help-10

[inland matrix]

Coz they can be in any interval in general

Oh

[ember frost]

ah youre damn right

i think i can take it from here

many thanks

[brazen gorge]

np

[ember frost]

.close

[obtuse pebble]

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[sick badge]

Given an M/M/1 queue with arrival rate lambda and service rate mu. If a customer arrives at a random time and the queue is empty, and we don't know if the server is busy, how do we calculate the probability that the server is busy?

[timid silo]

seems like a poisson process

and they want you to use an Erlang distribution or something

[sick badge]

Suppose arrival rate is 1 and service rate is 3 for system stability how would I calculate the probability the server is busy when the queue is empty?

[obtuse pebble]

@sick badge Has your question been resolved?

[obtuse pebble]

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[shy shoal]

was wondering on how to do this type of question for polynomial functions?

[jovial portal]

You can do some equations for a general polynomial of order 5 and slove recursivly

[high lily]

identify the zeros,
respective factors from factor theorem, then their power based on the shape of the curve around that zero
introduce scaling factor, determine that from the given g(0) = -2

[shy shoal]

1 -1 and -3

so my original equation is p(x) = (x+1) (x-1)^2 (x-3)^2

cause those tare the zeroes but idk

[jovial portal]

Looks good now input the g(0) = ... and you get the scaling factor

actually i thing it should be $p(x)= a \cdot (x-1) (x+1)^2(x+3)^2$

[warm shale]

Tobi

[high lily]

you're not applying factor theorem correctly

if $k$ is a zero, $(x\red{-}k)$ is a factor

[warm shale]

ℝαμΩℕωⅤ

[shy shoal]

so my original equation is not complete yet?

im not sure on what to do after that

[high lily]

its outright wrong

[shy shoal]

oh crap

[high lily]

because you didn't aply factor theorem properly

[shy shoal]

im not so sure what to do

i know my zeroes are

1 -1 and -3

[high lily]

do you know what factor theorem states?

[shy shoal]

not really

[high lily]

i made an extremely brief summary above

[shy shoal]

ok so if i take example of 1 as a zero

then x-1 is a factor

[high lily]

yes

and from the graph what's the multiplicity of that zero?

[shy shoal]

1

since it passes through

[high lily]

yeh, now move onto the next zero

[shy shoal]

so -1 as a zero

x+1 is a factor

and since it bounces the multiplicity should be 2

[high lily]

yes

[shy shoal]

so x+1^2

[high lily]

()

(x+1)^2

[shy shoal]

oh yeah mb

then

(x+3)^2

ohhh i see now

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

?

[high lily]

you'd want a scaling factor

[shy shoal]

what's that?

[high lily]

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

where a is to be determined

[shy shoal]

oh i see

[high lily]

what you currently have is a specific polynomial with those zeroes that doesn't necessarily satisfy the other condition

[shy shoal]

so we find a

but the y intercept isnt given

[high lily]

you're given g(0) = -2
don't know why they used different names here

[shy shoal]

oh but it said g(0) = -2

so do we replace p(x) with -2?

[shy shoal]

probably a mistake with

the reviewer

[high lily]

you replace x with 0 and p(x) with -2
in the same step

[shy shoal]

ohhh i see

so

0 = a(-2 - 1) (-2+1)^2 (-2+3)^2

[high lily]

no

you replaced things the other way

you replaced x with -2 and p(x) with 0

which isn't what you're supposed to do

[shy shoal]

oh so

-2 = (0-1) (0+1)^2 (0+3)^2

[high lily]

yes

[shy shoal]

oh yeah my bad since p(x) is 0

i get it

i got 2/9 as the a

so p(x) = 2/9 (x-1) (x+1)^2 (x+3)^2

[high lily]

yes

[shy shoal]

thats our complete equation?

[high lily]

yeh

[shy shoal]

thanks a lot!

[obtuse pebble]

@shy shoal Has your question been resolved?

[obtuse pebble]

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[blissful bane]

How do I find the integrating factor for a differential equation of the form M dx + N dy =0 to make it exact ?

[inland matrix]

Are M and N constants or functions?

[blissful bane]

functions of x and y

M(x,y) and N(x,y)

[inland matrix]

Oh

[obtuse pebble]

@blissful bane Has your question been resolved?

[obtuse pebble]

@blissful bane Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[sullen sparrow]

hi! i am studying vectorial subspaces and i dont understand why P3(x) is contained in P(x). isnt P3(x) greater than P(x)?

[pseudo swift]

alright

so what do you think P(x) and P3(x) is then ?

[fallow island]

by contained you mean it being a subspace?

[sullen sparrow]

polynomials

[sullen sparrow]

yes

[pseudo swift]

sure

that's not very specific though

that doesn't tell me what the difference between P(x) and P3(x) is

[sullen sparrow]

like

isnt p2(x) a polynomial like c + x + x^2 and p3(x) like c + x + x^2 + x^3?

[pseudo swift]

ok ok

well p2(x) has all polynomials of degree less than or equal to 2

p3(x) has all polynomials of degree less than or equal to 3

etc...

