help-42

[steady dome]

F(g(x))

[untold summit]

yes thats correct

what is g(f(x))

[steady dome]

Idk

X^2-3

[untold summit]

yes

ok lets go back to log(a-x)

log(a - x) = f(g(x))
you already identified g(x) = a - x

so we have log(a-x) = f(a-x)

so what does f(x) = ?

[steady dome]

Log x

[untold summit]

yes perfect

ok now we can use chain rule

f'(x) = ?

[steady dome]

1/x

[untold summit]

yep

g'(x) = ?

[steady dome]

-1

[untold summit]

yep

so d/dx log(a-x) = ?

[steady dome]

-1/(a-x)

[untold summit]

no

d/dx f(g(x)) = f'( g(x) ) g'(x)

[steady dome]

This s

[untold summit]

yes, do you understand why now

[steady dome]

Yes

[untold summit]

ok if d/dx log(a-x) = -1/(a - x) then what is int 1/(a-x) dx

[steady dome]

-loh(a-x)

[untold summit]

yep nice

only thing is when we integrate to a log you need absolute value signs

[steady dome]

Okok

[untold summit]

$\int \frac{1}{x} \dd x = \log |x| + C$

[potent lotus]

Acman

[steady dome]

$\int \frac{1}{-x} \dd x = \log |x| + C$

[potent lotus]

vamprem

[untold summit]

no

[steady dome]

Where is the negative sign

[untold summit]

it should be in front of log

[steady dome]

It should -logx

[untold summit]

yes

[steady dome]

Why showing wrong

[untold summit]

i dont understand what youre asking

[steady dome]

This

[untold summit]

yes i see that picture

i dont understand what youre problem is

the picture is not correct

[steady dome]

Where is -

[untold summit]

you didnt write it ....

[steady dome]

Yes

Int 1/-x == -log x

[untold summit]

yes

[steady dome]

Ok

I understand it now

And also figure out where I was wrong

Thanks

[untold summit]

nice

nws

[steady dome]

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[calm coral]

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[pure breach]

gm people

[unique jackal]

why’s this in a help channel though? :p

[pure breach]

i was typing

but got busy with soemthing else

so theres this question https://math.stackexchange.com/q/442654/1126252 thats in my part B workbook

its easy

but uhh i terribly misread this (somehow) as the following:
The integral part of the least positive value of a for which the maximum value of $4x^2 - 4ax + a^2 - 2a + 2=0$ on $\left[ 0,2 \right]$ is $3$ is

[potent lotus]

rak³en

[high lotus]

There's already help avaliable at the link.

[short prawn]

Is there a factoid for "Don't use polite greeting, just state the question already"?

[pure breach]

yes, but i just like to use a polite greeting

[pure breach]

anyhow so yeah this is the main question

what i have tried is complete bs

i have 0 idea what to do here

<@&286206848099549185>

[velvet osprey]

ok wait so

uhhh

what's that =0 doing there

@pure breach are you sure you meant to put the =0 there

or did you mean something like this

Let $f(a,x) = 4x^2 - 4ax + a^2 - 2a + 2$ and let $$A = { a > 0 : \max_{x \in [0,2]} f(a,x) = 3}.$$ Find $\floor{\min(A)}$.

[potent lotus]

Ann

[velvet osprey]

@pure breach is this what you meant

[pure breach]

by accident that happened

[pure breach]

and yes

[velvet osprey]

ok right

f is a quadratic in x

maybe we can rewrite it in a more convenient way

[pure breach]

errr i found the roots and they arennt nice

so maybe CTS? (completing the square)

[velvet osprey]

roots are useless

completing the square seems like the way to go

youll get (2x-a)^2 + g(a)

[pure breach]

yes

okay

O

OH

but how does that help

[velvet osprey]

you need to compare f(a,0), f(a,2) and f(a,a/2) (the latter only assuming a/2 falls in [0,2])

[pure breach]

thats for minimum no?

[pure breach]

but if it doesnt? we need that case too do we not

[velvet osprey]

then you dont

and only look at f(a,0) and f(a,2)

whichever is bigger

actually

you're right, x=a/2 will only ever give a local min

[pure breach]

oh and there cant be a middle value

[velvet osprey]

we're in fact guaranteed to not want that

so just f(a,0) and f(a,2) is what we need here

[pure breach]

no its a global min since its the vertex

[velvet osprey]

ok sure global

point is it's a minimum

[pure breach]

and we dont have to look at anything between 0 and 2 because quadratic is either increasing on decreasing if the vertex isnt there

and if it is there then we know 0 and 2 cant be the roots

so again its simply decreasing after 0 til vertex and increasing after 2 (actually vertex)

[pure breach]

yes

[velvet osprey]

.

this should be your takeaway

f(a,0) = a^2 - 2a + 2
f(a,2) = 16 - 8a + a^2 - 2a + 2 = a^2 - 10a + 18

[pure breach]

hmm i think f(a,0) > f(a,2) almost always?

well not almost always

solving that gives me

8a>16

a>2

and if its below it gives me a<2

wait no

[velvet osprey]

you want max{f(a,0), f(a,2)} = 3

[pure breach]

why tf am i doing this

just set it =3

yeah

[pure breach]

i just set f(a,0)=3 and f(a,2)=3 separately

then solve for a

and whichever gives me the max

[velvet osprey]

sure

[pure breach]

is the answer

right

[velvet osprey]

yes

dont forget a > 0 as well

[pure breach]

oh right good point

ty

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[tropic moth]

im struggling with point D

[fringe reef]

what is the height at D?

