help-39

[rancid steppe]

-16 +17 = 1 so don't we have left 1 +1 ?

[lofty berry]

the first 1 is multiplied by 2

2(x^2 - 8x + 16 + 1) + 1

[rancid steppe]

so we multiplied by 2 only the last term or what? i don't understand

[lofty berry]

do you agree with this:
2(x^2 - 8x + 16 - 16 + 17) = 2(x^2 - 8x + 16 + 1) + 1

2(x^2 - 8x + 16 + 1) + 1 = 2*(x^2 - 8x + 16) + 2*(1) + 1

[rancid steppe]

so we took out from the () the 1 we had left my doing 2(1) +1

[lofty berry]

yeah you need to distribute the 2

2(a + b) = 2a + 2b, but you did 2a + b

(here a = x^2 - 8x + 16 and b = 1)

[rancid steppe]

so we just distribute the 2 on the last term the 1

got it now

[lofty berry]

yup

[rancid steppe]

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[faint shore]

Is this correct?

[pearl pond]

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✅

[faint shore]

https://tenor.com/view/helpmepls-kuchnahiaatahelp-i-am-underwater-help-me-pls-uncle-gif-21956179

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.reopen

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✅

[faint shore]

.solved

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[exotic gale]

Can someone please help me on this question:

[stoic imp]

Why not plug -2

[exotic gale]

i did but if you continue, it ends up being a really large number

you can try for yourself

[warm current]

maybe try simplifying the fraction for f(x) first

[stoic imp]

Yeah simplify f(x) first please

[exotic gale]

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✅

[exotic gale]

how do you simplify this

[stoic imp]

Can someone check if I simplified correctly

[exotic gale]

is that the same software that the khan academy guy uses

[stoic imp]

Is samsung notes app 🤭

[pearl pond]

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[pastel raven]

can somebody help me understading this passage

in red

[quiet tendon]

if you consider the quantity (p-i)^3 as i goes from 1 to p-1, it's the exact same as the quantity i^3 as i goes from 1 to p-1

just the sum written in the opposite direction

meaning that i^3 + (p-i)^3, as i goes from 1 to p-1, is the same as twice i^3 as i goes from 1 to p-1

then all they've said is that ((p-i)_p)^3 is the same as (p - (i_p)^3)

since i_p is the remainder when dividing i by p, consider foiling out (p-i)^3 and seeing why when dividing by p, the remainder is the same as p - (i_p)^3

[pastel raven]

wdym by foiling out

[quiet tendon]

expanding

[pastel raven]

but the remainder is gonna be -i^3

[brave sluice]

i'm assuming the ^3 happens before the remainder operation

?

[pastel raven]

oh, it's non-negative..

idk

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[stoic imp]

Find a polynomial $P \in \mathbb{R}[x]$ of minimal degree such that it has as roots all the solutions of the equation
[
iz^2 + \overline{z} = \text{Re}(z)^2(1 + i)
]
and satisfies $P(2) = 60$.

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[pearl pond]

@stoic imp Has your question been resolved?

[cunning comet]

first step: solve for z.

[pearl pond]

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[brave sluice]

i'm implicitly using $(\phi(a)\phi(h))H'=\phi(a)(\phi(h)H')$

[jolly parrot]

Axe

[brave sluice]

is that ok or is it sloppy?

[thorny pulsar]

Yah proof is good. If you are worried about that part, you can say after that line: "because phi is a homomorphisms" or something like that.

[pearl pond]

@brave sluice Has your question been resolved?

[brave sluice]

thanks! 🙏

actually it's not the homomorphism property i'm worried about

if we define $xS={xs | s\in S}$ then $(xy)S=x(yS)$

[jolly parrot]

Axe

[brave sluice]

but rather this property

i'm probably overthinking this

i just realized my book uses notations like gHg^{-1} without a worry

[thorny pulsar]

hmmm I guess you can do it the long way if that makes you uncomfortable. But eventually you can skip this step: ahH=aHhH=aHeH=aeH=aH (abstract this example to your specific need)

[brave sluice]

ohh right

you've justified it with coset multiplication

i'll close this question then

thanks again

.solved

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[jolly parrot]

neon

neon

[glacial sequoia]

seems solvable

[pearl pond]

@stoic imp Has your question been resolved?

[stoic imp]

I understood until here

But idk how to continue

[pearl pond]

@stoic imp Has your question been resolved?

[glacial sequoia]

did you solve the system?

your solutions would be (0, 0), (1, 0), (3, -1), (0, -1)

[cunning comet]

how should y = 1 should be a solution for -y^2-y = 0?

[glacial sequoia]

mb fixed

[pearl pond]

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[dapper kraken]

status 1

[autumn trellis]

Hint: Use integrals to bound the sum

[warm current]

So basically, find the floor of A?

[pure rapids]

Pretty sure my teacher went through this before but i forgot

[autumn trellis]

,texsp ||$\frac{1}{\sqrt{x}}$ is decreasing so that $\int_{1}^{100}\frac{1}{\sqrt{x}}dx > A > \int_{2}^{101}\frac{1}{\sqrt{x}}dx$||

[pure rapids]

Ah

[jolly parrot]

4573r01d|)d357r0y3r 45²

[dapper kraken]

yea

[dapper kraken]

wtf

[pure rapids]

Its integral summation, good to use for bashing lol

[dapper kraken]

the intwgral of x^-1/2 is 2sqrt(x) right?

