help-23

[quasi bison]

ok right

what does 2/x^2 approach

[clear crest]

0

which means sin approaches 1 right?

[quasi bison]

what is sin(0)?

btw you put the ln in the wrong place

[clear crest]

nah still 0

[clear crest]

oh

[quasi bison]

your expression should be $x^3 \ln\paren{1 + \frac{1}{2x}} \ln\paren{1 + \sin\paren{\frac{2}{x^2}}}$

[flat frigate]

Ann

[quasi bison]

er

e^that

[clear crest]

bruuh right

so if sin(0) is still 0

then its ln(1)

which would still be 0

cuz e to the power of 0 is 1

[quasi bison]

yes

[clear crest]

and since im multiplying that with x^3 ln(1+1/2), it's still 0

which means the result would be e to the 0

which is 1?

[quasi bison]

??

i think you screwed up big time

you have an ∞*0 in here that you have to grapple with

also you missed an x

[clear crest]

nah wait i can't do that

[proper crypt]

You can't just take the limit in only one part of the expression

That's like saying the limit as x approaches infinity of x * 1/x is 0 because you get x * 0

Which is obviously not true

[clear crest]

yeah I'm absolutely clueless

I didn't even need to do that i just have more problems now

[safe radish]

@clear crest Has your question been resolved?

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[umbral swan]

set theory / limit Q

[umbral swan]

if I have some function f: A -> B

with A,B being subset of C (complex realm)

then any element of A is mapped onto B

now, when considering some convergent sequence a_n with a_n in A

if a_n converges within A, meaning lim(a_n) in A, then f(lim(a_n)) is also in B

since again it's just mapped

however, I got the intuition that if instead a_n converges on the boundary of A

then f(lim(a_n)) lies on the boundary of B or within B

is that correct & how could it be shown?

short version of the question:

f: A->B with A,B subsets of C (complex realm)
a_n convergent sequence with a_n ∈ A. lim(a_n) ∈ bound(A)
Is f(lim(a_n)) ∈ B∪bound(B)?

[safe radish]

@umbral swan Has your question been resolved?

[safe radish]

@umbral swan Has your question been resolved?

[safe radish]

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[lean otter]

magnitude of a = 3, magnitude of b = 1 magnitude of a-b=2, what is the magnitude of n=a-2b

[lean otter]

I'll try raising the n=a-2b ^2

[misty blade]

That's the right direction

[lean otter]

let's see where this gets me

do I have to use logic

as in since magnitude of a is 3 and magnitude of b is 1 the only way magnitude a-b=2 is if their angle is 0?

which then I can use with dot product to get a*b from the identity

[misty blade]

That's a good way

So you have the angle to be 0

Which means a•b is just the product of their magnitudes

[lean otter]

which is 3

so we have magnitude of n squared is 10-4*3

[misty blade]

So now magnitude of a-2b must be sqrt(a^2+4b^2-4a•b)

[lean otter]

oh wait

fucked up the math wait

|n|²=13-4ab

13-12

|n|²=1

|n|=1

[misty blade]

That's correct

[lean otter]

alright ty

.close

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[pseudo scroll]

$\sum_{k = 1}^{2023} ki^k$

[flat frigate]

jan Nejon

[pseudo scroll]

Is there no other way to do this other than expand it out and use AP summation formulae?

also why is my sigma small

[peak estuary]

why AP summation formula? this is more or less the derivative of the geometric sum. so you can use that

[pseudo scroll]

its not infinite

[peak estuary]

sum, not series

[pseudo scroll]

hmm

that was gonna be a followup question

if it was in infinite gp

the whole convergence thing

[peak estuary]

works out probably

I would need to look up the specific results but I know that it works out

[pseudo scroll]

$\frac{r^{n+1} - 1}{r - 1} = \sum_{k = 0}^n r^k$

hmm

[peak estuary]

not quite

[pseudo scroll]

did ii get it wrong

[peak estuary]

r^(n+1) and r^k

[pseudo scroll]

oh right

[peak estuary]

also starts at 0

[flat frigate]

jan Nejon

[pseudo scroll]

So I can just differentiate, multiply by r and plug in i?

[peak estuary]

yes

[pseudo scroll]

for the infinite case

does |x| < 1 still hold

[peak estuary]

yes

[pseudo scroll]

but now |x| is the complex modulus

alright

thanks

.close

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[pseudo scroll]

.reopen

[safe radish]

✅

[peak estuary]

well complex or real, either way <1

[pseudo scroll]

cool

.close

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[peak estuary]

express each number in (0,1] in binary

[restive parrot]

ive done that but what do i do next to show its equinumerus

[safe radish]

@restive parrot Has your question been resolved?

[junior smelt]

Each binary representation is equivalent to some element of ${0,1}^{\bN}$ - send each number in the decimal to its “cooordinate”

[flat frigate]

@junior smelt

[safe radish]

@restive parrot Has your question been resolved?

[restive parrot]

@junior smelt it doesnt matter if the functions in ${0,1}^{\mathbb{N}} are not bijection functions right

it doesnt matter if the functions in ${0,1}^{\mathbb{N}}$ are not bijection functions right

[flat frigate]

luke1

[restive parrot]

Rip

[junior smelt]

Backslash the curly brackets as well

[restive parrot]

Ye i forgor

[junior smelt]

But you want a function between ${ 0, 1}^{\bN}$ and the interval $(0,1]$ which is bijective

[flat frigate]

@junior smelt

[junior smelt]

You could represent elements in ${0,1}^{\bN}$ like $(x_1, x_2, \ldots)$, an infinite sequence where each term $x_i \in {0,1}$

[flat frigate]

@junior smelt

[safe radish]

@restive parrot Has your question been resolved?

