help-0

[fresh relic]

That makes sense

[tough hatch]

one-sided limit has either + or -

[fresh relic]

Ok

So how do we find the one sided ones

[tough hatch]

so you said this earlier

[fresh relic]

So the negative one is 3?

And the positive one is 4?

[tough hatch]

yes

tho i would not call them that.

[fresh relic]

Alright

For the next set of boxes the f(7) is 1 right?

[tough hatch]

so you understand this now?

and this?

[hexed aspen]

i need help with this too

anyone know how to do this?

[tough hatch]

@fresh relic

[fresh relic]

About to test it out

[tough hatch]

courtesy and common sense in chat, please

[fresh relic]

Can you please leave I’m being helped in here

[hexed aspen]

no

i need help too

[fresh relic]

Go to a different room

[tough hatch]

then you can ask on another channel

this channel is still occupied

[hexed aspen]

alright sorry

[tough hatch]

ty for understanding

[fresh relic]

But anyways @tough hatch for the next set of boxes f(7) = 1 right?

[tough hatch]

f(7) = 1

[fresh relic]

Sweet and then for the - and + they’re both 3 right?

[tough hatch]

yes

nice.

[fresh relic]

Alright and then the double limit is also 3?

[tough hatch]

thus the two sided limit is equal to ___

yes

[fresh relic]

So for 9

It’s all DNE right?

Because of the asymptote

[tough hatch]

not necessarily

any limit approaching x=9, one-sided or not, is indeed DNE

but it also happens that f is not defined at x=9

so yes, they are all DNE

[fresh relic]

It needs one of the dots for it to be defined right

[tough hatch]

if say, there was a shaded circle at x=9

then f(9) would exist

and the rest DNE

[tough hatch]

shaded circles, sure

[fresh relic]

Alright and then f(11) and the one sided limits and the double limit are all 3 right?

[tough hatch]

yes

[fresh relic]

For this one including the numbers in the table its just plugging them in right?

[tough hatch]

sure

[fresh relic]

If it is something divided by 0 it doesn’t exist right?

[tough hatch]

yes

[fresh relic]

So does this seem right?

[tight locust]

,rotate

[fresh relic]

?

[tough hatch]

the limit exists though

have u filled out the other tables?

do them first

[tight locust]

just use l'hospital's rule

[fresh relic]

Never heard of it

[tough hatch]

they likely don't know that.

for (a), that is not right

ln (1+0) = ?

ln 1, of course. but ln 1 is not equal to 1.

what exponent must e have in order to be equal to 1?

is essentially what ln 1 is asking

@fresh relic

[fresh relic]

What is e

[tough hatch]

it's a constant

[fresh relic]

For what

And where

[tight locust]

if you have a lim of the form 0/0 or inf/inf then you can take the limit of f'/g' and it will have the same value

pretty neat huh?

[ocean seal]

Results provided by WolframAlpha

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

Requested by ₀.₆₆₆...#0422

[tight locust]

,w e^ln(x)

[tough hatch]

this is e

[ocean seal]

Results provided by WolframAlpha

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

Requested by EndTimes#8053

[fresh relic]

I have absolutely no idea what that means

[tight locust]

ok. you know how sometimes you have to take the limit of a fraction?

[tough hatch]

don't bother trying to explain it to them

they are not there yet.

[fresh relic]

And I don’t see a single E in this problem

[tough hatch]

they don't cover derivatives yet

do you not know what ln 1 means?

[fresh relic]

I don’t

[tough hatch]

damn

better get used to them

[fresh relic]

You know the sad part is I took honors precalc in high school and got an A in it

[tough hatch]

you do not remember logarithms?

[fresh relic]

I don’t remember anything

[tough hatch]

damn

you need to brush up on that then

[fresh relic]

Since last year my academics went down the toilet

Now I’m a freshman in college

[fresh relic]

I can’t brush up if I don’t even know anything about it

[tough hatch]

yeah, you will have a hard time dealing with calculus and the above if your basics are not good enough

[fresh relic]

So like do I just put ln(1+0.5) in a calculator?

[tough hatch]

logarithms are not uncommon in precal, or in calculus

sure

you did not take any precalculus courses?

[fresh relic]

I did

Also when it says X what do I do

[tough hatch]

you were not taught about logarithms in your precal course, huh

that's just wrong

even a short review would've been helpful

where?

[boreal isle]

anyone know what my first step would be with this problem? It's asking for the difference quotient but the formula for the difference quotient seems to be different than what's here

this is review, i knew how to do this like 3-4 years ago

[tough hatch]

@fresh relic

[fresh relic]

we were i just dont remember

[tough hatch]

so you're done now?

[fresh relic]

no

i dont know what x means

[tough hatch]

the value of x

[fresh relic]

yeah its just an x

i dont know what to do

[tough hatch]

me either, there are a lot of x's here and you're not saying which one

[fresh relic]

any of them

it just says x

no X+1 or anything just x

[tough hatch]

then just copy the given column of x

[fresh relic]

?so just take 0.5 and move it over?

[tough hatch]

sure

[fresh relic]

whats the point of them even including that column?

[tough hatch]

who knows

maybe they are sadists

[alpine sable]

@tough hatch is this one still occupied

[fresh relic]

yes

[alpine sable]

Ok

[tough hatch]

are you done filling out the tables though? you haven't asked another question yet

[fresh relic]

I got this

Damnit I forgot about the 2nd table. This takes forever

Is this right though?