[pure rapids]

Think so

,w integral of 1/sqrt (x)

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by craze.cv

[dapper kraken]

so 18>A>2(sqrt101-sqrt2)

[pure rapids]

Yeah

[dapper kraken]

looks like 17 icl

[pure rapids]

Fr

[autumn trellis]

It is

,calc 2(sqrt(101)-sqrt(2))

[dapper kraken]

oh wait your deathcv lmao i just realized

[jolly parrot]

Result:

17.271324117496

[pure rapids]

lmfao

[dapper kraken]

alr thanks

.close

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[fathom saffron]

why is this wrong: $\sin^{-1} x = \tan^{-1} \frac{x}{\sqrt{1 - x^2}}$ Hence, taking tan on both sides, $\frac{\pm x}{\sqrt{1 - x^2}} = \frac{2x}{1-x^2}$

[jolly parrot]

SirGareth

[fathom saffron]

so either x = 0 or x^2 = -3 which implies only 1 solution

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[formal remnant]

I know how to solve it’s just I can’t find the maximum and minimum for the limits

[formal remnant]

Hi

[quasi sundial]

You have to build up the inequality from taking the maximum and minimum values of sin(npi/2)

[cunning comet]

whats the bounds for sin?

[formal remnant]

-1<= sin <= 1 (idk how to use the bot)

[quasi sundial]

You're close

[cunning comet]

use them instead of sin

[formal remnant]

Ok

Lemme try

Something like this?

[quasi sundial]

There you go

[formal remnant]

Yippee 👍

[cunning comet]

thats it

[formal remnant]

I’ll brb my mom is forcing me to cook lol

[cunning comet]

no need to hurry you should be done.

[formal remnant]

Ok I’m back

Ok I think I got it

Is that correct?

[quasi sundial]

Yes

Whenever you're working with the sandwich theorem look for a function whose bounds you're aware of and can be used for a neat inequality

[formal remnant]

Oki thx

.close

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[rancid skiff]

Hello, I'm looking for ressources about complex analysis used in calculating journey time in a city. My idea is to transform the urban plan into a polygon to reduce calculations of time travel using a car/public transportation/walking. Time travel calculation would be based on existing user constraints (e.g. using a car or walking according to temperature/humidity, total CO2 emissions, journey cost, journey time, carrying a suitcase compromises changing means of transport, etc.). I already got building/routes informations in the plan for the city i want to work on using QGIS. I'm looking for suggestions in order to complete this project

[pearl pond]

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.reopen

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✅

[rancid skiff]

up

[pearl pond]

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[pearl pond]

@rancid skiff Has your question been resolved?

[lunar knot]

It is not complex analysis but you may look for the shortest path problem, or the postman problem, over weighted graphs
Both are from graph theory / optimization / combinatorics

[pearl pond]

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[native coral]

no idea

[native coral]

answer key says something about monotonicity

i just don't understand what it means "call lp(f) the riemann sum for each f and P when we choose the left end-point for each subinterval"

like let's say this is the first rectangle

i'll actually make it clearer

[lofty berry]

it's giving you a different rule to form the terms of the sum

[lofty berry]

right

this is using the left endpoint

[native coral]

then would the second be this?

[lofty berry]

no

[native coral]

what would the second rectangle be

[lofty berry]

first of all it should be between x_1 and x_2

and it should have height f(x_1)

[native coral]

but what's the difference between that and the regular lower sum

[lofty berry]

lower sum chooses the minimum value

imagine you made a rectangle around [-1, 1]

lower sum would have height 5

(in your picture)

[native coral]

right

[lofty berry]

but you are onto something

[lofty berry]

related to this

[native coral]

i'm just confused on what it means by "we choose the left end-point for each subinterval"

[lofty berry]

it means for the interval [x1, x2], you chose the height f(x1)

for the interval [x2, x3], you choose the height f(x2)

so the left-end point of the interval is determining the rectangle height

unlike U_P and L_P, which "look at" the entire interval

like when you write out the Reimann sum, you have to choose a point in each interval to base the height of the rectangle

this rule says, use the left endpoint as the point

[native coral]

isn't x0 the left endpoint?

[lofty berry]

like

it's the left endpoint of the partition

what is a Reimann sum?

you take a partition of [a, b]

meaning $a = x_0 < x_1 < \dots < x_n = b$

[jolly parrot]

zkzach

[native coral]

yeah i have the definition right here

[lofty berry]

then you choose a special point $x_i^* \in [x_{i-1}, x_i]$ for each $i$

[jolly parrot]

zkzach

[native coral]

.rotate

/rotate

bruh

[vast wadi]

,rotate

[lofty berry]

exactly

so notice you wrote $x_i* \in [x_{i-1}, x_i]$

[jolly parrot]

zkzach

[native coral]

yes

[lofty berry]

so the left-endpoint rule is saying choose $x_i^* = x_{i-1}$

[jolly parrot]

zkzach

[lofty berry]

because $x_{i-1}$ is the left endpoint of the interval $[x_{i-1}, x_i]$

[native coral]

ah i see

alright i think i got it now

[lofty berry]

👍

[native coral]

ty

[lofty berry]

yw

[native coral]

.close

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[sterile turtle]

this is part a

for part b i set the integral up like

$\int_{4}^{8} \left( 1 - \frac{4}{x} \right) , dx$

[jolly parrot]

Sho

[sterile turtle]

and i got $4-4\ln{2}$

[jolly parrot]

Sho

[sterile turtle]

answers however say 4-ln2

what happened

did i set the integral up wrong?

