help-0

[surreal meadow]

w3 = u3 + v3 = (2u1 - 3u4) + (2v1 - 3v4)
yes

[dull lily]

right but since the OG equation uses x1,x2,x3,x4,x5

do i have to go through that process for all of the problems

[surreal meadow]

what do you mean

what process

[dull lily]

ok so basically

within the vector set w

we have cx1,x2,x3,x4,x5

wait hold on

right here we mean v + u is within w correct?

[surreal meadow]

no

that is what we're trying to prove

[dull lily]

right so we just set them equal to each other

[surreal meadow]

let me explain it and then you can confirm if you understand

[dull lily]

ok cool

thanks!

[surreal meadow]

we have a set W

which has elements of the form (x1,x2,...,x5)

and satisfying those conditions above

for example

the element (0,0,0,0,0) is in the set

[dull lily]

right

[surreal meadow]

because it satisfies the conditions above

what part b is asking of us

is to show that

if u and v are both in W

meaning if u and v satisfy the conditions to be in W

then u + v is in W

meaning

u + v satisfies the conditions to be in W

forget i said w = u + v

we're going to call it something different

say x = u + v

let me know if you understood that or not

[dull lily]

yeah let me just

process rq

lol

ok

so

ive made it all the way up to here

[dull lily]

.

[surreal meadow]

before we do that

[dull lily]

all the work makes sense

mhm

[surreal meadow]

can you confirm you understood everything past this

[dull lily]

yes

everything up to this point makes sense

[surreal meadow]

and after that i want to clarify that W and w

are different things

[dull lily]

right

[surreal meadow]

ok

so we have

w3 = u3 + v3 = (2u1 - 3u4) + (2v1 - 3v4)

[dull lily]

right

[surreal meadow]

so let's rearrange

w3 = (2u1 - 3u4) + (2v1 - 3v4) = (2u1 + 2v1) - (3u4 + 3v4)

[dull lily]

ok so u kind of just swapped the orders of things

im cool with that

[surreal meadow]

yes

w3 = (2u1 + 2v1) - (3u4 + 3v4) = 2(u1 + v1) - 3(u4 + v4)

[dull lily]

right and we factors

[surreal meadow]

and if you recall

w = <u1 + v1, ..., u4 + v4, u5 + v5>

[dull lily]

right

[surreal meadow]

so
w3 = 2(u1 + v1) - 3(u4 + v4) = 2w1 - 3w4

[dull lily]

lets cut out the middle man rq and you get w3 = 2w1-3w4

is it because we were able to get back to our condition

we proved that u + v vector are within w

?

[surreal meadow]

not yet

this is only 1 of the conditions

[dull lily]

i see

[surreal meadow]

we need to prove all conditions are satisfied by u + v

[dull lily]

right so we need to check the rest of the conditions

[surreal meadow]

yes

and its basically the exact same thing

[dull lily]

right so let me see if i can start us on this one

would we start with x2=3x1+4x5

then sub it out with u2 = 3u1 + 4u5

then same process for v

[surreal meadow]

take two vectors

u and v

same way we did before

and let w = u + v

[dull lily]

Basically right

[surreal meadow]

yup looks fine

ideally you'd include some comments on why each of those steps work

but yes

[dull lily]

the last condition looks kind of weird

right let me add that in

XD

[surreal meadow]

it's more of the same

just flip the equality

[dull lily]

uhhhh right ok

so

for this one

the 7 coefficient is throwing me off

[surreal meadow]

divide both sides by 7

or

maybe don't even do that

let's start with x = u + v

[dull lily]

right

[surreal meadow]

then x5 = u5 + v5, right?

[dull lily]

yep

[surreal meadow]

so 7x5 = 7u5 + 7v5

[dull lily]

right

[surreal meadow]

do the substituting stuff again

just forget the 7 is there on the left side

[dull lily]

7u5 = 2u3 +6u4

[surreal meadow]

yup

[dull lily]

ok cool

factored out you get 2 (u3+v3) + 6 (u4+v4)

[surreal meadow]

yup

[dull lily]

ok cool so

now we have all 3

satisfied

were chillin now right

[surreal meadow]

yes

[dull lily]

ok cool so by this crazy insane logic right

does that mean for part c, it would be x = uv

where u vector would be u = {u1,u2,u3,u4,u5}

and v vector would be u = {v1,v2,v3,v4,v5}

[surreal meadow]

no

[dull lily]

awe

[surreal meadow]

vector multiplication is not defined

[dull lily]

i was feeling super slick

[surreal meadow]

we want scalar multiplication

[dull lily]

oh wait youre right

so its just r as any numbre multiplied by u

where u = {u1,u2,u3,u4,u5}

so we have r which is any value

and u which is that massive value right

[surreal meadow]

r is any real number

u is a vector satisfying the conditions to be in W

we want to show ru = <r * u1, r * u2, ..., r * u5> is in W

meaning ru satisfies the conditions to be in W

[dull lily]

right

so can we start with

w = r * u(vector)

that would be an alright first step ye?

