help-36

[jagged root]

yueaa

[spare summit]

but the range depends on whether the poly is of odd or even degree

[jagged root]

i see i see

[spare summit]

if the poly is an odd degree poly, then the range is all real y

[jagged root]

odd as in example?

[spare summit]

if the poly is an even degree poly, then the range is all y greater than or equal to the min point, or less than or equal to the max point

[jagged root]

oh

i understand that

[spare summit]

odd as in having a variable term raised to an odd power as the highest exponent in the poly

that's the definition of degree anyway

[jagged root]

i see

[spare summit]

so x^3 + x + 1 is an odd-degree poly, for example

while x^4 + x^2 + 3 is an even-degree poly

[jagged root]

oHH

[spare summit]

anything else you'd like to ask?

[jagged root]

uh

I think i get it

ty for the help

[spare summit]

nps. do close the channel when you're done with .close

[jagged root]

.close

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[grand flare]

how do i do word problems like this

[deep condor]

is $\gamma$ the radius of the sphere?

[soft zealot]

حسیب ♥

[grand flare]

its r

but ye its changing at rate of 2cm/s

[deep condor]

oh that is an r, ok

[grand flare]

hm

[deep condor]

so, we know that the radius is increasing at a rate of 2 cm/s, and we want to find the rate of change in the volume

[grand flare]

when r=7

[deep condor]

let's say the radius is $r$ and the volume is $V$, what is the rate of change of $r$? and the rate of change of $V$?

[soft zealot]

حسیب ♥

[grand flare]

dr/dt=2

dv/dt=(dv/dr)*dr/dt

?

[deep condor]

this works, yes

now the thing we have to find is $\dv{V}{r}$, so let's find a formula with $V$ and $r$ in it, then differentiate $\dd r$

[soft zealot]

حسیب ♥

[grand flare]

V=4/3 pi r^3?

[deep condor]

now differentiate both sides with respect to r

[grand flare]

dv/dr=4/3 pi 3r^2

[deep condor]

then substitute that back into the chain rule equation here

[grand flare]

we get dv/dt=8 pi r^2

?

[deep condor]

technically you haven't subbed $r=7$cm in yet, so you have
$$\dv{V}{r} = 4\pi r^2 \dv{r}{t}$$ for any $r$

[soft zealot]

حسیب ♥

[grand flare]

now we js substitue r=7?

[deep condor]

yep

[grand flare]

we do it where

[grand flare]

here or

[grand flare]

here

[deep condor]

here, you subbed in dr/dt = 2 cm/s, but you didn't sub in r=7cm

so why not sub in all at once?

[deep condor]

but then you should have this step in between so it's clearer

[grand flare]

so dv/dt=8X22/7*7X7

[grand flare]

ohk

[deep condor]

$\pi$ is not exactly equal to $\frac{22}{7}$, so you can either write
$$\dv{V}{t} \doteq 8 \cdot \frac{22}{7} \cdot 7^2$$
or simply
$$\dv{V}{t} = 8\pi(7^2) = 392\pi$$

[soft zealot]

حسیب ♥

[grand flare]

but r=7 gives a sign to cancel the 7 in 22/7

thats why i thought to use it

[deep condor]

interesting way of thinking about it, because yeah teachers do like to make things cancel

but pi isn't 22/7

[grand flare]

ohk

ill keep that in mind

thanks

[deep condor]

you should write the "exact answer" (which is 392 pi) and then you can use 22/7 if you don't have a calculator on a test, for example

[grand flare]

ok

[deep condor]

so here are the general steps:

1. write out the chain rule [ dv/dt = (dv/dr)(dr/dt) ]
2. write the formula relating your variables
3. differentiate to find the unknown derivative (dv/dr in our case)
4. sub into the chain rule
5. sub the specific time (r=7)

do those steps make sense with what we did?

[grand flare]

ok ill try another qs and let u know if i face a problem

[grand flare]

yes they do

[deep condor]

ok, sounds like a plan :)

[grand flare]

one doubt
dx/dx^3

is it 1/3x^2

or 1/3*x^(-2/3)

[deep condor]

should be 1/(3x^2), dont forget

if you're ever confused, sub y=x^3

[grand flare]

so that makes it d(y^1/3)/dy

[deep condor]

then $\dv{x}{x^3} = \dv{(y^\frac{1}{3})}{y} = \dv{y} y^{\frac 13}$

[soft zealot]

حسیب ♥

[deep condor]

beat me to it :D

then differentiate and sub back in

[grand flare]

wdym eng is not my first lang 😢

[deep condor]

ah sorry, means that you did it faster than me

like beating someone in a race

[grand flare]

ahh ok

[grand flare]

alr

in this qs my solution is
dv/dt = 27
d a^3/dt=(d a^3/da) * (da/dt)
27=3a^2 * da/dt
now a is given as a=4 because v=64
27=3*4^2 da/dt
da/dt=9/16

now we need to find dA/dt where A=area=6a^2
d 6a^2/dt
=6* da^2/da *da/dt
=12a * 9/16
putting a=4 we get dA/dt=27

[final saddle]

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[grand flare]

?

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[plush haven]

Can anybody explain how this works

I don't understand why they decided to do U sub

I did it normally and got it right

[vital crag]

That's fine. Both options are valid

u subs of the form u = ax + b are pretty trivial

[plush haven]

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[ripe crater]

if im reading this right, i have to prove that

limn->infintity can = ca?

[loud sundial]

yes

[ripe crater]

do i have to make note for all the cases ot just prove it?

[loud sundial]

wdym by "cases"

[ripe crater]

so, my book says we have to consider 2 cases:

1. c = 0
2. c is not 0

but, like, do i have have to consider both of em?