[midnight haven]

sub ln 2 = 0.693

[sterile turtle]

?

[midnight haven]

u need area enclosed by curve and lines right

[sterile turtle]

yep

[midnight haven]

ln2 is a constant

[sterile turtle]

but where am I subbing that in??

[midnight haven]

in ur answer 4- 4ln2

[sterile turtle]

how does that make 4-4ln2 into 4-ln2

[midnight haven]

4 - 4ln2 = 4 - 4*0.693

now solve it

[sterile turtle]

solve what?

its a constant value

[midnight haven]

multiply and subract to find area

[sterile turtle]

so how does that equal 4-ln2

[sterile turtle]

please answer this question before i continue

[cosmic charm]

doesnt seem wrong tbh

[midnight haven]

u missed a 4

while integration

integral 4/x dx = 4* lnx

[sterile turtle]

im confused what the other dude is saying

[cosmic charm]

same

[sterile turtle]

so are the answers given to me just wrong?

[sterile turtle]

should be this right?

[midnight haven]

yup

[sterile turtle]

mk

thx for clarification

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[midnight haven]

hi

[midnight haven]

does anyone know what I did wrong in these 2 questions

[plush bramble]

,rotate

[plush bramble]

,calc 9^2

[jolly parrot]

Result:

81

[midnight haven]

[plush bramble]

You did 9^(1/2) = 81

[midnight haven]

🤦‍♂️

Thank you

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[gilded hollow]

$f(x,y,z) = sin(x¨2+y^2)+xyz$\
Determine all 2nd order partial derivatives

[jolly parrot]

Merineth 🇸🇪

[gilded hollow]

If i'm asked this. Do they mean..
f'_x
f'_y
f'_z
f''_xx
f''_xy
f''_xz
f'_yx
.....
f''_zz
?

[ember spear]

If direct or cross is not mentioned, I'd assume that

[gilded hollow]

ahh oki thanks

.close

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[tranquil sinew]

How can I solve this? $\lim_{x\to 1^-} \frac{\sin{\abs{(x-1)(x-2)}}}{x-1-1+x}$

[jolly parrot]

Shachar

[tranquil sinew]

I keep getting to -1/2 but it supposed to be -1

[quiet goblet]

the denominator can just be 2x-2

I suppose you simplified

[tranquil sinew]

yeye

it from the floor function

[quiet goblet]

so

|| is not absolute modulus?

but floor function?

[tranquil sinew]

the original is $\lim_{x\to 1} \frac{\sin{\abs{(x-1)(x-2)}}}{\floor{x}-1+x}$

[jolly parrot]

Shachar

[quiet goblet]

Oh

[tranquil sinew]

so I split it into 1+ and 1-

1+ is easy to compute

1- tho...

[quiet goblet]

I dont understand I think

[tranquil sinew]

wym

[quiet goblet]

[x] split into x-1

[tranquil sinew]

for left sided limit

and x for right sided

[quiet goblet]

Ohh

I see

[sharp smelt]

multiply and dvide by (x-2)?

[tranquil sinew]

what will it give me?

[quiet goblet]

Maybe begin with taking x-1=t?

[tranquil sinew]

the absulote value making sin(x)/x not very possible

[quiet goblet]

since you would want x tends to 0 in that

[tranquil sinew]

here it would "work for sin(x-1)(x-2)/(x-1)(x-2)"

[quiet goblet]

if x-1 = t, then x-2 will be t-1

[hollow pine]

.close

[tranquil sinew]

why

dont close me pls

[quiet goblet]

modulus sign is problematic

[tranquil sinew]

where is a modulus sign?

[quiet goblet]

inside sin

[tranquil sinew]

oh absulute

not really

[quiet goblet]

yea

how will you send modulus sign outside

[tranquil sinew]

it just take care of the problem of sin(-0)

you dont need

it doesnt makes any problems here

[sharp smelt]

Only mods can do that, don't worry

[tranquil sinew]

I mean maybe heine possible here, but I couldnt find two serieses that will work for my case

what is going on with the people here?

hi

go for any of the available channels pls

[fluid axle]

i can also close other chans, idk it's helpful or my yellow tho, it's not only mods

[tranquil sinew]

np

[spare lark]

You're able to close channels

[ember spear]

The problem arises because flr(x) = x-1

[tranquil sinew]

yup

[ember spear]

the floor function doesn't change in its locality

so, you can just say flr(x) = 0

and denominator is x-1

[tranquil sinew]

what does it mean doesnt change in it's locality

[ember spear]

Pardon me for my bad math language

when you change the input by a little, the output doesn't change

[tranquil sinew]

ye but am I allowed to assume x = 1 inside the limit?

isnt it makes a bit of limit in parts?

[sharp smelt]

I wanted to keep the secet so that people don't try to get helpful just to close channels

[ember spear]

Don't share the the secret ancient knowledge

Limit of products is equivalent of product of limits

the limit of a sum is the sum of the limits:

[tranquil sinew]

valid actually

so it's just long

😦

thanks

.close

[pearl pond]

Channel closed

Closed by @tranquil sinew

Use .reopen if this was a mistake.