[surreal meadow]

yes

[dull lily]

u vector being u1,u2,u3,u4,u5

can we say

w1 = r (u1)

w2 = r(u2)

[surreal meadow]

yes

[dull lily]

w3 = r(w3)

so on so fourth

[surreal meadow]

that is the definition of scalar multiplication

for R^n

[dull lily]

this part?

[surreal meadow]

do you know how to multiply a vector like
<1,2,-2> and a scalar like 5:
5 * <1,2,-2> = ...

[dull lily]

yeah so if its like

then it would be <5,10,-10>

right

[surreal meadow]

yes

scalar multiplication is multiplication component by component

so when i wrote

r * u = <r * u1, r * u2, r * u3, r * u4, r * u5>

[surreal meadow]

i was saying this stuff

[dull lily]

right ok

so this is right

then?

[dull lily]

.

[surreal meadow]

then what

[dull lily]

ok cool

then we go back to our conditions

find x3

sub out x3 for u

wait hold on i might have lost myself

lmao i was feeling really confident for a second

was i right up to this point

[dull lily]

.

wait no

[surreal meadow]

honestly dont know what you're saying

do you just want to walk through it together

[dull lily]

ok so

yeah sure

u win lol

[surreal meadow]

let's review the statement

we are assuming r is in R

and u is in W

we want to show

r * u is in W

let's call v = r * u

now what is v3?

v3 = r * (u3)

and since u is in W

we know that u3 = 2u1 - 3u4

so we can say

v3 = r * (2u1 - 3u4)

= r * 2u1 - r * 3u4

= 2 * (r * u1) - 3 * (r * u4)

[dull lily]

mhm

ok let me process really quick

[dull lily]

sorry but the variables are kind of messing me up

Am I somewhat close with this statement?

[surreal meadow]

i would prefer if you pointed out what confuses you

so i can explain

you are essentially writing a lot of statements which i don't know if you will actually understand

[dull lily]

Right here is where I’m getting lost

[surreal meadow]

review our starting assumptions

r is a real number

[dull lily]

Right

[surreal meadow]

and u is in the set W

[dull lily]

Mhm

[surreal meadow]

if u is in W

then u satisfies the conditions to be in W

these

[dull lily]

mhmright

[surreal meadow]

so u must satisfy
u3 = 2u1 - 3u4

[dull lily]

yup

ok cool

makes sense

i think i see it now

[dull lily]

so the next step would be x3 = r (2u1-3u4)

?

[surreal meadow]

yes

though i recommend you stick to one letter

either v or x

[dull lily]

gotcha

uhhh i have x written on my paper so ill use x

but i know x and v are interchangeable

so from this point

i distribute the r

[surreal meadow]

i want to make sure you understand that this isn't just a

assumption
step
step
step
step
answer

thing

why we are doing the steps should be clear to you

(to the point where we shouldn't be discussing each single step)

if it's not clear that's fine

[dull lily]

theyre clear for me nwo where if im understanding this properly

[surreal meadow]

but why we do stuff is more important than actually doing it

[dull lily]

im 90% sure im understanding everything so far

up to the point where we distribute r

[surreal meadow]

this is really all just algebra

[dull lily]

yeah, the computation is not too bad

its just the understanding

[surreal meadow]

associativity, distributivity, commutativity

etc

right

which is why im suggesting you read carefully and understand the algebra

and then analyze why we did what we did

[dull lily]

mhm, understood

[surreal meadow]

rather than just taking it 1 step at a time

it's about the big picture

[dull lily]

right

im basing everything right now off of this

[dull lily]

.

which now i can say i confidently understand after analyzing all the steps

i understand how b and c work now

my only question is

[dull lily]

the next step after this would be to use associative property and sub out values to get x3 = 2(x1) - 3(x4)

[surreal meadow]

yes

well

yes

[dull lily]

but why does this prove that u * scalar satisfies the description provided

[surreal meadow]

"the description provided" is the conditions to be in W

[dull lily]

ah ok

so because they are able to be subbed in and out of each other, they satisfy the conidion

condition*

[surreal meadow]

i dont understand your question

[dull lily]

Show that if ~u satisfies the description of W and r ∈ R, then r · ~u does as well

this is what the question asks

the process of proving that its the same is the same as part 2b

correct?

[surreal meadow]

in this case yes

what you should be thinking of is more general

[dull lily]

like what?

[surreal meadow]

how do i prove closure under scalar multiplication?

i take an arbitrary scalar r and arbitrary vector u in W, and show r * u is in W

this is the process

and how do you prove closure under vector addition?

you take two arbitrary vectors u,v in W

and show u + v is in W

[dull lily]

ah i see

[surreal meadow]

the specifics of this example are symbol pushing

you'll be doing it a lot probably

but that's the "whatever" part

the important part is understanding closure

[dull lily]

so instead of thinking of this as a 1 by 1 process, i should be memorizing the justifications of each step because in a wider picture

right

[surreal meadow]

you should understand how to prove:

• inclusion of the additive identity (0)
• closure under vector addition
• closure under scalar multiplication

[dull lily]

right and i do understand how to prove these now

but part d asks for the basis

so how would i find the basis of these properties?