[tropic moth]

hm is it zero?

[fringe reef]

yes

if you go a bit to the left it's +1 and if you go a bit to the right it's -1

so it cannot be a local maximum or minimum

[tropic moth]

right right

[fringe reef]

question now is whether it's a saddle point

[tropic moth]

can it be a saddle point?

[fringe reef]

you tell me :p

[tropic moth]

well Fx and Fy needs to be zero

[fringe reef]

look at what happens around D

in a small region

[fringe reef]

that's also true but idk if that's going to be particularly helpful

or actually yeah it's helpful

what is Fy at D?

[tropic moth]

how do we check fx fy values from a graph is where im confused at

[fringe reef]

hint: note the countour line on which D lies

[tropic moth]

itss on y line

[fringe reef]

yes but how did you know that the value at D is zero?

[tropic moth]

the y line has zeros on it lol

[fringe reef]

yes

so the value along the y line is the same

so Fy is?

[tropic moth]

oh its zero aswell

[fringe reef]

indeed

now let's check Fx

think about what happens if you go a bit to the right and a bit to the left of D

[tropic moth]

x line is in the 0 aswell but D isnt on the x line

[fringe reef]

yes you won't get an exact value

but we don't need it

[tropic moth]

it would change values

[fringe reef]

yes, in what way?

[tropic moth]

left it would get bigger right smaller

[fringe reef]

yes

[tropic moth]

so its like inbetween a slope

ooh thats not a saddle point then?

[fringe reef]

yep

Fx isn't 0

[tropic moth]

okkk that makes a lott of sense

thank uu

.close

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[knotty jay]

Need help, I will let you know what I've done so far

[knotty jay]

1. I added a^2(n+1) on both sides

Gimme a second

But answer seems to be off

[odd robin]

I think they want you to find the generating function for the recurrence and then plug in n=20

[knotty jay]

Like how

It's for high school math btw (Jee Advanced)

I don't know recurrence functions yet

[odd robin]

you can convert the above recurrence to a linear recurrence relation by setting b(n)=[a(n)]^2

Oh ok mb

[unkempt drift]

better learn them then

[knotty jay]

I have jee advanced in 8 days and it wouldn't be in my best interest to learn much right now...

[odd robin]

There's a shortcut for solving linear recurrences with constant coeffecients

If you want I could teach you

[unkempt drift]

characteristic polynomial first then ye

[knotty jay]

I did originally see the AI solution yea

But for say I don't do this and instead my method (which seems valid enough for me) my answer doesn't match

[odd robin]

It should ideally match, wait I'll check

[knotty jay]

for that 4 in gp series formula it ^19 which I took in the next step but wrote ^18 before

[flat quiver]

recursive functions grow at a rate similar to the golden ratio. find the coefficient of growth

[knotty jay]

yoooooo high school math man chill

[odd robin]

What's the answer, 21?

[flat quiver]

try the characteristic equation? mayhaps more intuitive

[odd robin]

Also you're not allowed calculators in JEE advanced, then how'll calculate the sqaure root of that large number?

[knotty jay]

I didn't I just ignored 7/2^19 and 1/2^20 cuz binomial expansion

and they gave fractional part anyway

[odd robin]

Hmm fair

[knotty jay]

Okay seems to me answer will be 21

but occording to me it's 39

[flat quiver]

there’s a nice component where by each term appears to be +-10 from a perfect growth of 4x from term to term

where if you just manually calculate first like 6-7 terms by hand you identify it

regardless you’ll realize each term is simply 2 less than a power of 2

this expansion takes the form of 2^(n+3) - 2 i think

[flat quiver]

the giveaway here is that the answer in a non calc exam is asking for a 2 base log (implying 2^x growth normally) and it’s a greatest integer function which means it’s probably modeled similarly to that growth

this means the answer is that the 20th term of this sequence will be 2 less than the 23rd term of the sequence, which means my intuition would guess the answer is 22

[odd robin]

Shouldn't there be a +14 at the end

And also shouldn't the alternating sum start from 4^20?

I did the same calculation as in this method and ended up with a(20)^2 = 2^42+2

[calm coral]

@knotty jay Has your question been resolved?

[knotty jay]

yea true

[knotty jay]

yea cuz I messed up signs I kept -14

[knotty jay]

aight mathematically I'll check it all

ty

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[halcyon crown]

4/b

[calm coral]

@halcyon crown Has your question been resolved?

[formal pulsar]

Do you mind translating it into English? I think that would best option for you get help.