[spare lark]

Lmao they'll just get ban

[pearl pond]

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[stiff mauve]

how is the lowerbound to x^2 + 3, 3?

[stiff mauve]

|x-2| < δ

if we say δ <= 1

then we have 1 < x < 3

1 < x^2 < 9

4 < x^2 + 3 < 12

shouldnt it be 4

[fringe raft]

3 is some lower bound

it probably doesn't matter how good it is

[stiff mauve]

is 4 also allowed

[fringe raft]

if you copy the proof from there but using the 4 instead, it should make sense too

it's just that proving 4 is a lower bound involves a bit more work

with 3 it's trivial

[stiff mauve]

ohh right

x^2 >= 0

thank you

[pearl pond]

@stiff mauve Has your question been resolved?

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[gilded hollow]

Find all the solutions to u(x,y) to the PDE (picture) which deepnds on r=sqrt(x^2+y^2). i.e, it is in the form u(x,y)=f(r)

[gilded hollow]

i assume i need to find find d^2u/dx^2 and d^2u/dy^2 first right?

Do i have to apply chain rule here? How do i know when chain rule is needed?

[pine jay]

Mostly whenever you have a substitution

So in this it would seem to be the r

Or rather f(r) that is

[gilded hollow]

$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} * \frac{\partial r}{\partial x} = \frac{\partial u}{\partial r} \frac{\partial \sqrt{x^2+y^2}}{\partial x} = \frac{\partial u}{\partial r} * \frac{x}{r}$

[jolly parrot]

Merineth 🇸🇪

[gilded hollow]

$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} *\frac{x}{r} \ \
\frac{\partial u}{\partial y} = \frac{\partial u}{\partial r} * \frac{y}{r}$

[pine jay]

Yes

[gilded hollow]

I think i managed to get the first order pd

[jolly parrot]

Merineth 🇸🇪

[gilded hollow]

i'm not entirely sure how to proceed here

do i do product rule on them?

[pine jay]

Yes

Or chain rule again possibly

Depending on how u do it

[leaden marsh]

Sorry for interrupting
Won't $u = \frac{x^{4} + y^{4}}{12}$

[gilded hollow]

$\frac{\partial^2 u}{\partial x^2} = (\frac{\partial u}{\partial r})' * \frac{x}{r} + \frac{\partial u}{\partial r} * (\frac{x}{r})'$

[jolly parrot]

@leaden marsh

Merineth 🇸🇪

[pine jay]

Are u suggesting an answer to their question??

[leaden marsh]

Yes

[pine jay]

They probably want to find that on their own

[leaden marsh]

Oh ok I'm sorry

[pine jay]

Also I’m pretty sure u(x,y) is not unique

[gilded hollow]

This seems right

I got

$\frac{\partial^2 u }{\partial x^2 } = \frac{\partial^2 u}{\partial r^2} * \frac{x}{r} + \frac{\partial u }{\partial r } * (\frac{1}{r}-\frac{x}{2r})$

[jolly parrot]

Merineth 🇸🇪

[pine jay]

You’re differentiating with respect to x I’m assuming?

[gilded hollow]

Yes

that should be u''_xx

My stomache also tells me u''_yy would be the exact same but replace x with y

$\frac{\partial^2 u }{\partial y^2 } = \frac{\partial^2 u}{\partial r^2} * \frac{y}{r} + \frac{\partial u }{\partial r } * (\frac{1}{r}-\frac{y}{2r})$

[jolly parrot]

Merineth 🇸🇪

[pine jay]

Well in that case you have to be a bit more careful, as the partial of u with respect to r when differentiated with respect to x is not the second partial derivative of u with respect to r

It might be more beneficial to use the suggested form of u(x,y), as f(r), to potentially reduce down to an ODE

[gilded hollow]

How would that work?

This is what i tried to do

[pine jay]

Well from the very start you could of maybe started with f

For example the partial derivative of u with respect to x

Can be written as $\partial /\partial x \big( f(r) \big)$

[jolly parrot]

Aslan

[pine jay]

Now this is just $f’(r)\frac{x}{r}$

[jolly parrot]

Aslan

[pine jay]

Almost as you had written

Now when taking the second derivative with respect to x, we do the same thing plus use the product rule

[gilded hollow]

wait so i take r = sqrt(x^2+y^2) and derive it twice?

[pine jay]

Well we can if we want just forget about $r$, it’s more of a short hand, really what we’re dealing with is $f(\sqrt{x^2+y^2})$

[jolly parrot]

Aslan

[pine jay]

So just differentiate this with respect to x and use the usual chain rule

Then for the second derivative

You use chain rule again

But with the product rule aswell since you’ve got that pesky factor

[gilded hollow]

So this is wrong?

oh those f' and f'' are derived with x and xx respectively

[pine jay]

It’s more of a preference I would say, using f here seems more intuitive to me atleast as it then seems to reduce down to a neat ODE

[gilded hollow]

ODE?

[pine jay]

A differential equation that depends on one variable

“Ordinary differential equation”

[gilded hollow]

I don't understand

[pine jay]

Once you’ve done the business for x

[gilded hollow]

is it wrong or right?

[pine jay]

By symmetry you’ll get the one for y

[pine jay]

Oh this?