[surreal meadow]

we can use row reduction

[dull lily]

rref

yeah, we are able to use a calc for this part

which im super grateful for

would step 1 be to put all the conditions as a matrix

and then calculate the rref

[surreal meadow]

that's what im typing up

yes

[dull lily]

cool!

[surreal meadow]

give me one minute

[dull lily]

your good!

[surreal meadow]

[
\begin{bmatrix} x_1\x_2\x_3\x_4\x_5\end{bmatrix} = \begin{bmatrix} x_1\3x_1 + 4x_5\2x_1 - 3x_4\x_4\\frac{2x_3+6x_4}{7}\end{bmatrix}
]

[ocean seal]

maximo

[dull lily]

OH like that matrix

ok i understand

[surreal meadow]

[
\begin{bmatrix} x_1\3x_1 + 4x_5\2x_1 - 3x_4\x_4\\frac{2x_3+6x_4}{7}\end{bmatrix} = x_1\begin{bmatrix} 1\3\2\0\0\end{bmatrix} + x_3 \begin{bmatrix} 0\0\0\0\\frac{2}{7}\end{bmatrix} + x_4\begin{bmatrix} 0\0\-3\1\\frac{6}{7}\end{bmatrix} + x_5\begin{bmatrix} 0\4\0\0\0\end{bmatrix}
]

[ocean seal]

maximo

[surreal meadow]

so we get the following matrix

[
\begin{bmatrix}
1 & 0&0&0&0\
3&0&0&0&4\
2&0&0&-3&0\
0&0&0&1&0\
0&0&\frac{2}{7}&\frac{6}{7}&0
\end{bmatrix}
]

[ocean seal]

maximo

[surreal meadow]

do you see how we got here?

[dull lily]

uhhh

[dull lily]

why did we skip x2?

[surreal meadow]

there's no x_2 in the original vector

it's basiacally x_2 * <0,0,0,0,0>

[dull lily]

ohhhh

os its there

but since its a 0 vector

we dont put it in

ok

yes

i get everything so far

[surreal meadow]

then you can reduce this to rref

and the pivot columns correspond to the columns which are basis vectors

in this original matrix

[dull lily]

right using the calculator

[surreal meadow]

you may run into an "issue" where you don't have integer values in the basis vectors

but you can just multiply each row by a scalar that will give you an integer

[dull lily]

im so confused

what the integer value in the basis vector means

[surreal meadow]

first lets discuss what you got

do you know what the pivot columns are

in that matrix

[dull lily]

x2

?

[surreal meadow]

no

[dull lily]

uhhh then no i do not

[surreal meadow]

lets talk about what a leading 1 is

a leading 1 is a "1" that appears at the left-most position on its row

you'll notice there's 4 leading 1s there

[dull lily]

right

r1-r4

all have leading 1's

[surreal meadow]

if your column has a leading 1, it's a pivot column

so the pivot columns are

c1

c3

c4

c5

[dull lily]

ohhhh

ok

got it

[surreal meadow]

so in our original matrix

[dull lily]

mhm

[surreal meadow]

the basis vectors correspond to columns 1, 3, 4, and 5

but columns 3 and 4

don't have integer values in every component

[dull lily]

can you elaborate on this

[surreal meadow]

is 2/7 an integer

[dull lily]

ohhhh ok

so in regards to our og column

we know that c1 c3 c4 c5 all have a leading one in them so they are the pivots

which we use to relate back to the OG matrix

[surreal meadow]

yes

[dull lily]

so the basis vectors are

c1 c3 c4 c5 of the OG matrix

?

[surreal meadow]

yes, but we want them to all have integer entries

[dull lily]

like

[surreal meadow]

look at c3

[dull lily]

the whole column has to have an integer

right

[surreal meadow]

is every entry

in that column

an integer?

[dull lily]

no

only 2/7

[surreal meadow]

so we multiply that column

by some scalar

so every entry is an integer

[dull lily]

but 0 x r is still

0

?

so it would only make the last value an integer

[surreal meadow]

we want every value to be an integer

[dull lily]

right

so the entries in c3 are {0,0,0,0,2/7}

no matter what we multiply by, the first four entries will always be 0

[surreal meadow]

yes

[dull lily]

so why would be multiply that column by some scalar

[surreal meadow]

.

[dull lily]

wait

im dumb

2/7 is not an integer

.-.

oops

so multiply that column by 7

[surreal meadow]

yes

[dull lily]

and the same thing with c4

[surreal meadow]

yes

[dull lily]

ok

so i did it

and now the matrizx

has all integers

or 0

[surreal meadow]

forget the matrix

write down the columns

[dull lily]

right

[surreal meadow]

those are the basis vectors

[dull lily]

so c1 {1,3,2,0,0}

c2 {0,0,0,0,0}

[surreal meadow]

dont write c2

c2 is not part of the basis

[dull lily]

c3 {0,0,0,0,2}

oh only ones with values huh

[surreal meadow]

only ones that were pivots

[dull lily]

is c3 right

or would it be 7c3 {0,0,0,0,2}

[surreal meadow]

you shouldnt be including the c1, 7c3, etc

you should just be writing these as vectors

like

[\begin{bmatrix} 1\3\2\0\0\end{bmatrix}]

[ocean seal]

maximo

[surreal meadow]

and ideally put them all in a set

[dull lily]

just like that?

