help-0

[glass lichen]

I doubt one exists if it needs to hold for all f: R to R

[little drum]

if it's a quadratic or a polynomial, it'd be nice if you're clear about it

or even a continuous function at that (@_@;)

[lone heart]

@upbeat kayak Has your question been resolved?

[upbeat kayak]

It's quadratic, forgot to mention this

sorry

For any quadratic function f(x)

[upbeat kayak]

nvm found a solution

x2<2<x1<-1<x2

a * f(2) < 0 and a * f(-1) < 0

[lone heart]

@upbeat kayak Has your question been resolved?

[lone heart]

Channel closed

Closed by @upbeat kayak

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.
• 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.

[regal crane]

.open

I need to make sure this isn’t flawed

[regal crane]

Someone told me there is something near the PQ part

[lone heart]

@regal crane Has your question been resolved?

[lone heart]

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.
• 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.

[white drum]

How do I do chain rule here

[distant flare]

why did you bring some stuff to the other side?

[pale kestrel]

thats fine.

You apply the chain rule like usual

[white drum]

We are deriving

[pale kestrel]

Derivative of the inside

multiplied by the thing derived as if the inside was just x

[white drum]

Which one is inside or outside

Ln

Or x

[pale kestrel]

?

what was the point of logging both sides

if you leave it in that form

why did you take the log of both sides?

[white drum]

To get sin out of the power

[pale kestrel]

Exactly, so your next step should be this

before differentiating both sides.

$$\ln(y-x^4-4^x) = \ln\left(x^{\sin(4x)}\right)$$

[ocean seal]

Shuri2060

[pale kestrel]

$$\ln(y-x^4-4^x) = \sin(4x)\ln x$$

[ocean seal]

Shuri2060

[pale kestrel]

$$\dv{x}\ln(y-x^4-4^x) = \dv{x}\left(\sin(4x)\ln x\right)$$

[ocean seal]

Shuri2060

[pale kestrel]

And then proceed.

[white drum]

Right?

[pale kestrel]

???

[white drum]

Did I do ln correctly

[pale kestrel]

of course they're not equal.

[white drum]

What?

[white drum]

That’s literally the equation

[white drum]

This is literally exactly what you wrote

[pale kestrel]

Think carefully agian.

[white drum]

I have no idea

[pale kestrel]

Then why is the LHS differentiated

[white drum]

What is lhs

[pale kestrel]

and the RHS the original thing

[white drum]

What is rhs

[pale kestrel]

left and right hand side

[white drum]

I have no idea what you’re trying to say

I can’t combine steps

[pale kestrel]

You differentiated the left

[white drum]

I’m working through this in steps

[pale kestrel]

and left the right untouched

The 2 sides are not equal

[white drum]

I just applied ln

[pale kestrel]

Look.

$$\dv{x}x^2 = \dv{x}x^2$$

[ocean seal]

Shuri2060

[pale kestrel]

$$2x = x^2$$

[ocean seal]

Shuri2060

[pale kestrel]

This is just plain wrong

[white drum]

I haven’t differentiated yet

I’m just applying the ln

[pale kestrel]

???

[white drum]

What is ln if the lhs then

[pale kestrel]

I don't understand

[white drum]

I am applying the ln to both sides

To get the exponents out of the exponents

Because that is what applying ln is

[pale kestrel]

I think you misunderstand what the log function does.

[white drum]

ln x^4 is 4x

[pale kestrel]

It's not.

[white drum]

What is it?

[pale kestrel]

$$\ln(y-x^4-4^x)$$

[ocean seal]

Shuri2060

[pale kestrel]

You have this thing on the left

And you're proposing that the ln can be split across all 3 terms?

1. That's not correct. Check your log rules

2. ln y isn't 1/y

3. ln x^4 isn't 4x

[white drum]

I have no idea what to do next

[pale kestrel]

I have no idea what you have done exactly

[white drum]

I have no idea what to do

[pale kestrel]

Ok slow down

Firstly let's get to the bottom of this

I agree with what you have done on the RHS

That is correct.

But the LHS is just algebraic nonsense

You've attempted apply log laws that don't exist

That half look like you're differentiating

===
Can you see the issue with what you have done, first of all?

Before we move onto what is the best approach to this

$$\ln(a+b)$$

[ocean seal]

Shuri2060

[pale kestrel]

This expression does not simplify in general.

There is no log law that does anything to this

[white drum]

okay

[pale kestrel]

Ok so backtracking

$$\dv{x}\ln(y-x^4-4^x) = \dv{x}\left(\sin(4x)\ln x\right)$$

[ocean seal]

Shuri2060

[pale kestrel]

You should be on this step --- which is differentiating both sides.

[white drum]

correct

[pale kestrel]

You can do things one step at a time

First attack the RHS

What rule can you do to differentiate this?

[white drum]

Chain rule

O=lnm

M=sini

i=4x

[pale kestrel]

🤔

I am unfamiliar with this particular methodology you seem to have for applying the chain rule...

