help-36

[hybrid heath]

nah before that, fam

[magic sparrow]

right we did

I was forgetting

[hybrid heath]

☁️

[magic sparrow]

this is probably an indication I should get more sleep

but let me try again one sec

[hybrid heath]

💤

[magic sparrow]

Okay

actually SWR

we did prove something similar

but it wasn't this

we proved for functions from Rn->R

not from R->R

[hybrid heath]

n=1

[magic sparrow]

the differentiability definition is different though

for R1

to any larger dimension

[hybrid heath]

Is it?

[magic sparrow]

yes

we have f(a+h)=f(a)+<c,h>+E(h) compared to
f(a+h)=f(a)+mh+E(h)

gradient

versus constant

[dull ravine]

How certain are you something is false here? It seems to me we have established that the limit of f approaches the same value from left and the right of 0, as we have defined E(h) that way

[magic sparrow]

I'm not certain there is anything false, I just feel like I remember the professor doing this proof in two parts, one for h->0- and one for h->0+ and I couldn't understand why that was necessary

[dull ravine]

It’s necessary most of the time, because we need to prove that the limit even exists for x ⇒ 0. But here we have already established that, so we don’t need to do it in two parts, at least not to my (admittedly very limited) knowledge

To my knowledge the limit only exists for x ⇒ 0 if the limit is the same for x ⇒ 0+ and x ⇒ 0-

But it’s very much possible that I’m wrong, so don’t take this at face value

Maybe someone else can (hopefully) confirm

[magic sparrow]

this is true

yah I feel like it is a nuanced issue here if there is one

@hybrid heath I think I

I've gathered my thoughts here

I have a different proof

presumably with no issue

[hybrid heath]

This was my only approach

(minus the =f' typo on line 2)

[magic sparrow]

Since f is differentiable at a,
$$f(a+h)=f(a)+mh+E(h)$$
$$\lim_{h\to 0} \frac{f(a+h)-f(a)-mh}{h}+\frac{E(h)}{h}=0$$
$$\implies \lim_{h\to 0} \frac{f(a+h)-f(a)-mh}{h}=0$$
$$\implies |f(a+h)-f(a)-mh| \to 0$$
And, since,
$$|f(a+h)-f(a)|=|f(a+h)-f(a)+mh-mh|\leq |f(a+h)-f(a)-mh|+|mh|$$
By the triangle inequality,
Since both of the terms on the left go to 0 as h to 0, then
$$|f(a+h)-f(a)|$$
also goes to 0 as h to 0 which implies that $f$ is continuous at $a$

[soft zealot]

Austin

[magic sparrow]

this should be better

latex cut off a small part at the triangle inequality line. should be an additional +|mh|

but as h->0 so does that

[dull ravine]

May I ask what m is?

[magic sparrow]

f'(a)!

[hybrid heath]

factorial?

[magic sparrow]

it is a suprisingly more convienent definition of the derivative that generalizes better to higher dimensions, while still being equivalent to your usual difference quotient definiton

in a small neighboorhood of f(a) (being f(a+h) for h->0) f can be approximated by a linear function: f(a)+mh, and an error term E(h) which dies off more quickly than h does as h to 0

so like essentially this tells us we can make a tangent line

with slope m

and you can rearrange this definition to get the difference quotient definiiton

HI Bungo!

[tiny gorge]

why do you need to divide by h on line 2, why not argue directly from the first equation

[magic sparrow]

we need to use the limit fact

[tiny gorge]

just use the general fact that if E(h)/h goes to zero, then E(h) goes to zero

as h->0

[magic sparrow]

Sure

but this works as presented still?

