help-13

[jaunty mural]

|g(x) - b| < delta_2

We WANT 0 < |g(x) - b| < delta_2

In order to use the last implication

[gentle lintel]

so we need to ensure g(x) ≠ b?

[jaunty mural]

yes.

[gentle lintel]

skjdjskdnfndp

this makes much more sense

[jaunty mural]

Condition 2 says g(x) =/= b for small enough epsilon

[gentle lintel]

then whatever mumbo jumbo we argued about yesterday

[jaunty mural]

Hence, we can use that final condition and combine

delta_2 and t are unused so I chuck those quantifiers

$$(\forall\varepsilon_2\in\bR^+)(\exists\delta_1\in\bR^+)(\forall x\in\bR)$$
$$(0<|x-a|<\delta_1\implies|f(g(x))-c|<\varepsilon_2)$$

[wraith dagger]

Shuri2060

[jaunty mural]

And that's exactly the statement that lim (x->a) f(g(x)) = c

Me not assuming the deltas/epsilons were the same were kinda necessary (it'd be just incorrect to assume they are)

[gentle lintel]

wait what

how can we combine

[gentle lintel]

we can’t sub t = g(x)

even with sufficient conditions

[jaunty mural]

I am combining by substituting t for g(x)

Because we know 0 < |g(x) - b| < delta_2

Hence that last implication applies for t = g(x)

[gentle lintel]

??

[jaunty mural]

That's NOT the same as what you were suggesting earlier

$$(\forall\varepsilon_2\in\bR^+)(\exists\delta_2\in\bR^+)(\forall t\in\bR)(0<|t-b|<\delta_2\implies|f(t)-c|<\varepsilon_2)$$

Look, we had this

[wraith dagger]

Shuri2060

[jaunty mural]

If I just substitute t=g(x) into the entire statement I get

$$(\forall\varepsilon_2\in\bR^+)(\exists\delta_2\in\bR^+)(\forall g(x)\in\bR)(0<|g(x)-b|<\delta_2\implies|f(g(x))-c|<\varepsilon_2)$$

[wraith dagger]

Shuri2060

[jaunty mural]

That's just wrong

'for all g(x) in R', blah blah

No.

Here I'm not replacing 't' everywhere with g(x)

I'm only doing it in that specific statement, NOT quantifiers as well.

@gentle lintel You're like suggesting something like this

$$(\forall x\in\bR^+)(-\sqrt{x} < \sqrt{x})$$

$$(\forall x^2\in\bR^+)(-\sqrt{x^2} < \sqrt{x^2})$$

It's just wrong

[gentle lintel]

wait how can we equate t and g(x) anyway

[wraith dagger]

Shuri2060

[gentle lintel]

aren’t they separate entities

[jaunty mural]

We have the 3rd statement

is true for all real t

g(x) is indeed real.

[gentle lintel]

oh yea

i cant think properly smh

[jaunty mural]

I can't make a proper example, but I hope you get what I meant

[gentle lintel]

yea so all this concludes all we need to know for change of variable shits?

[jaunty mural]

I'm not really changing variables.

[gentle lintel]

like the theorem

[jaunty mural]

I'm 'noticing' g(x) satisfies the criteria for t in the last statement

Yh I guess?

[gentle lintel]

and its intricacies

[jaunty mural]

ngl, we overcomplicated it way too much

Just a visual understanding is enough

[gentle lintel]

i’m just dumb lol

also another thing

when should i start analysis

[fallen heath]

wait so

did you get the proof?

[gentle lintel]

yea

[fallen heath]

for the theorem?

[gentle lintel]

i got both intuitive and algebraic understanding of that one fucking condition

and the entire theorem too

good for me i guess

[fallen heath]

Imma just save the chat (ヘ･_･)ヘ┳━┳

[gentle lintel]

after diff eqs?

or what

idk

[jaunty mural]

idk

no idea what u doing

If you can understand this quantifier whatever

And limits

You've already started analysis

[fallen heath]

Shuri, can you point out to all steps you used? to prove the theorem?

[jaunty mural]

I'll rewrite ok

[gentle lintel]

epsilon delta

and stuff

[jaunty mural]

just the screenshots

[gentle lintel]

lol

[crimson sedge]

So the main argument that @jaunty mural made was we needed 0<|g(x)-b|<epsilon for x close to a .For that we needed g(x) not equal to b close to a

[fallen heath]

Hmm.. thanks ^^

[jaunty mural]

First step is this. EDIT, I meant let epsilon_1 = delta_2

[wraith dagger]

Shuri2060

Shuri2060

Shuri2060

[jaunty mural]

Then the statements can be combined so. (I guess I really mean for x sufficiently close to a)

[wraith dagger]

Shuri2060

Shuri2060

Shuri2060

[jaunty mural]

Using the 3rd statement,

[wraith dagger]

Shuri2060

Shuri2060

[jaunty mural]

(getting rid of unused quantifiers)

[wraith dagger]

Shuri2060

Shuri2060

[jaunty mural]

Think that's it.

The statement was true for all epsilon

So picking out a specific epsilon = delta

is perfectly valid

[crimson sedge]

This proof is infact is really similar to showing if f,g are cts so is f compose g.Here we needed 0<|g(x)-b| but there g(x) can be equal to b as well I think

[crimson sedge]

Yes,it is a nice proof.

[gentle lintel]

this is why we need condition 2

[crimson sedge]

Yes, exactly

[jaunty mural]

The limit statement does not include a '0 <' for continuity

Because we require the function to equal its limit at that point

[gentle lintel]

one more thing

[fallen heath]

[gentle lintel]

is the ε-N the ε-δ for infinity limits?

[jaunty mural]

There is a different between sequences and functions

[gentle lintel]

limits of sequences are foreign to me

[jaunty mural]

It's the same idea

In analysis usually you do sequences first

[gentle lintel]

…, x > N ⇒ |f(x) - L| < ε

right

[fallen heath]

By this, we're basically fixing all epsilon 1,2 and delta 1,2?