[safe radish]

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[fallen thunder]

Hi, how would i do this?

[fallen thunder]

[glacial cairn]

Well X can't be more than 3, otherwise the carry would make it so the first column isn't Y

[solar hazel]

you can write e.g. YX7 = 100Y + 10x + 7

[glacial cairn]

In other words, there is no carry in the first column, so 6+X = 7 or 6+X+1 = 7

And the third column produces a carry, so it's 6+X+1 = 7

Or you just solve 100Y + 10X + 7 + 60 + Y = 100Y + 70 + X and find X and Y as integers between 0 and 9

[fallen thunder]

hmm and then how would this play out

YX7 = 100Y + 10X + 7
6Y = 60 + Y
so the sum is 100Y + 10X + 67 + Y

[fallen thunder]

hmm right, not sure if i'm slow 😭 but why do you knwo for sure the third column produces a carry?

[halcyon carbon]

X<7

[glacial cairn]

From 6+X = 7 or 6+X+1 = 7, X = 0 or X = 1, so to get 7+Y=X or 7+Y = 10+X you need 7+Y > 9

I'm so bad at explaining this

[solar hazel]

i think you'll get 9X + Y = 3, and that implies X = 0 (otherwise 9X > 3), which implies Y = 3

[fallen thunder]

Lol i think you guys are doing good with the explanation

It's embarassing i can't add hmm i'll keep thinking

cuz it doesn't fall together for me

[solar hazel]

100Y + 10X + 7 + 60 + Y = 100Y + 70 + X, right?

it is equivalent to 9X + Y = 3

[fallen thunder]

yes

oh lol i see

okay i'm blind and dumb 😭

tyy <33

now to understand nel's way as well

[halcyon carbon]

kappa and kanna

[solar hazel]

no kanna and i are the twinsies here

pure get a kuromi pfp to join us

[fallen thunder]

mad kuromi is the best

[halcyon carbon]

X+6 has to be less than 10, otherwise we get to carry over and the first column can’t be Y. So X<=3, but 7+Y > 3 for all Y, so 7+Y has to be 10,11,12, or 13, which produces a carry over, ie. X+6+1=7.

[glacial cairn]

Incredible Paint skills I know

[halcyon carbon]

Damn

[solar hazel]

[fallen thunder]

OWHHH @halcyon carbon @glacial cairn i see now!

sorry for caps

okay i think i get both ways now

[glacial cairn]

[fallen thunder]

[halcyon carbon]

[fallen thunder]

wait dumb question but after you surmised that x = 0 and y= 3 or x = 1 and y = 4

how did you affirm that it's the former pair

did you try them out?

[glacial cairn]

No, you get c = 1

(you get a carry)

[fallen thunder]

oh yeah y = 3 and y = 4 both produce a carry

so x has to be 0

[glacial cairn]

Then you just have 1 + X + 6 = 7

Yeah sorry the "7+Y = 0 or 7+Y = 1" should really be "7+Y = 10 or 7+Y = 11"

[fallen thunder]

yeah i get you're in mod 10 world ig lol

i think pure was in too

all of you guys are there except me

[solar hazel]

no pure doesn't know how to do this problem dw

[fallen thunder]

[solar hazel]

kappa is back

[zinc flume]

Is this solved?

[solar hazel]

yes

[zinc flume]

Cool

[solar hazel]

although pure still needs help understanding the solutions we gave

[safe radish]

@fallen thunder Has your question been resolved?

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[quartz wasp]

ok so i have a icocles trapezoid the sides are a a/sqrt3 a/2 and a/2 need to find the inscribed radius

[quasi bison]

this looks sus

do you have a pic of the question?

[quartz wasp]

thats wrong

im gonna tell you the problem rn

so i have a icocles trapezoid the trapezoids big side is a and the corner at the small side is 120 degrees i need to find the inscribed circles radius

@quasi bison

[quasi bison]

ok so then you don't have a picture, got it

[quartz wasp]

i do wait

Its suppost to be something like this if im not wrong

That 120 degrees i got bad writing

I need the circles area

[karmic hedge]

You can’t find it with just one length

[quartz wasp]

i dont need numbers

[karmic hedge]

Imagine an equilateral triangle with side length a

And draw a parallel to its base

That generates infinite trapezoids that satisfy your conditions

With different heights, and therefore different radii

[quartz wasp]

so what do i do?

This is the correct answer

@quasi bison

[quasi bison]

"here is a poorly stated problem, a shoddy diagram, and the correct answer. please reconstruct the problem and its solution from these data and explain it to me"

i dont rly want to do that

[quartz wasp]

ok can i just tell you another problem?

[quasi bison]

no

[quartz wasp]

ok

so we have a isosceles trapezoid

the big side is a

and the top corner is 120 degrees

i need to find the area of the circle

from that information

@quasi bison that is the problem i cant state it better

[quasi bison]

please stop pinging me.

[quartz wasp]

ok

sorry

@karmic hedge can you help me?