[sullen sparrow]

yeah yeah

[pseudo swift]

otherwise they're not vector spaces at all

and P(x) contains all polynomials

whatever their degree

[sullen sparrow]

hmm

[pseudo swift]

so why would P2(x) not be contained in P(x) in that case

[sullen sparrow]

so, p(x) is like R for real numbers?

[pseudo swift]

not sure what you're saying here

[sullen sparrow]

uhm

im still not sure what p(x)

like

its contains all polynomials, but like

its a universe of polynomials then??

[pseudo swift]

I guess

but like P3(x) would also be a 'universe' of polynomials if I understood you correctly, there are also polynomials inside of it

[sullen sparrow]

yes

[pseudo swift]

so maybe you should clarify what you mean by 'universe of polynomials'

okok

[sullen sparrow]

yeah like

the biggest one

i got it

idk how to verbalize my thoughts, but i got it

tysm

[pseudo swift]

but yeah P(x) is the biggest one indeed

you're welcome

[sullen sparrow]

.close

[obtuse pebble]

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[placid rose]

Can you help me understand why this is wrong?

[placid rose]

I was testing it and it says I’m wrong

[obtuse pebble]

@placid rose Has your question been resolved?

[placid rose]

no

[obtuse pebble]

@placid rose Has your question been resolved?

[placid rose]

No

Y’all don’t help

[placid rose]

Oh wait

Bro

I know I messed up but

It’s okay

I can close this

And nobody will remember

It will fade away into obscurity

And be gone 10000 messages away

.close

[obtuse pebble]

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[hoary idol]

right isosceles triangle there is a circle on it that circle toches 1 side and its center is at the hypotnus and the circle also toches the shatrp corner i need to find the radius

[obtuse pebble]

Please don't occupy multiple help channels.

[hoary idol]

.close

[obtuse pebble]

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[cinder ridge]

look at the cylinders part of my teacher's work

the formula for the cylinder is 2pi * r * h * delta x

i can kind of see why he didn't include delta x in the integral function

but in this case should we have included it or not?

[obtuse pebble]

@cinder ridge Has your question been resolved?

[cinder ridge]

he used the formula again and he made delta x magicaly disappear

[obtuse pebble]

@cinder ridge Has your question been resolved?

[obtuse pebble]

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[surreal mountain]

How do i approach this?

(The anser should be written as x + iy)

Im thinking de moivres

But its my first time using it

So im not sure what to do

[brazen gorge]

yes

convert to either cis form or polar form

each bracket

[surreal mountain]

Ok

That will take some time

Working on it

I want to rewrite the 25 as something smaller

But i have messed it up before

What is a good method for doing it?

[brazen gorge]

cos(25pi) = cos(pi)

same with sin

[surreal mountain]

But how would i write it out?

[brazen gorge]

not sure tbh

but do you understand why that is?

[surreal mountain]

I have a small clue

Both functions are periodical

By 2Pi

But how do i know its not = cos(0)?

Do i think of the closedt number possible to 25 (24) and the diffirence will be the answer?

25Pi - 24Pi = 1Pi

So cos(Pi)?

[surreal mountain]

How would you do it if it was cos(163Pi/6)?

[brazen gorge]

you subtract as many multiples of 2pi as possible

aka reduce mod 2pi

[obtuse pebble]

@surreal mountain Has your question been resolved?

[surreal mountain]

.close

[obtuse pebble]

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[glacial widget]

Pleaseeee help with this question.
Its intergration by parts. ill send the question and then my work. I cannot understand for the life of me whats going wrong lol

[glacial widget]

let me send the photos from my phone

so i did the box where i chose
e^-x as my intergration.
and 3x-4 as my dx

so that should be

d = 3
intergration = -e^-x

so using intergration by parts it should be

= (3x-4)(e^-x) - INTERGRATION OF: (3)(-e^-x)

multiplying inside the intergration before to simplify

its then

= (3x-4)(e^-x) - INTERGRATION OF: (-3e^-x)

then intergrating it its

should be

(3x-4)(e^-x)-(3e^-x)+C

but thats wrong. im so confused lol

any help is appreciated! pls ping me ❤️ thanks

[obtuse pebble]

@glacial widget Has your question been resolved?

[alpine oar]

@glacial widget can you try multiple answers?

does it have to be simplified a certain way or is any correct answer acceptable?

[glacial widget]

yeah

i can try any answers

[glacial widget]

any way

[alpine oar]

well okay for starters do you know the tabular method?

[glacial widget]

no

i was only taught this square thing

[alpine oar]

okay so you need to do integration by parts, you divide the the integral into two parts, the first part will be the bracket, and the second part will be the exponential function which is the (e^-x)

you select one to differentiate and one to integrate

ill show you on paper hold on

[glacial widget]

heres another example

i just dont understand wtf im doing wrong

like this is literlaly right

[alpine oar]

okay hold on let me try it

[glacial widget]

i just dont really undesrtand why youre suppsoed to multiply the stuff inside the intergration before intergrating

ok i solved the firstt one

i need the 2nd one now

thats hte one im really not understanindg

[alpine oar]

hey sorry im late

[glacial widget]

its ok!!