[jaunty mural]

$$\lim_{n\to\infty}a_n = L$$
$\iff$
$$(\forall\varepsilon\in\bR^+)(\exists N\in\bN)(\forall n\in\bN)(N\leq n\implies|a_n-L|<\varepsilon)$$

[wraith dagger]

Shuri2060

[jaunty mural]

(defn of limit for sequences)

[gentle lintel]

what

ok..?

[jaunty mural]

Using my previous example

f(x -> 0) : [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...] (limit is 0)

g(x -> 0) : [1, 0, -1, -0.5, 0, 0.5, 0.25, 0, -0.25, 0, ...] (limit is 0)

(fg)(x -> 0) : [0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, ...] (limit is undefined)

f(x -> lim g) : [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...] (limit is 1)

You give me an open interval around L

I can tell you a point in the sequence after which everything is in that interval

For g(x -> 0) : [1, 0, -1, -0.5, 0, 0.5, 0.25, 0, -0.25, 0, ...] (limit is 0)

If you tell me to get everything between (-0.26, 0.26)

I give you N = 7

[gentle lintel]

wait

does real analysis involve calc 3?

[jaunty mural]

🤷‍♂️

[gentle lintel]

rigorous definition of multi variable limit

or something

is this studied

or is it in 2 analysis courses

one for single var, one for multi var

[crimson sedge]

This is a handwritten version of @jaunty mural 's proof.

[jaunty mural]

Just realised I didn't substitute my deltas/epsilons properly

Forgot when rewriting the whole thing

[crimson sedge]

Oh

[jaunty mural]

Pretty sure I did the 1st time round

[fallen heath]

._. I was wondering why it looked scary

nvm I'll go about the whole discussion still anyways

including the visualization of limits

[jaunty mural]

Well the proof idea stands even if I didn't write it with the right letters lol

[fallen heath]

something new to learn

[crimson sedge]

Anyway,I got your argument from the chats .

[jaunty mural]

👌

[crimson sedge]

If you are familiar with showing composition of functions are cts then this is just an easy modification of that proof

[fallen heath]

no.. I mean the visualization thingy

this however should've been easy for me had I attended my analysis classes properly smh

[jaunty mural]

I think considering g as 'how you approach' is the easiest

[fallen heath]

or even did any of my analysis work last 3 semesters (ヘ･_･)ヘ┳━┳

[crimson sedge]

They won't teach you to visualise right 😅

Like they will teach you in Calculus

And tell you to forget that in real analysis

As we are supposed to write full rigorous proofs for everything

[fallen heath]

exactly

is why Visualization thingy was kinda cool n new for me :o

Anyways, nice to know your doubts been cleared @gentle lintel

[gentle lintel]

visual math is cool

[gentle lintel]

i’ve had worse, no worries lol

[crimson sedge]

You will love Algebraic topology then I think

[jaunty mural]

nvm I did substitute properly. It's just the 1st 'Let...' is off. Should be 'Let epsilon_1 = delta_2'

💤

[gentle lintel]

modelling himalayas

[jaunty mural]

alg topo is pretty good

[crimson sedge]

Yeah,it was like the first course for me where I was asked to visualise things and draw pictures instead of sometimes writting proofs 😅. Visualising was like a requirement there.

[gentle lintel]

i’ve had this channel on for like 14 hours lol

i should probably close it

.close

[cedar kiln]

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[fading socket]

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[cunning lava]

I don't know if this is the correct server to ask this question but it's about boolean algebra. I have X=ABC and I want to convert it into NAND. An example would be X=A + B = not(not(A)) + not(not(B)) = not(not(A)not(B))

[cunning lava]

I tried ABC = notnot(A)not(not(B))not(not(C)) but has no idea how to proceed.

[graceful karma]

I'd do not(not(not(not(ab))c))

As a final answer

Unless you're more interested in the process

[cunning lava]

Yes I would like the process, please.

[graceful karma]

I mean typically I'd just use the commonly known conversions between NAND, AND and OR

But if you wanted a visual process

[cunning lava]

Anything, I feel my brain is so pooped. 😫

[graceful karma]

I would say something like

ABC
=not(not(ABC))
=not(not(AB)+not(C))
=not(not(not(not(AB)+not(C))))
=not(not(not(not(AB))C))

I mean it's ugly and doesn't serve much purpose because it's easier to just directly convert it

But you can see a logical equivalence between the two ig

[cunning lava]

I see

Could you then remove two nots? so not(not(not(not(AB))C)) = not(not(AB)C)?

Or am I thinking wrong?

[graceful karma]

Well you could but you would then need a normal and gate

Assuming you only have access to NAND gates I think not(not(not(not(AB))C)) is the most simple

Oh I see what you mean

not(not(AB)C) wouldn't simplify to ABC I don't think

AB+not(C) would be as simple as you can get it

[cunning lava]

but then it would not be a NAND gate anymore right?

[graceful karma]

Yeah but I mean more in the terms of logical equivalence

Simplifying it just to see what it is without the nasty nand gate stuff

[cunning lava]

I am sure I am just being stupid. The problem I have is that NAND = NOT (A AND B), and I can't see that not(not(not(not(AB))C)) is a NAND, for example not(not(A)not(B)), I can apply the NAND to it and see that it is NAND.

[graceful karma]

For the single nots I'm assuming you're saying not(xx)

It's just way too long to write everything twice a bunch of times

[cunning lava]

yes

[graceful karma]

not(not(not(not(AB)not(AB))C) not(not(not(AB)not(AB))C)) is quite a bit less readable

Plus if you leave one nand input 0 it'll work as a not gate too

I can try to draw a schematic

[cunning lava]

That would help very much, Thanks!