[quasi yarrow]

can you give original problem?

[quartz wasp]

its in a diffrent language

you wouldnt get ir

it

[quasi yarrow]

what language?

[quartz wasp]

that is all thats given

[karmic hedge]

Yeah my bad the problem is right

Anyways it suffices to find the inradius of an equilateral with length a

Which is pretty easy so try it

[quartz wasp]

how do i find it?

is it a/2?

[quasi yarrow]

let b be the length of small high side, then the sides of trapezoid are (a+b)/2, the radius of the circle is equal to half the height of the trapezoid so you have to find the height which is easy with using trigonometry

[quartz wasp]

i dont rylly know trigonometry that well

@karmic hedge is it a sqrt3/6*s

[quasi yarrow]

radius is a*sqrt(3)/6, yes

[quartz wasp]

ok so now what

[quasi yarrow]

find the area of circle?..

do u know the formula?

[quartz wasp]

yes

okyy got it

thanks a lot

.close

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[vernal orbit]

a spring has natural length of 22cm if a force of 15N is required to keep it stretched to a length of 32cm, how much work is required to stretch it from 22cm to 40cm

[empty gyro]

!status

[safe radish]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[vernal orbit]

1

[empty gyro]

Do you know the force equation for a string?

[vernal orbit]

nope

[empty gyro]

That's a good first start then

err.. not string, spring

My bad

The force equation for a spring

[vernal orbit]

what is it?

[empty gyro]

https://www.google.com/search?q=force+equation+for+a+spring&rlz=1C1CHBF_enUS1011US1012&oq=force+equation+for+a+s&gs_lcrp=EgZjaHJvbWUqBwgAEAAYgAQyBwgAEAAYgAQyBggBEEUYOTIKCAIQABgPGBYYHjIKCAMQABgPGBYYHjIKCAQQABgPGBYYHjIICAUQABgWGB4yCAgGEAAYFhgeMggIBxAAGBYYHjIICAgQABgWGB4yCAgJEAAYFhgeqAIAsAIA&sourceid=chrome&ie=UTF-8

[vernal orbit]

is there any way to do this without hooke's law, like intuition

or is this necessary

[empty gyro]

You could intuitively derive Hooke's law I guess?

[vernal orbit]

what is x

[empty gyro]

Why are you trying to avoid Hooke's law?

[vernal orbit]

no reason just curious

since we didnt cover it

pretty weird

[empty gyro]

odd that you're given a spring equation without covering it

[vernal orbit]

yeah

so what does x mean

like in this case would it be 40?

[empty gyro]

https://en.wikipedia.org/wiki/Hooke's_law#Linear_springs

[vernal orbit]

so it is 18?

the question asks for the answer in joules

but doesnt this give the answer in newtons?

[empty gyro]

Depends on which displacement you are trying to find

[vernal orbit]

give me the full solution pls

[empty gyro]

!noans

[safe radish]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[vernal orbit]

a spring has natural length of 22cm if a force of 15N is required to keep it stretched to a length of 32cm, how much work is required to stretch it from 22cm to 40cm

oh

so heres my thought process

15N = 10cm*k

k = 3/2

[empty gyro]

Make sure you have the right units, but essentially yes

[vernal orbit]

F = 18cm* k = 18 * 3/2

oh

so F = 27N

?

[empty gyro]

This would be the force at the displacement of 18cm, but you want the energy required to displace it this far

[vernal orbit]

ohh

how would I calculate the energy instead of the force

ohh

wait

so joule is newton/meter?

or newton*meter

so 27N*.18cm=4.86J?

[frigid python]

what is the next step?

[empty gyro]

Not quite. Multiplying force and distance only works for comstant forces

[vernal orbit]

oh

integration?

[frigid python]

why do you need an integral]

[empty gyro]

You need to use the integral formula, yea

[frigid python]

why?

what does k represent

[empty gyro]

Read about Hooke's Law

[frigid python]

integral of 3/2 * x

[vernal orbit]

OHHHHHHHHHH

I GET IT

ty

[frigid python]

is x the distance

[vernal orbit]

integral of 0-22 of 3/2x

right?

.close

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[empty gyro]

wrong

[safe radish]

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[granite hollow]

!help how do i get the 6th time where it hits 17m

[safe radish]

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

[granite hollow]

desmos is giving me this

but idk how

[neat kiln]

6th time from what?

[granite hollow]

6th time where it hits 17m

from the start

which is at 1.5m

[granite hollow]

this is the equation

[neat kiln]

Why not solve the equation first

For y = 17

[quasi yarrow]

solve the equation y = 17 and find the root you want

[granite hollow]

i did

i got 41

but i did it with a diffrent equation

i wanna know how to get 41

with this one

i am getting 36

[quasi yarrow]

solve equation cos(pi*x/8) = -13/18

[granite hollow]

i did its 6.055

then how is the asnwer 41.945

[quasi yarrow]

it has infinitely many solutions, not only one

[granite hollow]

how

cuz its when does it hit the 6thm

when does the ferris wheel hit 17m the 6th time

so it should be 41.925

@quasi yarrow

[quasi yarrow]

solution of $cos(x) = a$ is $x = \pm \arccos(a) + 2\pi k, \quad k \in \mathbb{Z}$

[flat frigate]

