term?

i didnt think that was legal but that makes it much easier

so i get this and the limit is infinity right

okay sick

[lunar schooner]

okay the next question i also had something weird happen

what do i do when all the n terms cancle in the limit

[merry carbon]

You have a limit of a constant, which is just the same constant

[lunar schooner]

so its 1?

[merry carbon]

Well, be a bit careful with how you mean, $\lim_{n\to\infty} 1$ is 1 sure

[jolly parrot]

@merry carbon

[lunar schooner]

what do i do here when the limit approaches 0

[merry carbon]

Well, in that case, remember that you're trying to find the values of x such that the corresponding limit L is less than 1, and 0 is always less than 1

You converge for all real numbers in that case

(and finding what it actually converges to is not insanely difficult either!)

[lunar schooner]

how would i do that?

do i need to do it for the interval of convergence?

[merry carbon]

Well, you don't need to find what it converges to at any point, that was more of an interesting comment I thought to make

If you did, out of curiosity, want to find what it actually works out as, then if you know how e^x is defined, it's just a matter of manipulating what you have

(if I can algebra, it's e^{x - 1/9} - 1)

[merry carbon]

And also, as per before, as you get the limit being zero and know that it converges for all real numbers, you're pretty much already done, and don't need to test endpoints (and can't really even, the infinities aren't real numbers )

[lunar schooner]

it says that one is wrong but we just did all of them 😭

[merry carbon]

One of them is wrong

[merry carbon]

[lunar schooner]

what how?

[merry carbon]

For this one, if 8x - 9 is not 0 (so x is not 9/8), you found the limit to work out as infinity, and so there would not be any convergence

[lunar schooner]

but if the limit is infinity wouldnt the IoC be (-inf, inf)

[merry carbon]

Not quite

In fact it's more the other way around-

[merry carbon]

When the limit is 0, you get the interval of convergence as (-infty, infty)

[lunar schooner]

but theres no option for no convergence

[merry carbon]

If instead you get the limit as infinity, the only place you can get convergence is at "the single point" (where you'd have a series of just 0)

[merry carbon]

"if 8x - 9 is not 0 (so x is not 9/8)"

[lunar schooner]

okay

[merry carbon]

Happier now with which one it should be, and why?

[lunar schooner]

it should be d then

[merry carbon]

Also wait a moment too

[merry carbon]

Double check that (3) as well

[lunar schooner]

ya i def did something wrong there too

this was my work

[merry carbon]

Aha, I also wanted to make the comment before (but didn't get to), Be very careful here

What you ideally could do is to notice that this limit is "a limit of a constant", so is just |(x - 9)/9|

From which point, you want |(x - 9)/9| < 1, which I'm sure you can cook with

(in fact, you could even, in this case, get your answer "instantly" by noticing you basically have a geometric series, for which you may know exactly where they converge and diverge)

[lunar schooner]

wait where is the geometric series?

[merry carbon]

You effectively have a common ratio of (x - 9)/9, if you notice it...

[lunar schooner]

i fear i do not see how that is a common ratio...

[merry carbon]

What do you need to multiply $\qty(\frac{x - 9}9)^n$ by to get $\qty(\frac{x - 9}9)^{n + 1}$?

[jolly parrot]

@merry carbon

[lunar schooner]

you would need to multiply another (x-9)/9

[merry carbon]

Yep, do you see how that makes (x - 9) / 9 the common ratio of the series? It's what you multiply to get from one term in the series to the next one

[lunar schooner]

okay ya but i think that im still doing something wrong

[merry carbon]

The first one is correct, and you can simplify that and immediately see why you don't get convergence

[lunar schooner]

is it just 1?

[merry carbon]

The term you're taking the series of does simplify down to 1, yep

[merry carbon]

The second one you're doing is not correct, remember that you have absolute convergence when $\frac{\abs{x - 9}}9 < 1$, and make sure you rearrange that properly to realise the actual other endpoint you should be testing

[jolly parrot]

@merry carbon

[lunar schooner]

do u not multiply by 9 and then add 9?

[merry carbon]

Multiply by 9, sure, you get |x - 9| < 9 in that case

[lunar schooner]

and then i added 9 to get x by itself

[merry carbon]

Well, be very careful of how you mean, you first should really change what you have to something which doesn't have absolute values in it

[lunar schooner]

so would i need to change it to x+9?

[merry carbon]

Not like that

You do know how you're meant to change these, right?

[lunar schooner]

ummmm

[merry carbon]

I'll take that as a no naughty

Anyways, you may have seen it before, but saying that |a| < b is equivalent to -b < a < b

[lunar schooner]

yes i do know this part

[merry carbon]

That's what I wanted you to do here get this into -9 < x - 9 < 9 first

[lunar schooner]

ohhhh

[merry carbon]

Then you can happily add 9 to both sides

[lunar schooner]

okay

so its between 0 and 18

[merry carbon]

Correct, it's supposed to be x = 0 you test for that second one

[lunar schooner]

so then we get (-1)^n

[merry carbon]

We do, does that converge?