[calm coral]

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[umbral onyx]

Hello guys, Iam solving this diff equation. Iam trying to find the particular solution

[umbral onyx]

The problem is, that Iam not sure whether I should write *e^((4/3)x) or 4e^((4/3)*x) after the bracket

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@umbral onyx Has your question been resolved?

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[crystal breach]

can someone xeplain how do i find the gradient for this

[crystal breach]

its using the m= tan theta rule

[unreal isle]

what can you say about the blue angles

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[wet badge]

Hello!! So I have this assignment right? It's about practical maximum and minimum problem for derivatives. And I'm COMPLETELY stuck. Any help would be greatly appreciated

[wet badge]

(My main problem is that just don't really know how to start with solving B)

[dull wagon]

use the info on volume

you also haven't used y for your height

[primal smelt]

How do you calculate the volume

[dull wagon]

by applying the formula for volume of a prism/cuboid

[wet badge]

2x times x times y

[primal smelt]

And that equals?

[wet badge]

2x^2y

[primal smelt]

True, but look at the question

2x^2 y = ?

[wet badge]

Ah yeah

[primal smelt]

Now calculate the surface area

[wet badge]

oooooooh

I get it now

[wet badge]

yeah you substitute the y and then you just start moving stuff till it becomes the Area that matched the equation

thanks alot

.close

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[velvet osprey]

did GPT generate this problem for you

[calm coral]

@noble narwhal Has your question been resolved?

[noble narwhal]

no

it just rephrased it lol

<@&286206848099549185>

[civic dirge]

Problem is jibberish

[noble narwhal]

even question1?

q2 doesnt count

[noble narwhal]

even q1?

[velvet osprey]

well... hm

[noble narwhal]

solve in Q[X] the equation: (X² + 1)P + (X² - X + 1) Q = X - 1

like

we need to find all possible solutions

P and Q

[leaden thunder]

show the book you got the question from

[noble narwhal]

it's just a random homework exercise idk where they got this

why is it gibberish?

[leaden thunder]

screenshot the original page with the exercise

[noble narwhal]

ok

[velvet osprey]

i mean at least Q1 is sensible as-is in that the solution set isn't empty, but based on how Q1 plays out Q2 is a weird followup

[noble narwhal]

no you can ignore Q2

[noble narwhal]

i mean there's no screenshot it's just a written question

[leaden thunder]

then screenshot that

[noble narwhal]

but it's written in my notebook and its exacltly like how i wrote it here

qustion 2 is a mistake its not supposed to be here

[leaden thunder]

oh god

[noble narwhal]

i think the question makes sense

like

i feel like it's possible

but i dont know how to find the answer

[mortal orbit]

bezout couple for A and B and then multiply by X-1?

you know how to do euclid algorithm for polynomials?

[calm coral]

@noble narwhal Has your question been resolved?

[calm coral]

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[lethal heart]

for part b of this question after plugging into the equation i found 18xz^t^2-3xzt^2

but the scheme says 9xz^2t^2 im confused on how it came to this

should it not give 18 instead of 9 or is the second partial differentation not included for some reason?

<@&286206848099549185>

!status

[calm coral]

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

[lethal heart]

5

<@&286206848099549185>

[lethal heart]

<@&286206848099549185>

[calm coral]

@lethal heart Has your question been resolved?

[lethal heart]

<@&286206848099549185>

[lethal heart]

<@&286206848099549185>

[calm coral]

@lethal heart Has your question been resolved?

[calm coral]

@lethal heart Has your question been resolved?

[eternal shard]

3 x 3 = 9 the pdv wrt y is 0

[calm coral]

@lethal heart Has your question been resolved?

[calm coral]

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[potent lotus]

Rajvir

[calm coral]

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[idle jacinth]

If CP = x , then what x value would give us the maximum area? I tried but i think i need derivative, but you shouldn't use derivative here

[manic oracle]

you want to maximise the area of NOPM?

or ABC?

[idle jacinth]

oh of course NOPM

the rectangle

[rocky flower]

I’m pretty sure they want to maximise area of NOPM. Area of ABC is the same (90*70/2)

[idle jacinth]

first, we need to find the height of the rectangle right?

and i thought if we can use Thales' theorem, the one that Thales of Miletus used to calculate the height of the pyramids
like x/60 = OC/92.2 = OP/70 ?
or like (x/2)/30 = AN/76.2 = MN/70 ?

and then maybe putting everything equals everything something like that

[manic oracle]

the very fast solution is "fold the triangle onto the rectangle to show the rectangle is at most half the area of the triangle"

[idle jacinth]

hm

[manic oracle]

show A >= A', B >= B', C >= C' and D >= D'

[idle jacinth]

ok

[manic oracle]

the slower solution is "make an expression for the area of the rectangle in terms of x"

and the find what value of x maximises it

[idle jacinth]

and it's here where there's a problem since you can't use derivative in this problem

[manic oracle]

it's a quadratic

you can find maximum of quadratic conceptually

[idle jacinth]

ik it's quadraticc but can't seem to know how would i know the max without derivative, by completing the square and looking for the coordinates of the vertex?