I mean I don’t immediately see anything wrong it’s just different

For starters the f there is different

In our case we’re assuming f is the solution, but there it’s an auxiliary function

Hard to tell without words

I don’t like the notation for f’ tho, confusing

[fluid axle]

<@&268886789983436800> ad

[gilded hollow]

https://youtu.be/FSUCLGRVcXA?t=153
With the same reasoning he has in this video, can i say that i do

u -> r -> x and y

[fluid axle]

yeah the video is more complicated than what you're trying to do

this is fine

I'm not completely sure what you're trying to do in the last 3 lines

computing d/dx (x/r) maybe?

that's what it looks like at least

[gilded hollow]

yes

but like

If i just start with x

This part

If i want to derive this again

then i would have to apply the product rule to the rhs

Wouldn't that be correct?

[fluid axle]

yes and that's what you've done correctly indeed

[gilded hollow]

ooh you mean the last two rows?

or nvm

[fluid axle]

there's also an error in the last line if I interpret it correctly

this prime is an x-derivative right?

it's not an r-derivative

[gilded hollow]

Okayy hold on

[fluid axle]

I don't get why you have -1/r^2 then

[gilded hollow]

lets ignore my calculations at the end

i'll redo them properly

[fluid axle]

ok fine

[gilded hollow]

,rotate

[gilded hollow]

I believe this should be correct so far

[fluid axle]

yes

[gilded hollow]

Now i just have to apply the product rule appropriately

GImme a min

[fluid axle]

it is applied already

[gilded hollow]

Oh no i mean

i have to calculate it

I can't leave it as is, can i?

[fluid axle]

but there's some extra computation you can do

yea

[gilded hollow]

Right, that's what i assumed

[fluid axle]

the double derivatives you should try to simplify them

[gilded hollow]

Can you help me with the first product?

[fluid axle]

so d/dx (du/dr) and d/dx (x/r)

[gilded hollow]

d/dx (du/dr)

[fluid axle]

yea

[gilded hollow]

I'm not 100% sure what is done

du/dr * dr/dx?

[fluid axle]

well in the end you only want r-derivatives, that's why we did the chain rule in the first place

[fluid axle]

essentially that yes, you reapply the chain rule again

$\pdv{}{x}\left(\pdv{u}{r}\right) = \pdv{}{r} \pdv{u}{r} \cdot \pdv{r}{x}$

[jolly parrot]

aPlatypus

[fluid axle]

to derive wrt x, you derive wrt r then you multiply by dr/dx

it's still the chain rule, it's just that the inside function is more complicated than before

it's not just u or whatever, it's already a derivative of u inside

[gilded hollow]

I still get r-x/r^2

is it wrtong?

[fluid axle]

yeah

[gilded hollow]

...

[fluid axle]

again r is a function of x

[gilded hollow]

Then i have literally zero idea what to do

it's absolutely impossible to keep track of what i'm doing

[fluid axle]

so to compute (r^-1)' as you wrote before

it's the chain rule again

[gilded hollow]

so i have to

[fluid axle]

-r'/r^2 that's what you should get

[gilded hollow]

write out r?

ok

i redid it

for the 4th time

and i got

r-2x^2 / r^2

FMLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLDK AW' KPOAWD

never mind

redoing it

5th time

[jolly parrot]

aPlatypus

[gilded hollow]

ok

i got

(r^2-x^2)/(r^3)

[fluid axle]

yea

[gilded hollow]

what is this?

[fluid axle]

chain rule

[gilded hollow]

Like is it easier to keep track of stuff

if i write it as such+

[fluid axle]

to compute what you had trouble computing

[gilded hollow]

because i'm getting cancer writing it out

[fluid axle]

well if it helps you it's good

[gilded hollow]

trying to mentally keep track of everything is impossible

[fluid axle]

it's a pain in the ass yeah

at least we have paper

otherwise

[gilded hollow]

omg

i think i see what you did

it's so much easier?!

you did this?

[fluid axle]

gotta learn to love chain rule in this field of work

[fluid axle]

yes

[gilded hollow]

ok so we have derived it ONCE now?

xD

du/dx = (r^2-x^2)/r^3

?

So now i do it again?

[fluid axle]

and there's still the other weird derivative

[gilded hollow]

wrt y?

[fluid axle]

but here it is before your eyes

no

well wrt y just after yes

but it's very similar anyway

[gilded hollow]

shouldn't it be

r^2-y^2 / r^3

my stomache guess

<@&268886789983436800>

[fluid axle]

god they really want to shill their roblox crap today

[gilded hollow]

yeah indeed

[fluid axle]

yeah sure

I wasn't talking about that tho

[gilded hollow]

oh

but i'm halfway done

now i just have to derive it again

$\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x} (\frac{r^2-x^2}{r^3})$

[jolly parrot]

Merineth 🇸🇪

[gilded hollow]

Would this be correct?

[fluid axle]

wait what

[gilded hollow]

we found du/dx?

and du/dy

?

And now i want to find 2nd derivative

[fluid axle]

there was a typo I didn't catch

like this huge product rule is already your 2nd derivative

[gilded hollow]

ah right..

[fluid axle]

and now we managed to compute that red thing on the right

[gilded hollow]

i'm soooooooooo tired

[fluid axle]

now there's still the boxed thing to take care of

[gilded hollow]

Ok i think i know what yuou mean

one moment

[fluid axle]

which I already did, chain rule once again

[fluid axle]

^

[gilded hollow]

the last row

should be correct

oh wait

i have to expand the first term ?