But can you show me what your answer is

[white drum]

Alternatively product and chain rule but I don't understand why we cant do it all in one

Sure

[pale kestrel]

That doesn't look correct

[white drum]

I agree but idk why

[pale kestrel]

So you said apply the chain rule

But actually - that is impossible

with the chain rule I am familiar with

I was just wondering what your letters o m i were

[white drum]

Outer

Middle

Inner

[pale kestrel]

The chain rule is used to differentiate something like

f(g(x))

But we don't have that right now.

[white drum]

Okay that makes sense

[pale kestrel]

We have 2 things multiplied together

f(x)g(x)

Which means we first apply the product rule.

[white drum]

So product rule

4cos4x

I used chain rule to get g'

4cos4x is shown in the answer key at this step so I assume it is correct

[pale kestrel]

(fg)' = f'g + fg'

$$\dv{x}\ln(y-x^4-4^x) = \sin(4x)\dv{x}\ln x + (\ln x)\dv{x}\sin(4x)$$

[ocean seal]

Shuri2060

[pale kestrel]

You should get this after applying the product rule.

Next the chain rule is used to differentiate sin(4x), yes.

$$\dv{x}\ln(y-x^4-4^x) = \frac{1}{x}\sin(4x) + 4(\ln x)\cos(4x)$$

[ocean seal]

Shuri2060

[pale kestrel]

So you should get something like this, yes?

[white drum]

Still working on product rule

[pale kestrel]

You should try writing things one thing at a time

[pale kestrel]

Apply the product rule fully

writing it out like this

No need to think about what the differential of sin(4x) is yet.

[pale kestrel]

That is correct yes

But for more complex stuff that method of writing your working might not work out for you

[white drum]

How else would we find g’

[pale kestrel]

Also you are skimping out on writing d/dx

or '

Doesn't matter which you do - but don't miss it out

These 2 aren't equal

You differentiated the right, but left the left alone

Either you write it like this:

$$\dv{x}\ln(y-x^4-4^x) = \dv{x}\left(\sin(4x)\ln x\right)$$

[ocean seal]

Shuri2060

[pale kestrel]

$$(\ln(y-x^4-4^x))' = \left(\sin(4x)\ln x\right)'$$

[ocean seal]

Shuri2060

[pale kestrel]

Or you write it like this, it doesn't matter.

[pale kestrel]

perhaps you might prefer this because it's quicker, but the process is still the same.

I apply the product rule to the right hand side and get this

$$(\ln(y-x^4-4^x))' = (\sin(4x))'(\ln x)+(\sin(4x))(\ln x)'$$

(fg)' = f'g + fg'

[white drum]

Okay so we write the terms without actually deafferenting them yet

[pale kestrel]

yes.

[ocean seal]

Shuri2060

[pale kestrel]

It is a neater way of writing the working

Because if the expressions inside were complicated

You will find yourself lost I think

You look like you are differentating depth first

rather than breadth

=====
So alright, you did the right hand side.

$$\dv{x}\ln(y-x^4-4^x) = \frac{1}{x}\sin(4x) + 4(\ln x)\cos(4x)$$

[ocean seal]

Shuri2060

[pale kestrel]

Now you differentiate the left. Which rule to use?

[white drum]

chain rule ig

Or product

I don't see how it could be a product rule though

[pale kestrel]

well yes

it has to be chain rule

you have something of the form f(g(x))

(f(g(x)))' = f'(g(x))g'(x)

This is the version of the chain rule you are familiar with?

[white drum]

Yes

[pale kestrel]

So again, apply just this.

$$(\ln(y-x^4-4^x))' = \frac{1}{x}\sin(4x) + 4(\ln x)\cos(4x)$$

[ocean seal]

Shuri2060

[pale kestrel]

The inside function is a bit length, so usually people substitute something like u for it

[pale kestrel]

or I in your case

ok

[white drum]

I guess next steps are times both sides by the denominator

Plug in for y

simplify

Then move 4x^3-4^xln(4) to the other side?

[pale kestrel]

I haven't seen what you got after you applied chain rule

What does the LHS become

[pale kestrel]

i think you're missing brackets

ill check

$$\dv{x}\ln(y-x^4-4^x) = \frac{1}{x}\sin(4x) + 4(\ln x)\cos(4x)$$
Let $u = y-x^4-4^x$
$$\dv{x}\ln u = \frac{1}{x}\sin(4x) + 4(\ln x)\cos(4x)$$
$$\dv{u}{x}\dv{u}\ln(u) = \frac{1}{x}\sin(4x) + 4(\ln x)\cos(4x)$$
$$\dv{u}{x}\frac{1}{u} = \frac{1}{x}\sin(4x) + 4(\ln x)\cos(4x)$$
$$\frac{y'-4x^3-(\ln4)4^x}{y-x^4-4^x} = \frac{1}{x}\sin(4x) + 4(\ln x)\cos(4x)$$

[ocean seal]

Shuri2060

[pale kestrel]

ok no you're fine

Then like you said, plug in for y, and isolate y'

[white drum]

But why is the answer key different

Here is my whole process

Where did the ln inside the paranthesis come from

[pale kestrel]

I'm not sure how you got your final answer

[pale kestrel]

We agreed on this last line yes?

