help-49

[twilit field]

oh shit

[dreamy lichen]

how do you know that?

[subtle blaze]

shouldn't you have a fullstop after the set

[twilit field]

sorry

[subtle blaze]

then say ||suppose T is non-empty||

[twilit field]

oh, right

[dreamy lichen]

Ideally. And the next sentence should be (what frosst said)

[twilit field]

I've made that change

on my end

[dreamy lichen]

okay good, next thing is not necessary but i think its good to include it anyway. You can mention that you are using WOT. Like "By WOT, T has a smallest element, so let a be the smallest element of T."

[twilit field]

got it

[subtle blaze]

is that what WOT says

[dreamy lichen]

And you should expand on the => (a-1) ∈ S

[dreamy lichen]

WOT says that non-empty subset of Z+ has smallest element

and T is non-empty subset of Z+

[fleet moss]

i am still confused what we are even trying to prove here. if you want to say that this construction S is equivalent to Z+, we first need some definition of Z+ to compare it to.

this contradicts the definition of T
why does it contradict it?

[twilit field]

T is defined to consist of those elements that aren't in $S$, but I've shown that that's not true

[grand pond]

ƒ(Why am. I here)=I don't Know

[dreamy lichen]

basically, a is both element of S and T

[subtle blaze]

well i've just used Z+ as a scapegoat so he stops saying that T contains everything not in S

[dreamy lichen]

so thats how the contradiction arises

[dreamy lichen]

. You included the a =/= 1 too lately

also you can just generally expand on that implication and explain it in more detail

like a-1 isnt in T, lest it would be the smallest element

hence either a-1 isnt in Z+, or it's in S

[fleet moss]

well do we really know that a-1 is in S?

[dreamy lichen]

yes

[fleet moss]

how

[twilit field]

Sorry for making you spend so much of your time helping me on a simple proof .

[dreamy lichen]

a-1 isnt in T by minimality of a

hence a-1 is either not in Z+ or it's in S

(by definition of T)

the only way a-1 wouldnt be in Z+ is a-1 = 0 and a = 1, but 1 is in S, so 1 can't be in T and so it cant be minimal element of T

so a-1 must be in S

and then a must be in S by definition of S

all of that should be at least partially included in the proof btw

maybe not in that much detail, but it should

[subtle blaze]

when you do it for the first time you should write it in that detail and more

[dreamy lichen]

that's true too

[subtle blaze]

until you know you can do it, you should write it in that detail

if you're just going to handwave the parts you dont know then you might as well not be doing proofs

[twilit field]

got it, sorry

should I write it afresh ?

[fleet moss]

$x\not\in \mathbb{Z}^+\not\rightarrow x\in S$. for example 1.5 isnt in either set

[grand pond]

esca (@ with reply)

[dreamy lichen]

I never said it does

i used disjunctive syllogism or smth like that

idk how its called

a-1 is either not in Z+ or it's in S

I showed that a-1 is in Z+

hence its in S

[fleet moss]

how is a-1 in Z+

[dreamy lichen]

because A is in T and so in Z+

[subtle blaze]

small a

[dreamy lichen]

yeah

and so the only way for a-1 to not be in Z+ is a-1 = 0

[fleet moss]

a was in T, so a was in Z+. but how is a-1 in Z+

[subtle blaze]

you dont know

[dreamy lichen]

.

[subtle blaze]

until that

[dreamy lichen]

then i made argument that a cant be 1

since 1 is in S

and so 1 cant be in T

and a fortiori it cant be minimal element of T

[fleet moss]

i mean $a\neq 1\land a\in\mathbb{Z}^+\implies a-1\in\mathbb{Z}^+$ is literally the thing youre trying to prove, right?

[grand pond]

esca (@ with reply)

[subtle blaze]

no

we're trying to prove that S = Z+

[dreamy lichen]

Well I assume there are some properties about integers that are already established

and this one looks trivial enough

[fleet moss]

lay them out

[fleet moss]

so whats a baseline definition of Z+ we can work with

[twilit field]

the book says they are assumed to be given, we don't have to bother with them

[subtle blaze]

ok fine S contains all positive integers

[dreamy lichen]

not really

we are proving all positive integers is subset of S

[subtle blaze]

sorry yeah

[dreamy lichen]

We dont need to use von neumann numerals or anything

we just have some properties that we can use

i assume trivial arithmetical properties

which is what i used

and WOT

which is taken as axiom

in the book

[fleet moss]

what is an integer i guess should be my question

[dreamy lichen]

why care about that

we have some properties

[subtle blaze]

im pretty sure the exercise wants this part as the meat of the proof

[dreamy lichen]

and we dont care whether 2 = {∅, {∅}} or 2 = {{∅}}

[subtle blaze]

we're not looking to prove absolutely everything from scratch

[fleet moss]

so we take for granted that if a is a positive integer, then a+1 is also a positive integer

[twilit field]

yes, I guess

[fleet moss]

we also take for granted that 1 is a positive integer

[dreamy lichen]

yes

[fleet moss]

and we take for granted that a +b where b isnt an integer is not an integer

[dreamy lichen]

no?

1/2 + 1/2 = 1 which is an integer

[fleet moss]

a is still an integrt

[dreamy lichen]

okay, then yes i guess?

but this is already much less trivial than what i used

look, we just take some basic arithmetical properties for granted, and then we use WOT to derive new a base for new "proof technique", which is induction

I think we can assume that if a is in Z+, then a-1 is in Z+ or a-1 = 0

its equivalent to this btw:
if a is non-negative integer, then a+1 is positive integer

and that's just trivial

[twilit field]

Uh, so is anything still wrong with my proof?