[lunar schooner]

no both endponts diverge

[merry carbon]

Correct

(you could have of course been naughty and cheated by noticing what option C is, at which point you're happy )

[lunar schooner]

ya i did do that but i still need the practice of actually doing them 😅

[merry carbon]

I know, that's why I'm only saying it now, I wanted to be mean and make you actually do them and justify fully everything you do

[lunar schooner]

hahaha thx

do i need to figure out how to rewrite the bottom factorial in order to solve this? becasue this shit looks scary

[merry carbon]

It would help you quite a bit if you did, any ideas of how you could?

[lunar schooner]

ummm no this is hard to think about

why is it not just 4n! ?

[merry carbon]

Because 4n is the last term, you have a bunch of other stuff you're multiplying in

You have 4, 8, 12, ..., 4n that you have

Or, slightly differently, 4 * 1, 4 * 2, 4 * 3, ..., 4 * n

[lunar schooner]

but its not 4n!

[merry carbon]

It isn't 4n!, you're right (just seen your edit now as well, cheeky )

I'm sure it comes as no surprise to you that 1 * 2 * 3 * ... * n is n!, right?

[lunar schooner]

ya

[lunar schooner]

wait it is or isn't

[merry carbon]

You are right that it is not sorry

Alright, new question for you, what's 4 * 4 * 4 * ... * 4 equal to? (there being n appearances of 4 there)

[lunar schooner]

infinity?

[merry carbon]

Not quite, there aren't infinite 4's there

There are only n 4's that you have?

How about a slightly different question: what's 4 * 4? How about 4 * 4 * 4? Can you rewrite those?

[lunar schooner]

on is 4^2 4^3 and 4^n?

[merry carbon]

You got them all well done, you have 4^n you're working with

So you have that the product is effectively (1 * 2 * 3 * ... * n) * (4 * 4 * 4 * ... * 4), as you noted, the first is n!, the second is 4^n, I'm sure you can cook from there

[lunar schooner]

(4n!)^n?

or is it 4^n * n!

[merry carbon]

This one

[lunar schooner]

so i wrote it like this

[merry carbon]

Good, you can keep going

[lunar schooner]

sorry someone came into my work but im still working on it

[merry carbon]

Sure, just let me know when you're done

[lunar schooner]

this is where im at an the limit just goes to 1 right

[merry carbon]

The limit isn't 1, be careful (and also don't forget the absolute values )

[lunar schooner]

oops i added the absolute value signs but is it infinity?

[merry carbon]

It isn't

[lunar schooner]

what?!😩

[merry carbon]

Hint: you may wanna divide the numerator and the denominator of what you're taking the limit of by n

[lunar schooner]

omg is it 2/4?

[versed mica]

the limit?

[lunar schooner]

ya

[versed mica]

(2n+1)/(4n+4) —> 2/4 yes

notice that it’s just the ratio of the coefficients

for a rational function where the degree of the numerator is the same as the degree of the denominator you simply divide the coefficients of the highest degree terms

[lunar schooner]

i always forget that i can do this

[versed mica]

convince yourself of why this is true

i’ve showed you dividing by the highest degree before yes

[lunar schooner]

no it makes sense

im stuck here

[versed mica]

🤔

mixed fractions

[lunar schooner]

should i use a decimal?

[versed mica]

i’m not sure i follow what this is?

[versed mica]

this is the question yea

[lunar schooner]

ya

[versed mica]

what is this

oh i see what you did

i was like huh this converges everywhere? but yea i see what you did

ok so

$\frac{(n+1)^2(x-10)^{n+1}}{4^{n+1}(n+1)!} \cdot \frac{4^n n!}{n^2 (x-10)^n}$

[jolly parrot]

knief

[versed mica]

then this becomes

[lunar schooner]

im trying to plug in the end points of the interval

[versed mica]

$\frac{n+1}{4n^2} \cdot |x - 10|$

[jolly parrot]

knief

[versed mica]

which goes to zero

so the radius is infinite

[lunar schooner]

wait that does not look like what i did

[versed mica]

your mistake here was that (n+1)^2/n^2 ≠ 2n + 1

[versed mica]

because it isn’t

i’m assuming you did (n+1)^2 = n^2 + 2n + 1

then thought you could cancel the n^2

but no

you can not

[versed mica]

do you see how i got to this step

[versed mica]

from this

[lunar schooner]

oh

no

[versed mica]

you don’t see it?

ok i’ll write it out

[lunar schooner]

no i dont know how you did it

[versed mica]

you get this though right

[flint zenith]

Wtf

[versed mica]

<@&268886789983436800> get your free money

[flint zenith]

I’ve never seen a scambot in this server

[versed mica]

you must be new

[flint zenith]

😦

[versed mica]

@lunar schooner

[versed mica]

you get this?

