help-41

[tepid girder]

somewhere

I am so damn confused

[frozen horizon]

What is this integral

[tepid girder]

yash bro

I get it

but I have so many doubts

so it is advisable for me to do this tomorrow

bgn brother

💪

tysm for help tho

[teal oxide]

gn

no worries

[frozen horizon]

If you're going can you use .close

[amber wasp]

@tepid girder Has your question been resolved?

[tepid girder]

yh

[amber wasp]

Channel closed

Closed by @tepid girder

Use .reopen if this was a mistake.

[amber wasp]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[amber wasp]

Please don't occupy multiple help channels.

[ripe gull]

<@&268886789983436800>

[amber wasp]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[amber wasp]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[paper kindle]

i have no idea how to do this.. pls help

[tepid verge]

Write out as many facts about your two points (give them names) as you can

You can draw pictures

[paper kindle]

yea y^2 is 4x so x will be y^2/4 then i kept the value of x in two point formula.. and i got the m as y1 +y2 is 3 so the m is 4/3 after that i have no idea what to do

[tepid verge]

I don't know who m is

[tulip pumice]

Slope probably

[tepid verge]

So explicitly your two points are $(y_1^2/4, y_1)$ and $(y_2^2/4,y_2)$ right?

[paper kindle]

its slope

[grizzled pagoda]

Flip

[tepid verge]

Slope of the line defined by what two points, each of those and the origin?

[tulip pumice]

A chord will be considered untill it is inside the parabola right ?@tepid verge

[tepid verge]

Oh I see, sure

So it's the slope of the chord

What if the chord is vertical?

[tulip pumice]

Yeah

Would make sense

[paper kindle]

m is 4/3

the slope

[tepid verge]

I just don't see it, can you explain more? The slope of what, subject to what?

[paper kindle]

wait ill solve and send a pic

till what i have done

[tepid verge]

I would say the chord is always inside the parabola, because parabolas are convex

Or like, you know, the region containing the focus is

[paper kindle]

I have no idea like if this is how to proceed but this is how I did.. and idk what to do anymore and this is the slope of the chord from my side

@tepid verge

[tepid verge]

Oh interesting, ok so that is indeed the slope. I didn't expect it to be constant

So now you can designate one of y_1 or y_2 as your independent variable, representing one in terms of the other

Then you can define a line L, the chord of the two points, in terms of your single independent variable

And then, get the distance between the origin and L, and minimize it

[paper kindle]

k lemme try

the line isnt fixed cuz one of the y is variable so the 1st option is incorrect?

this is a multiple correct que btw

[tepid verge]

I assumed this was calculus

You have a function D(y), measuring the shortest distance between the origin and the line L(y), and you want to minimize D(y)

[paper kindle]

it could be solved by calculus but we haven't been taught that yet

[tepid verge]

Ok word, so unless if the distance function happens to be constant too, then I agree, it's not necessarily 9/20 because the line can change

[paper kindle]

yeaa

[tepid verge]

I mean really you could choose y_1 = 0, y_2 = 3

That line contains the origin, so the shortest possible distance is 0 after all

[paper kindle]

how do u know the line contains origin

pls explain

[tepid verge]

If y_1 = 0, then x_1 = y_1^2 / 4 = 0^2/4 = 0, so (x_1, y_1) = (0,0) is on the line

[paper kindle]

ohh

what abt option 2

how to see if thats correct or not @tepid verge

[tepid verge]

What is a normal?

[paper kindle]

A normal to a parabola is a straight line perpendicular to the tangent line at a specific point on the parabola, passing through that same point

[tepid verge]

Was that in your own words or copypasted?

[paper kindle]

mine

[tepid verge]

Ok, I had to ask

Well, how do you define the line normal to a parabola at an arbitrary point?

[paper kindle]

whats an arbitrary

[tepid verge]

arbitrary as in, its selection has no significance

I say R = (x_3, y_3) is a point on the parabola

with no further conditions, R is arbitrarily chosen

[paper kindle]

ohh?

[tepid verge]

It's not a math term

[paper kindle]

ohk

so how to proceed now? u have any idea for the 2nd option?

[tepid verge]

I already asked a question, how do you describe the line N normal to the parabola at a given point R?

[paper kindle]

its perpendicular

to the tangent

[tepid verge]

Can you give an equation for this line N?

[paper kindle]

at r

[paper kindle]

yes

[tepid verge]

Then do so for your two points and explain, if you could

[paper kindle]

its y= mx-2am-am^3

this is the slope eqation for normal

[tepid verge]

So is that also just a given equation or did you derive it at some point?

[paper kindle]

its a given

like its the standard noraml equation to parabola

[tepid verge]

Is it given when the parabola is of a particular form?

i.e. when does this hold true?

[paper kindle]

um

no

it hold tru verytime most prolly cuz its the standard slope form

of normal to parabola

[tepid verge]

y = ax^2, y = ax^2 + bx + c?

[paper kindle]

y = ax^2

only

not the second one

[tepid verge]

This parabola is different, this is x = y^2/4

[paper kindle]

no its y^2=4x which is the same thing as x = y^2/4

its in the form of y = ax^2

[tepid verge]

It's not exactly the same because the roles of x and y are switched

not ax^2, it's ay^2

so there might be some adjustments needed

[paper kindle]

yea sorry

mb

[amber wasp]

@paper kindle Has your question been resolved?

Channel closed

Closed by @paper kindle

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[hidden patrol]

Hello! I've been experimenting with procedural terrain generation algorithms, and while doing so, I stumbled upon a problem that I am probably not capable solving without some help.
Consider a Voronoi graph generated from sites on a square lattice. This graph would just be all square tiles, but if we apply some perturbation (random offset) to the sites, we get a much more chaotic looking Voronoi graph -- see first attached image.
Now, the question is, how much can the sites be perturbed, such that the sites keep their original neighbors? In other words, what shape is the boundary of positions where a site is guaranteed to neighbor another site that it is connected in the square lattice?

[hidden patrol]

Consider the following sites A, B and C. The yellow square would be the boundary of valid positions for the vertex X of the Voronoi graph. X cannot go below the bottom yellow line, because if there was another site D below A, it could cause the edge between B and A to vanish.

This applies to all other directions as well, hence the orange boundary being a square.

If I knew the position of B and C, I could compute the area where A could be placed -- here the darker regions correspond to those areas computed from the two remaining sites.

Using the fact that the areas should be symmetric, because the lattice could extend in all directions, I can take the intersection of rotational and reflective symmetries of each area:

And finally, I can superimpose the areas and intersect them, getting these black boundaries:

But these boundaries are not a general solution, they are computed from the current positions of the sites.
I would like to figure out the actual boundary, that will guarantee that when I place sites within them, the neighboring sites in the square lattice will remain neighbors in the Voronoi graph.

The computation of the orange Voronoi vertex X is a mapping from ℝ⁶ to ℝ², which makes this problem difficult.

Finally, here's an interactive tool I made to explore this problem: https://www.desmos.com/calculator/bdwvwknskw

[amber wasp]

@hidden patrol Has your question been resolved?

