help-36

[vapid torrent]

How did my username change

[formal trail]

your name was changed by the moderation team, to find out why and request a new one please DM @dusk parcel

[final saddle]

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[devout maple]

Hello, I was wondering how I can do this question. Im confused because there are some boxes bigger then others and i wouldnt know what to put for some corners (we need to use pascals triangle)

[abstract bramble]

treat the missing path as 0

[muted prairie]

For each point on the grid, count the number of paths from the top left to that point

Starting at the top left and moving down

[devout maple]

so would i just add 0+ whatever the other number is?

[abstract bramble]

yes

convince yourself this is valid with the simplest case

a single square with an edge removed

the idea is that it's 0 because the paths that COULD use it arent there anymore

so it's 0 paths from that point

[devout maple]

okay but then what would i add for this do i just do 3+0?

[abstract bramble]

yes

[devout maple]

ohhh okay

[abstract bramble]

wait

[devout maple]

thank you so much

oh

[abstract bramble]

i meant that as in when you add the paths together the one coming from the missing path is 0

you should have a 1 above that 0

then 3 + 1 0 = 3

don't skip cells when theres a path missing

[devout maple]

so i still put the 1 in between the two 1s up here

[abstract bramble]

yes

[devout maple]

okay

but i just treat it as 0

[abstract bramble]

when you add it to the corner below yea

[devout maple]

okay

thank you very much

.close

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[tight zenith]

HELLO can someone pls help me w these two questions

[tight zenith]

idk how to do proof by induction with diffentation especially to an unknown order of a differential

the second question i only need to understand the first part then i’ll be fine with the rest

[trail mango]

tell whoever wrote ts to learn latex

[gray vortex]

proof by induction is a method to prove things

[tight zenith]

edexcel exam board

[gray vortex]

it works the same as any other you've done

[tight zenith]

i don’t know how to apply it, let me show you how far i’ve gotten

[trail mango]

pretty good handwriting

[gray vortex]

ok what do you need to do to get information on d^(k+1)y/dx^(k+1)

[winter lava]

hint: d^(k+1)y/dx^(k+1) is the derivative of d^k y/dx^k

[cedar kraken]

well firstly what i'd do is set (3^k+1)/2 and (3^k-1)/2 to something like A and B, so it's easier to understand.

then prove the case for n=k+1, which is just an application of the product rule

[tight zenith]

yes i got it thankyou :)

it’s coming together

[tight zenith]

i’m too autistic for that

how about Q 11 part a

[cedar kraken]

but u seemed to have figured it out 👍

[cedar kraken]

split the odd and even terms?

[tight zenith]

i’ll attempt

[cedar kraken]

like for 2r, it's be even, and (-1)^even is 1, and same logic for odd

[tight zenith]

btw

this is the uk school system

what level of maths would you say this is?

[winter lava]

first year of university maybe?

[final saddle]

@tight zenith Has your question been resolved?

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[zenith furnace]

Would like help verifying if my proof is logically sound and rigorous.

Proof:
I define f(x) = cos(x) - x^3
Cos(x)andx^3 are continuous for all real x, thus continuity is satisfied
I choose interval [0, pi/2]

f(0) = 1 - 0, 1> 0
f(pi/2) = 0 - (pi/2)^3 = -pi^3/8, - pi^3/8 < 0

We get f(pi/2) < 0 < f(0), so L = 0 satisfies f(pi/2) < L < f(0)

By IVT, There exists some c in (0, pi/2) s.t. f(c) = 0.

cos(c) - c^3 = 0
let x = c
cos(x) - x^3 = 0
or cos(x) = x^3.
Thus there exists an x in Real Nums s.t. cos(x) = x^3

[rocky tusk]

looks good

[zenith furnace]

thanks again!

.close

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[vapid haven]

can i sub a.b =2 in that 2axb thing
then (a.b.a) x b
(|a|^2)(bxb) ------> 0.
c vector = -3b na
b.c = -3b^2 (cosalpha)

[vapid haven]

dk if im right tho

[shadow marlin]

No because it's wouldn't be a.b.a (which is not even defined) but (a.b)a (which is a vector.

Moreover, clearly 2axb is not always 0 so this couldn't have been true for all a and b

[vapid haven]

oh

ok ig got it

but how do i procceed now then

[shadow marlin]

Since we want to calculate sin alpha it makes sense to do bxc

The sin being squared hints at the length of bxc maybe

[final saddle]

@vapid haven Has your question been resolved?

[vapid haven]

hmm

[final saddle]

@vapid haven Has your question been resolved?

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[random vale]

I’m having trouble with this integral. Is it cyclic?

[barren hound]

yes

[night raft]

yeah it is, so a trick you can do is set I equal to the orginal integral

[barren hound]

(also be careful how you write things this looks like subtraction but you meant multiplication

)

[random vale]

Interesting 🤔. Is this one of those that gets manipulated algebraically?

[random vale]

Is this the correct integral that needs to be set equal to original?

[barren hound]

if youve seen any cyclic integrals they will have been exactly like this one tbh

[random vale]

This is my first one on the homework. We did it in class briefly, but I didn’t understand it 100%.

[barren hound]

i see

yes, it is

[random vale]

Should I factor out a negative before?

Ty for the help btw

[barren hound]

oh youre asking whether your work so far is correct, i misinterpreted

[random vale]

Yes

[barren hound]

hmm

why did we swap what parts u and v were taking

originally you had u = the exponential thing and v = the sine thing

then you swapped?

[random vale]

Oh I think that was an accident. I see. Second u is supposed to be the trig?

[barren hound]

well like

i think you dropped a factor of 2

maybe -2

and your result will be that the integral is equal to itself

[barren hound]

ah thats where the negative sign went

also if you want to make this mean multiplication you have to put parentheses around the second thing or else it means subtraction

[trail mango]

good handwriting

[random vale]

Thanks, if only I could write the right stuff 💀

I’m going to retry. Brb.

Do I need to set this equal to orig?