[surreal meadow]

these are basis vectors

your basis is a set of vectors

so
{<1,3,2,0,0>,<0,0,0,0,2>,...}

[dull lily]

right

i got it!

so finally last but not least

the dimensions

if i remember correctly

every matrix is A sub(mxn)

[surreal meadow]

the dimension is the number of basis vectors

that's a definition

[dull lily]

we have 4

basis vectors

so is dim(w) just 4

?

[surreal meadow]

yes

[vapid shuttle]

yes

[surreal meadow]

i concur

do you concur austin

[vapid shuttle]

I also do concur

I agree with the OP that the dim(w) is just 4

Do you agree that the dim(w) is equivalently 4 maximo?

[surreal meadow]

i believe i do

now this may come as a bit of a shock of a question

but austin

do you agree the dimension of the subspace is 4?

[vapid shuttle]

!help

[lone heart]

Please read #❓how-to-get-help

[vapid shuttle]

open an available channel buddy

[surreal meadow]

!nosols

[lone heart]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[vapid shuttle]

this one is occupied

[dull lily]

does dimension have anything to do with nullity?

[surreal meadow]

sort of

[vapid shuttle]

dimension no, rank yes

[dull lily]

cuz i know the dimension theorem talks a lot about how dimension of col and dimesnion of row space

ah thats the word

rank

[surreal meadow]

there is a nice result here though

which ties it back to dimension

[dull lily]

mhm

[surreal meadow]

every subspace of a fintie dimensional vector space is the image of a linear transformation

so the dimension is always the rank of some transformation

so there is a connection between dimension and nullity

[dull lily]

right

ok

everything makes sense and i feel a lot better now

thanks so much for your help!

[surreal meadow]

no problem

[vapid shuttle]

your welcome

[dull lily]

i really appreciate you taking the itme to help me!

[vapid shuttle]

anytime

[dull lily]

.close

[lone heart]

Channel closed

Closed by @dull lily

Use .reopen if this was a mistake.

[surreal meadow]

t!rep austinu

[quasi bramble]

Sorry, only verified users can use this command!

Command: Reputation
By using Tatsu Bot, you agree to follow our rules. View them with t!rules.

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[vapid shuttle]

t!rep

[quasi bramble]

Sorry, only verified users can use this command!

Command: Reputation
By using Tatsu Bot, you agree to follow our rules. View them with t!rules.

To get verified:

1. Check if your email is verified by Discord
2. Login to our Dashboard once.

[vapid shuttle]

we suck

[lone heart]

Available help channel!

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• Show your work, and if possible, explain where you are stuck.
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[surreal meadow]

do you have a question

[vale wigeon]

"yeah i want someone to hand me the answer"

[tardy stag]

this is just number crunching, compute AB and AC and verify that they're the same

[vapid shuttle]

verify it

[rose sigil]

[vale wigeon]

@balmy vector you there?

[balmy vector]

oh yeah

[alpine sable]

calculate AB and AC

that's all you need to do

[vale wigeon]

ok so what's troubling you with the problem

[balmy vector]

yeah thanks i got it!

[vapid shuttle]

go ahead and (.close) then

[lone heart]

@balmy vector Has your question been resolved?

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[true sandal]

\begin{align*}
\lim_{x \to 1} \frac{\sin(x-1)}{-(x-1)}
\end{align*}

[ocean seal]

casiofx991exz

[true sandal]

Guys how do I use the lim x to 0 of sint/t = 1 here

can I say

Let t = x - 1

t -> 0 <=> x -> 1

?

can I say that?

[rose sigil]

something like that

[true sandal]

😦

how to say it properly

[subtle birch]

yes

[left isle]

Let t=x-1,
Then as x -> 1, t -> 0.

so yes basically

[subtle birch]

x->1 implies t->0

[true sandal]

nice

[true sandal]

is this proper enough

I think <=> means equivalent

[rose sigil]

it’s not really proper

[left isle]

that specific arrow is not really used in this kind of case

[true sandal]

so I should just do this?

[left isle]

Yes

[true sandal]

thanks

I got another question

When proving limits don't exist in multivariable calculus, do I really have to check all the time if the thing I'm substituting has the point we're approaching

all the time?

Do I have to say something like, y = x passes through (0,0)

all the time?

just an example

[slate stone]

What does Sin stand for

[true sandal]

I will @ mods

[lone heart]

@true sandal Has your question been resolved?

[left isle]

are you talking about that second step? where you found that as x approaches 1, t approaches 0?