[alpine oar]

can you check the last answer and see if it will work, or does it need to be simplified

if it does, then I think I can explain it to you

[glacial widget]

lemme put it in

its gonna take me a bit to type with parenthese one sec lol

[alpine oar]

alr tyt

[glacial widget]

oops

wait

wrong e^x

[alpine oar]

its

e^4x

also put the parentheses around the whole thing except the C

[glacial widget]

my prof is gonna be sooo confused

LOL

[alpine oar]

uhh

the e^4x is supposed to be a common factor of the 3 fractions, is it written that way?

its not in the preview is it a bug

[glacial widget]

what

what does that mean

😭

[alpine oar]

you have 3 fractions right

[glacial widget]

yes

[alpine oar]

they're all supposed to be * e^4x

[glacial widget]

oh

yes

i added anotehr parnthesis

in that case its right

my bad

[alpine oar]

hmm

I wonder what the problem is idk

[glacial widget]

but i have no idea how u got to that lol

bc the way im taught is this stupid square theory

[alpine oar]

wait

is it correct?

[glacial widget]

uya

ya

[alpine oar]

OH OK NOW I CAN EXPLAIN

[glacial widget]

please dfo

Ll

[alpine oar]

okokok do you learn better by visuals

[glacial widget]

lol

[alpine oar]

should I open paint

[glacial widget]

NO LOL

the way i was taught is thi weird

square thign

[alpine oar]

no no I mean like you want me to type it or open paint and explain on it

[glacial widget]

forgive the bad handwriting but

[alpine oar]

its fine lol

[glacial widget]

you differentiate the algebric term and intergrate the e

and then

its

a diagnol multiplication

and then - intergration of the multiplied bottom two

[alpine oar]

yeah but this complicates stuff

[glacial widget]

😭

[alpine oar]

hold on

ill write it down

[glacial widget]

I appreciate your help but I have to show my work in the stupid box way. ill just go to the Ta's or something unfortuantely

[alpine oar]

this method is similar to it

[glacial widget]

i just do not understand why i am getting it wrong

even using the box way

my work is right

[alpine oar]

so the integration of the multiplied bottom two yeilds another box right?

[glacial widget]

uhh

what

[alpine oar]

when you start out

you do the box thing

its the product of the diagonal

and a new integral

[alpine oar]

this

[glacial widget]

yea

yes

[alpine oar]

okay so the new integration

is a new box

[glacial widget]

yes

[alpine oar]

did you do the new box

[glacial widget]

i dont

do new boxes

i just

writ in the formula

[alpine oar]

okay because the new box yeilds another box which is the last one in this case

but instead of boxes

you can write a table

it's better

let me show you

okay start by typing it like this

[glacial widget]

😭 i have no idea what im looking at

[alpine oar]

up top is the integration question

you draw a line splitting the page

[glacial widget]

heres an example of one i just got right

[alpine oar]

column on the left (D) stands for differentiate, column on the right (I) stands for integrate

[glacial widget]

(all by myself yay go me)

[alpine oar]

yay go you actually

[glacial widget]

LOL

[alpine oar]

okay did you write it

[glacial widget]

I dont understand waht that one is trityng to say though

[alpine oar]

wdym

[glacial widget]

like i dont get the + - thing

[alpine oar]

yeah thats part of it

when you write the table

you alternate between + and -

let me take you step by step hold on

you split the page

[alpine oar]

and put this

[glacial widget]

right

which is the box thing basically

[alpine oar]

yes

you add the + and -

start with the +

and alternate

you dont usually use them all, but add a bunch just to make sure

thats step two

got it?

[glacial widget]

ys

yes

[alpine oar]

okay now you have 2 parts of the integration

which should we differentiate and which should we integrate

[glacial widget]

well

LAE

Dx the x^2-7

[alpine oar]

alright

[glacial widget]

and intergrate e^4x

[alpine oar]

write it down and show me

[glacial widget]

where do i put them

on the

• thing

like just the next line

[alpine oar]

on the first +

[glacial widget]

OH

WAIT

CRAP

its

that rule is so stupid i hate it sm lol

[alpine oar]

okay now when you have a function that can reduce to 0 by differentiation

you keep going till you reach the 0

[glacial widget]

whaaat

😭

[alpine oar]

you have x^2 - 7 right

[glacial widget]

yes

[alpine oar]

when you keep differentiating that

whats the last product you're gonna get?

[glacial widget]

ohhh

2

[alpine oar]

and then?

[glacial widget]

0

[alpine oar]

yes

you reach the 0

[glacial widget]

write that on the next lines?