[dreamy lichen]

What's the current version of it?

this?

[twilit field]

this

yes

[dreamy lichen]

I dont like this part

you should explain in more detail why (a-1) is in S

start by (a-1) not being in T, by minimality of a

then use definition of T, exclude the possibility that (a-1) is not an integer and conclude that (a-1) is in S

[twilit field]

sorry, this is it

[fleet moss]

right and thats the thing we're trying to prove 😭

[dreamy lichen]

no it's not?

[dreamy lichen]

I see. You should assume that T is non-empty efore you use WOT

[twilit field]

Ok

other than that?

[dreamy lichen]

you repeat the same thing twice actually

delete the first one and it will be fine

and seperate your sentences

suppose T is non-empty let a be the smallest element of T (a-1) in S which by definition...

this should be 3 sentences

other than that, you should expand the proof there

explain why (a-1) should be in S

[twilit field]

ok, so that's basically as if $k\in S$ as is $k+1$

right

[grand pond]

ƒ(Why am. I here)=I don't Know

[dreamy lichen]

What do you mean?

[twilit field]

the explaination of this

[dreamy lichen]

Currently, you have that a is the smallest element of T. Use that and proof that (a-1) is in S

[twilit field]

by definition...

[dreamy lichen]

that explains that if (a-1) is in S, then a is in S

But what you need to explain is that if a is smalelst element of T, then (a-1) is in S

this is the big jump in your proof

it's correct, but the jump was too long

you skipped like solid 2 lines of proof

and not exactly trivial one

[fleet moss]

youre saying the only thing we know about Z+ is it has certain properties. and we're constructing S as some set that has exactly the same properties. then i dont understand whats left to prove

[dreamy lichen]

We are not constructing S

we have some set S that has the properties that it contains 1 and is closed under successor

and we are proving that it contains all positive integers

[fleet moss]

but all we know about Z+ is that it contains 1 and is closed under successor

[twilit field]

I'm really confused now, sorry

[dreamy lichen]

That's assuming a specific set theoretic construction

we dont always need to use the same construction with same properties

can we move this to DMs esca?

[fleet moss]

sure

[dreamy lichen]

at what point?

[fleet moss]

sorry physics rocks

[twilit field]

here

ok, so if $a-1$ is in S , as is $a$

[grand pond]

ƒ(Why am. I here)=I don't Know

[dreamy lichen]

I suppose this is supposed to be

"Let a be the smallest element of T. Then (a-1) is in S.

Is that right?

[twilit field]

yes

[dreamy lichen]

okay, that's not so trivial conclusion though

[twilit field]

so this approach is wrong?

[dreamy lichen]

that shows that if (a-1) is in S, then a is in S

but it doesnt show that if a is the smallest element of T, then (a-1) is in S

[twilit field]

by construction if an element isn't in T , it has to be in S

assuming it's an integer

[dreamy lichen]

exactly

and that's not so trivial

and you stuffed all of that in one single step

[twilit field]

ooh, ok

[dreamy lichen]

while it should be mentioned in the proof

[twilit field]

got it

[dreamy lichen]

you need to say that (a-1) is not in T, which follows from minimality of a

then you need to establish that (a-1) is a positive integer

and only then you can conclude that its in S

[twilit field]

oh, I have to do that too?

ok

[dreamy lichen]

you already partially did this part, though at wrong place

and partially

[twilit field]

so this shouls be near the beginning of the proof?

[dreamy lichen]

it should be mentioned somewhere during the process of proving that "if a is the smallest element of T, then (a-1) is in S"

meaning somewhere in between those 2 things

[twilit field]

ok

and thanks so much for all the help

[dreamy lichen]

np

[twilit field]

one minute

I'm lost

I think I'll just try this again later, sorry

is that fine?

[dreamy lichen]

yeah

[twilit field]

ok

sorry

[dreamy lichen]

np

[twilit field]

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[dreamy lichen]

you can finish it whenever you want, the progress is already quite good

the proof outline is correct

[twilit field]

👍

[last slate]

Can any1 help me with this

[midnight plank]

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[versed ridge]

for part b, i had $\int_{0}^{1} \int_{x^2-x+1}^{1} \int_{2-x-y}^{1-x^2} (x) dz dy dx$. I am wondering why this is wrong.

[grand pond]

Secret

[sweet wing]

do you have the drawing of the regions?

[versed ridge]

no but i could make one rq

here's what the region looks like

[midnight plank]

@versed ridge Has your question been resolved?

[versed ridge]

<@&286206848099549185>

[midnight plank]

@versed ridge Has your question been resolved?

[midnight plank]

@versed ridge Has your question been resolved?

[pearl idol]

That geogebra drawing looks terrifying lol

Lemme take a look at the problem

where did you get the bounds for y from? for example, couldn't y be 2? (if x=z=0)

[midnight plank]

@versed ridge Has your question been resolved?

[versed ridge]

so to find the bounds for x and y, i set the two surfaces equal to each other

and that gave me $2-x-y=1-x^2$

[grand pond]

Secret

[versed ridge]

which meant $y=1-x+x^2$

[grand pond]

Secret

[pearl idol]

That makes sense for the lower bound, but what about the upper bound of 1?