[white drum]

Yes

[pale kestrel]

$$\frac{y'-4x^3-(\ln4)4^x}{y-x^4-4^x} = \frac{1}{x}\sin(4x) + 4(\ln x)\cos(4x)$$

[ocean seal]

Shuri2060

[pale kestrel]

Next what was y...

$$y=x^4+4^x+x^{\sin(4x)}$$

[ocean seal]

Shuri2060

[white drum]

x^4+4^x+x^sin4x

yes

[pale kestrel]

$$\frac{y'-4x^3-(\ln4)4^x}{x^{\sin(4x)}} = \frac{1}{x}\sin(4x) + 4(\ln x)\cos(4x)$$

[ocean seal]

Shuri2060

[pale kestrel]

$$y'-4x^3-(\ln4)4^x = x^{\sin(4x)}\left(\frac{1}{x}\sin(4x) + 4(\ln x)\cos(4x)\right)$$

[ocean seal]

Shuri2060

[pale kestrel]

agreed?

Then move the stuff to the right and thats it more or less

[white drum]

where did the 4lnx come from

[pale kestrel]

it was always there

you just wrote it the other way round

ln x . 4cos x

[white drum]

How is that the same thing as 4lnx

[pale kestrel]

???

[white drum]

oh I see

multiplication is communative

Right?

[pale kestrel]

???

How.

Think you dropped brackets which need to be there.

[white drum]

What brackets

1/x*sin4x=sin4x/x

I have no idea how you got 4lnx*cos4x

In the product rule

f'g*fg'

f is lnx

g' is 4cos4x

Yeah I quit

It's been an hour

This is what happens when your parents send you to special ed school for the SOLE purpose of keeping you away from sex and drugs

Even when I was mentally and academically capable of regular school

.close

[lone heart]

Channel closed

Closed by @white drum

Use .reopen if this was a mistake.

[lone heart]

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.

[jade blaze]

.open

idk

[jade blaze]

yea im not sure exactly wat to do

is the top one supposed to be approaching below 7

guesss input wat isr eally close to the answer

[lone heart]

@jade blaze Has your question been resolved?

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[lone heart]

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.

[true geyser]

I feel like I'm doing this correctly but it isn't accepting my answer as correct

[glass lichen]

F isn't C

[true geyser]

Very cool I need to stop being dumb

.close

[lone heart]

Channel closed

Closed by @true geyser

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.
• 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.

[raw citrus]

I'm not sure how to go from here.

[raw citrus]

I'm supposed to assume there are two lines that go through P is perpendicular to line l. And then reach a contradiction using Property C

Nvm! I got it

.close

[lone heart]

Channel closed

Closed by @raw citrus

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.
• 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.

[alpine sable]

wow channel 0

[swift hemlock]

Come on now math is asking me questions

[pale kestrel]

whats u~

So we glue all the rationals together

but whats this u~ 🤔

uhhh the induced topology from this equivalence?

quotient topology i believe its called..

So Hausdorff --- forall x, y, exists neighborhoods around x, y that dont intersect

To show its not, you need a specific x, y as a counterexample --- any ideas?

@alpine sable

[alpine sable]

quotient topology

[alpine sable]

Yes!

[valid lake]

HI

[alpine sable]

Hmm, well, both Q and R are dense, yeah?

[valid lake]

i need a help

[alpine sable]

Make a channel, @valid lake.

[valid lake]

where ?

[pale kestrel]

well yes exactly.

[alpine sable]

Go to a channel that isn't taken. They're above this one.

[valid lake]

ok

[alpine sable]

So... Dedekind cuts? lol

[pale kestrel]

idk what they are but like...

if you pick any x y

pi

e

[alpine sable]

Ahh

Yeah.

[pale kestrel]

i think almost any pair will work

because all neighborhoods will contain a rational

[alpine sable]

Hmm...

Oh, so it contains the equivalence classes of both rationals and irrationals?

Or, am I missing something?

[pale kestrel]

what are the classes?

[alpine sable]

Any rationals where x - y is rational, right?

[pale kestrel]

thats the relation.

but the classes?

[alpine sable]

Hmm...

[alpine sable]

A rational minus a real?

[pale kestrel]

no

The classes are the sets

[alpine sable]

{Q + r} for r in R.

[pale kestrel]

no

$\sim_x:= {y\in\bR : x\sim y}$

I will use this notation.

So choose an x for me

and tell me its equivalence class

oh i made a typo

[ocean seal]

Shuri2060

[pale kestrel]

sry its this

so choose an x first.

Then tell me its equivalence class

[alpine sable]

Let's say x = 2.

[pale kestrel]

ok.

[alpine sable]

It has the equivalence class of... well... let's see. 2 + 1 is equivalent with this relation, right?

