help-36

[whole halo]

well if you need to go, we cant exactly help you with the question while youre away

[tranquil pine]

certainly

[whole halo]

linear combinations of what exactly

so youll need to .close the channel when you leave

[tranquil pine]

let's close it now i need to gtg asap

.close

[final saddle]

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[whole halo]

alr gl

tA + (1 - t)B is a straight line from A to B

consider that here, if you need a spoiler search up Jensen's inequality

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[warm python]

Can I have a hint

The lcm of the order of the elemnts must be 4

so I need $\phi(4)$

[soft zealot]

wai

[warm python]

right

[wheat mica]

So the combination of order must be (1,4), (4,1), (2.4) ....

[worldly mesa]

Z80000000 is cyclic so it has a unique cyclic subgroup of order 4

[warm python]

yup

that's it, right

[warm python]

same for Z_{40,0000}

[worldly mesa]

Yes, so you can embed Z4 x Z4 in a unique way in Z8000 .. x Z4000..

[warm python]

got it\

tq

.close

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[sharp wraith]

would i be right in saying in the case n = 10, there are no solution pairs, since y is rational, and x is rational, but root 10 / 10 is irrational, and you can't have a rational = rational x irrational

[frail moat]

yes

[sharp wraith]

thanks🙂

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[mental roost]

Context: Graph Theory, bipartite graphs, König's Theorem

[mental roost]

given is the graph above

i found a matching M, marked in orange

i also made partitions L and R, such that there are no edges (l,r), such that l in L and r in R

therefore this graph is bipartite i think

Next, I marked the exposed vertices in green

L* is the set of exposed vertices in L. Same for R*

The arrows on the graph where then created, such that all edges in the matchng M point from R to L. All edges in E\M (not in the matching) point from L to R.

In my script it then states the following

$C^:=(L\backslash S^)\cup (R\cap S^*)$

[soft zealot]

Martin

[mental roost]

and following König's Theorem: For any bipartite graph, the maximum size of a matching is equal to the minimum size of a vertex cover

I find C* to be {1,2,3,4,5,6,7,9,11} in this case, so |C*|=9
my matching however has 12 vertices.

isn't this a contradiction?

btw pls ping me when someone responds

[final saddle]

@mental roost Has your question been resolved?

[mental roost]

solved it!

my mistake was setting up S*

i thought that it asked for vertices that are reachable in one step

.close

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[glad kraken]

how did they go from $\frac{n^2}{n^3}$ to $\frac{n^3}{n^3}$

[soft zealot]

Mystic

[glad kraken]

in the 2nd/3rd line

[opal plinth]

,, \sum_{i=1}^n n^2 = n^3

[soft zealot]

Nel

[glad kraken]

what formula is this

[opal plinth]

,, \sum_{i=1}^n n^2 = n^2 \cdot \sum_{i=1}^n 1 = n^2 \cdot n = n^3

[soft zealot]

Nel

[glad kraken]

all i can remember is $\frac{n(n+1)(2n+1)}{6}$

[soft zealot]

Mystic

[opal plinth]

That's $\sum_{i=1}^n i^2$

[glad kraken]

doesnt this summation mean sum of first n squares

[soft zealot]

Nel

[glad kraken]

oh so that is i^2

what does n^2 even mean then

i^2 is just squaring and summing every element

[opal plinth]

n is the number of subintervals

[glad kraken]

what does this thing mean

[opal plinth]

The sum of n^2, n times

[glad kraken]

ahh k

alright one more thing

what if the question was about left hand endpoint

[glad kraken]

here it says right hand endpoint

[opal plinth]

It would be the sum from 0 to n-1 instead

[glad kraken]

for right hand its 1 to n, left hand its 0 to n-1,

and what about midpoint

[opal plinth]

It really just depends on how you define c_i

$c_i = i \Delta x = i/n$ for right-hand endpoint, so $c_1 = 1/n$ and $c_n = 1$ (the right sides of the first and last subintervals)

[soft zealot]

Nel

[opal plinth]

$c_i = (i-1) \Delta x = (i-1)/n$ for left-hand endpoint, so $c_1 = 0$ and $c_n = (n-1)/n$ (the left sides of the first and last subintervals)

[soft zealot]

Nel

[opal plinth]

$c_i = (i-1/2) \Delta x = (i-1/2)/n$ for midpoint, so $c_1 = 1/2 \cdot 1/n$ and $c_n = (n-1/2)/n$ (the midpoints of the first and last subintervals)

[soft zealot]

Nel

[glad kraken]

the riemann sum would become very tough then

if we use midpoint

[opal plinth]

I guess

Conceptually it's the same

Using midpoints means you need to sample the function at 2n points instead of at just n points

But since you're taking the limit anyway...

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[tranquil pine]

Consider the function f defined on [0,1] by (by the way (x,y) \in \mathbb{R}^2 and 0<x<y)

$t \rightarrow f(t) = \ln (tx +(1-t)y) - t ln(x) - (1-t)ln(y)$

By Studying f deduce that for all t $\in ]0,1[$ :

$t \ln(x) +(1-t)ln(y) \le \ln (tx + (1-t)y)$

Interpret the result geometrically

[soft zealot]

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

[tranquil pine]

can someone help i'm lost in useless calculations

in the question before this i proved this inequality with the mean value theorem

[final saddle]

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[final saddle]

@tranquil pine Has your question been resolved?

[final saddle]

@tranquil pine Has your question been resolved?

[final saddle]

@tranquil pine Has your question been resolved?

[soft zealot]

Civil Service Pigeon

[loud sundial]

L'inégalité souhaitée est équivalente à démontrer que $f(t) \geq 0$ et est très similaire aux approches standards pour la concavité. Ainsi, je vous encourage à considérer:
\begin{itemize}
\item Le signe de $f''(t)$
\item La valeur de $f(t)$ aux bornes de l'intervalle, à savoir en $t=0$ et $t=1$.
\end{itemize}

[soft zealot]

Civil Service Pigeon

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[steady vale]

hello!! i was wondering if someone can help brainstorm with me possible solutions for this? i dont want the answer right away, i wanna think about this

im stuck on this problem because if we pick a matrix like

1 0
0 0

acctually..

[desert mantle]

one possible option is to set A=(a,b; c,d), compute A^2 and then set enough of a,b,c,d=0 so everything in A^2 is 0

[steady vale]

ok i will give that a try; i think that when i was trying to solve the question, i forgot that matrix multiplication is multiplying rows and columns lol

you cant have something like

1 0
0 0
or anything where you just plug in positive numbers and zero only

the only combination that worked so far with me is picking a = 2 and then c = -2 and the rest zero so either multiplying by zero cancels out or -2(2) cancels out
0 2
-2 0

you can pick different numbers or maybe switch where you have #s and 0, like
2 0
0 -2

im wondering if theres another 2*2 matrix where thats possible

[desert mantle]

you can have matrices with only 0 and 1

[steady vale]

ok thank you, i need to think about this a bit more
actually
2 0
0 -2
doesnt work

--
is

0 -a
a 0
the only one that works?

