help-27

[hollow walrus]

how do you calculate that

.close

[devout snow]

Channel closed

Closed by @hollow walrus

Use .reopen if this was a mistake.

[radiant anvil]

actually

u can just look at the squares that can fit

and take the area of each

eg a 4 by 4 square would contribute 1 to 16 squares, so it effectively contributes 16 to the total sum. @hollow walrus

[hollow walrus]

.reopen

[devout snow]

✅

[hollow walrus]

so by this logic how would I use it?

[radiant anvil]

count the total number of each square size

then compute their total areas

eg for 1 by 1 squares, the total area would be 100(number) * 1(area) = 100

2 by 2 squares total area would be 9 * 9(number) * 4 (area) = idk

do it for all squares

[devout snow]

@hollow walrus Has your question been resolved?

Channel closed

Closed by @hollow walrus

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sonic arrow]

Not sure how they were able to get that transformation

[reef solstice]

i solved this ques by finding range

I got the ans but everywehere they used other method by finding minima of denominator

so i am correct ?

[sonic arrow]

aye bruh back off

[trail eagle]

!occupied

[devout snow]

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).

[chrome isle]

[chrome isle]

2x - x² -1 + 1 = 1 -(x-1) ²

So y = 2×2 ^ -(x-1) ²
So the graph of y will be just the graph of f(x) but having twice the slope and being shifted one place right by 1

[chrome isle]

Differentiate the function and do the second derivative test

[sonic arrow]

How does this correlate to 2^-x^2 being translated

I am no good at translations

ignore previous messages, why does the negative sign not do anything

[chrome isle]

In the function 2^-x² substitute x = x-1 you will get y/2

[chrome isle]

Huh which negative sign?

[wooden lance]

id just try all 4 options

lol

[sonic arrow]

1-(x-1)²

does the -1 before the square have no effect?

[devout snow]

@sonic arrow Has your question been resolved?

Channel closed

Closed by @sonic arrow

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[last egret]

guys im not getting

[last egret]

ivt thereom

[misty crest]

what are the choices

[misty crest]

so what do you think

[last egret]

i think that it is continous

[misty crest]

well yes it explicitly tells you that it is

[last egret]

i dont know if it can be applied

[misty crest]

"the function f(x) is continuous…"

[misty crest]

do you know what IVT is really telling you

[last egret]

i think it is something dealing with squeez thereom

[misty crest]

nope

nothing to do with squeeze theorem

[last egret]

is it if a point exists?

[misty crest]

consider a continuous function f and the interval (a,b), let f(a)=c and f(b)=d, then by IVT the function f takes on all values within (c,d)

essentially if the function is continuous then it has to go through every value in between the two y values at the endpoints in the interval between the two endpoints

there is no other way to go from c to d

then to go through every value between c and d

no matter the path you take

[last egret]

i think i get it

[misty crest]

since it’s continuous

[last egret]

so since it is continous it as to go trough these points right?

[misty crest]

mhm

no matter the path you take it has to go through the value of I

in this picture above

[last egret]

yes

[misty crest]

since f(a)<I<f(b)

so continuity is the only condition to apply IVT

[last egret]

so would the ivt be applied to all continosu limits?

[misty crest]

continuous limits?

what does that mean

[last egret]

the function

[misty crest]

continuous functions

but it’s more so continuous on whichever interval is in question

as in continuous on some interval

it can be continuous outside of an integral

but you can apply IVT on an interval for which the function is continuous

[last egret]

so how do we solve this

[misty crest]

we already did

we verified the condition

f is continuous on [-9,9]

thus we can apply IVT on that interval

[last egret]

there is mroe

[misty crest]

which you did not show

[misty crest]

i am not watching your screen

[last egret]

this pops up when you say it can be applied

[misty crest]

mhm so what is the interval

k is a value of y

and c is a value of x

[last egret]

would it be (6,2)?

[misty crest]

hmm i’m wondering if they want 6,2 or -3,6

because you can get both from the graph

they told you 6,2 are two points

but from that we can get -3,6

from -9,9

[last egret]

how did you get -3 6

did you subtrac?

[misty crest]

oh they said on -5,1

at the end of the question

so yea (6,2)

[misty crest]

well i just looks at the points for (-9,9)

f(-9)=-3

f(9)=6

but they only wanted it on the interval (-5,1)

[misty crest]

read the last line of the question

question sir

what is this for

[last egret]

-5 1?

[misty crest]

can you show the entire screen

[misty crest]

ok it’s just a homework

you’re good

[misty crest]

mhm

[last egret]

not a homework but a review for a unit test

[misty crest]

yea just not a test

[last egret]

no lol

[misty crest]

just wanted to make sure

[last egret]

how do we find the c

[misty crest]

it’s the x interval

c is some value of x in the interval

[last egret]

where do i find the x interavak

[misty crest]

it was given

they want you to apply IVT on (-5,1)

thus the interval is (-5,1)

[last egret]

so -5,1)?

[misty crest]

yes

[last egret]

so this means no f(c)?