I lost the 2* im adding it

[barren hound]

-# this still means subtraction... youd have to parenthesise the second thing, that is: cos(2t) (-e^-t)

[barren hound]

yes, that seems like a useful thing to do and its always allowed

[hushed bluff]

Are u finished?

[random vale]

Still working. Close.

Does this look okay?

[hushed bluff]

Correct

[random vale]

Thanks guys! Have a good one.

.close

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[twin pivot]

$\lim n\to +\infty \frac{\sin{(\pi xn)}}{\pi x}$

[soft zealot]

Goofy Joe

[twin pivot]

How is this limit calculated?

[loud phoenix]

!show

[final saddle]

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

[twin pivot]

Sin oscillates therefore should not exist

This Is my work

,w $\lim n\to +\infty \frac{\sin{(\pi xn)}}{\pi x}$

[soft zealot]

Failed to get a response from Wolfram Alpha.
If the problem persists, please contact support.

Feb 21, 2026, 11:59 AM

[loud phoenix]

That won't work here

[tranquil pine]

Try thinking sin(theta)/theta= 1

[loud phoenix]

$$-\frac{1}{\pi x} \le \frac{\sin(n \pi x)}{\pi x} \le \frac{1}{\pi x}$$

[soft zealot]

Ajay

[loud phoenix]

Because the lower and upper bounds are different numbers, the theorem cannot conclude that the limit exists

[twin pivot]

How?

[loud phoenix]

Squeeze theorem only works when lower and upper limits are the same

[twin pivot]

Yup

[loud phoenix]

It does not work in this case

[onyx peak]

Is it n -> infty or x -> infty?

[twin pivot]

n

[inland kettle]

yep

[onyx peak]

it depends on what x is, it does exist for some x

[loud phoenix]

Are you required to prove it does not exist formally?

[twin pivot]

I have to see it in the sense of distributions

[loud phoenix]

Could you elaborate?

[twin pivot]

Why?

[scenic pendant]

something similar happens

[loud phoenix]

Um, I think that's a tad advanced

more so than what the OP is asking for

[twin pivot]

What do you mean advanced?

[scenic pendant]

why?

[scenic pendant]

hes asking this

[loud phoenix]

To prove it wouldn't you need the inverse fourier transform?

[scenic pendant]

note that

$\int^{\infty}_{-\infty} \text{sinc}(xn) \mathrm dx = 1$

[soft zealot]

alee

[scenic pendant]

and $\int^{\infty}_{-\infty} \delta(x) dx = 1$

[soft zealot]

alee

[scenic pendant]

and when n -> inf

[scenic pendant]

this happens

[loud phoenix]

oh

[scenic pendant]

So “narrow × tall, constant area” => delta behavior

[loud phoenix]

I was definately overthinking this then

Oops

[scenic pendant]

wait

you can use fourier to say this

wait i think i made a mistake

[twin pivot]

Where did you get this from?

[scenic pendant]

here i mean n sinc(xn) dx

not only sinc(xn)!

[loud phoenix]

Let ${f_n}$ be a sequence of tempered distributions. Convergence to a limit $L$ is defined via the duality bracket with a test function $\phi \in \mathcal{S}(\mathbb{R})$:$$\lim_{n \to \infty} f_n = L \iff \lim_{n \to \infty} \langle f_n, \phi \rangle = \langle L, \phi \rangle, \quad \forall \phi \in \mathcal{S}(\mathbb{R})$$

[soft zealot]

Ajay

[scenic pendant]

[
X(f) = T,\mathrm{sinc}(fT) = \mathcal{F}\left{\Pi!\left(\frac{t}{T}\right)\right}
]

[soft zealot]

alee

[scenic pendant]

consider this

[loud phoenix]

Let $\text{rect}_n(\xi)$ be the indicator function on the interval $[-\frac{n}{2}, \frac{n}{2}]$. Its inverse Fourier transform is:$$\mathcal{F}^{-1}{ \text{rect}n(\xi) } = \int{-n/2}^{n/2} e^{i 2\pi x \xi} d\xi = \frac{\sin(n\pi x)}{\pi x}$$

[soft zealot]

Ajay

[loud phoenix]

So, $$\lim_{n \to \infty} \text{rect}_n(\xi) = 1(\xi)$$

[soft zealot]

Ajay

[loud phoenix]

and since $\mathcal{F}^{-1}$ is a continuous linear automorphism on $\mathcal{S}'(\mathbb{R})$

[soft zealot]

Ajay

[scenic pendant]

are you proving that $\mathcal{F}\left{1\right} = \delta(f)$ ?

[soft zealot]

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

[loud phoenix]

$$\lim_{n \to \infty} \frac{\sin(n\pi x)}{\pi x} = \mathcal{F}^{-1} { 1 }$$

[soft zealot]

Ajay

[loud phoenix]

Yes

by definition $\mathcal{F}{\delta} = 1$, which implies $\mathcal{F}^{-1}{1} = \delta$

[soft zealot]

Ajay

[loud phoenix]

So the limit equals $\delta(x)$

[soft zealot]

Ajay

[scenic pendant]

i have a question

[loud phoenix]

Ok

[scenic pendant]

since $1$ is not in $L^1$

[soft zealot]

alee

[loud phoenix]

Yes?

huh?

[scenic pendant]

i dont understand why we can find the FT by doing the limit

since 1 is not in L1

[loud phoenix]

I know that

[loud phoenix]

Yes, the standard integral formula for the Fourier transform fails

[scenic pendant]

yes

[loud phoenix]

That's why I umoved from functions to tempered distributions

[scenic pendant]

i mean why we can do lim T->inf rect(t/T) = 1 and lim T->inf sinc(Tf) = delta(f) and this implies

F{1} = delta(f)

i dont understand why we can do this

that is, is it a Fourier transform in the limit?

[loud phoenix]

Because the FT is a continuous operator on the space of tempered distributions

[scenic pendant]

and why is it "legal" to find a FT this way

[scenic pendant]

can i do another example ?

or i have to open new help ?