[fluid axle]

first of all you forgot the * x/r for the first term

and du/dr for the second

[gilded hollow]

what

ngl i'm super lost now

not sure what we are doiong

[fluid axle]

this is what your second derivative looks like, this is correct

[fluid axle]

why are the other parts of the product rule missing here

[gilded hollow]

that's right?

[fluid axle]

it's just copying what you have here really

[gilded hollow]

right ok

the second der for y

should be the same except with y instead of x

[fluid axle]

yea

[gilded hollow]

ok

final answer

now these two summed should equal x^2+y^2

I can't tell if they do

[fluid axle]

however what I've also been trying to tell you for the past 40mins is that you can simplify these two things

[gilded hollow]

:I

[fluid axle]

we want to get rid of the x and y derivatives

we just want r-derivatives in the end

and I've computed it for you

[fluid axle]

yes indeed

[gilded hollow]

how did you get that

like thought process

if i were to cover RHS and someone asked me what the rhs is i dont think i would've been able to reproduce it

[fluid axle]

I've tried to explain myself above also, but here's another explanation if you want

say Q is just some function of r

how would you compute dQ/dx using the chain rule ?

[gilded hollow]

q -> r -> x

dq/dr * dr/dx ?

[fluid axle]

we've done that a ton of times at this point, you'd get
$\pdv{Q}{x} = \dv{Q}{r} \pdv{r}{x}$

yeah

[jolly parrot]

aPlatypus

[fluid axle]

now

what if I told you Q was actually du/dr

[gilded hollow]

u -> r -> x
du/dr * dr/dx

[fluid axle]

why so

reread my argument

[fluid axle]

if you replace Q by du/dr in here, what do you get

[gilded hollow]

d(du/dr)/dx = d(du/dr)/dr * dr/dx

[fluid axle]

yea right

d(du/dr)/dr what is this then ?

[gilded hollow]

um

d^2/dr^2 ?

[fluid axle]

d^2 u/dr^2 yea

2nd deriv wrt to r of u

[gilded hollow]

i see

[fluid axle]

so $\pdv{}{x}\left(\pdv{u}{r}\right) = \pdv[2]{u}{r} \cdot \frac{x}{r}$

[jolly parrot]

aPlatypus

[fluid axle]

just replaced the dr/dx

[gilded hollow]

right

[fluid axle]

now you can plug that in here

[gilded hollow]

That seems right

[fluid axle]

almost

you forgot the extra x/r from before

[gilded hollow]

I did?

ahh

so it's

x^2 / r^2

and y^2/r^2

[fluid axle]

yes

[gilded hollow]

okokok

so now we have the full expression

these summed

should be x^2+y^2

[fluid axle]

yea

[gilded hollow]

picture above = x^2+y^2

how the fuck do i determine that HAHA

aaaaaaaah

this is insane

integrals?

[fluid axle]

there's a ton of stuff that simplfies first

with these x^2 and y^2 all around the place

try and combine the d^2u/dr^2 and du/dr terms together

[gilded hollow]

Ok i'll try one moment

like so?

[fluid axle]

right you can factor further

[gilded hollow]

I think that's all?

maybe the r^3 can be used to shorten the on the right term? Not sure

[fluid axle]

no

x^2+y^2

what do you think it is ?

that's the only real simplification here

[gilded hollow]

hmm

r^2 ?

since r = sqrt(x^2+y^2)

then r^2 = x^2+y^2

so the left term

should be

d^2u / dr^2

[fluid axle]

yes

and there's still something you can do on the right one

[gilded hollow]

replace r^2 with x^2+y^2

[fluid axle]

I mean you have -x^2 and -y^2

[gilded hollow]

that would leave x^2+y^2

in the numerator

[fluid axle]

yea

so 1/r du/dr

[gilded hollow]

yea!

hmm

d^2u / dr^2 + du/dr * 1/r = r^2 ?

[fluid axle]

yea

[gilded hollow]

d^2u/dr^2 + du/dr = r^3

i'm lowkey very tired rn

i can't even think

Way overdue breaktime

could you just explain how the last bit is done?

Do i just sub d^2u/dr^2 and du/dr with what we found before?

NO WAIT

I THINK I REMEMBER

[fluid axle]

it's just an ODE now

[gilded hollow]

It's this right

second order ODE

[fluid axle]

yea

[gilded hollow]

holy crap

that was long ago

I'll do it right after some food

not sure i remeember lmao

[fluid axle]

well i gtg lol

[gilded hollow]

hahah no worries

thank you so much

for real

<3

take care

.close

[pearl pond]

Channel closed

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[pearl pond]

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[gleaming arrow]

Hi there, Im trying to understand where the root comes in for this equation but I dont get it. For context, this equation is to calculate the surfrace area of 3d object using intergration

[gleaming arrow]

https://courses.lumenlearning.com/suny-openstax-calculus1/chapter/arc-length-of-a-curve-and-surface-area/#:~:text=Surface Area%3D∫ba,)%20)%202%20)%20d%20x%20.&text=Surface%20Area%3D%E2%88%ABdc,))2)dy.

I tried using this website to understand but got confused

[sharp vigil]

it is from the pythagorean theorem

[pearl pond]

@gleaming arrow Has your question been resolved?

Channel closed

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[errant fossil]

i need a little help understanding how to do this

[errant fossil]

idk what to do with x=6

[snow sail]

its perpendicular to the line x=6

how does it look

what does the line x=6 look like

what does that mean about f

[snow sail]

.