[misty crest]

so what do you think for the last part

remember what i said earlier

[misty crest]

^

[misty crest]

^

[last egret]

not exist

[misty crest]

why is that

so in this scenario

relating to what i said earlier

a=-5

b=1

c=f(a)=6

d=f(b)=2

from IVT

f takes on all values between 6 and 2 in the interval (-5,1)

so what does this mean for our question

[last egret]

oh so from 6,2

so 3 is inbetween

[misty crest]

indeed

[last egret]

if it was 2,6 would it be the same?

[misty crest]

same thing

doesn’t matter if it goes from lower to higher or higher to lower

[last egret]

i got it wrong

[misty crest]

since it’s continuous it takes on all the values in between

[last egret]

one more try left

[misty crest]

it is correct

[last egret]

should i submit it again?

[misty crest]

maybe they want

(2,6)

instead of (6,2)

[last egret]

yes it is 2,6

you are a genius

[misty crest]

i’m flattered

[last egret]

should i do one more and show you the steeos?

[misty crest]

the what

[last egret]

steps*

[misty crest]

steps?

yea

[misty crest]

you do this one on your own

i’ll correct you if you mess up

[last egret]

ok

so i know this thing is not contionus

because it didnt say it is continous

[misty crest]

well

[misty crest]

this isn’t true

you don’t "know" it’s not continuous

you just

"don’t know" IF it’s continuous

but yes since you don’t know if it’s continuous

[last egret]

so it would be this?

[misty crest]

then you can’t make the assumption and apply IVT

[last egret]

wait it is continosu since it gave the ivt

is this true?

[misty crest]

it never said IVT can be applied

they were just asking you what can you draw from IVT

you can’t draw anything from it since you don’t know if it’s continuous

[last egret]

so it would be not necessarily continous?

ok this si would i put but i go ti twrong

[misty crest]

what were the other options for there may not exist

oh wait no

you messed up the interval

(-3,0)

because it said value of c

thus value of x

which means x interval

[last egret]

yes i got it wrong

i am going to do another one

[misty crest]

even after my suggestion?

[last egret]

before

this is a new one

[misty crest]

you messed up the interval again

remember they gave you at the end, f(c)=k

[last egret]

so it would be (4,-2)?

[misty crest]

thus c is a value of x

and k is a value of y

so the interval corresponding to k should be a range

or y interval

and the opposite for c

[misty crest]

nope

[last egret]

ok let me do this again

[misty crest]

c is an x value between what and what

and k is a y value between what and what

[misty crest]

good but make sure that the k interval goes from smallest to largest because last time when we did it the other way the marked it wrong

the c interval is correct

so instead of (-2,-5)

it should be what

because -2>-5

[last egret]

so would it not exist?

[misty crest]

yes excellent

oh wait

no there may not

[misty crest]

my apologies

may not

[last egret]

how is it a there not be?

may not be*

[misty crest]

because i never said that -5 was the minimum

hold on let me draw

[last egret]

ok

[misty crest]

it can go like this

because all values between the two y values are still guaranteed

[last egret]

the left side is arrow right?

[misty crest]

what

i just drew x-y plane with a function

the two points

are the endpoints of the interval

but notice the IVT only guarantees values between the two functions

there may still be y values above or below the y interval

if it doesn’t specify that either of the endpoints are absolute minimums or maximums

[last egret]

i see

let me yty again

im doing another one i got it wrtong

[misty crest]

well it didn’t say f was continuous

if it doesn’t explicitly say it’s continuous

then you can’t apply IVT

[misty crest]

ok what do you think

[last egret]

this is a different one

im tring this one

[misty crest]

yes this is the answer

[last egret]

so i know this is continous since it is stated

[misty crest]

mhm

thus IVT can or can not be applied

what do you think

[last egret]

it can be applied since it is continous

the k value is 3 through 7

[misty crest]

nope

close

maybe a typo

[last egret]

oops 6

the formula for k is f(a) = k right?

[misty crest]

what is a

[last egret]

2

f(2) = k

[misty crest]

no

k is some arbitrary value of y in the interval

similarly c is some arbitrary value of x

in the given interval

[last egret]

i see

so it would be something lik ethis

?

for k right?

[misty crest]

k is between those

two lines

yes

if those are the bounds of our interval

[last egret]

yes

i see i think im gettin git

[misty crest]

so this is similar to the one before

[last egret]

i got it wrong

[misty crest]

is 3<2<7

[misty crest]

you submitted it before me

[misty crest]

^

[last egret]

so this is a may be not?

[misty crest]

mhm

think of 2 as k

when they say f(c)=2

can that be k?

if we said k is between 3 and 6

right

[last egret]

yes

[misty crest]

so IVT only holds for

k

[misty crest]

it can not be k

because 2 is not in between 3 and 6

[last egret]

so k coudlnt be 3 or 6 right?

[misty crest]

we stated that 3<k<6

well it could

but

they’re using open intervals

since they already told you the function takes those two values on

so to say it again would be redundant

by using an inclusive interval

[last egret]

"a value of cc in the interval left parenthesis, 2, comma, 7, right parenthesis(2,7) where f, of, c, equals, 2, .f(c)=2."

is 2 inside the interval

[misty crest]

2 is not in the interval (3,6)

[misty crest]

don’t confuse the x and y intervals

remember

c is an x value

which is the input

when you say f(c)=whatever

[last egret]

oh

[misty crest]

that’s a y value

[last egret]

it is using these

[misty crest]

the y value at x=c

[misty crest]

mhm

[last egret]

wait so it woudlnt exist right?