[graceful karma]

That should hypothetically work

It'd even fit on a 14 pin nand gate ic

[cunning lava]

Ah I get it now

[graceful karma]

Epic

[cunning lava]

I missed the not(not(AB)not(AB)), but now I get it. Thank you very much!

[graceful karma]

Oh

[cunning lava]

.close

[cedar kiln]

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[fossil silo]

Hi! I need to create a function to draw a curve between two points, with a variable that will define how smooth the curve is. So if the variable is low, it will be a normal arc, but if it's higher, it will be more like a logarithmic curve. I thought using logarithms would achieve this but if I understood them correctly, there's only one base that will give me a curve between two specific points, right?

I could do with just a name or resource I can look into to find out how to do this

[graceful karma]

Is there an application for the problem

To get a better idea of the curves purpose

[fossil silo]

It's a game where a character jumps, and I will have variables for the height of the jump (so, the Y of the second point), the time in which I want to reach that height, (which will be the X of the second point), and the magic variable that will define the sharpness of the curve. The first point will be Y=0 and X=0. It would make more sense to refer to T instead of X, maybe.

I'm coding this to practice these kinds of things, that's why I want to do it from scratch

[graceful karma]

So it'll always be like

"Upwards"

[fossil silo]

Yes, from 0 to Y which will always be positive

[graceful karma]

And the two points are always set?

Like if the jump has a smaller height it'll go just as far as far as the function cares

Just less curvy

[fossil silo]

Let me draw it, maybe it's clearer that way

[graceful karma]

Ok

[graceful karma]

Oh so it's not like x and y position?

[fossil silo]

No no, the actual X position of the character is independent

I'm using X here as the time

So I run the equation multiple times per second, inputting the time as X, and the magic variable which be a number that stays the same

And Y would be the output

[graceful karma]

Try sqrt(1-x^n)

[crimson sedge]

If you're working with conic sections, it might be worth looking into eccentricity

[fossil silo]

hmm for the second point being 1,1 that formula does exactly what I want

I can multiply the output to match the height and time I need, that'll work!

thanks!

.close

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[icy pendant]

i did all the others wiht a good understandning, but this one i put 3 and it says its wrong, but i am not sure what other number it would be

looking at the left side i see it bouncing back and forth but comes to 3

[austere wolf]

so

[deep trench]

keep in mind you're not searching for h(5) but the limit of h when x approaches 5 by the left side

[icy pendant]

oh i just tried it, its does not exist but how?

so because its going back and forth we do not have a precise number?

[austere wolf]

because by nature

yes kind of

similarly

check for graph sin (pi/x)

[icy pendant]

ah

[austere wolf]

it goes back and forth

it has close to infinite number of values

[icy pendant]

i see that makes sense

[austere wolf]

also if its 5 -ve

it comes from left

[icy pendant]

yes

ok that makes a lot of sense ty boss 🙏

[austere wolf]

no probelm

[icy pendant]

.close

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[crimson sedge]

any idea how to start

[cedar kiln]

@crimson sedge Has your question been resolved?

[dire geode]

just here to say this looks like an interesting problem but i have no idea

[tranquil oracle]

it's pretty standard LP stuff

are you doing stuff about linear programming?

[crimson sedge]

this was from a test

[tranquil oracle]

like past test?

[crimson sedge]

yea

[tranquil oracle]

step 1: Show that you can write A = pP+qQ+rR, where p+q+r=1

(in terms of vectors and stuff)

[crimson sedge]

nah idk how to do that as well

[tranquil oracle]

alright step 0: Show that for any point X on the line segment YZ, we can write X=yY+zZ for nonnegative y, z that sum to 1

[crimson sedge]

ah dont remember it, altho i've used it a lot

I'll check the proof online

[fallen heath]

OwO I can give a hint...

[fallen heath]

scary intuition

Just saying, perpendicular distance of a point from a line is given as: $D(A(m,n)) = \frac{|am+bn+c|}{\sqrt{a^2 + b^2}}$

(ヘ･_･)ヘ┳━┳ but I'd rather you figure why this and how this helps

[wraith dagger]

Ansh

[crimson sedge]

can we just use coordinate geometry
like if m and n are the ratio
then just multiply by unit vectors then adding them gives
X=nY/(n+m)+mZ/(n+m)

[fallen heath]

[cedar kiln]

@crimson sedge Has your question been resolved?

[fallen heath]

It should've been... but idk if Arkos read my hint or not

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[clever jewel]

is it possible to define a function that has the set of integers as the domain, but the rationals/reals as the image?

[dire geode]

What do you think

[cedar kiln]

@clever jewel Has your question been resolved?

[clever jewel]

that it's not possible, but I'm still curious

[dire geode]

How do you construct the rationals?

Try thinking of how you can get some rationals with just a single function of the natural numbers

Being as simple as possible

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[crimson sedge]

i did understand it geometrically, but idk how to write the proof

[fallen heath]

Uhh umm

perpendicular distance of a point from a line is given as: $D(A(m,n)) = \frac{|am+bn+c|}{\sqrt{a^2 + b^2}}$

[wraith dagger]

Ansh

[fallen heath]

which in this case would be

$D(A(m,n)) = \frac{|f(A)|}{\sqrt{a^2 + b^2}}$

[wraith dagger]

Ansh

[fallen heath]

draw a random graph and try to show the result geometrically or using coordinate geometry using the above idea

You just have to show now that
D(P) < max{D(A), D(B), D(C)}

[crimson sedge]

yea that is what i failed at
idk how to show that for any line theres always a vertex who's perp distance >=any point on the figure

[fallen heath]

ehh

draw lines parallel to the given line

passing through the three vertices

and the given point P

You'll have 4 parallel lines + given line

now show that no matter whatever the placement of the given line is, there's always a line in these 4 parallel lines constructed that lies the farthest

and that's not the one passing through P

:o

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

ah i see now
thanks for ur help

[fallen heath]

Lol GL

[crimson sedge]

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[real tide]

Hi, i have a problem : i have a space, and 3 points, O, A, B. I know the coordinates of the 3 points. A and B represent 2 corners of that cube, and O is that cube's center. Here's a representation i made in blender :

[real tide]

Now, i need to get about 20 points on every edge of that cube in orange

How to do that ?