Alisia

[granite hollow]

that makes no sense

on the graph it hits 17m multiple times

but the 6th time is only once

[quasi yarrow]

show full graph

[granite hollow]

wdmy show full grpah

[neat kiln]

Exactly

It's never ending

[quasi yarrow]

show full graph that you're investigating

[granite hollow]

here

its not a graph that i am investigating

its a equation

[quasi yarrow]

with 6'th root and axes pls

[granite hollow]

whats that

someone got 87

which makes no sense

.close

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[upbeat needle]

short, question about the basis of a an infinite vector space. so for finite vector spaces of the form F^N where F is a field and N is finite i know that the set of all indicator functions is a basis of the vector space and that the numbers of elements in the basis is the dimension and the largest possible set of linearly independent vectors and that the basis is maximally independent. now my question is, given an infinitely large vector space, is the set of characteristic functions still a basis and can i still assume that adding an element to that set makes it linearly dependent?

[steep lily]

what do you mean by set of characteristic functions?

what I'm interpreting that as initially isn't linearly independent

[upbeat needle]

indicator functions if that is the correct term? sorry if my terminology is unclear it's not my language of study. what i mean is the function X_S : T -> {0, 1} with X(T) = 1 if t is in S and 0 otherwise where S is a subset of T and iirc the set of all indicator functions of the subsets with 1 element of N is supposed to be a basis for F^N

[steep lily]

unfortunately, that's not spanning

(characteristic functions is the right word, you just didn't clarify you meant subsets with one element)

we only allow finite linear combinations when discussing spanning

because we aren't guaranteed convergence if we allowed infinite sums in the general case e.g. Q^N

I think the general proof of the existence of a basis for all vector spaces relies on the axiom of choice, so I'm uncertain how explicit you can get with a basis when you have an infinite dimensional space

[upbeat needle]

yeah

[steep lily]

as for the second bit, adding an element to a basis does make a linearly dependent set regardless of the size of the basis, because the basis was spanning before

[upbeat needle]

hm that is what i find confusing because if the set in question is still a basis for an infinite set, aren't there elements that we can never construct with finite linear combinations

[steep lily]

I'm saying the set you gave isn't a basis, but if you had another set and it was a basis then that would hold

[upbeat needle]

ah

i see

thanks

.close

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[icy lance]

57

.close

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✅

[trim swan]

<@&268886789983436800>

thank you

.close

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[livid crow]

for questions like this, when plugging in x and y and the result is undefined, whats the rule for that?

[livid crow]

i remember learning about tangent lines but was it just a/0 is vertical tangent, 0/a is a horizontal tangent?

also for this question, i got 2 tangent lines (1 vertical and 1 normal one) but is that correct

since i got 2 x values

[rough storm]

!showwork

[safe radish]

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

[livid crow]

nvm i got it (ty)

.close

[safe radish]

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[merry saffron]

how do I solve b?

[misty blade]

Looks like a test

[merry saffron]

It is

Previous year

[royal ferry]

can you use $\cos 2x = 1 - 2\sin^2x$

[flat frigate]

LF

[royal ferry]

@merry saffron

[merry saffron]

But it’s cos4x

How would I rewrite it to make it cos2x

[royal ferry]

let y=2x

so now its cos 2y

and you can use it

[merry saffron]

I’m confused

[royal ferry]

what is confusing you6

[merry saffron]

How would it look like

How can we spilt up cos4x

[royal ferry]

Let $\theta = 2x$. Then, $\cos 4x = \cos 2 \theta = 1 - 2\sin^2 \theta = 1 - 2\sin^2 2x$

[flat frigate]

LF

[terse lichen]

sum identities

[royal ferry]

also

[terse lichen]

cos(4x) = cos(2x + 2x)

[merry saffron]

Tried that

[terse lichen]

and what did you get

[terse lichen]

you should get (cos(2x))^2 - (sin(2x))^2 yes?

[safe radish]

@merry saffron Has your question been resolved?

[merry saffron]

Yes

[terse lichen]

ok now whats cos(2x)

out of the three identities for cos(2x), which one do you think is more suitable

[merry saffron]

Cos^2 - sin^2 ?

[terse lichen]

yep

and whats sin(2x)

[safe radish]

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[merry saffron]

2sinxcosx

[terse lichen]

now sub and expand

and you get what you want

[merry saffron]

Alright thanks

[safe radish]

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[rich sinew]

Find the point on the graph of the function ( f(x) = x^2 ), that has the least distance to the line ( y = x - 2 ).

[flat frigate]

mjd

[rich sinew]

Hello. I used the distance formula to get the distance d=sqrt((x^2-x+2)^2). I derived it to find the minimum distance,equated it to zero, and got the point 1/2. I plug it back into the d formula and i got 7/4. Does this seem valid/ is there a better way to do it?

Also, is there a tool online I can use to find the answer?

[potent seal]

You could find the tangent to the parabola having the same slope as that of the line

The distance between the tangent and the line would be the least one

[rich sinew]

Alright thank you. I’ll try it out and see if I get the same answer as my method

[wise shell]

I have another method I think

From distance formula

[rich sinew]

What is it

[wise shell]

Wait lemme write it

[rich sinew]

Take ur time

[wise shell]

Nvm it's wrong

[rich sinew]

It’s all good

[potent seal]

Did you get the same ans?