[loud phoenix]

You can open a new help channel

[scenic pendant]

i did

[final saddle]

@twin pivot Has your question been resolved?

[twin pivot]

Thanks

.close

[final saddle]

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[copper roost]

hello ive got this function

[copper roost]

so basically

im checking out the symetry

and an e function cant be symetric right

[strange sparrow]

there are many kinds of symmetry, what symmetry are you referring to here?

[copper roost]

to the y axis or the other one

i mean like when a normal function has x^ even numbers

then its symetric to the y axis yk

but how does this apply to here

[final saddle]

@copper roost Has your question been resolved?

[copper roost]

<@&286206848099549185>

[strange sparrow]

the y axis is not an axis of symmetry for f_a, for any a ≠ 0

in order for the y axis to be a line of symmetry the function must be even, ie. satisfy f(x) = f(-x)

but it is easily seen that f(x) and f(y) have opposite signs whenever x and y have opposite signs

[copper roost]

ohh yhh

[copper roost]

but generally an e function isnt symetrica right?

[strange sparrow]

what do you mean by "an e function"

an exponential function?

[copper roost]

yeah

[strange sparrow]

a^x is not an even function except when a = 0 and a = 1

[copper roost]

btw how do i solve for x = 0

hate this shit

whenever a parameter is involved i dont know what to do

[spring crow]

Wdym

[copper roost]

the zeropoints

[spring crow]

Zeros of the function?

[copper roost]

yeah

so we know e ^-ax cant have any right

so we do like 2 parts

and then a^2 * x

idk how tod o that

[spring crow]

X =0

[copper roost]

yeah but

how i do that

[spring crow]

A squared is multiplied by X

[copper roost]

yes

so we divide by a square

so x = a ^2

so can i just write for x1(a^2/y) ?

[spring crow]

If you divide by a square it is gonna be 0 I think don't get what your question is

a^2 is a constant so it is just gonna be x=0

[copper roost]

omg 🤦‍♂️

right

ty

i forgot that its connected to x

[copper roost]

so its x(0/0) right

the ex^a part doesnt have any x = 0 right

[spring crow]

Yeah it can't be equal to zero

[final saddle]

@copper roost Has your question been resolved?

[copper roost]

.close

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[keen hare]

is my substitution method insane or did i do this right

[vale mist]

right

[keen hare]

dope

.close

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[sinful tapir]

Where would i go to send in a potential discovery i made

[sinful tapir]

I think ive found something

[mint orbit]

probably a university

[shell condor]

in #math-discussion maybe, there's not a place specificably I think

[trail mango]

[thorny canyon]

is this abt the super super root thing

[trail mango]

nowhere

[lilac bison]

you should tell me it

i will tell the relevant people

[mint orbit]

be nice

maybe they are more inspired than you

[sinful tapir]

Se where we go from here chat

Enemy combatant

Enemy combatanant chat

I might make a video start email youtube people

[lilac bison]

what

[sinful tapir]

From mind your decissions or something

[trail mango]

have you read their messages over the past day

[trail mango]

you can put it on vixra

[sinful tapir]

Wdym by this

[lilac bison]

wait im lost, was this the guy who sent the videos in the help channel yesterday?

[sinful tapir]

My messedges have been very thought provoking thak you

[shell condor]

what topic is your potential discovery in?

[sinful tapir]

Ill take my thoughts elsewhwere ig im not wanted here

Turns head away

[sonic kelp]

I wanted to ask that tii

[sinful tapir]

Dramatically

Ha

[sonic kelp]

yo stay

[sinful tapir]

I cant beleive this

[shell condor]

god

[sinful tapir]

Good heavens

[lilac bison]

if i had gif perms, i would send dies of cringe

[sinful tapir]

Lmao

[thorny canyon]

she claims she found a generalized definition of tetration/pentation to fractional powers

[sinful tapir]

So vixra

Im just joking guys

We allowed to joke a little

[lilac bison]

YEAH, in the discussions channel

[sinful tapir]

I know i was getting butthurt a little here and there

But you guys are okay lmao

[sonic kelp]

generalised definition?

are you sure it doesnt exist already?

[thorny canyon]

it doesn't

[sinful tapir]

I found it for roots and logs too

[shell condor]

Well, if you've got to ask a mathematical problem you're stuck in, then you can ask, otherwise I'd say move to a discussion channel

[sinful tapir]

For the inverse ops past tetration

[sonic kelp]

as in?

[sinful tapir]

Vixra sounds cool

Ill prolly try there

[lilac bison]

chat if this is legit, we will be noted down in history as the opposers..

[sinful tapir]

And ill prolly try making a youtube vid and email to people

[sonic kelp]

im not opposing broo

im just curious

[sinful tapir]

Like mid your decissions

[lilac bison]

well i am.

[sonic kelp]

i wanna know what they did

[sinful tapir]

Or numberphile

Or maybe veritassium

[sonic kelp]

do it

[thorny canyon]

😔

[sonic kelp]

i wouldnt go there tho

[sinful tapir]

Dm

Clos

.close

[final saddle]

Channel closed

Closed by @sinful tapir

Use .reopen if this was a mistake.

[sinful tapir]

I can tell you in dms

[coarse horizon]

hi?

wow I just came at the time where the chat died

interesting reason to leave;-;

[final saddle]

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Remember:
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[jagged root]

!help

[final saddle]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[lilac bison]

whats your question mate

[jagged root]

I need some help with definitons

of domain and range

conceptual*

[lilac bison]

okay

what do you not understand

[jagged root]

the restrictions

when there is like a square root for example in an equation

Im struggling w the restrictions

should i give u an example?

[lilac bison]

so when dealing with the set of real numbers, we know that complex numbers arent defined in the set of real numbers

[jagged root]

oh they arent?