[cedar kiln]

@real tide Has your question been resolved?

[real tide]

[lone grove]

I am confused by this question.

[real tide]

Hum ...

[lone grove]

Perhaps you could detail the question exactly as it is.

[real tide]

It isn't an academic problem or anything i have to answer
It's for a programming personal project

[cedar kiln]

@real tide Has your question been resolved?

[real tide]

I give up

[cedar kiln]

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[noble garnet]

I've got a problem stating "There are 13 apples and 17 oranges. What is the chance of making a group of 4 where there are at least two apples?" I've tried doing $\frac{\binom{13}{2}*\binom{11}{2}}{\binom{30}{4}}$ , however the result is wrong. The correct way to do it is to multiply all possibilities in the numerator (i.e. AAOO, AAAO and AAAA). Why does my way not work?

[wraith dagger]

mteXD

[jaunty mural]

what is 11c2

[noble garnet]

• there is a typo, i meant 28c2 instead of 11c2

[jaunty mural]

then there is double counting

i think

i will choose 2 apples, then 2 apples

this is double counted for all possibilities

[noble garnet]

i don't quite get this...

if i divide the numerator by 2 it still comes out wrong

[jaunty mural]

uve not double counted every combination

when there are 3 or 4 apples you count these combos multiple times

[noble garnet]

ow i see...

kind of

ok but is there any shorter way to do it than to multiply all the possibilities (13c4 + 13c3 *17 + 13c2 * 17c2 )

or is this the only way?

[cedar kiln]

@noble garnet Has your question been resolved?

[cedar kiln]

@noble garnet Has your question been resolved?

[noble garnet]

ok i did some thinking i've got it now thx

[cedar kiln]

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[viscid anvil]

hello, i have a stats question that i need help with

[viscid anvil]

would the answer simply be 30x29x29?

[jaunty mural]

🤔

combinatorics, not stats

[jaunty mural]

I think I agree but 🤔

if we have 0 1 2

[viscid anvil]

i am having doubts bc if u pick the 1st and 3rd number first, it would be 30 x ? x 30

they didnt specify which numbers get picked first?

[jaunty mural]

010
012
020
021

101
102
120
121
...

[jaunty mural]

it shouldnt answer in final answer

so if its 012, 3x2x2

[jaunty mural]

If you consider this, then there are multiple cases

same, not same.

[viscid anvil]

so i am just going to assume the simplest case and go from left to right

ok then i guess it should be 30x29x29

[jaunty mural]

yh

[viscid anvil]

👌 thanks

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[midnight fractal]

i need help with implicit differentiation

[midnight fractal]

is my operation in step 4 valid?

i think i effed up

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[real bone]

the x from the numerator never gets canceled so i dont see how the leading term would not approach infinity

[real bone]

i started by multiplying by sqrt(5x^2 + 7) / sqrt(5x^2 + 7)

[pallid arrow]

what happens if you divide both the numerator and denominator by x?

[real bone]

yes I did that next

so (5x^2)/4x*sqrt(5x) then divide by x got me 5x/4sqrt(5)

[pallid arrow]

sqrt(5x^2)

not sqrt(5x)

[real bone]

oh

[pallid arrow]

you dont have to multiply sqrt(5x^2 + 7) / sqrt(5x^2 + 7) btw

[real bone]

5x/4*sqrt(5x)

do i just start by dividing by x?

[pallid arrow]

yes you can just start by dividing by x

[real bone]

ok

il try that next

but if I finish with simplifying

[pallid arrow]

sqrt(5x^2) not sqrt(5x)

[real bone]

after dividing by x

[pallid arrow]

4x * sqrt(5x^2) / x = 4 * sqrt(5x^2)

$\frac{4x \cdot \sqrt{5x^2}}{x} = 4 \cdot \sqrt{5x^2}$

[wraith dagger]

xdk1235

[real bone]

i thought you had to do /sqrt x within the square root

[pallid arrow]

$\frac{4 \cdot 3 \cdot \sqrt{5 \cdot 3^2}}{3} = 4 \cdot \sqrt{5 \cdot 3^2}$

[wraith dagger]

xdk1235

[pallid arrow]

same thing

you only get rid of one factor

[real bone]

so if i started out by dividing by x

then i would get sqrt(5x)/4

[pallid arrow]

$\sqrt{x^2} / x$ is not $\sqrt{x}$

[wraith dagger]

xdk1235

[pallid arrow]

$\sqrt{x^2} = x$ so $\frac{\sqrt{x^2}}{x} = 1$

[wraith dagger]

xdk1235

[real bone]

oh

[pallid arrow]

eh that is slightly wrong ig

$\frac{\sqrt{x^2}}{x} = \frac{\sqrt{x^2}}{\sqrt{x^2}} = 1$

[wraith dagger]

xdk1235

[pallid arrow]

but you know what i mean

[real bone]

they specifically tought me that sqrt(x^2)/x approaching infinity you have to think about x as sqrt x which would mean sqrt(x^2)/sqrt(x) = sqrt(x)

oh

no you are right

i missed the square

so sqrt(5)/4

[pallid arrow]

yes

[real bone]

thank you

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[twin rose]

hey

[twin rose]

need help w g h and f

how is this done

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[graceful sable]

ABCD is a parallelogram and BC is produced to a point Q such that AD = QC.If AQ intersects DC at P, show that ar(BPC)= ar(DPQ) (figure given below)
i didnt get how to solve this question

[little void]

Idk if im gonna be able to help, but @graceful sable where is Q

its the bottom right

[graceful sable]

oh yeah sorry its the bottom point i didnt take full screenshot

[little void]

Ok

i dont remember geometry stuff too well

https://www.teachoo.com/9791/2968/Ex-9.4--4-Optional/category/Ex-9.4-Optional/

but im pretty syre i found this exact question on the internet

[graceful sable]

i thinks this is the same question

thanks

!