[manic oracle]

yeah

[idle jacinth]

ok

thanks

i guess that's what i needed to be clarified

i'm gonna work on it from here

[stoic pagoda]

there would be no need to complete the square btw

as in, not fully to get the vertex form

just -b/2a is needed

[idle jacinth]

oh ok

i guess i'll close this chanel noow

.close

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[warm stone]

help

[warm stone]

Suppose ( X_i ) are independent ( N(\mu, \sigma^2) ) random variables for ( i = 1, 2, \dotsc, n ).

Will
[
\frac{ \bar{X} - \mu }{ S / \sqrt{n} }
]
always be ( t_{n-1} ) distributed?

[potent lotus]

Beluga

[velvet osprey]

wouldnt it be NORMALLY distributed??

or wait

what's S

[warm stone]

Sample standard deviation

[velvet osprey]

ah hm. nevermind then

[warm stone]

If it was $\sigma$ then it would be standard normally distributed and not just normally distributed😉

[potent lotus]

Beluga

[calm coral]

@warm stone Has your question been resolved?

[warm stone]

<@&286206848099549185>

[calm coral]

@warm stone Has your question been resolved?

[calm coral]

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[empty dune]

This problem is impossible, right? D must not include 1 or 3, but if this is the case, then you cannot add two rectangular regions together to get set R.

[calm coral]

@empty dune Has your question been resolved?

[calm coral]

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[dapper bramble]

A minimal spanning tree with no injective edge function is still unique if the duplicate wheighted edges with unique vertices either both close cycles (all can be added) or neither closes cycles (all can ne added)

[dapper bramble]

Is my assumption correct?

[calm coral]

@dapper bramble Has your question been resolved?

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Use .reopen if this was a mistake.

[calm coral]

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• 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>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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[fickle flume]

i still dont get what went wrong here

or what the actual answer should be

[fickle flume]

here

the negative sign

was my attempt at correcting for the absolute value of cosx when it is negative

so i replaced cosx with -cosx inside, i just moved out the -1 outside the integral

when i asked wolfram alpha it said the volume should be 2pi^2

that makes me think i've gone wrong in many places

because i dont understand how i can rid of the -4pi in my result without changing the 2pi^2 as well

i might have found what i did wrong

verifying on paper rn

nvm still stuck

[old falcon]

I'm sorry whats the original problem here

[old falcon]

this integral near the bottom?

[fickle flume]

yes sorry, it's to rotate cos(x) around the y-axis, im using the shellmethod

0<=x<=pi

i think i found what went wrong, at least changing it gives me the right answer

[old falcon]

alright

[fickle flume]

this should be -1 right?

ignoring the c i mean

[old falcon]

why is there a c

[fickle flume]

idk im bad at integrals and idk when it appears tbh

[old falcon]

isnt this a definite integral

[fickle flume]

it is but i thought there was still a c

but they just took eachother out

[old falcon]

definite integrals dont get a C

[fickle flume]

ah ok

WELL ok except for that

[old falcon]

yea, this is -1

cos pi = -1

cos pi/2 = 0

[fickle flume]

yes ok and when i calculated it i put it as 1

now the -1 and +1 cancel eachother

and i get the wolframalpha answer of 2pi^2

this might be a dumb question

but why dont definite integrals get a c?

i get that they are called definite

so in that sense they should be sure of themselves

no flimsy c

[old falcon]

+C shows up when for antiderivatives

when theres multiply answers

"family of solutions"

definite integrals where they exist are just a value

not a function

well, i guess im being pretty general here

but thats the loose idea

[fickle flume]

i think that's good for me

and for initial value problems

that's when the c is important

that's what i'm solving for

in an initial value problem

[old falcon]

I feel like those would usually show up in a differential equation?

[fickle flume]

oh yeah

[old falcon]

those usually have indefinite integrals, youre looking for antiderivatives

[fickle flume]

first order and second order

when do i get multiple cs?

[old falcon]

so for sure you'll pick up a C or so somewhere

[fickle flume]

in the second order one

dont i get like a c_1 and c_2 somehow

[old falcon]

it can happen in different ways

yea, if you wind up with two LI solutions

then any linear combination of them is also a solution

[fickle flume]

what does LI mean?

[old falcon]

i guess im introducing too much theory

like for y'' = -y

we really find two solutions

so each one gets a constant

we are maybe getting off track

[fickle flume]

i only watched org. chem tutor video on it a long time ago, i remember there were 3 different cases

and i think it was for second order differential equations

[old falcon]

hmm maybe youre thinking of harmonic oscillator

[old falcon]

were definitely off track now haha

[old falcon]

yea, we can manufacture this being necessary

[fickle flume]

we're on-track in the sense that i need to learn this stuff too, off-track to the original question, i basically started a new track

idk if that is allowed

[old falcon]

idk how much you wanna just talk about random stuff tho

yea i dont think it really matters

[fickle flume]

for me this isn't random it is on my exam

[old falcon]

you get how we get r?