[scarlet sequoia]

Have you tried your example?

[steady vale]

i tried

0 -1
1 0
and if im not mistaken, it should work

[desert mantle]

it doesnt

[steady vale]

oh?

how?

[scarlet sequoia]

Can you show your computation

[steady vale]

okay

[scarlet sequoia]

What is this matrix squared

[steady vale]

ill send a picture

youre right

it doesnt work

i dont know why i thought the -1 * 1 can just cancel out

[scarlet sequoia]

As a hint

Don't try to think about canceling things out

Just try to find a configuration where all coefficients in A² are 0+0

[steady vale]

thank you! ill try again

[scarlet sequoia]

(Ofc there are other matrices such that A² = 0 that don't follow this type, but they are a bit harder to find)

[steady vale]

youre right, i tried with a very simple example

0 0
a 0
where a = 1
and it did work

[scarlet sequoia]

Yep

[steady vale]

can we just move a anywhere and we will get A^2 = 0?

[scarlet sequoia]

So technically you've already solved the exercise, if you pick two different values of a

[steady vale]

thats true

[scarlet sequoia]

Try to think about if it's true for every position

[steady vale]

okay

0 a
0 0
where a = 1 works

but if you try to position 'a' anywhere in the main diagonal, it doesnt work

a 0
0 0
doesnt work

0 0
0 a
doesnt work

[final saddle]

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[gritty flume]

can i get help w this

[gritty flume]

im suppose to find the volume and surface area

but i cant find the surface area

[icy current]

!showyourwork

[final saddle]

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

[gritty flume]

ik the formula is 4pir^2

but it doesn’t work when I divide it all by 2

since its not a full sphere

[warped herald]

r is half of 5 m

[gritty flume]

2.5

[warped herald]

then plug r in formulas

and divide by 2

make sure you notice the units

[gleaming halo]

there is a circle too?

[gritty flume]

the answer they got ,reference is 58.90

i got 39.26

[warped herald]

for volume or surface area?

[gritty flume]

nvm thanks

surface area

[warped herald]

solved it?

[gritty flume]

yuh

the other person made me notice the circle

[icy current]

!done

[final saddle]

If you are done with this channel, please mark your problem as solved by typing .close

[gritty flume]

.close

[final saddle]

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[gritty flume]

.reopen

[final saddle]

✅ Original question: #help-36 message

[gritty flume]

i have a question

for the surface area of this

they didn’t use the whole formula for the cone

[frail moat]

elaborate

[gritty flume]

the cone formula is for surface area = (pi x r^2)+(pi x r x s)

but they didn’t use the first part

I don’t understand why

[icy current]

That's the TSA formula

[frail moat]

that's because they curved surface area of cone, here the bottom part of cone is not considered

as we have semisphere attached

[gritty flume]

so no one for the circle/

?

[frail moat]

yh you can say the bottom part is now inside the 3d figure and not on surface

[gritty flume]

does tht only work for all shapes when calculating the surface area

or is it also for the volume

[pliant shore]

not for the volume

no volume disappears when you combine two shapes together

for the surface area, you should always be thinking: "which parts are on the outside?"
then you'll figure it out

[gritty flume]

ic

tysm

.close

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[shrewd wren]

hi, i need help with letter (b).

[shrewd wren]

thanks in advance!

<@&286206848099549185>

[vital crag]

!15m

[final saddle]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[shrewd wren]

oh i apologize

[coarse aurora]

Which question?

[shrewd wren]

letter (b)

i have done this

i just tried to visualize

[coarse aurora]

Have you tried integrating?

[shrewd wren]

no

i am not sure what to integrate

which area

[coarse aurora]

So you want two integrals. One is to get the area between P and the curve until 4 (from the intersection til 4) the other is for the area between Q and the curve from 4 till their intersection

In general when two curves are given in the plane by continuous functions say f and g. The area between them on [a, b] is $\int_a^b |f(x) - g(x)|dx$

[soft zealot]

Beth

[coarse aurora]

But notice that the curve sits above both lines here? This means the absolute value simplifies

[shrewd wren]

so these two areas?

[coarse aurora]

The areas below

[shrewd wren]

oh yess

[shrewd wren]

ohh

[coarse aurora]

You’ve drawn your diagram nicely so it looks obvious (although of course it still needs to be shown)

[shrewd wren]

these

okiee

thanks

[shrewd wren]

but wouldnt this be difficult?

cause i dont have the function of the tangent

[coarse aurora]

Ahhh. Well you can get it. If you want to get the tangent of a differentiable function at a point, it’s $y = f’(a)(x - a) + f(a)$. I would suggest trying to geometrically intuit why, but this gives you your tangents

[soft zealot]

Beth

[shrewd wren]

ohh okai

thanks

[coarse aurora]

For the record, this is called a linearization

[shrewd wren]

okie

[shrewd wren]

is this correct?

[coarse aurora]

Looks good. Now you just have to do the next step

[shrewd wren]

yayy

okie

i will do that now

[shrewd wren]

so this would be the area for the first one

we can already see that the area doesn’t depend on m and n

which we want to show

[shrewd wren]

or we want to show it indirectly

we want to show that the area is equal regardless of the values of m and n

ohh i made a mistake i believe

[coarse aurora]

You have. It’s kind of a small one though

[shrewd wren]

shouldnt i find the integral from 2 to 4?

ohh

so i have to do it all over again?

[coarse aurora]

I would suggest finding the antiderivatives of the curve, P and Q so that it just becomes substitution

[shrewd wren]

oh i did that actually

[coarse aurora]

Since you seemed to have found the antiderivative, no. Thankfully you’ve done lots of the work

[shrewd wren]

yesss

[shrewd wren]

but now the answer i got is wrong

[coarse aurora]

For Q it is 4 6

[shrewd wren]

but this is for P

[shrewd wren]

but yes

[coarse aurora]

Because The peach region is from 4 to 6

[shrewd wren]

yes

but now i tried to find the pink region

[shrewd wren]

but i have done a mistake

cause the area should be 8/3

for both regions

[final saddle]

@shrewd wren Has your question been resolved?

[shrewd wren]

@coarse aurora hope its ok i pinged you

<@&286206848099549185>

[coarse aurora]

Hey

Sorry was busy. I am now in front of a desk. Try simplifying $f(x) - ( f’(a)(x - a) + f(a))$ symbolically

[soft zealot]

Beth

[coarse aurora]

So just hold $a$ constant and simplify this expression
Hint: you should get $(x - a)^2$ at the end and get something very symmetrical as a final step

[soft zealot]

Beth

[shrewd wren]

hii

[shrewd wren]

no worries

[shrewd wren]

but why f(x)-h(x)

[coarse aurora]

Well we are taking the difference between f and a linearization of f at a point in any case. So what can we say about it generally?