[amber wasp]

@hidden patrol Has your question been resolved?

[amber wasp]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[hidden patrol]

.reopen

[amber wasp]

✅ Original question: #help-41 message

[hidden patrol]

Update: I have no proof but I have a hunch this is the solution. It's too perfect not to be it.

.close

[amber wasp]

Channel closed

Closed by @hidden patrol

Use .reopen if this was a mistake.

[blazing spindle]

"Can someone please tag the bot or send a link for the 'Bartle Real Analysis' solutions PDF? I can't find it in the search

[amber wasp]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[round zodiac]

Could anyone help me with this problem?

[round zodiac]

I'm stuck in part a, I've been trying to see something like f(x+a)=f(x) somewhere by going from one integral to the other

but I'm gettng that integral of f(x+b) from 0 to a is equal to integral of f(x) from b to b+a

but the period is a

not b

and cannot use change of variable or sorts, just fundamental theorem of calculus and definition

[terse rivet]

split $\int_b^{b + a} f = \int_b^a f + \int_a^{b + a} f$

[round zodiac]

oh right, I'm so dumb

.close

[amber wasp]

Channel closed

Closed by @round zodiac

Use .reopen if this was a mistake.

[leaden stone]

i am dumb

[amber wasp]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cedar tundra]

the lesson is about finding the area of a triangle with this method
I'm wondering how the j became positive or this is all an error

[fluid ridge]

it's probably an error

[cedar tundra]

.close

[amber wasp]

Channel closed

Closed by @cedar tundra

Use .reopen if this was a mistake.

[cedar tundra]

thx

[amber wasp]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[keen pawn]

I'm kind of confused here

The likelihood function is

$\prod_{i=1}^{n} (1- \theta)^{x_i-1} \theta= \theta^{n} (1- \theta) ^{\sum){i=1}^{n} x_i -n}$

[grizzled pagoda]

wai

[keen pawn]

And sure , we can maximise its log

$n \log(\theta) + (\sum_{i=1}^{n} x_i -n ) \ln (1- \theta)$

[grizzled pagoda]

wai

[keen pawn]

so $\frac{n}{ \theta} = \frac{ \sum_{i=1}^{n} x_i -n}{ 1- \theta}$ is needed

[grizzled pagoda]

wai

[keen pawn]

so $\theta= \frac{n} { \sum_{i=1}^{n} x_i}$

[grizzled pagoda]

wai

[keen pawn]

unless I' tripping?

but I don't see what this has to do with the factorisatioon theomr

Like this suggests \sum x_i -n is sufficient too, but?

[vernal bobcat]

Isn’t this just sum of the xi

Or just the sample mean

[vernal bobcat]

At this point you have your model

What does the factorisation lemma say

@keen pawn

[keen pawn]

oops

[keen pawn]

hi, yes

[keen pawn]

if I can write the likelihood function as h(x) g(\theta, T(x)), Tx is a sufficient statistic

[vernal bobcat]

Okay

So $\theta^n \exp((\sum x_i -n)\ln(1-\theta))$

[keen pawn]

sure

[grizzled pagoda]

frosst

[vernal bobcat]

Oh no that’s not right is it

How do you pull the 1- θ out

Exp first?

[keen pawn]

where

[keen pawn]

here?

I took the log

[vernal bobcat]

Yeah

Yeah but there’s a way to turn a^b into something * e^b

I’m pretty sure

I’m so dead rn I can’t think

lna * e^b maybe

[keen pawn]

that doesn't sound right

[vernal bobcat]

ae^b?

[keen pawn]

lemme just verify

one minute

[vernal bobcat]

Hold up a^b = exp(ln(a^b)) = exp(blna)

Okay good

[keen pawn]

$e^{ b( \ln(a)-1)}$

[vernal bobcat]

This was right

[grizzled pagoda]

wai

[vernal bobcat]

Where did this -1 come from

[keen pawn]

a^b=xe^b

bln(a)= b+ ln(x)

[vernal bobcat]

Okay whatever it doesn’t matter

[keen pawn]

ln(x)=. b(ln(a)-1)

[keen pawn]

so you're saying this is the neyman factorisation

[vernal bobcat]

This guy is clearly $h(\theta)g(\theta, T(x))$

[grizzled pagoda]

frosst

[vernal bobcat]

With $T(x) = \bar x$

[grizzled pagoda]

frosst

[keen pawn]

sure, what did I do wrong here though

[vernal bobcat]

Why did you differentiate and set to zero

That’s for part 2

First you need to show it’s sufficient

[keen pawn]

okay, so overline{X} is a sufficient stat

[vernal bobcat]

Oh why do they want to maximise the log

[keen pawn]

I thought I'd first maximise the log

yeah

[vernal bobcat]

Maybe it’s just giving you some hints

[vernal bobcat]

This hints towards the mle of θ is a function of xbar

[vernal bobcat]

A minimal one even

[keen pawn]

it does yeah, which is why I was trying to figure out what I did wrong

because I'm getting 1/overline{X}

[vernal bobcat]

Oh there’s also a theorem for this

https://stats.stackexchange.com/questions/253568/maximum-likelihood-estimator-that-is-not-a-function-of-a-sufficient-statistic

Your mle becomes a function of a sufficient statistic

So you can see mle of θ is a function of xbar

[keen pawn]

I see

[vernal bobcat]

That’s like the hint to be oh look for the xbar for factorising

[keen pawn]

got it

thanks so much !

.close

[amber wasp]

Channel closed

Closed by @keen pawn

Use .reopen if this was a mistake.

[amber wasp]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vapid portal]

Hello

[vapid portal]

i had a problem in

how do you integrate

tan inverse (1-x/1+x)

like integral of tan^-1(1-x/1+x)

[tepid verge]

all you really do is use some of the integration tricks you've stored

here's what I thought: the derivative of arctan(-) is algebraic, so why not try ibp?

[magic lily]

lol cbse board?

put $x = tan{\theta}$

[grizzled pagoda]

parthisjoking

[magic lily]

the inside becomes
$\tan({\frac{\pi}{4}-\theta})$

[grizzled pagoda]

parthisjoking

[magic lily]

its ez after that

[magic lily]

if u dont see this happening lmk

[amber wasp]

@vapid portal Has your question been resolved?

[vapid portal]

LMFAO yes howd you know

[magic lily]

same set

[vapid portal]

OHHH SHI
What was the answer for this

How was your paper bro??