like, we assume a form of the solution

normally you have something like $y'' + by' +ay = 0$

[potent lotus]

jan Niku

[fickle flume]

no tbh i dont get how we get r, when i tried this on my first go at this exam (which i failed) i just tried to memorise

[old falcon]

and you assume that $y=e^{rx}$

[potent lotus]

jan Niku

[fickle flume]

because i didnt start studying early

[old falcon]

so, we want to find r. If we can find r, then we have the solution, its here

[old falcon]

you plug it into the diff eq here

we get $e^{rx} \qty( r^2 + ar + b) = 0$

[potent lotus]

jan Niku

[old falcon]

its nice that e^rx is never 0; we can just divide both sides by e^(rx) and we get something really familiar

[old falcon]

$r^2 + ar + b =0$ like you have here

[potent lotus]

jan Niku

[old falcon]

kind of

[fickle flume]

is it okay if i ask you to slow down?

[old falcon]

yea

sorry

[fickle flume]

i dont know how to put it in a non-rude way

np

sorry for not saying earlier

[old falcon]

lol its all good

[fickle flume]

im with you here

i can assume that y equals that

i dont really know why we are assuming that

[old falcon]

well, lets guess this form

[fickle flume]

yes sure lets

and then

[old falcon]

and see if we can find solutions

if we can, then nice

if not, then okay, try something else

[fickle flume]

we plugged it in and got this, i dont see how we would get this

[old falcon]

okay

can you find $y' = \dv x \qty( e^{rx} )$

[potent lotus]

jan Niku

[fickle flume]

yes re^(rx)

i think

[old falcon]

how about $y'' = \dv[2] x \qty( e^{rx})$

[potent lotus]

jan Niku

[fickle flume]

i think i see how we get it now, but im bad at this so

[fickle flume]

let me try this one still

[old falcon]

nah its good to go through step by step

[fickle flume]

i think it is r^2e^(rx)

[old falcon]

yea, thats right

[fickle flume]

then we break out e^(rx)

because every term has e^(rx)

[old falcon]

so $y' = re^{rx}$ and $y'' = r^2 e^{rx}$

[potent lotus]

jan Niku

[old falcon]

you can substitute them into $y'' +ay' + by = 0$

[fickle flume]

let me do this on paper

actually

[old falcon]

sure

[fickle flume]

and confirm we get what yo usaid there

[potent lotus]

jan Niku

[fickle flume]

okay yes

sorry im very slow

i see that we get that now

[old falcon]

so good to $e^{rx} \qty( r^2 + ar +b) = 0$?

[potent lotus]

jan Niku

[fickle flume]

yes

that makes sense

[old falcon]

and okay to divide by e^(rx)?

[fickle flume]

yes because we have 0 on RHS

so that gives us freedom

[old falcon]

its more like

we might be worried that e^(rx) is 0 someplace

[fickle flume]

mm and yes but i get that since we have that base

it can never be 0

it will always be some positive number

[old falcon]

well as long as r is some real number

e^x is either flat, or explodes, or shrinks

[fickle flume]

unless ^i can turn it neagtive

i dont remember what ^i does

[old falcon]

yea if r is imaginary we have some concerns

[fickle flume]

rotates or something

[old falcon]

but, lets hope its not

[fickle flume]

ok!

[old falcon]

so now were just solving $r^2 + ar + b = 0$

[potent lotus]

jan Niku

[fickle flume]

oh yeah lol just a quadratic

[old falcon]

yea

[old falcon]

which we still need to be sure of this

so we gotta make sure that a^2 - 4b > 0

[fickle flume]

if r^2 has a coefficient

[old falcon]

we can just divide it away

[fickle flume]

oh yeah

and also im not really used to that 4b version of the quadratic formula

[old falcon]

we can change the letters

i was worried to use c

i didnt want to introduce more issues

[fickle flume]

lol!

[old falcon]

but sure

$r^2 + br + c = 0$ is probably more familiar

[potent lotus]

jan Niku

jan Niku

[fickle flume]

the one im familiar with uses p and q

p for b and q for c

[old falcon]

[fickle flume]

let me take photo

[old falcon]

we can speak in general too

[old falcon]

okay

[potent lotus]

jan Niku

jan Niku

[old falcon]

so, no imaginary solutions

[fickle flume]

damn i dont understand that

[fickle flume]

like i dont get why we want this

[old falcon]

because we'd like real solutions for r

[fickle flume]

for you, you see that and go, we dont get imaginary solutions

for me, i dont see why we would get either real or imaginary

how do you determine that

also very nice cat picture

on your profile

[potent lotus]

jan Niku

[old falcon]

looking at this we have a worry about the square root

[fickle flume]

right if it becomes negative

[old falcon]

normal anxiety is if we are taking a square root of a negative number

[fickle flume]

i get we get solutons that aren't purely real

[old falcon]

so, we just want to make sure we arent taking the square root of something negative

if thats something we can make sure of, then theres nothing else that can make it imaginary

so we just enforce p^2-4q > 0

[fickle flume]

ok

idk how to enforce that

but let's say it is non-negative

[old falcon]

we actually dont get to choose

[fickle flume]

oh

[old falcon]

its given by the problem

but, we can talk about cases where its true

[fickle flume]

oh wait that's what you meant

by this?

