help-28

[sinful wasp]

so i would do the equation and whatever i got for m1 will be the same as m13?

[hard veldt]

yea

[sinful wasp]

alr

i got 40

how do i start this?

?

[hard veldt]

What is the measure of $\angle ADC$?

[glossy valve]

Dork9399

[sinful wasp]

128?

ohhh

wait

is the answer 52?

like

[hard veldt]

yea

[sinful wasp]

alr

how would i start

[hard veldt]

what is the measure of $\angle ABC$?

[glossy valve]

Dork9399

[sinful wasp]

57

[hard veldt]

What is the measure of CBD?

[sinful wasp]

123

[hard veldt]

CBD, not CBE

[sinful wasp]

uh

[hard veldt]

its an isosceles triangle with a vertex angle of 90

what does the make the other two angles?

[sinful wasp]

im not sure

[hard veldt]

The angles of the triangle are 90, x, and x. What does that make x?

[sinful wasp]

45?

[hard veldt]

yes

[sinful wasp]

so b=45?

[hard veldt]

So ABC = 57, CBD = 45. What does that make EBD?

[sinful wasp]

78

[hard veldt]

So what is BDE?

[sinful wasp]

12?

[hard veldt]

yay!

[sinful wasp]

thats all

.close

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[sinful wasp]

ty

.close

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[charred geode]

can I cancel differentials ?
for example: if dx/dt = dy/dt does that mean that dy=dx?

[muted summit]

uhhh... kinda
whats the context

[hybrid nexus]

idt that makes sense

[muted summit]

because, dy/dx is its own thing,

But you can do that trick when doing integrals

[sacred sparrow]

no that'd mean the derivative of an arbitrary function is 1.

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[proven spade]

hello

[proven spade]

i need some urgent help!

[gritty rose]

!15m

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[arctic void]

Bruh

[proven spade]

?

[arctic void]

Do u not understand reduction and enlargement

💀

[proven spade]

not really, but thats why im asking

[proven spade]

<@&268886789983436800>

[arctic void]

<@&268886789983436800> this can't be a doubt

[proven spade]

bruh im just asking a damn question

[arctic void]

Rules state u have to try first

Pls share what u have tried

[proven spade]

I DID!

[arctic void]

.

[proven spade]

i just did a whole page of other work

i dont need to show you what I completed your not even a helper

<@&286206848099549185> its been 15 mins

[devout dawn]

Who need help?

[proven spade]

i do

some guy interjected into my question

he was really rude

he was juding me for not understand a question

[devout dawn]

I am very sorry for him, if we understand everything it is impossible to learn

So what is your question?

[pale otter]

kirito is a helper.

but for the question, if enlargement makes an image bigger, and reduction makes it smaller, then how do you think the small triangle b in question one must be scaled to make the big triangle A?

[proven spade]

this is really disrespectful honestly.

[proven spade]

.

[proven spade]

so the first one would be enlargment and reduction is the 2nd one?

[pale otter]

good, you got it

[proven spade]

so basiclly englargment is bigger and smaller is reduction

[pale otter]

yeah

[arctic void]

Nice✅

[devout dawn]

That is

[arctic void]

But u didn't share yr 2 page working

[proven spade]

okay and I have one more question

[arctic void]

.

[pale otter]

yeah

[pale otter]

yeah

[proven spade]

its asking to find the scale factor out of these two

but im not sure how to find it

is it the bigger one?

[devout dawn]

To find the scaling factor, divide the lengths of the corresponding sides

[proven spade]

so divide every single side?

so like 40/20 and 8/6?

[devout dawn]

For example in the figure on the right, the scaling factor is 2

The corresponding sides of the two similar figures

[proven spade]

for each?

[devout dawn]

The larger figure is obtained by multiplying each side of the smaller figure by 2.

8/4=2

6/3=2

[proven spade]

where do you get the 4 from

but its asking me find the scale factor

out of the 2

[devout dawn]

The scaling factor is the number by which you multiply the sides of one figure to obtain the other

[pale otter]

you see how the figures are congruent? (they look the same, with the same ratios between side lengths)

I think factoring could help:
try factoring the side lengths on each figure ans see what factors match up

[proven spade]

so would it be the 2nd one?

[devout dawn]

48/40

[devout dawn]

On the left side

8/4 On the right side

[proven spade]

sso would the correct on be on the left?

[full forum]

@proven spade Has your question been resolved?

[keen mango]

It looks like the question is asking for a number, not for you to choose a shape

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[twin wolf]

So I graphed f(x) in desmos and the sign testing holds up when I check the sign from (1, inf) it’s always increasing and from (-inf,-1) it’s always decreasing.

But what if I have a function that changes concavity somewhere towards -inf and + inf? Will the sign testing still hold up

[gritty rose]

What question are you using the sign test for

[twin wolf]

well i dont have to use it here but i just wanted to try and see if i could curve sketch it

all i had to do was find the max or min points and find the stationary point

im just curious if i was to use the sign test to curve sketch

would the sign test catch whether there would be a change in concavity in a later interval

like for example i know that this function will always be concave up from (1,inf) but what if it changed somewhere during that interval ( ik it doesnt but just asking)

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[twin wolf]

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[vapid spindle]

Find the smallest possible integer n, such thar n! is divisble by 2⁵×3⁵×5⁷

Is the answer 35 correct?
My reasoning was since we need 2, 3 five times in the factorial and 5 seven times we can go with five and find the number with five seven times.