[tough hatch]

i am not going to check that myself

u can recheck yourself using a calculator

[fresh relic]

I did use a calculator

You crazy you think I’d do that without one?

[jaunty egret]

1.642857142/1
now multiply the denominator to a number in such a way that the numerator becomes whole
and then multiply the numerator and denominator by 3

[tough hatch]

recheck

[fresh relic]

What?

[tough hatch]

yes and it's big bang all over again

[fresh relic]

Oh honestly it took too long for the first one I can’t with this anymore I’ll just do the - now

[tough hatch]

yep, they are definitely sadists

and possibly masochists

[fresh relic]

theyre dicks

[jaunty egret]

wut

[fresh relic]

Ok @tough hatch I finished both tables

I did C and it just says error

[tough hatch]

what value does the function approach, based on the tables alone?

[fierce ivy]

Hello Guys, this is an exercise of my differential calculus, do you know how to answer? I have no idea.

[fresh relic]

1

[fresh relic]

Please leave

[tough hatch]

so that is ur answer

[fresh relic]

I think

as the values get smaller in the first and larger in the second it comes to 1

[tough hatch]

this is what is asked of you in c.

[fierce ivy]

alright, sorry for not asking yet

[fresh relic]

so it is 1?

[tough hatch]

the first table showcases the values of x approaching 0, from the right side of 0

the second showcases the values of x approaching 0, from the left side of 0

[fresh relic]

so 1 right?

you still around? @tough hatch

[tough hatch]

hmmmm...

[tough hatch]

this isn't a question

[fresh relic]

but am I picking the right thing

like see in the far right as you go down it comes to 1

thats what ur referring to right?

[tough hatch]

yes

the limit is indeed equal to 1

[fresh relic]

Do you know how to do these?

[tough hatch]

which?

[fresh relic]

Both of them

[tough hatch]

well

you know that the limit is 1

[tough hatch]

which eventually becomes equal to 1 at x=0

but what does it mean for a fraction to be equal to 1?

numerator ? denominator

[fresh relic]

=

[tough hatch]

yes

ln(1+x) is the numerator

x is the denominator

[fresh relic]

So for D just put that?

[fiery berry]

there are certain number of kids in a school. 20% take cars. 3/5 of the remaining students take bus. 612 walk

[tough hatch]

up to you.

[fresh relic]

but I don’t wanna get it wrong

how do you know what to put?

[tough hatch]

i have given an explanation above.

[fresh relic]

But why did you say up to you

[tough hatch]

it's up to you how you write your answer

[alpine sable]

is this free?

@fresh relic

[fresh relic]

well how would you write it?

[fresh relic]

yes

[tough hatch]

i just explained it above

lol

[alpine sable]

ok

Wait so I can ask a question @fresh relic

if ur done sorry

[fresh relic]

ok so do you know how to do D?

[fresh relic]

im sorry man im having anuerisms with this i still have more questions

[tough hatch]

didn't i just say it above?

[fresh relic]

or I am sorry I meant E

[tough hatch]

ok

why not use a graphing tool

like Desmos

or Geogebra

[fresh relic]

Ok

[alpine sable]

@fresh relic u good?

[shy trout]

help?

pls

[keen shell]

local seed = userid*day + userid*offset
is there anyway 2 seeds can be the same if userid is unique, offset is a a number ascending from 1, and day is the amount of days that has passed since 1970 (unix)?

[devout bronze]

Let the coordinate C be (x,y)

find the distance from A to C and B to C

but since C is on the x axis

y = 0

equate the two distances together

and solve for x

[jolly stone]

yes although very very very unlikely

[shy trout]

ty

[keen shell]

like... how unlikely? less than %1

[jolly stone]

depending on your size of userid, day, offset

[keen shell]

sowwy for the dumb question im trying to program and im having a brainfart

[keen shell]

an example mihgt be 1231453, 100, 2

[jolly stone]

how small and large can the number of days be?

100 days so still year 1970

[keen shell]

anywhere from

[jovial iris]

explain me linear

[jolly stone]

oh ok so 0 to around 5-digit number

[jovial iris]

i little bit dont understand

[keen shell]

it could be any number above 18k

although it probably wont ever be

[jolly stone]

how about id?

oh btw i think way less than 1%, just not sure the magnitude

[keen shell]

any number between 0 and 1 billion

unique to each player

[jolly stone]

oh now that might have problems with small id number

[keen shell]

i think the earliest ids that are active are 10000+

[jolly stone]

i see, so assuming id from 10k to 1b

id of 10k-100k will probably cause some problems, though pretty unlikely

how about offset?

[keen shell]

offset is just how many times the player buys the "next day" pass

so 0 - inf although its not likely to be over 1k

[jolly stone]

so also unique to players?

[keen shell]

nope, almost certain that offset is not unique

[jolly stone]

ok that might be a problem

[keen shell]

day is also not unique

[jolly stone]

best thing i can think of is, try with your current db

if collision even occurs on day 1 of your db, your seed is pretty bad

so basically your seed should be totally random and equally distributed right?

[keen shell]

yes, it should be different for each player

[jolly stone]

have you tried using any well known hash functions?

if that's your goal of generating random numbers

[keen shell]

it has to be constant, thats the only thing

[jolly stone]

what does "constant" mean here?