[misty crest]

not necessarily

it could exist

but isn’t guaranteed

because it could just draw a straight line between the points

and it would never go below to 2

[last egret]

when can it be it cannot exist?

[misty crest]

assuming IVT can already be applied?

[last egret]

yes

[misty crest]

only if they state either of the two endpoints are absolute minimums or absolute maximums on the interval and the value in question is below the min or above the max

that would not be possible

it’s contradictory

[last egret]

ok lets do one

i think i can get this right

i got it correct

[misty crest]

nice

[last egret]

how did i get this wrong?

[misty crest]

you said therefore there exists

while also saying it can not be applied

if it can not be applied then you can’t assert that there exists…

[last egret]

so it may not exist

[misty crest]

there may exist may not

🤷🏼‍♂️

[last egret]

thank you

you are a legend man

[misty crest]

thank you pingu

[last egret]

can i griend regques tyou

you are a legend

i cannot express this enough

[misty crest]

no need

i hope you do well

on your exam

[last egret]

thank you

how did uou learn this muhc calclus?

[misty crest]

you’re welcome

[misty crest]

books

youtube

practice

[last egret]

what do you think i should search for ivt

[misty crest]

khan academy is always a great resource

use the ap calculus courses on there

free

[last egret]

do you think chemistry tutor is good

[misty crest]

they have practice exercises and videos

[misty crest]

yea he’s not bad

he’s sort of like khan academy

i’ve watched him before

not as of late but

yea was good back in the day

[last egret]

so khan academy is the main?

[misty crest]

some of his videos i didn’t like

[misty crest]

yes i’d definitely recommend khan academy

[last egret]

what are some books?

[misty crest]

stewart’s calculus was great

larsons book is great

[last egret]

is it good for beginnrse?

[misty crest]

mhm

single variable calculus

when you start doing derivatives and integrals i’d recommend chris mcmullens calculus exercises

he has a workbook

[last egret]

is it free online?

[misty crest]

it may be

[last egret]

i see

[misty crest]

i bought it for cheap

it’s a workbook so it’s probably like 15 dollars

[last egret]

i am starting derriatives soon

[misty crest]

it’s great though

great exercises

[last egret]

so do you think it is good to start?

[misty crest]

clean and concise explanations

[misty crest]

it’s how i started

[last egret]

what are the main things for calculus i heard it is limits deravitives

and some other thighs

[misty crest]

i had that and some book by thompson

[last egret]

have you haerd the book calculus made easy by thmpson?

[misty crest]

hmm i’ll look it up

nope

[chrome isle]

It's a useless book

[misty crest]

that’s not the one i read

i first read calculus for they practical man

but i wouldn’t recommend that one

[chrome isle]

Altho it teaches u the intuitive aspect of calc it won't teach u how to use it

[last egret]

i see so the mcallien thing?

[misty crest]

some of the answers to the exercises were incorrect

[misty crest]

mcmullen is a quick workbook

i definitely recommend

[last egret]

i see do you know any good youtube videos for ivt right now?

[misty crest]

you’ll want more resources as it’s not a textbook with theorems and such

[last egret]

i see

[misty crest]

i’m sure if you searched IVT on youtube you’d find some

khan academy has on

they post on youtube as well

i’m sure professor leonard has one

mit ocw is great

[chrome isle]

Il suggest to use calc by stewart Or thomson if ur beggner for learning how to use it
And 3blue1browns series of essence of calc for the intuitive reasoning

[misty crest]

i didn’t use them for single variable calc but still

[misty crest]

this as well

essence of calculus is essential

nice video series

great animations

to give an intuitive approach

[last egret]

i will look into these book

[misty crest]

just type pdf at the end

you can find them online

[chrome isle]

U can go for spivak Or apostol once ur done with calc once but not if ur starting since they are too rigorous

[last egret]

i always have roblem the formulas

like f(a) something similar to that

[misty crest]

yea definitely don’t do spivak right now lol

[last egret]

i just cannot apply it

[chrome isle]

Yup the best math youTuber

[misty crest]

him or bprp

bprp is more fun

i guess

but 3b1b is more educational

[last egret]

bbrp

[misty crest]

blackpenredpen

[last egret]

i will look it up

[chrome isle]

I tried to read it once Im still traumatized by it till this day

[misty crest]

spivak is advanced calculus

[chrome isle]

And dont go for apostol too it's even harder than spivak

[chrome isle]

It's more analysis based il try it again once I finish my calc 2 n 3 lol

[misty crest]

it is analysis really

you haven’t even done calc2?

why are you reading spivak

[chrome isle]

tru lol

[misty crest]

😭😭

[chrome isle]

Have I'm revising

[chrome isle]

It was suggested by someone he said it's for beggingners

[misty crest]

anyways do you have any other questions @last egret

if so then type .close

[devout snow]

@last egret Has your question been resolved?

[devout snow]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[warm compass]

It seems as tho this is the wrong answer:

[warm compass]

where did I go wrong?

and what would be a more efficient way to solve for x here?

[lofty crescent]

start by multiplying both sides by (x+2)(x+3)
and we get x^2+2x=3
then factor into
x^2+2x-3=0
(x+3)(x-1)=0
x=-3
x=1

[warm compass]

dang, so the crossing was incorrect?