[true sandal]

nah

1 sec

\begin{align*}
\lim_{(x,y) \to (0,0)} \frac{5xy^3}{x^2+2y^6}
\end{align*}

[ocean seal]

casiofx991exz

[true sandal]

this for example

let's say I do

Limit along x = 0

Do I always have to point out if x has (0,0)

Because writing sentences take tiem

I am slow at that

[left isle]

assuming you do a substitution like z=xy+1, then you would plug in (0,0) and say: as (x,y) -> 0, z -> 1

and then z -> 1 would go under the limit of your new equation

sorry i meant to put 1's in the example. i changed it

[true sandal]

yeah

[slender marten]

You can often come up with a family of paths such that the limit changes for different members of the family. For example, x = ky^3, to show the limit doesn't exist.

[lone heart]

@true sandal Has your question been resolved?

[lone heart]

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[safe tartan]

how do you do ii).

Im guessing i have to construct a differential equation but not sure how

[wind cloak]

Solid of revolution

Oh wait 2

Shit

Uh just differentiate the function V(u)

[lone heart]

@safe tartan Has your question been resolved?

[safe tartan]

but its repescts time

[wind cloak]

yes so differentiate wrt time

[lone heart]

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[regal remnant]

I need some help with this, I know that triangles ABC, CEB and BDE are similar, and triangle BDC is similar to CEA. Im not sure how to continue from here tho

[flat lava]

Let BE=x

ED is 4 since BC=BE

x/4=9/x

Solve for x

[regal remnant]

Damn thats easier than i thought

ok ty

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[remote flint]

Can you explain the second last line to the last line of the solution

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<@&286206848099549185>

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<@&286206848099549185>

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[lapis furnace]

(√12+2)(√12-2)-(√3-2)^2=
1-4√3 <- is that answer i got
but i was wrong andi t said 1+4√3 is correct
why?

[lapis furnace]

i was doing (√12+2)(√12-2)-(√3-2)^2=
then i got 1-4√3
cuz (√12+2)(√12-2) can be solved by the x^2-a^2 equation
and (√3-2)^2 can be solved by the x^2-2ax+a^2 equation
which makes (√12)^2-2^2-(3-4√3+4)
then thats 12-4-7-4√3 cuz (3-4√3+4) is the same thing as 3+4-4√3
but then i was wrong and it said the answer is 1+4√3
why?

[ornate ginkgo]

then thats 12-4-7-4√3 cuz (3-4√3+4) is the same thing as 3+4-4√3

Bracket

Inside the bracket it's negative

[wary stream]

which makes (√12)^2-2^2-(3-4√3+4)
then thats 12-4-7-4√3 cuz (3-4√3+4) is the same thing as 3+4-4√3
You forget to distribute the negative to -4sqrt(3)

[lapis furnace]

wait let me process it

[lapis furnace]

OHHH

so the -(√3-2)^2's first minus also affects the -4√3

i forgot about that

[lapis furnace]

ty!!!!

ty @ornate ginkgo @wary stream

i understand what i got wrong now

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[ocean seal]

10001016114

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[indigo coral]

how to show that when $V$ and $W$ are vector spaces with respect to the field $F$ that the tensor product is equal to the span of all elementary tensors so: $V \otimes_F W = span{v \otimes w | v \in V, w \in W }$ by only using the universal property of the tensor product. I know how to show that both sets are isomorphic, since they have the same universal property, but I don’t know how to show that they are the same set.

[ocean seal]

10001016114

[worn fox]

They are not the same set in general, the tensor product is only defined up to (unique) isomorphism

[indigo coral]

Yeah I see rn that I wrote the set wrong I mean the span of the elementary tensors is equal to the tensor product. How can I show this to be true

[worn fox]

I know you meant span, and my message still holds

[indigo coral]

I know that the universal property is defined up to unique isomorphism, but my prof said that one could also proof this fact by using the universal property

[worn fox]

If you've already shown that the span of elementary tensors is a tensor product then there's nothing else to do

[indigo coral]

But is this enough to conclude that it’s the same set as $V \otimes W$?

[ocean seal]

10001016114

[indigo coral]

I know that it’s a tensor product, but why exactly this one

[worn fox]

No, and it doesn't have to be the same set

There are many tensor products, they are all isomorphic and not the same sets

[indigo coral]

ah that’s what you mean

The notation of being the same is being isomorphic for tensor products

But they also are the same set, because we can form a basis of these (atleast one can prove that)

So one should be able to prove that they are the same set or not?

[worn fox]

A tensor product is just anything satisfying the universal property

This need not be the set of the span of elementary tensors

But it will be isomorphic to it

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[vapid steppe]

is there a lhopitals rule for multivariable limits

[alpine sable]

No, it only goes for single-variable limits only

[slender marten]

There is a version.

The notation is a bit weird.

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[dire star]

how am I supposed to find the asymptotes if i dont have h(x)?

[gleaming ridge]

You're not asked to find the asymptote

Just guess which of the graphs

[dire star]

how else would i identify them

[gleaming ridge]

I mean, look at A and B

How do they differ?

[dire star]

a goes up to y=2 and b only goes up to y=0

[gleaming ridge]

Yesss, for A, f(0)=2 and for B, f(0)=0

One of them cannot be possible

Look at your numerator!