And 2 + 4?

[pale kestrel]

well yes

all of those are in the class

[alpine sable]

Right.

So... 2 + {anything that makes this rational}

[pale kestrel]

uhh

[alpine sable]

And then choose something like pi.

[pale kestrel]

i dont think 2+ is relevant

[alpine sable]

Oh?

[pale kestrel]

oh wait aaaaaa

ok i misread your original question

no wonder lol

[alpine sable]

dang

[pale kestrel]

I think theres a typo there lol

[alpine sable]

That's why I'm so lost.

[pale kestrel]

x ~ y iff x - y in Q

yes?

$$\sim_x:= {y\in\bR : x\sim y}$$
$$= {y\in\bR : x-y\in\bQ}$$

[ocean seal]

Shuri2060

[pale kestrel]

So indeed this observation is correct

[alpine sable]

Got it,

[pale kestrel]

$$\sim_x = x+\bQ$$

[ocean seal]

Shuri2060

[pale kestrel]

So if we have x, y that contradict the Hausdorff statement

We should be able to always find an equivalence class which is in the intersection of any neighborhoods of x and y

And like you said, this argument will probably hinge on denseness

[alpine sable]

Okay.

[pale kestrel]

I think it is possible to show this algebraically. Not just handwave

[alpine sable]

So, since Q is dense, there cannot be a neighbourhood for q in Q containing only q!

[pale kestrel]

Yes the denseness intuition is important

but I think you dont need to use it in your proof

Well at least for what I have in mind

[alpine sable]

Oh?

Honestly, I think it would work, no?

[pale kestrel]

I think if you pick epsilon_1 and epsilon_2

[alpine sable]

Ah.

Rigorous.

[pale kestrel]

$$(x-\varepsilon_1, x+ \varepsilon_1)\cap (y-\varepsilon_2, y+\varepsilon_2) $$

[ocean seal]

Shuri2060

[pale kestrel]

yes so you prove this is non-empty

maybe its not so easy

Note I refer to the equivalence classes here

maybe I should use open ball, not interval but yes

[alpine sable]

This makes sense.

[pale kestrel]

Actually if I use this notation and refer to x and y as the representatives

you want to show there is a number from each interval

which is a rational number in difference

$$a\in(x-\varepsilon_1, x+ \varepsilon_1)$$
$$b\in(y-\varepsilon_2, y+\varepsilon_2) $$
$$a-b\in\bQ$$

[ocean seal]

Shuri2060

[alpine sable]

Now, use denseness to prove it is nonempty?

[pale kestrel]

You could do it that way, also

Maybe you have to use denseness either way, I wasnt 100% sure

Perhaps instead of denseness, you can consider ceil(1/min(e1, e2))

but hmmm this doesnt quite get you there...

Well its up to you which argument you want to try 🙂

[alpine sable]

I think I can try using denseness from here.

[alpine sable]

Could I ask one more question?

It's my last problem and it seems quite complex.

[pale kestrel]

👀

[alpine sable]

Here it goes:

(S1 x I) / ~ is homeomorphic to {x in R^2 | ||x|| <= 1} when ~ is defined s.t. (x, y) ~ (s, t) iff xt = ys.

[pale kestrel]

S1 x [0, 1]

visually, what does this look like?

[alpine sable]

A disc, yeah?

Like a ball of dimension two.

[pale kestrel]

im not sure...

I had a different image

[alpine sable]

It's a circle over the entirely interval [0, 1].

Oh?

[pale kestrel]

surely a ball of dimension 2 is IxI

[alpine sable]

Hmm...

[pale kestrel]

wait

so the set

{x in R^2, |x| =< 1}

this is clearly the disc in R^2

===
S1 x [0, 1]

But this, isnt

we're meant to have
(S1 x [0, 1])/~ is homeomorphic to the disc

[alpine sable]

ohh

hmm

[alpine sable]

Yeah.

I can't even visualize the eq relation.

[pale kestrel]

yes thats why i asked for

S1 x [0, 1]

just this first.

[alpine sable]

unit circle cross [0, 1]...

Isn't it just the...

@pale kestrel I think I have it.

(S1 x I)/~ is compact and the unit disc is Hausdorff so a function f mapping the former to the latter defined using the quotient topology makes a homeomorphism.

[pale kestrel]

Well the answer I do not know.

But you should try visualizing what this quotient is, regardless.

[alpine sable]

I just see it as the disc. :/

Sadly.

[pale kestrel]

S1 x [0, 1]

whats this first of all

[alpine sable]

a circle paired with a point on that interval

[pale kestrel]

geometrically?

[alpine sable]

Or rather, a point on the circle paired with a point on the interval.

[alpine sable]

Hmm, circle cross line segment...

Cylinder? :p

[pale kestrel]

yes a cylinder without the ends

hollow tube

[alpine sable]

:0

omg

[pale kestrel]

Now the equiv

look at that equation carefully

[alpine sable]

Yes.