[little void]

I dont know how much it will help you lol

[graceful sable]

yeah i found the solution

thanks

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[proven oak]

how to draw the graph?

[jaunty mural]

'Hence'. You need to use part (a) to help you.

[crimson sedge]

and is this a exam or smth

[proven oak]

(a) only asks to prove the statement i dont think its usable in (b)

[proven oak]

no its an exercise but we took the questions from other states' tests

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[crimson sedge]

What is the weight of 15.625 cubic feet of aluminum

[jaunty mural]

that's a job for wolfram

,w hi

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Shuri2060#9525

[crimson sedge]

??

ok

[jaunty mural]

type your query in lol

[crimson sedge]

holy-

Alright thanks

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[cloud venture]

I am more wanting to understand the why for this Question.

I am finding the Domain without graphing the function, I have gotten the answer and to make sure I am correct, I went to look at the answers, and its x <= 9/2 but I am not sure why its <=

[cloud venture]

I have gotten down to 9/2, but I don't know why its x <= 9/2

[dire geode]

Can you find other numbers in the domain?

[cloud venture]

Its only 9/2 because I took the 'function' out of the sqrt and make it equal to zero, then got down to x = 9/2 but thats as far as it goes for me

[tropic oxide]

the domain consists of those values of x for which your function is defined

[cloud venture]

I know that, I am just wondering why I have x = 9/2

and not x <= 9/2

[tropic oxide]

and you want the stuff inside the sqrt to not be equal to zero but rather be greater than or equal to zero.

[cloud venture]

right

so how does it switch to <

[tropic oxide]

$9-2x \geq 0$ is what you should be starting with

[wraith dagger]

Ann

[cloud venture]

Ohh thats where I messed up

Do you swap the sign when you divide by -?

Cause it ends up as -2x >= -9

I'm more wondering at which point within the problem does it switch to a <=

[dire geode]

You do switch, but it's also easier to just add 2x to both sides

[cloud venture]

Right, okay, thanks for the clarification I was pretty confused on how it ended up as a <

But thanks

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[tidal crow]

I don't get how this problem works

[tidal crow]

15 cubic centimeters per second doesn't give me anything I could use in the volume formula (outside of setting v to 15, which doesn't really seem useful), so I don't know what else to do

[violet rapids]

Related rates

Differentiate the volume equation with respect to time

[tidal crow]

dV/dt is given in the problem if I'm not mistaken

[violet rapids]

Yes

So you need to differentiate the volume equation with respect to time (dt)

[tidal crow]

dV/dt is differentiating the volume equation with respect to time if I'm not mistaken

[violet rapids]

Yes

i hate latex

[wraith dagger]

sills

[violet rapids]

@tidal crow

[azure cosmos]

Volume of sphere..?

[tidal crow]

That formula's in the problem
dV/dt is 15 cubic centimeters per second
This is the information we know
I don't know how to find dr/dt from dV/dt, because 15 cubic centimeters per second doesn't give me anything I can use in the volume formula

[violet rapids]

@tidal crow https://www.youtube.com/watch?v=7tjarK_7cOQ

dt = the rate at which time changes

[tidal crow]

dV/dt=8/3pi*r^2
I don't see how this helps me

[violet rapids]

You didn't differentiate correctly

After doing power rule, you need to add dr/dt

wait wtf

how do you do this is if it doesn't give you r

[tidal crow]

lmao

[violet rapids]

You can't solve if it doesn't give you r

[tidal crow]

I have no idea either

[violet rapids]

If it said "find the rate at which the radius is changing when the radius is 3" then you could solve it

$\frac{dV}{dt} = 4\pi\frac{dr}{dt}r^2$

[wraith dagger]

sills

[tidal crow]

Well, if I'm being honest, I googled it, and it does have a solution
It doesn't really make sense to me, but I could give you the solution they got if you want to look

[violet rapids]

What's the solution?

[tidal crow]

No idea if this is right but

[violet rapids]

Oh

yeah I forgot you can solve in terms of things for that

[violet rapids]

But that's how you differentiate

[wraith dagger]

sills

[tidal crow]

Think I know where to go from here, gonna keep this open but I think I know

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[warm crown]

Hello

[warm crown]

I need help with q3

I’ve proved (x+2) is a factor

Dunno how to continue from there on

[short socket]

try polynomial long division

and then apply the quadratic formula

[warm crown]

Aight gimme a sec I’ll try it

Thanks for da suggestion

[short socket]

nw

[warm crown]

Ok thx I got this so far

[short socket]

yes looks correct

[warm crown]

But how do I apply the quad formula with an unknown tho?

[short socket]

it works just as well with an unknown

since you only need to find a range of possible answers

[warm crown]

Oh ye you’re right

Man why didn’t I think of that lol

Thanks fellow Dave fan 🙃

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[short socket]

nw glad to help :)

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[halcyon veldt]

I don't understand what a) is asking for

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[sage merlin]

how do you solve 2 3 and 4 my teacher explanation was horrible

[livid hound]

you could set up a general quadratic function
f(n) = an^2 + bn + c
and then solve a system of equations

[sage merlin]

i dont understand

[dire geode]

How did you do 1?

[obsidian coral]

Probably from the teacher's "horrible" explanation

[sage merlin]

no actually from a youtube tut

[stable talon]

you're given that's it's a quadratic, and that f(1) = a+b+c = 3, f(2) = 4a+2b+c = 12, f(3) = 9a+3b+c = 27, f(4) = 16a+4b+c = 48
so you have 4 linear equations in 3 variables - which you can solve

[obsidian coral]

Using the info Ramonov gave, you have the general form of the quadratic. You also know the nth term. IE n = 1, you have 6. That's like a coordinate point. You need just three of them to plug in, you'll have 3 equations, 3 unknowns, solve for each unknown using elimination/substitution

[wanton glacier]

btw in your f(2), where are 4 and 2 from?