Or is 30 the answer?
Since there's two fives in 25?

[full forum]

@vapid spindle Has your question been resolved?

[sacred yarrow]

30 is correct

[sacred yarrow]

There are eight 5 in 35!
5
10
15
20
25
30
35

It’s excessive

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[raven mauve]

hi

[raven mauve]

can someone see if i am correct please?

[raven cloud]

if the 2 after square root of 5 is a power yes ur correct

[raven mauve]

oh shoot my bad i didnt see

i thought it was a power

so in that case

its just 2 square root 5?

[raven cloud]

its weird how its behind the square root lol why would not they just write 2*(square root of 5)

im guessing its a power

[raven mauve]

ye thats what i thought

ye possibly a mistake

is this correct

wait oops

[raven cloud]

u gotta show that 45 is equal to that not that 3 square root of 5 is equal to that

square root of 45= square root of 9 times by square root of 5

then square root of 9 is 3

and its 3 times square root of 5

[raven mauve]

sorry like this?

[raven cloud]

ya

[raven mauve]

i messed up my bad

i keep making small mistakes

[raven mauve]

so questions like this

u just have to prove it?

[raven cloud]

yea

got anything else? im bored cause no one answered my questions in the help channel

i could help

[raven mauve]

this one is the last one

[raven mauve]

ohhhh

[raven cloud]

devide by 2 so its easier to solve

gonna be

sqrt(x)=8

[raven mauve]

just 8?

[raven cloud]

then square both sides

[raven mauve]

wait what

[raven cloud]

basically

if u devide both sides by 2

nothing changes

but it becomes more simple

easier to solve

[raven mauve]

so final answer is 8?

[raven cloud]

no

u gotta find x

[raven cloud]

not sqrt(x)

[vernal gale]

pythagoras

[raven cloud]

this channel is taken go to #help-5

[raven mauve]

ohhh wsit

[vernal gale]

ok

[raven mauve]

so x = 8

pretty much?

[raven cloud]

no

when u devide both sides by 2

what do u get

?

[raven mauve]

sorry im confused

as hell

what do u mean divided by both side

this?

[raven cloud]

yes

then

u gotta find x

[raven mauve]

so then u do 8-1?

waist no

wait

ye im confused

[opal isle]

10 = 7 + 3 <==> x7 both sides is 7(10) = 7(7+3)

2sqrtx = 16

÷2 both sides

sqrtx = 8

[raven cloud]

if u wanan get x from sqrt(x)

u gotta square it

x=8^2

x=64

[raven mauve]

huh

wait i think i got it

OHHH wait ye i got it

ty guys

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[hoary bluff]

hey can someone help me with a.

[hoary bluff]

im confused what its asking me to do

[hot herald]

long division

[hoary bluff]

yes but

like

what divided by what

well i know

but

if i do will it turn into the form a/x-1

[raven cloud]

do u know how to devide that thing by x-1?

[hot herald]

yeh, do the division

[hoary bluff]

yeah

[raven cloud]

do it

[hoary bluff]

okay ill do it and come back

brb

[hoary bluff]

okay im back I got x^2 + 3x + 8 + 9/x-1

[raven cloud]

thats it

[hoary bluff]

huh

[raven cloud]

x^2+3x+8 is P(x)

[hoary bluff]

oh

[raven cloud]

and a is 9

[hoary bluff]

so its just weird wording for solving long division

uurrrgghh

thank u tho

haha

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[vast rover]

Could I get some help with question 1 please? I have done the base case but how do I make use of the orbit stabiliser theorem (or its corollary) in the inductive step?

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@vast rover Has your question been resolved?

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@vast rover Has your question been resolved?

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@vast rover Has your question been resolved?

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[desert musk]

"Assume that f is uniformly continuous on (-∞, 0]. Also, assume that f is uniformly continuous on [0, ∞). Show that f is uniformly continuous on the entire ℝ."

[desert musk]

how do i show this

[sacred sparrow]

use continuity definition

to show that f is continuous at 0.

[manic token]

try to start from f（0）

[sacred sparrow]

then f is continuous on all R

[desert musk]

yeah, i suppose i only need to show it is on f(0)

[sacred sparrow]

well all other values have already been assumed to be true

[desert musk]

where both x and y are 0

or only one of them

[manic token]

x=0

[desert musk]

what about y

[manic token]

y=y

🙂

it doesnt matter

[sacred sparrow]

you don't know what f(0) is, f is an arbitrary function

[quaint prawn]

Uniform continuity tells you for any epsilon there is a delta that works for all x in those two intervals

Take the minimum of the two deltas you are guaranteed

And rejoice

[desert musk]

what exactly is the y then

[sacred sparrow]

you don't know? y = f(0)

[desert musk]

i thought we had x and y because of mutiple dimensions

[manic token]

it doesnt matter and you dont know it

[quaint prawn]

It doesn’t matter and you don’t know it

Just feel like it should be said again

[manic token]

yeah lol

[quaint prawn]

@desert musk

[manic token]

lim x=0+ ,f(x)=f(0)

lim x=0- ,f(x)=f(0)

em

[quaint prawn]

There’s no need for limits either

[manic token]

sorry

im not doing well

[sacred sparrow]

nw, help is always appreciated. The left/right side limit approach would work if it were continuity instead of uniform continuity :)

@desert musk what have you written so far?

