help-36

[fringe vault]

where does the

how does it expand to that

[edgy mauve]

Huh

Wait

So (x-2) is a

2(sqrt(2)) is b

U hv it in a²-b² form

U can break it down to (a+b)(a-b)

Cant u

[fringe vault]

but wheres the

-4-4 from

[grizzled lily]

He added and subtracted 4 like it never existed so you won’t use general formula

[edgy mauve]

See u did it urself here

Here

So i thought u would get it

[crimson socket]

Why don't you just use the quadtaric equation to solve for x?

[fringe vault]

i jst followed a formula 😢 i dont get it

[fringe vault]

already did

[crimson socket]

oh

[fringe vault]

i need to shoe it with factoring

[edgy mauve]

I'll explain

So (x-2)² has x²-4x+4

Do we understand that

[fringe vault]

mhm

[edgy mauve]

But the guven eqn has -4 instead of +4

Ryt?

[fringe vault]

yea

[edgy mauve]

So what we want to do is
We want to for a whole square which would be
(x-2)² here

And to get that we will put a +4 in the eqn

But +4 wasnt in the eqn initially so we need to neutralize it

That's y i added another -4 so that
+4-4=0

Let me know if u dont understand

[fringe vault]

so like that?

in a way

[edgy mauve]

The -4 shouldnt be in the bracket

Cuz -4 isnt a part of (x-2)²

[fringe vault]

ohh

how did you get the part afte (x-2)^2 -8?

[edgy mauve]

-4-4=-8 ryt?

[fringe vault]

yes

how did you get the part after it tho

[edgy mauve]

Sqrt 8 gives u what

[edgy mauve]

I m on it

[fringe vault]

2(sqrt2)

[edgy mauve]

Exactly

So when u square 2(sqrt(2)) ,u get 8

Am i wrong?

[fringe vault]

yes

why is it a b tho

like why do you do that

[edgy mauve]

Okay so do u understand how i got what i got

[edgy mauve]

Umm i js used b to denote it to make u understand

We wanted to factorize didnt we

[fringe vault]

oh QAIT i get it

oops

[edgy mauve]

Do u?

[fringe vault]

no

why is the -2(sqrt2) to the power of 2

[languid robin]

:p

[dapper hull]

absolute cinema

[edgy mauve]

(2sqrt(2))^2
Gives u 8

So to bring 8 in squared form

I used that

We need a²-b² to work out here

[fringe vault]

why does (x-2)^2 expand to that

[languid robin]

this is the only channel with problem i can do

[edgy mauve]

Take over

[languid robin]

what the question

[languid robin]

until this step?

[fringe vault]

this is what i have with the help of

victimizeer

this is what i understand

[edgy mauve]

Um (2root(2) )^2

[languid robin]

em

[edgy mauve]

2nd last line

[ripe jewel]

this isn't called factoring btw

it's called completing the square

[edgy mauve]

We couldn't find other ways

[fringe vault]

i already did completing the square

[edgy mauve]

We did quadratic

[languid robin]

second line isnt it x^2-4x+4-8

[edgy mauve]

Did completing square too

[languid robin]

💀

[ripe jewel]

so what are you doing if not completing the square

because this isn't factoring

[languid robin]

second line wrong.

-4 can become +4-8

[edgy mauve]

And she phrased it wrong

[fringe vault]

wait so idk what im doing 😭 i wa sjust going alon

[edgy mauve]

How would u do it
We dk other methods

[ripe jewel]

this looks like completing the square to me

[languid robin]

oh ok

[edgy mauve]

Help

[ripe jewel]

what are you trying to do

[languid robin]

i understood

[edgy mauve]

.

[languid robin]

the use a^2-b^2

[edgy mauve]

5his is the main question

[languid robin]

oh

4 methods..

[ripe jewel]

well it depends on which of the 4 methods youve done and which you havent

[edgy mauve]

This is how i tried

[edgy mauve]

Quadratic done

Compelting square done

[languid robin]

done with quadratic .>.

factoring done?

[ripe jewel]

huh

[languid robin]

well the bot only pined a "hwllo"

[ripe jewel]

it's not factorable though

[languid robin]

then whats wrong with the question

[fringe vault]

should i just say that because i wqs thinking that too

[languid robin]

💀

[ripe jewel]

well you can factor it using the quadratic formula

but that really is the same answer as usijng the quadratic formula

[edgy mauve]

Yeah

[languid robin]

em

[ripe jewel]

but thats what I would do

write it as (x-r1)(x-r2)=0

[edgy mauve]

Yeah not through middle split up

[edgy mauve]

That's what i tried

[languid robin]

no need to be =0 isnt it

like

$x^2-4x=4$

[soft zealot]

Question mark

[ripe jewel]

if it's not = 0 then factoring is pointless

[languid robin]

ha

[ripe jewel]

if you say (x-r1)(x-r2)=0 you can deduce that x=r1 or x=r2

if you say that (x-r1)(x-r2)=5 you can't say that x=5+r1 or anything like that

[languid robin]

:p oh ye x is not integer

[ripe jewel]

i would just use the quadratic formula first

for the quadratic formula answer

[edgy mauve]

It's done

Already

[ripe jewel]

and then for factoring use the values from the quadratic formula

[languid robin]

graphically done? :p

[edgy mauve]

Proceed with factoring then

[ripe jewel]

that's the only way you can "factor" it practically speaking

unless you want to start fucking around with

pq=-4, p+q=-4 and non integer solutions

which is not easy

[edgy mauve]

Nah

I mean doesnt a²-b² consider factoring

Like it still does make sense

Tho similar to completing squares

[languid robin]

we need to draw out the graphic using hand??