[keen shell]

e.g if a player rejoins and the function to get their seed is run again

it has to be the same as when they left

with the exception of the offset or day changing

[jolly stone]

when day or offset changes, your seed must change also?

hash functions should also handle that, using the day and offset as input also

[keen shell]

okie, ill look into that

[jolly stone]

[midnight girder]

could anyone explain where to start here?

[pastel schooner]

I just asked this today lmao. Ill explain

So we know that f(x) must be contuous

as stated in the question

So that only leaves 9 as a possible discontinuous.

Also note that 6x - 2 = f(9) = -7x + b

cause for it to be continuous it must f(x) must have the same limit value as x goes to 9 from both directions.

You plug in 9 for x

6(9) - 2

which is 52

and -7(9)+b

-63 +b

so 52 = -63 + b

+63 on both sides

b = 115

if @crisp iron can see if I am right.

[wary stream]

FYI this is a person

Not the helper ping

@Helpers

Is what you want

[pastel schooner]

ah I see thank you

is my explanation right?

[crisp grove]

[pastel schooner]

Dont want to give them the wrong info lol

[wooden cloud]

can someone help me w this?

this is a partial fractions question

[crisp grove]

if you already know what to do, why are you asking?

[wooden cloud]

i know what the method is called i'm just unsure of how to proceed

i split up the fraction and set up my A + B equation

but one of the variables couls be either positive or negative and im not sure what to do when that happens

[kindred warren]

this is just $(x^3 - x)^{-1}$

[ocean seal]

Tra-Guy

[kindred warren]

and you can use chain rule to antidifferentiate

[rigid smelt]

[stark lantern]

lol

[crisp grove]

[jolly stone]

can you show your current work?

[rigid smelt]

wouldnt it be great to have a reverse chain rule

[kindred warren]

wait shit

i thougth we doing dy/dx

wopes

[jolly stone]

[wooden cloud]

IS THERE ONE

[rigid smelt]

technically there is, its just not general

and it only works for composition of linear functions, like 1/(2x+5000)

[crisp grove]

[wooden cloud]

@jolly stone

[crisp grove]

you have to take $\frac{Bx+C}{x^2-1}$ instead of $\frac{B}{x^2-1}$

[ocean seal]

Ryuzaki

[wooden cloud]

huh

[crisp grove]

alos

also*

(x^2-1) = (x-1)(x+1), you can factorize again

$\frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$

[ocean seal]

Ryuzaki

[wooden cloud]

https://tenor.com/view/wojak-gif-21991855

[jolly stone]

you can split x^2-1 = (x-1)(x+1)

ok

soz for afk

[wooden cloud]

okayyy i never knew that you could do that

i think i'd benefit more from learning how and why this method works

like when it's to be used

[jolly stone]

[wooden cloud]

what do i do from here?

i guess OH

pluggin in zero would get rid of b and c

ok hold on give me a sec

[midnight girder]

@pastel schooner agh i did what you did but i did it backwards and plugged in 9 after solving for b whoops

that should be right i hope?

115

it was!!
thank you!

[wooden cloud]

@jolly stone am i heading in the right direction?

[jolly stone]

yep

the rest is trivial and is left as an exercise to readers

[wooden cloud]

[jolly stone]

don't forget it's ln|x+1| and ln|x-1|

[wooden cloud]

okee thank you

[jolly stone]

[wooden cloud]

[harsh belfry]

I could be approaching the problem incorrectly.

But wouldn't it be impossible to have a continuous function given these rules?

[tough hatch]

the function is continuous if the (existent) limit at x=a is equal to the value of g(x) at x=a

[harsh belfry]

Give me moment to wrap my small brain around what you said.

haha

so is a 10?

[tough hatch]

do you see what to do now?

[harsh belfry]

I need to find the value of a in context of the top function plugging in 10 for a

[tough hatch]

no.

[harsh belfry]

Alright, I'm very sorry but I am confused then.

[harsh belfry]

I think I'm grasping it alittle more now.

But what would my first step be?

[tough hatch]

we are taking the limit of something divided by one of its factors

x-a is a factor of x^2-a^2

because x^2-a^2 = ?

[harsh belfry]

right?

[tough hatch]

yes

now what?

[harsh belfry]

I must cancel out the two x-as

[tough hatch]

yes, and this works because we are taking the limit anyway

[tough hatch]

yes

so what is the limit now?

and what is it equal to?

(again)

[tough hatch]

no no

i am asking

what the limit of x+a as x approaches a is

[placid zinc]

It's important to keep the lim{x → a} when you're working on paper

Obv that's kinda tough here

[harsh belfry]

I'll be honest I'm very confused.

[harsh belfry]

ah

a+a

[tough hatch]

this becomes pretty straightforward now

yes.

which is just?

?????

[harsh belfry]

oops sorry

[tough hatch]

a + a

not (a)(a)

[harsh belfry]

I didnt mean to write that sorry!

[tough hatch]

so what is it

[harsh belfry]

a+a is equal to a+a

[tough hatch]

jesus.

[harsh belfry]

Im not sure how else to write that

[tough hatch]

so 1+1 is just 1+1

i see

[harsh belfry]

2a

[tough hatch]

yes.

but what was the original limit equal to?

in the first place

[harsh belfry]

the original limit was x?

[tough hatch]

nope.

[tough hatch]

what is this then?

[harsh belfry]

the limit of x as it approaches a is 10

[tough hatch]

thats bad wording

the limit of g(x) as x approaches a is 10

so 2a = ?

[harsh belfry]

20

[tough hatch]

huh?