The simplifying turns out to make life much easier

[coarse aurora]

Like. Write integral sign maybe 4 times at most

[shrewd wren]

oh i meant why f(x)-h(x) and not h(x)-f(x)

[shrewd wren]

silly mee, i could have simplified here

but i did that

and still got the wrong answer

[coarse aurora]

Must be an arithmetic mistake. Things are correct up to line 3. I suggest looking at it and seeing if you can factorise?

[coarse aurora]

Recall this. Somehow you should get a nice factorisation

The same would go for the other integral

[shrewd wren]

(x-2)^2

right?

x^2-4x+4 = (x-2)^2

[coarse aurora]

Now let u = x - 2, du = dx. And the bounds of integration change too. With 2 mapping to 0 and 4 mapping to 2

Then you should get something with u^2 only in your integral

For f and Q, you do a very similar thing

[shrewd wren]

but what do you mean by "With 2 mapping to 0 and 4 mapping to 2"?

[coarse aurora]

Then you wouldn’t even need to evaluate the integrals technically speaking to determine if they are equal

[coarse aurora]

Ahh, let me clarify. $\int_a^b dx$ becomes $\int_{a - 2}^{b-2} du$

[soft zealot]

Beth

[shrewd wren]

oh i dont understand

wouldnt a = 2 and b = 4

for the area between h(x) and f(x)

[coarse aurora]

Nope. Because $u(a) = a - 2$ and $u(b) = b - 2$

[soft zealot]

Beth

[coarse aurora]

When you do a u substitution into a definite integral, you apply what you substitute onto the bounds of integration

[shrewd wren]

oh

[final saddle]

@shrewd wren Has your question been resolved?

[merry crypt]

How you did the latex ?

[final saddle]

@shrewd wren Has your question been resolved?

[burnt pond]

Struggling with 2A

it’s not an integration thing, this is ch4 applications of derivatives

no clue how to find the time traveled if the velocity is changing

[vital crag]

!occupied

[final saddle]

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).

[shrewd wren]

hm i still don’t understand why this is wrong

<@&286206848099549185>

[loud sundial]

sign error

[final saddle]

@shrewd wren Has your question been resolved?

[final saddle]

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[gritty flume]

Could i get help

[celest crane]

1a is just a variant of the Angles of Intersecting Secants Theorem.

[gritty flume]

huh

[celest crane]

https://www.mathsisfun.com/geometry/circle-intersect-secants-angle.html

[gritty flume]

i still don’t get it

[final saddle]

@gritty flume Has your question been resolved?

[timber basin]

For these questions, it is useful to remember that the tangent to a circle is at a 90 degree angle from the origin to the point. Hence, we can say that angle OBP is 90 deg in 1a, for example.
Thereafter, you can see that line PB and PA are equal, though not explicitly given in the problem. Knowing that, we also know that the other two angles in triangle ABP are equal, and so you can use the fact that the sum of angles in a triangle is 180, subtracting angle P.

Finally, now you know that OBP is 90 deg, and that ABP is (180-68)/2 deg. that means that angle OBA is 90-56 = 34 deg. Since OBA and OAB are equal, we get that angle AOB (or a) can be solved for with 180-(2*34) = 112

this is the sort of reasoning you can use if you don't want to use the theorems, it is doable and gives you a deeper understanding, but often times it is faster to apply the theorems.

b should be possible to do, remembering that the PAO is 90deg tho.

@gritty flume makes sense?

[gritty flume]

but how

i kinda understand

but the angle at a and b are 90 yes

but how do they go into the triangle

[timber basin]

So the angle at A and B can be split into two parts

you have both OBA and ABP which together add to OBP

and therefore, you have 90 = ABP + OBA

[gritty flume]

I don’t get it

[timber basin]

so if you know one, you can calculate the other.

[gritty flume]

why are they 90

[timber basin]

that's the interesting part, which is a bit tricky

[gritty flume]

in only know when tangents meet they are equal

[timber basin]

the tangent is defined as being basically intersecting the circle at exactly one point

and because there is only that outermost point, the line has to be exactly perpendicular

and it is 90 degrees.

[gritty flume]

oki i understand tht

[timber basin]

nice.

So when you have the triangles with a known angle, like in 1a, you can find the other two as well

because we know that they are equal

and so, you can say that the unknown angles, which are equal, each denoted by x

satisfy the equation: theta + 2x = 180

180 - theta = 2x

[gritty flume]

wait, what if i cut the 68 into two parts and find the a

[timber basin]

and we know theta = 68

thing is, we'll ultimately find that 68 + a = 180, by the theorem that guy mentioned

and this is basically the working towards that

[gritty flume]

where did the 2x comes rom

[timber basin]

i get you, there are three angles of a triangle, and they all sum to 180 deg. Critically, we know that the unknown angles are equal.

that is why there is 2x instead of something else

this is only because we are dealing with an isoceles triangle

[gritty flume]

is the equal angles the WHOLE a

[timber basin]

well, yeah actually

sum them up and they equal a

[gritty flume]

liked his?

like

[timber basin]

a = 112

so i think you got something mixed up in the start there

Nice spotting the triangles

[gritty flume]

wouldn’t the a be divided into two parts

find one part and x by 2

[timber basin]

but the unknown angle would be equal to a/2

so your triangle would be a 90-34-a/2 triangle

[gritty flume]

idk tht formula

[timber basin]

that's alright

but you drew the triangle

and that triangle is almost correct

[gritty flume]

ic

[timber basin]

you just have to label the corner where you had "a" "a/2" instead

[gritty flume]

so i should use a different variable for the divided a

[timber basin]

ideally i would do that yes

[gritty flume]

oki oki

ty ty

[timber basin]

this is how i would solve it

notice that the equal angles have dashes through

[gritty flume]

ic ic

ty ty

im gonna take tht

can i

[timber basin]

maybe try doing b?

[gritty flume]

ok wait

atp i only know the formal but not how to solve them

[timber basin]

try recognizing which angles are equal

sketch it out

For b, OAB = OBA

so you could use the same trick

[gritty flume]

maybe a and b

honestly idk im guessing

ik an angle in a circumference is 90

but tht one is 132

[timber basin]

the first thing i do is draw the triangle

because we know that it is an isoceles triangle

[gritty flume]

is every triangle like tht and isosceles

[timber basin]

so 180-132=2x

more or less yes

if the lengths of the legs are equal in length, then yes

often times you can recongnize these because they are in a circle problem

[gritty flume]

how do ik their equal

they didn’t gibe the lie tht

[timber basin]

the construction took away some parts that might be useful for recognizing it

[gritty flume]

i got 24

[timber basin]

yes, x = 24

correct

but b = 66

[gritty flume]

ooh so to find be

its 24 +b =90

right

[timber basin]

yes!