[pale kestrel]

Theres a few ways to view it

Do you recognise it in any way?

(S1 x I) / ~ is homeomorphic to {x in R^2 | ||x|| <= 1} when ~ is defined s.t. (x, y) ~ (s, t) iff xt = ys.

[alpine sable]

Oh, you're locking it to a plane.

[pale kestrel]

??

[alpine sable]

nvm

It does kind of remind me of Euclid's gcd formula.

[pale kestrel]

no

if we had a b c d

instead of x y s t

ad - bc = 0

[alpine sable]

Oh, it looks more familiar now.

OH

RATIONALS

[pale kestrel]

?

[alpine sable]

a/b = c/d

[pale kestrel]

you can do that yes...

[alpine sable]

equivalence classes

[pale kestrel]

not rationals though

x/y = s/t

treat the points as vectors

the vectors must be multiples of each other

or if we take ad - bc = 0, this a 2x2 determinant. The determinant is 0 means the columns or rows are linearly dependent

[alpine sable]

Right, okay.

[pale kestrel]

(S1 x I) / ~ is homeomorphic to {x in R^2 | ||x|| <= 1} when ~ is defined s.t. (x, y) ~ (s, t) iff xt = ys.

[alpine sable]

Oh, of course. From linear algebra.

[pale kestrel]

Actually now I think about it I'm confused

How is S1 defined

[alpine sable]

The unit circle?

[pale kestrel]

yes in terms of numbers

I can't tell what the x and s are meant to belong to

[alpine sable]

Ah. Good point.

[pale kestrel]

[0, 1] but you connect the end points???

But this makes no sense to me...

[alpine sable]

Maybe????

tbh

[pale kestrel]

look at the original relation

[alpine sable]

I found a hint on stack exchange and went from there.

[pale kestrel]

if x = y = 0

[alpine sable]

And it's totally different from what you're saying.

But obviously there are different ways of looking at it.

[pale kestrel]

Notice (0, 0) is related to every other point. This creates a single class. This surely cannot be right

[alpine sable]

But it's an argument from compactness and Hausdorfness using different equivalence classes.

[alpine sable]

yeah...

[pale kestrel]

Is this the exact original question?

[alpine sable]

@pale kestrel here

[pale kestrel]

yes i just found this

i understand

x and s are vectors

the equation tx = ys is a vector equation

[alpine sable]

Oh?

[pale kestrel]

This now makes a lot more sense

x and s live in R^2

on the unit circle

t and y are scalars in [0,1]

$$x, s\in{x\in\bR^2 : |x| = 1}$$

[ocean seal]

Shuri2060

[pale kestrel]

$$t,y\in[0, 1]$$

[ocean seal]

Shuri2060

[alpine sable]

Hmm

Interesting

[pale kestrel]

So now look at that equivalence relation again

and it makes a lot more sense

(S1 x I) / ~ is homeomorphic to {x in R^2 | ||x|| <= 1} when ~ is defined s.t. (x, y) ~ (s, t) iff xt = ys.

since x and s are vectors, there are only a few occasions when this equality will hold

My bad for not noticing earlier 💤

$$x\in{x\in\bR^2 : |x| = 1}$$

$$y\in[0,1 ]$$

Think about what xy must belong to

[ocean seal]

Shuri2060

Shuri2060

[pale kestrel]

x lives on the unit circle

but then y scales the length of x between 0 and 1

If you like, x is the angle. y is the length

(x, y) is exactly polar form.

The equivalence relation handles the point at 0

[alpine sable]

👀

[alpine sable]

unit disc lol

[lone heart]

@alpine sable Has your question been resolved?

Channel closed

Closed by @slim pulsar

Use .reopen if this was a mistake.

[lone heart]

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.

[heady prairie]

is it normal that in the way i defined a plane, i only render it from "a corner" (which is my sample point and not beyond it) heres what i mean

[heady prairie]

my plane is supposed to take up entire

floor

but its only doing so from that one corner

heres how i defined its intersection

[lone heart]

@heady prairie Has your question been resolved?

[heady prairie]

.close

[lone heart]

Channel closed

Closed by @heady prairie

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.
• 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.

[alpine sable]

what is natural logarithm of -1 ?

[alpine sable]

<@&286206848099549185>

[abstract fractal]

You need to wait 15 minutes before tagging

[alpine sable]

its an emergency

i cant find help online about it

[abstract fractal]

Are we only considering real numbers?

[alpine sable]

yes

[abstract fractal]

Picture the graph of e^x in your head. Is there any value of x that makes the function negative?

[alpine sable]

e^(-3)?

hmm

no?

so it doesnt exist?

[abstract fractal]

It doesn't exist. No real value of x gives a negative answer

[alpine sable]

thanks

[alpine sable]

.close

[lone heart]

Channel closed

Closed by @smoky socket

Use .reopen if this was a mistake.

[little drum]

but like Dio said, no real works

[lone heart]

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.

[little drum]

,rotate

[finite flax]

well...