[quaint prawn]

Because f is uniformly continuous on (-inf, 0]
For all epsilon > 0 there exists a delta1 > 0 such that for all x if
|x1-x2| < delta1 then |f(x1)-f(x2)|<epsilon

[quaint prawn]

Now apply the same definition on the other side of the interval

[quaint prawn]

The line about x<0<y is unnecessary

[manic token]

another??

[desert musk]

what?

i just never really feel done when i do epislon thing

you just write down the defintion and say now its proven

[sacred sparrow]

mind showing what you have?

I'd split it into:
Let x,y e R. we want to show that
|x-y|<delta ===> |f(x)-f(y)|<eps

[desert musk]

yes

[sacred sparrow]

if x,y both in the first interval, then it's true since delta ≤ delta1

[desert musk]

but i never feel like nice i got it when i do these sort of questions

you just write down the defintion and say "now it holds"

[sacred sparrow]

and x,y both in the second interval then it's true since delta ≤ delta2

[sacred sparrow]

hm, well have you shown that it actually holds?

[desert musk]

see im never really sure if i have

i dont have a picture in my head of what im doign when i do this

[sacred sparrow]

k a sec I'll switch to pc

here we know that f is uniformly continuous for all R\{0}

now if we let x to be in the first interval and y to be in the second, we'll need to show the same

[desert musk]

it is correct x and y just a points

that are some distance away from each other less than epislon

just like in the normal continuity but than y is instead of a?

[quaint prawn]

This is not correct

You have the quantifies backwards

[sacred sparrow]

oops I switched delta epsi

[quaint prawn]

That’s not the only issue

We have uniform continuity

There’s still an issue

there exists delta, such that for all x, y

Not the other way around

[sacred sparrow]

what am I doing lol

[desert musk]

can you draw me a visual graph of it

[quaint prawn]

We don’t know what f is

So no

[desert musk]

ill rephrase

an abitrary graph

[sacred sparrow]

it'll be more helpful if you get used to the abstraction

[quaint prawn]

I don’t think it’d be useful here

[quaint prawn]

Now take the minimum of these two deltas

If. |x-y| < that then what can you say?

Was telling them

[desert musk]

i tried nevertheless

[sacred sparrow]

yea the eps-delta rectangle is what you can visually infer

[desert musk]

then draw from x and y

up to the points on the graph

i think for my graph it doesnt even hold

but still

why are we letting it be the minimum of delta

?

[desert musk]

here

[sacred sparrow]

you'll see once we regard the third case

x in the first interval and y in the second

[desert musk]

then i wont learn

[quaint prawn]

Because then it works for both intervals

[desert musk]

i need to be able to guess the answer myself

[quaint prawn]

Say one delta is 2

[desert musk]

what is the rational behind it

[quaint prawn]

And one delta is 5

If we say, let our big delta be < 2

[sacred sparrow]

since delta <= delta1 and delta <= delta2, delta applies for the uniform continuity of both intervals

[quaint prawn]

Then we know we get both conditions

[desert musk]

why is your "big" delta smaller than delta 2

big doesnt imply being the biggest?

[sacred sparrow]

we defined delta = min(delta1, delta2)

I can also add it into the first two trivial cases iyw

[desert musk]

so you just the smallest delta of your two deltas

which ever one that is

so we write delta = min(delta1,delta2) now i am sure that i've picked the smallest delta. Why is that important?

[quaint prawn]

This is going nowhere

We need to clean up our ideas and work

[desert musk]

i am learning

[quaint prawn]

Mav

[desert musk]

if you can answer questions

[quaint prawn]

You are trying to show uniform continuity on (-inf, inf)

Write out what you must prove

[desert musk]

i just need to show it for f(0)

the rest has already been given

[quaint prawn]

That’s not important

[desert musk]

but im still asking you it is important to pick the smallest delta

[quaint prawn]

Yes it is important

[desert musk]

and now why?

[quaint prawn]

[desert musk]

that f is uniformly continous at x =0

[quaint prawn]

No

[desert musk]

how is that not it

[quaint prawn]

And that’s doesn’t have any quantifiers

I want you to write out formally

What it means for f to be uniformly continuous on (-inf, inf)

As that is what we are trying to prove

[desert musk]

For a function f to be uniformly continuous on (-∞, ∞), it means that for any ε > 0, there exists a δ > 0 such that for all x, y in (-∞, ∞), if |x - y| < δ, then |f(x) - f(y)| < ε, regardless of how large or small x and y are

[quaint prawn]

Yes and the last part doesn’t need to be included

[quaint prawn]

The idea is that If |x-y| < min(delta1, delta2), then
|x-y| < delta1
And
|x-y| < delta2

So then you can consider cases of x,y

[desert musk]

why is that true

[quaint prawn]

What

if x < min (5,7)
x < 5
And
x < 7

[desert musk]

why

it just says x < 5

[quaint prawn]

?