[gritty flume]

tysm person

[timber basin]

also, this is kinda interesting, because this problem set more or less illustrates that the angles have this relationship to eachother

notice that b is exactly half of 132

[gritty flume]

ohh

wait

[timber basin]

and for the 1a, notice that a + 68 is exactly 180

[gritty flume]

i need to also learn how to write the explanation tho

[timber basin]

you'll get it with some more practice 🙂

[gritty flume]

tysm

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[dense garnet]

I'm trying to show $C_c(X)$ is separable, where $X=\mathbb{R}^d$. We know that $C_c(X)\subset C_0(X)$, so if I can show the latter is separable, so will $C_c(X)$ be. The Stone-Weierstrass theorem for locally compact Hausdorff spaces I'm familiar with is:\

Let $X$ be a noncompact LCH space. If $\mathcal{A}$ is a closed subalgebra of $C_0(X,\mathbb{R})=C_0(X)\cap C(X,\mathbb{R})$ that separates points, then either $\mathcal{A}=C_0(X,\mathbb{R})$ or $\mathcal{A}={f\in C_0(X,\mathbb{R}):f(x_0)=0}$ for some $x_0\in X$.

[soft zealot]

psie

psie

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[radiant carbon]

for b

how is the root to this equation very small?

the graphs meet at x = 0

[cursive bough]

bc you drew the line incorrectly

2cos(4x) is at (0,2), not (0,1)

[tired walrus]

not at exactly x=0 they don't!

[radiant carbon]

mark scheme:

[radiant carbon]

okok

[tired walrus]

misaligned tbh

their 10x+1 line is off a few pixels to the left

[radiant carbon]

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[spiral olive]

Can anyone help me? I am trying but like i really dont get it, with the table of variations i am showing, i want for example to get the table of variation of u( v(x) ) but the problem is i dont understand the I1 (interval 1) or I2 (interval 2) or idk what, what i think when i read the theorem : for me it means that u function is like the g function, and v function is like the f function, which means : u : x [0 ; 3] which is I2 -> u(x) and v : x [0 ; 3] which is I1 -> v(x) [1 ; 3] which is NOT I2, but the problem is not stopping here, when i see the correction i notice that the table of variations of u( v(x) ) is not strictly increasing or decreasing, its not easy to explain but like i simply dont get it, idk if im missing a point with the interval im saying but i dont get

if you want, here is the correction of the table of variation of u( v(x) ) so you can maybe understand better

and yet, even with the correction, i dont get what the theorem means

[halcyon ether]

v(n) and u(n) are not monotonic function in [0,3]

The theorem states that it's true for f(x) and g(x) are strictly monotonic

[spiral olive]

By the way if anyone need to see a little bit more clear of what i tried

[spiral olive]

What does monotonic means actually?

Dont get it

Does it mean that like its only increasing or only decreasing on a certain interval?

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@spiral olive Has your question been resolved?

[proper dagger]

yes.

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[radiant carbon]

why doesn't
s = 16.8
u =
v = -19
a = -10
t =

work in v^2 = u^2 + 2as

19^2 - 2(-10)(16.8) = u^2 means u = sqrt(697)

which is not 5

[tepid bluff]

Displacement is downwards

Meaning s is negative

19^2-2(-10)(-16.8)=u^2 gives u=sqrt(25)

[radiant carbon]

okay

now doing the second part

-44 = -10t

t = 4.4s

but t = 2.4s in the mark scheme

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[dense garnet]

Consider the collection $\mathcal{E}$ of functions $\mathbb{R}^d\ni x\mapsto e^{-|x-q|^2}$, where $q\in\mathbb{Q}^d$ and $|\cdot|$ is the Euclidean norm. We do not include the constant $1$ function. $\mathcal{E}$ is a countable family of functions.
Consider the $\mathbb{Q}$-algebra that $\mathcal{E}$ generates, that is, linear combinations of products of elements in $\mathcal{E}$ with coefficients from $\mathbb{Q}$. Is the closure of this $\mathbb{Q}$-algebra a subalgebra of $C(\mathbb{R}^d,\mathbb{R})$, i.e. a real vector subspace, closed also under point-wise multiplication?\

I tried to verify this, but I got overwhelmed by the workload of things to check. My end goal is to apply the Stone-Weierstrass theorem for LCH spaces to some closure of a collection of functions that is countable (and this closure being an algebra), to prove $C_0(\mathbb{R}^d)$ and hence $C_c(\mathbb{R}^d)$ is separable.

[soft zealot]

psie

[midnight inlet]

Suppose you took a pair of elements in the closure

Say, $f,g$, since they’re in the closure you can get sequences $f_n, g_n \in \mathcal E$ such that $\Vert f_n - f\Vert\to 0$ and same for $g_n\to g$

[soft zealot]

Sharp, Archon of Severing

[midnight inlet]

Can you show $\Vert f_n g_n - fg\Vert\to 0$ and $r_n \to r \implies \Vert r_n f_n - rf\Vert \to 0$

[soft zealot]

Sharp, Archon of Severing

[midnight inlet]

Once you get those two, do you see how to proceed?

Note that none of this relies on basically any of our assumptions besides that you’re closing this normed algebra and that our operations are nicely continuous

(And being metric spaces with uniform continuity makes things much nicer)

[dense garnet]

Yeah, thanks. A key thing is though that our topology is that of uniform convergence on compact sets, right? So to verify that c f, where c is real and f is an element in the closure I need to consider sup_{x in K} where K is a compact set.

[midnight inlet]

Your topology is in fact just outright uniform convergence

And normed

Verify this, but clearly \sup |f(x)| < \infty for any f in C_0

Turns out it gives you a norm

And it’s exactly the topology you have here

[dense garnet]

Hmm, ok. I thought for LCH spaces we always consider the topology of uniform convergence on compact sets. (Since R^d is LCH.)

[midnight inlet]

So, I also want to point out that this isn’t “key” either, the real bit is continuity

[midnight inlet]

It vanishes at infinity, prove C_0(R^d, R) is a Banach space in this norm and that it’s the same topology

[dense garnet]

Oh boy. Ok. 🥲 But there would in principle be nothing wrong by choosing the topology of uniform convergence, right?

[midnight inlet]

They are the same thing

[dense garnet]

Sorry, meant the one on compact sets.

[midnight inlet]

Yea

They’re the same thing

When you’re dealing with C_0 here

Clearly it’s the same with C_c right?

[dense garnet]

Yes, makes sense. 👍

[midnight inlet]

The norm makes it easier by letting you throw things like these, but you can just as easily only ask about convergent sequences or nets or such here

But that’s more troublesome so it’s nice to say you don’t need to

[dense garnet]

Ok. I struggled with verifying the closure is closed under scalar multiplication (real coefficients). That took me some time.

[midnight inlet]

This is essentially just verifying that scalar and algebra multiplication are continuous btw

[dense garnet]

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[twilit jetty]

How do i shade the water

[onyx peak]

wdym by shading it?

makign it all blue?

y <= [your water level curve]

[twilit jetty]

ill try that one out

yeah

it works

thanks

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[onyx peak]

np, btw it can be used to color other stuff as well

e.g. if u wanated to make the boat red, you could do
boat floor <= y <= boat ceiling

[twilit jetty]

wait

how

[onyx peak]

oh, did u draw it by hand? Or is it a formula?

[twilit jetty]

its not working

[onyx peak]

If its drawn by hand, then it probably wont work

[twilit jetty]

formula

[onyx peak]

can u send the formula?