[harsh belfry]

10+10

right?

[tough hatch]

2a is exactly the limit of g(x) as x approaches a

and you know that the limit of g(x) as x approaches a is 10

so 2a = ?

2a = something

something = 10

[tight locust]

(x-a)(x+a)/(x-a) = (x+a)

lim x->a (x+a) = 10

[harsh belfry]

10a?

[tough hatch]

2a = something

something = 10

[harsh belfry]

2a is 10.... ah

[tough hatch]

this something is the limit of g(x) as x approaches 10

yes.

[tight locust]

the limit of a continuous function is just its value there.

f(x)=(x+a) is continuous over R

[harsh belfry]

so the function of g(z) as x approaches a is equal to 10

the function is equal to 2a

2a=10

[tight locust]

so lim x->a (x+a) = (a+a)

[tough hatch]

yes.

[harsh belfry]

and 2a=a+a=10

a=5?

[tough hatch]

oh totally not.

jk

[harsh belfry]

I was about to just show myself out

you crushed me with that joke

oh god I just had the hardest time understanding this but I get it now

thanks for bearing with me and not giving up

Im really hard to work with I know

[tough hatch]

there are others worse than you

[harsh belfry]

I'm sorry

[rotund kraken]

Hello, our professor gave us a problem that wasn't taught to us

We really don't know what to do

[wind bane]

@rotund kraken substituting x = tan^2(u), dx = 2tan(u) sec^2(u) du seems to be like a good idea

[crystal crag]

or simply, taking x common from the bracket does the job

[thorn sundial]

do you guys know this

[tight locust]

let s = 14

then find the least positive solution to t

presumably you want the t value when the ball is ascending to 14 feet

not when it is coming back down

[ivory kiln]

does anyone have chegg here ?

[thorn sundial]

not me

[dark granite]

I can only speak for myself.....no, I don't have chegg

[ivory kiln]

aww, thanks man. i just dont know how to solve my mechanics for deformable bodies assignment

it's kinda off math topic more of physics with axial load and shear stress problems. so i really dont think some one can answer those kind of questions here. but thanks anyway

[ionic jewel]

try physics server

definitely can answer things there

until you pass undergrad level work at least it's unlikely you won't be able to find someone who can help

[ivory kiln]

thanks :>

[jagged plover]

the mode here is small m, and i'm unsure as to how we got that from this pdf

from what i know the mode is equal to the x value of the maximum point of a pdf, but because this pdf is a linear function the single derivative is a constant, and the double derivative is just 0

so i'm not sure how to work this out

[zealous flame]

heyo i need help dm me if you could help me i would appreciate it if you could help me

[jagged plover]

<@&286206848099549185>

[jagged plover]

pinged for this question

[dark granite]

wait, what grade are you in?

[hollow cypress]

any constructions of dependent paths with the help of: #discussion message

three transports in a trenchcoat
i don't think channel's latex could help me any further.

[grim raft]

Can someone help me out with this question?

[dark granite]

dependents path of what? I got lost after clicking the link

[dark granite]

here's a small but super helpful skill to learn....try explaining to yourself what that statement means in plain english. Also, your question is a question better suited for the Analysis channel.

[hollow cypress]

if you're familiar with Agda, PathP

[dark granite]

oh hmm. Sorry I'm not familiar with that

[alpine sable]

hey is this channel busy right now?

[dark granite]

kind of, is your question quick?

[alpine sable]

Kinda

So, I have a doubt about combinations with repetition

[dark granite]

okay

that's my forte, so I can answer your question

[alpine sable]

Can't I think of that as like

say we have to choose 8 things from 3 options

then 1st object has 3 options

second has 3 options

3rd has 3 options

and so on

3 options as in, it can be red, it can be blue, it can be green (say)

and I have 3^8 ways to choose them?

Apparently I am wrong https://www.cs.sfu.ca/~ggbaker/zju/math/perm-comb-more.html from this source

[dark granite]

choose 8 things from 3 options?

[alpine sable]

Yeah say I have to choose 8 dogs from 3 types of dogs that kind of stuff

[dark granite]

okay

now I'm following

[alpine sable]

the first dog can be one of the three (say, A, B, C)

[dark granite]

no because ABC is really the same as BCA and so forth

[alpine sable]

wait lemme consume what you said

[dark granite]

it all really depends if you want order to matter

[alpine sable]

order doesn't matter here

[dark granite]

okay, then 3^8 ain't it

[alpine sable]

https://www.cs.sfu.ca/~ggbaker/zju/math/perm-comb-more.html this is what I am following, I kinda need some help understanding why is it so

[dark granite]

because 3^8 would count ABC as different from BCA for instance

[alpine sable]

yes,

but

oh

Oh I get it

thanks

[dark granite]

yea you need to use stars and bars kinda idea aka indentical balls into distinguishable boxes

[alpine sable]

just another small question, why do we separate it in the example they have

this step (starred in red)

[dark granite]

this separation is how we distinguish between the varieties

[alpine sable]

okay

so basically to make sure order doesn't matter

am I correct?

i dont think I am correct

[dark granite]

yea so the order of the dividers and dots matters, but in reality this gives us the unorderdness we want because we are talking about unordered selection with replacement

[alpine sable]

oh

thanks a lot

[dark granite]

You're welcome!

[tender star]

how do I solve this?

I really need help

[hollow cypress]

does the definition PathP : (A : I → Type) → A i0 → A i1 → Type, where I is the Interval type help?