[desert musk]

5 is min of 5 and 7

[quaint prawn]

yeah and if x<5

It’s certainly less than 7??

Since 5<7

[desert musk]

oh nvm

well ok

so now we are saying that it is smaller than the smallest delta then it is smaller than all deltas?

[quaint prawn]

Smaller than the 2 that we are given yes

So consider an arbitrary |x-y| with x, y in (-inf , inf)

If |x-y| is less than our minimum chosen delta

Well if x, and y are in the same interval then we’re already done

Because we’re given uniform continuity in each

So only considering now WLOG x<0<y which I think was written above

[desert musk]

WLOG?

[quaint prawn]

Without loss of generality

[desert musk]

what does that mean

[quaint prawn]

It doesn’t matter if it’s y < 0 < x

Or x < 0 < y

Basically since the variables don’t matter here

We can only consider one case as they are the same

So without loss of generality means we can ignore the other case, it’s the same

[desert musk]

so

if 3 fruits either apple or banana at least two of them will be the same

and it doesnt matter if i take the case of the banana or apple

an example for WLOG @sacred sparrow

[sacred sparrow]

just wondering why you said 3 fruits

[quaint prawn]

Well if the qualities of them being an apple or banana don’t matter

Sure

[sacred sparrow]

you just want to express that x,y e R but x and y are in the different intervals

[desert musk]

WLOG just means that it could have been y < 0 < x

right?

[quaint prawn]

Yes

It’s like saying, given two distinct numbers a and b

We can say WLOG that a<b

Right because otherwise b<a

And we can just relabel

[desert musk]

alright

[sacred sparrow]

I'll put the solution for the case you currently write as spoiler here

[desert musk]

arh

smart triangle inequality

[sacred sparrow]

ys you almost always use estimates

to show |f(x)-f(y)|<eps

[desert musk]

idk it just feels weird doing these sort of things

[sacred sparrow]

yeah it may feel overly strict and formal at first

[desert musk]

im just giving myself what i need to solve it by for exmaple saying less than episolon/2

[sacred sparrow]

only once you accept the freezing you'll start liking the cold water :3

[desert musk]

feels like solving a physics problem and just given yourself the information you need

[sacred sparrow]

yop

you do these steps in reverse

above I first used the triangle inequality

to see that I need eps/2

I didn't know it beforehand

(well I partially did, but let's assume I didn't)

often you first rearrange |f(x)-f(y)| to find out what estimates you actually need

[desert musk]

and we knew that |f(x)-f(0)| is less than epislon epsilon/2

because we said it was for any x and y

which also includes y = 0?

[sacred sparrow]

in (-inf, 0]

x and 0 are in that interval

that's why I chose f(0)

because 0 is in both intervals

[desert musk]

yes, but the idea its that it has to be smaller for all x and y in the interval

which will also be 0

smaller than epislon/2 that is

[sacred sparrow]

ys

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[cursive nacelle]

Is this a quiz

[marsh summit]

!status

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What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[misty lynx]

no its hw

but i just got it

thanks tho

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[royal pawn]

Not really sure how to do this. Any suggestions?

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[nimble crane]

small confusions i had when revisiting this proof:

1. why is the sum going to be direct?
2. what does it mean when it says "by uniqueness on 0 \in V"?

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@nimble crane Has your question been resolved?

[wide star]

I think uniqueness refers to the direct sum

bad wording

Any vector in a direct sum can be written uniquely as a sum of vectors

Each term of the sum comes from one of the spaces in ur direct sum

(Up to order)

[nimble crane]

hm okay, why does that imply that T(xi) = 0?

[wide star]

Tx=0

Now expand x out using the direct sum decomp

[nimble crane]

ah

[wide star]

T-invariance tells u each term is in the cortesponding space from the direct sum

By uniqueness since Tx=0 each term in the sum for Tx ends up being 0

[nimble crane]

got it

now for Q1...

why is the intersection of all those kernels going to be 0?

[wide star]

There's several ways to pose being a direct sum so perhaps there is a better way to think about it?

Maybe I am misunderstanding. I'm still trying to think that out myself tbh

[nimble crane]

hmm

[wide star]

I know from friedbergs linalg book that V=R(T)+N(T) implies V=R(T) oplus N(T). It is an exercise there at least.

Your direct sum is really a sum of restrictions of T to subspaces so maybe chaining some silly facts like that together might work.

[wide star]

In general for T from V to V where W is a T-invariant subspace you can show N(T restricted to W) is N(T) intersected with W.

This was also an exercise from friedberg.

[wide star]

Scratch this. This is useless nvm.

Could the null space of say T restricted to V_1 be in V_1 by T-invariance maybe?

Oh I think it's literally just intersection bs

V_1 and V_2 already only intersect at 0

Further intersections won't change that.

Same idea should generalize to the whole direct sum I think

[nimble crane]

so (N(T) intersect V1) intersect (N(T) intersect V2) is also 0?