[twilit jetty]

its quite a bit, want the link?

[onyx peak]

sure

[twilit jetty]

https://www.desmos.com/calculator/n5cuqzciqr

LMAO THE THUMBNAIL 😭

the formulas are a bit messy, i only just started getting the hang of this

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[twilit jetty]

@onyx peak sorry to ping, it js closed the channel

[onyx peak]

https://www.desmos.com/calculator/fnlmb4qnao

its gonna take some time since your boat is defined in a lot of pieces, but you can fill it by bunch of inequalities like that one above

and u can mess with the opacity in settings btw

[twilit jetty]

ah ok

thankss

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[austere nymph]

does anyone get what my teacher is on about she assigned this as homework. i’ve gotten 10.000 but I don’t think that would be right or it wouldn’t say to 3d.p

The triangle with vertices
A (1,3), B(-2, 4) and C(3, - 5) is not right-angled.
Changing only one of these six coordinates can result in ABC being the vertices of a right-angled triangle.
Find (to 3d.p.) the area of the smallest right-angled triangle that can be created in this way.

[next thorn]

what does "3d.p." mean?

[austere nymph]

decimal places

might just be a uk thing

[next thorn]

I also got 10! by changing C's y coordinate... altho im not sure if that makes the smallest possible area 😔

Im not sure how to prove it either

[final saddle]

@austere nymph Has your question been resolved?

[next thorn]

yup! sorry I cant really prove if that's the smallest...

it will be extremely annoying if i were to keep guessing by changing the coordinates of the other points

especially because the question didnt state that the coordinates for the vertices have to be integers

unless... theres calculus in this 🤷‍♂️ good luck!

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[spice tangle]

I’m stuck on this question on my homework in which I’m differentiating the equation x^2+4y^2=7+3xy
This image shows the work I’ve done but according to AP Classroom my goal is dy/dx = (3y-2x)/(8y-3x)
can someone point out what I’ve been doing wrong? I got this same process three times now and can’t tell where a change would be important

[spice tangle]

This is Calculus >> Deriving implicit functions

[leaden moon]

$x^2 + 4y^2 = 7 + 3xy$?

[soft zealot]

1 divided by 0 equals Infinity

[leaden moon]

try to make a quadratic of $y$

[soft zealot]

1 divided by 0 equals Infinity

[spice tangle]

Yes

A quadratic? Hmm

[leaden moon]

something like $ay^2 + by + c$ where $a, b, c$ can contain $x$

[soft zealot]

1 divided by 0 equals Infinity

[spice tangle]

Before differentiation?

[leaden moon]

ye

[spice tangle]

I’ll try it, hold on

[leaden moon]

do it in your draft

[halcyon ether]

When take differential of 3xy you have to apply product rule

[spice tangle]

OH

[halcyon ether]

It's not 3dy/dx

[spice tangle]

that’s been my issue with so many of these I forgot about the product rule

Thank you Hunter 😭

[halcyon ether]

I got u

[spice tangle]

3$xy$ turns into $3(x)’(y)+3x(y)’$ right

[soft zealot]

@ON | Flappy the Turd

[halcyon ether]

Yes

[spice tangle]

Ok cool

I’ll get back once I’m done with whether it worked

Oh yeah I can already see this working

Thank you so much! I’ll make sure none of the other questions’ issues are product rules either in case I need to open another question in a moment

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[tranquil pine]

hi i have a question

[tranquil pine]

x here is supposed to be the vector written as the linear combination

but then when they do the matrix representation of the system they treat x like its the coefficient matrix, why??

[final saddle]

@tranquil pine Has your question been resolved?

[tranquil pine]

<@&286206848099549185>

can anyone help

much appreciated

<@&286206848099549185>

[hollow niche]

wait wdym treat x like its the coefficient matrix?

[tranquil pine]

mb I thought i sent it but here

[x]_B

[hollow niche]

[x]_B is just the coordinate vector of x in the standard basis B AKA x itself,

[tranquil pine]

Yeah but in the system it’s just the vector

in the matrix representation of the same system it’s interpreted now as the coefficient matrix???

howcome

[hollow niche]

im still a little confused about what you're confused about, can you point out this 'coefficient matrix'?

the 3x3 matrix, aka P, that I see is the change of basis matrix from B' to B

[tranquil pine]

the coefficient matirx is the constants that you use to linearly combine a basis to form some vector x

but when we construct the system of equations, that comes from expressing x as a linear combination of a non standard basis

so x here is what? A vector

but as soon as they expressed the system as a Ax=b, b here which is x became interpreted as the coefficient matrix?

i mean sure you did use it as a coefficient matrix when expressing x via the standard basis, but in the system, its just x

and you are representing the system

sooo

yea

[hollow niche]

im guessing it's taught differently around the world, but at my college, a coefficient matrix is a matrix of all the coefficients of a linear system

[tranquil pine]

are you down to vc? I can explain better

[hollow niche]

sry i cant vc

[tranquil pine]

its okay but

[tranquil pine]

Okay so look at the system

[1 2 -1]^T is x right

see?

[hollow niche]

yeah

[tranquil pine]

Okay is that a coefficient matrix [x]_B too??

[hollow niche]

yeah, since B is the standard basis

idk if calling it a coefficient matrix is correct though, since it's a coordinate vector/matrix

[tranquil pine]

but how

the matrix representation Ax=b represents the system

in the system x is just x

Its just the vector generated by the linear combination

[hollow niche]

right, but [x]_B is how you represent x in terms of the standard basis

the vectors of the standard basis form the columns for the identity matrix. so when you solve for I[x]_B = x, you get [x]_B = x

[tranquil pine]

How

like

1,2,-1 is the result vector

in the system

and Ax=b is supposed to represent the system

system says 1,2,-1 is the spanned vector

Ax=b says no, now its the coefficients c1,c2,and c3 that spanned 1,2,-1 via the standard basis

and that happens here

[hollow niche]

The coefficients c1, c2, and c3 correspond to the vector [c_1 c_2 c_3]^T which is [x]_B'. the coefficients c1 c2 c3 times the basis B' results in [1, 2, -1]^T

I hope it makes sense

[tranquil pine]

here I’ll explain better:

[x]_B’ = the vector column of the ** coefficients ** used to multiply by a non standard basis to get x

[x]_B = the vector column of the ** coefficients ** used to multiply by the standard basis to get x

Looking at our system, the right hand side is x

Now when we represented our system via Ax=b, b should be x, right?? And x is the spanned vector.

Yet in the screen shot, they label the x column as [x]_B, which we established what it represents above

So why did x become [x]_B if Ax=b is supposed to represent the system and the system treats the right hand side as x and not as [x]_B

[hollow niche]

yeah so you are right that b in this case is x. the reason they are representing x as [x]_B is to show the change of basis from B' to B. to convert from the standard basis to the standard basis, you multiply by the identity matrix, so x is [x]_B.