[magic lily]

i did the 1/x^1/2 + x^1/3

[magic lily]

100 hopefully

[vapid portal]

oohh that one

how to do that

[vapid portal]

Ayy lessgoo

[magic lily]

put x = t⁶

[vapid portal]

Oohhh

F

[magic lily]

yes

[hot lagoon]

Another method is using arctan a - arctanb as arctan of a-b/1+ab and then take a and b as 1 and x

[magic lily]

or

[vapid portal]

Ohhh

Bhai mai to kuch bhi krke aayega isme 😭

[magic lily]

this

[vapid portal]

ilate laga diya or shit

[vapid portal]

i was thinking aisa hi hoga

[magic lily]

its ok , it was kinda tuff

[vapid portal]

but idk click hi nahi hua

[vapid portal]

Bhai indefinite wala tough tha

[magic lily]

definite was ez

[vapid portal]

Vo sin^2026 ka ans π/2 tha na

[vapid portal]

Okieee

[magic lily]

everything was ez

[vapid portal]

Kasam se

[magic lily]

i didnt have that

[vapid portal]

Aayein

Konsa tha

Tera

[magic lily]

regionwise diff hota h paper

set3

[vapid portal]

Ohh

Mera set 1

Nahi set 3

Sorry

Mera bhi

Abe par kaise

[amber wasp]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[quartz saddle]

In my linear algebra book (Local university) they are not having such topics...i don't know why

[quartz saddle]

Can anyone tell me which book contains these topics?

[pallid canopy]

linear algebra book: https://textbooks.math.gatech.edu/ila/overview.html

use index

e.g.

[quartz saddle]

Thank you very much

.close

[amber wasp]

Channel closed

Closed by @quartz saddle

Use .reopen if this was a mistake.

[amber wasp]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[surreal vigil]

Is there a good way to approximate the piecewise (z<0 : x else y) with polynomials, other than lagrange?

[amber wasp]

Please don't occupy multiple help channels.

[surreal vigil]

completely different question ):

[indigo cloud]

well there are lots of different ways to compute the interpolating polynomial but it will still end up the same anyway

fundamentally the piecewise probably isnt continuous which will fuck up your approximation anyway

[surreal vigil]

Given it’s continuous and so is its derivative

But higher degrees no

I can’t taylor expand bc it’s a piecewise but I’d prefer a nonsampling method if at all possible

[shy kraken]

Why cant u taylor expand?

[indigo cloud]

often chebychev nodes are optimal for interpolation

[surreal vigil]

because taylor expansion relies on d derivatives, but piecewises break that. so the approximations using the 2nd order derivative and so on will break completely at the piecewise switch

and I need this to be approximated solely with additions, subtracts, multiplications and divisions, hence a polynomial

[indigo cloud]

sounds like xy

[shy kraken]

Can u send the problem maybe that helpd

[indigo cloud]

divisions are stronger than just polynomial

there are things like pade approximations but I dont know how well those do here either

[surreal vigil]

ideally as few divisions as possible** since each division requires an LU of a matrix of functions

[indigo cloud]

the heck are you doing

[surreal vigil]

it's a personal project, I am working with a system of nonlinear ODEs that I am putting through spectral transforms, which means I have nonlinear functions of weighted sums of basis vectors, and need to approximate those as nonlinear functions times each basis vector separately so I can turn it into a system of equations

unfortunately piecewise functions have been a problem since it seems I must sample, and I really want a neater way than that

[shy kraken]

I see i see

[shy kraken]

Maybe try other type of functions

Sry for bad engliscj

[surreal vigil]

I do not have a choice in the system of equations I am transforming lol

and I'm using simplified models to boot

I expect the real models are significantly worse

the piecewises come from transistor equations in this case

[indigo cloud]

btw spline interpolation is also generally good

wait, that's just piecewise again

nvm

[shy kraken]

How far are u into this project

[surreal vigil]

wdym?

as far as I know, everything is feasible / I know methods / I have seen methods for resolving every problem I know of in implementation

technically I can approximate piecewises with lagrange

it's just a really messy solution and I really want something better

but yes in the bases I'm working with, only addition, subtraction, multiplication and division are defined between my vectors

so e^(vector) needs to be approximate through 1 + vector + vector^2/2! + ...

though I can improve that by extracting the mean from my vector, and expanding around that instead of 0 (for various reasons, that does work)

but yeah I checked, spline won't work because on evaluation, I won't be able to know which spline to use (by piecewise) as I won't have real values, but spectral coefficients instead (which corresponds to not being able to split the output into basis vectors)

padé won't work because piecewise can't be approximate by power series, but I should keep that in mind for functions that can

[surreal vigil]

oh you meant in context with real models — circuit simulation doesn't typically use real models all the time, you use simplified models for standard simulation (standard parts work very hard to correspond to standard behavior), then real models when doing delicate stuff, ie when you're doing ASIC design with analog transistors that's when the models are basically maxwell's laws + numerical interpolations from measured data, instead of simplified piecewises

"simplified" models like that I am using are good enough most of the time

[surreal vigil]

ah ok, that would help answer #help-7｜zen1thxyz for me then

[amber wasp]

@surreal vigil Has your question been resolved?

[amber wasp]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[amber wasp]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lime ingot]

can i get help man

[lime ingot]

i have no clue whats goin on here at all

[hidden narwhal]

a or b

[lime ingot]

can someone explain this to me intituitively

everything

[hidden narwhal]

so you havent solved this yet?

[lime ingot]

we went from finding tangent planes to this all of a sudden and for the life of me im tryuna understand the correlation here

[lime ingot]

no not at all ive only found partial derivatives but i have no idea why we need them either

[hidden narwhal]

$$dV = \frac{\partial V}{\partial r} dr + \frac{\partial V}{\partial h} dh$$ this?

[grizzled pagoda]

Minhh

[lime ingot]

not dV just dV/dr and dV/dh

i have no idea what the intuition is tbh

<@&286206848099549185>

[hidden narwhal]

well this means if you want to build a cylinder, you have to add "stuff" in two directions, which are outward (radius) and upward (height)

[pastel salmon]

Yes ?

Sorry can’t help, too advanced

[hidden narwhal]

imagine if you have a finished soda can, if you want to make it thicker you wrap around it a layer, that area of that side is the circumference 2 pi*r times the height h

while the thickness is dr

and you know that the intuition here is volume = thickness multiply area

but if you want to make the can taller, you add a thin disk both to the top and bottom

the thickness here is now dh and area of top of circle is pir^2

thats how i imagine of, idk if this helps

[amber wasp]

@lime ingot Has your question been resolved?

[lime ingot]

is this what you mean

[strong pulsar]

reminder!

[hidden narwhal]

Yes

Thank you for reminding us lunar

[amber wasp]

@lime ingot Has your question been resolved?

[amber wasp]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wanton charm]

could someone help me with this question?

[winter badge]

work out the first 6

and see what happens...

[deep edge]

6 ?

[winter badge]

well I mean youll see it at 5

[deep edge]

I would’ve said 5

Yeah

[coral wigeon]

compute as many as you need to before you see what’s going on

[winter badge]

^

[wanton charm]

like the 6th term?

[wanton charm]

aight

so after n! where n > 4 all compute to 1

[winter badge]

=4...

but yes

[wanton charm]

ok thanks

[deep edge]

Yup

[wanton charm]

so final is i + 95

[deep edge]

Yup

[wanton charm]

thanks

.close

[amber wasp]

Channel closed

Closed by @wanton charm

Use .reopen if this was a mistake.

[amber wasp]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lime ingot]

can i get help for this bruh

[lime ingot]

how does this relate to tangent planes

why do we care about the total differential here

[amber wasp]

@lime ingot Has your question been resolved?