Where did you get that from

[versed ridge]

i think i figured that if $z=1-x^2$ and we are bounded by the plane $z = 0$, $-1 \leq x \leq 1$ but we only really consider $0 \leq x \leq 1$. Now, looking at the second surface, we have $z=2-x-y$. If $z \geq 0$, then $x+y<=2$ and i think i plugged in $x=1$ into this inequality.

[grand pond]

Secret

[versed ridge]

tho ig i should plug in x=0 if i were to find the upper bound for y

[pearl idol]

yeah

I think your upper bound for y should probably depend on x

[versed ridge]

oh so 2-x

ok yea that makes sense

thank you!

.close

[midnight plank]

Channel closed

Closed by @versed ridge

Use .reopen if this was a mistake.

[pearl idol]

Wait does 2-x actually work lol I haven't checked

[versed ridge]

imma try it out and lyk

[pearl idol]

kk good luck!

[versed ridge]

ok this is the revised integral and it gives me the wrong answer. Do you see anything else wrong with it?

[pearl idol]

.reopen

[midnight plank]

✅

[pearl idol]

So my question is

Are you sure you're looking at the right region

Because the bounds you've set make me think you're looking at this region

Between the blue parabola and the red plane, and so on

But it seems like that region would just continue on forever to the right

Lemme try plotting it

what I'd assume is you're trying to find this region, under both the plane and the paraboloid?

In that case, you should have that x+y+z ≤ 2 and x^2 + z ≤ 1

[versed ridge]

ye i think the one that is bounded should be the region

[pearl idol]

Before, you were assuming, I think, that x+y+z ≥ 2 in order to find lower bounds for y and z

[versed ridge]

ye

[pearl idol]

The inequality should be the other way around though

So that we can have a bounded region

[versed ridge]

wait so if the region is under the plane

then would it be less than or equal to the plane

[pearl idol]

yea

[versed ridge]

i see

ok i'll try this out

[pearl idol]

So you've got
0 ≤ x
0 ≤ y
0 ≤ z
x + y + z ≤ 2
x^2 + z ≤ 1

[versed ridge]

ye

[pearl idol]

I would prob do the outside integral 0 ≤ z ≤ 1

Although, not really sure

[versed ridge]

wait wouldn't it be z between 0 and 2

[pearl idol]

No bc x^2 + z ≤ 1

[versed ridge]

oh yea you take the lesser bound

[pearl idol]

Yeah this is tricky

[versed ridge]

we can do y as innermost

and then go from 0 and 2-x-z

and x goes from 0 to 2-z

does that make sense?

[pearl idol]

Then you're not really using x^2 + z ≤ 1, are you

[versed ridge]

ye ig i have to take cases for x

i have two possible inequalities $0 \leq x \leq (2-z)$ and $0 \leq x \leq \sqrt{1-z}$

[grand pond]

Secret

[midnight plank]

@versed ridge Has your question been resolved?

[pearl idol]

0 ≤ x ≤ 1
0 ≤ z ≤ 1 - x^2
0 ≤ y ≤ 2-x-z

Do these bounds work? @versed ridge

They feel like they should

[versed ridge]

let me check

yep that's right

[pearl idol]

Awesome

[versed ridge]

how did you get there?

[pearl idol]

So, I noticed that if you look at this picture, the y-axis is the only one where the bounds are not piecewise

y always goes from 0 to the plane

So I figured that one should go on the inside of the integral

Once we've got that covered, that ensures we're always under the plane

And then we just need to deal with the x and z being inside the parabola

You can either do this like
0 ≤ x ≤ 1
0 ≤ z ≤ 1-x^2
or I think
0 ≤ z ≤ 1
0 ≤ x ≤ sqrt(1-z)
should work fine too

And then that solves it

[versed ridge]

ah i see

[pearl idol]

So like, even though the region of integration is somewhat complicated, if you project it onto the x-z plane, it's actually really simple

(just the area under a parabola)

[versed ridge]

ig you should be able to do it with x and z as innermost too, right? it would just involve taking two different cases

[pearl idol]

I think it would be a lot more complicated than this way

You can try it

[versed ridge]

i see either way, thanks a lot, man!

appreciate it

[pearl idol]

you're welcome!

[versed ridge]

.close

[midnight plank]

Channel closed

Closed by @versed ridge

Use .reopen if this was a mistake.

[midnight plank]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[tidal turret]

let $f(x) = x^2 \ln x$ $\\1)$ how many solutions does the equation $f(x) = \frac{-1}{6}$ have $\\$2) for which values of $k$ the equation $f(x) = k$ has exactly one solution

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

first one is doable

,, \frac{-1}{6} = x^2 \ln x

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

well

[frank wolf]

Try sketching the graph

[twilit field]

I don't think you actually need to solve it

try finding the number of turning points and the concavity of the graph

[tidal turret]

how to sketch the x^ 2 lnx doe

ok so lets calculate the critical points and the points of inflection, no?

wdym concavity

how do I find concavity and convexity of this

?

[twilit field]

I don't meanu numerically

[tidal turret]

,, x^2 \ln x \ \frac{d}{dx} \left(x^2 \ln x\right)

[twilit field]

I mean is it concave up or concave down

or neitger

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

neitger?