[lofty crescent]

not sure but you solve these problems by canceling out the denominators of both sides usually

lemme see what happened

[warm compass]

ok

[lofty crescent]

oh yeah it is the cross multiplying

[warm compass]

alright, I'll watch out for it

[mystic scarab]

<@&268886789983436800>

[warm compass]

ty ty

.close

[devout snow]

Channel closed

Closed by @warm compass

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[balmy oracle]

hi

[glossy basalt]

hi

[balmy oracle]

How does this not equal 52

ABE + EBC = 180

so 5r - 10 = 180

190/5 = 38

r=38

38(2)-20 is 52

[silk zenith]

what are m and r?

just variables?

[balmy oracle]

uh

wheres m and r

r is a variable

[silk zenith]

it says m<EBC

[balmy oracle]

M is angle

basically

so the angle of EBC is 3r+10

[silk zenith]

ohk

[balmy oracle]

so that means since BA and BC are opposite rays ABE and EBC add up to 180

so use the variables that means 3r+10 + 2r-20 = 180

[silk zenith]

yup

[balmy oracle]

Okay and it says it's wrong

After I plug it in

For ABE which is congruent to EBF

wait im so dumb

38 x 2 is not 72

🙄

76-20 is 56

I stg if this is the answer

yep im done

i just wasted 30 minutes for a dumb multiplication error

[silk zenith]

this is wrong

[balmy oracle]

ik

[silk zenith]

yup

[balmy oracle]

its 56

I literally waited for 30 mins

Okay homework time is over

Im done for now

[silk zenith]

im sorry, I kind of went to another chat because I saw someone else take over here

💀

[balmy oracle]

your fine im just going to do my spanish hw now

geometry can wait

[silk zenith]

gl lol

[amber umbra]

Did you get assistance yet? I would be happy to help you out.

[balmy oracle]

no it has nothing to do with that

I just wasted a lot of time

for no reason

[amber umbra]

Got it. If you need anything, feel free.

[devout snow]

@balmy oracle Has your question been resolved?

[silk zenith]

ye, pretty ssure the question has been solved

[devout snow]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dire hatch]

4. In kindergarten, Viktoria and Emil each have a glass of juice with exactly the same amount of juice in each glass. However, Viktoria is not entirely happy because Emil's juice contains twice as much sugar as hers. The glasses are not so full that they cannot pour a little from one into the other, and smart as she is, Viktoria convinces Emil to play the following game: She pours $\frac{1}{9}$ of her juice into Emil's glass, asks him to stir it well, and then pour the same amount of juice back into her glass so that they again have the same amount of juice.

The mixing procedure described above is repeated several times. Let $x_n$ and $y_n$ be the amount of sugar in the glasses of Viktoria and Emil, respectively, after the procedure has been performed $n$ times.
a) Show that

$$
\begin{aligned}
x_{n+1} & =0.9x_n + 0.1y_n \
y_{n+1} & =0.1x_n + 0.9y_n
\end{aligned}
$$

b) Let $M$ be the matrix such that $\binom{x_{n+1}}{y_{n+1}} = M\binom{x_n}{y_n}$. Find the eigenvalues and eigenvectors of $M$. \
c) Write $\binom{2}{4}$ as a linear combination of eigenvectors of $M$, and find $M^n\binom{2}{4}$. \
d) How many times must the mixing procedure be performed for the ratio between the sugar content in Viktoria's juice and Emil's juice to be at least $0.95$?

[woven radish]

Michael

[dire hatch]

Need assistance on d)

[devout snow]

@dire hatch Has your question been resolved?

[dire hatch]

<@&286206848099549185>

Apologize for a lot of text.

[hot steeple]

i don't know which direction the ratio is meant to go in

so idk if it is x_n = 0.95 y_n

or the other way around

[dire hatch]

Viktoria = x_n, Emil = y_n

so x_n / y_n = 0.95

[hot steeple]

but the idea is you want to find n such that x_n = 0.95 y_n

or y_n = 0.95 x_n

[dire hatch]

yeah

but how do I do that...

[hot steeple]

or more

[dire hatch]

do I need to insert one of the previous equations?

[hot steeple]

you start with some x_0 and y_0

some sugar content values

[dire hatch]

So I should just make some values up?

[hot steeple]

they need to fulfil the condition that emil has twice viktorias

[dire hatch]

Yep

[hot steeple]

but other than that it's irrelevant how much exactly

as you want the ratio

and if you pick some value that is twice the intended value

for example let x_0 and y_0 be the intended values

then x_0 = 0.5 y_0

and you choose twice those values

since the condition needs to be met

you can just divide that out

that multiple

anyway

pick some numbers

[dire hatch]

yeah

[hot steeple]

like 2 and 4

and apply the 'mixing matrix'

[dire hatch]

Ahh

[hot steeple]

until instead of x_0 = 0.5 y_0 you get x_0 >= 0.95 y_0

[dire hatch]

and then I can use the result from c

[hot steeple]

and you want to find n

oh sorry

it's x_n >= 0.95 y_n

not _0

[dire hatch]

np, I understood that you meant n

alright then I got it. Thanks!