[rich sinew]

Yea I did I got the same x value

And by using distance formula same answer

So thank you

[potent seal]

Oh alr

Np

[rich sinew]

.close

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[tiny gorge]

this optimization problem

i dont know what equations to link the variables

[fossil shell]

draw a picture

[merry sleet]

call x the angle BOC with O being the center of the circle

then write the time taken by the woman

if she goes to B from A with the boat and then walks from B to C along the circle

depending on the value of x

or you can call x AOB

whatever

[tiny gorge]

ah ok i will try that

[merry sleet]

then you can differentiate

wrt x

and find the minimum

[safe radish]

@tiny gorge Has your question been resolved?

[tiny gorge]

what would be the relationship between AB and BC?

because i have the equation $t = \frac{d_1}{2} + \frac{d_2}{4}$

[flat frigate]

marco.py

[tiny gorge]

d1 is AB and d2 is BC

[safe radish]

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[tiny gorge]

.reopen

[safe radish]

✅

[tiny gorge]

is x = to 2theta?

ah ok i think it is

[safe radish]

@tiny gorge Has your question been resolved?

[tiny gorge]

ah ok it works

thank you

.close

[safe radish]

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[merry saffron]

,,sin 2 x = -root 3 /2

[flat frigate]

Hanz

[merry saffron]

question on my test

idk if I solved it right

I just did 2sinxcosx = -root3/2

and I did 2sin(2pi/3)cos(2pi/3) = -root3/2

which kinda proved that x = 2pi/3

but idk

[potent seal]

What was your reason behind putting x=2pi/3?

[safe radish]

@merry saffron Has your question been resolved?

[merry saffron]

I knew it would equal 2pi/3

How would you solve this tho

[safe radish]

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[safe radish]

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[quasi bison]

comparison with 1/n

or limit comparison with 1/n ig

[wet wigeon]

$$\log (1+x) \leq x$$ for all $ x\ > -1$

[flat frigate]

Cyrenux

[wet wigeon]

Right, it currently doesnt help, but there is also left side of this inequality

That might help?

$$ 1 - \frac{1}{1+x} \leq \log (1+x) \leq x$$ for $ x > 0$

[flat frigate]

Cyrenux

[tiny wraith]

Log rules make it telescoping

[wet wigeon]

$$ \log (a/b) = /log (a) - /log (b)$$

I think you havent learnt limit comparasion test yet

That could be why

Yeah, you take limits on that one

[safe radish]

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[raw pivot]

$\int \sec (x) dx$

[raw pivot]

How can I solve this

Im not sure where to start here

[flat frigate]

tobi

[raw pivot]

.close

[safe radish]

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[plucky elk]

Use alternating series test instead

[rough storm]

you can turn what you have also into the form $\left(1+\f{1}{n}\right)^{n}$

[flat frigate]

Moosey

[rough storm]

inside the logarithm

[atomic spruce]

random tangent your username scares me..

[plucky elk]

Ah shit

Yea thankfully moosey can read unlike me

I just looked at your work

[rough storm]

I think I misread too

ononono

wait

[atomic spruce]

simplify the log expression

log(n+1)/log(n) is what

its been awhile since i did horrendous convergence tests

wait, its not being subtracted

[junior smelt]

Note that $\log(n + 1) = \log(n) + \log\pqty{1 + \frac1n}$

[flat frigate]

@junior smelt

[rough storm]

AH

there it is

[junior smelt]

(as you can factor n + 1 = n(1 + 1/n), makes life a bit nicer!)

[atomic spruce]

we love doing a little cheeky factoring

[rough storm]

and that makes it all nice :)

[flat frigate]

Mycobacterium

[rough storm]

you can move the n into the log(1+1/n) as a power

then just apply limit to that to turn it into a nice constant :)

[junior smelt]

Alt: factor as
[
\abs{ \pqty{\frac{n}{n+1}} \pqty{ 1 + \frac{ \log\pqty{1 + \frac1n} }{ \log(n) } } }
]

[flat frigate]

@junior smelt

[junior smelt]

Both of those factors, when you work on them, end up going to 1, hopefully clearly

[rough storm]

hmm...

[atomic spruce]

i cannot believe i broke several math laws to reach that answer

[rough storm]

lmao

[safe radish]

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[quiet plume]

The ratio test is inconclusive if the limit is 1.

If it's >1 (including infinity) then it should diverge.

[untold sky]

that limit is certainly not 1

[quiet plume]

If it's <1 it should converge absolutely.

Well it can not exist in other ways

Think about what the test actually shows

If it's >1, then it means that a given term tends to be greater than the last in magnitude.

So it makes sense that it would make the series diverge.

[safe radish]

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[tame aurora]

i've used the exponential growth formula and compound interest rate formula to solve this, both with different answers. thoughts?

[tame aurora]

two ways i did this

For a calculus assignment. Haven't come across a formula that explicitly calculates GDP rate. I assumed we had to use principles from our topics and the first thought i came up with was the exponential growth function

[safe radish]

@tame aurora Has your question been resolved?

[tame aurora]

<@&286206848099549185>

[safe radish]

@tame aurora Has your question been resolved?