[loud sundial]

um

yes?

because |c|>0 doesn't hold for c=0

am I missing smt

[ripe crater]

hmm alright alright

makes sense

[loud sundial]

[final saddle]

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[ripe crater]

|ca_n - ca| < epsilon
|c| |a_n - a| < epsilon
a_n - a < epsilon/c
a_n < epsilon/c + a

is this right?

and then do it again, but at step 2 replace a_n with epsilon/c + a

[cobalt sonnet]

yo

someone helpppp

[loud sundial]

mhm for the start but don't drop absolute values

[loud sundial]

!occupied

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[ripe crater]

wdym?

[loud sundial]

line 2 to line 3

why did you drop the absolute value

[ripe crater]

i dont have a why, i just dropped it to move the variables

[loud sundial]

????

what

$$|c| |a_n-a|<\varepsilon \implies |a_n-a|<\frac{\varepsilon}{c}$$

[soft zealot]

Civil Service Pigeon

[loud sundial]

the original limit definition has absolute values anyway

so it's not like you need to have no absolute values in the end

I should also say that you shouldn't start with $|ca_n-ca|<\varepsilon$ since that's assuming what you're trying to prove

[soft zealot]

Civil Service Pigeon

[loud sundial]

but when you write it in reverse, it's fine

I thought it was obvious at first, but perhaps it deserves a quick remark

idk

[ripe crater]

then what do i start with?

[loud sundial]

with the given condition?

[ripe crater]

the way i do is:
Let E > 0
then that step

[ripe crater]

yeah with the given condition, or just in general

my way of doing this is:

Let E > 0
| f(x) - L| < E

then whatever e i end up getting
Let delta/N (dependong on the question) and then resolvoing it, but replacing it

[loud sundial]

You're given $\lim \limits_{n \to \infty} a_n=a$ to start with that. You want to \textit{end} at $\lim \limits_{n \to \infty} ca_n=ca$.

[soft zealot]

Civil Service Pigeon

[loud sundial]

the overall methodology and rules of how to write proofs doesn't change just because what you're writing the proof about changes

[ripe crater]

im lost on what ur trying to say

[loud sundial]

You want to prove that
$$\lim_{n \to \infty} a_n=a \implies \lim_{n \to \infty} ca_n=ca.$$
When you prove $A \implies B$, it's common to start with $A$ and do stuff to end up at $B$. (Granted this isn't the only way to do this - contrapositive and other methods exist for example.) What you \textit{should not do} is start at $B$ and end up at $A$ - that would prove $B \implies A$ rather than $A \implies B$. (You may recognise this as proving the \textit{converse} of $A \implies B$ instead.) A common example of why this is an issue is because you can do stuff like
$$1=2 \implies 1 \cdot 0=2 \cdot 0=0=0$$
which allows you to end up with a true statement even if you started with a false one (\textit{vacuous truth}).

[soft zealot]

Civil Service Pigeon

[steady locust]

hes saying to do this in reverse basically

🙂

that is what the question asks

[ripe crater]

so, what i did is... right? im not going b to a?

[ripe crater]

reverse how?

[steady locust]

you start with $\lim \limits_{n \to \infty} a_n=a$, and you want to show $\lim_{n \to \infty} ca_n=ca$.

you did it the other way around

which is wrong

[warm python]

Lemme try , you assume $\lim_{n \to \infty} a_n =a$ is true. From this you wish to show $\lim_{n \to \infty} ca_n = ca$

[soft zealot]

wai

c2b7_97705

[ripe crater]

how do i bring in C if i start from a_n = a?

[steady locust]

if |a_n - a| < epsilon, then |c||a_n - a| < |c| * epsilon right

you are just multiplying both sides of the inequality by a positive number

[ripe crater]

oh

[final saddle]

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[halcyon roost]

DiffEQ,
In the most recent homework two of the problems seemingly don’t appear in my notes really at all. I write everything that I said and done, so I have no idea how to solve these questions. My friends in the class had no idea either
I’ve attached my notes to this message as well

[halcyon roost]

Just a way to get started would probably be ideal, especially using concepts from my notes, but any help is greatly appreciated

[vital crag]

You want someone to read your notes to help you get started?

That's the whole point of taking notes

[halcyon roost]

I mean, Ive read my notes, Im not sure what part applies to these questions

I wrote them too ig lmao

[trail mango]

good handwriting

[halcyon roost]

haha thanks

[vital crag]

What part of the instructions in the question is confusing you

[halcyon roost]

yeah so, neither couchys theorem, superposition principle, or the wronskian really apply to verifying two parameter families of functions of nonhomogenous DEs

Superposition deals with homogenous

[vital crag]

Well do you know what the question is asking

[lilac bison]

How do you make such neat notes and not burnout

[halcyon roost]

Im gonna be honest, not entirely?

[vital crag]

Yea do that first before reading your notes

[halcyon roost]

bro it can be tiring, 75 minutes of straight writing, after class im exhausted

[halcyon roost]

ok fair enough

[vital crag]

It's just telling you to plug the solution given into the DE and show that it's true

[halcyon roost]

what does two parameter mean in this context?

[vital crag]

$c_1$ and $c_2$ are called parameters

[soft zealot]

riemann

[halcyon roost]

ah ok that makes more sense

[lilac bison]

Aren't we just supposed to plug in the values?

[halcyon roost]

Yep

trying that now

[lilac bison]

Yuh

Getting started with differential equations?

[halcyon roost]

I guess yeah, we finished first order DEs now were onto higher order DEs

Just plugging it in worked btw

Thanks @vital crag @lilac bison

[lilac bison]

I didn't do anything lol

[halcyon roost]

you sort of did

[vital crag]

[halcyon roost]

oh also the other benefit is that friends depend on me for notes

aight thanks guys

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[fleet briar]

I am not able to understand which arc makes what angle

Shouldn't be angle POR = 200°?