[wide star]

Yah

Lemme dbl check iterated direct sum stuff

[nimble crane]

apologies, but that doesnt seem obvious to me

[wide star]

It is a subset of V_1 intersect V_2 which u know is {0}

[nimble crane]

ah

got it

that makes sense

[wide star]

It also trivially contains 0 (so is a superset of {0}).

[nimble crane]

okay, then this seems to work

ty

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[glossy valve]

4C

[hallow walrus]

I think you mean $e^{cx}$

[glossy valve]

kheerii

4C

[hallow walrus]

Ah, ok

Oh, well y=0 is only a solution when the equation has no independent x terms or constants

Why not

Yeah, only when there are no independent x terms or constants

Just try putting y=0 into any random DE

[glossy valve]

4C
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[hallow walrus]

It is, because there are no independent x terms or constants

[rare dock]

what even is the question

i read that

[hallow walrus]

As in, each x term(if any) is multiplied to some derivative of y

What?

y=0 can only be a solution if BOTH of the conditions are true

[hallow walrus]

Yeah, so this satisfies the conditions I said

How…

There’s no constant

And there’s no independent x term

It’s multiplied to the y’ though

Something like y’+2y”y+3x=0 for example

This can’t have y=0 as a solution

Or y”y+3y’+1=0

Yeah, which is why I said those two conditions

Lmao

Not sure how to go about the rest of the question

It implies y=0 is a solution, unless y is in the denominator

Yeah, and that y or any of its derivatives is not in the denominator

Unless im missing something

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[twin wolf]

I don’t understand what’s going on here

[torn jolt]

sin(x) is the integral of cos(x)

it's a substitution

[umbral dome]

so (in a slightly different form) we have
[ \int u v' , dx = uv - \int u' v , dx]
where $u$ and $v$ are both functions of x

[glossy valve]

cloud

[umbral dome]

(which is essentially the power rule in reverse)

so then the strategy is to split the integral into a product of two functions, one of which you differentiate, and one of which you integrate, to get a new and simpler integral back

[twin wolf]

So wait

[twin wolf]

Why can’t we let v = cosx?

Why does dv have to be cosx?

[umbral dome]

because our original integral must have one thing we call "u" and one thing we call "dv"

[twin wolf]

Is dv a function?

Or like a derivative operator

[umbral dome]

i wrote an equivalent definition up here, "dv" is really just [ \odv{v}{x} dx] where $v$ is a function of $x$

[glossy valve]

cloud

[rugged void]

"dv" is basically the integrand for a function "v"

e.g. how if you want to integrate f(x) = say, x^2, you integrate x^2 dx not just x^2

it's basically a function times dx, except that the function is a derivative of another function, which we call v

[twin wolf]

Hm.

I think the notation is confusing me the most with this

[rugged void]

e.g. if v = sin x, dv would be (dv/dx) times dx = cos x dx

[twin wolf]

if v = sin(x) then dv/dx = sin(x)

[rugged void]

$$v=\sin x$$
$$\frac{\dd v}{\dd x}\dd x=\dd v=\cos\left(x\right)\dd x$$

[twin wolf]

Wait why is dv/dx being multiplied with another dx

[rugged void]

because dv = (dv/dx) * dx

the dx's cancel out

[glossy valve]

ren

[twin wolf]

wait if the dx’s cancel out then why isn’t there still a dx next to cosx?

Wouldn’t it just be dv = cosx

[rugged void]

no

dv/dx = cos x

(dv/dx) * dx = cos (x) *dx = dv

[twin wolf]

then you rearrange to solve for dv and get cos*dx?

[rugged void]

you mean cos(x)dx? yes

[twin wolf]

yes

This is hurting my brain

[rugged void]

look

you understand the concept of integrals and derivatives right

[twin wolf]

yes

[rugged void]

the underlying logic behind the formula for a derivative

like why the formula is as it is

??

[twin wolf]

yes I think so

[rugged void]

okay

dx = h, then?

[twin wolf]

What

What is h

[rugged void]

one second

[glossy valve]

ren

[rugged void]

this is the formula for a derivative, correct?

[twin wolf]

yes

[rugged void]

however, this can also be written as df/dx

correct?

[rugged void]

you get what df and dx mean right

[twin wolf]

Yeah tiny difference in function f over tiny difference in x

[rugged void]

yes

so h is the tiny difference in x aka dx right

[twin wolf]

yes

[rugged void]

well

then dv = v' times dx

and they cancel out

bc integration variable

and bc it's a LIMIT not bc dx = 0

[twin wolf]

Right..

So uh how does this work with the ibp

[rugged void]

lemme show you

actually, wait one second

[glossy valve]

ren

[rugged void]

this is the product rule: (uv)' = u'v+uv'

got it?

@twin wolf

[twin wolf]

yea

[rugged void]

okay

multiplying by dx, we get
$$\dd (uv) = v \cdot \dd u + u \cdot \dd v$$

[glossy valve]

ren

[rugged void]

okay??