[twilit field]

,w graph x^2ln(x)

[grand pond]

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[twilit field]

notice it;s concave up

on (0,1)

[tidal turret]

sure

what about the derivatives

[twilit field]

what's the derivative of x^2ln(x)

[tidal turret]

,align x^2 \ln x = \ \frac{d}{dx} \left(x^2 \ln x\right) &= \frac{d}{dx} \left[x^2\right]\ln(x) + (x^2)\frac{d}{dx}\left[\ln(x)\right] \ &= 2x \ln(x) + \frac{x^2}{x} \ &= 2x \ln(x) + x

[twilit field]

pretty sure that's wrong

what's the derivative of ln(x)

yes

this is right

now notice it's 0 at a point between 0 and 1

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

ln(x) in here x > 0 strict

[last slate]

ye so e^(-1/2) is the exttema

[tidal turret]

wdym

[last slate]

factor x out

in the derivative

[tidal turret]

,, 0 = 2x \ln(x) + x \ 0 = x(2 \ln(x) + 1)

[last slate]

ye

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

so either x = 0 or
2ln(x) + 1 = 0

[last slate]

no

x=0 isnt possible

[tidal turret]

okay

why doe

[last slate]

cuz f(0) doesnt exist

[tidal turret]

okay

sure

[last slate]

ye

[tidal turret]

then $2 \ln(x) + 1 = 0$

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

thats why e^-1/2

[last slate]

ye

u can ignore x in the derivative

so the function is ez to plot

[tidal turret]

,w solve 2\ln(x) + 1 = 0

[grand pond]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

e^-1/2

now what

[last slate]

can you draw 2lnx + 1

if you can ur gonna realize that f(x) is gonna look like a quadratic

with vertex at x = e^(-1/2)

[tidal turret]

how do I draw it doe

[last slate]

you dont know how lnx looks like?

[tidal turret]

,w plot ln(x)

[grand pond]

Results provided by WolframAlpha

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[last slate]

ye

[tidal turret]

so what?

can you draw 2lnx + 1
how

[last slate]

it looks pretty much the same

you dont have to be precise when using derivatives to check where the function is increasing/decreasing

thats why i ignored x in the first place

cuz x>0

[tidal turret]

sure

now what

[last slate]

u did it?

[tidal turret]

no I still cannot

can you draw 2lnx + 1
how

[last slate]

if you cant do it you can check where 2lnx+1 is positive n negativd

[tidal turret]

how did you do it

I want to know the deepfriedpack way of doing stuff

[last slate]

its simple

[last slate]

u do this but change the root

so this

[tidal turret]

here root is 1?

[last slate]

ye but its now 2lnx+1

[tidal turret]

I mean in lnx

damn

got it got it

[last slate]

yeah but were doing 2lnx+1

[tidal turret]

sure

now what

,calc e^(-1/2)

[grand pond]

Result:

0.60653065971263

[last slate]

u now know where the function is increasing or decreasif

[tidal turret]

exactly

(e^-1/2, +infty) increasing

mmm

still confused

need more hints

[last slate]

gimme a sec

[tidal turret]

sure, I will scroll tiktok real quick

[last slate]

which one

[tidal turret]

unsure how to answer 1 and 2 from my original question

I am getting confused all over again

I need more handholding

[last slate]

can u plot fx from this

[tidal turret]

fx?

x^2 ln(x)

how would I

do that

[last slate]

u know this

so

something like this

[tidal turret]

(0, e^-1/2) is decreasing

[last slate]

this is enough to solve those 2 questions btw

[tidal turret]

(e^-1/2, infty) increasing

[last slate]

u dont nesd that much detail

[tidal turret]

what is that

[last slate]

f(x)

[last slate]

look

[tidal turret]

how

need more hints

[last slate]

i just used the definition of decreasing n increasing

[tidal turret]

okay

I get it now

now what?

[last slate]

whats f(e^(-1/2))

[tidal turret]

tangent zero

[last slate]

not the derivative

the actual function

[tidal turret]

,calc e^(-1)*-1/2

[grand pond]

Result:

-0.18393972058572

[last slate]

ye now compare that to -1/6

[tidal turret]

,calc -1/6

[grand pond]

Result:

-0.16666666666667

[last slate]

so -1/6 abit bigger

[tidal turret]

,calc -1/5.5

[last slate]

than f(e^(-1/2))

mb

[grand pond]

Result:

-0.18181818181818

[tidal turret]

its not 1/6 doe

[last slate]

what

[tidal turret]

,w 1/sqrt(e) = 1/6

[last slate]

-1/6 is abit bigger than f(e^(-1/2))

[grand pond]

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[tidal turret]

slightly, agreed

[last slate]

im not saying f(e^(-1/2)) is -1/6

[tidal turret]

okay

now what

[last slate]

so can u tell how many roots there are in f(x) = -1/6

[tidal turret]

1

[last slate]

why

[last slate]

ye how many intersections are there

[tidal turret]

two

[last slate]

yeah

try the second one now

[tidal turret]

how?

,, x^2 \ln x = k

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

this ?