[alpine oar]

that's your first stop

[glacial widget]

• 0

OMG STOP DISCORD

[alpine oar]

you stop when you reach 0

[glacial widget]

minus -

0

[alpine oar]

yeah you keep differentiating

you have 2x now

its gonna be 2

[glacial widget]

omg but intergrating is gonna take forever

[alpine oar]

then 0

[glacial widget]

for the

next lines

[alpine oar]

no not that long

you just divide by 4 each time

do that and show me

[glacial widget]

whaaat

dnt u have to susbtituion rule

and stuff

[alpine oar]

hmm?

[glacial widget]

its gonna taek forever

(1/4)e^4x

[alpine oar]

what substitution rule

[glacial widget]

to intergrate that

uhh

u = 4x

[alpine oar]

to integrate that

[glacial widget]

u' = 4

dx = 1/4 du

that one

[alpine oar]

uhh

okay what I do is

I put the e^4x

and then divide by the differentiation of the power

which is 4

[glacial widget]

☠️

😭

i think i may have to go to the TA because I think learning a whole other way of this is gonna harm me for my finals

like if my professor wants to see this stupid box method

i got all other 8 questions its just this one really messing me up

[alpine oar]

well yeah if it's gonna harm your finals to do it this way then yeah maybe not

but it's just so much easier

[glacial widget]

I understand :(. and i respect u sm for trying to help me but i gotta go to the TA to learn the way he wants us to learn it which is so.. ugh

tysm ! tho

[alpine oar]

oh no problem at all

[glacial widget]

u gave me the right answer which led me in teh right direction so

i appreicate it sm

[alpine oar]

I can link you a video in case you wanna learn the tabular method

[glacial widget]

Lol ill remember it before calc 2

[alpine oar]

alright

[glacial widget]

hopefully

[alpine oar]

have a good day/night!

[glacial widget]

ty1

how do i mark as done

[alpine oar]

.close

[glacial widget]

.close

[obtuse pebble]

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[obtuse pebble]

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[burnt ruin]

I need help with mathcad. I'm trying to plot a piecewise function and have the function peg out if it exceeds a max val or goes below a min. I have the piecewise itself working but I can't figure out why my other program isn't working

[obtuse pebble]

@burnt ruin Has your question been resolved?

[obtuse pebble]

@burnt ruin Has your question been resolved?

[obtuse pebble]

@burnt ruin Has your question been resolved?

[burnt ruin]

Bump

[burnt ruin]

figured it out

range var was wrong

[obtuse pebble]

@burnt ruin Has your question been resolved?

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[coarse thicket]

How do I tell if the coefficients of the sine or cosine term in a fourier series are zero or non zero?

[gilded needle]

can you give an example?

[coarse thicket]

f(x) = 0 for 0 <= x <= 3, 1 for 3 <= x < 6 where f(x) = f(x+6)

By looking at the graph of periods of a function similar to this how would I be able to tell the above?

[gilded needle]

one thing to observe is that if you subtracted a constant 1/2 from this function, it would be an odd function

therefore all the cosine coefficients (excluding n=0) would be zero

so the fourier series would be of the form 1/2 + sum(n=1 to infinity) b_n sin(omega n) for an appropriate choice of omega

[coarse thicket]

hmm i think i get it

what about a function that looks like 4 for 0 <= x < 3, or -4 for 3 <= x < 6 same f(x)=f(x+6)

[gilded needle]

same deal except the constant term would be 0 instead of 1/2

[coarse thicket]

why is constant term 0 here?

or wait i see it nevermind

thank you

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[obtuse pebble]

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[tidal relic]

how do i do this

[tidal relic]

i only know u sub rule rn

[fathom flicker]

The binomial theorem is all you need here

[high lily]

id ibp this

[fathom flicker]

but maybe it'd be better to usub u=x+8 or ibp

(then u-8=x)

[tidal relic]

whats ibp

[fathom flicker]

integration by parts

but you likely haven't learned that yet

[tidal relic]

oh

[tidal relic]

OH

wait

do i then do du - 8 = dx

[timid silo]

Be sigma and do

Pascal’s triangle

And distribute the x

😈

The do some power rules

That would go lit

In all seriousness I think setting u = x+8 gets you off and running

Wait

No I don’t think it does

[tidal relic]

i got to the integral of x(u)^8 dx and then

integral of x(u)^8 -8 du

idk how to get the X out

[timid silo]

Yea

You don’t know

Integration by parts

Correct?

(u-8)(u)^8du

Hmm

I wonder what we can with that

Perhaps multiply out the u^8

And go from there :P

[tidal relic]

ive literally only been taught u sub

ahhhhhhhhhh

[timid silo]

Lmk when u get it

I think I have a rough idea of the answer

[brave bramble]

Continue with this

[timid silo]

He got it

[tidal relic]

integral of x(u)^8 -8 du?

[brave bramble]

Oh you might be missing that
x = u - 8

You can't leave an x in there after your u-sub

[tidal relic]

OHHH

(u-8)(u)^8 -8 du?