[lilac bison]

so we avoid having a negative number inside a suqare root

[lilac bison]

they are indeed not

[jagged root]

that would be the restriction?

for domain or range tho

[lilac bison]

precisely

[jagged root]

thats where i get confused

[lilac bison]

think of it like this

you have a function call it a machine

[jagged root]

input output

right

[royal gust]

You just need to memorize a few rules. For example, you can't have a negative under a square root

[lilac bison]

you give the machine some input, thats your domain, the machine does its magic, and gives an output

thats your range

[jagged root]

oh

[jagged root]

yea

ok

i think i get it now

ty

!close

[lilac bison]

its with a dot

not an exclamation mark

[jagged root]

.close

[final saddle]

Channel closed

Closed by @jagged root

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[jagged root]

ty

[final saddle]

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[vagrant knot]

Hi

[vagrant knot]

Can u tell me what I did wrong here

[shell condor]

must have been the wind

[vagrant knot]

.

Bruh

[shell condor]

looks fine

what's the answer?

[vagrant knot]

Do I just flip

[nocturne cape]

putting minus exponents is correct but why

[vagrant knot]

Because I have to follow the rules

[shell condor]

true, you're making things complicated for yourself

[nocturne cape]

u can also simplify the two minuses

[royal gust]

What you've got on the page is correct.

Are you asked to do something with this?

[vagrant knot]

Simplify

Is there an easier way to do that

The negative is kinda annnoying

[nocturne cape]

yeah
the numerator and the denominator both being negative can be simplified, so you get a positive fraction

as for the minus exponents, a minus just means u can replace it with the inverse
so x^-1 becomes 1/x since 1/x is the inverse of x

y^-4 can be written as 1/y^4, since 1/y^4 is the inverse of y^4
does that make sense?

(and vice versa
1/x^-1 is x since the inverse of 1/x is x)

so u can use this rule to simplify and remove the minuses in the exponents

what you did isn't "wrong", it's correct.
but usually when u need to simplify you need to exhaust everything and keep going until you can't anymore

idk if you have to get the ^2 inside tho, seems cluttery

[shell condor]

basically $x^n = \frac{1}{x^{-n}}$

[soft zealot]

doctorstrangejr

[shell condor]

(for reference)

[nocturne cape]

i forgot this bot existed

[vagrant knot]

Yeah I know what rule, but cant I just flip them to the bottom to make them positive?

[nocturne cape]

yeah u can
only the things with a minus exponent yes
so the 3 for example stays up

[nocturne cape]

yeah

now get the 2 that's outside inside

and u got a simplified fraction gg

[vagrant knot]

Thank you so much

It’s very helpful

[nocturne cape]

what was the thing that troubled you?

where did you get stuck exactly?

[vagrant knot]

The negative, it bothers me, I didn’t simplify them

I might need your help again on the challenging ones haha

[nocturne cape]

it's a good idea to deal with them as early as possible sometimes, so they don't drag on and make stuff more complicated or unreadable
having to keep track of them can make you cause more errors

[final saddle]

@vagrant knot Has your question been resolved?

[vagrant knot]

Hi rat, 54 is kinda weird, can u explain the 3rd step to me

There is a negative and I have to subtract

[nocturne cape]

yeah so its simplified already

[vagrant knot]

Yeah I searched it up

[nocturne cape]

you can't remove the negative anymore

[vagrant knot]

But I didn’t really get it

[nocturne cape]

ah okay

so basically you have 3 minuses in the fraction, 2 of them nullify each other leaving you with only one
the fraction is negative, you can't remove the minus anymore
so sign wise, it's simplified

[vagrant knot]

Oh sorry I didn’t say the question I need help with

It’s 54

[nocturne cape]

yeah

isn't 54 the fraction?

[vagrant knot]

Yeah

[vagrant knot]

Oh i see, so I just do y^1 - y^8, which is y^-7?

[nocturne cape]

ye
$\frac{-y}{-8y^4.-5xy^4}$
the denomintaor simplifies itself very easily to 40xy^8
is that okay so far?

[soft zealot]

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

[nocturne cape]

yeah

[vagrant knot]

Yep

[nocturne cape]

what u do is normal subtraction
you take the bigger of the exponents, and subtract from it the smaller one

so if u had y^4/y^3
you subtract 4 from 3,
leaving you with y

if it's y^3/y^4, ud get 1/y

[vagrant knot]

Got it, thanks again

[final saddle]

@vagrant knot Has your question been resolved?

[final saddle]

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[earnest shuttle]

is this right?

[old quarry]

No

You didn't use chain rule correctly

[earnest shuttle]

i just used the sinx = cosx rule and cosx = -sinx

[old quarry]

y=f(g(x))
dy/dx= f'(g(x)) INTO g'(x)

[earnest shuttle]

oh shi

is this right for the first part?

[old quarry]

Yes

[earnest shuttle]

wait is it supposed to be let u = sintheta or u = theta?

[old quarry]

sin theta

[earnest shuttle]

cuz i remember there was a question like tan theta cubed and they let u = theta cubed

[old quarry]

u=theta would make no difference

[earnest shuttle]

true that

it was like tan(x^3)

[old quarry]

Yeah to make it easier u can use x^3=u

[earnest shuttle]

is this right yoda?

[old quarry]

Yeah

[earnest shuttle]

alr thank you master yoda

i deeply appreciate your mentoring

its an honour to be your student

.close

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[copper roost]

hello so

i dont understand what they mean by multiplicity of zeros

[abstract bramble]

multiplicity of the roots

[copper roost]

uhh

[copper roost]

wait so wdym

[abstract bramble]

the roots of x^2 are 0 and 0 so we say 0 is a root of multiplicity 2

[copper roost]

so

i calculated 3rd root of 2a

then i got a multiplicity of 3

?

[abstract bramble]

no multiplicity is simply how many duplicates of a root there are

[copper roost]

so

there are none

here

[abstract bramble]

this is a quartic so it should have 4 "roots"

[copper roost]

ohhh

so

now i need to do if a is bigger than 0 and stuff

[opal plinth]

If that's the actual exercise then you didn't copy f_a(x) correctly

[copper roost]

i didnt

oh my god

thanks

[final saddle]

@copper roost Has your question been resolved?