[last slate]

ye

[tidal turret]

how do I do that

[last slate]

y = k

k = -1/6 had two roots

[tidal turret]

okay

[last slate]

what should k be so that those 2 have 1 intersection

[tidal turret]

how would I know that

the intersection has to be at the turning point

at the vertex of the parabola

[last slate]

possible ks

actually nvm its a range

[tidal turret]

how do I do that

[last slate]

is there a condition that says k <0

[tidal turret]

yeah definetely

x^2 is postiive

,w range of ln(x)

[grand pond]

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[tidal turret]

wait, log can output negative aswell?

didnt knew that

[last slate]

in the og question

[last slate]

this

[tidal turret]

let $f(x) = x^2 \ln x$ $\\1)$ how many solutions does the equation $f(x) = \frac{-1}{6}$ have $\\$2) for which values of $k$ the equation $f(x) = k$ has exactly one solution

wdym

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

I dont get it

wdym

@robust cloak

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

Can someone help my fatass

[last slate]

I'm here

[hollow oyster]

I think you have to use the W Lambert function

Like let lnx^2=u

Then x^2=e^u

[hollow oyster]

And in f(x)=-1/6 multiply 2 on both sides

[hollow oyster]

Do you now know what to do

[upper crag]

lambert w won't help, it's not in most curriculums

[hollow oyster]

Oh I see

But it will give the solution

[upper crag]

some calculus will help, basically 1 solution occurs at the min of f if we think of this k value as a vertical translation

[hollow oyster]

I check in Wolfram also

Then what do you suggest mero

[upper crag]

this

[hollow oyster]

I'm not sure if I understand that

[upper crag]

what part

[hollow oyster]

The assumption that k is a vertical translation

[upper crag]

https://www.desmos.com/calculator/j3enmiyens

maybe this is convincing

[hollow oyster]

Also how can we be sure there's only 1 solution

[upper crag]

move k so that the minima lies on the x-axis

[hollow oyster]

Ok i see

[upper crag]

also these are increasing functions on the domain [0,infty)

ok cool I'll let you work on it now

[midnight plank]

@tidal turret Has your question been resolved?

[hollow oyster]

Mero my dumb brain is not getting what to do T_T

[upper crag]

well ok first off, were you able to solve for the x value of the minima?

[hollow oyster]

1?

[upper crag]

nope

f(x)=x^2ln(x)

now the minima occurs when f'(x)=0

[hollow oyster]

e^-1/2

[upper crag]

good

how did you get that can you show your work

that was too quick like you're just plugging this into wolfram alpha 😛

that's no fun

well regardless once you have that, then plugging e^{-1/2} into f will be the height of the minima

[last slate]

!showwork

[midnight plank]

Show your work, and if possible, explain where you are stuck.

[last slate]

nice eyes drawing

[hollow oyster]

Lol

[upper crag]

eh?

2xln(x)+x=0

since x !=0 isn't in our domain, you divide it out

2ln(x)+1 = 0

[last slate]

,rotate

[hollow oyster]

Ik i didn't think that time

[upper crag]

oh ok cool

[last slate]

was this just so that we can see the eye drawing more better 😭

[upper crag]

so back to this point where we left off

[last slate]

yeah lets lock in fr

[upper crag]

lmao

[upper crag]

👁️ 👁‍🗨

[last slate]

👁️ 🧿

[last slate]

bro has math equations for a mouth 💀 💀

[hollow oyster]

I got e^-1/2 as min

[upper crag]

specifically the x value of the min

we want the y value of the min since that's how much we need to translate the graph by so that it's the only point tha ttouches the x-axis

[hollow oyster]

1/2e

[upper crag]

I think it's -1/(2e) unless you mean the translation amount up

[hollow oyster]

Thus k=1/2e

[hollow oyster]

Yes I meant that

[upper crag]

then k=-1/(2e)

[hollow oyster]

My bad

[upper crag]

all good

[hollow oyster]

Well how to do the first w/o W Lambert function?

[upper crag]

same sorta calculus stuff

[hollow oyster]

But it is different

Man i have to revise my clac

Well it is my hw for the week

Bye mero

[upper crag]

nite nite

[last slate]

k>=0 also works though

[last slate]

bye byee

the answer should be a range

[hollow oyster]

I ain't going anywhere tho

[last slate]

wdym, you said bye?

[hollow oyster]

I'm just not gonna slove this

[last slate]

ohh it's alr bro, you tried your best anyways

[hollow oyster]

So "bye"

[hollow oyster]

I got the answer for the first but mero said it will not be accepted in some curriculum

[last slate]

mhmm

[hollow oyster]

But I clearly have no idea how to do with out that method

[last slate]

it's fine bro

[midnight plank]

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[bleak pier]

hints please

[sage helm]

what is an "intermediate" value

[deep vine]

over what interval?

intermediate value theorem requires an interval

[bleak pier]

I do not know

[bleak pier]

yes

and we do then derivative

[deep vine]

but what is the interval

[bleak pier]

as we can see it is not given

[sage helm]

Half the questions you bring are incomplete in some way

Do you just need us to confirm that there's nothing to be done?

[bleak pier]

see

[midnight plank]

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[bleak pier]

Hints please

[sage helm]

Do you have expressions for the volume and surface area

You need a function for the volume in terms of your favorite conical parameter

[midnight plank]

@bleak pier Has your question been resolved?

[bleak pier]

.cl;soee

.close

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[midnight plank]

@flint heron Has your question been resolved?

[flint heron]

<@&286206848099549185>

[midnight plank]

@flint heron Has your question been resolved?

[iron pasture]

you seem to have solved the problem already.

p is congurent to 1 mod 4 only if p-1 is divisible by 4

[flint heron]

Yes, I am not sure why that is. Should I learn about Legendre symbols?