[lunar schooner]

ya!

[versed mica]

ok

[opal lantern]

sigh

[versed mica]

lmao

more free money

[lunar schooner]

okay actually im about to get off work at 2 so if we do t finish im going to have tp try again when i get back home

[versed mica]

\begin{align*}
\frac{(n+1)^2(x-10)^{n+1}}{4^{n+1}(n+1)!} \cdot \frac{4^n n!}{n^2 (x-10)^n} &= \frac{\cancel{(n+1)}(n+1)\cancel{(x-10)^n}(x-10)}{4 \cdot \cancel{4^n (n+1)n!}} \cdot \frac{\cancel{4^n n!}}{n^2\cancel{(x-10)^n}} \
&= \frac{n+1}{4n^2} \cdot (x-10)
\end{align*}

[jolly parrot]

knief

[versed mica]

can you see it now

ugh

[lunar schooner]

okay ya and then u take the limit of (n+1)/(4n^2)

[versed mica]

which is

[lunar schooner]

0

[versed mica]

which means what for the radius

remember L = 1/R

[lunar schooner]

i thought that L was this 0<L<1

[versed mica]

well

we have 0 * |x -10| and we want this to be less than 1 right

but it’s always less than 1

regardless of what x is

it’s zero

so the radius is..

[lunar schooner]

is the radius 0 then?

[versed mica]

radius zero means it converges now where except the center

this isn’t true

[versed mica]

that’s why i said rhis

[versed mica]

^

L = 0 = 1/R

what’s R

[lunar schooner]

infinity?

[versed mica]

btw

when you see a factorial in the bottom

it’s like a dead giveaway

[lunar schooner]

its a dead giveaway that it will be infinity?

[versed mica]

that it will converge everywhere

[lunar schooner]

huh okay

[versed mica]

this isn’t always true of course

but

usually a good insight to have

[lunar schooner]

ya ill have to def do the work on the exam anyway

[versed mica]

it outgrows the exponentials a^n and the powers of n

n^k

[versed mica]

mhm

it’s usually because you’ll have a residual factor of (n+1) from the ratio test whereas the power terms like n^2 or n^k for some k will have limit 1 and the exponentials like 4^n or more generally b^n will just have a factor of b

like here we had a factor of 4

but that doesn’t do anything for the limit

[versed mica]

this n+1 in the denominator however does change things

and it usually means the limit goes to zero

does this make sense

[lunar schooner]

ya i think so

thanks

[versed mica]

you’re welcome

[lunar schooner]

.close

[pearl pond]

Channel closed

Closed by @lunar schooner

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[brave sluice]

for 13 can i assume that "a" is nonzero?

[cursive wraith]

0 is a divisor of 0

[brave sluice]

oh

my book says if a and b are two nonzero elements of a ring R such that ab=0, then a and b are divisors of 0

but wikipedia agrees with what you said

thanks 🙏

.solved

[pearl pond]

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[vagrant ingot]

Could someone please help me with the problem in the attached image?

[sacred fern]

i think you need to design a finite automaton or a grammar in JFLAP format that accepts the valid strings? im not sure sry

[pearl pond]

@vagrant ingot Has your question been resolved?

[vagrant ingot]

yes

[sacred fern]

okay 👍

[pearl pond]

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[final zinc]

how to get help channel TWT

[final zinc]

oh

i see

i was wondering how would u get the base and height of a triangle with only the area given

[pearl pond]

@final zinc Has your question been resolved?

[final zinc]

norpe

[ember patio]

use ratio

first consider BOC and ABD

since $h_{BOD} = h_{DOC}$

[jolly parrot]

C

[ember patio]

This mean BD : DC = 2 : 6
Same as DO : OA = 2 : 3

use any of this

To figure out |ADC|

[final zinc]

im not sure i get it

how do i use the ratio in finding out the missing area?

[ember patio]

do you know that if there height is equal
It’s mean ratio of area = ratio of Base right?

[final zinc]

no, i did not know that TWT

[ember patio]

If you have 2 triangle that there height is equal ( h1 = h2 )
Assume
base triangle 1 = a and
base triangle 2 = b
Now consider ratio of area of triangle 1 : 2
ah/2 : bh/2 = a : b

now back to your question

[final zinc]

is it possible to get the base and height of the triangle with just the area given?

[ember patio]

The question want you to find area not the height

[final zinc]

yes, i was thinking of finding the base and height of AOB, OBD, and ODC

and then yknow the 1/2 * b * h

will that work?