[lime ingot]

<@&286206848099549185>

[granite kiln]

id imagine

this has nothing to do with tangent planes

and we do not care about the total derivative

[lime ingot]

do you have a different way of doing it

because in reality it lowkey does

[granite kiln]

isnt it just like

replace r and h with 1.05r and 1.05h

and compare that to the volume of the cylinder

[lime ingot]

well i dont know man thats the point

[granite kiln]

🤷‍♂️

[fierce linden]

youve asked about total differential a few times. if u have some would u mind me giving a mini lecture on its derivation and application?

[lime ingot]

the total derivative is basically the linear approximatuion of (delta)z, as (delta)y and (delta)x approach 0 of some initial point on a given curve,

(delta)z can be found by finding its change along the curve parallel to one axis + the other axis

and apparently thats the partial derivatives

[fierce linden]

thats correct but i can explain how it relates to errors if ur interested

[lime ingot]

yes please

id appreciate it

[fierce linden]

for notational convenience lets go back to singlevar. at the end ill just outright state the multivar analog of our results bc its proof is similar

we have a function $f:\bR\to\bR$ and an input $x\in\bR$ with output $f(x)$. suppose that we measure the input with error $h$. when we try to record $f(x)$, this error causes us to instead record $f(x+h)$, so we want a formula that expresses the induced error in the output
$$f(x+h)-f(x)=\text{ error term that is accurate for small }h$$
equivalently
$$f(x+h)=f(x)+\text{ error term that is accurate for small }h$$

[grizzled pagoda]

ロケット・ジャンプ

[fierce linden]

@lime ingot all clear?

[lime ingot]

i guess?

so how would this apply here

[fierce linden]

sry ive only just started. im going very slowly bc itll get wild

[lime ingot]

no worries

[fierce linden]

the thing is what i just wrote is unrigorous. to be rigorous ill use little o notation. are u possibly familiar with it?

@lime ingot

[lime ingot]

o notation?

[fierce linden]

'little o notation'

[lime ingot]

ohh

yes

like x_0

?

yeah

[fierce linden]

no

ok ill assume u dont know and define it

[lime ingot]

okay

[fierce linden]

given functions $F$ and $G$ and a point $h_0$, we say "$F(h)$ is little o of $G(h)$ as $h\to h_0$" if
$$\lim_{h\to h_0}\frac{F(h)}{G(h)}=0$$
when this is true we can alternatively write
$$F(h)=o(G(h)) \text{ as }h\to h_0$$
the bottom equality is not literal equality. its a way to say that $F$ is in a "class" of functions, and that class is defined by vanishing when we divide by $G$ and take the limit $h\to h_0$

[lime ingot]

oh okay

[grizzled pagoda]

ロケット・ジャンプ

[fierce linden]

@lime ingot clear?

[lime ingot]

sure

[fierce linden]

sry but i need to take another detour before rigorously rewriting my intro

[lime ingot]

its okay but

can i tell you where im at right now with the geometric interpreation so you know what to write

[fierce linden]

when i mention a linear function, it doesnt mean $g(x)=Ax+B$ from high school. it means a function $g$ that "respects" addition and scalar multiplication of its inputs. the former means $g(a+b)=g(a)+g(b)$ for all inputs $a$ and $b$. the latter means $g(ca)=cg(a)$ for all scalars $c$ and inputs $a$

[grizzled pagoda]

ロケット・ジャンプ

[fierce linden]

yep thats the correct visual

and the boxed result will follow from my derivation

[lime ingot]

so far i understand the labels except for one thing which is the total derivative being expressed as the partials in the x and y direction

[fierce linden]

it will come once i state the multivar analog of my results

[lime ingot]

okay

[fierce linden]

now ill state the rigorous version of my goal

unrigorously we want
$$f(x+h)=f(x)+\text{ error term that is accurate for small }h$$
rigorously we define a linear function $df(x):\bR\to\bR$ (the $(x)$ in the function name denotes that its "attached" to the input $x$) such that
$$f(x+h)=f(x)+df(x)(h)+o(h) \text{ as }h\to0$$
we desire this so that we can informally write
$$f(x+h)\approx f(x)+df(x)(h) \text{ for small }h$$
our goal is to find an explicit formula for $df(x)(h)$ for all $h$, so that the entire equation above (called the linear approximation of $f$ at $x$) is made explicit

@lime ingot clear?

[grizzled pagoda]

ロケット・ジャンプ

[lime ingot]

im sorry i hate to make you right all that just for me to not get it but its not clicking tbh with you

for now is there like a simple way to understand this because i have a test tmrw so i think the best option is to let go of understanding the concept for now

not the best thing but i have no other choice tbh

[fierce linden]

ik its not visual friendly, its just total differentials rigorously come from bashing real analysis

for now if u just bear with me and blindly accept the above goal of linear approximation then i guarantee the rest is mostly algebra and standard calc

[lime ingot]

alright then

[fierce linden]

ill recap the goal that youve blindly accepted. attached to $x$ we define a function $df(x):\bR\to\bR$. we impose that $df(x)$ is linear and that it satisfies
$$f(x+h)=f(x)+df(x)(h)+o(h) \text{ as }h\to0$$
our goal is to find an explict formula for $df(x)(h)$ for all $h$

[grizzled pagoda]

ロケット・ジャンプ

[lime ingot]

alright 😭

[fierce linden]

now my final detour.. is a relatively straightforward result from linear algebra (im aware youve probably not taken it)

the only linear functions $\bR\to\bR$ are constant*h. in other words if $g:\bR\to\bR$ is linear then there exists a constant $c$ such that $g(h)=ch$ for all $h$

[grizzled pagoda]

ロケット・ジャンプ

[fierce linden]

this fact tells us that there exists a constant $c$ such that $df(x)(h)=ch$ for all $h$. we are just left to find $c$

[grizzled pagoda]

ロケット・ジャンプ

[fierce linden]

now lets do some algebra and calc

plug in the above fact to get $f(x+h)=f(x)+ch+o(h)$

[grizzled pagoda]

ロケット・ジャンプ

[fierce linden]

$f(x+h)-f(x)=ch+o(h)$

[grizzled pagoda]

ロケット・ジャンプ

[fierce linden]

$\frac{f(x+h)-f(x)}h=c+\frac{o(h)}h$

[grizzled pagoda]

ロケット・ジャンプ

[fierce linden]

take the limit $h\to0$. recall the definition of little o notation: the $o(h)$ (as $h\to0$) term means that it vanishes if we divide it by $h$ then let $h\to0$
$$\lim_{h\to0}\frac{f(x+h)-f(x)}h=c$$

[grizzled pagoda]

ロケット・ジャンプ

[lime ingot]

im sorry but it doesnt seem like im understanding at all 😭

the problem is im completely lost when it comes to this

[fierce linden]

is there any confusion in the algebra?