[dark granite]

it doesn't unfortunately. Thanks for trying to help me understand what you meant tho.

[dark granite]

what grade are ya in?

[hollow cypress]

if you're familiar with any category theorists here that might understand the query, may you inquire to them for aid?

homotopy foundations.. etc

[dark granite]

ohhh. Yea I'm not too familiar with category theory. Maybe you could try asking in that channel tho

[lost violet]

how can i check if $1,x,cos(x),sin(x)$ are linearly independent

[ocean seal]

alef0

[lost violet]

i make the wronksian and gives me 1

then are linearly independent right?

[tender star]

7th grade

[placid zinc]

Yes

[alpine sable]

I'm 14 in 10th

[dark granite]

so you're 12?

[grim raft]

Thanks I would have posted there but I am not able to access that channel somehow

[dark granite]

you can visit #get-advanced-access for those roles

[grim raft]

Got it. Thanks again!

[dark granite]

You're welcome!

[vale wigeon]

this looks kinda tough, ngl

[dark granite]

this seems like a math comp problem. Maybe that channel might be better.

[alpine sable]

Yeah sorry

[dark granite]

all good.

This channel is open for questions

[alpine sable]

👍 Got it

[toxic elk]

lmao wth is this

Work hard man

you got this

[dark granite]

how old are ya?

[vale wigeon]

bruh

[dark granite]

<@&268886789983436800> see above

[tame falcon]

Thanks.

[tender star]

13

Was 12 a couple months ago

[vale wigeon]

wasn't addressed to you, prem

[dark granite]

lmaoooo

it was

I asked him a while back

[tender star]

^

[dark granite]

it's funny he only replied now after the other guy got banned

but oh well lmaoo

[vale wigeon]

oh

[dark granite]

kinda funny you brought this up to hahaha

[tender star]

well good thing I didn't join last year when I was a illegal lmao

[daring roost]

can someone help me with a chemistry question?

[vale wigeon]

you can certainly try your luck with it

[daring roost]

Can someone tell me how to answer these sort of questions?

[vale wigeon]

oh so this is stoichiometry

[daring roost]

or is this too offtopic

[vale wigeon]

...wait is this a timed test/exam

[daring roost]

no

[vale wigeon]

okay

[daring roost]

not a timed test

[vale wigeon]

calculate everything in moles first

[daring roost]

ok

[vale wigeon]

,w CuO molar mass

fuck no i meant ,w

[ocean seal]

Results provided by WolframAlpha

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

Requested by Ann#0413

[vale wigeon]

there

[daring roost]

ok

is that it

[vale wigeon]

wym

[daring roost]

if its just finding molar mass i can already do that

[vale wigeon]

no it's not just finding the molar mass

calculate how many moles of CuO went in

[daring roost]

ok

[vale wigeon]

that'll also be how many moles of H2 will be needed

[daring roost]

so do their moles have to be the same

i think i did it

@vale wigeon is the answer B then

[vale wigeon]

seems like it

[daring roost]

the second one

like 1.01

ok

thanks for help

appreciate it.

[frank grotto]

$$ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}n}{(n+1)2^{n-1}} $$

[ocean seal]

NikolaJ

[frank grotto]

How can i find the sum of this series?

We did power series so I assume I have to find a power series that can become this for a certain value of x but I am no sure how to find it here

or if that is the correct approach

[vale wigeon]

x=-1/2 seems like a fine candidate? esp if you reindex the series to start with 0

you have sum[n=1,infty] n/(n+1) x^(n-1)

[frank grotto]

thanks, I'll try that

I assume I should then either find the integral or the derivative of it?

[vale wigeon]

...maybe? no idea what's best to do here ngl

[frank grotto]

is this valid?

$$ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}n}{(n+1)2^{n-1}} = 2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}n}{(n+1)2^{n}} | x = \frac{1}{2} $$

[ocean seal]

NikolaJ

[vale wigeon]

....no, not as written

[frank grotto]

why?

because I wrote $2^{n-1} = 2^n * 2^-1$ and then just extracted that out of the sum

[ocean seal]

NikolaJ

[vale wigeon]

but then x is nowhere to be seen in there

...

i don't have enough energy to fix this

[fast mantle]

is this channel free?

[frank grotto]

$$ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}n}{(n+1)2^{n-1}} = 2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}n}{(n+1)2^{n}} => 2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}n}{(n+1)} \frac{1}{2^n} => 2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}n}{(n+1)}x^n $$

[ocean seal]

NikolaJ

$$ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}n}{(n+1)2^{n-1}} = 2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}n}{(n+1)2^{n}} => 2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}n}{(n+1)} \frac{1}{2^n} => 2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}n}{(n+1)}x^n $$

[frank grotto]

here is the last part that cut off

$$ 2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}n}{(n+1)}x^n $$

[ocean seal]

NikolaJ

[alpine sable]

hello everyone, looking for some feedback regarding some questions i have about "Powers of the imaginary unit" specifically simplifying.

[jade birch]

Okay

[stuck jolt]

Why is there an x ?

You can't add a constant to this series, it cannot be valid

[frank grotto]

that x was to turn it into a power series

[alpine sable]

so when we are simplifying let say i^15 we would find the number that is a multiple of 4 (or closest to the multiple of 4) which in this cause would be 12, then we would do (i^3)^4 which would then become -i, we then multiple that -i by the reminder which is going to be i^3 which finally would = i (-i * -i would then = i) have i got this right?