[ember patio]

What’s your ‘b’ here mean

[final zinc]

base

[final zinc]

would this be possible?

[ember patio]

It doesn’t matter what there base height are, we use only ratio to figure out this question

[final zinc]

oh

i dont know how to deal with ratios TWT

[ember patio]

It’s mean
2y : 3y = 6 : |AOC|
2/3 = 6/|AOC|

[pearl pond]

@final zinc Has your question been resolved?

[pearl pond]

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[torpid mountain]

Can anyone help me please

[torpid mountain]

I thought of like different sizes

Like 15x25x35

And then find somehow how to arrange it

I’m a bit stuck on how to rearrange

[toxic lichen]

well let's see

let's say the target dimensions for the brick are x, y and z

[torpid mountain]

Yeah

[toxic lichen]

then the current ones are x+5, y+5 and z+5

but we also know we can cut one of those in half to get an x by y by z brick

we can also assume x <= y <= z here

[torpid mountain]

Ok

[pearl pond]

@torpid mountain Has your question been resolved?

[toxic lichen]

if you assume the longest dimension is the one to be cut in half you can get it almost immediately

[pearl pond]

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[hardy totem]

Hello, I would like help with this question. My main point of confusion is the definition. It says that the l(n) function is a linear function such that l(n) = p(n), however p(n) is not a linear function? How are you supposed to sketch this graph?

[toxic lichen]

$\ell_n$ is a linear function whose graph passes thru the points $(n,P(n))$ and $(n+1, P(n+1))$

[jolly parrot]

Ann

[hardy totem]

im new to this server, what is that supposed to mean?

[toxic lichen]

we have a bot that can render LaTeX code from discord messages

handy for writing things down with proper math notation

[toxic lichen]

look at the image that the bot sent here in response to my message

[hardy totem]

oh interesting, I have never worked with LaTeX code.

[pearl pond]

@hardy totem Has your question been resolved?

[pearl pond]

@hardy totem Has your question been resolved?

[pearl pond]

@hardy totem Has your question been resolved?

[pearl pond]

@hardy totem Has your question been resolved?

[pearl pond]

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[earnest mortar]

Can anyone see why my method is not correct?

I understand that they use the moment around the x axis in conjunction with the FR x the y distance

But I do not see why my method wouldnt give the correct answer

MEMO ANSWER ( x; y) = (3; 3)

[pearl pond]

@earnest mortar Has your question been resolved?

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@earnest mortar Has your question been resolved?

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[low slate]

Find summation of series who General term is this ( use telescoping)

[low slate]

.close

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[tardy anvil]

Im a little confused because I know order matters

[tardy anvil]

So would I say like AB = AB, then multiply by the inverse of B?

[toxic lichen]

AB = AB

you can find A by multiplying AB with B^-1 yes but you have to be careful with the order.

[tardy anvil]

well yea just as a starting point 😭 idk how else I would do it

so would it be A = B^-1 (AB)

[ivory swallow]

remember that matrix multiplication is associative

[toxic lichen]

nope

[tardy anvil]

oh so B^-1(AB) is the same as (AB) B^-1 ?

[toxic lichen]

no, you're mixing up associative w commutative

commutative would mean you can switch shit around
associative means you can rebracket

A = ABB^-1 is correct

[tardy anvil]

ok im just a lil confused because I was doing a similar problem and was told like when Im multiply something by an inverse, it always goes to the front

[ivory swallow]

not always. you want the inverse of B to be next to B so they cancel out

[tardy anvil]

oh alright I think then it was just moved to the front because for that question I was taking the inverse of A

so it would be A = ABB^-1 makes sense

.close

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[dusky willow]

I don’t know what to do, since after the transformation it’s not gonna be a circle any more idk how to like write the equation for that image

[ivory swallow]

expand

[dusky willow]

Oh does it work sir

[ivory swallow]

should be an ellipse

[dusky willow]

Like the shape after transform isn’t circle thou

[dusky willow]

Idk the equation for ellipse and I assume it won’t be like = radius^2 thou Isk

But I will try

It is different from the answer

The answer is 20x^2 bla bla bla

[pearl pond]

@dusky willow Has your question been resolved?

[dusky willow]

Or is this unsolvable

[feral sedge]

If I have a point (x', y') which satisfies the original equation

And I want to make a new equation which is satisfied by the point (2x'+3y', 2x'+4y')

The way to do that isn't to replace every x with 2x+3y and every y with 2x+4y

just like how if I want to shift a graph to the right by 3 I don't replace every x with x+3

[dusky willow]

True . So how do u think I should solve this question?

[feral sedge]

Usually you want to apply the opposite of whatever transformation you want to your variables

[dusky willow]

(Btw by what u said do u mean like I am changing every point in the circle to the location after transformed?)