[brave bramble]

Wait where's the -8 from

[tidal relic]

du-8=dx

or is that wrong

is du=dx

okay

cuz i used x=u-8 on accident

and it confused me

[brave bramble]

∫ (u - 8)u⁸ du

[tidal relic]

i see

[brave bramble]

Which is easier to get than the original, for sure

[tidal relic]

so now

do i distribute

[timid silo]

Yes

That’s what I said earlier

[tidal relic]

thank u

and everyone else

thanks

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[dull void]

how you solve this derivative

$e^{x^2}$

[warm shale]

odokawa

[dull void]

if you could provide me a rule or something it would be great

i have the rule

[sage geode]

Chain rule

[dull void]

WHAAT

why?

[sage geode]

Because you have a composition of two functions

[dull void]

there is a rule which say

ok

so it would be

$e^{x^2} . 2x$

@sage geode if not let me know

[sage geode]

What's the derivative of x^2?

[dull void]

2x

[sage geode]

Right, so it's $e^{x^2} \cdot 2x$

[warm shale]

odokawa

A Lonely Bean

[dull void]

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[dull void]

.reopen

[obtuse pebble]

✅

[dull void]

@sage geode what about if we have

$e^{x^2} . 2x$

[warm shale]

odokawa

[dull void]

and now we have to derivate it again

[sage geode]

Product rule and chain rule

[obtuse pebble]

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[pure nymph]

,rotate

[pure nymph]

Guys how do I solve this?

,rotate

[obtuse pebble]

@pure nymph Has your question been resolved?

[obtuse pebble]

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[timid silo]

What's the derivative of this function?

[timid silo]

I think it's -5

But according to my book, it should be 5

give us some information on what the letters are

are P_K and P,_D independent variables from each other?

yeah

it's an economic function

and I need to take the first derivative of the function

okay when you say "find the derivative of the this function" you are being ambiguous since you have to specify with respect to what also

alr, sorry

it could be with respect to P_K or P_D

hmm

if it is with respect to P_D, you treat P_K as constant and the answer is indeed 5

okay yes it is with respect to P_D then

treat P_K as a constant and differentiate

I get it

so if it would be /aP_K

it would have been -10

yes

alright

thank you a lot

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[mellow olive]

I don't really understand where to start/how I can use double formula

[mellow olive]

just never see this layout of question before

[shut lagoon]

Use the fact that sin(π-x) = sin(x).

Start from sin(2π/5) and apply the double angle formula.

[mellow olive]

wait what? do i spliy the 2pi/5 and go pi/5 + pi/5?

💀 OH WAIT

my bad

ik what u mean now

like pi/5 is theta

@shut lagoon

so im left with 2sin(pi/5) cos (pi/5)

whats next

[shut lagoon]

Try the identity I gave you earlier.

[shut lagoon]

This

[obtuse pebble]

@mellow olive Has your question been resolved?

[mellow olive]

bro idk ngl

idk how to use t he identity

😭

im just not seeing it

[shut lagoon]

What is pi - pi/5?

[mellow olive]

pi4/5

OH

dang

💀

[mellow olive]

is there also an identity like that for cosine?

[shut lagoon]

Hum

Maybe

cos(x) = cos(2pi - x)

[mellow olive]

where x is pi/5 right?

[shut lagoon]

Yes that will give you the first identity

[mellow olive]

what i don't really get is why you couldnt just stay with 2 sin pi/5 cos pi/5

[shut lagoon]

Because it asks to show that sin(2pi/5) = 2cos(pi/5)sin(4pi/5)

[mellow olive]

yes but if u didt know the idetity the u wouldnt rlly know what to do right

well at least in the textbook i didnt see that identity

oh wait i mean

not really in that unit

but yeah i get what u mean

thanks 😄

can i try part b and show u my work

[mellow olive]

b is just cos^2(3pi/7) - sin^2(3pi/7) tho i dont know what identity to use for cosine

@shut lagoon i thought that this was the identity so shouldn't the 4pi/7 be a negative and not a positive?

[shut lagoon]

It works with that one since it's squared

So the sign doesn't matter

[mellow olive]

oh alrihgt thanks bro

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[hollow igloo]

$x + \frac{\pi}{6} = 3x,$
$x = \frac{\pi}{12}$

[hollow igloo]

but what about the others solutions?

this functiosn have different phase how do i calculate all of their solutions then?

[warm shale]

Yanek

[cunning breach]

Two things:

1. What is the domain?
2. What angle is equivalent to cos(x)?

[hollow igloo]

domain is R

angle equalivent to cos(x) would be any $x + 2\pi k, k \in Z$

[warm shale]

Yanek

[hollow igloo]

@cunning breach

but cos(3x) has different phase

so how do i calculate this

[cunning breach]

This is one set of solutions, think about quadrants.

[timid silo]

Hi, can you please explain how to solve a problem like this? Step by step?

The call centre of a certain company receives an average of 30 calls per hour. The calls are independent of each other. 60% of the people who call the centre are men. The 100th person,
who calls the center gets a special reward from the company. Using CLV, calculate
the probability that

1. there are at most 75 men among the 150 people calling the center,
2. the waiting time for the caller(s) to receive the reward is at least 3 hours.

[cunning breach]

Sin, cos and tan have two quadrants with equivalent values.