[dire star]

how can I test the value without the denominator though

what if its negative or something

would b not be possible because of dividing by 0

[gleaming ridge]

B is saying f(0)=0

And to have a rational function = 0

The numerator must be = 0

Our numerator at x=0 is not 0

So, f(0) is not equal to 0

[dire star]

oh ok

thank you

[gleaming ridge]

wait

A and C remain

Show us C

i cannot see it D:

[dire star]

f(0)=0 for all of them

[gleaming ridge]

Oh, so it's A

The only possible one

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[wispy ice]

Good Afternoon mathematical ppl. In the photo, I didn't understand this until now. They didn't express 3 hm then 6 dam, they took 364 meters. Because it is on a scale or the metric system
Like 1 m 67 cm is = 167 cm
ALL is right??

[limpid spade]

whats 2 dam 5m and 7dm in meters

[hoary blaze]

anyone need help?

[limpid spade]

no

[hoary blaze]

ok

[rancid trail]

The creator of this table was on some drugs

It took me 1 minute of pure staring to understand what was going on

[limpid spade]

???

[hoary blaze]

itz toooooo easy

[rancid trail]

I have never seen a table like that before

[vale wigeon]

this is never an appropriate thing to say

anyway so it looks like each row in this table is a single length measurement but given in this maximally broken down form

[rancid trail]

Go check out other channels if you don't have a job

[vale wigeon]

but... @wispy ice can you repeat what your issue is with this problem?

[grand oasis]

So 1m 67 cm is basically 1m + 67 cm.

[wispy ice]

the understanding of what thy wanted me to express because the numbers are broken up
But thank you all for your help and this community

[limpid spade]

you got 25,7 m for second one?

[grand oasis]

Convert it to the "whole" form rather than the "split" form would make it easier for you to understand.

For instance 1m 67cm = 1m + 67cm = 100cm + 67cm = 167cm

[wispy ice]

25.7m

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[steep latch]

Why have I been given a function with one line while every other YouTube video I find about Fourier series showing two lines of the function

[honest hawk]

your function is correct :
x² in the interval -3/2 to 3/2 , it is 3 length
then in the second line, you have 3-period
so its good

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[slender gull]

Is that all that's given?
Nothing about differentiability and such?

[tawny crown]

By hit and trial f(x)=x² comes out to be one such function..

[steep latch]

Sorry for late reply! Thank you for explaining

[slender gull]

There's others as well, guessing will only get us so far.
Besides something like f(x) = 0 works as well, nothing sophisticated to think about.

[tawny crown]

True

One thing we know is the function is going to be even for sure

[slender gull]

Yeah.

[lone heart]

@daring totem Has your question been resolved?

[daring totem]

both f(x) = 0 and f(x) = x^2 satisfy the functional equation, idk about differentiability

Nothing too rigurous

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[daring totem]

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[foggy lava]

does anyone know how to tell if function is continous?

[honest hawk]

hi,

[foggy lava]

hey

[honest hawk]

you check on the bounds of the interval

so in x= 0 and in x= 2

no ned moiren because 2+t is continous on R, 2-t too

in x = 0 in first interval you calculate f(0)

[foggy lava]

I don't really know what those meant

[honest hawk]

and you compare to f(0) in the second interval

[foggy lava]

I haven't done ned moiren

[honest hawk]

in [-2;0] you have f(0) = 2 + 0 = 2

Do the calculation in [0;2]

[foggy lava]

2-0 = 2 as well

but how is that continous

[honest hawk]

so f is continous in x = 0

[foggy lava]

so by continous it means that they connect?

[honest hawk]

on a graph, it means they connect yes

[foggy lava]

but why does -2 and 2 matter then

[honest hawk]

you can draw the curve of f without lifting the pencil from your paper

now you check in x = 2 : the bound between the second intervall and the 3rd interval

[foggy lava]

f(2) = 0

f(-2) = 0

[honest hawk]

yes

[foggy lava]

but that doesn't show they conenct

[honest hawk]

yes it is continuous

because for x = 2 : before the 2 and after the 2 they connect in image 0

for x = -2 it is the same

[foggy lava]

image 0?

[honest hawk]

draw the cruve you will understand better

[foggy lava]

i have drawn it

but those two points do not connect back to each other

they are the outer bounds

[honest hawk]

https://i.gyazo.com/d42b5407236edd7a259610a0fc302dce.png

continuous in x=-2, x= 0 , x= 2

[wicked fjord]

This doesn’t make sense can someone explain

[winter light]

Show your drawing, maybe there's something wrong in it

[foggy lava]

oh ok i sorta get how its continous

the 0 goes on forever

but what if it doesnt have 0

for otherwise

[honest hawk]

it is said f(x)= o otherwise

🙂

[winter light]

Then it would be continuous only at x = 0

[foggy lava]

so if i wanted to say the function was continous at t

it would just be if the bounds of each function connect

[winter light]

What puzzles you in particular??

[wicked fjord]

A and b on the picture I sent

[honest hawk]

when the function is defined by intervals, you look for continuity at the bounds of the intervals

[wicked fjord]

They didn’t really do any math but just got an answer

[winter light]

Not mathem. rigorous, but that's the idea yeah

[winter light]

But where precisely?