[stable talon]

2^2 and 2

I'm subbing n = 2

[sage merlin]

my brain cant handle these tips

[wanton glacier]

oh

[sage merlin]

please simplify

[stable talon]

sub n = 1, 2, 3, 4 into a general quadratic an^2+bn+c
set each one of these quadratics to the nth term in the sequence you're given (hence why we stop at 4 cause you're only given 4 terms)
this gives you 4 linear equations in a, b, c
then solve for a, b, c using standard methods to solve simultanious linear equations

[obsidian coral]

Example: problem 2. General form $$f(n) = An^2 + Bn + C$$. First term is 3, also written as (1, 3), where it is (n, f(n))
If n = 1, and f(1) = 3, then
$$3 = A(1)^2 + B(1) + C$$

[wraith dagger]

dldh06

[obsidian coral]

Follow for the next ones, until you get 3 equations

Solve for the unknowns

[wanton glacier]

and what are a, b, and c?

[obsidian coral]

This isn't your question that you asked, so why are you questioning the process?

[wanton glacier]

simplifying it for that guy

[stable talon]

it's perfectly reasonable to question the process

[stable talon]

a b and c are what you're trying to find

they're the coefficients of the quadratic that give you that sequence

[sage merlin]

im questioning the process the way you are trying to help me is to complicated for my brain im only 14 i dont know half the terms you guys are using

[stable talon]

no I guarantee you know all of them

just read it back slowly and follow the process step by step

[livid hound]

you should know what a quadratic is
at this stage you should also know about basic substitution

and you should also know what a sequence is

[sage merlin]

ik what a quadratic is just not about substitution

[sage merlin]

well not what you guys are doing

[livid hound]

how do you not know about substitution

[sage merlin]

dont know

[obsidian coral]

I found this, maybe it will help you out, slightly difference process then what we're showing, but still gets to the proper answer
https://youtu.be/5NY6CpB2Mog?t=420

[livid hound]

what is the value of
x + 2 when x is 4?

[sage merlin]

its due in an hour ive been stressing with it for last couple days so i resorted to this server to help me

[livid hound]

are you able to do

what is the value of
x + 2 when x is 4?

[sage merlin]

ye

[wanton glacier]

what is it?

[sage merlin]

6???

[wanton glacier]

yeah

[livid hound]

so you do know substitution

[stable talon]

ok then you know subsitution

[sage merlin]

ok

[livid hound]

start with (whether you use f(n) or t_n is up to personal preference)
f(n) = t_n = an^2 + bn + c

and substitute n=1

and tell me what you get

[sage merlin]

what an^2 +bn+c mean

[wanton glacier]

here, try this

[sage merlin]

i got that far

i have so far 3a + b=12

[livid hound]

an^2 + bn + c
is a general quadratic

[sage merlin]

ok

[livid hound]

where a,b,c constants you want to determine

start with
t_n = an^2 + bn + c
substitute in n=1 and tell me what you get

this is not supposed to be a trick request

[sage merlin]

sorry i cant answer i have to go and submit my homework in

[wanton glacier]

here are 3 steps you can do:

1. 2a = second difference
2. 3a+b = U2 - U1
3. a+b+c = U1

*U2 and U1 are 12 and 3 in number 2

@sage merlin should be pretty easy now

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[haughty warren]

Did I solve it correctly?

[cedar kiln]

@haughty warren Has your question been resolved?

[kindred dock]

Ohh my bad

My bad

[haughty warren]

I just wanted to do $\frac{2}{5} \frac{5}{2} 5x^{2}=2x^{2}$

[haughty warren]

Oh

[wraith dagger]

JSGF

[haughty warren]

Well the 5/2 goes out

It may have to do with the trig sub

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[sharp kettle]

Hello, I am a physics fist semester and have some difficulties in Analysis and linear algebra. Can someone give me some tips and resources to understand these topics better?

[obsidian coral]

I just looked up resources on Google

[upper abyss]

3b1b does a good series on applied lin alg, a great way to intuit the ideas

Analysis is tough, keep at it.

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[random finch]

I am trying to find a horizontal parabola equation. I know the vertex (-9.5,0) and a point of the parabola at (-3.6,5.963), another point is at (-3.6,-5.963)

[random finch]

I have tried to input this into the y^2 = 4ax equation to find my a but with no luck.

So currently my equation to find 'a' looks like: 5.963^2 = 4a(-3.6+9.5)

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[velvet egret]

Are all sequences that converge to a limit cauchy?

[dawn jetty]

yea

@velvet egret interesting site :)

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[gentle lintel]

hi

[gentle lintel]

.close

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[crimson sedge]

can i get a nudge with factoring this expression: 7x^3-7x?

[gentle lintel]

do you know factor by gcf?

[crimson sedge]

this is what i got when i tried to factor: 7x(x^2-x)

it didn't look right and i feel lost

yes i know factor by gcf

[gentle lintel]

it doesn’t

you factored 7x wrongly

[crimson sedge]

can i get another nudge? im still lost,

[gentle lintel]

if you expand this

you get

7x³ - 7x²

[fallen heath]

chromium don't occupy many channel

[gentle lintel]

note that factor is just reverse expand

[gentle lintel]

none are being used

that one i closed

somehow still has my name

idfk why

[fallen heath]

XD jk

[crimson sedge]

x(7x^2-7) ?

this looks better

[gentle lintel]

yea

but still not done

[cedar kiln]

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[crimson sedge]

Im not seeing it

[crimson sedge]

im lost again

job looks finished to me

[gentle lintel]

7x³ - 7x right

do you know factoring = reverse expanding?