[fringe vault]

i think so

[languid robin]

:0 brain ded

[edgy mauve]

Okay so what do we do

Continue with a²-b² or what

[languid robin]

:p wait wait

maybe try this

:p this is completing the square

[fringe vault]

i already did

[languid robin]

$x^2-4x-4=0$

[soft zealot]

Question mark

[languid robin]

sorry ok

what happen when u apple

apply

[edgy mauve]

This is what she got

[edgy mauve]

Oh

Sigh

Messy ah situation

[fringe vault]

does this make sense

at all

i cant make sense of it

probs cz i havejtslept tho so my brains runningnonly on stuff i know

[edgy mauve]

Like u cant get a 0 as a product if at least one of the numbers u r multiplying isnt 0

Cuz 1×1=1

1×0=0

So no matter what u multiply if one of the number isnt 0

U cant get 0 as a product

[fringe vault]

im going to bed ive been on this questionf or like 2 horus now

[edgy mauve]

Sure

[fringe vault]

ill try to figure it out in the morning]

thabks for the help

[languid robin]

:p its 11.43AM here

[fringe vault]

PM here

[final saddle]

@fringe vault Has your question been resolved?

[tired walrus]

.close (OP said she's off to bed)

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[remote vine]

Hello,
with numbers 7 and 15,
summarizing them at any individual amount,
prove we can represent any Natural number bigger than a certain number
What should I learn to solve this?

[cedar obsidian]

number theory i believe

[stable copper]

hello

whats up

i am new here]

[worthy wren]

peano axioms

and if by any number u wanna include reals then u'd get into the construction of reals and well-ordering principle

[wheat lark]

i need help f (x) lim->2 42π^2/42-72^27 (62-4^9)

[worthy wren]

!occupied

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[iron mist]

Kinda sounds like Chicken McNugget Theorem

[worthy wren]

oh i properly read it

can u post the full question

there should be more hypothesis otherwise it is trivial since 7 spans R

[sturdy cypress]

it's cuz you can add 15s until you're over and then you replace 15 with 7+7 to go back to where you want

but you can't go below 7

[iron mist]

7+7=14 not 15

[sturdy cypress]

yes, so your number is smaller

so you can go backwards and forwards and reach any number

there's nothing to learn

you need to learn induction to write a formal proof i suppose

oh ok, it's not 7, it's much higher

i can't solve it ok

it's like, 120 maybe

[remote vine]

it's from a high school additional student book, and it is short in text so that is full version. Could you say more about the triviality of it, I don't get it why?

[sturdy cypress]

there's no name for what you need to learn, it's just some subset of number theory without a name

[iron mist]

If you are allowed to use the Chicken McNugget theorem (yes that is a real thing, google it), then it’s over instantly

Otherwise, you have to break up the naturals into a bunch of cases and then find a formula for each case (as well as exceptions thereof)

[remote vine]

Thank you guys

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[narrow river]

I dont understand this, especially the last part about the difference between mathematical incorrectness vs process incorrectness

[final saddle]

Please don't occupy multiple help channels.

[sudden comet]

what that last part means is what rule of math does this break as opposed to there being a mistake in the process

[narrow river]

is it that they ignored the terms with y in the second integrand? because idk why they would have done that

i mean i also think the first step is wrong, but after that i dont know why theyre ignoring y terms

[sudden comet]

there's that, but there's also integration with a seperate variable than what the integral is being taken with respect to i believe

[teal glen]

Ignoring terms with y leads to incomplete integrations and destroys the potential function construction of exact differential equations

[narrow river]

I agree. but can you really just integrate without separating the variables and call it a solution? that doesnt seem right either

my thinking is that by integrating without separated variables you are ignoring the impact of x and y in the differential equation for the left hand and right hand side, respectively

[sudden comet]

nope, for instance take $\frac{dy}{dx} = \frac{y}{x}$, normally you would need to solve it linearly using $e^{\int P(X) dx}$, but if you did it the way the problem does you would get $x dy = y dx \implies xy = yx$ which is obviously wrong

[soft zealot]

orange20063

[narrow river]

yea

but even if you were to do it in that wrong way why would you ignore y terms in the rhs integration

[sudden comet]

I honestly couldn’t tell you

But since we’re not working with partial derivatives, we can’t integrate like this

[final saddle]

@narrow river Has your question been resolved?