[errant fossil]

oops sorry

[errant fossil]

6/1

?

[snow sail]

no i mean what does it look like

,w graph x=6

[jolly parrot]

Results provided by WolframAlpha

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Requested by janniku

[errant fossil]

ooh

[snow sail]

its vertical right, up and down

so if f is perpendicular to this vertical line, how does it look

[errant fossil]

itll be horizontal

[snow sail]

yea

so, no slope

well slope = 0

so f=mx+b, but m=0

must be f=b

can you find the equation?

[errant fossil]

would b = 5?

[snow sail]

yea i think so

so f=5

,w graph x=6 and y=5

[jolly parrot]

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[errant fossil]

so the -1 in this is useless

[snow sail]

we dont need it if we solve it this way no

you could use it if you solve another way

[errant fossil]

ooh

[snow sail]

it doesnt change the answer

[errant fossil]

can u explain how to do it another way?

out of curiosity

[snow sail]

you can do point slope

so use y-5 = m(x+1)

but then you realize that the slope must be 0

y-5 = 0 * (x+1)

y-5=0

[errant fossil]

so 0=(x+1) is just 0

[snow sail]

yup

[errant fossil]

i seee i see

[snow sail]

so we ""used"" -1

it doesnt change hte answer

[errant fossil]

yeep

thank you very much

[snow sail]

[errant fossil]

.close

[pearl pond]

Channel closed

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[young widget]

hi can someone check this please

[blazing notch]

I think it's good

It def has 3 mins

[young widget]

okay thankss

.close

[pearl pond]

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[sharp smelt]

I'm trying to determine if this statement is true or false : If every proper subsequence of $(x_n) $converges, then $(x_n)$ converges as well.
\
I feel examining the contrapositive will be more prudent here
\
If $(x_n)$ diverges, then there exists a proper subsequence of $(x_n)$ that diverges. This statement is trivially true if we consider the sub- sequence $(x_i), )i \geq 2)$. Thus if every proper subseqence of $(x_n)$ converges, then $(x_n)$ converges as well

[jolly parrot]

math_rocks

[brave sluice]

can i just leave off the first term and call that a proper subsequence?

[sharp smelt]

i think so

thats what i did

[brave sluice]

oh sorry i didn't fully read it

[sharp smelt]

so does this work?

[brave sluice]

the contrapositive isn't as clear to me

i'd rather say every proper subsequence converges => x_i i>=2 converges => x_n converges

the contrapositive should work too though

[sharp smelt]

Got it

thanks

One more question

If $(x_n)$ contains a divergent subsequence, then $(x_n)$ diverges. To prove this , I was thinking contrapositive + bolzano-weierstrass

[jolly parrot]

math_rocks

[brave sluice]

i don't think you need that theorem

but the contrapositive makes sense to me

[sharp smelt]

If $(x_n)$ converges then all sub- sequences of $(x_n)$ converge. By defn, $\froall \varepsilon >0, \exists N \in \N$ st $\froall n>N$ , $\abs{a_n-a}< \varepsilon$. From this it follows that every sub-seqeunce converges too

[jolly parrot]

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

[brave sluice]

if x_n converges, then all the terms after some N are close to the limit, so certainly the terms after N that are contained in the subsequence are also close to the limit

[sharp smelt]

so something like this

rigght

[brave sluice]

yeah

you should probably use \implies after n>N

[sharp smelt]

could probably use a little refinement

[brave sluice]

there could be more details but i'm not exactly sure how to index a subsequence

like if you have a sequence a1 a2 a3...
and you take the odd terms, do you say
a1 a3 a5...
or do you have to re-index so it's
b1 b2 b3...

[sharp smelt]

hmm, yeah

.close

[pearl pond]

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[sharp smelt]

thanks

[pearl pond]

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[sharp smelt]

If (x_n) is monotone and contains a convergent sub -sequence than (x_n) converges. I was thinking $a_n =-n. a_{n-1}=-n$ as a counter examples

[jolly parrot]

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

[sharp smelt]

nvm,. that doesn't converge

$a_{2n}=\sum_{i=1}^{n}\frac{1}{n} ;a_{2n+1} = \frac{1}{n^2}$

[jolly parrot]

math_rocks

[ocean hornet]

not monotone?

also wait, do you mean $a_{2n} = \sum^n_{i=1} \frac{1}{i}$

[jolly parrot]

Bair

[sharp smelt]

yes

my bad

[plush bramble]

Where's the subsequence

[sharp smelt]

the subsequence a_{2n+1} converges

but not the entire seqeunce

[ocean hornet]

_ _

[sharp smelt]

why not

[plush bramble]

How is 2n+1 a subsequence of 2n?

[ocean hornet]

because the odd terms are increasing and the even terms are decreasing?

[sharp smelt]

I mean the subsequence of all the odd terms

[plush bramble]

Where's the sequence then

[cunning comet]

the whole sequence has to be monotone, not only subsequencies.

[sharp smelt]

I see

okay

thanks

[inland ivy]

what makes you think this isn't true?