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[worn needle]

I'm stuck on how I got this answer wrong, as it stated the answer is 919.9 . I thought I was doing the steps correctly, and I'm not sure what went wrong when I was trying to solve it The answer that I got was 785.52, and when rounded to the nearest 10th it was 785.5

[robust mulch]

why is r equal to 3

[worn needle]

Because of the semisphere

[robust mulch]

what is the diameter of the hemisphere

[worn needle]

I'm not sure the question included the diameter, as for the whole question is in the photo on the computer

Like the whole #9 is on the computer, that's the whole problem it gave me to solve

[robust mulch]

the hemisphere goes across the top of the box

so you should know the distance across the top

[worn needle]

The question did not give that to me. Can you please help me see what I did wrong?

Because I showed how I did it on paper, and the computer said it was wrong

[barren hound]

you got r = 3 somehow which doesn't make sense; the diameter of the hemisphere is 9 yd

[worn needle]

Can you show me a visual..?

[worn needle]

Yeah, the circle. Half of it?

And the 9yd

[barren hound]

the red line is 9 yd

[worn needle]

So the hemisphere is also 9 yd?

[barren hound]

the diameter of the hemisphere is 9 yd

because its flush with the edges

[worn needle]

Yeah, so the hemisphere is also 9yd. But how do I get 919.9 yd^3 as my answer

[barren hound]

the diameter of the hemisphere is 9 yd

so the radius of the hemisphere is 9/2 yd

or 4.5 yd

[worn needle]

So I would plug that into the calculator

Because I'm still confused on how you're explaining this to me

Sorry, I just need detailed explaining..

[lone talon]

What’s the formula for your model

Just the raw formula

No numbers

it’s the formula of the cube + formula of the hemisphere

[worn needle]

Everything I was given was basically on the computer and I had to solve it out. I'm just still confused how it's because explained and how to actually plug it in correctly 💔

[lone talon]

$x^3 + \frac{2}{3} \pi r^3$

[soft zealot]

zain

[lone talon]

what’s r and x

once you have those you just plug it in

should get the answer

[robust mulch]

the main issue is we are trying to show how to find r

[lone talon]

Ah

[worn needle]

Oh.. so it's 9/2 but why is it that? Since I did 1/2•4/3

Because the sphere volume formula was 4/3 πr^3

[robust mulch]

why did you pick r as 3

[worn needle]

Oh, I see what I did wrong

Thx for making me see my mistake, I was kinda slow to notice

.close

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[round cedar]

i got -3 but calculator says -2

[round cedar]

any small hint will be cool

(whenever i got ״0/0״ i did lhopital rule)

[hushed bluff]

But x -> 1 so x - 1 -> 0 and ln(x) = ln(x - 1 + 1) ~ x - 1

[radiant igloo]

completely forgor about f(x)^g(x)= e^(g(x)(f(x)-1))

[radiant igloo]

all correct

[final saddle]

@round cedar Has your question been resolved?

[onyx peak]

ln(x) = x-1 wasnt enough here

[hushed bluff]

No its not

[round cedar]

wym

[hushed bluff]

No

I reply that guy

[round cedar]

howcome you got -2

[round cedar]

y tho

they both tend to 0 when x--->1

[onyx peak]

-2 is the correct result, your result was incorrect because the ln(x) = x-1 approximation wasnt sufficient here

[onyx peak]

oh, then why not approximate it by 0?

it also -> 0

[radiant igloo]

so i was correctly incorrect

[onyx peak]

you'd need to use the 2nd order approximation here i believe

[round cedar]

plz explain further?

when can i use lnx~ x-1 , and when i cannot do so?

and what other alternatives i have..

[hushed bluff]

e^xlnx ≠ lnx

[onyx peak]

well, you can lhop it

[hushed bluff]

I think so

I mean

[round cedar]

oh you mean before using tricks

[onyx peak]

ideally, you should always write lnx ~ x-1+O(x^2)

[onyx peak]

you could also notice that the denominator is 2nd degree, because lnx - (x - 1) ~ x - 1 + O(x^2) - (x-1) = O(x^2)

that means that you have to make your numerator expansion 2nd degree as well

[gloomy river]

You can get away with using 1st degree for the numerator if you do everything with Taylor expansions, interestingly

[onyx peak]

uh

how would u do the denominator

[gloomy river]

You have to do 2nd degree for the denominator

lnx = (x-1) - 1/2 (x-1)^2 + O((x-1)^3)

[onyx peak]

i dont think you can get away with that

[gloomy river]

Well, technically you are still taking the 2nd degree expansion of e^(xlnx)

But a 1st degree expansion of lnx suffices for that

[onyx peak]

$\frac{e^{x\left(x-1\right)}-x}{-1/2(x-1)^{2}}$

this?

[soft zealot]

MathIsAlwaysRight

[onyx peak]

that -> -3 already

[round cedar]

those degrees yall talking about is related to this "O(x^2)" thing?

[gloomy river]

um

[round cedar]

cuz i aint sure they taught us this

[onyx peak]

that O(x^2) means that there is a missing x^2 term and further in the approximation

[round cedar]

so the degree is 2? because of x^2

[onyx peak]

oh and i shouldve written O((x-1)^2)

[onyx peak]

yes

[onyx peak]

If your approximation worked, you'd expect those O terms to become insignificant in the limit (so e.g. if there was + O((x-1)^2)) in the term, it would be fine, since that -> 0). But if it was e.g. O((x-1)^2) / (x-1)^2, then it wouldnt tend to 0 any anymore, so you'd have to make your approximation better

[gloomy river]

Oh right nvm, I took Taylor of x^x around 0 instead of 1

[onyx peak]

that O((x-1)^2) thing just lets you work with the 2nd degree approximation, without bothering to compute the coefficient

and you hope that it will vanish later

which in this case it wouldnt

[round cedar]

sadly i still cant get -2

um

nvm

found the error lol

@gloomy river @onyx peak

i think thats cool, thx

.close

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[quartz vigil]

<@&286206848099549185>

[final saddle]

@quartz vigil Has your question been resolved?