[wicked fjord]

A)

[winter light]

Where is x² - 4 ≥ 0?

[wicked fjord]

Yes

[winter light]

I mean, solve that inequality 😅

[foggy lava]

if i wanted to prove the function was continous for my maths exam, how would i write it btw?

[winter light]

With limits

[foggy lava]

since you said it was not mathematically rigourous tho

[honest hawk]

btw the exercice you had is not rigourous

they put t = 0 in both intervall, it is a fault

[foggy lava]

yeah but i just want to know the correct thing to say in my exam

[winter light]

Here ( https://en.wikipedia.org/wiki/Continuous_function ) you can find it, I'm not able to write the formulas in a good looking manner

[honest hawk]

you first say that t --> 0 and t --> 2+t and t --> 2-t are continuous on R so they are contin ous on each intervall

then
you said you will check the bound of each intervall

[foggy lava]

yeah but in notation

[honest hawk]

for t = -2 : you calculate f(-2) = 2 + (-2) = 0
and you calculate
lim( t --> -2 , 0 ) = 0

then f is continous in t = - 2

[foggy lava]

lim f(t-) t->0 = lim f(t+) t-> 0

[honest hawk]

definition is :
f is continuous in a
<==>
lim (t-->a , f(t) ) = f(a)

[foggy lava]

ok i think i get it then

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[mortal river]

Not sure where to start with u substitution for this problem

[mortal river]

Could someone point me in the direction?

[mortal trellis]

when in doubt just u-sub the annoying part and see what you get

[barren portal]

the substitution here is slightly different

set u= 1+x so du= dx and x=u-1

[mortal trellis]

well I would argue the sum inside the sqrt is the annoying part

[lone heart]

@mortal river Has your question been resolved?

[mortal river]

You’re saying sub out the entire radical? Or just the 1+x term?

[mortal trellis]

doc gave you the sub

[mortal river]

Oh okay

[\int\frac{u-1}{u} du]
[\int u-1 *\frac{1}{u} du]

[ocean seal]

dopediscorduser

[mortal river]

Or could you do this?

[mortal trellis]

how did you get rid of the sqrt?

[mortal river]

[\int\frac{u-1}{u} du]
[\int \frac{u}{u}\frac{-1}{u} du]

[ocean seal]

dopediscorduser

[mortal river]

Oops

I didn’t I just forgot it

[mortal trellis]

I figured

[mortal river]

[\int\frac{u-1}{\sqrt{u}} du]

[\int\frac{u-1}{\sqrt{u}} * \frac{\sqrt{u}}{\sqrt{u}} du]

[\int\frac{u \sqrt{u} - \sqrt{u}}{u} du]

[ocean seal]

dopediscorduser

[mortal river]

Maybe something like this?

[mortal trellis]

dont overcomplicate it

just pull the fraction apart

simplify exponents

[mortal river]

Pull apart like this?

[mortal trellis]

u/u -1/u

ah wait sqrt

u/sqrtu -1/sqrtu

[mortal river]

Ah gotcha

Could you subtract the exponents of the u/sqrt u term?

Like 1-(-1/2)?

[\frac{u}{\sqrt{u}} - \frac{1}{\sqrt{u}}]
[u^1 * u^{-1/2} - u^{-1/2}]
[u^{-1/2} - u^{-1/2}]

[ocean seal]

dopediscorduser

[mortal river]

Would this be correct? Just in terms of simplifying?

[hollow jackal]

$1 - 1/2 \neq -1/2$

[ocean seal]

compusomnia

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[vivid tusk]

I don’t know how the answer to this equation was reached

2sq5 over 5

[foggy current]

You mean how $\frac{2/3}{\sqrt{5}/3}$ is related to $\frac{2\sqrt{5}}{5}$?

[ocean seal]

math_is_fun

[vivid tusk]

Yes

[foggy current]

First multiply the numer. and denomin. by 3

Do you know how to rationalize fractions?

[vivid tusk]

Apparently I have forgotten

[foggy current]

Heres a quick recap

$\frac{1}{\sqrt{c}-\sqrt{d}}=\frac{\sqrt{c}+\sqrt{d}}{c-d}$

[ocean seal]

math_is_fun

[foggy current]

In this case, c=5 and d=0

[vivid tusk]

I will have to look into this again. I don’t really remember learning that

[foggy current]

Maybe you haven't

[vivid tusk]

I’m in precalc learning about trig right now so that’s what all of this is

[foggy current]

In your question you can just multiply the numer. and denomin. by sqrt(5)

[vivid tusk]

Maybe it was somewhere today but I forgot

Thank you for the help

.close

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[main lantern]

Chi(z) is the function takes value 1 if |z| <= R, 0 otherwise

How to estimate the heat flux at |z| = R?

reference about taylor series of average around Ball or on the shell

d is the dimension, here d = 3

The relative estimate I got is depending on 1/R which is very large. But it should be a percentage which should be very small

My answer didn't depend on f(t) which is impossible, I must make it too large

the initial condition is T(r,0) = 0

which is initial flat

[alpine sable]

care to show your work?