[frank grotto]

then find a function representation of that power series and plug 1/2 back into it

[stuck jolt]

I see what you did now

[stuck jolt]

No

Here you would have

i^12xi^3

So its just -i

[frank grotto]

is that the correct approach then?

[velvet condor]

apparently im trying to find a graph that looks like this curve....... any idea what type of equation could it be?(except from ecliptic ones)

[stuck jolt]

I am actually not sure

[velvet condor]

hmm

[alpine sable]

is this not i? as -i * -i would therefore = i

[stuck jolt]

I haven't looked at these series in a while

[velvet condor]

ill just wait then....

[stuck jolt]

I don't get what you are saying

i^15=i^12xi^3

That just simplifies to 1xi^3

and i^3 is -i

So the answer is -i

[alpine sable]

yeah so i was right lmao

[stuck jolt]

No

[alpine sable]

wait hold on i need to double check

[stuck jolt]

You say the answer is i

It isn't

The answer is -i

[alpine sable]

hold on i am just taking a look at my notes, only started this yesterday

[alpine sable]

hey thanks man you are right just need to double check how to get there, thanks!

[alpine sable]

also i have another question when we are trying to simplify whatever to the ^i are we always looking for a multiple of 4?

[stuck jolt]

Yh, it is often the case

Even later on in advanced math, you look for powers of 4 w complex numbers

[alpine sable]

right okay interesting oh and final question, you know how with i^15 we found our multiple of 4 which was 12, and therfore we have to multiple our reminder can the reminder be bigger than 3 for example?

like what if our remainder was like 15 or something or would that not happen?

[mystic ravine]

hii

i am new here

hope you dont mind

[alpine sable]

go to general category-
you ask doubts here

[mystic ravine]

okk thx

[civic galleon]

can you help me? my study questions

[stuck jolt]

Well, in the case of i^15, there is no remainder

I don't really know what you mean by remainder

Usually a remainder comes w long division, but in the case of your questions there is none of that

[alpine sable]

the remainder being 3 because 4 goes into 15 3 times with a remainder of 3

[stuck jolt]

There cannot be a remainder of 3

We discussed this already

i^3 is -i

I don't understand how you get the remainder of 3

[tight sphinx]

does root 2 times root 2= root 2? Or just 2?

[velvet condor]

bruh

its 2

${\sqrt{2}}^2=2$

[ocean seal]

Minh Baka

[halcyon matrix]

Can someone please tell me what constitutes as a small angle?

[small trellis]

Less than 90 degrees I would assume

[jagged imp]

as in in the context of the small angle approximations? Depends on what you're okay with for error

at less than 10-15 degrees, small angles are probably good enough for most things

[small trellis]

I wonder if he's talking about an acute angle or not

[jagged imp]

i doubt they're talking about acute angles

[small trellis]

But if there's a separate term that is "small angle" that I'm not familiar with then I should probably stop talking

[jagged imp]

but its just speculation until they reply

[halcyon matrix]

Yes, small angle approximations

@jagged imp

[small trellis]

"more typically, saying 'small angle approximation' typically means θ≪1, where θ is in radians; this can be rephrased in degrees as θ≪57∘"

I found this on google

[halcyon matrix]

Ah, so less than 1 radian/less than 57 degrees?

[jagged imp]

uh that could be a pretty big error

at 1 radian

[small trellis]

Depends on what measure you're using

[halcyon matrix]

I can't remember what the exact question was but I remember in an exam a few months ago there was something like "Is 278 degrees appropriate for small angle approximation or not?" and apparently the answer was "yes" so is there something to do with a period/cycle??

Or is there only one range for small angles

[jagged imp]

the reason is that the real angle you're worried about there is 8 degrees

which is absolutely fit for small angle

[placid zinc]

If you're trying to keep something at 270, then 278 is a small error

[jagged imp]

270+8=278. If you know sin(8 deg) (using small angle, mayhaps), you can figure out cos(278 deg)

and vice versa

using trig identities

[halcyon matrix]

I see. Then what doesn't constitute as a small angle?

[jagged imp]

I mean 278 doesnt constitute as a small angle

but it is fit for the small angle approximation

but seriously, something like sin(45 degrees)

would not be very well estimated using small angle

if you're more than 15 degrees from a multiple of 90

you're probably not gonna have much luck

[halcyon matrix]

I think I get it...?

Why from a multiple of 90

[jagged imp]

since If I have the trig functions evaluated at x, I can figure out the trig functions at x+90deg , x+180deg, x+270 deg etc

for instance, sin(45)=-cos(90+45)

[halcyon matrix]

sin(90-45)?

[jagged imp]

sorry typo

typoed twice, it is correct now lol

so if i want to evaluate cos(135), and I know sin(45), I can work that out

[halcyon matrix]

But 135 is more than 15 degrees away from a multiple of 90

[jagged imp]

Replace 45 with 5

what I'm saying is that using that principle I used to evaluate cos(135), you could evaluate cos(95), using small angle to get sin(5)

[halcyon matrix]

OK I see

Thanks Luffy pfp

[jagged imp]

yw

[placid zinc]

I mean small angle is as good as x - sin(x). For a balancing bot, this is pretty precise even for pretty large angles. For a mars rover, you might never want to use this

[slow girder]

Anyone wanna help me?

[slow girder]

<@&286206848099549185>

[alpine sable]

can someone help with c

[glass lichen]

did you do b?

also is it an assessment?