.close

[devout snow]

Channel closed

Closed by @dire hatch

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[plush flax]

The text above this problem says “determine whether the function is continuous at the given values of c”

[plush flax]

I honestly dont know how to start problem 2 all🥲

[plush flax]

See i bet thats really helpful but i still just dont understand for some reason

I think bc there are so many terms im just getting really stuck

[devout snow]

@plush flax Has your question been resolved?

[tawny pewter]

draw a graph of the function and think about it and\or see if you get 1 when plugging in -1 in -x-1 (do the same for the other points c=0 and c=1 according to the function definition in each part)

[devout snow]

@plush flax Has your question been resolved?

[devout snow]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[ancient condor]

Hello, how are you, could someone help me? Does anyone know how to solve this step by step?

[restive river]

integration by parts

[devout snow]

@ancient condor Has your question been resolved?

[ancient condor]

I got this, sorry for the handwriting, sorry if it is not very understandable, I would like to know if what I did is right before evaluating the limits

At the bottom left there is an x^2 but you can't see it, sorry again

<@&286206848099549185>

[devout snow]

@ancient condor Has your question been resolved?

[devout snow]

@ancient condor Has your question been resolved?

[devout snow]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[chrome nymph]

I proved this to be equal to x^n - y^n. Is there a way to translate it to the more traditional form which is on the second screenshot?

[devout snow]

Please don't occupy multiple help channels.

@chrome nymph Has your question been resolved?

[devout snow]

@chrome nymph Has your question been resolved?

[devout snow]

@chrome nymph Has your question been resolved?

[devout snow]

@chrome nymph Has your question been resolved?

[kind tiger]

since i>0 you can write (x-y)^i as (x-y) (x-y)^(i-1) and factor out the (x-y) from the sum

that would be the first step

then you would have to do another binomial expansion on (x-y)^(i-1) giving you a nested sum

[chrome nymph]

I see that but I don't know how to go from that sum to the traditionally used one

[kind tiger]

im not sure i understand what youre fully trying to do

the nice thing about finding these two different representations if that you can equate the coefficients individually

giving you some perhaps interesting or otherwise useful results

[chrome nymph]

I'm just trying to go from my form to the traditional form

[kind tiger]

\begin{align*}
x^2 - y^2 &= \sum_{i=1}^n {n \choose i} (x-y)^i y^{n-i} \
&= (x-y) \sum_{i=1}^n {n \choose i} (x-y)^{i-1} y^{n-i} \
&= (x-y) \sum_{i=0}^{n-1} {n \choose i+1} (x-y)^i y^{n-i-1} \
&= (x-y) \sum_{i=0}^{n-1} {n \choose i+1} \sum_{j=0}^i {i \choose j} x^j (-y)^{i-j} y^{n-i-1} \
&= (x-y) \sum_{i=0}^{n-1} \sum_{j=0}^i {n \choose i+1} {i \choose j} (-1)^{i-j} x^j y^{n-j-1}
\end{align*}

[kind tiger]

ok so youre ok up to this?

[chrome nymph]

looking through

[woven radish]

Acman

[kind tiger]

missed a -1 whoops !

[chrome nymph]

It will take a bit of time for me to understand

lines 1 and 2 of 4 look good

Line 3 of 4 looks good too

where did the (-y) go?

[kind tiger]

$(-y)^{i-j} y^{n-i-1} = (-1)^{i-j} y^{i-j} y^{n-i-1} = (-1)^{i-j} y^{n-j-1}$

[woven radish]

Acman

[chrome nymph]

Okay

So far looks good

[kind tiger]

ok the next step is going to be to change the order of summation, have you seen anything like that before

[chrome nymph]

Nope

Or probably I have but years ago

[kind tiger]

this table shows which pairs of (i, j) are summed over

do you see how this is working

the other sum goes from 0 to n-1, these are the rows

then the inside sum is j which goes from 0 to i

so for instance the term where i = 0 and j = 1 is not included because when i=0 the inside sum is j=0 to i=0

changing the order of summation is exactly what it sounds like, instead of doing i first we rewrite it we sum j first

we do this so that we can group all the coefficients of e.g. x^2 y^(n-3) together

[chrome nymph]

Seems relatively straightforward

[kind tiger]

ok cool so we can look at this chart to see that the new summation will be [\sum_{j=0}^{n-1} \sum_{i=j}^{n-1}]

[woven radish]

Acman

[kind tiger]

does that make sense ?

[chrome nymph]

You sure the second sum will go up to n-1?

[kind tiger]

yep

look at the chart

summing i first is like going across the rows

summing j first is like going down the columns

and every column includes i=n-1

[chrome nymph]

Oh yeah, I was trying to go in the opposite direction but the result would be the same

Continue please

[kind tiger]

\begin{align*}
x^n - y^n &= (x-y) \sum_{i=0}^{n-1} \sum_{j=0}^i {n \choose i+1} {i \choose j} (-1)^{i-j} x^j y^{n-j-1} \
&= (x-y) \sum_{j=0}^{n-1} \sum_{i=j}^{n-1} {n \choose i+1} {i \choose j} (-1)^{i-j} x^j y^{n-j-1} \
&= (x-y) \sum_{j=0}^{n-1} x^j y^{n-j-1} \sum_{i=j}^{n-1} {n \choose i+1} {i \choose i-j} (-1)^{i-j} \
&= (x-y) \sum_{j=0}^{n-1} x^j y^{n-j-1} \sum_{i=0}^{n-j-1} {n \choose i+j+1} {i+j \choose i} (-1)^i
\end{align*}