[twin wolf]

I’m just trying to wrap my head around the treating the derivative operator like it’s a normal fraction

[rugged void]

xD

i thought that's what we always did lol

the "d" here is just to represent infinitesimal change

[twin wolf]

dy/dx = y/x

[rugged void]

no...

dy = infinitesimal change in y

dx = infinitesimal change in x

the "d" represents change
it's not MULTIPLIED

think of the d as a function

[twin wolf]

You cancel the d’s

[rugged void]

d(y)/d(x)

d(t) = infinitesimal change in t.

does that help you understand??

[twin wolf]

ok

[rugged void]

does it

[twin wolf]

i think so

Maybe

[rugged void]

okay

[rugged void]

integrating, this, we get:

[glossy valve]

ren

[twin wolf]

yes

[rugged void]

got it?

[glossy valve]

ren

[twin wolf]

Why do we want to solve for udv?

[rugged void]

because that's a product

like you can't integrate x cos x normally

or cot^2 x

[twin wolf]

Isn’t uv the product

[rugged void]

no, u dv is

we're trying to integrate u dv... uv is sitting there

@twin wolf FOCUS

[twin wolf]

okay

I’m just confused like

If we want to integrate a product

of uv

[rugged void]

no

if we want to integrate uv

[twin wolf]

Why integrate only u with respect to dv

[rugged void]

we'll have to do a double integral

[rugged void]

U and V are both functions of "x"
we don't integrate u with respect to dv; we substitute functions

e.g. for x cos x dx, we set dv = cos x dx and u = x
then, x sin x - int(sin x dx) = int(x cos x dx)

@twin wolf

look i don't seem to be able to explain this to you, sorry

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[brittle parcel]

I'm having trouble understanding this statement "Vectors, colinear with a line, make 1D real linear space"

[brittle parcel]

Ik why they create a real linear space

the rest of the statement is "On the line exists a single non-zero vector colinear with the line. All +1 vectors are linearly dependent"

and it just says "There, 1D"

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@brittle parcel Has your question been resolved?

[sacred sparrow]

is the dimensionality of the space unclear?

why it must be 1D

[brittle parcel]

I don't know either

I'm having trouble fully comprehending this

[sacred sparrow]

ok, consider an arbitrary line in some vector space, R², R³ etc.

now which vectors are colinear with that line?

well colinear geometrically interpreted is the same as being parallel to it

so all vectors that 'go' in the same direction as that line are colinear to it

more specifically, if that line goes through the origin

then only all the vectors exactly on that line are colinear to the line

since they are the only vectors parallel to the line

the line could be elsewhere in the space of course

but the vectors colinear to it just form the same line, moved to the origin

so you can imagine that the set of colinear vectors to a line form a line on their own which is parallel to the other line.

and since a line is one-dimensional, that can be considered a 1D real linear space

because all vectors within that set

fulfill the properties of a linear vector space (addition, multiplication, laws, etc.)

@brittle parcel hope that made it visually a bit clearer

[brittle parcel]

ones that are parallel or lying on the same line in said space

I think?

[sacred sparrow]

only parallel

[brittle parcel]

well if they lie on the same line, every line is parallel to itself

[sacred sparrow]

if the line goes through the origin

then all vectors parallel to it

are the line itself

[brittle parcel]

right

[sacred sparrow]

eg if we were to vis it in 2D:

this vector would be colinear to the line

this one as well

[brittle parcel]

riiight

I see

[sacred sparrow]

or these three

now let me draw all vectors colinear to it as points:

which is just another line

since there are infinitely many vectors colinear to the blue line

[brittle parcel]

I see, I see

[sacred sparrow]

in essence:

the red line = the blue line just moved

and a line that goes through the origin is a 1D linear vector space

you are going to see the same for other objects

a plane of vectors that goes through the origin is a 2D linear vector space

[brittle parcel]

right, ty

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[clever sand]

I've got a little problem I'm trying to solve for my own fun which probably has been solved centuries ago but still
what I'm trying to do is make a """"function"""" that takes an actual function "f" and an angle "alpha" and rotate f's graph by alpha radians
first thing I tried to tackle is making a function that takes a point and outputs a point alpha degrees clockwise to that point

I managed to do it without much trouble except for a few parts: when the x of the point xor the y of the point is negative then as you move through the values of alpha the point follows an ellipse instead of a circle, and more importantly: I don't know if it's even possible to make a desmos function that takes a point, let alone another function as its input

this here is what I've done so far

[clever sand]

I don't know if there's much of a better way to solve this, there probably is, but what I did is make a function g that takes a and b, or the x and y of the point, for the x take the arccos of its x/hypthenuse to get the angle, add alpha to it then use cos to reverse it and get the ratio from the wanted x to the hypothenuse, then just multiplied by the hypothenuse to get the final x. I did the same for the y but with arcsin and sin instead of cos

I'm curious to see how the shape of the ellipse changes as I change the parameters now too, I never really learned any of this officially so I don't really know why the ellipse popped out there.

ping me if someone's interested

[sacred sparrow]

lovely

I once tried to create a function that has a fixed distance to another function, so that for any function, I can find a function whose values all have a distance of >=1 to all values of the other function

but that was awfully difficult and restrictive

in coooontrast to your question

which is quickly solvable :) @clever sand

but we'll have to slightly redefine it

[clever sand]

I guessed that it's easily solvable
I just didn't know how so I challenged myself

[clever sand]

sure, how so?