But I know it's wrong

[strange sparrow]

why do you think POR = 200?

[heavy stump]

nice one, are you trying to find that angle x?

[fleet briar]

Angle subtended by an arc at the centre of the circle is twice the

Angle subtended by it

[fleet briar]

Yez

[strange sparrow]

hint: construct a cyclic quadliteratal involving PQR and another arbitrary point on the circumference

[fleet briar]

At any point of circle

[fleet briar]

Yea then?,
I just want to know how to figure out that which arc makes/subtends what angle

It's so confusing when reflex angles come

[strange sparrow]

Okay, say you have picked a point S on the circumference, say on the lower half of the circle (for easier visualisation)

what is PSR?

[delicate sorrel]

huh

[fleet briar]

80

[strange sparrow]

right, can you find POR then?

[fleet briar]

Oh yea

[fleet briar]

But I wanna know what was my mistake in previous?

[strange sparrow]

What was your previous reasoning

[fleet briar]

Angle por = 200

Seeing that other angle 100

[strange sparrow]

it's an incorrect application of the theorem

it's just not the right shape

[fleet briar]

Don't those two angles got subtended by same arc?

If not pls explain

[heavy stump]

yeah 160 at the center makes it 10 for x, nice one 0lante

[fleet briar]

0lante?

[strange sparrow]

it is not applicable when the point on the circumference lies on the arc

[fleet briar]

Arc that is subtended?

[strange sparrow]

sorry, just arc

[fleet briar]

So that 100° angle is subtended by which arc?

[strange sparrow]

it is subtended by the "outside" arc

[fleet briar]

And Angle POR is subtended by inside arc?

[strange sparrow]

they are subtended by different arcs

if you apply the theorem with the lime arc then it tells you reflex POR = 200 so POR = 160

which is another way to arrive at the same conclusion

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[primal lava]

Could you solve this problem? They said wrong

A. 4/7

B. 5/7

[leaden moon]

!noans

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[leaden moon]

First question is wrong because the parts aren't equal

[primal lava]

So I need to make equal

[leaden moon]

There's a big shaded part that looks like $\frac14$

[soft zealot]

1 divided by 0 equals Infinity

[primal lava]

I got it

So my answer is wrong

[leaden moon]

And the remaing part has 3 parts shaded out of 6 equal parts

Note that these 6 parts represent $\frac{3}{4}$

[soft zealot]

1 divided by 0 equals Infinity

[primal lava]

Thank you

.close

[leaden moon]

Try again

[primal lava]

.close

[leaden moon]

Don't edit 😭

[primal lava]

Sorry

.close

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[tough pike]

I don’t really get how I’m meant to do this question,
If Z=1+I Is a zero of the polynomial z^3 +az^2 +bz +10 -6i
Needing to find, a and b, where a and b are real

[lilac bison]

z = 1 + i?

[strange sparrow]

plug z = 1 + i into the polynomial, you will get 0

[lilac bison]

if thats the case, since its a zero of the polynomial, plugging that in, will result the equation to become zero

from there you have some equating to do

[strange sparrow]

from this you can equate the real and imaginary parts to form a set of equations

[tough pike]

That makes so much sense ty

[tough pike]

That makes so much sense ty

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[steady wraith]

just want to double check if im right or not because i got a slightly different answer but also it seems like the people who made the mark scheme are engineers and wrote g = 10

[steady wraith]

might as well just say e^2 = pi^2 = g = 10 by that point 💀

oh wait

it literally says

to use g = 10 which again feels stupid because this is maths not engineering

mb

.solved

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[slow crypt]

Hello, could someone help me to get started on this?

[vital crag]

do you know how to draw the right triangle?

[lilac bison]

oop

[vital crag]

if so then use

,tex .sohcahtoa

[soft zealot]

riemann

[lilac bison]

how did you type it that fast? are you goated or smth?

[slow crypt]

Will the hyp be the square root of 3?

[abstract bramble]

no

[slow crypt]

Ah I'm getting my triangles confused

[abstract bramble]

what triangles

use pythagoreans

[slow crypt]

I thought it was one of the special ones from first glance

It's the square root of 5

[vital crag]

the hypotenuse is sqrt(5) yes

[slow crypt]

Ah I got it

Lol panicked a little bit my b

thank you

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[hardy jackal]

im so lost, would i not be able to do question 5 like 4?

[hardy jackal]

where am i going wrong

[abstract bramble]

uh why are you using -3 and -1

[barren hound]

i dont understand what's happening in the blue and white but the red purple and yellow is correct

[abstract bramble]

do it like question 4

[hardy jackal]

well its to find a point on the graph i could use as a reference

thats how my teacher showed us

[hardy jackal]

what numbers would work

[abstract bramble]

2?

i mean

-2

youre trying to find the limit at -2

[hardy jackal]

but doesnt it have to be less than -2 for the first one

my teacher did it like this

[barren hound]

we care about the behaviour as it approaches -2

so think like -2.000000000000000000000001

[hardy jackal]

why didnt my teacher just sub in 1 for her example, since the limit approaces 1?

[barren hound]

good question

[abstract bramble]

no idea

[barren hound]

thats what i woudl have done

plotting it at -1 is like vaguely helpful for drawing its behaviour

but you know how to draw a line lol

[hardy jackal]

thats true

i dont want her to dock marks for doing it another way since shes super strict about that

idk what to do

[barren hound]

ok well uh... go ahead and plot those points but theyll not tell you anything useful about its behaviour at -2

[hardy jackal]

i mean those point dont even work after putting it in desmos

[barren hound]

what do you mean by "those points dont even work"?

the graph should go through them

[hardy jackal]

wait ur right i mixed them up

i switched the linear line with the quadratic

[hardy jackal]

also

for thise one

wouldnt i add arrow heads at the ends (where the blue dots are )

[heavy stump]

since both sides give 0 at -2 it's actually continuous there

[barren hound]

sure

[hardy jackal]

sure

[final saddle]

@hardy jackal Has your question been resolved?