[lime ingot]

no no its the concept

i just dont understand why / how tangent planes, are supposed to relate to a freakin error,

[fierce linden]

im very close to stating the results

[lime ingot]

the book sucks, my professor sucks

they dont even explain this stuff

@fierce linden can i ask something

[fierce linden]

?

ill wait for ur question. in the meantime ill state the results

our equation says $f'(x)=c$. we finally have an explicit formula
$$df(x)(h)=f'(x)h$$

[grizzled pagoda]

ロケット・ジャンプ

[fierce linden]

our results for singlevar:
\begin{enumerate}
\item formula for the total differential $df(x)$ of $f$ at $x$:
$$df(x)(h)=f'(x)h$$
\item formula for error in output caused by error in input:
$$f(x+h)=f(x)+f'(x)h+o(h) \text{ as } h\to0$$
equivalently
$$f(x+h)-f(x)=f'(x)h+o(h) \text{ as } h\to0$$
\item linear approximation of error:
$$f(x+h)\approx f(x)+f'(x)h\text{ for small } h$$
equivalently
$$f(x+h)-f(x)\approx f'(x)h\text{ for small } h$$
\end{enumerate}

[grizzled pagoda]

ロケット・ジャンプ

[fierce linden]

now ill state the multivar analog of the results. their derivation is similar to the singlevar case. note that in the multivar case $x$ and $h$ are vectors and written as column vectors whenever necessary.
\begin{enumerate}
\item formula for the total differential $df(x)$ of $f$ at $x$:
$$df(x)(h)=Jf(x)h$$
where $Jf(x)$ is the jacobian matrix of $f$ evaluated at $x$. \tbf{note the right side is a matrix vector product}
\item formula for error in output caused by error in input:
$$f(x+h)=f(x)+Jf(x)h+o(\norm{h}) \text{ as } h\to0$$
equivalently
$$f(x+h)-f(x)=Jf(x)h+o(\norm{h}) \text{ as } h\to0$$
\item linear approximation of error:
$$f(x+h)\approx f(x)+Jf(x)h\text{ for small } h$$
equivalently
$$f(x+h)-f(x)\approx Jf(x)h\text{ for small } h$$
\end{enumerate}

[grizzled pagoda]

ロケット・ジャンプ

[lime ingot]

okay

is there any chance you could explain this to me like youd do to a 5 year old

like these

[fierce linden]

in the context of ur problem the volume is a scalar function $V(r,h)$, so its jacobian matrix is
$$JV(r,h)=\m{\pdv{V}{r}&\pdv{V}{h}}$$
unfortunately we recast $h$ as a scalar representing height. we will rewrite the vector $h$ in our results as $\m{dr\dh}$. result (1) says what youve been after the whole time,
$$dV(r,h)(dr,dh)=\m{\pdv{V}{r}&\pdv{V}{h}}\m{dr\dh}=\pdv{V}{r}dr+\pdv{V}{h}dh$$

[grizzled pagoda]

ロケット・ジャンプ

[lime ingot]

yeah no im sorry this made it worse

idek what a jobean matrix is

all i know is how to find a tangent plane equation given a multi variable function

and thats literally ALL i know and can do regarding this material

thats it

[fierce linden]

for a scalar function its just the matrix of partial derivatives (just a row vector)

if the function has multiple components then its more complicated but we dont worry about it

[lime ingot]

@fierce linden do you think i can take this approach to solve these kinds of problems?

[fierce linden]

thats the full story of how total differential relates to errors. ik u havent absorbed too much of it but i invite u to try after the exam is done. for now all u need is to apply the results to solve ur problem

plug in the function $V(r,h)$ to result (3) to get the linear approximation of error
$$V(r+dr,h+dh)-V(r,h)\approx\pdv{V}{r}dr+\pdv{V}{h}dh \text{ for small }dr,dh$$

[grizzled pagoda]

ロケット・ジャンプ

[lime ingot]

tbh my test is tmrw and im pulling an alnighter and im probably gonna need a memory based method to go about this, do you have any recommendations?

like given these kinds of problems, im pretty sure theres a memorization way to go about this or something right

like plugging in stuff

[fierce linden]

i already provided the results. u just need to plug in the given function

ur problem says estimate error in the volume. that means plug in V(r,h) into the linear approximation given in result (3) as ive done above

[lime ingot]

but it shows you ended up with a combination of partial derivatives

[fierce linden]

did u expect any different?

its consistent with the boxed equation in ur tangent plane picture

[lime ingot]

its just not making sense man

[fierce linden]

lets say we're looking at a cylinder so $V(r,h)=\pi^2rh$. plugging into my approximation,
$$\text{error in volume}\approx\pi^2hdr+\pi^2rdh\text{ for small }dr,dh$$

[grizzled pagoda]

ロケット・ジャンプ

[fierce linden]

lastly plug in r and h along with the errors dr and dh

[lime ingot]

yes okay so

is that for question 4?

[fierce linden]

4 and 5a bc they both ask for error estimates

[lime ingot]

yes

but what about h and r

[fierce linden]

weird thing is 4 doesnt give values of r and h

4 and 5 have different shapes so this is what we apply

https://cdn.discordapp.com/attachments/908078493655502919/1483395129149624441/150414892225134592.png?ex=69ba6ede&is=69b91d5e&hm=724f88ba422d41d0957d55dec3dc82bb7b005f94ab193978e29f2d8506811e1f&

if it wasnt clear, the whole left side is the error in volume

[lime ingot]

yes i see that tbh

uh

what abt r and h

[fierce linden]

weird thing is 4 doesnt give values of r and h

[lime ingot]

yeah so what do we do about it

but the thing is in the homework there was an answer

[fierce linden]

my most generous interpretation of "maximum possible" is to maximize the error estimate as r and h vary over all values

[lime ingot]

i dont think ive ever not understood a topic in math like this

any advice for the rest?

🤣

[fierce linden]

this is quite dumb bc the error blows up as r and h get larger..

5b is the exact same..

it asks for error estimate

plug function $R(R_1,R_2)$ into result (3). it will look like
$$\text{error in }R\approx\dots\text{ for small }dR_1,dR_2$$

[grizzled pagoda]

ロケット・ジャンプ

[fierce linden]

its late and i need to sleep so let me finish off

in case u dont go back to reread my derivation, at least memorize the results

https://cdn.discordapp.com/attachments/908078493655502919/1483390626123878400/150414892225134592.png?ex=69ba6aad&is=69b9192d&hm=cd6335c7f67d035d92750734806933d1c3ed2c922541505800441392381ca455&

for every problem about error estimates plug the given function into result (3). if ur not sure how, mimic the examples i did above

[lime ingot]

ill try my best

thanks alot for your help

[fierce linden]

@lime ingot in theory now u know the full story of total differential, errors, and linear approximation. i did warn it would be wild. i encourage u after the exam to read it back to try understanding as much as u can, and feel free to ask me follow up questions on any part

have a good one, gl on the exam

[amber wasp]

@lime ingot Has your question been resolved?

[lime ingot]

you too, thanks alot

[amber wasp]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lime ingot]

can someone help me understand differentials and approximation

[lime ingot]

like conceptually

if someone can help then ill let you know where im stuck on exactly

[amber wasp]

@lime ingot Has your question been resolved?

[amber wasp]

@lime ingot Has your question been resolved?