[velvet egret]

thx

[crimson sedge]

if that means factoring = reversing a polynomial then yeah i suppose

a bit

[gentle lintel]

so you first factor by gcf

[gentle lintel]

what’s the gcf of 7x³ - 7x?

[crimson sedge]

x

7x*

[gentle lintel]

no

yea

7x

so what do you get after factoring 7x out?

[crimson sedge]

7x(x^2-1)

[gentle lintel]

yes

one more step

[crimson sedge]

what was that

[gentle lintel]

do you know the difference of squares?

[crimson sedge]

a little

[gentle lintel]

what does it state

[crimson sedge]

a^b-b^2 = (a +b) (a-b) i think

[gentle lintel]

no

so in this diagram

we wanna find the area of grey

do you see how a² - b² and (a + b) (a - b) are equivalent expressions?

[crimson sedge]

yes

[gentle lintel]

so algebraically, they’re equal

a² - b² = (a + b) (a - b)

difference of squares yey!!!!

[gentle lintel]

it’s helpful back here

we can factorise x² - 1

[crimson sedge]

how

[gentle lintel]

use this

[crimson sedge]

(x^2+1) (x^2-1) ？

[stable falcon]

Without the squares

[crimson sedge]

（x+1）（x-1）

so 7x³ - 7x = x+1）（x-1）

？

[gentle lintel]

no

you forgot the 7x

[crimson sedge]

（7x+1）（7x-1)?

[gentle lintel]

uhh no

remember how you got 7x (x² - 1)?

we know x² - 1 = (x + 1) (x - 1)

[gentle lintel]

so we put it back here

[crimson sedge]

(7x + 1） （x^2-1)?

[gentle lintel]

…

no

7x (x² - 1) = 7x (something)

[gentle lintel]

we directly apply this result

[crimson sedge]

ooooh

7x ( x+1) ( x-1)?

[gentle lintel]

yes

good

[crimson sedge]

on hood cuh imma juh become a rapper hit my partna nba youngboy up

thanks anyways jitt

actually gangsta before i get locked up from shooting a block and give it all up let me get help with one more

5x^4 - 80

this is where i got to: 5(x^4-16 = 5(x^2 +4) (x^2 -4) = 5(x +2) (x-2)

[gentle lintel]

almost correct

the last part you missed something

[crimson sedge]

5(x+2) (x-2)^3?

wtf idk what i just typed

nah bruh i need a push in the right direction im stucker than a milf in a dryer

oh wait i see

its 5(x ^2+4) (x^2 -4)

yes

right @gentle lintel

[gentle lintel]

no

[gentle lintel]

5 (x + 2) (x - 2) is almost correct

but you forgot something lol

[crimson sedge]

what i forget?

oh

5(x^4-16) 5(x=2)(x-2)?

omg i think thats the answer

i wont go rob anyone if thats it

ill quit the gang life

@gentle lintel

[gentle lintel]

?????

[gentle lintel]

what even is 5 (x = 2) (x - 2)?

[crimson sedge]

oh i meant x+2

[gentle lintel]

still no

you factor (x⁴ - 16) into 3 things through difference of squares

can you figure out what those are

[crimson sedge]

x^2+4 x^2-4

i cant make it make sense

where da 5 go?

@gentle lintel

[gentle lintel]

we omit that for a sec

[gentle lintel]

which is..?

that’s two things

[crimson sedge]

5

[gentle lintel]

uhh no

currently we’ve got 5 (x⁴ - 16)

[crimson sedge]

yea

[gentle lintel]

our focus is x⁴ - 16

now this can be split into 3 things

can you find them

[crimson sedge]

whattt

how

i can only see two

what do you even mean by split into 3?

@gentle lintel

[gentle lintel]

this is (x² + 4) (x² - 4) right

one of these you can further factorise

[crimson sedge]

why only 1?

if you factorise one then it would be ((x² + 4)(x-2) right?

@gentle lintel

[gentle lintel]

wdym

one of these you can apply difference of squares again

one of these you can’t

so only one can be further factorised

[crimson sedge]

uh

not getting it

oh wait

(x^2+4) (x+2)(x-2)?

@gentle lintel

[gentle lintel]

yes

so final answer 5 (x + 2) (x - 2) (x² + 4)

[crimson sedge]

all that?

wtf

imma go rob my teacher

[restive forge]

Hey @crimson sedge if your question has been resolved, type .close to free up the channel for other questions

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

Channel closed

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Use .reopen if this was a mistake.

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[crimson sedge]

Had a problem exactly like this few minutes ago, and have another one. Tried to use the same method but don't see it going anywhere (factoring both numerator & denominator)

[crimson sedge]

I don't have the definitive answer to it

[kindred storm]

@crimson sedge Well, 65536 is 2¹⁶ and 32768 is 2¹⁵.

[dawn jetty]

are you allowed a calculator ?

[crimson sedge]

no

that's what I'm going to ask now

consider I don't have the info about that

[dawn jetty]

are you expected to know that?

[crimson sedge]

that 65536 is 2^16

[dawn jetty]

that 65536 is 2^16?

[crimson sedge]

no

that's the only way?

[dawn jetty]

doubt it

[crimson sedge]

can't I somehow factor or agroup the numbers?

and simplify if possible?

because of course now that I know that 65536 is 2^16 and possibly all the numbers follow the same pattern it's easier

but certainly I wouldn't know that in a exam that I have 3 hours to complete

[wraith dagger]

Chai T. Rex

[kindred storm]

Well, it's a bit cut off.

[crimson sedge]

that's what I thought

it's okay

I was discussing with my friend that sent me this question

ty

[kindred storm]

No problem.

[crimson sedge]

after all I just need to complete the simple algebra?

confirming before I close

[kindred storm]

Yes, there are probably tricks to make it easier as well.

You'll get an integer times x.