[quartz vigil]

<@&286206848099549185>

[white venture]

!15min

[final saddle]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[white venture]

!show

[final saddle]

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

[quartz vigil]

I don’t get this at all

Like I m unable to understand

so when switch is closed capacitor acts like a short circuit

Letting current pass through it

infinitely?

I m so confused ugh

but then after a while it acts as an open circuit

so we ignore the capacitor wire

as if it doesn’t exist?

will u help

🙁

[white venture]

Sry, I don't know this stuff. I was asking for more info to make it easier for a different helper who does know it.

Nobody can help without first understanding where you're stuck after all.

[quartz vigil]

oh

[true void]

yes, initially - capacitor basically acts like a wire, and once fully charged - acts as an open circuit so u can ignore it.

[quartz vigil]

can u explain me how u would solve

the circuit

then

using this info

I m stuck at this part

@true void

Manish bhaiyya

[true void]

Initially, replace the capacitor by a wire

[quartz vigil]

Okay

[true void]

so u basically have R1,R3 in parallel and R2,R4 in parallel right

[quartz vigil]

HOW DO WE FOGURE THAT OUT

[echo crown]

Hmm

[quartz vigil]

like in this square

how do I know what’s parallel to what

[echo crown]

It’s easy after the capacitor is charged

Just disregard the capacitor and you get two parallel branches

[quartz vigil]

Manish bhayya

[echo crown]

I think nodal analysis can be used for the case when it is being charged but not sure

[quartz vigil]

@true void

pls bhaiyya help kardein

[true void]

idk hindi lol

So once you replace capacitor by wire, you can see R1,R3 ends have same potentials right

[echo crown]

Idk if you can take it as such in the first case

[true void]

why not

[echo crown]

Oh yeah when it has just started charging it acts as a straight wire

[true void]

Yes

@quartz vigil

are u following

[echo crown]

Alright then makes sense

It’s easy from there

[true void]

Yea but he seems to have a problem in finding that they are in parallel

[quartz vigil]

Hey

[quartz vigil]

Exactly

I don’t get that

how to find what’s in parallel

[true void]

resistors connected across same potential are in parallel, so here u have R1 and R3 connected across P and Q

and so are R2 and R4

across Q and the bottom point

[quartz vigil]

Basically we find the place where current is being split

?

Okay I get part a now

I got voltage across R1 and R3 as 4.94 and R2 and R4 as 7.06

@true void how about b

[true void]

For b

like i said you can just remove the capacitor

since it is an open circuit

so now, R1,R2 and R3,R4 are in series

[echo crown]

In general things are said to be parallel with both of their ends are joined which is true for these two wires when the Capacitor starts charging

[quartz vigil]

R1 is in series with R2

R3 with R4?

[true void]

yep

[quartz vigil]

Is my answers for a correct

[true void]

idk i didnt calculate

[quartz vigil]

can u pls

🥹

[true void]

yes they are 👍🏻

[quartz vigil]

Voltage across R1 is 4 V
Voltage across R2 is 8 V
Voltage across r3/r4 is 6 V

@true void b?

pls

[true void]

one sec

yep

[quartz vigil]

r u sure

[true void]

right

[quartz vigil]

Thanks bro

[true void]

Np

you can do .close

[quartz vigil]

@true void another question

[true void]

Ok post it

i mightve to leave in a few min tho

[quartz vigil]

I think I m messing up the right hand rule

and the internet has 1000 different versions

[true void]

You want to find the direction of force on the wire?

@quartz vigil

[quartz vigil]

Yeah

[true void]

you know the formula is I(l x B) right

l is in the same direction as current flow

So place your right hand along the direction of l and curl it towards B.

Your thumb will point towards the dirn of force

[quartz vigil]

My first finger

is current or magnetic field?

[true void]

All of your fingers along current direction, then curl your fingers towards B

right hand only

[quartz vigil]

Huh

Isn’t it cross product

[true void]

*make sure your thumb is perpendicular to other fingers.

[true void]

idk the rhtr version i learnt is this

yes it is cross product

[quartz vigil]

thumb is force

first finger and middle finger I confused them

[true void]

like you said there are multiple versions and the one im talking about does not invlove first finger and middle finger pointing towards different directions

[final saddle]

@quartz vigil Has your question been resolved?

[final saddle]

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[left trail]

can someone explain what they mean by (4+h) feet at its ends

[thorny canyon]

that's a frustum yeah

basically, the diameter of the top part is 4 feet

diameter of the bottom is (4+h) feet

cuz h is given to be positive

[left trail]

ok. I see. also what does he mean by "if the axis coincides with that of the log"

[thorny canyon]

the frustum is the log

and you're trying to cut a cylinder out of it

[thorny canyon]

this means that the axis of the cylinder is the same as the axis of the log

[left trail]

so if the blue are the axis. then something like this?

[thorny canyon]

the vertical blue is the axis

for a cone we usually don't take horizontal stuff as axii

[left trail]

ok so all there saying is that they are on the same axis meaning the cone is crooked or anyting like that

[thorny canyon]

means the point from where u measure the radius of the cylinder is the same as the point from where you measure the radius of the frustum

[left trail]

ok. thx

.solved

[final saddle]

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[kindred bloom]

I dont understand demorgans law at all how do I do this

[opal plinth]

You don't need to understand it to apply it (though you should try to understand it anyway)

[kindred bloom]

i know but what i mean is the 1st one is P^Q why isnt it P ^ ~Q

i also need to understand it because I have a no note exam on thursday

[opal plinth]

You have the answer key?