[whole halo]

the mistake is here

"ignoring any terms" cant work

for one, you can see the y^2 - 6xy on the right completely disappear which is pretty bold

ignoring terms only work if y doesnt depend on x, such as if x and y are both functions of t

here, y depends on x because y is a function of x

if you had f(x)^2 - 6 x f(x) - 3x^2, you wouldnt integrate it by treating the f(x) as constant

[tulip coyote]

(for what it's worth, I interpret their "process incorrectness" to mean that of a stated step, they made a mistake in doing it, as opposed to one of the stated steps being incorrect of itself)

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[errant basin]

in proving [0,1)x[0,1) is homeomorphic to [0,1)x[0,1], line 3 paragraph 2 "clearly"

[errant basin]

i would love it if somebody gicves a motivation for the proof, or rather, construction

[tired walrus]

so the very rough visual motivation is that [0,1) × [0,1) has two "open sides" while [0,1] × [0,1) has one "open side"

and the goal of the homeo is to "stretch" that one open side into two

[errant basin]

uhh

okay???

uhh

[errant basin]

how can you strecth something that doesnt exist.

[tired walrus]

i did say it was very rough and informal

but heres another way to think about it

in terms of existent things

X has 3 "closed" sides and Y has 2

so you need to "compress" those 3 sides into 2

while keeping everything else continuous

id probably do it differently than them tbh

[errant basin]

what would be your solution.

[tired walrus]

mmmm idk if i can describe it in full verbally

but basically im thinking of fixing the center (1/2, 1/2) in place and then mapping the lines joining the center to the corners roughly like so:

[errant basin]

oh, thats more intuitive;

[desert mantle]

I have yet another way of thinking about it: blow out the sides so that the square gets converted into a disk. so part of the boundary of the disk is included and part isnt. and then by stretching/squishing certain sectors of the disk its imo very intuitive that you can convert between 1/4th not included and 1/2th not included

[tired walrus]

true

which is basically the same as what i showed just now only without the inflation

[desert mantle]

yeah

but the added symmetry of the disk helps a bit imo

[errant basin]

wait, what is a homeomorphism between square and circle.

oh

okok

i think i can formalise that?

thanks a lot that rly helped, some of the answers in this pdf is bit "strange"

.closed

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[zenith cedar]

2 points A and B are chosen above the x axis. 2 points P and R lie on the x axis and Q lies on the line ax+by+c=0 above the x axis. Minimize the path APQRB
P Q R are variable points and A B are fixed

[zenith cedar]

#help-46 message

[long thorn]

Try to use principle of reflection

[zenith cedar]

I have used that I am not sure if this gave the minimum path

There was a possibility of no minimum in the previous discussion

[long thorn]

Did you use P and R to lie on the line segments?

[zenith cedar]

I don't think so

[final saddle]

@zenith cedar Has your question been resolved?

[ebon agate]

This is the figure to keep in mind?

[zenith cedar]

yes

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[zenith cedar]

.reopen

[final saddle]

✅

[zenith cedar]

there was also another part to this where the angle of reflection equals angle of incidence abd we need to minimize path ABPQR

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@zenith cedar Has your question been resolved?

[sullen gale]

wdym?

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[solar crest]

I’m so confused and stuck

[solar crest]

Ignore the last. 2 lines lol

I ended up returning to where I started…,..

[golden slate]

L’Hopital’s rule?

[terse crypt]

holy shit

derivative defn

sub with n := x - pi/2

etc

[solar crest]

Not allowed

[slender basin]

tan(2h+pi) is simply tan(2h) then just write it as sin/cos and then use limit sin(x)/x

[solar crest]

Thta wat I did ….

[solar crest]

Oh just rewrite as sin and cos?

[terse crypt]

[ \lim_{n \to 0} \frac{\tan(2n)}{n} = \lim_{n \to 0} \frac{\sin(2n)}{n} \cdot \frac{1}{\cos(2n)}]

[soft zealot]

k

[terse crypt]

thisi s what ViNton is saying

[thin cloud]

I remember some one asked this problem recently

[terse crypt]

this is my method

[ \lim_{x \to \pi/2} \frac{\tan(2x) - \tan(2(\pi/2))}{x - \pi/2} = \dv{x} (\tan(2x)) \big|_{x = \frac{\pi}{2}} = 2\sec^2(2\left(\frac{\pi}{2})\right)]

[solar crest]

I think it was me

[slate dove]

аааа задроты

[thin cloud]

Can you just multiply both by 2 and get $\lim_{h \to 0} \frac{2tan(2x)}{2x}$

[soft zealot]

Alexis_Fx

k

[terse crypt]

or this

[solar crest]

No dedication fees

[solar crest]

How does that’ll help

[terse crypt]

[ \lim_{x \to 0} \frac{\tan(x)}{x} = 1]

[soft zealot]

k

[thin cloud]

that x should be h

[terse crypt]

it should be noted that as ${x \to 0}$, ${\sin x \approx \tan x \approx x}$

[soft zealot]

k

[terse crypt]

anyhow

[slender basin]

or you can manually get it to sin(2x)/2x to use one of important? limits (idk how they called in eng)

[terse crypt]

U can use:

1. sin(x)/x -> 1
2. tan(x)/x -> 1
3. derivative defn
4. l'hopital
5. approx

[terse crypt]

does it even have a name

[slender basin]

in my language yes literally called 'wonderful limits' xD

[thin cloud]

If you haven't been taught sin(x)/x , prove it would be a problem

@solar crest Have you learnt this yet?

[solar crest]

no…………..