I have it

but I don't know if I can explain it

b.c. it's slightly different for every one you do

but you want (pretty much every time)

to scale the input linearly with some offset

so basically

x = at+b

in this case,

x = 2t -1

y = 1-x^2 = 1-(2t-1)^2

we need the input to go FROM -1 to 1,

TO 0 to 1

yes

I don't know about that

should reduce to a single answer

well I mean

there should be known way to do this

ok ok

what is the length of the segment of the x axis from -1 to 1?

and you get why I am asking that so far right

just bear with me

so we need to make is so the distance from 0 to 1 (which is 1) will cover length 2

hence, we scale t by 2

that's 'a'

but we need to START at -1

so we offset by adding -1

that's b

x = 2t - 1

we just did it together!

we need to start with a distance from 0 to 1

and make it so we can cover the distance from -1 to 1 (which is 2)

that's why we scale t by 2

and we offset it by -1 so we can start at -1

instead of 0

boom.

yes

the parameterization is f(x,y) = <some function of t, another function of t>

y is already a function of x

and x is a function of t now

thus...

?

??

???

...

yup.

every time.

it usually is

the harder ones begin parametric

far as I know

in school anyway

ok

how about another quadratic

let's do one that has roots at -1 and 3

no, roots

zeros

oh, yes

it would

(x+1)(x-3) = x^2 - 2x + 3

and let's say we want a parameterization that takes t in [0,1] to x in [-1,3]

do that the same we just did the other

that's the domain

part of the domain, I mean

yup

f(x(t),y(t)) = <4t-1, and whatever y(x(t)) simplifies to :P>

t is always 0 to 1...pretty much

normalized

k

yes

ok

don't

I'm about to get run out a teaching credential program b.c. I choke making my writeups

I've been watching HBO MAX for 3 hours

I hate this

I want to die

more than I want to go back to that classroom--oh look, math

no

middle school

but I was trained for hs

shit prep first semester

no the kids aren't the problem

what's the math problem

ok, so normalized, t is in [0,1] as usual

and where do you need to take it?

uh-huh...

let me ask you, are we good where we start?

at 0

?

focus on my questsions

t in [0,1] ---> theta in [0, pi/2]

so...what do you need for a and b?

theta=at+b

whoops, yes

read again

and tell me

b.c. it's the same

t in [0,1] ---> theta in [0, pi/2]

yes

no

lol

a = pi/2

ok, one more time, more clearly

the distance from 0 to pi/2

is pi/2

that's 'a'

and you don't need an offset

right, because x starts at 0

theta

my bad

from earlier

oh....

hang on

I'll just pencil these out rq

and take a screenshot

ok well that first part worked

just fine

really

what is the right answer for part a?

I see it

hm...

f(r,t)

ugh, let's get in a room

and get out Desmos

yes

I think math voice has screenshar

e

I mean

advanced mathematics

I'm there now

I'm there right now

it's under VOICE

might need to be expanded

really?

[little drum]

$(x, y) = (\sin 2\theta \cos \theta, \sin 2\theta \sin \theta)$

[little drum]

isn't polar basically parametrization??

Also, for 'b' isn't the interval for "t" just doubled

[lone heart]

@alpine sable Has your question been resolved?

[little drum]

Your curve 👀

[little drum]

First petal :o

Did you maybe misunderstand the question?

<sin 2t, t> <--- -that's not how polar works lol

[lone heart]

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.
• 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.

[swift shoal]

How to solve?

[alpine sable]

then find the value of y?

[swift shoal]

yes

[vague coral]

by making y the subject and solve

[swift shoal]

Don't laugh

[vague coral]

yeah you just invented algebra

[alpine sable]

just a sec grabs out copy

[vague coral]

its all wrong tho

[swift shoal]

:/

Can someone solve

and send

[vague coral]

solve it yourself, we can only show you how to

[swift shoal]

ok

[vague coral]

use this fact :
$\frac{a}{b} = \frac{c}{d}$ means that $ad = bc$

[ocean seal]

Herels

[vague coral]

obviously b and d different of 0

[swift shoal]

is my first step correct? @vague coral

[alpine sable]

yea

its easy

take the denominator

[swift shoal]

.

[alpine sable]

(1-3y)(4-y)=4-y-12y+3y^2 right?

or

4-13y+3y^2

simplify

the numerator and denominator

and

then use what she said

[vague coral]

wrong

[swift shoal]

ok

[alpine sable]

u can rewrite as

(1-3y)(4-y)=(2-3y)(1-y)

if thats what easier for u?

[vague coral]

I mean are you sure you know the basic rules of algebra ? @swift shoal

[swift shoal]

no

[alpine sable]

uh

i am sus on the age

lmao

[vague coral]

go learn them first before doing that thing

[swift shoal]

ok

[alpine sable]

ya

[swift shoal]

.close

[lone heart]

Channel closed

Closed by @swift shoal

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.
• 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.