[tranquil pine]

i don’t understand bro ngl

are you saying they decided to express the result b suddenly as the coefficient vector column without clarifying that they decided to not interpret b as the right hand side of the equation anymore

[hollow niche]

wydm?

[tranquil pine]

Idk how to explain at this point 😂

[hollow niche]

yeah sorry 😭

yeah idk how else to explain tbh, but 3blue1brown's made a great video on it
https://youtu.be/P2LTAUO1TdA

[tranquil pine]

thank u bro

i appreciate your help regardless thanks

[tranquil pine]

No worries

[tranquil pine]

btw I still don’t understand how any of this is a “tranforation”

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@tranquil pine Has your question been resolved?

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[tranquil pine]

.reopen

[final saddle]

✅ Original question: #help-36 message

[tranquil pine]

i need help

[main mirage]

Hi! What's S?

Is S the standard basis

[tranquil pine]

i assume so yeah

theres a lot of intuition i don’t understand if you can help me out i really would appreciate it

coordinate matrix = scalars used to linearly combine a set of basis to express some vector

right?

[main mirage]

Yep

[main mirage]

Recalling from our discussion long ago, remember that a basis represents the 'building blocks' of a vector space. In other words, a coordinate in the basis B means that (a, b, c) in the B basis refers to:

• a units in the (1, 0, 0) direction
• b units in the (-3, 1, 0) direction
• c units in the (1, -2, 1) direction.
  That means to convert the basis from B to S, you need to figure out how much each 'direction' in B goes in a 'direction' in S.

There's a couple of ways people explain it, but this is my way: The strategy going from one basis to another is to look at how each basis vector gets transformed with respect to the other basis.

Here's an example:
Take the point (1, 0, 0). How would you represent it in the standard basis?

[tranquil pine]

you need to find c1 c2 and c3 : c1(1,0,0) + c2(0,1,0) + c3(0,0,1) = (3,-1,4) right?

[tranquil pine]

(1,0,0) is already standard right

[main mirage]

Yeahh but (3, -1, 4) is in terms of the basis B, so you should really write it as [(3, -1, 4)]_B

[main mirage]

Perfect! So (1, 0, 0) gets mapped to itself. In other words, (1, 0, 0) -> (1, 0, 0).

[tranquil pine]

yes

[main mirage]

How about (-3, 1, 0)?

[tranquil pine]

using the standard basis to describe this particular vector span would be a linear combination of a standard basis to describe this vector so:

c1(0,1,0) + c2(1,0,0) = (-3,1,0)

[main mirage]

Good (you should technically include the (0, 0, 1) vector too)

[tranquil pine]

but (-3,1,0) lives in a 2D space right

[main mirage]

So would you agree that (-3, 1, 0) = -3(1, 0, 0) + 1(0, 1, 0) + 0(0, 0, 1)?

[tranquil pine]

yes

i do

[main mirage]

Well it's technically still a 3D vector, just that it lies on the XY plane.

[main mirage]

Right, so here's the thing.
When we move one direction in the b direction in [(a, b, c)]_B, we are essentially moving -3 in the x direction, 1 in the y direction and 0 in the z direction, agreed?

[tranquil pine]

wdym by in the b direction

[main mirage]

Like

The middle component

The vector [(a, b, c)]_B is a vector according to the B basis

[tranquil pine]

yes

[main mirage]

So whenever we move one unit in the middle of the vector, i.e [(a, b, c)]_B

We are actually moving -3 in the x direciton, 1 in the y direction and 0 in the z direction

[tranquil pine]

but thats just moving in all components no?

i don’t understand tbh

[main mirage]

wait are you by any chance able to vc

[tranquil pine]

yes please

of course dude ive been waiting for so long 😂😂😂

one minute

[main mirage]

lol

how do you vc in this server

[pliant shore]

you can't VC for maths

there's only one Minecraft VC for active+

[main mirage]

Damn

[gentle zephyr]

You still need help or not

[tranquil pine]

here or in dms?

[tranquil pine]

oh

[tranquil pine]

yes

of course

[main mirage]

DMs if you don't mind

[gentle zephyr]

Ok let's do this asap

[main mirage]

o

[gentle zephyr]

Which one, a), B) or c)

@tranquil pine

[tranquil pine]

a

i mean all but starting with a yeah

[gentle zephyr]

The matrix that changes coordinates from B to standard is just the basis vectors of B placed as columns

That's all for a)

@tranquil pine

For example, if you have two basis, B1 and B2

[tranquil pine]

whats the intuition tho

[gentle zephyr]

You know linear transformations?

[tranquil pine]

I don’t think so

[gentle zephyr]

The matrix that changes coordinates from B ={b1,b2,b3} to C = {c1,c2,c3}, is the matrix that has columns (b1)_C, (b2)(C), (b3)(C), the intuition is that, when you multiply a 3x3 matrix with a 1x3 vector, you are essentially making linear combinations with the basis vectors

[covert mountain]

hiii

[gentle zephyr]

You are reconstructing that vector in terms of the basis you have, for example when B-> std you are transforming B coordinates to standard, because the matrix that has columns as the basis vectors of B when you multiply it by the coordinates of a vector in B you are telling the matrix the coefficients of the linear combination of how to form that vector, and the result of that multiplication is in standard basis

[tranquil pine]

is this what you are referring to?

[gentle zephyr]

Yes it all boils down to linear combination and solving systems of equations

[gentle zephyr]

In this case you are trying to find the coordinates of (1,2,-1) in the B' basis

(1,2,-1) is in standard basis

And you want to go from standard to B'

Simpler alternative would be to find the change of basis matrix that goes from std to B'

Aka, the inverse of B' -> std

,w {{1,0,2},{0,-1,3},{1,2,-5}}^(-1) * {{1},{2},{-1}}

[soft zealot]

Results provided by WolframAlpha

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Requested by renatoxbz

[tranquil pine]

whats the point of doing it like this instead of just solving the system

[gentle zephyr]

When you see change of basis of linear transformations maybe it would make more sense of why we use change of basis matrices

For now, just see it as a quick way of finding coordinates of a vector with respect to some basis

[tranquil pine]

thats the only way i see it rn

[gentle zephyr]

Later I think it will be fruitful when you get to more complicated math

[tranquil pine]

but is there a specific reason we represent it as Ax=b ?

like the system

[gentle zephyr]

Wdym?