[terse crypt]

this is the mistake

[solar crest]

Huh

[terse crypt]

oh shi

nvm

[solar crest]

yea so cooked

[thin cloud]

$\lim_{x \to 0} \frac{sin(x)}{x}=1$

[soft zealot]

Alexis_Fx

[terse crypt]

there is no mistake on that part

[solar crest]

So far so good?

ok here

[terse crypt]

but in the end

u forgot this exists

[solar crest]

how is that even true OMF

[terse crypt]

squeeze theorem

u will learn soon

🙂

anyway

so u've gotten here

[solar crest]

I learnt it

[terse crypt]

what

[solar crest]

but not in this format

[terse crypt]

now thats a surprise 😭

[solar crest]

Bro I rushed it and self learnt it son

[terse crypt]

but u are allowed to use this right?

[solar crest]

So*

[terse crypt]

ahh

makes sense

[solar crest]

Nope…

[terse crypt]

i get ya

how abt

derivative defn?

2 cats

[thin cloud]

If you prove it, you will be allowed to use it right? 😂

[solar crest]

huh wdym by this

[terse crypt]

[solar crest]

LOL

[solar crest]

I guess,.l

Is there another way

😭

[terse crypt]

[ f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}]

[soft zealot]

k

[solar crest]

Without having to do a whole proof

[terse crypt]

can we use this

[solar crest]

Oh

Lining definition

Limit

Ues

[terse crypt]

YAY

uve learned derivative, right?

[solar crest]

Yep

[terse crypt]

cool

now

[ \lim_{x \to \frac{\pi}{2}} \frac{\tan(2x) - 0}{x-\frac{\pi}{2}} = \lim_{x \to \frac{\pi}{2}} \frac{\tan(2x) - \tan\left(2\cdot \frac{\pi}{2}\right)}{x-\frac{\pi}{2}} ]

[soft zealot]

k

[terse crypt]

do u agree with this

[solar crest]

Ye

[terse crypt]

can u find f(x) and a

from this format

[solar crest]

a=pi/2?

f(x) is 2x?

[terse crypt]

f(x) is wrong

look again

a is right tho

[ebon agate]

Wow wtf 😭

[final saddle]

@solar crest Has your question been resolved?

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[lethal tapir]

Is the best way to approach these problems by plugging in numbers?

[vivid walrus]

yes

sometimes its not like when its goes to infinity

[lethal tapir]

ok thank you

.close

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[lethal tapir]

How does f(x) exist as x appraches 2? Also, how can you tell if the 2 functions are equal?

[stuck juniper]

Wym by the first question

[lethal tapir]

Its telling me that f(x) does exist as x approaches 2. I dont know how because there's a hole and the other point isn't going anywhere

at x=2

[stuck juniper]

This looks like a removable discontinuity

[inner acorn]

at x=2 there is a solid point at (2,1)

which means that f(2) exists

[tardy void]

The limit does exist though

[inner acorn]

however when you APPROACH f(2)

it approaches 0.5

[stuck juniper]

lim is dne

But the discontinuity is removed

Due to the piecewise

[inner acorn]

nonono

[tardy void]

Huh

[inner acorn]

the limit is still there if

[stuck juniper]

Oh its there

[inner acorn]

lim x->2+ = lim x->2-

[stuck juniper]

?maybe im wring then

[inner acorn]

and this function has that property

both the limit and the value of f(2) exists

[stuck juniper]

Its not equal to the function value at that point tho

[inner acorn]

they are just different values

[inner acorn]

yeah that is allowed

you gotta understand that

f(2) is not lim x->2 f(x)

they are different as one is the exact value and the other is the value it APPROACHES when getting close to x=2

[lethal tapir]

So f(2) is the closed point and the limit is the hole?

[inner acorn]

exactly

[lethal tapir]

Wait, so holes can exist?

[inner acorn]

it is not that the hole exists

but as you go INFINITLY close to x=2

[tardy void]

Yes they can

[inner acorn]

it GET infinitly close to the hole

but when you actualy touch x=2

it jumps to the solid point

[tardy void]

Have you dealt with greatest integer functions

[inner acorn]

for these limits, you gotta watch out, as the limit only exist if lim x->2+ = lim x->2-

in the picture shown tho, it is v ery clear that lim x->2+=0.5 and lim x->2-=0.5

therefore lim x->2=0.5

[lethal tapir]

How did you get 0.5

specifically for 2-?

[stuck juniper]

Plug it into the given function

[tardy void]

It’s indeterminate

You can’t directly

[stuck juniper]

U can convert it

[tardy void]

Yeah

But that’s where the removable discontinuity comes from

[stuck juniper]

Should be quite straightforward

Factorise x from numerator

And use difference of squares

[tardy void]

Does op know about asymptotes and removable discontinuities

[thin cloud]

Should wait for OP to say something first

[stuck juniper]

Probably should given that she is given the graph and stuff

[lethal tapir]

so you get (x/x+2) then you plug in 2?

[stuck juniper]

Yes

[lethal tapir]

yes

But how do know that 2 appraching to the right and appraching to the left are the same from that

[tardy void]

Look at the graph

[lethal tapir]

ok thanks

.close

[final saddle]

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[torn summit]

i need help at solving y''+a(x)y'+b(x)y=c(x)

[torn summit]

i saw earlier this:

y''-4y=0

(y''-2y')+2(y'-2y)=0

and then

z=y'-2y

and from there its easy

but this is one case, and they did it very easily.

now i want to generalize it.