[hardy jackal]

can someone check these over:

[gritty drift]

First is good

Second is good. Small note, for continuity it is not sufficient for the right and left side limits to match. Additionally the function value at that point must also equal the limit.

If the left & right side limits match -> limit exists.

If limit exists AND f(a) is equal to the limit at a -> continuity

[hardy jackal]

would this be better: the graph is continuous at x=−2 because the left andd right limit equal 0, and the function value f(−2) is also 0

[gritty drift]

yes

[hardy jackal]

would everything else be fine?

[gritty drift]

yes

[hardy jackal]

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[paper thorn]

How do I solve this? Might be a dumbass question but it's late and can't figure it out

[paper thorn]

Do I substitute

[muted prairie]

What does the lightning bolt mean?

is that a 5

[paper thorn]

4

√x-2 + ³√4-x = 2

[vital crag]

are you supposed to find just 1 solution or all of them?

you could guess a solution by using 1 + 1 = 2

[zinc maple]

thats a crude guess

it would assume √(x-2) = ³√(4-x)

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[trail shale]

What did I do wrong?

[final tangle]

you didn't mark the angle of elevation correctly

angles of elevation (and depression) are angles made with the horizontal

[trail shale]

What do you mean

[trail shale]

Oh ok

Is this always true

[final tangle]

yes, its how these terms are defined

[trail shale]

Ok

But I have one question

My teacher told me to use alternate interior angles to “move” the angle of depression so it is in the triangle but that would make it the same angle as the angle of elevation so how does that work

[loud phoenix]

Because you will usually be given either the height or length

[final tangle]

alternate angles, parallel lines

[loud phoenix]

By moving it inside you can use your trig

[final tangle]

the above angles of depression and elevation are congruent by that property
(since the horizontals are parallel to each other)

[trail shale]

Oh ok I think I understand thank you

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[left trail]

For number 21 can someone tell me if I am missing something. I feel that what I found for c is not right but I can't find any errors

Also for the green right triangle we have phi for one of the angles and then one of the smaller right triangles has alpha as an angle. In this setup alpha can never be greater than phi right?

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[left trail]

.open

.reopen

[final saddle]

✅ Original question: #help-36 message

@left trail Has your question been resolved?

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[chrome ermine]

57

[steady locust]

are you supposed to take the derivative?

what does the question say to do

[olive lantern]

what to do?

[chrome ermine]

d/dx

derivitave

sorry

[olive lantern]

arcsin(x) has a direct result

[chrome ermine]

for that one

i got

[olive lantern]

for the next one use the multiplication rule

[chrome ermine]

2/sqrt(1-x^2/16)

[chrome ermine]

i did but i tihnk i did it wrong

[olive lantern]

can u show what u did?

[chrome ermine]

yes

-1/2(-x^2(16-x^2)+sqrt(16-x^2))

[olive lantern]

small error

it should be $-1/2(-x^2/sqrt(16-x^2)+sqrt(16-x^2))$

[chrome ermine]

what is wrong in mine

[olive lantern]

u did the chain rule wrong

[chrome ermine]

its 1/2(sqrt(16-x^2)) * -2x no?

[soft zealot]

Hardik

[olive lantern]

yes but u are diffentiating sqrt(16-x^2) by d(16-x^2)

[chrome ermine]

oh i forgot the -1

on the power rule part

[olive lantern]

so it becomes $1/((16-x^2)^(1/2)) multiplied by d(16-x^2)/dx$

omg

i dont know how to use that bot sorry

[chrome ermine]

its okay

[olive lantern]

u also forgot too diff sqrt 16-x^2

somehow u managed to square it

[chrome ermine]

okay im going to do the whole product rule again and see what i get

[olive lantern]

ok

ill try out the bot

$d(ab)/d(x) for a,b are functions of x is a'(x)b(x)+b'(x)ax$

bruh

[chrome ermine]

x* 1/2 (16-x^2)^-1/2 * -2x + sqrt(16-x^2)

thats what i got

[olive lantern]

yes

[chrome ermine]

okay

[olive lantern]

thats correct for the second part

[chrome ermine]

ok

ok and i simplify

and get

x^2/2sqrt(16-x^2) - sqrt(16-x^2)/2

[olive lantern]

yes

[chrome ermine]

can i simplify more or is that it

[olive lantern]

if u want u can make a common denominator

that will cancel out the x^2

[chrome ermine]

i see it

okay i understand

okay thanks you

[olive lantern]

dont forget

to add arc sin

deiff

diff

[chrome ermine]

ya

got it

ty

.close

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[ripe crater]

so... to solve this:
subseqeucne 1: k^2
subsequence 2: k

the limit for subsequence 1 is 1
for sub2 limit is 1/n which translates to 0

now to use the epsilon-delta/n
how would i find epsilon?

[trail mango]

can you explain your subsequences a little more

[ripe crater]

ie... off... ah... eee... my bad k^2

[trail mango]

unless i misunderstand, they do not converge (which is kinda a good thing)

[ripe crater]

nono, my bad, i wrote it wrong

[trail mango]

subsequence 2: k
what does this mean?

is that a different sequence than f?

[ripe crater]

no, the 1/n

so if the subsequence 1 i.e 1 if n is a perfect square is k^2
then subsequence 1 i.e 1/n otherwise can be k

or am i missing something?