[tranquil pine]

see how they represented the system as a Ax = b

Whats the point?

why not just do elemination on the equations

like the normal guassian way

[gentle zephyr]

You have (1,2,-1) and want to find (1,2,-1)_B' = (X,Y,Z)
By definition of coordinates wrt to a basis, that is
(1,2,-1) = Xb1 + Yb2 + Zb3 where B'={b1,B2,b3}

[gentle zephyr]

They used Gaussian when they say (matrix) bla bla The solution of this system is

[tranquil pine]

yeah but i can just do that without writing the system as Ax = b

[gentle zephyr]

The idea is to show you where this change of basis matrix comes from

[tranquil pine]

i see

[gentle zephyr]

C1,C2,C3 are the coordinates of (1,2,-1) in the basis B'

Aka (1,2,-1)_B' = (C1,C2,C3)

[tranquil pine]

can i aks something else

[gentle zephyr]

This is clearly showing that the Ax = b thing is doing that A is the matrix that changes coordinates from B' to std

You give it C1,C2, C3 that is

The coordinates that you are looking for your vector (1,2,-1) in B'

And it gives you the (1,2,-1) in standard

You solve for C1,C2,C3 and you find (1,2,-1)_B'

[gentle zephyr]

What

[tranquil pine]

see here in the picture,:

[x]_B’ = the vector column of the coefficients used to multiply by a non standard basis to get x

[x]_B = the vector column of the coefficients used to multiply by the standard basis to get x

Looking at our system, the right hand side is x

Now when we represented our system via Ax=b, b should be x, right?? And x is the spanned vector.

Yet in the screen shot, they label the x column as [x]_B, which we established what it represents above
So why did x become [x]_B if Ax=b is supposed to represent the system and the system treats the right hand side as x and not as [x]_B

i get that [1 2 -1]^T is also [x]_B but thats not what the system of equations was saying

the system of equations was representing (1,2,-1) as just the resultant vector

which is in the Ax=b correspondence thats b

But theyre saying b = [1 2 -1]^T = [x]_B

which the system didn’t interpret it like that

[gentle zephyr]

It's still the same thing

[mighty lantern]

Question

[gentle zephyr]

(1,2,-1) = c1(1,0,0) + c2(0,1,0) + c3(0,0,1)

Where (c1,c2,c3) = (1,2,-1)

[1,2,-1]_B = (1,2,-1)

[gentle zephyr]

Can you elaborate?

B is the standard basis

[tranquil pine]

yes but

in the system of equations

Which is the same as Ax = b

1 2 -1 is just the resultant vector

so why interpret it as [x]_B when it isnt the case in the system if equations

[gentle zephyr]

They are trying to show you it's also possible to do it not only from basis B' to standard but also B' to D where D is another basis

In this case D was just the standard basis, but you can also have change of basis that goes from one nonstandard basis to another non standard basis

[gentle zephyr]

Let's agree that x = [x]_B when B is standard basis, so still saying same thing in my perspective

We can also do it from basis B' to another basis D

[tranquil pine]

sure but

[gentle zephyr]

Where D is non standard

[tranquil pine]

yeah i don’t understand this

would it be the same exact representation?

[gentle zephyr]

No

The book gave a bad example

Because

[tranquil pine]

agreed

[gentle zephyr]

If you have a basis B = {b1,b2,b3} and basis D = {d1,d2,d3}, then the matrix that changes coordinatew from B to D is the basis vectors from B written wrt D, so the columns would be (b1)_D, (b2)_D, (b3)_D and this is harder to encapsulate in a system of equations little and simple example

in the last example that they gave it worked since the coordinates of the basis vectors of B' in the standard basis B are still the same, but if you use a non standard basis for B, the last example falls apart

[tranquil pine]

do you not recommend we write it out algebraically then as a matrix?

[gentle zephyr]

up to you really, but in linear algebra you try to prefer turning things into matrix form rather than keep it in system of equations form

also, at some point you start making good use of the matrix form factor

[tranquil pine]

@gentle zephyr i have a question, so I noticed that in this section they solved Ax = b ⇒ x = A^-1 * b to solve for the coordinates

but in other places such as solving for the nullspace of a row space, they didn’t take any inverse and solve like that

so can you do this way in any system of equations?

meaning can you rewrite any system then solve by taking inverse on both sides

[gentle zephyr]

nullspace of a row space?

[tranquil pine]

Nullspace

[gentle zephyr]

only if A its invertible, in this case, ALL change of basis matrices are invertible

[gentle zephyr]

well, solving for the nullspace is solving for Ax = 0

[gentle zephyr]

they do that because change of basis matrices are always 100% invertible

theres two ways as to see why they are always invertible

one is, the determinant is always non zero for a matrix made out of basis vectors (they are linearly independent)

also, forgot to mention, but when the determinant of a square matrix is zero we have a nontrivial kernel/the dimension of the nullspace is not 0

second is, my favorite, but when you get to linear transformations you will see that the change of basis matrices are the matrix representation of the identity operator acting on different basis vectors and the identity linear transformation is an isomorphism (bijective linear transformation, which of course has a invertible matrix representation)

[gentle zephyr]

only when the A in Ax = b, only when A is invertible

if say for example you are solving for the nullspace or the rowspace

and solve it like that, by using the inverse of A

you will be assuming A is made out of linearly independent rows and columns

and by that logic since A is invertible then nullspace(A) is empty

[tranquil pine]

but it is though

the row space does contain linearly independent vectors

[gentle zephyr]

the thing is, you know the concept of dimension?

dimension is the number of vectors a basis has

so for example, say that you are solving for the rowspace

and say you find a basis for the rowspace, that is made out of linearly independent vectors

then, if your matrix A is square, say for example a matrix 3x3

theres two possibilities

either the dimension of the rowspace is 3

or the dimension of the rowspace is < 3

[tranquil pine]

could be 2 or 1

yes

[gentle zephyr]

in those cases

A is not invertible

but if the rank is 3, then the 3x3 matrix A is invertinle

A is only invertible if its a square full rank matrix, where rank(A) = dim(Col(A)) = dim(Row(A)) where row and col means rowspace and colspace

[tranquil pine]

true because REF ⇒ a row with all zeros (in case the dimension is less than 3) which ⇒ 0 det(A)

[tranquil pine]

in this case can you actually solve in Ax = 0 for x?

[gentle zephyr]

you will get a trivial kernel/nullspace ={0}

[tribal ocean]

it's x=0 since there is only one solution if A is invertible

[tranquil pine]

good point

[tranquil pine]

but the null space for a 2D space is a line

how can a line be 0

say your square matrix has 2 independent vectors in R^3

wait thats not investable

[tribal ocean]

Then it's not invertible yes

[tranquil pine]

but wai t3 independent vectors in R^3 would have a 0 nullspace?

[tribal ocean]

yes when they're the rows/cols of a matrix A

[tranquil pine]

ohh

[tribal ocean]

If you know bijections, being invertible means you can go between x and Ax freely since its a pairwise correspondence

it maps R^n to R^n

And there is a matrix A^-1 that reverses the operation of A

[tranquil pine]

i see

i didnt get to that section yet tbh

[tribal ocean]

That's fine, this stuff is basically thinking about what happens when you try to go from one type of dimensional space to another

If you go to a higher dimensional space from a lower one, you can't cover every single point

[tranquil pine]

i see

[tranquil pine]

by the way, why is the graph stretching like that

My prof didnt teach us that

Ive seen this before too

but idk what it means

similar with determinants if I remember correctly

[tribal ocean]

If you think about the 2d grid you are familiar with, its usually drawn with squares right? your axes are just the lines for the vectors (1,0) and (0,1) passing through the origin

those vectors are called the standard basis vectors since you can think of every vector in R2 to be a combination of them

for example (3,2) is just going 3 units right and 2 up, so 3(1,0)+2(0,1)

But that's because we assumed that everything is made of (1,0) and (0,1)

When you change the basis you're basically changing your perspective of the grid

if you assumed everything is made of (1,0) and (1,2), then to get to the same point on the plane you would have to take different steps

These are the steps for the standard basis

If we use a different step system we should draw each basic square in the shape of our new steps instead, so hence the basic unit goes from the square to this parallelogram

This helps us count by hand how many of the (1, 0)s and (1, 2)s you need to follow to get to a point in our new grid

[final saddle]

@tranquil pine Has your question been resolved?