This topic wasn't covered in my textbook

[iron pasture]

One sec let me check some stuff

[flint heron]

Thank you!

[iron pasture]

It seems like you need to use Legendre symbols for this. Is this problem from your textbook?

[flint heron]

Not directly from it. I will look this up and try to figure out and come back if I get stuck. Thank you

.close

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[summer ravine]

solve the following system of equal for real x,y,z,;
x+y-z = 4
x^2-y^2+z^2 = -4
xyz = 6

[summer ravine]

.close

[midnight plank]

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[summer ravine]

.reopen

[midnight plank]

✅

[summer ravine]

.close

[midnight plank]

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[summer ravine]

nvm

.reopen

[midnight plank]

✅

[summer ravine]

help me solve this please

x+y-z = 4
x^2-y^2+z^2 = -4
xyz = 6

real solutions

helpers

nvm i got it.

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[steady trail]

Guys, but as I'm practicing finding the maxima and minima of $f(x,y)=\arcsin(x^2-y-1)$ I get some derivatives which seem infinite. Is there a way to do it faster, like parameterization?

[grand pond]

card

[steady trail]

Domain D:
${(x,y) \in \Bbb R^2 : 0 \leq x^2-y \leq 2}$

[grand pond]

card

[midnight plank]

@steady trail Has your question been resolved?

[steady trail]

<@&286206848099549185>

[junior flower]

i have a very fast way
for any t, we have -pi/2 <= arcsin(t) <= pi/2, right?

[steady trail]

Yes

[junior flower]

so now just find a pair (x,y) that makes arcsin(x^2 - y - 1) = -pi/2 and another pair that makes arcsin(x^2 - y - 1) = pi/2

this shows the min and max are -pi/2 and pi/2 respectively

[steady trail]

but what gives us the certainty that it takes on those values

[junior flower]

we can write the pairs explicitly

[steady trail]

for Extreme value theorem ?

[junior flower]

no

[steady trail]

I mean how do we know that the function certainly has a maximum and minimum

[junior flower]

from this and the fact that arcsin(x^2 - y - 1) attains the values -pi/2 and pi/2

[steady trail]

Okay

[steady trail]

I do not know how to do it

[junior flower]

if you want arcsin(x^2 - y - 1) = -pi/2, you want x^2 - y - 1 = -1

pick any x,y that satisfy that equation

[steady trail]

So when x^2-y=0

x=0, y=0

[junior flower]

yep that works

[steady trail]

But couldn't there be others now?

[junior flower]

other x,y that make arcsin(x^2 - y - 1) = -pi/2?

[steady trail]

x=1, y=1

[junior flower]

yea but that doesn't matter

[steady trail]

Why?

[junior flower]

we have arcsin(x^2 - y - 1) >= -pi/2 for any x,y, and there exist x,y with arcsin(x^2 - y - 1) = -pi/2

that shows the min is -pi/2

think about the minimum value of sin

it's -1

and we have sin(x) = -1 for lots of different x's but that doesn't matter, the min is -1

[steady trail]

Indeed

So max

x^2-y=2

x=-1 , y=-1

[junior flower]

yep that works

[steady trail]

But in fact, seeing the domain, setting x=0 and y=0, solving the equation I find 0≤0

And setting x=-1 and y=-1 i find 2≤2

Here I find precisely the extremes of the domain

[junior flower]

sure

that's expected since arcsin is an increasing function

[steady trail]

Last question: but what does it mean to parameterize this function

[junior flower]

wait no that's not what i meant

[steady trail]

Okay

[junior flower]

nvm i can't explain this well

[steady trail]

Maybe with an example ?

[junior flower]

well i can just give my stupid explanation again. the graph of f consists of points of form (x, y, arcsin(x^2-y-1)) where (x,y) is in D. call this set of points G
paramaterizing f means making some real variable functions a,b,c and a set A where [the points of form (a(t), b(t), c(t)) for t in A] is G

so you can get a point on the graph of f by just taking one real number t and looking at the point (a(t), b(t), c(t))

this is how i think about it but it's maybe too abstract

<@&268886789983436800>

[visual tiger]

<@&268886789983436800>, no advertising especially in help channels

[junior flower]

(spammed in lots of channels)

[steady trail]

Thank you so much

[steady trail]

.close

[midnight plank]

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[whole lantern]

????????????????????

[midnight plank]

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[sonic dragon]

it doesnt work

[old storm]

Why the hell did u close your previous one

[lusty cave]

lol

[old storm]

Shush ur just as guilty

[sonic dragon]

can some one just please help me

[lusty cave]

🤐

[violet storm]

maybe the answer form expects the answer fed to it in some specific way, sounds like a pain with this problem. maybe worth trying to factor out the e^ part out of both factors

as far as doing it correctly, seems ok to me

[old storm]

Wait

Your answer is right

Wtf

[sonic dragon]

thank you

yea electonic maths things arnt well suited for this i guess

[old storm]

Hmm

Remove one pair of brackets

From the second part

[violet storm]

try to factor this out, i think oftentimes they prefer factored versions of answers?

[old storm]

Also the brackets

You have a pair of useless brackets

[violet storm]

oh wait you forgot a parenthesis here, this will do it

[old storm]

Or that!

[sonic dragon]

whats a parentisis?

[violet storm]

( )

...

[sonic dragon]

oh thanks

[old storm]

[old storm]

I loved this reaction

[sonic dragon]

im english we call them brakets

[sonic dragon]

so?