[dawn jetty]

I somehow doubt that it's the method

won;t it getg too convoluted?

or will it actually simplify

[crimson sedge]

but considering

that I don't know the info that Chai said before

it's apparently the way

like I had a problem that had 3^x

I don't know log

[dawn jetty]

well

hmm

[kindred storm]

It does simplify, but there may be an easier way that I haven't seen.

[dawn jetty]

the constants will cancel

oh

yea

that might actually work

[crimson sedge]

so the only way was to consider 3^x = y

[dawn jetty]

actually I think thi sis the way

the trick is just seeins that the constants will cancel right away

but also, given that the answer is in the form of 2^16
you're probably expected to know that 65536= 2^16

[crimson sedge]

so just slowly improving the equation?

[crimson sedge]

but using this way wouldn't it come as powers?

well, I'm going to try it right now

[dawn jetty]

Oh wow

it actually wounds up being rather nice

[kindred storm]

You can probably expand easily with the binomial coefficients and then do a long division of the polynomials.

[dawn jetty]

I don't think you have to

you can see that everything besides x^3 and x^2 will cancel each other out

then again... can see is not always the best argument

[crimson sedge]

1 question

why did you consider x as 32768?

[kindred storm]

To avoid fractions with x/2 in the denominator if I chose x = 65536.

As to why I'd choose either 32768 or 65536, that's to make the things you add to it very simple.

[crimson sedge]

oh I see, so divide by 2

okay

oooooh now I understood completely why (2x - 3) and all others

you choose a base, in this case 32768 and adding or subtracting depending on the number

I see, didn't thought about that

and another question

I could if I wanted consider x as 32767, right?

[kindred storm]

You could if you wanted to, but since the answers all have 2¹⁶ in them, it's not going to be pretty.

[crimson sedge]

so the requirement for choosing it as 32678

as the 2^16

if I wouldn't notice that but still saw the pattern and consider some number as x how would I think for choosing the number 32678?

is there there another thinking process on it?

[kindred storm]

Well, you notice that the answer has powers of 2 in it.

Also you notice that the problem's numbers have powers of 2 in them.

[crimson sedge]

so I would go by "numbers divisible by 2"

I see

[kindred storm]

Like the top has 65536 and the bottom 32768.

[crimson sedge]

I see it

[kindred storm]

Not just even, but a power of two.

If you use 32767 instead, you'll get a(x + 1) as the result.

Instead of ax.

It just changes the result of the polynomial long division.

[crimson sedge]

well I'm interested on how I would continue if I did that way

[kindred storm]

Well, you'd expand top and bottom.

Then you'd do polynomial division or factor both top and bottom and cancel.

[crimson sedge]

but still I haven't solved considering x = 32768 so I will stop for a few minutes before that yet

[crimson sedge]

uuuuh I think I got it

I've written all the replacement of the equation as x

could I immediately cancel some numbers?

[kindred storm]

If they're factors of the entire top or bottom.

If they're just factors of some term on the top or bottom, no.

[crimson sedge]

actually I found a strange thing

why is there (x - 1)x + x(x + 1) ?

you would multiply it twice?

[kindred storm]

Because you had 32767 × 32768 + 32768 × 32769.

[crimson sedge]

ooooooh sorry

[kindred storm]

Yes, because you had x twice in the original.

[crimson sedge]

yeah makes totally sense

alright will give you feedback in minutes

[kindred storm]

OK.

[crimson sedge]

so on the denominator I thought of doing the \multiplications

x^2 - 5x + 6 (but as a result it came out as quadratic equation and I don't think this is the way you're expecting me to)

and I thought of too

[kindred storm]

No, that's the correct expansion of (x - 3)(x - 2).

[crimson sedge]

Yes

but it's not the best way to go, right?

[kindred storm]

No, it's fine.

You can speed it up slightly.

[crimson sedge]

forgot the comma

sorry

could I cancel the factors like

(2x - 3)^3

/ (x - 3)

certainly no, is that correct?

[kindred storm]

No, because those aren't factors of the ENTIRE top or ENTIRE bottom.

[wraith dagger]

Chai T. Rex

Chai T. Rex

[crimson sedge]

I see but can it be a + b / a + c

[kindred storm]

Only when it's a factor of the entire top and entire bottom can you cancel. Factor means multiplied by the whole thing.

[crimson sedge]

and I can cancel them?

[kindred storm]

No, because it's added to the entire top and entire bottom.

But it's not multiplied by the rest of the top and multiplied by the rest of the bottom.

[crimson sedge]

because I thought you can't cancel until there are + or -

[kindred storm]

No, you can't cancel unless you're multiplying something with the rest of the top and the rest of the bottom.

[wraith dagger]

Chai T. Rex

[kindred storm]

That has a multiplied by the entire rest of the top, right?

[crimson sedge]

in this case

yes

I can cancel a

[kindred storm]

Also it has a multiplied by the entire rest of the bottom.

[crimson sedge]

b + c / d

[kindred storm]

Yes, if it's multiplied by the entire rest of the top and the entire rest of the bottom, you can cancel.

But it has to be multiplied by the rest, not added to the rest.

[crimson sedge]

so that's what I should be searching in my equation now?

[kindred storm]

No, first expand the entire top and bottom.

The top has a trick.

The top is a bunch of terms that are (x + a)³ for some a (the a for the x³ term is 0).

So, expand (x + a)³ and then use that to do the hard work.

[crimson sedge]

expand both (x - a)^3 & (x + a)^3?

[kindred storm]

No, just (x + a)³.

You can make a negative for the subtractions.

[crimson sedge]

by expand you mean

(x + a)^3 = (x + a)(x + a) -> quadratic equation multiplied by (x + a)

?

[kindred storm]

Yes, you can do it that way.

[crimson sedge]

Is the most efficient way?

[kindred storm]

Yes, because you do the hard work of multiplying all that out once.