[hollow igloo]

yes

cos have 1 and 4 quadrant the same

or x = -x

[cunning breach]

Excatly, so you need the negative one.

[hollow igloo]

$x + \frac{\pi}{6} = -3x$ but what about the phase?

[warm shale]

Yanek

[hollow igloo]

this gives 2 solutions?

so know this is just $\frac{pi}{12} + 2\pik , k \in Z$? and the other solution with 2pik?

[warm shale]

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

[cunning breach]

What do you mean by phrase? This gives x = pi/12 + k*pi, and so does the other set of solutions.

[hollow igloo]

so it doesn't matter that cos(3x) has phase of $\frac{2\pi k}{3}?$

[warm shale]

Yanek

[cunning breach]

$$cos (x+\frac{\pi}{6}) = cos(3x)$$ $$=> x+\frac{pi}{6}=\pm 3x + k*2\pi$$

[warm shale]

Good

[hollow igloo]

i see

ok so when i have 2 functions f(x) = f(y) that have different phase i should take only the bigger one?

[cunning breach]

I'm not quite following here.

[hollow igloo]

$cos(3(x + \frac{2\pi k}{3}))$

[warm shale]

Yanek

[hollow igloo]

@cunning breach would it work the same with lets say cos(2x) = cos(3x + pi/6)?

2x = -+3x + pi/6 + 2pik?

[cunning breach]

It dosen't matter which side you put the $\pm$, you will result the same.

[warm shale]

Good

[cunning breach]

Try cos(2x) = cos(3x)

[hollow igloo]

i was talking about the phase if it should be something different than 2pik there

i just not sure

but i quess you are right

[cunning breach]

2pi is the same as 360 degrees, which is a full circle.

cos(0) = cos(2pi) = cos(4pi). If this is what you was wondering.

[hollow igloo]

yes, i know that, but the thing is the phase of cos(2x) is different than cos(x) and cos(3x)

[cunning breach]

You mind explaning what "phrase" is? I'm not native english, so I'm unfamiliar with some words.

[hollow igloo]

https://en.wikipedia.org/wiki/Phase_(waves)

im not native too so dont worry

wolfram alpha shows one solution that we didn't have

https://www.wolframalpha.com/input?i=cos(x+%2B+pi%2F6)+%3D+cos(3x)

@cunning breach if you mind looking at that

[cunning breach]

Still don't get, if that's period, then it indeed would change, but you start with 2pi, since that a circle, else you need for a other persons explanation.
Speaking of wolframes three solutions, this might caused by + = +, + = -, - = +. You could try out.

[hollow igloo]

ok

thanks a lot for you time

[cunning breach]

Welcome.

[obtuse pebble]

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[next wagon]

Hello, if the limite of a sequence function is +oo, does it imply that it’s inferior limit is +oo ?

[fossil crag]

If the lim exists then lim = liminf = limsup ;)

[next wagon]

J’osais pas te redemander des questions alala je t’aime tellement

Ah d’accord donc + et -oo inclu

Merci beaucoup ❤️

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[marble jay]

can someone explain what the max(S_x,S_y)/min(S_x,S_y) means????

[stark root]

Can anyone can solve this sequence problem?

[wanton hull]

!occupied

[obtuse pebble]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[stark root]

Sorry

[obtuse pebble]

@marble jay Has your question been resolved?

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[maiden sage]

Hello, I really need your help I have been struggeling 2 hours on this. So I have a task that I need to complete for school. This is working with statistics of population and demographics.

So my problem:

I have found that the population in Malaysia is 32698,1 in 2022 (number 000') and that it is expected to reach 33379,5 in 2023.

In the population you have citizens and non-citizens. We have 30198,2 citizens and 2500,0 non-citizens in 2022. In 2023 it is expected to be 30415,3 citizens and 2964,2 non-citizens.

So that is a total increase of 2,083913132567% from 2022 -> 2023.
And for citizens an increase of 0,718917021544% from 2022 -> 2023.
And for non-citizens an increase of 18,568% from 2022 -> 2023.

So now if I calculate total population with a population growth of 2,08...%. Or I calculate the citizens and non-citizens with an population growth of 0.71% and 18,5% individually and add them up I get something completely different from the total.

How is this possible please help!

[obtuse pebble]

@maiden sage Has your question been resolved?

[obtuse pebble]

@maiden sage Has your question been resolved?

[obtuse pebble]

@maiden sage Has your question been resolved?

[obtuse pebble]

@maiden sage Has your question been resolved?

[obtuse pebble]

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[autumn delta]

can anyone help me with this

[autumn delta]

i dont get it at all

[lone echo]

post the instructions

[autumn delta]

my ad

bad

i just did the top function = -2 but i dont understand why it works

oh hold on

nvm

.close

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[autumn delta]

i dont get this question

[oak aurora]

Do you want to try writing out the IVT theorem first?

[autumn delta]

my bad it stands for intermediate value theorem
if f is a continious function on [a,b] and n is a number between f(a) and f(b), then there exists a number c such that f(c) = n

i graphed it on desmos it's not a continious function?