[kindred bloom]

Nope i used ai

[opal plinth]

I'm so tired of people saying that I don't even know how to respond

[kindred bloom]

I dont have any other options we didnt go over a single example in class

litterally the only thing i can do

its tragic

[opal plinth]

You have search engines that can give you hundreds of pages explaining these things

You have Wikipedia

You also have this server, of course

In no circumstance would I use AI to explain or check maths if I were a student

[kindred bloom]

If i had an answer key id use that its right most of the time and I dont have the time to research anything

[opal plinth]

,, A \to B \equiv \neg A \lor B

[soft zealot]

Nel

[opal plinth]

This is not De Morgan's law, it's just an equivalence that you should remember

[kindred bloom]

Okay

[opal plinth]

Now for (a), you have P -> not Q

What does that give you if you use this equivalence?

[kindred bloom]

not P / Q

/

\ /

[opal plinth]

Don't type \ on Discord unless you really want to (you can type \\ to show \)

Use words

So that's not P or Q, right?

[kindred bloom]

yes

[opal plinth]

Ok, and so then you have the negation of that

That's where you use De Morgan's

[kindred bloom]

Alright?

[opal plinth]

Basically you distribute the negation and switch between "or" and "and"

[kindred bloom]

so it would be P and not Q?

[opal plinth]

Yes, if you didn't make a mistake in the previous step

,, A \to \neg B \equiv \neg A \lor \neg B

[soft zealot]

Nel

[opal plinth]

This is exactly the same equivalence as before

Just the B from the first one became a "not B"

[kindred bloom]

Okay?

[opal plinth]

Well that's exactly the same form as P -> not Q

[kindred bloom]

So is this another demorgans law?

[opal plinth]

No

$A \to B \equiv \neg A \lor B$ is like the definition of implication

[soft zealot]

Nel

[opal plinth]

In your case, $A = P$, and $B = \neg Q$

[soft zealot]

Nel

[opal plinth]

So it becomes $P \to \neg Q \equiv \neg P \lor \neg Q$

[soft zealot]

Nel

[opal plinth]

Does that not make sense to you?

[kindred bloom]

No it doesnt make sense sorry

[opal plinth]

Can you point to what exactly doesn't make sense? I can't really think of a more basic explanation

Is it the variable substitution?

[kindred bloom]

Okay I dont really know what your tryign to describe like from my pov youre telling me the equivalences and I just dont know what im getting out of that liek why is that important, I just watched a video on demoragns law it only takes place when a not is behind parentheses with an and or an or symbol so i understand that now but I dont know how to convert the regular ones

[opal plinth]

What regular ones

[kindred bloom]

like in the case of this problem where there isnt an and or an or we are translating its a if then

So translating an if then its called an implication?

[opal plinth]

Can you tell me what the -> symbol means for you? How is it defined?

[kindred bloom]

if then what do you mean by defined?

P->Q

[opal plinth]

Have you learned truth tables?

[kindred bloom]

its conditional and only false if the p is true and the q is false

[kindred bloom]

yeah

[opal plinth]

Alright, so P -> Q is an expression that evaluates to true or false, specifically it's always true unless P is true and Q is false, which is represented by the truth table of ->

[kindred bloom]

yeah

[opal plinth]

That -> symbol represents an operation (or operator), just like + represents the addition operation

That operation is called a material implication, or a material conditional

You said conditional just now, so I'm guessing that's how you learned it

[kindred bloom]

Ive only heard it as conditional yeah

[opal plinth]

Implication means the same thing

[kindred bloom]

okay

[opal plinth]

As you can see, in LaTeX there's this symbol called "implies":

,, \implies

[soft zealot]

Nel

[kindred bloom]

yeah

[opal plinth]

It has two lines but it's the same as -> in this context

"implies" comes from "implication"

So anyway, when you have an implication, you can rewrite it using only the operations "or", "and", and "not"

[opal plinth]

This way

[kindred bloom]

so it still means the same thing but you can use or and and not to change it?

[opal plinth]

Both expressions are equivalent, they have the same truth table

[kindred bloom]

Okay

[opal plinth]

Btw I hope you understood that this symbol means "is equivalent to":

,, \equiv

[soft zealot]

Nel

[kindred bloom]

yeah ik that luckily lol

[opal plinth]

Alright

So the exercise is to rewrite the statements in such a way that the negation symbol only appears before variables and not before subexpressions in parentheses

Well in general it's going to be a lot easier to get rid of any implication symbol before doing anything else

[kindred bloom]

wait okay i think im understanding

So not (P and Q) is wait so the not is making the statement false correct?

[opal plinth]

No, whether a statement is true or false in general depends on the truth values of the variables

[kindred bloom]

oh

[opal plinth]

What the not operator does is invert the truth value

So "not true" is "false" and "not false" is "true"

[kindred bloom]

okay

[opal plinth]

"not" is one of the only two boolean operators with one argument (the other being the identity)

Pretty much all the other operators have two arguments (one of the left, one on the right)

You can think of them as functions if that's easier for you

[kindred bloom]

okay

[opal plinth]

Like the function "and" is defined by:

• and(true, true) = true
• otherwise, and(x, y) = false

But we usually write it in between the arguments instead

[kindred bloom]

yeah

[opal plinth]

So...

,, A \to B \equiv \neg A \lor B

[soft zealot]

Nel

[opal plinth]

For (a), since you have $\neg(P \to \neg Q)$, substitute $P$ for $A$ and $\neg Q$ for $B$ in this equivalence

[soft zealot]

Nel

[opal plinth]

That gives you $\neg(\neg P \lor \neg Q)$

[soft zealot]

Nel

[opal plinth]

Then apply De Morgan's: $\neg(A \lor B) \equiv \neg A \land \neg B$

[soft zealot]

Nel

[opal plinth]

Again you need to substitute the variables correctly

[kindred bloom]

So you Have to turn the statement from -> to and or or before you even start it?

[opal plinth]

In this case yes, in general it's just easier to handle expressions that only contain "and", "or", and "not"

[kindred bloom]

I guess im having trouble translating it at this step then

[opal plinth]

In this case you're forced to if you want to apply De Morgan's afterward, which you need

[opal plinth]

Which step? Implication or De Morgan's?