[vital crag]

There's no generalized formula

All depends on the form of c(x)

[torn summit]

why so?

i mean, can you make some integral games with it somehow

to get something... closed form?

[vital crag]

Sure try it yourself and see

[torn summit]

i will.

[ebon agate]

There's a way to solve if a,b,c are constant functions like this case

[ripe jewel]

in general

if you have an ode of the form

P(Dt)[y]=0

where P(Dt) is a polynomial in the derivative operator

you can try factoring

[ebon agate]

Yeah

Ah mb

[torn summit]

i dont accept failure.

[ebon agate]

Just wait

[ripe jewel]

@torn summit do you understand my notation

[torn summit]

P(x) is polynomial

whats Dt

eulers derivative?

[ripe jewel]

the derivative wrt t

yes

so like

[torn summit]

why 3 notations :-;

[ripe jewel]

because they have different connotations

you can have something like

[torn summit]

where is D used?

i saw dy/dx

for example

[ripe jewel]

(Dt²-2Dt-3id)[y]=0

[torn summit]

y' sin y= cos x

[ripe jewel]

now lol

what i just wrote would be awkward with other notations

[torn summit]

so deriving around t?

as in, d/dt?

[ebon agate]

Is this the same as y" - 2y' - 3?

[ripe jewel]

yes but from a different perspective

[ebon agate]

Meaning?

Of diff perspective

[ripe jewel]

meaning you are now considering objects such as

D²-2D-3I

this has a polynomial form

but its not a function of numbers

[ebon agate]

Ah ok

[ripe jewel]

its a function of functions

[ebon agate]

Ah got it

[torn summit]

now tell me

Dy^2+2Dy+1

can i

(Dy+1)^2?

[ebon agate]

+y*

Ig

[ripe jewel]

(D+I)²[y] is a better notation

[ebon agate]

Not 1

[ripe jewel]

or just (D+I)²y

[torn summit]

huh.

can i sqrt it then?

[ripe jewel]

nooooooo

illegal in 49 states

[ebon agate]

There was a video on half derivatives that I saw 😭

Like 2 years back

[ripe jewel]

but you can do this

first solve (D+I)y=z

then solve (D+I)z=0

[torn summit]

🤔

📝 🤔

[ebon agate]

https://youtu.be/2dwQUUDt5Is?si=fFHg2sVkVXMzQeJy

[torn summit]

but sqrt d/sqrt dy

it sounds so wrong

[ebon agate]

That's a square root?

[torn summit]

its wrong, so wrong

[ripe jewel]

there is such a thing as fractional function composition

but ive never seen it actually useful

[torn summit]

how so?

[rocky tusk]

fractional calculus

[torn summit]

sounds useless tbh

[ripe jewel]

like see if there exists a function f such that fof=cos

[ebon agate]

I thought this was very gimmicky in its definition and everything

But who am I to judge

[ripe jewel]

thats my impression of it

[rocky tusk]

word

[torn summit]

yes...

[rocky tusk]

it’s niche

[torn summit]

does this f have a use?

all we know is that f(f(x))=cos x

[ripe jewel]

however

[torn summit]

and i dont see a use for that.

[ripe jewel]

square roots of matrix operators exist often enough to be useful

but theyre generally not unique

[torn summit]

so, what you are telling me

you can find, for some reason (you want to find it for some unknown use)

the function f, such that f(f(x))=g(x)

can wolfram alpha even solve these things?

bc this sounds VERY niche

[tired walrus]

cause it is

and maybe WA can't

[torn summit]

i tried giving it

f(f(x))=x^4

its easy to guess it is f(x)=x^2

it doesnt understand anything

it does function analysis on x^4, which is not what i wanted.

[tired walrus]

well then i guess WA can't do it. cry about it ig lol

[ripe jewel]

i said for matrix opetators specifically

[torn summit]

anyways, we are swerving here, i still want to solve y''+a(x)y'+b(x)y=c(x)

i just finished the first book, about first order ODEs

slope fields and all, it makes sense tbh

the second book is second order ODEs, its far fatter than the first

the y'' stuff is weird.

also, @ripe jewel i love your ideas for solving it

but i dont think my uni accepts D as an existant thing

i remember coming to it

and seeing y'=y

[ripe jewel]

ty but what i said is not a novel idea its just not usually presented exactly that way

[torn summit]

and saying, lets just divide by y, integrate

get ln y=c+x

[ripe jewel]

usually its done using an Ansatz and writing a polynomial in r instead of in Dt

[torn summit]

im 2 pages in

the first problem was

y''-4y=0

doing here the trick of y'/y, wont work

we dont have here difference of 1, we have here 2

[ripe jewel]

yeah so the usual approach is to write

[torn summit]

so y''/y, is useless.

[ripe jewel]

consider the polynomial r²-4

and skip the presentation of a polynomial in Dt

[torn summit]

well, i thank you for this idea

its amazing

[ripe jewel]

[torn summit]

but i dont think its ok to use

[ripe jewel]

[torn summit]

the uni didnt teach it yet

and teachers are this way.

[ripe jewel]

[torn summit]

if you do 2+1 as 1+2, teachers get mad.

because you do it "wrong"

even tho differential equations have a trillion ways to solve

[torn summit]

im sorry, i dont want to lower my grade

being smarty pants in my first course isnt an idea.

i needed to do calculus first, and iirc also matrices

but we didnt yet get to those.