[feral sedge]

Yes

Probably

[dusky willow]

Oh

Nah

I was like

Checking the radius of the thingie after transformed

[dusky willow]

That’s sounds like a very good tactic

But idk what shape it will transform to

So idk what equation to use ig

[dusky willow]

However I assume it’s gonna become a ellipse

But ngl I have no idea how to apply this equation lol

[nimble socket]

yes since it is not equal transformation

i will try to help

one second

[dusky willow]

Thanks !

[nimble socket]

ah you know what might help

parametrising x^2+y^2=4

[dusky willow]

(Wait let me get out of toilet)

[nimble socket]

skibidi toilet?

[dusky willow]

Real toilet

[dusky willow]

What’s parametrising

[nimble socket]

oh

ok

😦

[dusky willow]

Sorry

[nimble socket]

no nono its fine

[dusky willow]

thanks

[nimble socket]

ah wait

[nimble socket]

like here

what you want to do is "opposite" it

so try to "opposite" the matrix you have

[dusky willow]

Okay

[dusky willow]

With this formulae ? 😭

Oki

I get how the ellipse formulae works now! Let me try

Nuh uh

[dusky willow]

For the ellipse formulae does b^2 = -a^2?

Cuz it’s just a reflect

[nimble socket]

no

rather

nevermind this will probably overcomplicate it

[feral sedge]

well you have some matrix transformation

and you want to do the opposite of it

do you know of anything which might work like a matrix transofrmation in reverse?

[dusky willow]

Oh

Inverse it?

[feral sedge]

yes

[dusky willow]

Ohhh

Let me try

[dusky willow]

I am so cooked 😭

What is that 😭

[feral sedge]

that

is not how you multiply matrices

[dusky willow]

uh

I am cooked

[pearl pond]

@dusky willow Has your question been resolved?

[pearl pond]

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[dusky willow]

Not sure how my answer is wrong

[pearl pond]

@dusky willow Has your question been resolved?

[dusky willow]

<@&286206848099549185> sorry for ping but I am confused

[dense jasper]

]

,w -6+5m=-m^2

[jolly parrot]

Results provided by WolframAlpha

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Requested by civil_service_pigeon

[dusky willow]

lmfao silly me

Thank you so much

Have a lovely day or night

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[dusky willow]

.close

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[midnight haven]

In Newtonian mechanics, is it possible to axiomatize the concept of energy as an arbitrary scalar quantity X that remains constant over time and satisfies certain conservation laws (like dX/dt = 0 when forces are conservative and ΔX = ∫Fnon-conservative dr)? Imo, defining KE as 1/2 mv^2 seems somewhat that is justified by trial and error or as an empirical fact. What I would like to do is find that any candidate X that matches the idea of what energy should be must take the form X = 1/2 mv^2 + U(r) or something similar, not the other way around (so not showing that this definition for energy magically satisfies those conservation laws)

[pearl pond]

@midnight haven Has your question been resolved?

[sharp vigil]

you can mathematically prove that:

• the line integral of the net force acting on a particle is given by 1/2 m(v_2^2 - v_1^2)
• since line integrals are additive, this work integral can be split into the integral of each force separately
• by definition, the conservative forces are the ones whose line integrals are force-independent
• a theorem in vector calculus says that the line integral of a conservative vector field is given by the difference in a potential function

so from the math we have the work energy theorem:
[ W \coloneq \int \vb F \cdot \odif{\vb r} = \frac 12 m \Delta(v^2) ]
additivity:
[ W = \int \vb F \cdot \odif{\vb r} = \int (\vb F_c + \vb F_{nc}) \cdot \odif{\vb r} = W_c + W_{nc} ]
For conservative forces $F_c = \grad \phi$ and $W_c = \Delta \phi$. summing up we have
[ W_{nc} + \Delta \phi = \frac 12 m \Delta(v^2) ]
then it seems like a natural choice to define $U \coloneq - \phi$ and [ E \coloneq \frac 12 m v^2 + U ]

[jolly parrot]

cloud

[pearl pond]

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[hazy ember]

$\cot\theta=\frac{\cos\theta}{\sin\theta}$

[jolly parrot]

reverileo

[hazy ember]

when does $\cos\theta=0$?

[jolly parrot]

reverileo

[hazy ember]

this is better

[hazy ember]

think about it this way

you got it

$\arccot(0)$ is essentially asking for which $\theta$ does $\cot\theta=0$

[jolly parrot]

reverileo

[hazy ember]

sin you don't care because if the numerator is zero, then the whole thing is zero

well except when the demoninator equals 0 too, but clearly sin(90) isn't zero

$\cot 90^\circ=\frac{\cos 90^\circ}{\sin 90^\circ}$

[jolly parrot]

reverileo

[hazy ember]

no

in fact there's no $\theta$ such that $\sin\theta=\cos\theta=0$ simultaneously

[jolly parrot]

reverileo

[hazy ember]

like arctan(0)?