[old storm]

Why do you have 5he outer brackets

[violet storm]

sorry i should have known to use UK english given your username and profile pic : )

[old storm]

Remoce them

[sonic dragon]

or

[old storm]

Bottom one

[sonic dragon]

ayy it was right

nice

thank you

[violet storm]

sure...the outer bracket isn't really necessary

but i guess doesn't matter

[sonic dragon]

i really dont like simplifying

thank you

🫡

.clsoe

.clsoe

.close

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[thorny glen]

Anyone have an idea of how solve this integral

[old storm]

Write sin theta as u

[rigid mirage]

u = sinx

du = cosx dx

so
you do

[vocal inlet]

where do u see x

[rigid mirage]

x as theta so i dont need to write theta

[violet storm]

easier to type in discord than theta 😛

[rigid mirage]

:)

[vocal inlet]

damn these mathematicians they see x everywhere

[vocal inlet]

theta is theta.

[rigid mirage]

I'd hardly call myself a mathematician

so

so

[old storm]

What about a weeb

[rigid mirage]

[violet storm]

$\int_0^{\f\pi 2}\cos(\sin\theta),d\theta=\int_0^{\f\pi 2}\cos(\sin x),dx$

[grand pond]

∫oosh (lemonsaurus appreciator)

[violet storm]

just rename the variable :p

[old storm]

There is a hentai joke here somewhere with the x but for the life of me I cant think of it

[rigid mirage]

i mean

with doing u = sinx

it still won't work

[thorny glen]

I got it from this integral when x is sin(theta), I’m not sure if this integral is easily solved, if not it’s fine but would like to know the actual solution

[violet storm]

did you use a trig sub?

[thorny glen]

Unless I messed up somewhere in the bounds

Yes

[old storm]

You lost me

[violet storm]

you used x = sin t ?

[thorny glen]

Yes

[violet storm]

seems like what you would do

[old storm]

I wonder if you should just polar this

[violet storm]

looks like a pretty standard trig sub integral

[vocal inlet]

trigonometric substition?

okay nevermind

[thorny glen]

The cosine is tripping me up

[old storm]

Does that to all of us sometimes

[thorny glen]

Is there a way to turn cos(sin(x)) into something

[vocal inlet]

like the weeb above said

u = sin x

[thorny glen]

That would just reverse the substitution though

Btw this isn't a homework problem just made the integral up and try to solve it

[violet storm]

maybe you just stumbled upon a hard integral then

kinda easy to do with integrals :p

but attempting it with trig sub seems like a good thought at least

[violet storm]

this isn't so easy to simplify because there's no obvious geometrical interpretation, sin spits out some geometric length and now you're feeding it to cos as an angle, so it's just kinda random blarb

[thorny glen]

It came from this integral, I only made it this far

.close

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[last slate]

HELO

[last slate]

must be 8?

wait

would it be half the length

i forgot about this not going to lie

any ideas?

.close

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[bleak pier]

hints please

[olive matrix]

just evaluate it

order of operations

[bleak pier]

what?

[olive matrix]

BODMAS or whatever

start with the exponents

[bleak pier]

it is going to infiity

[olive matrix]

is it? it doesn't look like it to me

[fathom tendon]

i guess theres just a hidden ... at the right

[bleak pier]

i guess yes

[olive matrix]

you guess?

[fathom tendon]

do you have the original problem

[olive matrix]

girlie you're the one who posted the question

[bleak pier]

acutally NO

[fathom tendon]

Peculiar

[bleak pier]

my answer by a formula will be e^2-e

[worthy wing]

Post the original question

[bleak pier]

as I said I dont have the original

someone asked me and i asked here

[worthy wing]

Your question doesn’t make sense

[bleak pier]

.

[worthy wing]

Lack of context and visibility

[bleak pier]

i am saying it is tending to infinity

[humble torrent]

ok so the sum $\sum_{k\geq1} \frac{2^{k-1}}{k!}$ ?

[grand pond]

Aslan

[bleak pier]

no

[zealous schooner]

This isn’t even the same sum

[bleak pier]

right

[humble torrent]

Then a hint would be to maybe make use of the fact that $e^x=\sum_{k\geq 0}\frac{x^k}{k!}$

[bleak pier]

2^3-1

[zealous schooner]

What makes you think this is wrong

[grand pond]

Aslan

[bleak pier]

i am just asking

[humble torrent]

?

[zealous schooner]

Looks good to me

[bleak pier]

i see

thanks

i want to know the process

[humble torrent]

wym

[bleak pier]

i used a formula only

[zealous schooner]

The sum in question is $$\sum_{n=0}^{\infty}\left(\frac1{n!}\sum_{r=0}^{n-1}a^r\right)$$

[grand pond]

kheerii

[zealous schooner]

You can use the formula for the sum of a GP

To simplify the inside summation

And then use the identity $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$

[grand pond]

kheerii

[zealous schooner]

Not quite, there’s one term missing

[humble torrent]

oh nvm theres a sum

in the numerator

God damn

[zealous schooner]

Indeed

[midnight plank]

@bleak pier Has your question been resolved?

[midnight plank]

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[midnight flame]

show that ^ for positive integers n

[humble torrent]

how much progress have you made?