[sacred sparrow]

with the essence of linear algebra it becomes simpler

when viewing it from an analysis standpoint it'll seem difficult

[sacred sparrow]

k instead of seeking a function

that takes a function and converts it into a new one

[clever sand]

k?

[sacred sparrow]

we'll instead let our initial function be given as an equation

an arbitrary equation involving x and y

[clever sand]

unfortunately, I barely have basic terminology of linear algebra in my mind so it might be difficult to explain how

[sacred sparrow]

no problem

[clever sand]

sure, go on

[sacred sparrow]

now, the graph for that equation is all points [x y] for which the equation is true

and we want to rotate each of these points [x y] by some angle a

which we can express in linear algebra:

as a product of a vector with a rotation matrix

meaning Rotation * Coordinates = Rotated_Coordinates

[clever sand]

as I said
I haven't got the faintest clue what vector multiplication is
I mean I do understand what you are doing though
ok interesting

[sacred sparrow]

dw, I'll explain it in a different way in a sec

[clever sand]

ok

[sacred sparrow]

you may accept that for now :D

the matrix multiplication yields us: x' = x*cos(a)-y*sin(a)

and y' = x*sin(a)+y*cos(a)

now why would that have any use?

well if you now take some arbitrary function, say ```
x+y = 5

which is just some random line

we can substitute the x with the rotated coordinate x'

and y with y'

x*cos(a)-y*sin(a)+x*sin(a)+y*cos(a) = 5

and now, we have the rotated graph by an angle a!

[clever sand]

is that how you calculate a rotation usually?

[sacred sparrow]

ys in R²

[clever sand]

is the way I did it fine though

how would you write it using the method I used?

[sacred sparrow]

well the issue will be that you can't necessarily rotate most types of functions

whereas the principle of "let's just rotate every single point instead"

is applicable for arbitrary functions

they don't have to be continuous or anything

sry gtg

[clever sand]

what I did was a function that takes a specific point and outputs the rotated point, could you just use that function and do the matrix thingy using that (sorry I don't know exactly how that works so I'd like a clarification if that won't work) [I'm talking about the first line in the picture]

[sacred sparrow]

but if you want to know why these are the rotation values

you can search for "point rotation in 2D or R²"

[clever sand]

ok thanks :)
how do I implement that in desmos if that's possible? I doubt it supports stuff like that

just curious

[sacred sparrow]

only in part, because it restricts our function outputs

it'll only work for continuous functions and they aren't allowed to rotate far enough to potentially have 2 outputs

[clever sand]

anyway bye, I guess I'm gonna learn linear algebra now from random sources because I'm intrigued

thanks

[sacred sparrow]

so the method works, yes, but it'll only cover a small sample of possible functions

[sacred sparrow]

it won't require algebra btw

it's just sin & cos

for rotating an arbitrary point

nywys wish thee good luck, like the experimental spirit

🦇

[clever sand]

I'm more intrigued as to what the hell matrix multiplication is and vectors and all that
I've been hearing it thrown around in a variety of contexts so I'm curious as to what it holds

cya

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[willow sundial]

I need help solving this integral

[velvet sedge]

a u-sub makes this become a little bit nicer integration by parts

[willow sundial]

what u-sub should i use?

[glad hound]

I thought to try (x-pi/2), but that doesn't take care of cos(nx)

[velvet sedge]

i think either way you'll need to use two integration by parts

[glad hound]

save the u-sub for later, and go with integration by parts first

Remember that pi is not a variable, and it can be moved outside with the 16

(the pi in the denominator... the one in parentheses is stuck for the moment)

[willow sundial]

thx

ill try solving it with that

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[vast haven]

I need help solving this frq question for precalc

<@&286206848099549185>

[neon fulcrum]

https://www.take-a-screenshot.org

[vast haven]

I can't use discord on that computer so I had to take pic from another device

<@&286206848099549185>

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[strong pawn]

How do we find (rs)^11 in D6 hexagon without brute force? r is rotation ccw, and s is flip across x axis.

[torn jolt]

Split it

[fathom saddle]

The dihedral groups obey:
r⁻¹s = sr

[torn jolt]

(rs)^6+5

[strong pawn]

rsrsrsrsrsrsrsrsrsrsrs

[fathom saddle]

You can pass r through any s, just take the inverse when you do

[torn jolt]

Bruh

Do you want to get reported

[strong pawn]

I split it like you asked

Wdym

[spice knot]

they split it for you

[torn jolt]

@fathom saddle Can't you split it?

[strong pawn]

I wasnt looking

i was counting my rs's

[spice knot]

rs^(6+5)

[strong pawn]

on the keyboard

mb

[spice knot]

lmao

"rsrsrsrsrsrsrsrsrsrsrs" never gets old

now split it like this:

[strong pawn]

Ok ya i get it

[spice knot]

rsrsrsrsrsrs rsrsrsrsrs

[strong pawn]

so u can write rs ^6 + 5 right

[spice knot]

make sure its (rs)^(6 + 5)

dont forget parentheses

rest assured this is the first time youve seen that

[strong pawn]

whats that supposed to do now

[torn jolt]

@strong pawn Since rs^6 and rs^5 are consecutive we can simply it to e (rs)

[strong pawn]

????????????