[vernal bobcat]

it's saying that it's approximately the same

but if we look at first order approximation it's equal

we only look at linear approximations of the surface (so only up to first order)

[lime ingot]

im confused on the part where delta z is calculated as a combination of partiual derivatgives

whats the geometric logic behind that

[vernal bobcat]

do you agree that ∆z = ∆x + ∆y?

well, ∆z = ∆z_x + ∆z_y

[agile sun]

i think it won't make sense without a little linear algebra

you are basically picking a basis on the tangent plane and using linearity of Df(x,y) to reduce the computation of Df(x,y)(Δz) to a linear combination of Df(x,y)(Δx) and Df(x,y)(Δy)

[amber wasp]

@lime ingot Has your question been resolved?

[amber wasp]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[orchid patrol]

how can I obtain 10a? I have $-\frac{\sin^2(x)\cos(x)}{3} - \frac{2}{3}\cos x$ but I don't see what sub I can do

[grizzled pagoda]

BigBen

[pallid canopy]

use d/dx sin(x)= cos(x)

[spiral zealot]

What's the result in exercise 7 and 8

[orchid patrol]

im not following

[stuck flint]

u=sinx

[pallid canopy]

did you not learn derivatives before integrals?

if not that's fine, just surprising

[orchid patrol]

I didn't read your message properly

your sayign derivte sin is cos

[pallid canopy]

d/dx represents the derivative yes

[orchid patrol]

I was viewing it as d/dx*sinx

[orchid patrol]

just misread

but that is using u sub. the exercise is trying to tell me to use 7 or 8 and from the second part of 8 I have what I wrote but I can't see what trig subs to do from there

I tried sin^2 = 1-cos^2 x

[pallid canopy]

well yea i suggested that before you showed what 7 and 8 were

i suppose you have to integration by parts on sin^2(x) * cos(x) again using 7 to integrate sin^2(x)

from 7

alternatively, you can also use 8 with n=3

[orchid patrol]

with 8 and n = 3 I can only see how to simplify to $-\cos x -\frac{\cos^3 x}{3}$

[grizzled pagoda]

BigBen

[pallid canopy]

simplify what to that?

[orchid patrol]

so we have $\int \sin^3 x dx = - \frac{\sin^2 x \cos x}{3} + \frac{2}{3} \int \sin x dx$

[grizzled pagoda]

BigBen

[orchid patrol]

I turned that into this

[pallid canopy]

are you not able to integrate sin(x) ?

[orchid patrol]

I did thats where I had the $-\frac{2}{3}\cos x$

[grizzled pagoda]

BigBen

[pallid canopy]

,w int sin^3(x) dx

[grizzled pagoda]

Results provided by WolframAlpha

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

Requested by rie.mann

[pallid canopy]

did you try using sin^2 = 1 - cos^2

[orchid patrol]

ye thats how I turned the other fraction into $\frac{-\cos x - \cos^3 x}{3}$

[grizzled pagoda]

BigBen

[orchid patrol]

these two things lead me to this

[pallid canopy]

,tex .double angle

[grizzled pagoda]

pi_day

[pallid canopy]

maybe try the 2nd formula to reduce the power. there may be other trig identites you'll need

[stuck flint]

I think according to exercise 7 and 8 the book want them to solve it through IBP

[orchid patrol]

ok let me see what I can do

[orchid patrol]

I don't know what IBP is

[stuck flint]

Integration by part

[orchid patrol]

ye but we don't have any intergrals anymore

[pallid canopy]

yea this is incorrect somewhere

,w -1/3 *( cos(x) - cos^3(x)) - 1/12(cos(3x) - 9cos(x))

[grizzled pagoda]

Results provided by WolframAlpha

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

Requested by rie.mann

[pallid canopy]

since that's ^ not a constant

[orchid patrol]

isn't this different than what we had?

[pallid canopy]

this was the last time you had a correct step. show your entire work in one or two messages to get to an expression without integrals

[pallid canopy]

your answer

[pallid canopy]

the correct answer

if your antiderivative was correct, it would differ from wolfram's by a constant

[orchid patrol]

This is all the steps I had

[pallid canopy]

sign error here

-(a-b) = -a + b

[orchid patrol]

Thanks

The second part for the last line looks like cos(3x) just missing the sin parts

I got it but I had to go backwards

.solved

[amber wasp]

Channel closed

Closed by @orchid patrol

Use .reopen if this was a mistake.

[amber wasp]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cyan marlin]

Given f is a scalar function and v is a vector function ,

To prove :-

d(f r)/dt = f dr/dt + r df/dt

See image *

Is the proof correct .

[cyan marlin]

The step:
∆t-> 0 f(t+∆t) ≈ f(t)
Is what I am unsure about

[shrewd elm]

it's valid, but you need justification

that statement is equivalent to saying that f is continuous at t

[cyan marlin]

Continuity?

[shrewd elm]

yep

[cyan marlin]

Then , it's fully Rigorous?

[shrewd elm]

up to the point where the screenshot ends, yea

you have to explain why f is continuous at t, though

[cyan marlin]

Okay , understood

Thank u

[shrewd elm]

yw

[cyan marlin]

.close

[amber wasp]

Channel closed

Closed by @cyan marlin

Use .reopen if this was a mistake.

[amber wasp]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[stable kelp]

can someone help me find the currents of every resistor in this circuit? i got these equations from kirchoff but got wrong values when i tried solving it
not even sure if they are right, and there should be an easier way to do this right?

[amber wasp]

@stable kelp Has your question been resolved?

[shadow stump]

the fact that none of the equations have any negative signs is suspicious

[astral umbra]

uh which way are the I1, I2, and other Is

[stable kelp]

the last 2 are the nodes between three resistors so the sum should be 0
i did 4 loops clockwise

[astral umbra]

well those two i can understand but i think you need to determine the direction of the currents for the first three as @cloud said

[amber wasp]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[eager pebble]

can someone help please? Sequences

[pallid canopy]

Wut

[pallid canopy]

!original

[amber wasp]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[sinful sapphire]

any two could be true

[amber wasp]

@eager pebble Has your question been resolved?

[winter badge]

I'm assuming this is 1 geometric series and you just want to find which terms make sense. It's good to start off by recognising that these are powers of 2. Then you will see why frowny's statement is true and it's actually 2 geometric sequences that are contained in the answers here

[winter badge]

Is this new information from the question?

[eager pebble]

Yes sorry i didnt see the full question 😭

[winter badge]

well because its geometric

you know to go from the 6th to the 7th term you multiply by r

so to go from the 6th to add term you multiply by r twice i.e. $r^2$

[grizzled pagoda]

Nyxzore

[winter badge]

knowing that you can work out what r is

[eager pebble]

ah i seee

[eager pebble]

Whats r used for

[winter badge]

the ratio

,w geometric series

[eager pebble]

is rhere a follow up formula for this

[winter badge]

well you know any geometric sequence is
${ar^n}$

if i had 3 9 27 ...

my r value is 3

cuz to go from one term to the next i am multiplying by 3

so to get the 4th term id just say 27*3

and so on

[grizzled pagoda]

Nyxzore

[eager pebble]

Ouhh whats a

[winter badge]

guess

[eager pebble]

is it the first term

[winter badge]

well done

[eager pebble]

but we dont know term 1

[winter badge]

we dont need to

I just need to know how to traverse terms

lets say i can you a random sequence

3, ..., 12, 24, 48

[eager pebble]

6

[winter badge]

exactly!