[oak aurora]

Right so the question here is telling you that f(0)<10 and f(4)>10, so if we blindly apply IVT, we should have that there exists x in (0,4) such that h(x)=10

[oak aurora]

yep

so the point of the question is to make sure you know that IVT is only true for continuous function

the function your working with here isn

[autumn delta]

Ohhhh okay thanks

.close

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[thin thistle]

here we need to find f’(x)

[thin thistle]

and as i understand to do that, we need to confirm if it’s continuous then check for differentiation

i get how we proved its continuous

then we moved on to check for the differentiation of

1. x≠0
2. x=0

and again, we pretty much differentiated xsin(1/x) for the first term

[oak aurora]

are you sure lim xsin(1/x)=0?

[thin thistle]

but i dont get what we did for the second term

[thin thistle]

by the squeezing theorem, yea

[verbal prism]

the second term, as in the differential for 0?

[thin thistle]

the differential for when x=0

yes

[verbal prism]

graph x = 0

what is the slope?

[oak aurora]

oh oops, confused myself

[thin thistle]

undefined

[verbal prism]

you can't differentiate it because of that

the top one is product rule, the dne is because it is undefined

DNE = Does Not Exist

[thin thistle]

but i meant about the take “because” part

that went into the technicalities

[verbal prism]

ah, okay

[thin thistle]

why did we use the limit as x approaches 0

and why did we put in it the evaluation of the first term

[verbal prism]

the 'because' part where they do the limit has a limit because we are dividing by 0

we can't divide by 0, and there will be a hole there

BUT if it approaches on both sides, it exists and we can pretend there is no hole

the -1/x(cos1/x) makes it necessary to do the limit there

[thin thistle]

well okay but then again

[thin thistle]

yea why

i mean why did we put it inside

[verbal prism]

ah, inside where, inside the cosine and sine?

sorry, I am confused as to what you are asking

[oak aurora]

the terms are within the limit

so we he doesn’t have to consider x=0

but anyway what is the definition of f’(0) @thin thistle

like in the most general sense, given some f(x)

[thin thistle]

limx -> 0 f(x) - f(0)/x - 0

[oak aurora]

is that the same as your expression here?

[thin thistle]

are u asking me because ur confused or are u asking me to get me to arrive to a conclusion

because im getting more and more confused

[oak aurora]

the latter

[oak aurora]

or more directly im saying that the f’(0) here is not the same as the definition of f’(0)

[thin thistle]

to clear the confusion, my question is,

when we want to check for the differentiation in x=0, why are we using the limit? and why are we asking about the limit of that very specific term

[oak aurora]

because definition of f’(0) involves the limit

but at the same time note that the limit your taking is not f’(0).

[thin thistle]

what is it then

if thats not it

[oak aurora]

lim… xsin(1/x)/x

[thin thistle]

but why

if we’re talking about when x=0

how is that any relevant

[oak aurora]

Because the definition of f’(0) is about the limit of f(x)-f(0)/x-0 as x approaches 0

As x approaches 0, x=/=0

so f(x) is still just xsin(1/x)

[verbal prism]

lusgnol you're confusing me too and I am in cal 3 🥵

[thin thistle]

😭😭😭

[oak aurora]

Your disagreeing with me on the definition of f’(0)?

[verbal prism]

no, I am disagreeing with you on whether it is relevant to their question

[thin thistle]

wait let me say something

and correct me if im wrong

[verbal prism]

they are just trying to find domain of something and confused as to why the limit is needed, I think

[oak aurora]

it is relevant because the f’(0) they have written is completely wrong

[thin thistle]

the first term refers to the derivative when x≠0

well when x≠0, x is just a constant

and the derivative of any constant is 0

so we’re asking for the limit of that derivative?

to check for 0

which clearly doesnt exist in that manner

[verbal prism]

i am explaining this from a physics-applied-math viewpoint and not a pure-math-knowledge viewpoint BUT

this is piecewise because that hole gets in our way

[thin thistle]

yea

[verbal prism]

so we fill in the hole with the value it approaches using a piecewise function

HOWEVER the value that we use to fill it in was (likely, I haven't checked) acquired by taking the limit of f(x) approaching 0 when x!=0

the f'(x) ALSO is the same function, but there still is a hole there. we can use that function to find the value of the hole and fill it in

[oak aurora]

and why exactly are you filling in the hole?