[alpine sable]

we often know e to be defined as limit upto infnity of $(1+1/n)^n$ right? thats the place where it comes from now take another series in math $2^infnity=0$ yes there is a very quick way on how u get that u can see on google basically $2* 2 * 2 *...=r$ so 2r=r thus r = 0 but in the e formula lets just say (1+1/n) when n=infinity is equal to c a constant and n=infinity so c^infinity as we know is 0 does that mean e=0

[alpine sable]

sorry i had some problems with how to format it

[tacit arch]

yes you've proved e=0. feel free to .close

[alpine sable]

☠️ I strongly belive that I got wrong somewhere

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

<@&286206848099549185>

[glad sluice]

is this a genuine question?

if so... 2 * 2 * 2 * ... = r, r can be undefined

so you can't do the normal arithmetic on it

[alpine sable]

Wait

Imma show u

https://youtu.be/leFep9yt3JY

What this video is saying at

Time 6:09

I mean that

[glad sluice]

that is not possible in the real number system

i mean you could create a system where that is true

[alpine sable]

Dude there are hundreds of series solved with the same logic

Power tower golden ratio

[glad sluice]

yeah but they're not necessarily convergent

[alpine sable]

Dude

Then r u

Saying

Ramajuan as wrong

Here

1+2+3+4+...=-1/12

[glad sluice]

no that is analytical continuation

[alpine sable]

I mean thats also divergent

[glad sluice]

there is a system where that is true

[glad sluice]

yes, it is divergent in the "normal" number system

[alpine sable]

Ok then golden ratio is not true?

[glad sluice]

and that is not true in the "normal" number system

[alpine sable]

What is not normal in

2* 2 * ....n

[glad sluice]

this

[alpine sable]

What?

[glad sluice]

nothing nvm

[glad sluice]

hm?

that's clearly divergent

[alpine sable]

Yes

But many series

Are divergent

Still we compute that?

Like 1+2+3+.....

[glad sluice]

that is analytical continuation, you're using formulae out of bounds, bounds which they're supposed to be used under

you could also say with the same logic
1 + 2 + 4 + 8 + ... = 1/(1-2) = -1

but the infinite geometric progression sum formula

only holds true for -1 < r < 1

so you're technically "misusing" the formula

[alpine sable]

So like

Outside of real analysis

This still has sense?

[glad sluice]

maybe

[alpine sable]

Hmmm

[glad sluice]

I may be wrong here though

[alpine sable]

Hmm ok

I actually am kinda confused

Because

Stuff like that is seen

Again in

x^x^x^x... Series

Anyway

I think I have been going wrong here

That (1+1/infnity)

Is not a real number

[glad sluice]

yeah infinity is a weird concept

[alpine sable]

Yea

We can't just hold it as a constant

.close

[lone heart]

Channel closed

Closed by @vale aspen

Use .reopen if this was a mistake.

[lone heart]

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.

[grave hamlet]

how is this ?

[grave hamlet]

lets say probability of 1st event is 0.20 then others are 0

how it will goes to zero ?

[vale sapphire]

Say your first event is A_1

Then $\sum^\infty_{i=2} \mathbb{P}(A_i) = 0$ because $\mathbb{P}(A_i) = 0$ for all $i\ge 2$

[ocean seal]

Syst3ms

[vale sapphire]

What is changing is the starting index

You make the summation start later @grave hamlet

Although, this isn't true for all choices of A_i

Heck, if you take P(A_i)=1, the sum isn't even finite for any given n

[grave hamlet]

aaah so the summation starts at infinity ?

because i = n

still confused

so is it like saying after some point prob is zero ?

because there is just infinite of them

and prob(all events) has to be finite measure

so yeah tail sum of convergent series is just zero

right ? @vale sapphire

[vale sapphire]

The summation starts at n and goes to infinity

And at point the probability is not necessarily zero, but it is true that the tail sum of a convergent series tends towards zero

[grave hamlet]

could you please tell me why its zero ?

[vale sapphire]

That's a limit

[ocean seal]

Syst3ms

Syst3ms

[vale sapphire]

You're removing the first n terms from the infinity series

[ocean seal]

Syst3ms

Syst3ms

Syst3ms

[vale sapphire]

If you keep removing more and more terms from the start of a convergent series, it will tend to 0

[grave hamlet]

yup exactly

thanks man

@vale sapphire

also this

how we take limit

i mean how it transformed to limit

[vale sapphire]

This is only true if the B_n are decreasing with regards to inclusion

[grave hamlet]

also note that Bns are nested decreasing

[grave hamlet]

could you please tell me the intuition here ?

[vale sapphire]

Well, what is being said is that if all of the events imply each other successively, then the limit of their probabilities is the probability of all of them happening at once

[grave hamlet]

yeah i think i got this

[vale sapphire]

For a fixed $N$, what is $\bigcap_{n=1}^N B_n$ ?

[ocean seal]

Syst3ms

[grave hamlet]

so the last bn are intersecting with all previous bns. so the prob of that intersection is just lim n goes to infinity

right ?