[safe radish]

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[cyan dune]

"Show that if n is not prime, 2^n - 1 is not prime"

[cyan dune]

How would I go about proving this?

wait nevermind i got it

.close

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[inner current]

i have f(x) = [x-1] + [x-2] in x E [0,3]

[.] is GIF

[inner current]

and i have to find the points of non differentiability

so i rewrote f(x) as

2 [x] - 3

and we know, [x] is ND at all integers

so we get 0,1,2,3 as the answer

but 0 isnt included in the answer

why

ping if answering, thank you

[robust river]

[x] = n such that n <= x and n is an integer, right?

[inner current]

yes yes

Floor function

[3.8] = 3

[robust river]

alright

[inner current]

how

😞

[robust river]

it's correct

one sec

[robust river]

that's because of the domain

[inner current]

then why is 3 included

[drowsy karma]

the floor x function always has jumps - points of discontinuity at integral points

[robust river]

A function is differentiable on a segment [a, b] if it's differentiable on an interval (a, b), has right derivative at a and left derivative at b

[robust river]

Because there's no left-derivative at x=3

[inner current]

yes i am aware

[inner current]

huh

no right derivative right?

because 3- is in domain

[drowsy karma]

and points of discontinuity are also points of non-differentiability 🙂

[inner current]

youre right but then

why is 3 included

3 is also an extreme point

we cant check at 3+ or 0-

[robust river]

$\lim_{h \rightarrow 0+0} \frac{f(3) - f(3-h)}{h} = \lim_{h \rightarrow 0+0} \frac{3 - 1}{h} = \infty$

[flat frigate]

EQUENOS

[robust river]

no left derivative

[inner current]

...

how is that

if the function is from 0 to 3

then left side of 0 is not included and right side of 3 is not included

whats even the point of calculating derivative of 3-

[robust river]

No, the zero should be included then

[inner current]

man

youre seriously confusing me rn 😭

[robust river]

$\lim_{h \rightarrow 0+0} \frac{f(h) - f(0)}{h} = \lim_{h \rightarrow 0+0} \frac{0 - 0}{h} = 0$

[flat frigate]

EQUENOS

[inner current]

i dont get it sorry,

[robust river]

So there's a right derivative at 0

[inner current]

yes ofc

there is

it is differentiable at 0+

the problem is at 0 but then 0 isnt included

[robust river]

Again, let's look at this definition

[inner current]

a function is differentiable in a segment if

it is continuous

has no vertical tangents

does not oscillate

[inner current]

like i get this

[robust river]

I'm saying that 0 is the point of differentiability because we live in [0, 3]

We "can't" approach 0 from the left

In topology this is called an induced topology

[inner current]

yes

this is correct

[robust river]

So that's why 0 is not in the answer

[inner current]

then 3 has to be too

we just dont apporach it from right side

[robust river]

Approach 3 from the left

We can approach it from the left

[inner current]

yes

and we can approach zero from the right

[robust river]

This shows the problem of 3

[inner current]

what does that change

but 3 is in the answer

[robust river]

Any sequence approaching 3 from the left is less then 3 and therefore the floor of that sequence is <= 2
While floor(3) = 3

[robust river]

Yes, because it's not the point of differentiability

[drowsy karma]

but this is left-hand differentiability, not general differentiability

[inner current]

but why does that matter

[inner current]

what

man

i dont wanna be rude

but you seriously cannot explain

like i appreciate your help

i really really do

but youve just made it worse 😭

[robust river]

The discussion got kinda chaotic

[inner current]

yeah i agree mb

i just dont get why 0 is removed while 3 is not, you cant just remove ONE of them and blame it on the domain

either remove both of remove none

[robust river]

Ok let's to the following thing

[inner current]

this is from desmos

[robust river]

Try calculating

1. The right derivative at 0
2. The left derivative at 3
   And share your results with me

[inner current]

for right derivative at x = 0

we have f(x+h) - f(x) / h

so we have, 2[x+h] - 3 at x = 0, which will give us - 3

and f(0) = 2[0] - 3

is this wrong?

[robust river]

This is correct

[inner current]

so we have

-3 - (-3) / h

which is just 0/h

since numerator is perfectly 0 we have 0

[robust river]

splendid

[inner current]

or did i fuck it up

☠️

[drowsy karma]

but you must compute derivative from both sides, using two limits

[inner current]

but it isnt in the domain

how do i come from the left side

[robust river]

You proved that the right derivative at 0 is 0
Now let's calculate the left derivative at 3

[inner current]

if my graph starts from 0

[robust river]

We can't compute the left limit at 0 in an induced topology on [0, 3]

[drowsy karma]

ofc, but it must be written x->0+

[inner current]

theres no point calculating these limits because we already know where the function is non differentiable

these limits UNLESS outside the domain

MUST exist

because the only thing in f(x) causing issue is [x]

which we know is ND at integers ONLY

[inner current]

for exmaple i cant calculate 3+ or 0-

[drowsy karma]

Equenos is right ab out differentiabilty on [a,b]

[drowsy karma]

he wrote earlier that a function is differentiable on [a,b] when it is differentiable on (a,b), right-differentiable at x = a and left-differentiable at x = b. And now, for me, even though it is differentiable on an [a,b] , then it is not differentiable at x = b, similarly if x = a,

[inner current]

man, fancy definitions

[inner current]

look, graph continious

graph no kink

graph no oscillates

graph has slope

graph differentiate

[drowsy karma]

fo rme, it is clear 0, 1, 2, 3, are points of non-differentiabiltiy

[robust river]