[woven radish]

Acman

[kind tiger]

but notice that the result we are trying to show is that
[x^n-y^n = (x-y) \sum_{j=0}^{n-1} x^j y^{n-j-1}]
so all we need to do from here to finish is show that
[\sum_{i=0}^{n-j-1} {n \choose i+j+1} {i+j \choose i} (-1)^i = 1 \quad \text{for } j=0, \dots, n-1]

[woven radish]

Acman

[kind tiger]

this is what i was talking about here
we could treat that last equation as a new identify we found via this method

im not sure how you would otherwise prove that identity, probably can be done but i was never good at combi haha

[chrome nymph]

Same, same

[kind tiger]

there are bunch of identities here https://en.wikipedia.org/wiki/Binomial_coefficient#Identities_involving_binomial_coefficients

likely the key in one (or a combination of multiple) of them

[chrome nymph]

I just wonder why we can't get rid of -1

[kind tiger]

wdym

[chrome nymph]

We are supposed to have only pluses without minuses

[kind tiger]

yeah we do

the (-1)^i is in the inside sum not the ouside

[chrome nymph]

Yeah

But doesn't it mean there will be minuses

[kind tiger]

not necessarily

4 - 3 + 2 - 1 = 1

[chrome nymph]

We are not working with specific numbers here

[kind tiger]

no that was just an example

[chrome nymph]

It should be true with just the variables too

[kind tiger]

my point is the fact that we have an alternating sum as the coefficient does not necessarily mean the coefficients will be altnerating themselves

look at this

the inner sum is the coefficient each x^j y^(n-j-1) term

like you say, we want them all to be 1

[chrome nymph]

According to vardenmonde's identity

Not that it gives much

[kind tiger]

i dont think you have applied that identity correctly

[chrome nymph]

I made k=i, r=2i + j + 1

[restive river]

correct but that is combinatorics

like this n+i+j C 2i+j+1

[chrome nymph]

Oh

Yeah

[kind tiger]

the issue is vandermonde is about a SUM of binom coefs

and since our sum is over i there should be no is left in the final expression

since its the summing variable

i dont think vandermondes can be applied to our sum

[chrome nymph]

Okay I don't have the mental strength to deal with it right now. Unless someone else is interested in it I'm going to close the topic

.close

[devout snow]

Channel closed

Closed by @chrome nymph

Use .reopen if this was a mistake.

[mild sorrel]

he

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[chrome isle]

What's the formal proof of Riemann series theorem

[static ember]

elaborate

[upper schooner]

The one about being able to rearrange conditionally convergent series to whatever you want?

[chrome isle]

Ahh yes

[chrome isle]

I seem to understand all of it
Now how can I prove that the partial sums of the given sequence have superior y and inferior x

Sorry for not stating my question correctly

Hello anyone

[static ember]

I think it all comes down to pn and qn being null series, even though their series diverge. So the amount you overshoot y when adding up the pn and the amount you undershoot x when subtracting the qn has to converge to zero

[chrome isle]

Why? Why can't it Ossicilate and never equal the converging value

[static ember]

what I dont really get in this proof is the need for these sequences xn and yn, why not just take constant x and y

it does oscillate

thats why we take limsup and liminf

[chrome isle]

they are just trying to show the idea of overshoot and undershoot using it

[static ember]

but each time the oscillation reaches x or y, it hits these numbers closer and closer

[chrome isle]

Uh? But then the series won't ever actually converge to the value we want

[chrome isle]

How can we show it hits closer and closer?

[chrome isle]

Yea but the rearrangement actually attains that value somewhere

At infinity

[static ember]

yeah we only want it to have a limit superior and a limit inferior, not necessarily a limit

[static ember]

because pn and qn are null sequences

pn -> 0 for n -> inf

[chrome isle]

Why can't pn approach infinity and qn approach - infinity

[static ember]

because the original series converges

so an is a null sequence

thus pn and qn as subsequences too

[chrome isle]

Can u elaborate I'm a little dumb

[static ember]

you know what conditionally convergent means?

[chrome isle]

Yea that the series converges but the series formed by the mod of corresponding terms diverges

[chrome isle]

Also will in this case lim inf be equal to limsup

[static ember]

and for a series to converge, the sequence of summands has to be a null sequence

[chrome isle]

Ok

[static ember]

only if x = y

[chrome isle]

If x isn't equal to y then it's going to be a series that alternates between them never reaching a finite no?

[static ember]

yeah

[chrome isle]

Ok so the rearrangement of the original series may or may not be converging depending on lim inf and sup

I think I get it now thanks alot @static ember

.close

[devout snow]

Channel closed

Closed by @chrome isle

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[topaz beacon]

!nopdf

[devout snow]

Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

[chrome isle]

Can u send some photos or smthng else can't really open the link

What's the name of the book?

I've heard it's name so il suggest to go with it since I don't know how ur proffesor teaches
Also teaching sequence series after limits is wierd cause we were taught that in calc 2

[devout snow]

@light hazel Has your question been resolved?