[midnight flame]

um i tried it and then i think i did something completely wrong

so barelyt any

lemme give it one more go before asking sos

[humble torrent]

Yes

[midnight flame]

yeah so i'm unsure of what do to with RHS

[humble torrent]

my idea is to start with the LHS

by using the binominal theorem

[midnight flame]

i started with LHS and got 2 cos (cos + isin)

[humble torrent]

so set say x = cos(..) +isin(..)

[midnight flame]

binomials yeahh

[humble torrent]

then you want to expand (1+x)^n

you get $\sum_{k=0}^n \binom{n}{k}x^k$

[grand pond]

Aslan

[humble torrent]

see if this helps

maybe writing x as e^i(2pi/n) instead

[midnight flame]

nah haven't done euler's yet

meant to be solved w demoirves and binomials i reckon

[humble torrent]

ok so ignore eulers

if youre allowed demoivers it deosnt really matter

i used it as shorthand really here

but yeah try using that

then you "distribute" k into the arguments of cos and sin

if that even helps

[midnight flame]

ok thankss

.close

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[twilit field]

$\frac{\left(1\cdot2\cdot3\dots2n\right)}{2^nn!}=\frac{2^n\left(1\cdot1\right)\left(2\cdot2\right)\left(3\cdot3\right)\dots}{2^{n\ }n!}$

[twilit field]

is this right?

[grand pond]

ƒ(Why am. I here)=I don't Know

[deep vine]

no

[hollow oyster]

What are you trying to prove

[twilit field]

one minute

$\frac{(2n)!}{n! \cdot 2^n}$ is an integer

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

Induction would probably be enough here

now that I think of it

[deep vine]

just prove it directly

[twilit field]

👍 let me think of a way

[burnt flame]

think of double factorial*shifted double factorial

[deep vine]

(2n)!/(n! 2^n) = [ (n+1) (n+2) ... (2n) ] / 2^n
top has exactly n even numbers, so at least n factors of 2

[twilit field]

wait, doesn't the top have n/2 even numbers?

I reached until there too

but was stuck thereafter

[deep vine]

yeah mb

to be more specific

n/4 numbers divisible by 2 only, n/4 numbers divisible by 4

that should do it

[twilit field]

hmm

yeah

makes sense

thanks

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[twilit field]

.reopen

[midnight plank]

✅

[burnt flame]

each additional term adds a factor of 2

just as 2^n does

[twilit field]

got it

thank you

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[midnight plank]

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[twilit field]

is this right?

[twilit field]

I can justify myself if need be

[agile lichen]

yes but don't forget to prove the case n = 1

[twilit field]

I'll justify myself, just a minute

[subtle blaze]

$\frac{1}{1^2} + \frac{1}{2^2} \textcolor{red}{+} ... \textcolor{red}{+}\frac{1}{n^2}$

[grand pond]

Frosst

[subtle blaze]

[twilit field]

oh

sorry

[subtle blaze]

why do you say this

it doesn't seem very obvious why it's true

[agile lichen]

he's doing induction

[subtle blaze]

i know he is

[twilit field]

ok, not too sure of how I'd justify it

it just feels right

[dreamy lichen]

"We will prove it by induction on n:"
Base case:
blah blah blah blah. Base case works.

Inductive step:
Assume 1 / 1^2 + 1 / 2^2 + ... + 1 / n^2 <= 2 - 1/n. [Insert inductive step here]

mention that you are using induction

[twilit field]

ah, ok

[dreamy lichen]

seperate base case and inductive step

and thats pretty much all you will need to do

[twilit field]

so what I'm doing is overkill

[dreamy lichen]

no, it's unclear

you are doing induction

but in a very unclear way

[twilit field]

got it

thanks

[agile lichen]

write up the whole proof and then post it here

[twilit field]

👍

done, let me TeX it

just 2 minutes

To prove : $\sum_{i=1}^{n} \frac{1}{i^2} \leq 2 - \frac{1}{n}$
\
base case:- $1 \leq 2-1$
\
inductive hypothesis :- $\sum_{i=1}^{n} \frac{1}{i^2} \leq 2 - \frac{1}{n}$
\
we now prove that the hypothesis holds true for the (i+1)th case
\
$\sum_{i=1}^{n} \frac{1}{i^2} + \frac{1}{(n+1)^2} \leq 2- \frac{1}{n+1}$

but $\sum_{i=1}^{n} \frac{1}{i^2} \leq 2 - \frac{1}{n}$
so saying
\
\
\
$2- \frac{1}{n}+ \frac{1}{(n+1} \leq 2 - \frac{1}{n+1}$ isn't wrong
\
\
solving this inequality we get this is true for all $n\in \mathbb{N}$
\
\
thus our hypothesis is true
\end{document}

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

I think I've made a errors when translating to TeX

oops

is this better

[dreamy lichen]

so saying ___ isn't wrong

I'd definitely replace this with something like
"so it suffices to show that ___"

Also, I feel like you are trying to avoid doing the proof that inequality, so instead you just said that it suffices to solve the inequality

[twilit field]

I did the inequality in my notebook

I got $n \neq n+1$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

wait

my bad

$n \leq n+1$

[dreamy lichen]

\leq?

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

which shows it's true for all n

[dreamy lichen]

okay, but in a proof you'd usually write that in one line

yk what i mean

[twilit field]

yeah

[dreamy lichen]

it's good to learn to "translate" between solving an inequation and doing that one-line proof

[twilit field]

got it

is the rest of the proof fine?