[final saddle]

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[granite tapir]

I wrote a foundations in Lean4, and have the axioms, operations, but this seems difficult, because I don't know these concepts yet:
field of quotients of an integral domain
Lie algebra of a Lie group

if they are functors then they are arrows
field_of_quoitents : 𝒞 ⟶ 𝒟

lie_algebra : 𝒞 ⟶ 𝒟

[tired walrus]

how much ring theory do you know

in particular do you know what an integral domain is

if you dont then uhhh im sorry to say this but this exercise is simply inaccessible to you yet.

[granite tapir]

Thanks @Ann, I will look these terms up

F : IntDom ⟶ Field
G : LieGrp ⟶ LieAlg

[tired walrus]

field of quotients is probably better called field of fractions imo bc it is a construction that almost directly mimics the construction of the rationals from the integers

and THAT is sth you're surely familiar with from school

[tranquil pine]

interesting

i mean i dont see how that would be helpful

or useful

but thats maybe cause i dont know much about it yet

[final saddle]

@granite tapir Has your question been resolved?

[granite tapir]

That's awesome insight, thanks @tired walrus , seems to makes it more accessible

So it's a field of things p/q, p, q integers?

[granite tapir]

a functor maps objects and arrows of a category to those of another category

IntDom has objects integral domains, and arrows ring homomorphisms between them
Field has objects fields and arrows field homomorphisms

Frac : IntDom → Field

R = ℤ, f(n) = 3n
fraction 2/3 ∈ Frac(ℤ)
Frac(f)(2/3) = 3·2 / 3 = 6/3 = 2
object ℤ → Frac(ℤ)=ℚ,
ℤ ↦ ℚ
arrow f → Frac(f)
f : ℤ → ℤ, e.g., f(n) = 2n ↦ Frac(f) : ℚ → ℚ, Frac(f)(p/q) = 2p/q

to show in general still seems difficult in my experience, maybe simple?

how to prove:
objects: R ↦ Frac(R)
arrows: f : R → S ↦ Frac(f)(p/q) = f(p)/f(q)
id preserved: Frac(id_R) = id_{Frac(R)}
composition preserved: Frac(g ∘ f) = Frac(g) ∘ Frac(f)

[granite tapir]

a field is an integral domain, but not always converseley is this relevant?

[worldly mesa]

Not very relevant

From every integral you can construct is fraction field

You need to show for very homomorphism of integral domains it induces a homomorphism on the fraction fields, and that it defines a functor

[granite tapir]

from Survey of Algebra - Mac Lane / Birkhoff
this seems relevant also

"The field of quotients Q(D) is a field for any integral domain D."

[worldly mesa]

Yeah

Pretty much by definition its a field

[granite tapir]

"Any integral domain D can be embedded isomorphically in a field Q(D), each element of which is a quotient of two elements of D." this is like the functor

[worldly mesa]

The homomorphism part is what you are really missing

[granite tapir]

Meaning how arrows are mapped?

[worldly mesa]

Yeah

But I am trying to remove abstraction

[granite tapir]

the composition of things in the integral domain matches those of the field of quotients

integer addition is same as rational number addition

+(a,b) ⟶ + a/b ? I don't yet understand it

a ↦ a/1
a+b ⟼ a/1 + b/1

[worldly mesa]

Yeah

[granite tapir]

f : ℤ → ℚ,
f(n) = n/1
a, b ∈ ℤ
f(a + b) = (a + b)/1 = a/1 + b/1 = f(a) + f(b)

[worldly mesa]

This is the localization map

You can use it to define the functor

Given g: A --> B, think how to define the induces morphism on a/1

[granite tapir]

add      : a + b ↦ a/1 + b/1 = (a+b)/1
mul      : a·b ↦ a/1 · b/1 = (a·b)/1
add_unit : 0 ↦ 0/1
mul_unit : 1 ↦ 1/1
add_inv  : -a ↦ -a/1
mul_inv  : a/b ↦ b/a, a ≠ 0

[granite tapir]

g: A → B in IntDom lifts to Frac(g): Frac(A) ⟶ Frac(B) in Field

objects map:
A ↦ Frac(A), B ↦ Frac(B)
arrows map:
g: A → B ↦ Frac(g): Frac(A) → Frac(B), defined by
Frac(g)(a/b) = g(a)/g(b)

is this local global pattern?

local maybe because pick specific things and operations with mapsto, but functor is global for it maps forall things and every map

which is maybe the whole point of the proof, show there is a global map

I appreciate your kind help @tired walrus and @worldly mesa
ty / thx

[final saddle]

@granite tapir Has your question been resolved?

[final saddle]

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[thin night]

I need help with my math studies. About finding a value of sin cos tan

[woven ledge]

could you be more specific?

[thin night]

This is the problem that I’m trying to solve

[woven ledge]

,rccw

[pliant shore]

which question exactly?

[thin night]

243

[pliant shore]

oh okay the other things were just identities

[woven ledge]

what does that say

[pliant shore]

can you draw theta = 7/6 pi on the unit circle?

then you should be able to convert it back to an angle such that 0 <= theta <= pi/2

(you might need to change the - sign into +)

[thin night]

For the following θ, find the values ​​of sinθ, cosθ, and tanθ.

[thin night]

Unit circle?

[woven ledge]

ok, do you know sin (pi/6)

[thin night]

I don’t think so-

[woven ledge]

hm

do you know that sin (x + pi) = -sin x

[thin night]

Yeah

[woven ledge]

right, so how can we use this to conver sin (7pi/6) to sin of some angle in the first quadrant

[pliant shore]

that's your issue

you should know that sin(0), sin(30), sin(45), sin(60), sin(90) are 0, 1/2, sqrt(2) / 2, sqrt(3) / 2, 1 respectively

[thin night]

Sin0=y/r?

[thin night]

Oh you mean that I didn’t know what it meant in English

[pliant shore]

I don't know

seemed like you didn't know it in any language

[woven ledge]

did you learn these values of sin in radians or degrees

[thin night]

Yeah but I was sleeping during those classes- well I mean I did study today about it during the day and did some few problems about it

Kind of understand what it is from YouTube videos

[pliant shore]

yes, then you'll also need the unit circle definition

the unit circle has r = 1, by definition

so sin theta = y/r = y
cos theta = x/r = x

[thin night]

Ohh

[pliant shore]

then you need to compare these angles 180 degrees apart

one of them is actually -q

below the origin is -q