[torn jolt]

rs^(6+5) -> rs^6 * rs^5

[strong pawn]

Ya so now we got rsrsrsrsrsrs and rsrsrsrsrs

and yeah thats the same as rsrsrsrsrsrsrsrsrsrsrs

[torn jolt]

Yes

Shut up

[strong pawn]

then how tf do we get e (rs)

[torn jolt]

Do you know that r^6 = e

[strong pawn]

Yeah

[torn jolt]

s^2 = e

[strong pawn]

Cuz if u rotate it 6 times u get trhe same thing

And if u flip it twice u get the same thing

But then doesn't it change because they are not together

They are separate

rsrsrsrsrsrs

[torn jolt]

Oh

What I did was r(sr)^4

becomes r(sr(sr)^3)

That simplifies to r(s^2r^2)

@strong pawn Now what can be evaluated

[strong pawn]

where is this from

i dontget it

so rn we goet (rs)^11 = (rs)^(6+5) = rs^6 * rs^5 = rsrsrsrsrsrs * rsrsrsrsrs

[torn jolt]

Oh sorry I got it from rs^5

Yeah but we know r^6 = e already

So now we only have to worry about rs^5

[strong pawn]

(rs)^5 = rsrsrsrsrs != r(sr)^4 = r srsrsrsr

[torn jolt]

rsrsrsrsr

[fathom saddle]

So this is a rule I like to remember, useful for breaking down dihedral groups. In this case, we can use it to show rsrs = e

[torn jolt]

Ok you know what kaynex help you

Because I give up

I am retire

[strong pawn]

thanks bro

"pass r thoguh any s"

[torn jolt]

For retiring?

[strong pawn]

that sounds wack

[torn jolt]

Dam u hate me a lot

[strong pawn]

no i was being sarcastic

nah dw

everything is forgotten

jk

you suck

jk ^2 *

[strong pawn]

is it ok if u elaborate?

i dont get ur "pass r thoguh any s"

i do know a theorem

r^i s^j r^k s^l = r^(i-k) s^(j+l) if j = 1, and r^(i+k) s^l if j=0

but tahts only for rsrs

[fathom saddle]

That's just true in any dihedral group. rs = sr⁻¹

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[quick fox]

Can someone help me figure out what I'm missing for this problem?

[quick fox]

This was my attempt at a solution, but 0.436 m/s was wrong.

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@quick fox Has your question been resolved?

[quick fox]

<@&286206848099549185>

[thin flint]

you integrated with respect to V not t

[quick fox]

I don't have time in this problem so how would I solve if I used V = -gt? What would my new bounds be?

[thin flint]

you would just add -g to the result

and keep the rest

new bounds would be 0 and t

t being the time at which the velocity is measured

[quick fox]

Do you think there's a +C involved here where the mass of the raindrop at t = 0 is the initial mass? Or is that already accounted for?

[thin flint]

the + C is related to indefinite integrals

specifically because there is no starting point

you could have a starting time

t0

[quick fox]

Alright, thank you!

.close

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[wanton basin]

What logarithm rules are being referred to here?

or arctangent rules?

Ok i have made signifcant progress

But I am still struggling with the bottom portion

[gritty talon]

from the second = sign to the third equal sign, it looks like ln 2 = (1/2) ln x

how would you solve for x?

[wanton basin]

maybe multiply the 1/2 over

so 1/2ln(2) = ln(x)

and then

exponentiate both sides

so e^1/2*ln(2) = x

is what I thought

[gritty talon]

i think you have to multiply by 2 to get rid of the (1/2)

[wanton basin]

oops

you are right

lol

so ln(4) = x

what about the bottom 2 though?

[gritty talon]

actually, i think this is still wrong

2ln(2) = ln (x)

e^(2ln(2)) = x

[wanton basin]

then bring the 2 in

[gritty talon]

x = 2^2 = 4

[wanton basin]

its ln(2^2)

ah

e^2ln(2) = x

is 4 = x?

[gritty talon]

ah no because remember exponent rules: x^(yz) = x^y * x^z

[gritty talon]

yes

[wanton basin]

I messed up my order

It was this

the 1/2 ln(blank/5) refers to what though?

[gritty talon]

can you find out a way to combine all of the logs in the question?

hint: use log rules

[wanton basin]

is it just a subtraction

then division

so 8/5

?

[gritty talon]

yeah so 8 should go in the box

[wanton basin]

ok thanks

that answers my questions, I see what they were doing now

the way its ordered with all the steps disjointed makes it hard for me to parse

[gritty talon]

yeah tbh i kinda dislike those problems too

[gritty talon]

np!

[wanton basin]

.close

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[flat phoenix]

can someone help me rq with this word problem i dont know how to start it off

[flat phoenix]

im not too sure how to start this off

[exotic mason]

0.9^3*807