@main lantern

[main lantern]

The second image is my progress

[alpine sable]

this?

[main lantern]

Yes

But the answer is too loose

[alpine sable]

right, its quite compact hah

[main lantern]

I am trying to get a better bound or estimation

I didn't even make it an upper bound

[nocturne hemlock]

@main lantern what r you studing for

[alpine sable]

right

[main lantern]

It is an engineering problem

[alpine sable]

thats thermodynamics

[nocturne hemlock]

thought so

[alpine sable]

the heat flux of a point on an obvect a point in time

[main lantern]

So in the space, there is a sphere with one material and the surronding is another

There is a thing called interfacial thermal resistance that is proportional to the heat flux at the contacting boundary

I am trying to estimate the resistance by estimating the heat flux

[alpine sable]

yeah, the layer between the two materials

[alpine sable]

so using the taylor series to determine the function between the point of contact?

sorry if my english doesnt make sense, its not my first language

[main lantern]

That is what I am doing

[alpine sable]

that's too broad though, no?

[main lantern]

it is too inacurrate

the problem starts from estimate this integral where T is the solution of the heat equation

I claim that it is bigger than 4/3 pi R^3 min Delta T

But that makes the magnitude so big

[alpine sable]

well yeah, but i dont see another way you can do it

[main lantern]

Do you think my estimation of Delta T(0,t) is good or bad

[alpine sable]

i would do it like that

i think the problem is with the taylor series

shouldt you use the chebyshev approximation to get a more accurate result?

since you're using polynomials to determine a function

[main lantern]

which chebyshev

[alpine sable]

1st kind

[main lantern]

I think the problem is here

this estimation has a remainder at order of R^2

[alpine sable]

thats the approximation of delta T?

[main lantern]

Which will shrink rapidly for small R

Yes

[alpine sable]

i think i see the error

if its that well

youre going to feel stupid when you figure it out

[main lantern]

Hmmmm i still didnt see it

[alpine sable]

did you solve for r?

R

[main lantern]

what do you mean

T(R,t) \approx T(0,t) + R^2/6 Delta T

• o(R^4)

[alpine sable]

oh nvm didnt see that

and the problem youre having is that the answer is too broad?

[main lantern]

2/RG is going to blow up to infinity if R to 0

[alpine sable]

yeah

[main lantern]

that is kind of impossible

[alpine sable]

R cant be 0

i dont see the problem

[main lantern]

Or maybe I am right

But that is percentage

The temperature drop cannot go down to negative

[alpine sable]

well actually

depends

[main lantern]

if 2/RG is > 1, then the outside will be negative

This equation is modeling there is a heat source at the ball

it is heating up the surronding

[alpine sable]

if the heat flux is getting "absorbed" by a material with a better heat conduction coefficient

or am i reading into it wrong

[main lantern]

I am thinking my answer is wrong

It is not realistic

because there exists materials that G is relative small compare to 1/R

[alpine sable]

yeah

[main lantern]

The solution will be very concave

so this is very loose

[alpine sable]

yeah, thats too loose

[alpine sable]

what engineering program are you in?

[main lantern]

in reality the temperature drop is not big

ECE

[alpine sable]

electrical?

damn, i did mechanical

[main lantern]

LOL ECE is board

It is not a typical engineering program lol

[alpine sable]

oh?

[main lantern]

I think a lot of professor in ECE are just doing convex optimization

or machine learning

[alpine sable]

well, isnt that hardware issue going to resolve itself with quantum computing?

[main lantern]

I know nothing about it

Well another confusion is I think I should use f(t) which is the heat source somehow

but I didnt use it at all

Which is weird

[alpine sable]

well, how would you use it

[main lantern]

I dont know, but intuitively they should be related

[alpine sable]

so heat loss?

wouldnt that make it overly complex?

[main lantern]

I dont know tbh

I have completely no idea

[alpine sable]

Q= dKA(T 1 −T 2 )
​

[main lantern]

Maybe I should post this to stack exchange

[alpine sable]

yeah haha

as much as id like to help, this isnt my domain

[main lantern]

Or maybe I should post it in ode and pde channel

[alpine sable]

i have a rough idea of what you're doing but i dont know the specifics of it, sorry

[main lantern]

It's fine

Nice chat with you

[alpine sable]

you too

good luck

[main lantern]

.close

[lone heart]

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[scenic wing]

How do you know what angles to use as points on the x axis

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@scenic wing Has your question been resolved?

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[dark crag]

Not sure the answer to this question.

[slender quiver]

what did you try

[dark crag]

Adding 3 to both sides then seeing the slope. And the question is looking for a line that is not parallel to that line so I was looking for something without the same slope but it was kinda hard to even put those answer in a form that I could properly see the slope.

[sour dove]

so what's something that we know about parallel lines? What do they all share in common?

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@dark crag Has your question been resolved?

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[vapid shuttle]

Don't close and reopen a new channel just to ask the same question

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@proud plank Has your question been resolved?

[proud plank]

My discord was bugging so I had to

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@proud plank Has your question been resolved?

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