$\tan\theta=\frac{\sin\theta}{\cos\theta}$, so you would find $\theta$ such that $\sin\theta=0$

[jolly parrot]

reverileo

[hazy ember]

on a calculator you have a button for inverse trig functions

oh are you trying to find $\arctan\infty$?

[jolly parrot]

reverileo

[small mango]

its n

its n over 2

[limber pivot]

You can't divide by zero

[small mango]

i meant n over 2

mb

idk i solved it

its n over 2

[hazy ember]

you should clarify that n represents pi

[small mango]

ok

is that it

naw

[limber pivot]

I mean what model is your calc

[small mango]

i dont feel like typing it

ngl

[limber pivot]

If you want to know it on your calc only

[hazy ember]

for which $\theta$ does $\tan\theta$ become undefined?

[jolly parrot]

reverileo

[hazy ember]

can you answer this question?

can you find $\theta$

[jolly parrot]

reverileo

[small mango]

yes maybe

it will take a second

[hazy ember]

yes

that's it

yes

exactly

so $\arctan(1/0)=90^\circ$

[jolly parrot]

reverileo

[small mango]

i know i deleted my 90 degrees

i did not sa

[hazy ember]

it's improper notation but

[small mango]

say

im pretty sure

it is 0,1

[pearl pond]

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[prime bane]

I dont get how 3(ii) was done in the solution

[nimble socket]

let me explain

let me cook

🔥

[prime bane]

Sure

[nimble socket]

so $\qty(\frac{3}{x^2}+x)^8=\sum_{k=0}^8\binom{8}{k}x^{n-k}\qty(\frac{3}{x^2})^k$ agree?

[jolly parrot]

vengeance

[prime bane]

What the flip is the e

I never learn that E symbol

But thr rest i agree

[nimble socket]

just sigma notation

anyways

[prime bane]

so $\qty(\frac{3}{x^2}+x)^8=\binom{8}{k}x^{n-k}\qty(\frac{3}{x^2})^k$ agree to this

[nimble socket]

not correct but ok

so we want when $x^{8-k}\cdot\frac{1}{x^{2k}}=1$

[jolly parrot]

vengeance

[nimble socket]

Or in other words, when power=0

[prime bane]

Hmm

I dont understand

Why = 1 instead of 0

They asked independent of x

Which means there should be no x term

In other words x^0

[prime bane]

ya

[nimble socket]

eh abuse of notation

[prime bane]

but that statement does not mean power=0right?

[prime bane]

wdym

[nimble socket]

well actually now i am sus

accidentally gaslit myself

[prime bane]

THINK HARDER BUDDY

LOCK IN

[nimble socket]

ah i get it

sorry

we only dealth with the first one

we need to consider the whole, (3/x^2+x)^8(5-2x)

so if we add that in, where do you think the coeff of x will be independent

[prime bane]

whats this now

[nimble socket]

wait a minute

bro you did part 1 alr

?!

[prime bane]

yes

[nimble socket]

dead gievaway

[prime bane]

i dont see the thing

[prime bane]

HELP ME SEE THE THING

[nimble socket]

ok so

[prime bane]

i just dont understand why they do =-1 instead of 0

[nimble socket]

we have an extra x term

[prime bane]

yes

[nimble socket]

now if we nicely distribute we get a nice form

[prime bane]

its not so nice on my end

so messy distribution

[nimble socket]

namely $x^{3r-16}\cdot x=x^{3r-16+1}=x^{3r-15}$

[jolly parrot]

vengeance

[prime bane]

its 2x not x.....

[prime bane]

also you need to multiple with the other terms

[nimble socket]

i know but I am showing you how it would be multiplied with x

therefore r=5 makes it independent

[prime bane]

ya i know indeces law

[prime bane]

huh

[nimble socket]

3(5)-15=0

[prime bane]

oh

i see

wait why did you put-15

i could just say 3x-18

and just make x=6

[prime bane]

i have a feeling your just as lost as me😭😭

[nimble socket]

no i get it now

[nimble socket]

-16+1=-15

[prime bane]

why you put+1 in the first place

thats what i dont get

if i know that i know the rest

[nimble socket]

because indice law!

x^(a+b)=x^a*x^b

[prime bane]

this thing

why did you multiple by x

instead of 2x

!