[vale sapphire]

For a fixed N

N doesn't move

I get your drift, but let's go a tad slower, shall we?

[grave hamlet]

yeah

[vale sapphire]

Hint : which event is incuded in all of the others?

[grave hamlet]

BN

if it is decreasing

[vale sapphire]

Correct

[grave hamlet]

Bns

[vale sapphire]

So in short $\bigcap_{n=1}^N B_n = B_N$

[ocean seal]

Syst3ms

[vale sapphire]

Wait a minute

I don't actually recall how this is proved, give me a sec

[grave hamlet]

intuitively its proved

hehe

[vale sapphire]

Okay, I had wholly forgotten the actual proof

Have you studied the case where the sequence is increasing and you're taking the union instead?

[grave hamlet]

nope

does that matter now ?

[vale sapphire]

Yeah, because the proof goes by proving that first case, and then using a complement to get the intersection case

[grave hamlet]

aah okay

[grave hamlet]

i think iam gonna trust my intuition for now

okay ?

[vale sapphire]

And the first proof actually uses one of the probability axioms

Although, the proof isn't complicated, i can show it

But it uses a trick that's not obvious if you've never seen it

Take some increasing sequence of events A_n

[grave hamlet]

yeah i think its enough

[vale sapphire]

And set B_0=A_0 and B_n=A_n \ A_(n-1) for all n ≥ 1

What you can notice is that $\bigsqcup_{i=0}^n B_n = \bigcup_{i=0}^n A_n$

[ocean seal]

Syst3ms

[vale sapphire]

But there is something very special about the B_n, and that's denoted by the square union symbol

The union is disjoint

And one of the axioms of a probability functions only works for a disjoint union of events

That's why we have to define the B_n that way, it's to get disjoint events

Hence, $\mathbb{P}(\bigcup_{i=0}^\infty A_i) = \mathbb{P}(\bigsqcup_{i=0}^\infty B_i) = \sum^\infty_{i=0} \mathbb{P}(B_i) = \lim_{n\to\infty} \sum^n_{i=0} \mathbb{P}(B_i) = \lim_{n\to\infty} \mathbb{P}(\bigsqcup_{i=0}^n B_i) = \lim_{n\to\infty} \mathbb{P}(\bigcup_{i=0}^n A_i) = \lim_{n\to\infty} \mathbb{P}(A_n)$

[ocean seal]

Syst3ms

[vale sapphire]

The second and fourth equalities use the additivity of probability with a disjoint union, and the last equality uses the fact that the sequence is increasing

This switch to a disjoint union is subtle

[grave hamlet]

yupp

[vale sapphire]

But now that you've seen this, hopefully you'll have an easier time with the intersection case by taking the complement

[grave hamlet]

yep thanks @vale sapphire

[tall flume]

Wtf is thatttt

Damn

Integrals i guess

[vale sapphire]

No integrals here

Both of those symbols are unions

One is written slightly differently to remark that the union is disjoint

[lone heart]

@grave hamlet Has your question been resolved?

Channel closed

Closed by @grave hamlet

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.
• 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.

[alpine sable]

.help

[lone heart]

Commands:
clopen: .close, .reopen
factoids: .tag
help: .help

Type .help <command name> for more info on a command.

[alpine sable]

.reopen

f:(0,infinity ) -> R f(x)=1/x(x+1)

[alpine sable]

?

[swift hemlock]

whats the ques

[robust dagger]

wait who are you asking?

this isnt in my name

[alpine sable]

what do you mean by ques?

[coral moss]

Question

[swift hemlock]

what are u asking mentioning this?

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[lone heart]

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.

[dull lily]

can someone help me through this

[covert agate]

which

[dull lily]

uhhh i guess i need all 3 parts

but just start me off with a and see if I can do b and c on my onw?

so so far I know y = 3rad 5x^2-7

so far so good?

[covert agate]

are you stuck on a

so far

[dull lily]

yeah

[covert agate]

hmm

[dull lily]

rn im on a

[covert agate]

do you know how implicit differentiation works

[dull lily]

i had the idea to chain rule this

and find the inner

outter

and outtest

[covert agate]

$\frac{\dd}{\dd x} y^2 = 2yy'$

[ocean seal]

Chromium

[covert agate]

you understand this right?

[dull lily]

uhhh

the derivative of y^2 = 2y^1

where is the y' coming from again?

[covert agate]

you know what the quantity $\frac{dy}{dx}$ means right?

[ocean seal]

Chromium

[dull lily]

yeah

find the derivative of

right?

[covert agate]

hmm not really

[dull lily]

what does it mean .-.

[covert agate]

$dy, dx$ represent small changes in $x, y$

[ocean seal]

Chromium

[dull lily]

can u define

small changes

[covert agate]

do you know why the first principles is the way it is

[dull lily]

no

the rate of change one then no

i just know it is what it is I never knew the reasoning

[sudden hinge]

well do you know what a derivative represents?

[dull lily]

change of slope on anywhere in a graph

right