like, this is the problem

[old falcon]

im just trying to be clear which solutions were talking about

[fickle flume]

you are manufacturing this "solution-case"

[old falcon]

these kind of methods where you assume a form of the solution and go blundering forward

usually involve a few loops of having to introduce additional restrictions

and then you end up with cases

[fickle flume]

mmm you're using some very big words now

i'm starting to shake in my boots

[old falcon]

im getting sleepy lol

[fickle flume]

lol np ty for putting up with me

[old falcon]

i dont usually talk like this i dont think

anyways

what were we doing

in general i mean

[fickle flume]

this

oh in general

[old falcon]

sure, yea

[fickle flume]

i was just curious about 2nd order differential equations

[old falcon]

so lets say we get two distinct solutions for r

then youre done

[fickle flume]

ok so

i did all the steps we went through

and

the quadratic

at the end

and got two real solutions for r

is that what we're saying happened in this case?

[old falcon]

yea

if we just get real solutions for r

[fickle flume]

yes

[old falcon]

theres two things that can happen

one is that we get two distinct r

[fickle flume]

oh

or a double root

?

[old falcon]

yea

[fickle flume]

ok

[old falcon]

the double root is a bit of a concern

say theres one root r1

we might want to write $y = c_1 e^{r_1 x} + c_2 e^{r_1 x}$

[potent lotus]

jan Niku

[old falcon]

but thats just $y = (c_1 + c_2) e^{r_1 x}$

[potent lotus]

jan Niku

[old falcon]

we've only found one solution

[fickle flume]

that doesnt seem good

[old falcon]

we just make up another one

[fickle flume]

okay!

lol

[old falcon]

$y = c_1 e^{r_1 x} + c_2 x e^{r_1 x}$

[potent lotus]

jan Niku

[fickle flume]

oh yeah i remember one of my tutors

tryign to remember why that x was there

why the hell is it there, we made it up

we needed one more

[old falcon]

its a way of manufacturing another solution

[fickle flume]

because we had two unknowns?

but the solution works

so it wasnt a random manufacturing

[old falcon]

this is differential equations course?

if you finish it you'll be able to convince yourself

[fickle flume]

the course is called single variable analysis

but im pretty sure it is just calc 2

[old falcon]

you will probably see techniques that allow you to do things like this if you are talking about second order differential equations

but its a long story

[fickle flume]

unfortunately this course is over

[old falcon]

oh

[fickle flume]

so no more lectures for it, at least not until after summer

i need to do re-exam

if i fail in june i will go in august

[old falcon]

hmm

well, its not too involved to work out this second solution thing

[fickle flume]

if nothing else i can ask my math tutors about that technique, if you are feeling tired/want to do something else

dont be afraid too

[old falcon]

but it is probably a severe detour

[fickle flume]

i wouldnt midn

mind

im actually curious about this stuff

unlike from when i first took the course

like just understanding

[old falcon]

we found one solution, right

[fickle flume]

why that x would appear

[old falcon]

$y_1 = c_1 e^{r_1 x}$

[fickle flume]

if we found the same solution twice

[potent lotus]

jan Niku

[old falcon]

but $y_2 = c_2 e^{r_1 x}$ is not another solution

[potent lotus]

jan Niku

[fickle flume]

y_2 = y_1

?

[old falcon]

c1 is arbitrary right

some constant

so $y_1 + y_2 = (c_1 + c_2) e^{r_1 x}$ is really just $y_1$ again

[potent lotus]

jan Niku

[old falcon]

its some arbitrary constant times e^r1 x

[fickle flume]

like c_1 + c_2 = c_arbitrary

[old falcon]

yea

[fickle flume]

yes im okay with that, im just trying to remember

uhh

where we got the c_1 and c_2

we went over

we just went a little fast im gonna scroll up

[old falcon]

theres not much to remember

say you have some y that solves the diff eq

try showing that cy solves it, too

well that'd be (cy)'' + p(cy)' + q(cy)

but derivatives are linear, the constant can come out

c (y)'' + pc(y)' + qc(y)

do you see what happens?

[fickle flume]

sorry im trying to follow

[old falcon]

sorry im prolly going too fast again

[fickle flume]

im just not rock solid

i think most people would be able to follow you just fine

[old falcon]

the important part is that, if y is a solution to some differential equation

then so is cy

some constant multiple of y

[fickle flume]

c is just scaling y

[old falcon]

yea

and its not crazy to show or some wild theorem

you can just plug cy directly in

factor out the c from each term

and thats it

so, thats why c

if y is a solution, so is cy, for any c

[fickle flume]

there are mutliple reasons why im having a hard time following, but i think the biggest one is just, we had different c's

for other situations

or wait

arbitrary c

was additions of non-arbitrary cs?

or

idk

[old falcon]

we write c_1 and c_2 and such but its really not that important, theyre just arbitrary

[fickle flume]

ok they are all arbitrary

[old falcon]

yea and if you get more comfort with this stuff you'll be able to feel out when theyre necessary

wanna keep going?