[torn summit]

also, what do you mean here

i dont think you can treat y^(n) as y^n

tho, im stupid, so enlighten me.

[ripe jewel]

https://en.m.wikipedia.org/wiki/Characteristic_equation_(calculus)

its more or less the same thing as writing P(D)

more of a different perspective than a different approach

[torn summit]

and how do you solve it?

i did the substitution trick

where you say

[ripe jewel]

you assume the answer is a linear combination of functions exp(rt), where r is a root of the characteristic equation

[torn summit]

y''-2y=z

[torn summit]

mhm.

we do use exponents

but i like solving, not assuming

[ripe jewel]

the WP page explains it better than i can

[torn summit]

and it looks like honestly gibberish

and btw, i still cant use it

the trick they used here

[ripe jewel]

i hear that, but, {e^(rt)}_r in C is a very powerful basis for these types of problems

you took lin alg yeah?

[torn summit]

uhhhhhhh

nope i did not

i jumped straight from highschool stuff to this

[ripe jewel]

oh

[torn summit]

as i said

[torn summit]

consider tomato is a member of stupid.

i learn fast, but i need the why of things

thats why i learn slower than others

and took 2500h to finish highschool math

[ripe jewel]

i thought Israeli math courses used English books

[torn summit]

they do...

if i bought them

in english

[ripe jewel]

oh

[torn summit]

its in hebrew, because the official language is hebrew

[ripe jewel]

im fluent, i am aware

well, proficient

[torn summit]

and ben gurion, first prime minister, dictated (quite literally) that hebrew is THE ONLY language

[torn summit]

at hebrew?

[ripe jewel]

level dalet proficiency

[torn summit]

huh.

well, good enough i guess

thats hebrew that i dont understand

like mathematical english

but asimptota, you know its an asymptote

we didnt invent our own language

[torn summit]

either way

they say here that they did a so called trick

with the (y''-2y')+2(y'-2y)

but this seems so primitive and limited

@ripe jewel i dont know how it would work with, say, y''+e^x y'+2x sin x y=0

this method is so limited, i cant see it ever being used

[ripe jewel]

im no expert in diffeqs

but i know most of them cant ve solved explicitly

[torn summit]

ish yes

eh if only.

may i ping helpers?

[torn summit]

<@&286206848099549185>

[shrewd nexus]

Depends on a, b, c

You cant find a general solution to it

[torn summit]

:-;

[shrewd nexus]

Do you need help with a specific case?

[torn summit]

i cant take that as an answer man

[torn summit]

well

[shrewd nexus]

Well it is how it is you know

[torn summit]

they did here the y''-4y=0

and they did it easily

[shrewd nexus]

Depending on a,b,c, some of these difeqs are literally unsolvable

provably unsolvable

[torn summit]

(y''-2y')+2(y'-2y)

and then z=y'-2y

and you solved it.

but this is a stupid solution

[shrewd nexus]

not me, but ye

[torn summit]

because this works for a handful of solutions

its not useful and brings me nowhere

[shrewd nexus]

also doing it like that is a bit convoluted lol

but sure

[shrewd nexus]

Well thats how difeqs are man

[torn summit]

its such a stupid solution

[shrewd nexus]

we cant do much about it

[torn summit]

in first order i can

we define some A(x)

and integrate a(x)

[shrewd nexus]

you cant solve it generally

[torn summit]

in the form of

y'+y a(x)=b(x)

then

[y e^A(x)]'=b(x)

and thats easy.

y=B(x) e^-A(x)

@shrewd nexus but this doesnt work for second order!

btw, i have zero background on linear algebra, and calculus i barely know.

[shrewd nexus]

you might not be able to integrate tho, so that solution doesnt really work does it

[torn summit]

well, thats true.

[shrewd nexus]

even first order

[torn summit]

but more than this theres nothing

[shrewd nexus]

You cant say much in the general sense

[torn summit]

i want to reach some form of an integral

[shrewd nexus]

yeah ofc

[smoky egret]

if there is no general solution for first order diff eqs then why do you expect a general solution for any other order

[shrewd nexus]

but some integrals cant be expressed as a closed form of known functions

[ebon agate]

I actually tried the exact same thing in school dng

[torn summit]

it is some solution

y=B(x) e^-A(x)

[smoky egret]

there are some ways to deal with second order linear homogeneous diff eqs for example

[shrewd nexus]

B and A might not exist...

[smoky egret]

but thats too specific

[torn summit]

thats something at least.

[shrewd nexus]

eh not really tho but ok

[torn summit]

i can use a million methods to go around the integral

and i want my problem to be this integral

[smoky egret]

idk what you mean by this

[torn summit]

taylor works

[smoky egret]

but if you mean that you can solve the integral to get an elementary function then the answer is no

[smoky egret]

well using power series is a thing but i am not sure if you would be satisfied with a power series as your y

[torn summit]

i prefer one unknown function

rather than two

[ebon agate]

Ig he just wants the solution of the differential in integral form

To solve is his problem

?

[torn summit]

i havent seen how they solve it, but the y''-4y=0 trick was honestly useless

[torn summit]

yes.

i want some y=f(x)

or x=f(y)

[ebon agate]

I mean did you get what gfauxpas said?