[trail mango]

so you mean the subsequence f(k_1), f(k_2), ... where (k_i) is the sequence of non perfect square integers i guess

you're not really describing the subsequences well but you have the right idea i think

[ripe crater]

yes yes

[ripe crater]

i.. sorry, im not good with knowing the exact terminology of writing stuff

[trail mango]

it's ok

so what facts about sequences and subsequences and limits are we allowed here? i'm not sure what you are requesting still

now to use the epsilon-delta/n
how would i find epsilon?
like what are you talking about here?

are we allowed standard facts about sequences like... subsequence converges to X implies ambient sequence converges to X, and a sequence has at most 1 limit point

[ripe crater]

so, basically to prove it doesnt exist we are using the contradiction,
we say
the limit exist and then prove that, when that cant be done we say it doesnt exist

so, to do that, we need some value of epsilon

[trail mango]

ok if no standard facts at all and definitions only...
we want to show there is no L for which there is an epsilon such that for all N, there is an n >= N with |f(n) - L| >= epsilon

or if you insist on contradiction, sure we can say an L exists

[ripe crater]

yeah

so, how do i find the epsilon which i have to use?

[trail mango]

there are many you could pick

1/2 should do just fine

[ripe crater]

but how, how did we come to the conclusion that 1/2 can worl

[trail mango]

well there are subsequences converging to 0 and 1. so the limit kinda needs to be one of those (if we just imagine there is a limit for now)

if L = 0, then there are infinitely many n's such that f(n) = 1, so we'll be able to find one such that |f(n) - L| >= 1/2

if L = 1, similar thing

we can find large n such that f(n) is near 0 and hence |f(n) - L| >= 1/2

[final saddle]

@ripe crater Has your question been resolved?

[ripe crater]

ah gotcha

alright alright makes sense thank you

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[rustic wedge]

HALLO
so last month apparently 3b1b posted this puzzle:
consider a ladybug starting at 12 on a clock. At any momment, it travels 1 step clockwise or anti-clockwise with equal probability (of 1/2). Every time it reaches a new number, the number is coloured red. What is the probability that the last number to be coloured is 6?

[rustic wedge]

now i just came across this

and i have an idea

the issue is i am missing something because of which i am not getting the right answer

Consider the following setup:
Let S be a permutation of the set {1,..,12} representing the order in which the numbers are coloured

clearly, the first element is fixed to be 12

meaning 11! elements in our sample space

.close

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[rustic wedge]

i am so stupid

why did i need to write this out to make sense of it 😭

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[twin pivot]

What Is the Quasi-Fourier Transform?

[final saddle]

@twin pivot Has your question been resolved?

[runic needle]

no clue

[stone wagon]

context? where did you see that

[sturdy cypress]

q is quick

[stone wagon]

quickuasi fourier transform

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[old quarry]

how to integrate this

tried converting to sec and tan

but was not getting anything

[final saddle]

@old quarry Has your question been resolved?

[wicked obsidian]

$$1 - 5\cos^2 x = (\sin^2 x + \cos^2 x) - 5\cos^2 x = \sin^2 x - 4\cos^2 x$$

[soft zealot]

akeanti

[wicked obsidian]

looks like if u simplity the numerator like this and separate both fractions u will be left with a simple derivative

[runic tulip]

Multiply and divide by sin
Then let t = cosx

[runic tulip]

Tho it might not work

[old quarry]

ill get 1/sin^3xcos^2x

and ill have to integrate cosec^5x

which in itself is a kinda long process

[wicked obsidian]

$$I = \int \left( \frac{1}{\sin^3 x \cos^2 x} - \frac{4}{\sin^5 x} \right) dx$$

[soft zealot]

akeanti

[old quarry]

yes

[wicked obsidian]

what is inside looks simillar to the derivative of $$f(x) = \sin^{-4} x \cos^{-1} x$$

[soft zealot]

akeanti

[old quarry]

i get something very ugly

[old quarry]

that is not easy to see at all

[wicked obsidian]

oh well

[old quarry]

yea but like u just kinda got to there

[wicked obsidian]

yea xD

[stark minnow]

why not bring cos^2x to numerator as sec^2x and let tanx=t

does that not work ?

[old quarry]

yes but the problem is power 5

ull have to multiply and divide by cos^5 to get that tan term

and cos^5 bring to the numerator n write as sec^5

which ull have to break as sec^2 sec^2 sec

the sec^2 can be written as 1+tan^2 but what will u do about the sec

[stark minnow]

yeah i got that

was seeing if it is solvable further

[stark minnow]

nvm this method works brilliantly

[old quarry]

that is not a method they just got the answer and differentiated that..

[wicked obsidian]

it looks also possible to be solved using t=cosx and then using ostrogradsky integration method to simplify both fractions

[old quarry]

what is the ostrogradsky integration method 💀

[wicked obsidian]

smt to help with all those fractions

[old quarry]

uhh thats a bit too complicated for me

[candid pulsar]

yeah

||thats the answer||

a reasonable approach is to IBP on $\int \frac{\sec^2(x)}{\sin^5(x)} dx - \int \frac{5}{\sin^5(x)} dx$

[final saddle]

@old quarry Has your question been resolved?

[old quarry]

uhh sorry?

ibp on which integral

cosec^5x ?

or both

[candid pulsar]

first one

sec^2(x)/sin^5(x)

[old quarry]

o but how will that help?

ohh wait

Ohhhhhhhhh

it cancels

thats crazy

thanks bro

.close

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[cedar obsidian]

!status

[final saddle]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[thorny canyon]

find the intersections first

can be done via substitution

[loud sundial]

,w x-2y-1=0, y^2=4x

[soft zealot]

Failed to get a response from Wolfram Alpha.
If the problem persists, please contact support.

Feb 19, 2026, 4:44 PM

[loud sundial]

the intersections are ugly though

@cedar obsidian

You're better off parameterising the parabola by $(t^2, 2t)$

[soft zealot]

Civil Service Pigeon

[loud sundial]

since this allows you to ||do Vieta's on everything instead||

[thorny canyon]

oh that's smart

[cedar obsidian]

yes true

[cedar obsidian]

like i form the biquadratic equation right?