The question is asking for the derivative of the piecewise function

[verbal prism]

well we want to know if there is anything there, so we take the derivative and the limit of that derivative as we approach the hole

[verbal prism]

we can plug 0 in here, to find the value of the hole

but we know that won't work because of the -1/x, we would be dividing by zero

[thin thistle]

which brings me to my second question

[verbal prism]

BUT even if there is no hole, taking the limit will get us the value at that point

[thin thistle]

in that limit

sin1/x is dne

as x approaches 0

i get it

but why is -1/xcos1/x = 0

using the squeeze theorem
-1<= cos1/x <=1

1/x <= -1/xcos1/x <= -1/x

shouldnt the limit for 1/x and -1/x be dne

[verbal prism]

good point

the limit of cos(x) is DNE because it doesn't approach anything

the limit of -1/x is negative infinity because the divisor is so small it makes the number on top huge

[thin thistle]

and 1/x is infinity

the math isnt mathing

[verbal prism]

ok the limit of 1/x approaches infinity so cos(1/x) should be DNE to my humble knowledge

the fact that it is multiplied by -1/x shouldn't matter, DNE * infinity should be DNE

lemme graph this

[thin thistle]

i think ill just take it as a rule that the derivative of 0 is dne and call it a day

my head hurts 😭

[verbal prism]

well yeah the derivative of x = any constant is DNE because of the slope

you can remember it because of the slope

[thin thistle]

huh

the derivative of any constant is just 0

no?

[verbal prism]

my calculus 1 education at my school is actually garbage

but

what I was taught is that they don't have a slope

so you can just say it's 0

but maybe I should not be peddling that around

sorry if its y=0 then its a horizontal line

so it is no slope

but if its x = 0

then it is a vertical line

which makes more sense when you do 3D calculus

so its 0 because the constant will always be y = 0 for you

i am sorry for confusing you 😭

it has been a while since cal 1

[thin thistle]

lmao its alright

thanks for ur time

:D

[verbal prism]

you're good

are you still confused on the original part?

[thin thistle]

no i think i get it now

tysm :D

[verbal prism]

honestly not sure why its approaching 0 and not DNE but maybe squeeze theorem will help

you're welcome 👍

im sorry lusgnol you are very good at math but my simple physicist brain can only math that makes physical sense to me

[oak aurora]

Because it DNE and his doing something very wrong in his original expression which is why i keep talking about his expression for f’(0)

[verbal prism]

🤔

that looks like what he had

[oak aurora]

im talking about the cos expression

and i dont want to argue with you but your definition of filling holes to find the derivative is just not how derivatives are defined

[verbal prism]

not what i said

you fill holes to find the piecewise function's values at each point where the function itself does not suffice

works well enough for me in applied math

[oak aurora]

Is his final expression correct?

That’s my question to you

his f’(0)

[oak aurora]

but this is again not how derivatives are defined

derivatives refer to the slope

you can fill in such that the piecewise function can be continuous

[verbal prism]

I am not talking about the derivative, just the hole

the derivative is the function in question there

but I meant that for any function where we have holes we want to check out

also his f'(0) should be undefined there i think

and his derivative is right

[oak aurora]

im talking about the expression itself

the lim_{xapproaches 0} -1/xcos(1/x)+sin(1/x)

is this f’(0)?

[thin thistle]

i thought it was

[verbal prism]

it's f'(x) as x approaches 0

but in lazy wording

[oak aurora]

it is not

[verbal prism]

hey man it's good enough for my professors

[oak aurora]

f’(0)=limit as x approaches 0 of f(x)-f(0)/(x-0)

which should be (xsin(1/x)-0)/(x-0)

[thin thistle]

u mean f(x) - f(0) / (x-0)?

[oak aurora]

yes

[thin thistle]

wells thats the definition yes

but what would u do after that

[oak aurora]

and so we have f’(0)=limit of sin(1/x) as x approaches 0

not whatever your expression is there

Likely for you calculus class, you can assume that limit of sin(1/x) as x approaches 0 is undefined

[thin thistle]

which is xsin(1/x)/xsin(1/x)

meaning undefined

right?

[oak aurora]

where did you get this?

[thin thistle]

since that cancels out to 0/0

[thin thistle]

its this

but substitute x with xsin(1/x)

[thin thistle]

oh right

i divided by f(x) instead of x

but yeah i get what ur trying to say

but the limit in the picture is sent isnt talking about the definition

its using another logic

[oak aurora]

yes your f’(0) comes from nowhere..

it’s possible for function to be differentiable everywhere but the derivative to not be continuous

your trying to test for continuity of derivative at 0 but that’s simply not the same as finding f’(0)

[thin thistle]

that is if the derivative of any constant is 0, and if the derivative of any constant thats not 0 in that function -1/x * cos(1/x) + sin(1/x)

then the derivative of 0 would be the limit of -1…

[oak aurora]

you should not be caring about the derivative of constants

because the f’(0) is about the behaviour of the function as it approaches 0

[thin thistle]

u can’t differentiate a point if its not continuous

[oak aurora]

i’m talking about the continuity of the derivative

not the continuity of the original function

[oak aurora]

yes this is correct for the original function

the final expression you have in your original working is about the continuity of the derivative right?

[thin thistle]

well i understand my professor’s way but jts a pain in the ass, ur definition made it click easier

now i can use the same way of thinking for all similar scenarios

[thin thistle]

the final expression is not about the continuity

[oak aurora]

i would be very shocked if this is your professor’s way.

[thin thistle]

its about the differentiation of x=0

[oak aurora]

yes and i’m trying to tell you that your final expression is not the derivative of 0.