Channel closed

Closed by @light hazel

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[thin inlet]

try drawing a picture :3

this is what the scenario would look like. you want the green area

[devout snow]

@flint zodiac Has your question been resolved?

[devout snow]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[molten spindle]

why doenst it work? at the end i get 0 instead of 14pi

[dense lynx]

your first equality is incorrect

[molten spindle]

was the complexifycation incorrect?

how

[dense lynx]

you rewrote sin(3x) / sin(x) as e^(3ix) / e^(ix), which simplifies to e^(2ix), or cos(2x) + i * sin(2x)
as you can see, what you ended up with is nothing like what you started with

to be precise, euler's formula states that e^(ix) = cos(x) + i * sin(x), which you understand to imply sin(x) = Im(e^(ix))

your issue comes from overextending this formula

to properly complexify this integral, you would use the identity sin(x) = (e^(ix) + e^(-ix))/(2i)

[molten spindle]

woah, i didnt know this formula existed

thanks

[dense lynx]

(here's the typical exponential forms of trigonometric expressions)

[molten spindle]

.close

[devout snow]

Channel closed

Closed by @molten spindle

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[faint halo]

Been trying this problem for awhile now.

like other problems i've inserted x into the function i can post my results

F(1) = 4.83
F(0.5) = 2.10
F(0.05) = 0.41

F(0.01) = 0.041
F(0.005) = 0.021
F(0.0005) = 0.0041

to my understanding the numbers should be really close to the answer but they're all over the place

[devout snow]

@faint halo Has your question been resolved?

[faint halo]

<@&286206848099549185>

[ember pumice]

I think you've done most of the work here. Looking at the values of f(x) can you guess the limit ?

[faint halo]

0?

0.1?

[ember pumice]

You are already below 0.1 at x = 0.0005

So 0 would be a better guess

[faint halo]

oh true

[ember pumice]

(and the actual limit)

[faint halo]

oh you're right!

Ty ty!

[ember pumice]

No worries, good luck

[faint halo]

.close

[devout snow]

Channel closed

Closed by @faint halo

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[pine patrol]

Would I just use these limits for 22c question I dont rlly know how to do the right limits when there isnt any lines on the right that go to the holes

[pine patrol]

And also for 23 how would I describe the right/left limits

[devout snow]

@pine patrol Has your question been resolved?

[pine patrol]

<@&286206848099549185>

[restive river]

ok for 23

i assume they want right/left limits to converge on a point, preferably 0?

[pine patrol]

Im not sure but probably

So would I just say what the y is when x is 0 for the right and left sides @restive river

[restive river]

you would say that the limit as x approaches 0 from the right is 0

basically, i think?

based on what 23 looks like

[open vale]

is there any other descriptions for 23??

[restive river]

oh nvm

[pine patrol]

Ya that sounds right, but for 22 how would I do the right side for the open circle

[restive river]

wait wait

b says that the limits of the end behavior

[pine patrol]

No

[restive river]

the limits dont converge on a point

they are to describe the end behavior of postive and negative infinity

thats what it is

[pine patrol]

Im kinda confused

[restive river]

read b.

[pine patrol]

But thats just for determining the end behaviors

Which is inf and -inf

[restive river]

and for 23

which has no discontinuities

you can only do end behavior

[pine patrol]

So it would be DNE

ok thank you, what about the right side for 22

On the -2 hole

[restive river]

the limits for 23 arent DNE

they have values they go to

mainly

[pine patrol]

So what would I put then

-inf and pos inf

[restive river]

right side goes down forever (which is an unlimited number in negatives)

left side does the opposite

i hope you got the hint

[pine patrol]

Ya I think I understand 23

What abt 22

[restive river]

22 has a lot of discontinuities, therefore, a. will be having some answers for 22

[pine patrol]

My teacher said there should be 4 limits

Dont I just do the left and right sides of both holes

[hard wagon]

Help plz

[pine patrol]

<@&286206848099549185>

[open vale]

hey

what's the pro?

@pine patrol

[pine patrol]

22

@open vale

[pine patrol]

<@&286206848099549185>

[devout snow]

@pine patrol Has your question been resolved?

[devout snow]

@pine patrol Has your question been resolved?

Channel closed

Closed by @pine patrol

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[iron siren]

How do you do gaussian elimination on a 5x4 or 4x5 matrix what???

[wicked turtle]

same way you do it in general

[smoky nimbus]

The same way you would for a square matrix

[iron siren]

so the end result should be something like this??

dots = numbers

[smoky nimbus]

That, or you can have row of zeros possibly

[iron siren]

if i have a row of zeros post gaussian wouldnt that make it near impossible to solve for 5 variables

[wicked turtle]

not if it's all zeros

if you have something like 0 0 0 0 0 | 1 then that means there's no solution

[iron siren]

isee

ok ill throw myself at the problem and brb rq

ok this is hard

could I get some help starting the process?

[wicked turtle]

first, you want the entries in the first column, rows 2 through 4, to be 0

[iron siren]

so rather than go row by row look at the columns of 0s that need to occur?

[wicked turtle]

yea you want to clear one column at a time

to get into upper triangular form

[iron siren]

like thius?

WAIT

cant then you do this??