[loud sundial]

um

oh

um

where would that come from

[cedar obsidian]

its given in the solution and i thought that the solution is messy by that algebra

[thorny canyon]

uhm there's no biquadratic involved

[loud sundial]

I see a quartic, but it's not biquadratic

slip of the tongue?

but this looks mostly like what I intended

[cedar obsidian]

t^4 + ... the 4th isn't called biquadratic?

am i forgetting smh

[thorny canyon]

synonyms

[loud sundial]

well eh

idk

the less common usage ig

[thorny canyon]

like biquadratic is sortof a blanket term

any eqn of deg 4

[loud sundial]

meh I'm not in the mood to debate archaic terminology rn

I'll leave y'all to it since my contribution is basically done

the rest is algebra bash as seen above

[cedar obsidian]

.close

[final saddle]

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[cedar obsidian]

oh its still open

nice

.reopen

[final saddle]

✅ Original question: #help-36 message

[cedar obsidian]

solution is weird

it directly says that both are diameters of the circle

it means every chord is a diameter to a circle which is not true

[opal plinth]

No, it means every pair of chords that bisect each other are diameters

[cedar obsidian]

ooh shit

[final saddle]

@cedar obsidian Has your question been resolved?

[cedar obsidian]

.

one more problem i need help with is this ^

[opal plinth]

Have you tried anything?

[tiny kraken]

||equilateral triangle|| and ||GP Sum||

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[cedar obsidian]

hello

[cedar obsidian]

no ideas coming to my head :((

[worldly spruce]

Hello

[worldly spruce]

This?

[worldly spruce]

Oh this

[cedar obsidian]

it's this

[worldly spruce]

Yeah

[opal plinth]

ABC is a special triangle. What kind?

[worldly spruce]

I don't like triangles, bye

Best of luck though!

[tiny kraken]

reasonable reaction

[cedar obsidian]

seems isosceles triangle

[opal plinth]

Don't play any 3D video game

[opal plinth]

Right, which sides are equal?

[cedar obsidian]

BC & AC

[opal plinth]

Ok and can you find the area of AHB?

Or just the coordinates of H

[cedar obsidian]

area found is 16/root3

AHB^

[opal plinth]

Ok not sure how you got that but it's correct

Now what do you know about centroids

[cedar obsidian]

coordinates of H and then determinant formula for area

[opal plinth]

Idk what the determinant formula is

[cedar obsidian]

concurrence point of medians

it divides median in 2:1

[opal plinth]

Yes

And ABC is isosceles, so where are all these centroids G_i

[cedar obsidian]

on the median or the altitude passing from H

[opal plinth]

Exactly

So what is the area of AG_1B compared to that of AHB

And then that of AG_2B compared to that of AG_1B...

[tiny kraken]

Can't we just say it's a equilateral triangle so H is also the centroid.... and the area is a^2sqrt{3}/4...

[opal plinth]

It would be 1/2 * 8 * 4sqrt(3)/3 = 16sqrt(3)/3

[cedar obsidian]

yes

its not coming in format on (sqrt3*a^2/)/4

[tiny kraken]

what are you referring to

[opal plinth]

Area of AHB

[tiny kraken]

ah, I meant the area of ABC

yeah divide that by 3 since H is centroid yeah

[opal plinth]

Right sure

[opal plinth]

@cedar obsidian ?

[cedar obsidian]

it should be half right?

[opal plinth]

No

Same base, but how do the heights compare?

[opal plinth]

.

What's the total length of the median when you say the centroid divides it in 2:1?

[cedar obsidian]

3

[opal plinth]

Right, so...

[cedar obsidian]

area is 3 times

[opal plinth]

Of what

[cedar obsidian]

the area of AG_1B is 1/3 times compared to that of AHB

[opal plinth]

No

G_1 is inside AHB, how can AG_1B be bigger than AHB?

[cedar obsidian]

right?

[opal plinth]

Yeah ok, 1/3

[cedar obsidian]

mb for that

[opal plinth]

So can you express the sum now?

[cedar obsidian]

[
\frac{16}{\sqrt{3}} + \frac{16}{3\sqrt{3}} + \frac{16}{9\sqrt{3}} + \cdots
]

[soft zealot]

BlackidoZΣ

[opal plinth]

We are not including AHB

[cedar obsidian]

oh yes sigma notation is for G_I

[opal plinth]

Yes

[cedar obsidian]

[
\frac{16}{3\sqrt{3}} + \frac{16}{9\sqrt{3}} + \frac{16}{27\sqrt{3}} + \cdots + \infty
]

[opal plinth]

Can you express this using sigma notation?

[soft zealot]

BlackidoZΣ

[cedar obsidian]

yes

[opal plinth]

Don't write +inf at the end that's nonsense

[cedar obsidian]

but at the end we want to find the infinite sum right a(r^n-1)/(r-1)

then why not directly do it

[opal plinth]

It's an infinite sum, not a sum where the last term is "infinity"

[cedar obsidian]

i am facing problem to convert it into sigma notation

i need to practice that again

[opal plinth]

You divide by 3 each time

That means you multiply by 1/3

What does repeated multiplication look like?

[cedar obsidian]

1/3^n

[opal plinth]

Right

And then the 16/sqrt(3) stays constant

[cedar obsidian]

oh got it

[
\frac{16}{\sqrt{3}} \sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^n
]

[soft zealot]

BlackidoZΣ

[cedar obsidian]

[
\sum_{n=1}^{\infty} \frac{16}{\sqrt{3}}\left(\frac{1}{3}\right)^n
]

[soft zealot]

BlackidoZΣ

[cedar obsidian]

now?