[torn summit]

which of his words

[ebon agate]

Characteristic equations to solve these kinds of problems

[torn summit]

i also cant say "lets try ..."

i want to solve

not guess

[ebon agate]

Where A(x),B(x) and C(x) are constants

[torn summit]

because for the y''-4y=0 problem i guess it is 0, and you cant tell me no.

[torn summit]

if they are

what can you do, to solve it?

[ebon agate]

suppose you have like

y" - 6y' + 5y = 0

[torn summit]

a(x) isnt reliant on x, nor is b, or c

[torn summit]

looks good.

ok better

how do you?

[ebon agate]

This is always going to be in the form of

Ae^r1*t + Be^r2t

And r1,r2 will be the roots of a characteristic equation

[torn summit]

yes.

[ebon agate]

You will make the characteristic equation by taking this equation

But treating double differential as squaring

[torn summit]

wait whar

wait wait

where squaring from

[ebon agate]

y" is analagous to r^2

y' is analagous to r

y is analagous to 1

[torn summit]

huh.

so quadratic equation?

[ebon agate]

Yes

[torn summit]

well, that sounds too easy.

[ebon agate]

Double differential will turn into a quadratic

Same way if it was a triple differential

It would turn into a cubic

[shrewd nexus]

Second order difeqs with constant coefficients are super common in physics if you take that you will see some

[torn summit]

huh. ok.

[ebon agate]

Find the roots of this quadratic equation now

r^2 - 6r + 5 =0

The roots of this equation will be r1 and r2

[shrewd nexus]

You will see if you take it, you will learn to solve them in the general sense very quickly

[torn summit]

6 plusmin sqrt 26 div 2

3 plusmin sqrt(13/2)

[ebon agate]

Ig it turns a bit complicated if there are complex roots (trig fxns) and repeated roots

[shrewd nexus]

sqrt (16) not 26

[torn summit]

mb not 2ac

[shrewd nexus]

ye algood

[torn summit]

sorry, cosine theorem

[ebon agate]

Try now for an equation like

y" - 5y' + 6y = 0

[torn summit]

wait wait

so y''=r^2?

[ebon agate]

Ohk

Yeah

[torn summit]

and y^(n) is r^n?

[ebon agate]

I mean equal wont be a good word

[ebon agate]

Yes

[torn summit]

but this seems so absurd

[ebon agate]

Oh well

[torn summit]

how so?

[ebon agate]

Once you find the solution to this differential, put it back in, you'll see it how it works

It's pretty neat

[torn summit]

can you show me how you reached from y''+a(x)y'+b(x)y=c(x) to a solution with r?

[torn summit]

so if say, r=5

[ebon agate]

I just know when a(x),b(x) are constants and c(x) is preferrably 0

Very specific case

[torn summit]

so y=5?

[ebon agate]

No no

Y = Ae^5x

[ebon agate]

.

[torn summit]

so y=A e^(rx)

whats this given A?

[ebon agate]

A is the arbitary constant

Yk how n degree differentials have n arbitary constants

[shrewd nexus]

A and B are found with the conditions of the problem

[ebon agate]

Same thing

[shrewd nexus]

like y(0) = 0 for example

or whatever

[ebon agate]

Initial value conditions

[torn summit]

so... c?

[ebon agate]

Yes

Basically

[torn summit]

integration constant.

[ebon agate]

Yupp

Ig when you are integrating, you are solving a diffrential

[torn summit]

ok.

no no, you helped me a lot.

thank you

[ebon agate]

Welcome

[torn summit]

im now reaching home at last

and can open the book to see the methods

im really confused yet still

but this one is helpful.

ok i see

so, they have here a problem, harder than before

y''=(1-y')/x

now they did here, since theres no y, z=y'

[shrewd nexus]

yes

[torn summit]

and here things get worse.

[shrewd nexus]

and then you can solve like you were saying earlier

[torn summit]

y'' = (y')^2/y(y-1)

[torn summit]

yes, first order is somewhat east

easy

it makes sense.

as much as integrals can.

@shrewd nexus for you to know

i never did anything except d/dx

with this notation

i didnt get f(y) dy=g(x) dx

[shrewd nexus]

?

I didnt catch that

[shrewd nexus]

Do you need help with this is that it?

[torn summit]

this one

heres what they do

[shrewd nexus]

ok

[torn summit]

z=y'

dz/dx=z^2/y(y-1)

yes... ok...

works i guess.

dz/dy = dz/dx dx/dy

x'=1/y'=1/z

uh huh.

z dz/dy=dz/dx

[shrewd nexus]

Its hard to follow like this 😅

[torn summit]

man

thats all i have

[shrewd nexus]

What is this line

[torn summit]

its written

exactly

x'(y)=1/y'(x)

=1/z

[shrewd nexus]

oh

ye

weird stuff they are doing

[torn summit]

yeah you telling me.

oh god

let me quote what they say

to cancel complications, we use the same letter z to mark two different functions:

z(x) and z(y)=z(x(y))

sorry

[shrewd nexus]

they are doing some weird stuff man tbh

[torn summit]

for example, z(x)=y'=3x^2

end of quote.

[torn summit]

yes, dont tell me

im perfectly aware.