Like this: $\lim_{h \rightarrow 0+} \frac{f(x) - f(x-h)}{h}$

[flat frigate]

EQUENOS

[inner current]

but

there

is no left side

it is NOT in the domain

[inner current]

yes but then

but

hm

[drowsy karma]

0 and 3 belongs to the domain

[robust river]

At x=3? both x and x-h are in the domain because h>0

[inner current]

two answers

[inner current]

but

i dont care about the left side of 3

we already established the function is ND at 3

clearly it has to be D at EVERYWHERE else

except the other integers

[robust river]

Except 0 is the point of differentiability if the domain is [0, 3]

[inner current]

but why

[drowsy karma]

i disagree

[inner current]

dont give me a fancy definition

show me the graph

explain

[drowsy karma]

0 is a pont of right differentiability

[inner current]

show the beauty of maths not fancy english

[drowsy karma]

but not generl diffeetiability

[inner current]

yes

the problem is

if i say the domain is

(0,3)

would 0 and 3 still be differentiable

[robust river]

it's also a point of general differentiability because of how the induced topology on [0, 3] works

[inner current]

no right

[robust river]

You can't approach 0 from the left if there's no space on the left of 0

[inner current]

true yes

i agree

[robust river]

Except for a trivial sequence x_n=0

[inner current]

that makes no sense

[robust river]

ok forget about the trivial sequence

[inner current]

i am in school man, show me graphs and maths

show me where the problem is graphically

or tell me why without fancy english

we either include both 0 and 3 or none of them

[robust river]

My point is that at the edges of our segment one-sided differentiability becomes general differentiability

[inner current]

and what does that change

[robust river]

ok, one sec

[drowsy karma]

the derivative at x = 0 exists only on the right side - the right-hand derivative, so we cannot say that x = 0 is the point of differentiability, similarly x = 3,, but from left side

[inner current]

are you saying we include 0

[inner current]

yes

so we either exclude both or none

[drowsy karma]

i exclude 0, an di exclude 3

[inner current]

hm

[drowsy karma]

0, 1, 2, 3, are all non -diffenrentiabiltuy points

[inner current]

so this is the same as domain being (0,3)

[drowsy karma]

yes different is

that yo ca say

[inner current]

different ?

[drowsy karma]

that f is differentiale on [0,1), (1,2),(2,3,] but that is another deifntion = ddefintion of diffeerntiability on interval

but if we discuss

not intervals but points

i conider that 0 and 3 must be excluded

[inner current]

see we cant talk about intervals now

All of us have established the function to be ND at x E integers

the domain is [0,3]

so the question is of boundary points

[drowsy karma]

yes

[inner current]

0 and 3

[drowsy karma]

i exclude 0 and 3

[inner current]

because?

[drowsy karma]

beucae derivative, is only oen side on those points

[inner current]

yes but you see

the graph begins at 0

[drowsy karma]

tjhere is also a thoerem:

a function is differentiable at x = a if it is left-differentiable and right-differentiable

at x = a

and here, we cant

talk

[inner current]

hm

[drowsy karma]

abotu it

if it coem sto x = 0 and x = 3

[inner current]

so 0 and pi are points of ND

[drowsy karma]

if you get restricted sinx

on 0, pi open

then all out of 0,pi doe snot exist for us

we jsut do not think about the rest

[inner current]

0- and pi+ DNE

[drowsy karma]

0- and Pi+ does not exist for me

in yoru example of sine

[inner current]

my intelligence limits my understanding of this graph

please explain

[drowsy karma]

desmos is sure nice tool but we cant take it as a proof in my opinion

[robust river]

When we consider the right limit at an integer point, we get a horizontal line (derivative = 0)
When we consider the left limit at an integer point, we get a vertical line (derivative = inf)

[robust river]

he asked me

[drowsy karma]

ok

[robust river]

I already prooved it formally earlier

[drowsy karma]

form right side fo rx = 0 and let side of x = 3

but not generally

[robust river]

x=0 is the point of differentiability though
You see, we can't approach it from the left because the domain doesn't extend to the left.

[drowsy karma]

i know

[inner current]

then

[robust river]

Your statement holds only if x=0 has a neighbourhood inside the domain

[drowsy karma]

i am not talkign ab left

[inner current]

you must say the same for 3

[drowsy karma]

but you msut talk ab right siee of x = 0

[inner current]

how can you give this explaination about the left side of 0 but not about right side of 3

we either include both or none

[drowsy karma]

$f'{+}\left( 0 \right)=\lim{\Delta\text{}x \to 0^{+}} \frac{f\left( 0+\Delta\text{}x \right)-f\left( 0 \right)}{\Delta\text{ }x}$

[flat frigate]

Joanna Angel

[drowsy karma]

hence f is not differentuiable at x = 0, god

onyl right

again

a function is differentiable at x = a if it is left-differentiable and right-differentiable

since given function is not differentiable at x = 0, nor x = 3

[inner current]

someone needs to math stack exchange this

[drowsy karma]

lol

[inner current]

so the guy who i am learning from did it wrong

hm

[robust river]

You're right - we can't approach 3 from the right side. But the difference here is that if we approach it from the left we get an infinite limit, not zero. Therefore x is not left differentiable.

[robust river]

It's incorrect to talk about the left derivative of f at zero because it's the left edge of the domain

[drowsy karma]

who did it ?