[alpine sable]

na homework

[alpine sable]

ye

i got

(x- 1/2) ^2 - 17/4

[glass lichen]

yes, so now the quadratic is in vertex form

so read off the vertex

[alpine sable]

idk what that means

[glass lichen]

completing the square puts a quadratic in vertex form

[alpine sable]

ye

[glass lichen]

so.. tell me what the vertex is

[alpine sable]

im not american so we dont use them words

oh u mean the intercepts?

[glass lichen]

No, I mean what I said

[alpine sable]

💀

[glass lichen]

the turning point

[alpine sable]

ohhh oh ye

[glass lichen]

the min/max point, whatever

[alpine sable]

i forgot lol been a while

its 1/2

[glass lichen]

that's the x component, yes

I asked for a point though

[alpine sable]

one sec

[glass lichen]

$y=a(x-h)^2+k$ has vertex $(h,k)$

[ocean seal]

Mosh

[alpine sable]

oh

i subbed x= 1/2 into it

to get y

i got - 17/4 lol

ur ways easier tho

[glass lichen]

yeah.. so (1/2,-17/4) is the minimum point

then just read off the axis of symmetry

[alpine sable]

so the line of symettry is where x= 1/2

?

[glass lichen]

yes

x=1/2 is the line of symmetry

[alpine sable]

ty

can someone help with this too

i got this so far

now idk what to do next

[median tendon]

Maybe try to get rid of the fractions

[alpine sable]

let me try expand

[median tendon]

You now have same denominator

Combine, then multiply denominator to other side

[alpine sable]

okay ima do that

[past prism]

would anybody be able to help at #help-6?

[alpine sable]

i got

do i use quadratic formula

oh nvm

im dum

2x +2 = 0 so 2x =-2

so x= -1

and x =2

can someone help with this

[crystal crag]

take x^(1/3) as some variable say, "t" and solve the quadratic in terms of "t"

[alpine sable]

oh ye okay thanks

ive forgot all this stuff since summer holidays

[crystal crag]

np

[alpine sable]

i got x =1

or

x = cube root of 16

,w x^(2/3) - 17[x^(1/3)] +16=0

[ocean seal]

Tra-Guy

[alpine sable]

,w cube root 16

[ocean seal]

Results provided by WolframAlpha

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

Requested by şaka şaka#6731

[alpine sable]

i dont get why it’s giving answer as 4096

stop im using this wtf

[sour peak]

ohh ok

I thought you're done sorry

[ocean seal]

Results provided by WolframAlpha

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

Requested by şaka şaka#6731

[alpine sable]

whys it saying x=4096

[kindred warren]

because it is one of the solutions

[crystal crag]

bcoz its a solution, as Tra-Guy said

[indigo warren]

hey guys, how would you attack this?

the answer is ||64 + 31 + 61||

could you pls explain where we got those numbers?

thanks :)

[alpine sable]

ye ik but i got root 16

cube root 16

[crystal crag]

no

you will get x^(1/3)=16

[alpine sable]

ye

[crystal crag]

so now for x, you need to cube both the sides

and not cube root

[alpine sable]

oh ye

thanks

[indigo warren]

<@&286206848099549185>

[dark cave]

Is that correct ?
And here is modulas and argument gonna be 1 ????

<@&286206848099549185>

[iron loom]

need help wwith the formula please

<@&286206848099549185>

[vapid oak]

does the question specify that point P has to lie on the line AB?

[silver dew]

can somebody tell my what is the answer pls?

[iron loom]

not entirely sure

do i need to use the section formula?

[vapid oak]

yes

[iron loom]

can you help with with an example with a?

ill try my best to understand so@i can do the b and c

[vapid oak]

yes of course

so I'm going to write out the section formula and you tell me if you have learnt it with different pronumerals

[sand pawn]

Can someone tell if the following is true?

[iron loom]

okay!

[vapid oak]

could you please go onto another channel? we are currently solving a problem here. maybe go onto channel 1

[daring roost]

when it says 192 of magnesium do i multiply that by 2 because in the formula magnesium has 2 or is 192 both of them already

[vapid oak]

same applies to you

[vapid oak]

.

$P(x, y) = (\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}$

[ocean seal]

Omfish

[vapid oak]

where m and n are the ratios of the side lengths

[vapid oak]

sorry it took a while. im just replying so you get a ping to see what ive written

[daring roost]

what

[vapid oak]

this. When a channel is in use, its best to go into an empty channel to ask questions

[daring roost]

what do you mean by '.'

[vapid oak]

i was just replying to your message for you to see

[daring roost]

ik

[wary stream]

That person is ping replying to your message, trying to tell you that this channel is currently busy so ask in a not busy channel

[daring roost]

so which one is correct

[iron loom]

its fine its fine

[daring roost]

is it free now

[vapid oak]

we are still in the process of answering the question so no. Maybe go onto channel 8.

[vapid oak]

$P(x, y) = (\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}$

[ocean seal]

Omfish

[vapid oak]

is this what youve learnt?