[eager pebble]

so is r 2

[winter badge]

yes

knowing r can get you anywhere

[eager pebble]

okay so 1024/(2r) is 256 ?

[winter badge]

i see the confusion

well we know the 6th is 256 so lets write it like this
{..., ..., ..., ..., ..., 256, ..., 1024}

3, ..., 12, 24, 48 looking back at this

you saw that r=2

to go from 12->48

i'd need to times by r twice

i.e. 12.r.r

[eager pebble]

so r^2

[winter badge]

ayep

[eager pebble]

Thank you

.close

[amber wasp]

Channel closed

Closed by @eager pebble

Use .reopen if this was a mistake.

[amber wasp]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[solemn escarp]

Determine all monotone functions $f : \mathbb{R} \to \mathbb{R}$ such that [f(x +f(y)) = f(x) + f(y) ] for all real $x,y$

[solemn escarp]

well clearly f(f(x)) = f(x) + f(0) for all x but idk how to continue

[pseudo crescent]

f(x + f(y)) = f(x) + f(y) looks quite interesting,
f(x + k) = f(x) + k (where k = f(y) for any y you pick)

[solemn escarp]

oh wait i forgot a part

[grizzled pagoda]

Copter

[pseudo crescent]

oh that might change some stuff

[solemn escarp]

monotone means injectivity right

[pseudo crescent]

depends

it might not be strictly monotone

montonicity includes constant sections too usually

[solemn escarp]

oh okay

suppose that we have f(a)=f(b)
then f(a+f(y)) = f(b+f(y))

if f(y) non const then a=b

[pseudo crescent]

if f(y) non const then a=b
Wdym?

why?

[solemn escarp]

if f(y) nonconstant then (a+f(y),b+f(y)) is some interval adm we have for all y so we can vary y

i think

[pseudo crescent]

maybe f(y) just doesnt vary as much as you would hope it would

i just managed to find quite a non-trivial solution, which is not injective (not the x=0 one)

[solemn escarp]

oh

🫩

[pseudo crescent]

f(x + k) = f(x) + k (where k = f(y) for any y you pick)

This is probably the most interesting thing

the most obvious functions with this property is probably f(x) = x+c

then there are some heavily discontinuous (but not monotone examples) and also some discontinuous but still monotone examples

try finding some of those discontinuous and still monotone exampels

f(x + k) = f(x) + k (where k = f(y) for any y you pick)
This basically says
f(x+k) = f(x) + k for all k in ran(f) btw

[solemn escarp]

hmmm

how would i use monotone for this though

[pseudo crescent]

im not yet sure how to do the formal proof, but the montonicity is there just to remove some very ugly very discontinuous sols

f(x+k) = f(x) + k for all k in ran(f) btw
Try just constructing something with this property

oh i think i know how to do the formal proof now

do you know how to do it, assuming f is injective?

[solemn escarp]

yeah

but there are cases where it is not injective?

[pseudo crescent]

yeah, there are such cases

what does it mean if a function isnt injective but is monotone

what can you say about such function

[solemn escarp]

there is some interval where it is constant

[pseudo crescent]

yep, its constant on some interval

so lets suppose that on some interval [a,b], f(x) = c

f(x+k) = f(x) + k for all k in ran(f) btw
Draw it and use this to get some values outside of [a,b] and see how f starts to look like

[solemn escarp]

linear?

[pseudo crescent]

well, its constant somewhere so it cant be exactly linear

[solemn escarp]

i mean its linear outside of (a,b)

[pseudo crescent]

Not necessarily

For example if on [1,2) f(x) = 1, it would force e.g. f(2.5) = f(1.5 +1) = 2, similarly f(2.1) = 2...

[solemn escarp]

would it be constant of the same interval

[pseudo crescent]

Mhm

In this specific case, itd result in floor(x)

[solemn escarp]

i see

[pseudo crescent]

So these floor-like functions are another class of solutions

Now youll probably have to go through to casework on the constant intervals (for example if f is 0 on the interval, we arent able zo get much info from it) and see where the floor-likeness is actually forced

[fierce linden]

i think we can do a lil algebra

f(x+k)-(x+k)=f(x)-x

so g(x)=f(x)-x is k periodic. then use f monotone to get some conditions on g

[pseudo crescent]

That was my og idea, but i didnt find the monotonicity condition very useful

It could be used with continuity or to construct some very ugly sols tho

[fierce linden]

||g periodic kills f decreasing||

||then f increasing gives a nice concrete condition on g||

[amber wasp]

@solemn escarp Has your question been resolved?

[pseudo crescent]

honestly the class of floor-like solutions is so big that its quite difficult just to describe it, for example
f(x) = 1 on [0, 1/3)
f(x) = 2 on [1/3, 1)
f(x+n) = f(x) + n for n in Z
works too, and its not a plain floor
Describing the g functions as rokketo suggested might be easier

||You can pick any y and then H := {ny | n in Z} will be the range. Now pick an arbitrary monotone function f' mapping [0, y) -> H and extend it by the rule f(x + y) = f(x) + y. That gives you f: R -> H, which is a valid sol.||

There are more sols than I expected

They mightve intended "monotone" to mean "strictly monotone"

[fierce linden]

u know ur cooked when even helpfuls misspell ur name

[fierce linden]

this doesnt really affect much, at least from what i see in my solution

[amber wasp]

@solemn escarp Has your question been resolved?

[amber wasp]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[amber wasp]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[amber wasp]

Please don't occupy multiple help channels.

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[small copper]

<@&268886789983436800>

[amber wasp]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[languid nest]

Help me with 31 and 32

[small copper]

What have you tried?

[shy kraken]

dude

<@&268886789983436800>

poor guy

[languid nest]

What

[languid nest]

I have attached

[shy kraken]

there was a gambling ad guy a gain

[languid nest]

Someone help

[small copper]

Do you have any ideas how to continue?

[languid nest]

I just know it’s gonna be like a vase

But I can’t draw

[small copper]

Here you have all the necessary to 31.

What we have in x and y?

[languid nest]

That’s what I m confused about

I don’t know what to draw

[small copper]

The traces at y = k are circles that get smaller until they become a point at y = 1. This tells you that the figure has a peak on the y-axis and opens up like a cone.

Right?

[languid nest]

I can’t visualise

[small copper]

You can draw the axes to visualise it better.

Using our values.

[languid nest]

which values

We just have k

[small copper]

It's just a way of saying any number. Try substituting it with any number.

[languid nest]

oh I still don’t get it

[small copper]

Did you draw it?

[languid nest]

I m confused

so I just skipped that one

I m now stuck on another similar problem

[small copper]

Send it here. And state what you tried or what ideas you have. 🙂

[small copper]

Don’t worry with practice and drawing you will get it.