[fickle flume]

yes so

that part

we could break otu e^(r_1 * x)

[old falcon]

you can see why this doesnt work out

[fickle flume]

and after that

like okay wait so

we only got one r

and r is found

by that stage

and then, with these problems, often an initial value

thing

so i plug in x and y

[old falcon]

sometimes

[fickle flume]

but like you say

c_1 and c_2

i cant solve for them

that way

like two unknowns

still

and one equation?

[old falcon]

you can solve for one constant given one piece of information

[fickle flume]

but i dont get how another x would help tbh

[old falcon]

well, we dont have this weird solution combining problem

[fickle flume]

am i wrong in saying that

we cant solve for the two unknown c's

because we have one equation

and two unknowns

is that a bad way to understand

why it is a problem

[old falcon]

its maybe more correct to not think of the constants as unsolved

they're there because were trying to include all solutions

if were given more information about the situation, then we can pick out individual solutions

[fickle flume]

that is how im thinking

like

if we are given a point

an x and a y

then we cant solve for the c's

right?

[old falcon]

you can solve for one

[old falcon]

oh, here, well

we havent found all the solutions in what you posted just now

theres really only one constant

so that we can "solve" with one piece of information

[fickle flume]

damn i dont really get that part about there only really being one constant, you did explain earlier that c just scales y, but in all these problems there is a c_1 and c_2 right?

[old falcon]

its because we can rewrite it as $(c_1+c_2) e^{r_1 x}$

[potent lotus]

jan Niku

[old falcon]

c_1 + c_2 is just an arbitrary constant, a single one

[fickle flume]

okay yes and that becomes some constant

that's true

[old falcon]

yea

so, theres really only one constant here

[fickle flume]

but then what is the problem

with the double root

if we can solve for that big boy constant anyway?

[old falcon]

theory tells us that our equation should have multiple solutions

[fickle flume]

and it does have multiple, it is just the same one twice

[old falcon]

one twice is just one

[old falcon]

this actually is not that interesting and is just algebra

once you have as many pieces of info as you have constants, you just get a system of equations out

[old falcon]

this doesnt work because of the combining thing weve been doing, like here

you cant write one solution as two

because it will always end up just being equal to the one solution by itself

[fickle flume]

im not saying this to talk down on myself

but i swear i have a lacking sense of logic

let me paint you a picture and i wont be too long

i see i get a double root, and i think, ok, we only have one solution here

my mind doesnt wander off and think

"i should try to manufacture another"

i accept the one i got

and im happy with it

yeah thats it

i dont understand the need for this 2nd solution

you said

theory

theory tells us we should have 2

[old falcon]

were using the work second pretty loosely

its probably more accurate to say second half

[fickle flume]

is it just negative?

or something?

i dont know

[old falcon]

you can get it in a few ways

or, you can just remember to multiply by x

its not too hard to derive it

[fickle flume]

i kinda wanna know where that x comes from

[old falcon]

you'd just say $y_1 = c_1 e^{r_1 x}$

[potent lotus]

jan Niku

[old falcon]

and let's assume that $y_2 = u_1 (x) y_1$

[potent lotus]

jan Niku

[old falcon]

same thing as normal, we plug this into the ODE

do you want to do it?

theres some product rule type stuff

[fickle flume]

im very slow if there is the product rule

[old falcon]

or we can just look at what happens

[fickle flume]

i would prefer to look at it

if that's ok

like in practice i will probably just remember to multiply by x

but i wanna see where it came from

[old falcon]

we sub it into the diff eq

what we get is $u''_1y_1 + 2u'_1y_1' + u_1 y''_1 + p(u_1'y_1 + u_1 y_1') + q(u_1 y_1)$

[potent lotus]

jan Niku

[old falcon]

man algebra

[fickle flume]

ok now i feel bad i didnt know it would be this gruesome

[old falcon]

the trick is to spot the original ode in here

[fickle flume]

i can accept after doing the product rule correctly

[old falcon]

since we know $y_1 '' + py_1 ' + qy_1 = 0$

[potent lotus]

jan Niku

[fickle flume]

maybe stuff cancel

[fickle flume]

and yeah we get something that looks kinda like that

[old falcon]

we end up with $2u_1 ' y_1' + u_1''y_1 + pu_1' y_1=0$

[fickle flume]

and that's what y_2 is equal to?

[old falcon]

sorry, its equal to 0

since were assuming y2 is a solution

[potent lotus]

jan Niku

[old falcon]

this is easier than it looks

we know y1

its just a first order equation

we gettt

$u_1 ' \qty( 2 y_1 ' + py_1) + u_1 '' y_1 = 0$

[potent lotus]

jan Niku

[old falcon]

idk is this helpful

i feel like were in the weeds

[fickle flume]

i feel like im in the weeds but im basically choosing the weeds over just accepting a random x teleporting into an equation for, what is to me, no good reason

[old falcon]

okay

we can solve this

[old falcon]

$\frac{ u_1' (2 y'_1 + py_1) }{ -y_1 } = u_1 ''$

[potent lotus]

jan Niku

[old falcon]

this is seperable

[fickle flume]

oh yeah so we can get y stuff on one side

and u stuff on another

[old falcon]

yea

and we pretend that this is really a diff eq for u'

[fickle flume]

but we have u''

does that not matter?