[nimble socket]

coeff of x doesn't matter here

all we need to consider is x when getting r

[prime bane]

oh

oh

[nimble socket]

then we take coeffs and all that

once we find r value

[prime bane]

oh

then i get it

[nimble socket]

yeah

[prime bane]

tysm

[nimble socket]

see ezpz

[prime bane]

.close

[pearl pond]

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[prime bramble]

I need some help with this problem

[prime bramble]

here is what I've written so far

my idea was to compute \omega \wedge d\eta and d\omega \wedge \eta, compare the two, and hopefully get to a point where they're equal, up to a constant multiple

that constant would be c, and I'd be done

[prime bramble]

but uhh, this wedge product between those two sums is really ugly and I'm not sure if I either

a) have the wrong approach and need to rethink this; or
b) need to just bite the bullet and work through the wedge

in the case of b), I can't figure out how to do it

it's just overwhelming me

so I'm kinda looking for somebody to assist there, if possible

[feral sedge]

Is there a nice expression for dω^η + ω^dη

[prime bramble]

heck if I know

[feral sedge]

There should be, right?

[prime bramble]

uhh, is there?

if there is, I was certainly not made aware of it

[feral sedge]

d(ω^η)?

[prime bramble]

is that true?

[feral sedge]

hmm

Ah

[prime bramble]

aha

[feral sedge]

It depends on if omega is an even form or an odd form

[prime bramble]

this

[feral sedge]

Yeah this

But this is sufficient to solve too

[prime bramble]

hmm

okay, I can work with this

thanks for bringing it to my attention moni

I'll try again now

[feral sedge]

np

[prime bramble]

.close

[pearl pond]

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[pearl pond]

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[gray bison]

cscx

[pearl pond]

@jade moss Has your question been resolved?

[pearl pond]

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[brittle urchin]

Is there any server like this one to help solve doubts of physics and chemistry? With similar channels for individual doubts like this one

[tall flint]

#old-network has a collection of other servers

[brittle urchin]

But there we can't share our doubts like this with various channels

Like there we have to post our doubts and then they remain unanswered for days

In forums

[drowsy adder]

I think you can send physics here but chemistry becomes too unrelated

[brittle urchin]

Won't I get kicked?

[drowsy adder]

No lmao

[brittle urchin]

That's great then

[brittle urchin]

Tysm

.close

[pearl pond]

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[pearl pond]

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[radiant fable]

Plz can you help me understand trigonometry

[pearl pond]

Please don't occupy multiple help channels.

[clear niche]

Sure, and please close your previous help channel if you’re finished with it

[radiant fable]

I'm not yet done with it

. close

[oak ivy]

its .close

[pearl pond]

@radiant fable Has your question been resolved?

[pearl pond]

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[rustic zealot]

Been trying this for probably 40 minutes. I know its not strictly maths but physics is similar enough haha anyway help ASAP would be appreciated@!I!!

[oak ivy]

They haven't given any sin values right?

[plush moss]

Note the angles at both normals are 90 degrees

[rustic zealot]

wdym? Probs not thats all the question has

[plush moss]

You guys can use calculator right

[rustic zealot]

Hmm yeah ive been basing it off of that. Geometry has never been my strong suit

Yes

[oak ivy]

Because I thought of snell's law

[rustic zealot]

yeah thats what we are supposed tod o

I can calculate the first refraction

[oak ivy]

n1sini =n2sinr

yeah

[rustic zealot]

but im struggling to find the incident angle for the second refraction

thats all if i can get that then i could do the question

[oak ivy]

then we use it again?

oh ok

Can you find deviation angle

[rustic zealot]

deviation angle?

[oak ivy]

angle of deviation is it possible to find it

[rustic zealot]

Dont think ive learnt about that

[oak ivy]

um

[oak ivy]

A+D=i+e

[plush moss]

First can you find this angle

[rustic zealot]

or we possibly use a different term

[rustic zealot]

yeah thats around 50 degrees

Cause angle of refraction for that one is 40.7

[plush moss]

Yea

[rustic zealot]

so itd be 49.3 i suppose

[plush moss]

After that find this

[oak ivy]

Is that deviation angle?

[rustic zealot]

now thats where teh struggle begins aight lemme give it a go

[oak ivy]

If it is question is done

[rustic zealot]

oh wait

Its just a triangle

[oak ivy]

Yep

[rustic zealot]

Goodness me

[oak ivy]

I think that is deviation right ?

[rustic zealot]

70.7

My goodness

well now i can do it

hahahaha

[oak ivy]

Ok

[rustic zealot]

How funny is taht

i spent so long on it

[plush moss]

Yea exactly

[rustic zealot]

But it was so easy all along

Tghanks guys

Thats all

[pearl pond]

@rustic zealot Has your question been resolved?

[pearl pond]

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[cinder bane]

I was wondering if it is possible to know many subsequences of a certain number contains an even digit based on the number of even digits it contains.
Like for example the number 123456, it has a total number of subsequences (discarding empty subsequence) of 2^6 - 1 = 63.
Is there a fast way to do so?