[languid nest]

okay

Drawing this

[small copper]

What ideas you have. You need only to draw it, right?

[languid nest]

yeah but I m so confused

when I have to draw in 3 d

Like on paper

What’s the directrix gonna be

U there?

[small copper]

The hyperbolic paraboloid is a ruled surface. This means you can draw it by joining points between two principal lines (or curves).

[languid nest]

huh?

but we do need

two hyperbolas

on each surface

[small copper]

Remember that in a hyperbolic paraboloid, the parabolas are not the structure, but the result of cutting the figure.

[languid nest]

Can u explain me how you draw on paper I feel clueless

[languid nest]

Then how to draw it

[small copper]

First, draw a parabola opening upwards in the yz-plane and a parabola opening downwards in the xy-plane, both intersecting at the origin. Right?

Now you can know how to conclude?

[languid nest]

I m confused

Won’t it be hyoerbolas

Can u draw it i m so confused

[small copper]

The parabolas are the skeleton you draw first, and the hyperbolas are the walls that connect them to form the saddle.

[small copper]

Don’t worry.

[small copper]

I can’t give you the answer sorry.

🙂

[languid nest]

Yeah I wanna do it myself too

Lemme try

[small copper]

Try it.

If you made mistakes I will correct you, don’t worry.

[native stag]

is this thomas?

[languid nest]

@small copper hey

I don’t know how we make the y = -x^2

[stoic valley]

Hello bhai

[languid nest]

Han bhai

yar

Yeh hyperbolic paraboloid

ne

halat kharab ki huwi hai bhai

rona araha hai

[stoic valley]

Arey koi na bs all is well boldo🌚

Abhi kaunsa 31 kar rhe ho na?

[small copper]

Good!

[stoic valley]

Acha theek ye banda kar dega badhiya se help

acha hai ye wala banda usse hi smjh lo

[small copper]

Just draw a parabola that starts at the center and opens flat along the floor (the xy-plane), pointing towards the negative y-axis.

[languid nest]

is the one I drew yz

@small copper

[small copper]

Better.

[small copper]

You have to draw that same inverted U but starting from where all the axes intersect.

It’s the only think.

[languid nest]

Why so

[small copper]

Because when y=0 x=0. Right?

[amber wasp]

@languid nest Has your question been resolved?

[languid nest]

Got it

Thanks

[amber wasp]

Channel closed

Closed by @languid nest

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[clever hull]

I'm doing ellipses and hyperbole in high school. I need some help, i know how to do the basics, but when it gets even a lil bit harder i get stuck.

[clever hull]

this is the excercise where im stuck

[echo pivot]

Whats the question

[clever hull]

I need help to start this excercise. I can do the last part, like drawing it on the graphic, but i just need to find Centre, a, b, c and e.

[clever hull]

I need some help gettin there

[echo pivot]

Aight , so do you know if its an hyperbole or an ellipse?

[clever hull]

help*

[echo pivot]

Asking to see how much you much you know currently

[clever hull]

I understand this is and hyperbole, since there is the - in between.

[echo pivot]

So the basic form of an hyperboal is, in this case,
(y - β)²/b² - (x-α)²/a² = 1

So, on the right hand side, you need to get 1

Then the center would be (α,β)

And a² would the opposite of the coefficient thats with x². In this case, itd be 9/16

Youd get something like this, meaning β = 2 and α = 0

And 1/b² = 1/9

And 1/a² = 16/9

[amber wasp]

@clever hull Has your question been resolved?

[clever hull]

mh, so the centre is (0;2)

but shouldnt the a be 3 just like the b

Cuz in hyperboles and ellipses i calculate the a and the b by doin the square root of whats under the fraction

[clever hull]

like this

[echo pivot]

Not quite, you dont square root just whats under the fraction, you need to root the entire coefficient

[clever hull]

sorry if i explain myself bad, english is not my first lenguage. if u need some clarification, im herw

[clever hull]

oh alright

[echo pivot]

It worked for that cuz there wasnt anythinf multiplied in the numerator

[clever hull]

but this is at the power ²

[echo pivot]

Yes

[clever hull]

so if i do the square root

it comes out right?

[echo pivot]

Yeah, youd get 1/a = 4/3

So a = 3/4

[clever hull]

OHHHHH ALRIGHT

tysm man, i really appreciate it❤️‍🩹

[amber wasp]

Channel closed

Closed by @clever hull

Use .reopen if this was a mistake.

[echo pivot]

[amber wasp]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[rapid plaza]

hey guys, how am I supposed to do this. Im in adv func so I dont know any calculus

[round rune]

multiply by conjugate/conjugate

[stoic valley]

Have you tried taking conjugates?

[rapid plaza]

multiply the root by the cojugate?

cuz isnt that usally used when you have a fraction

or do I make it into a fraction

[stoic valley]

by root i hope you mean the whole numerator right

[rapid plaza]

yea

[stoic valley]

Okay go ahead then

[rapid plaza]

ok I got to 3x+7/ (sqrt(x^2 + 3x + 7) + x)

[stoic valley]

Okay nice

Now leaving the Question aside for a moment

what is a/x when x is tending to infinty?

(a is a definite constant)

[rapid plaza]

decreasing?

[jagged zephyr]

Noo

[stoic valley]

Nope i am asking the value of the limit

[jagged zephyr]

What is 1/infinity= (tending to ...)

[rapid plaza]

ummmm aprouach 0?

[jagged zephyr]

Yeeee

[stoic valley]

Okay good!

[jagged zephyr]

So you have to make smthing like that

[rapid plaza]

what do you mean make something like that

oh cuz roots cant be negative

[jagged zephyr]

Make 1/infinity type so u get 0

[rapid plaza]

and denom cant be 0

[rapid plaza]

type?

[jagged zephyr]

Umm u need pic ?

[rapid plaza]

ok wait first what does geometrical meaning is that the graph?

[stoic valley]

Can you convert all the terms in the expression into smth like a/x or a/(x^2) or a constant itself?

[jagged zephyr]

Idk I can help to solve questions but idk what that ?

[stoic valley]

not required to analyze graphically here

[rapid plaza]

so the top is 3/x + 7/x^2

or 3 + 7/x

[stoic valley]

uh?

Oh you mean numerator

[woeful sable]

technically DNE

[stoic valley]

Yeah!

[stoic valley]

well not exaclty as you could say its tending to zero itself

[stoic valley]

Oh well then what did you do for this?

[rapid plaza]

wait can ygs check this can I based it off this

[stoic valley]

I am assuming you divided the numerator by x?

[stoic valley]

This is correct but this wont lead you anywhere!

[rapid plaza]

ad then I multpy by the conjugate

[stoic valley]

Sure you could do that but its better to multiply by conjugate before divding by x

[woeful sable]

ahh my bad thought you were talking about 1/x when x approaches 0

[stoic valley]

Ah😅

[rapid plaza]

wdym also the conjugate is just the same equation but +1 instead of -1 right

[stoic valley]

Nope look carefully!

[stoic valley]

here you are getting a form of inf x 0

[rapid plaza]

ok wait lemme restart

[rapid plaza]

ill do this