[kindred bloom]

Implication its the one turning it from -> to and or or

[opal plinth]

Right well that's something you need to memorize

[kindred bloom]

Okay i have it written down for now

[opal plinth]

(or understand why it's true so you can derive it again)

Like you said yourself A -> B is only false if A is true and B is false, right?

[kindred bloom]

yes

[opal plinth]

So it's always true when A is false, and also always true when B is true

[kindred bloom]

mhm

[opal plinth]

,, A \to B \equiv \neg(A \land \neg B) \equiv \neg A \lor B

[soft zealot]

Nel

[opal plinth]

You can memorize it like this too

The first step is "A -> B is false when A is true and B is false"

The second step is De Morgan's

[kindred bloom]

okay so this is the rule I need to follow?

[opal plinth]

To simplify an expression that contains an implication, yes, it's almost always a good idea to do that

[kindred bloom]

okay so with mine not(P ->Q) since its different then the rule one how would i start to translate it

[opal plinth]

You mean not(P -> not Q) ?

[kindred bloom]

Oh yeah that my bad

[opal plinth]

If I give you f(x, y) = 2x + 3y, can you give me the expression for f(2x, -y) ?

[kindred bloom]

no..

[opal plinth]

Well, you may want to review highschool algebra / functions

You substitute a variable for another

In the case of not(P -> not Q), at first we are only interested in the inner expression P -> not Q

The "rule" is "A -> B" is equivalent to "not A or B"

"P" matches with "A", and "not Q" matches with "B"

So the rule becomes "P -> Q" is equivalent to "not P or not Q"

[kindred bloom]

okay

[opal plinth]

It's literally just replacing all instances of "A" with "P" and all instance of "B" with "not Q"

[kindred bloom]

okay

[opal plinth]

A and B are variables, but they can represent whole expressions

In this case the only expression that's not directly a variable is "not Q" but in the next question, (b), you will have to deal with two bigger expressions

Just think of A and B as boxes if you will, then you put whatever is on the left of the implication in box A and whatever is on the right in box B

Then you write the equivalence, and finally you simplify

[kindred bloom]

Is there a part I ignore first? like do i just take the inside convert that then apply the not?

i understand the conversions but my example is different then the rule you showed me thats where im stuck theres an abnormality that i have no idea how to solve

[opal plinth]

Not sure what you mean

You can apply rules in any order

[kindred bloom]

okay so as a real life example it feels like your teaching me to swing a baseball bat when in my problem i need to swing a hammer its both swinging but the process is completly different

ive never solved something like this i need examples or steps to take in order to know what to do

[opal plinth]

I mean... I'm not going to give you a full course on this topic

[kindred bloom]

yes i know its just I dont know what to do next to me this is doing something ive never done before

[opal plinth]

You have at least three more questions to do from what you posted

Just try to work them out, and if you're stuck, open a channel explaining exactly where you're stuck

I have to go, but I'm sure other helpers will step in if you close this channel and open a new one

[kindred bloom]

okay thank you for helping

.close

[final saddle]

Channel closed

Closed by @kindred bloom

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

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Remember:
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[mighty grotto]

can someone help me with this pls i dont think my teacher taught how to solve this mathematically

[mighty grotto]

i went on desmos and stuff to plot it out and like do it but idk how i do now

<@&286206848099549185>

[tender pollen]

what properties does a perpendicular bisector have?

[mighty grotto]

it forms a right angle i think

[tender pollen]

and the other?

[mighty grotto]

theyre equal

equal distance

[tender pollen]

yes, so how could you represent that graphically?

[mighty grotto]

i could draw a line down the middle

from 4,6 to 2,2

[tender pollen]

and what other line could you draw

[mighty grotto]

from -2,4 to 6,0

[tender pollen]

and what must be true for them to be perpendicular?

[mighty grotto]

that it forms a right angle

[tender pollen]

and how could they be perpendicular?

(hint: ||slopes||)

[mighty grotto]

im lost now

[tender pollen]

what's confusing?

[mighty grotto]

only slope i remember is y = mx+b or something

[tender pollen]

ok, if two lines are perpendicular, what can you say about their slopes?

[mighty grotto]

that their slopes are equal?

[tender pollen]

no

something else...

should i tell you?

[mighty grotto]

yes please i didnt learn this

[tender pollen]

ok

basically

if two lines are perpendicular...

||then their slopes, a and b, satisfy ab = -1||

so what can you do?

[mighty grotto]

multiply?

[tender pollen]

well actually wait

did you prove JOE is isosceles?

[mighty grotto]

i kinda just assumed it was so no

[tender pollen]

ok, how could you prove it's isosceles?

(hint: ||distance formula||)

[mighty grotto]

no idea what that is 😔

but it looks equal

the two sides

[tender pollen]

wait, you don't know the distance formula?

[mighty grotto]

no

[tender pollen]

what do you know?

[mighty grotto]

algebra 2 and sin cos tan

and the sin theorem

last time we were in class (friday before this one) we learned about 3d shapes

[tender pollen]

ok, i feel like you should know the distance formula...

[mighty grotto]

yeah we didnt learn that at all

[tender pollen]

anyways, the distance formula says:

the distance between $(a,b)$ and $(c,d)$ is $\sqrt{(a-b)^2+(c-d)^2}$

[soft zealot]

nadat12

[tender pollen]

that's the euclidean distance

(proof: it's a corollary of the pythagorean theorem)

[mighty grotto]

the only word i understood was the pythagorean theorem

[tender pollen]

the distance formula is basically the pythagorean theorem

[mighty grotto]

sorry math isnt my strongest suit i hope thisnt a bother

[tender pollen]

just in cartesian coordinates

[mighty grotto]

so i would swap about abcd for the coordinates?

[tender pollen]

yep

and then show that two of the lengths are equal

[mighty grotto]

wait but what coordinates go where

[tender pollen]

it doesn't matter

the distance formula can be used vice versa*

that is, swapping the points doesn't matter

[mighty grotto]

so $\sqrt{(4 - 6)^2+(2+2)^2}$