[opal plinth]

Now it's a geometric series

[cedar obsidian]

infnite gp right

[opal plinth]

Yes

Just be careful with the starting index

[cedar obsidian]

getting 8/root3

option c

wow its correct

[opal plinth]

[cedar obsidian]

when that question suddenly come in exam

altho at home i couldn't recall that property of median bymself

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@cedar obsidian Has your question been resolved?

[final saddle]

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[karmic glen]

I am tasked with finding the supremum and infimum of this set

[karmic glen]

I need hints

this is first analysis course homework at my uni

[] denotes the floor function

trivial inequalities on the floor dont yield anything

useful

[lime crest]

{n√2} is dense in [0,1] because √2 is irrational

[final saddle]

@karmic glen Has your question been resolved?

[karmic glen]

nice, this takes care of the supremum

what about the infimum

[cerulean epoch]

i think this was one of the first questions i asked in this server

[final saddle]

Channel closed

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[karmic glen]

.reopen

[final saddle]

✅ Original question: #help-36 message

[tiny gorge]

simulating it numerically for say n = 1 to 1 million and checking the min result is interesting

chances are you'll be able to recognize what the number is, and that gives you something to aim for

[karmic glen]

okay

ill try that

[cerulean epoch]

you can show that inf ≥ ||1/2✓2|| using the fact that ||2, 3 are not quadratic residues mod 4|| but i'm not sure how you would show this bound is tight

[karmic glen]

0.353553

idk what this number is its prolly related to sqrt(2)

[karmic glen]

its not:(

[karmic glen]

oh

mb

sorry

[cerulean epoch]

actually i think you can get to tightness using the ideas of this proof

[scarlet sequoia]

The usual idea is to ||take an approximation p/q ≈ sqrt(2) through pell equation, and look at a_q||

[final saddle]

@karmic glen Has your question been resolved?

[karmic glen]

$m = \lfloor n \sqrt{2} \rfloor$

We have $n(n\sqrt{2} - m) = n \frac{2n^2-m^2}{n \sqrt{2} + m}.$

Since $ n\sqrt{2} > m. $ We get $2n^2 - m^2 > 0$. Because m and n are integers this means
$2n^2 - m^2 \geq 1$.
$$a_n = \frac{n(2n^2 - m^2)}{n\sqrt{2} + m} \geq \frac{n \cdot 1}{n\sqrt{2} + m} = \frac{1}{\sqrt{2} + \frac{m}{n}}$$
Since $m = \lfloor n\sqrt{2} \rfloor$, we know that $m < n\sqrt{2}$, which implies $\frac{m}{n} < \sqrt{2}$. Therefore:
$$\frac{1}{\sqrt{2} + \frac{m}{n}} > \frac{1}{\sqrt{2} + \sqrt{2}} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$$

[soft zealot]

k

[karmic glen]

okay I got the bound thank you

[karmic glen]

okay this seems like the key insight for the infimum thats pretty cool

imma look into this I didnt know you could do this

oh and thank you also

.close

[final saddle]

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[alpine shadow]

Help

[deep condor]

,rcw

[alpine shadow]

B isn't the ans

[deep condor]

so what have you tried?

[alpine shadow]

Uh so I tried like solving the line and the parabola tgt to get sm rel bw y and m or x and m and finding the triangles area but I'm getting confused in like when I make the quadratic eqn (after solving parabola and the line) what do I do after it

[final saddle]

@alpine shadow Has your question been resolved?

[floral stream]

if u have a phone, you could send pictures of what uve done

i can help look for errors or give hints

[final saddle]

@alpine shadow Has your question been resolved?

[alpine shadow]

Sure just a sec

,rcw

[trail mango]

bad handwriting

[alpine shadow]

😭

[final saddle]

@alpine shadow Has your question been resolved?

[naive spoke]

m is such that maxima is attained find the value of t for maxima and put it in the eq with m

as F lies on y = mx+c asw

[alpine shadow]

Nvm I got it tysm

.close

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[heavy plaza]

Given $g:\mathbb{Z}/39\mathbb{Z}\to\mathbb{Z}/54\mathbb{Z}$ is a ring homomorphism, find all values of $g(1)$.

[soft zealot]

lyric

[heavy plaza]

so im trying to find all possible homomorphisms here, but i think i need help in my understanding

what i have so far is that since we need $g(0)=0$, $g$ must have the form $g:x\mapsto m\cdot x + 54\cdot k$ for some $m\in\mathbb{Z}$ and for any $k\in\mathbb{Z}$, so we can say $g:x\mapsto m\cdot x$

[soft zealot]

lyric

[heavy plaza]

so like trying $m=1$, im trying to see if thats a homomorphism. we require $g(x+y)=g(x)+g(y)$. so if we let $x=y=30$, how would we define $g(x+y)=g(60)$?

[soft zealot]

lyric

[heavy plaza]

since g is defined only on $\mathbb{Z}/39\mathbb{Z}$?

[soft zealot]

lyric

[heavy plaza]

guess it makes sense we'd take the input mod 39, so thatd show g(60)=g(21)=21, g(60)=g(30)+g(30)=30+30=60=6 so m=1 doesnt work

if im right so far, how would i proceed in finding all possible m?

[pure sorrel]

hello

im new

[heavy plaza]

[grim nebula]

your ring homs aren't required to send 1 to 1?

[heavy plaza]

why would they?

[grim nebula]

err

that's part of the definition of a unital ring homomorphism

commonly called just a ring homomorphism

[strange sparrow]

maybe it's actually a rng homomorphism

[grim nebula]

because people often deal with unital rings only

anyway

since 1 generates Z/39Z, you only need to worry about defining g(1)

since for example g(2) = g(1) + g(1)

the one requirement is that 0 = g(0) = g(39 * 1) = 39 * g(1)

[heavy plaza]

assuming g(39)=39 ?

oh wait no gotchu

[heavy plaza]

how are you sure this is the only requirement?