[iron loom]

yeah!

but the prof didnt explain

how to use the formula ;-;

just gave it to us

[daring roost]

is this multivariable calculus

[vapid oak]

ok so basically you have two points given to you in the question, A and B with coordinates

[iron loom]

uhuh

[vapid oak]

so A(4, 3) and B(5, 1) these are x1, y1, x2 and y2. So in this example x1=4 y1=3 x2=5 and y2=1

[iron loom]

mhm

[vapid oak]

and then you want to split the length AB into a ratio of 1/3 so the length of AP will be 1/3 and therefore the rest of the line must be 2/3. so your m factor is 1/3 and your n factor os 2/3

so subbing that it

$P(x, y) = (\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n} = (\frac{\frac{5}{3}+\frac{8}{3}}{1} , \frac{\frac{1}{3}+\frac{6}{3}}{1}$

[ocean seal]

Omfish

[iron loom]

mhm

[vapid oak]

so the point is $(\frac{13}{3} , \frac{7}{3})$

[ocean seal]

Omfish

[iron loom]

OOOOOO

ok i get it know thank you @vapid oak

[alpine sable]

How would u approach question 5

I’m clueless 😥

[vapid oak]

do you remember the rule that loga+logb = log(ab)?

try and use that rule in reverse...

[alpine sable]

Oh ok

What do i do next

[devout sigil]

Simplify a and x

[wary stream]

I don't think that's the correct use of log properties

[vapid oak]

it is correct, however not going to get you to the solution

[devout sigil]

It is if u write it in a fraction

[wary stream]

I'm used to fractions, not division signs, my bad

[vapid oak]

$log_a(a^2 \sqrt{x}) - log_a(x^2 \sqrt{a}) = log_a(a^2) + log_a(\sqrt{x}) - log_a(x^2) - log_a(\sqrt{a})$

[devout sigil]

It would be simpler to just simplify the powers of a and x before splitting them up again

[ocean seal]

Omfish

[alpine sable]

/

Like that

[vapid oak]

that is correct, however remember to expand the brackets, so that last sign is a negative not a plus

[alpine sable]

Uh

What brackets?

[vapid oak]

sorry i mean that the negative sign carries out

so the entire second term of the original problem was subtracted, so when you expand that term make sure each piece is minused too

[alpine sable]

Ohh I got it

Am I in the right direction?

[vapid oak]

yes exactly

perfect!

[alpine sable]

I’m guessing it would b

Alr I got it

TY!

[vapid oak]

yay!

[alpine sable]

@vapid oak I have 1 more question

[vapid oak]

ofc

[vapid oak]

ahh this is a nice question

[alpine sable]

It’s says to give exact question

[vapid oak]

maybe before I help you, what do you notice about the exponents?

[alpine sable]

It’s has an e

[vague coral]

exponent*

not base

[vapid oak]

sorry, what do you notice about the number in the power

or index you may call it

[alpine sable]

Oh

[vague coral]

no

[alpine sable]

The power of x

[vapid oak]

if I write this out, does it give you an idea:

$(e^x)^2-2e^x=0$

[ocean seal]

Omfish

[alpine sable]

I remember I used to ln( both sides to solve for it.
But I forgot how to do it

[vapid oak]

i dont think ln is necessary. Do you understand what I've done here and why it is true?

[alpine sable]

We have to do it with factor theorum specifically not with long division.

[alpine sable]

Ye

[vapid oak]

please go onto channel 7 or another empty channel

[alpine sable]

If an exponent has an exponent then they multiply

Ok sure

[vapid oak]

ok so what do you notice about this equation?

what does it look like?

[devout sigil]

Maybe replace e^x with y can help u

[vague coral]

That sentence makes no sense, but I guess that you have the idea xd

[glass lichen]

inaccurate phrasing

[alpine sable]

Ok cmon

[glass lichen]

$a^{b^c}\neq (a^b)^c$

[ocean seal]

Mosh

[vapid oak]

$(e^x)^2-2e^x=0$ Let e^x = u, then u^2-2u=0$

[ocean seal]

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

[silver dew]

somebody can tell me how simplificate this?

[vapid oak]

please go onto an empty channel to ask this. this channel is currently in use

[alpine sable]

I got u= 0, u =2

[vapid oak]

yea exactly

so now subsitute back in. u = e^x = 0 or 2

[vapid oak]

you want to apply ln to both sides

so if e^x = 0 we get x=ln0 which is undefined

e^x = 2 we get x = ln2 which is our answer!

[alpine sable]

What do u so after this

[vapid oak]

apply log laws

so you bring the exponent to the front of the log.

x * lne = ln2

but lne = 1

so x = ln2

[alpine sable]

Oh then ln(e) become 1

Oh ok I got it

Thank you so much 😊

[vapid oak]

my pleasure!

[wary stream]

Sounds like a person who works at Chick fil A(if you know what that is)

[tawny lion]

hey

I am confused on where to start

[fringe ember]

@tawny lion try using induction

[tawny lion]

how would I do that?

[fringe ember]

what would s(n+1) be when you have s(n)

use the conditions given in the prolem

[tawny lion]

but my question is regarding i)_

[alpine sable]

Sure

Find S1 then express S2 with S1

[tawny lion]

I'm having trouble finding that

could you help me?

[alpine sable]

Let's see

You have a interest rate of I so our ammount after a month is P+P*I/100

Then there is the repayment so the amount is reduced by R

[tawny lion]

right

[alpine sable]

So you have S1 the ammount for the first month

[tawny lion]

it would be $P(1+\frac{I}{100})^t - R$ ?

[alpine sable]

Then you take S1 as the initial ammount to find S2

[ocean seal]

kurama

[alpine sable]

what is t

[tawny lion]

t is 1

how do I find it for 2 though

do I just input 2?

[alpine sable]

no