[sharp smelt]

just following my nose

[inland ivy]

or is the exercise to give a counterexample

[ocean hornet]

i think it is true because the sequnce must be monotone and bounded

[sharp smelt]

prove or give a counter example

[inland ivy]

yeah it does seem to be true

[cunning comet]

your nose is wrong, the statement is true,

[sharp smelt]

I see, okay

why have I been blocked , ThM

I think contrapositive would work well here then

if (x_n) doesn't converge , then (x_n) is not monotone or does not contain a converegent subseqeunce

[inland ivy]

Let ${x_{n_k}}{k=1}^\infty$ be the convergent subsequence of some sequence ${x_n}$, then for all $\varepsilon>0$ there exists $N>0$ such that $\abs{x{n_k}-L}<\varepsilon$ for all $k>N$, where $L$ is the limit of $x_{n_k}$

[jolly parrot]

kheerii

[inland ivy]

I think you can use monotonicity to prove the theorem you want

[cunning comet]

wdym?

[sharp smelt]

I can't react to your messages

[inland ivy]

you want to prove that for all $\varepsilon>0$ there exists $N>0$ such that $\abs{x_n-L}<\varepsilon$ for all $n>N$

[jolly parrot]

kheerii

[sharp smelt]

okay, now I can

thanks

Yeah, okay. I think that follows almost directly from the defn of convergence, no

[inland ivy]

no

[sharp smelt]

if ${x_{nk}}$ is a convergent subseqence, then for if $n>N$, for some $N in. \N$, then $\abs{x_{nk} -L}< \varepsilon$

[jolly parrot]

math_rocks

[karmic fern]

The idea here is that the convergent subsequence gives you the limit L, and you can use its elements to bound / squish the elements of the parent sequence against L

which is exactly what you need to do to prove convergence

and you can do this squishing because you know the sequence is monotone

WLOG you can probably just treat monotone sequences going up

[sharp smelt]

hmm

Why do I have to do that

[karmic fern]

you don't have to do anything

[sharp smelt]

Isn't this sufficient

[inland ivy]

that's just a statement

it doesn't prove anything

[sharp smelt]

right, I got confused

my bad

I think contrapositive would work here: if $(x_n)$ doen't converge, then $(x_n)$ is not monotone or does not contain a convergent subsequence. From the contrapositive of the monotone convergence theorm, we can conclude that $(x_n)$ is not monotone.

[jolly parrot]

math_rocks

[sharp smelt]

Good so far?

[inland ivy]

what would the contrapositive of the monotone convergence theorem be

[sharp smelt]

if (x_n) doesn't converge than it's not monotone or not bounded

[inland ivy]

are you sure?

[sharp smelt]

yes

[inland ivy]

no

if (x_n) doesn't converge then it's not BOTH monotone and bounded

it can still be any one of them

[sharp smelt]

Oh

wait, what

The monotone convergence theorm is : If a seqeunce is montone and bounded than it converges

[steel dagger]

Isnt (-1)^n a counter example to this

[inland ivy]

ah I misread your statement

your statement is fine too

[steel dagger]

Bounded but not monotone

[inland ivy]

yes and it doesn't converge

so it's not both monotone and bounded, it's only one of those things

[inland ivy]

how do you know (x_n) is bounded?

[sharp smelt]

I don't know

[sharp smelt]

It's an or statement, so either one works

[inland ivy]

what?

[steel dagger]

Doesnt every real sequence have a convergent subsequence

Instead of using contrapositive you can prove that

[sharp smelt]

The seqeunce of naturals doesn't

Consider the sequence 1,2,\dots

no subseqeunce of it converges

[steel dagger]

Ah i see

Every bounded real sequence has a convergent subsequence

[sharp smelt]

I think contrapositive would work here: if $(x_n)$ doen't converge, then $(x_n)$ is not monotone or does not contain a convergent subsequence. From the contrapositive of the monotone convergence theorm, we can conclude that $(x_n)$ is not monotone. OR. from the contrapositive of the monotone convergence theorem, we can conclude that $(x_n)$ is not bounded, and hence doesn't have a convergent subseqeunce .. Thus if $(x_n)$ doesn't convegre then $(x_n)$is not monotone or not bounded

[inland ivy]

from the contrapositive you can conclude that (x_n) is EITHER not monotone or not bounded

you can't conclude both

the direct proof is the easiest method here imo

[inland ivy]

start from this and use the condition of monotonicity

[inland ivy]

to prove this

[sharp smelt]

that sounds a lot harder to me, can I please have advice on contrapositive, or should I do a direct proof

[sharp smelt]

yeah

[inland ivy]

if you've proven the monotone convergence theorem this is very similar to that

[jolly parrot]

math_rocks

[sharp smelt]

How

I have some personal work to attend to gtg

thanks for the help

.close

[pearl pond]

Channel closed

Closed by @sharp smelt

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[dapper kraken]

i extended BD by AD to get E such that AD=DE

that was my only idea tho D: i got a somewhat simmilar q a while back with this trick, i did a bit of angle chasing but didnt get anywherw

[pearl pond]

@dapper kraken Has your question been resolved?

[pearl pond]

@dapper kraken Has your question been resolved?

[rigid lava]

try using angle bisector theorem

[errant fable]

solution simple with trig

with geo however...

[rigid lava]

i can even begin to understand this with geo

angle bisector theorem wasnt useful

[errant fable]

i think i solved

[rigid lava]

present ur soltion

[errant fable]

answer is 100 degrees

[errant fable]

EBC is langleys triangle

[dapper kraken]

whar

[dapper kraken]

https://en.m.wikipedia.org/wiki/Langley's_Adventitious_Angles
this??

[errant fable]

yep