[inner current]

i am getting 2/h

[drowsy karma]

i said ab rigth derivative

at x = 0

$f'{+}\left( 0 \right)=\lim{\Delta\text{}x \to 0^{+}} \frac{f\left( 0+\Delta\text{}x \right)-f\left( 0 \right)}{\Delta\text{ }x}$

[flat frigate]

Joanna Angel

[inner current]

this is brain hurting

[drowsy karma]

not about left side

[robust river]

I think I know what's the source of confusion here

[inner current]

its me

[drowsy karma]

that is a thoerem, digt in literature

[inner current]

if that wasnt very obvious

There is a second method of solving this question

i will try the other way

[drowsy karma]

We have established here that the derivative at x = 0 is only right-handed, and at x = 3 it is left-handed, so we spend so much time on these points that they are only points of one-sided differentiability but not of differentiability.

[robust river]

Let's say that [a, b] is the domain of function f. Let c be a point from (a, b). We all agree that f is differentiable at x=c if it has both left and right derivatives at c and they're equal. Now what should we do with a, b?

I say that f is differentiable at x=a if it has the right derivative at a. (because the left derivative is not even defined at a)
Joana says that f is never differentiable at segment edges.

[drowsy karma]

yes 🙂

[robust river]

that's why we ended up being so confused

[drowsy karma]

yes )

and htat is ok in maths because

no President decides ab it )

that is consensus

in other words, we have definitional differences, and these are issues that are often conventional, unfortunately.

[robust river]

Yeah, some definition details are conventional

After all, one can specify which derivative mean

[drowsy karma]

i agree

[inner current]

i calculated LHD and RHD at x = 0

and yes i did it without caring about domain

LHD = 4/h and RHD = 6/h

[robust river]

RHD = 0
LHD = inf

You forgot that h->0-

[inner current]

i did use it

like

LHD, f(x) - f(x-h) / - h

RHD, f(x+h) - f(x) / h

f(x+h) = 3

urgh

i will just ask it as a doubt idk this shit sucks

thank you for so much time

.close

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[azure quiver]

I understand that i have to factor out the denominators

but then after that im confused on what to do

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@azure quiver Has your question been resolved?

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[gritty pier]

Partial Fraction Decomposition. Which method is the correct one?

[gritty pier]

One method is from my teacher, and the other one is from yt 💀

both are different, but which one is it now

(dont mind the typo + should be -)

OK just realised theyre just the same 💀

.close

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[narrow vault]

can someone explain me how to differentiate something like this or how itss called such th`at I can watch a video on it?

[plucky elk]

Have you taken multivariate calculus

[narrow vault]

yes last year

[plucky elk]

Use chain rule

https://tutorial.math.lamar.edu/classes/calciii/chainrule.aspx

[narrow vault]

seems like I am missing one c to satisfy the condition

meaning the c's on the RHD are one power less than on the LHS

[narrow vault]

or actually the answer to this question is false. so did I do the derrivatives correctly?

lmk when you have time riemann 🙏

[plucky elk]

Oh you're right. The question is wrong

https://math.libretexts.org/Bookshelves/Differential_Equations/Elementary_Differential_Equations_with_Boundary_Value_Problems_(Trench)/12%3A_Fourier_Solutions_of_Partial_Differential_Equations/12.02%3A_The_Wave_Equation

[plucky elk]

Should be c^2 at the end

[narrow vault]

the question is a true or false thing

but thanks. If what I wrote is correct then I got it

.close

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[tiny gorge]

,rotate

[tiny gorge]

im just stuck trying to find the range of r

what should be the maximum for r?

[low vector]

you have the perimeter which is 2r+2x+pi*r and you know its equal to 18

so you can write x in terms of r

[tiny gorge]

yeah

[low vector]

do that

[tiny gorge]

$x = \frac{18-r(\pi+2)}{2}$

[flat frigate]

marco.py

[tiny gorge]

yes this

[low vector]

yes

[tiny gorge]

is that the max?

[low vector]

no

thats just the side length of the rectangle

[tiny gorge]

yes

[low vector]

such that the total perimeter is 18

[tiny gorge]

yes thats the height of the rectangle

[low vector]

we can also calculate the area of the shape

[tiny gorge]

yeah its $A = \pixr^3$

[low vector]

are you sure?

[flat frigate]

marco.py

[low vector]

the area is the sum of the area of hte rectangle and the area of the semicircle, correct?

[tiny gorge]

yeah

[low vector]

so how are you getting an r^3?

[tiny gorge]

ah im multiplying the areas

[low vector]

:)

[tiny gorge]

ok so $A = \frac{4rx+\pi*r^2}{2}$

[flat frigate]

marco.py

[tiny gorge]

yeah

so the range of r would be

[low vector]

yes

we now have the area

how do we get the maximum of the area

[tiny gorge]

we find the critical points of A

by differentiating

but don't we first need to find the possible values of r?

[low vector]

so what is the derivative of A

[low vector]

we do that later

[tiny gorge]

ok

[low vector]

and with respect to what do we differentiate

[tiny gorge]

r

so i have to replace x with r

[low vector]

technically no

[tiny gorge]

well replace x with $\frac{18-r(\pi+2)}{2}$

[flat frigate]

marco.py

[low vector]

you can use the product rule

remember that [f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)