[cinder bane]

The aprouch I am using to determine how many subsequences have an even number is by dividing the subsequences according to size then for each size I determine how many of them have an even number.
The upper number has 3 even digits and 3 odd digits.
So for the size of 1 subsequences, the number containing even digits = 6C1 - 3 = 3
for size of 2: 6C2 - 3x2 = 9
for size of 3: 6C3 - 3x2x1 = 14
and all the other sizes must have an even number

[limpid lily]

By subsequence, do you include 1256? Does "contains an even number" mean that the subsequence itself is an even number?

[cinder bane]

i meant even digits

[autumn fossil]

I think that you could calculate how many subsequences are made out purely of odd digits

that should be fairly easy given any number

let's do e.g.

1112211211

this has sequence of 3 odds, 2 odds and 2 odds

[cinder bane]

That is a big number to consider as an example

but ok

[autumn fossil]

doesnt really matter

[autumn fossil]

Can you see how there are 3 groups of odd numbers? Every odd subsequence must be part of one of those groups

[cinder bane]

by the way why would determining if the number have an odd digit be easier than the number have an even digit

[autumn fossil]

I said purely odd digits

meaning no even digits

it's easier to count subsequences with no even digits, than subsequences with at least one even digit

[cinder bane]

oh ok

[cinder bane]

ok

[autumn fossil]

Can you now think of a way to count those purely-odd subseqeunces?

[cinder bane]

considering them as a number and counting the subsequences it has?
like in this case there are 3 + 2 + 2 = 7 odd digits
so it should be 2^7 - 1 = 127. Right?

[autumn fossil]

almost, but this overcounts

oh wait

you dont requre your sequences to be connected, right?

[cinder bane]

yes

[autumn fossil]

for some reason i was thinking of connected ones

[autumn fossil]

Yeah, then this would be fine

[cinder bane]

isn't subsequence mean that you delete stuff from the sequence while keeping the order

[autumn fossil]

Yeah, it does

[cinder bane]

oh ok

[autumn fossil]

i dont know why i was thinking of connected ones

anyway, this method still works and your result is correct

[cinder bane]

I thought that you would account for repetition of subsequences. So I was about to tell you that I don't care much about the repetion

[autumn fossil]

Fortunately, accounting for that would be kinda hard

Anyway we have 2^oddDigits - 1 purely odd subsequences

and 2^totalDigits - 1 (nonempty) subsequences in total

subtracting those, we get 2^totalDigits - 2^oddDigits subsequences containing an even number

[cinder bane]

Yeah, got it

[autumn fossil]

we could also factor out 2^oddDigits and get:
2^oddDigits(2^evenDigits - 1) as an alternative form

[cinder bane]

Ok, Thanks for the help

[pearl pond]

@cinder bane Has your question been resolved?

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[livid bolt]

How to do question d?

[tardy reef]

rate of decrease = derivative

so you know, differentiate V wrt time, and evaluate at V=8volts

[livid bolt]

what’s wrt

[surreal coral]

with respect to

[livid bolt]

is this right

[surreal coral]

where did the 14 go

[livid bolt]

um…..

idk. I was thinking it disappears

because 14^0

do I have to use the product rule

[surreal coral]

derivative of Ae^kt = kAe^kt

you can show that using the chain rule

[surreal coral]

why did you add the 1(e^kt) for

[livid bolt]

uhh…. I thought that’s how u do product rule

[surreal coral]

you're not using product rule

[livid bolt]

ohh

derivative of 14 is 0

[surreal coral]

or an easier way to think about it is that 14 is just a constant so you just need to differentiate the e^kt

and then multiply back the 14 after

[livid bolt]

ok I got that

is that right now

[surreal coral]

no need to write the + 0(e^kt)

but 14ke^kt is correct

[livid bolt]

hey I do that because it’s easier

in my brain

):

[surreal coral]

as long as it makes sense to you

now the questions asking "at what rate was the voltage decreasing at the instant it reached 8 volts"

so you found the rate of change of voltage

[livid bolt]

yea

[surreal coral]

do you have any ideas on where to go from here

[livid bolt]

um

I do

I find the time it takes to reach 8 volts and sub it in

[surreal coral]

mhm

that's right

[livid bolt]

they did it an easier way

how do i do their way

[tardy reef]

Its the same way. Just that they recognized that the derivative kAe^kt contains Ae^kt which is V

[livid bolt]

did they use this thing

oh

ok surely i dont need that

that formula is too hard

[livid bolt]

how do they know that

[tardy reef]

you can just see it. Like its right there in the expression...

kAe^kt = k * (Ae^kt)

[livid bolt]

how is Ae^kt = v

oh

[tardy reef]

its given

[livid bolt]

ohzzx

its in the question...

how come i have a 14 tho?

and they dont