[wicked turtle]

yea

you're not quite done though

[iron siren]

same goes for R3 though...

[wicked turtle]

^

[iron siren]

then you'd end up with 4 0s again

all i know know is that x5 = 3

[wicked turtle]

the last two rows are the same

so subtract row 3 from row 4 to finish

[iron siren]

can i just

combine them to make the matrix smaller?

[wicked turtle]

you could, but the "standard" RREF should be the same size as the original

the last row will just be all zeros in this case

so it doesn't contribute anything to the solution

[iron siren]

can an RREF be a REF?

[wicked turtle]

yea, RREF is just a special case of REF

i meant REF here actually

[wicked turtle]

yep

[iron siren]

but then how can i solve for each variable individually

there's too many of them

or od i have to keep going to make the 2nd row just 2 vairables

[wicked turtle]

well you'll have two "free" variables that can be anything

namely x4 and x2

[iron siren]

wait waht

[wicked turtle]

as you said, x5=3

[iron siren]

yes

[wicked turtle]

then the 2nd row tells you:
x3 + x4 + 2x5 = 0

plug in x5=3

that becomes:

x3 + x4 = -6

[iron siren]

6

[wicked turtle]

so you can let x4 be anything

[iron siren]

but then theres an infinitely wide range of what they can be

[wicked turtle]

yea there will be infinitely many solutions

because you have fewer equations than unknowns

like for example if i just gave you one equation with two unknowns, say:
x1 + x2 = 5

there are infinitely many solutions to that, right?

[iron siren]

yes

[wicked turtle]

same idea here

[iron siren]

so when it says solve the linear system.

does that mean just simplify it as best as possible?

[wicked turtle]

yea in this case it means you'll express x1, x3 and x5 as functions of x2 and x4

(x5 is just a constant, which is a trivial function)

[iron siren]

so what do i have to do after getting to the matrix i just did?

repost)

[wicked turtle]

start at the bottom and go up

x5 = 3

then

x3 + x4 + 2x5 = 0, so
x3 + x4 = -6

thus:

x3 = -x4 - 6

[iron siren]

oh.

[wicked turtle]

finally the top row

[iron siren]

so im now just writing the "solution"

using the variable names

[wicked turtle]

x1 = -x2 - x3 -x4 - x5 + 1

[iron siren]

due to there being infinently many solns

[wicked turtle]

plug in your expressions for x3 and x5

[wicked turtle]

yep

[iron siren]

ok!

i think i can try the next question on my own and see what i can do

if there's more variables than there are eqns

then its not a concrete answer

[wicked turtle]

yea, there will either be infinitely many solutions or no solutions

(when there are more variables than equations)

[iron siren]

tysm ❤️

ill be back in case i need help

:3

.close

[devout snow]

Channel closed

Closed by @iron siren

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vernal burrow]

Can anyone please tell me if this is correct i have no idea why both 32,000 net profit

[devout snow]

@vernal burrow Has your question been resolved?

[vague frigate]

Addition mistake in manufacturing top sheet

[vernal burrow]

why is it mistake

[vernal burrow]

manufacturing is decrease to 1,500,000

everything in the table is fixed costs

COGS is variables costs

[vague frigate]

Hmmm... Well just focus on the year 2 columns on both, total cost for the two sheets doesn't seem right...

Oh the COGS... The cogs are what balances the total production costs

[vernal burrow]

ohhh so it is correct then?

Cuz fixed cost manufacturing gets lowered but COGS increases so it cancels out i think

[vague frigate]

Yes sir

[vernal burrow]

Thank you brother

.close

[devout snow]

Channel closed

Closed by @vernal burrow

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[humble sapphire]

Is this correct

[devout snow]

@humble sapphire Has your question been resolved?

[humble sapphire]

Hello?

[devout snow]

@humble sapphire Has your question been resolved?

Channel closed

Closed by @humble sapphire

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[heavy current]

okay, I am going insane

[heavy current]

is it true that $\coprod_{y \in Y} X$ is a discrete space?

[woven radish]

higher!

[heavy current]

or am I wasting my time trying to prove something that is false?

I'm really hoping it's true, otherwise my proof idea for part of this exercise is gonna fall flat on its face

like, to prove it is discrete, it suffices to prove that every singleton in the disjoint union is open

so if ${(x,y)}$ is a singleton in the $y$th copy of $X$, then ${(x,y)} \cap X = \emptyset$ for every other copy of $X$ by the definition of the disjoint union

[woven radish]

higher!

[heavy current]

and \emptyset is open in X always

but if we intersect ${(x,y)}$ with its own copy of $X$, then we have problems

[woven radish]

higher!

[heavy current]

the intersection is just ${(x,y)}$ itself, so now I need to show that this singleton is open in the $y$th copy of $X$

I want to somehow use the fact that Y is discrete to do this, but I cannot figure out how for the life of me

does anybody have an idea, or a suggestion?

or you can tell me that the disjoint union space is actually not discrete and that I'm wasting my time

I just need a direction to go in

[woven radish]

higher!

[dire forge]

Why do you think it's true

Have you tried an example yet

[heavy current]

because it would make my life highly convenient

then I can say that ${\text{Id}|}X: \coprod{y \in Y} \to X$ is continuous

[dire forge]

^

[woven radish]

higher!