[shrewd nexus]

hmm

[torn summit]

let me write it down

[shrewd nexus]

My instinct is to notice that y'' / y'^2 = d/dx(1/y')
Therefore your difeq is essentially:
d/dx(1/y') = 1/y(y-1) but im not sure how to continue for now

[torn summit]

ok

i got here something

@shrewd nexus i can now just throw there the integral

[shrewd nexus]

mhm

Shouldnt the second to last line be dz/dx and not dz/dy?

[torn summit]

no, because z is defined by y

z(y)=y'

[shrewd nexus]

ohhh like that

right

👍

[final saddle]

@torn summit Has your question been resolved?

[slate dove]

пацаны скоко 142646 + 245 срочнгоооо

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[ionic gate]

Continuity means the delta doesn't depend on the point c we choose to work with. Or is that uniform continuity?

[blissful meadow]

That's uniform continuity. Do note that uniform continuity describes the behaviour of a function on a set, whereas continuity discusses behaviour at a point.

[ionic gate]

So, for continuity here, are we just evaluating specific points to understand the behavior?

[blissful meadow]

Yes.
Informally, continuity ensures there are no jumps in your function at a point, which you can think of as infinitely steep ascents/descents at a certain point.
Uniform continuity is stronger in the sense that it also forbids very quick ascents/descents, instead of just straight up jumps.

[final saddle]

@ionic gate Has your question been resolved?

[ionic gate]

I see.

ANother question I have is about proving the that there exists a point c that lies in a function:

For continuity.

What would be the logical starting point?

[ripe jewel]

see if g has a root

[blissful meadow]

I don't think the path is to use continuity, at least not directly.

Use the function g you're given. Is it continuous? See what happens when you plug in key points, and then try to make use of fun theorems that arise when dealing with continuous functions on closed intervals ||think intermediate value theorem||

[ionic gate]

5.3.7 Bolzano’s Intermediate Value Theorem Let I be an interval and let f : I — R be continuous on I. Ifa,b € Land ifk € R satisfies f(a) < k < f(b), then there exists a point c € I between a and b such that f(c) = k.

[faint edge]

Try it with g

[ionic gate]

Could I take f(0) < k < f(1)?

[ripe jewel]

euro symbol for set membership...

[ionic gate]

Blame the book publisher

[ripe jewel]

What's L?

[ionic gate]

L is the limit that we are trying to evaluate.

We don't know yet. The problem doesn't say.

[ripe jewel]

"if a, b are in the limit" doesn't make sense

[faint edge]

I think L is a typo for I?

Did you copy paste this?

Sometimes formatting gets fucked up when you copy paste latex

That's probably why set membership became the euro symbol too.

$\in \text{ vs €}$

[soft zealot]

DootDooter

[blissful meadow]

brexit

[faint edge]

You have g from the hint in your problem. g is defined on a particular interval. Do you see any possible a,b values you can try in that interval?

[ionic gate]

0, 1/2, 1

[faint edge]

Yeah that'll work.

What happens if g is nonzero between 0 and 1/2?

The intermediate value theorem tells you something nice in that case.

[ionic gate]

There exists a point where g(c) = k

[ripe jewel]

whats k

and where is c

[ionic gate]

A real number s.t g(a) < k < g(b)

[faint edge]

Yeah. So you see we are letting a=0, b=1/2?

[ionic gate]

What about that exactly?

[faint edge]

You can plug those in and look at g(0) and g(1/2).

[ripe jewel]

why do you assume g(a)<g(b)?

[faint edge]

It doesn't matter.

Can you see anything about how the signs of g(0) and g(1/2) are related?

[ionic gate]

What?

[faint edge]

g(0)= f(0)-f(1/2) right?

But g(1/2)=???

[ionic gate]

What?

[faint edge]

I'll tell you more if you write out what you think g(1/2) is.

[ionic gate]

g(1/2) = ??

[faint edge]

Use the definition they gave you here for g. Plug in 1/2.

[ionic gate]

g(1/2) = f(1/2) - f(1)

[faint edge]

Yeah and they told you f(1)=f(0) so g(1/2)=f(1/2)-f(0)

My calculation a sec ago was g(0)=f(0)-f(1/2)

You see that these have opposite signs?

[ionic gate]

Yeah

[faint edge]

Okay, so if we assume between 0 and 1/2 g is never zero, then either g(0) < 0 < g(1/2) or g(1/2) < 0 < g(0).

What's the problem with this?

[ionic gate]

It can only be an our, I guess:

[faint edge]

Yeah that's a special case of this theorem.

Intermediate value theorem tells you that there would have to be a point c somewhere between 0 and 1/2 where g(c)=0

Which would be a contradiction.

So, g has to be zero between 0 and 1/2 somewhere

There's several ways to phrase the proof I was trying to walk you through. The main pt is that applying intermediate value theorem (or your 5.3.5) to g will force you into getting the c you need by considering the possible signs g can have at 0 and 1/2.

[ionic gate]

I see. What does the antipodal mean here for this question?

[faint edge]

Ah oops

Sorry if you have a pt x the antipodal pt should be -x

Like, if you were on a circle at a pt x the antipode of x would be on the opposite side of the circle

[ionic gate]

This is confusing. I am skipping this question