[grim nebula]

well

think about the ring homomorphism requirements

g(x + y) = g(x) + g(y)

but since every element of Z/39Z is just 1 + ... + 1 a bunch of times

then this will always hold once you decide what g(1) is and then define g(1 + ... + 1) = g(1) + ... + g(1)

similarly for g(xy) = g(x)g(y)

hmmm

i might be lying actually

im definitely lying

g(1^n) = g(1)^n

clearly i never work with non-unital rings

[heavy plaza]

these rings aren't unital?

[grim nebula]

nah

[heavy plaza]

1?

[grim nebula]

1?

wdym

[heavy plaza]

is 1 not a multiplicative neutral element in both?

[grim nebula]

thats the definition of a unital ring

but not the definition of a unital ring homomorphism

[inland kettle]

unital homomorphism is what snow's talking abt i think

[grim nebula]

when you define an object in mathematics, you should also define their morphisms at the same time

so when i talk about unital rings, i really also mean with homomorphisms forced to send 1 to 1

[heavy plaza]

ohh k i got u

[grim nebula]

anyway

[grim nebula]

you need this additional requirement that is usually automatic for unital rings

or just like

g(1^2) = g(1)^2 is enough to guarantee that

[heavy plaza]

oh k i can see that

so you're saying any solution x=g(1) to x^2=x mod 54 would give a valid definition of a homomorphism?

[grim nebula]

yes, unless i'm lying

well

wait

39x = 0 is the other one i said initially

[heavy plaza]

alr, so to define a valid homomorphism, it suffices to define g(1) from the fact that addition is linear. then from the fact that multiplication is linear, we have the conditions that g(1)=g(1)^n for all n, and 39*g(1)=0.

g(1)=0 satisfies both, g(1) satisfies the first but not the second, and there are no other solutions to the first

does that sound right?

[grim nebula]

did you miss something in the second part of the sentence

[final saddle]

@heavy plaza Has your question been resolved?

[heavy plaza]

i guess ;-;

[grim nebula]

18^2 mod 54 is 0

you need x^2 = x mod 54

[heavy plaza]

ok there has to be a better way of solving this

x^2=x mod 54 => x^2 - x - 54k = 0 => x = 1/2 +- 1/2 * sqrt(1 + 54k) => 1+54k is square => im just plugging in values of k to find square numbers

[grim nebula]

uh

39x = 0 mod 54 => 13x = 0 mod 18

13 and 18 are coprime so x = 0 mod 18

that gives 0, 18, 36 as candidates

you check their squares and only 0 works

[heavy plaza]

u absolute legend

just tried it on a similar problem (Z/32Z -> Z/100Z) and it's correct

[grim nebula]

nice

[heavy plaza]

my only issue still is thinking to define g by its action on 1, but from there i can now see how you'd derive those conditions

thanks so much <333333

[grim nebula]

if you've worked with vector spaces or abelian groups, its essentially the same idea

when you want to define a homomorphism of these algebraic structures, it suffices to say where the generators go

subject to the relations they satisfy

[heavy plaza]

ah i see, i'll keep that in mind when working with them then

[grim nebula]

indeed

[final saddle]

@heavy plaza Has your question been resolved?

[final saddle]

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[final saddle]

Available help channel!

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[raven hawk]

i am lost here

i dont know how to approach

[abstract bramble]

how many are in the large circle

[raven hawk]

80

[ocean marlin]

ok so

there are 3 circles

and two types of balls

@raven hawk u still there?

[raven hawk]

yes

[ocean marlin]

cool

[ocean marlin]

first you have to figure out what you are looking for

Can u tell me that?

[raven hawk]

footballs

[ocean marlin]

ok

how many types of balls are there?

[raven hawk]

2

[ocean marlin]

Good, basketballs and footballs

so the entire shipment contains balls

meaning if a ball is not a basket ball, then it has to be a?

[raven hawk]

a football

[ocean marlin]

Good

[raven hawk]

interesting

[ocean marlin]

now lets take a look at the diagram

[raven hawk]

ive never seen a venn diaram like this one

[ocean marlin]

theres a circle that includes basketballs right?

[raven hawk]

yeah, the not basketballs is fooballs

[ocean marlin]

correct

so how many footballs are there

[raven hawk]

60

[ocean marlin]

good

now there is a large section right

[raven hawk]

yes, 40 large footballs

[ocean marlin]

good

thats your first answer

now if you have a total of 60 footballs

and 40 are large

how many are small

[raven hawk]

20

[ocean marlin]

good

now can u find out how many large basketballs there are?

[raven hawk]

40

[ocean marlin]

Exactly

now the questions asks large basketballs to small footballs

so that would be?

[raven hawk]

2:1

[ocean marlin]

Great

thats the second question done

[raven hawk]

thanks

[ocean marlin]

Np man

if u have any other questions, u can ask them here, or if ur done just type .close

[raven hawk]

all good thanks for the help

.close

[final saddle]

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[final saddle]

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[left belfry]

hi

[left belfry]

3+3

please help me

I am very bad at math

[night raft]

why it's 6, but you shouldn't use the help channels to troll

[left belfry]

3+3x9-2x88/3a

my math is linear

[loud sundial]

<@&268886789983436800> troll

[left belfry]

algebra

[formal trail]

@left belfry don't abuse the help channels like this.

.close

[final saddle]

Channel closed

Closed by @formal trail

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[vapid torrent]

What if hes not trolling ☠️

Some people struggle in math

I know people in 6th grade who struggle with 2x+6=16

[night raft]

I'd say that's different to asking 3 + 3 since it's basic arithmetic. But ofc people with dyscalculia exist