[old falcon]

so $u_1 ' = \int \frac{ u_1' (2 y'_1 + py_1) }{ -y_1 } \dd x$

[potent lotus]

jan Niku

[fickle flume]

should the RHS not say u''?

[old falcon]

bleh maybe its easier with an extra step

say u_1' = v

[fickle flume]

wait wait wait!!

[old falcon]

haha

[fickle flume]

plz answer that

[fickle flume]

;~;

[old falcon]

rhs?

the integral?

[fickle flume]

this!

[old falcon]

oh wait maybe i did screw this up

[fickle flume]

what happened to u''

[old falcon]

hold on

well here lets do it like this

let u_1 ' = v

$\frac{ v (2 y'_1 + py_1) }{ -y_1 } = v'$

[potent lotus]

jan Niku

$\frac{ v (2 y'_1 + py_1) }{ -y_1 } = v'$

[old falcon]

so v' = u''

[fickle flume]

oh god ok

[old falcon]

$\frac{ v (2 y'_1 + py_1) }{ -y_1 } = \dv{v}{x}$

[fickle flume]

ok yes that's seperable still

[potent lotus]

jan Niku

[old falcon]

$\int \frac{ 2y_1 ' + py_1 }{ -y_1 } \dd x = \int \frac{ \dd v }{ v}$

[potent lotus]

jan Niku

[old falcon]

classic separable business

[fickle flume]

how do we get dy now

[old falcon]

the left hand integral?

you can try to work it out, if you want

it just ends up being $\int -(2r_1+p)$

[potent lotus]

jan Niku

[fickle flume]

okay o_o

[old falcon]

so

$u' = v = e^{-\int (2r_1 + p) \dd x }$

[potent lotus]

jan Niku

[fickle flume]

sure

that is integrating factor?

[old falcon]

i feel like something went wrong

anyways

[old falcon]

it looks like that, huh

[fickle flume]

i like the way it looks you made a really nice integrating factor and i think you did a very good job of it

[old falcon]

man i have a horrible feeling something went wrong

look im super tired

i can help maybe at another time

i think im not going to be super helpful right now

sorry to have wasted some of your time

[fickle flume]

ok it is no problem i hope you didnt feel forced to stay here

no it is ok

[old falcon]

nah i just am maybe not going to be too uhh

you know

[fickle flume]

i dont feel it was wasted

[old falcon]

another time

[fickle flume]

one

last thing

oh nvm

!!

it is ok

byee hope yo uget nice sleep

.close

[calm coral]

Channel closed

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Use .reopen if this was a mistake.

[calm coral]

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[frozen rampart]

hello

[frozen rampart]

i need hint on how to do this

I will translate

P is a point on 9 point circle of triangle ABC, connect AP and draw the perpendicular line to AP at P which meets BC extended at Q. Let X be on PQ such that XA is perpendicular to AQ. If H is the orthocenter of ABC, and D and M are the midpoints of BC and AQ respectively provs that's HX perpendicular to DM.

btw the diagram has extra points labelled and im gonna use them

[tough prism]

is this an oly problem or are the chinese kids too cracked 🤔

[frozen rampart]

So firstly DG is diameter of nine point circle as AH is height and thus AH perpendicular to BC and thus AK perpendicular to BC, since GKD is 90 then DG is diameter

[frozen rampart]

i am a chinese kid and this is an oly problem

[tough prism]

ah i see

[frozen rampart]

"原题是基本的几何题" 😭 一点也不基本

theres a lot to use

but i didnt want diagram to be too cluttered

i need help changing XH perpendicular DM into smth nicer

progress so far, i think 最后结论 might be useful

15m

<@&286206848099549185>

我等

Another thing I got was APCQ concyclic with M center

i have some stuff to use w.r.t the 9 pt circle

But the center seems not important in this question

am stuck

倒角 time

I also got that triangle XAP is similar to GDP since (in angles) XAP = AQP (similar) = PKA (AQKP conyclic) = PDG (PGKD concyclic)

And XPA = DPG = 90

essentially this

this isnt my first diagram btw the first one was smth i attempted for 1h and then gave up

(it was messy af)

anyone?

[calm coral]

@frozen rampart Has your question been resolved?

[calm coral]

@frozen rampart Has your question been resolved?

[frozen rampart]

anyone?

@kindred folio

hs dont bash this pls

[calm coral]

@frozen rampart Has your question been resolved?

[calm coral]

Channel closed

Closed by @frozen rampart

Use .reopen if this was a mistake.

Channel closed

Closed by @frozen rampart

Use .reopen if this was a mistake.

[frozen rampart]

.reopen

[calm coral]

✅

[frozen rampart]

Okay

[full quarry]

dude ngl i have no clue how to prove it

[frozen rampart]

sad