[tiny gorge]

so f(a+h) = f(a) + mh + E(h)
right hand side goes to f(a) as h->0
therefore the limit of the left hand side exists and equals f(a)

[magic sparrow]

hmm maybe you have a better way with words than I do

XD

[tiny gorge]

sure, but you detour into "divide by h" land and then have to multiply by h later, what's the point

[magic sparrow]

True

the proof just added a few steps for no reason

I see

[tiny gorge]

you can confine the fiddling to the E(h) argument

$E(h) = \frac{E(h)}{h} h$

[soft zealot]

Bungo

[tiny gorge]

rhs goes to zero, hence so does the left

[magic sparrow]

yes

I'll do it that way

TY

[tiny gorge]

btw (just scrolled back a bit) i see nothing here that would call for having to argue with one-sided limits

unless f only has one-sided derivatives for some reason

[magic sparrow]

Maybe I am remembering a different proof then where he made that remark

Seems like I've constructed like 4 different proofs of differentiability => continuity at this point

oh well can't hurt

[tiny gorge]

like SWR said, if you already have a proof for R^n then it should apply for n=1 as well

but no harm doing n=1 directly

[magic sparrow]

I did notice that

the gradient just becomes mh

[tiny gorge]

yea

[magic sparrow]

brain fart

[tiny gorge]

n=1 is hard

[magic sparrow]

not too bad once I got everything organized now

[dull ravine]

Is it possible to reason that f can only be differentiable at x if for any h>0 there is a value for f(x+i) for some i<h, which is pretty much the definition of continuity?

[magic sparrow]

I don't think so

[dull ravine]

Nvm I think I mixed up the definition of continuity

[magic sparrow]

yeah

[dull ravine]

It’s too late lol

[magic sparrow]

and a singular point can be differentiable

[dull ravine]

Yeah it’s 2 am and it’s noticeable lol

[magic sparrow]

I think an example of this is like ( I might be wrong) but f(x,y) = 1 if xy=0 and =0 elsewhere

then the partials along x and y to (0,0) both exist and are continuous

so f(x,y) is differentiable at (0,0)

[tiny gorge]

that's not differentiable

[magic sparrow]

no?

I am misremembering then

[dull ravine]

This is too advanced for me lol

[tiny gorge]

no, if it were, then its directional derivative at (0,0) would be 0 in every direction

since directionals are a linear combination of the partials

[magic sparrow]

oh right the partials have to be continuous in a neighboorhood of (0,0)

not just along a line

that is not a ball

[tiny gorge]

yea

[magic sparrow]

there's probably some kind of example though

[tiny gorge]

it's not hard to construct examples in R^1

take sin(1/x) times appropriate powers of x, typically

oh wait that does the opposite of what you're saying

differentiable everywhere except 0

if you want differentiable only at zero try multiplying the characteristic function of the rationals by appropriate powers of x

[magic sparrow]

yes I've done a problem like this before

The completed proof

Checks out?

[tiny gorge]

yea looks fine

[magic sparrow]

ty bungo and SWR and others

I was wondering if someone could check my work on this.

By the MVT, for all a,b in R, there exists a c in R s.t
f(a)-f(b)=f'(c)(a-b)

and since f'(c)<=|f'(c)|<=C

f(a)-f(b)<=C(a-b)

so f is lipschitz continuous

which implies f is uniform continuous

qed

[tiny gorge]

all correct except it should be |f(a)-f(b)| <= C|a-b|

[magic sparrow]

where did the abs value bars come from

[tiny gorge]

it's more like, how did you get C into that inequality without using abs values

and anyway, to prove uniform continuity you need to bound |f(a)-f(b)|, not f(a)-f(b) since the latter can potentially be large negative

[magic sparrow]

yeah the statement for lipschitz continuity involves absvals

so they should be here

but MVT does not give us them

so I'm wondering how we pick them up

[tiny gorge]

just take abs vals of both sides

of f(a)-f(b)=f'(c)(a-b)

[magic sparrow]

ah

Bungo I am not sure if I told you this yet

but that hellish problem from my midterm

no one got any credit for it, he kept the scores, and then assigned it to our HW for this week

😭

[tiny gorge]

harsh

you don't get to simply accept defeat and lose points on the midterm, you have to actually solve it haha

[magic sparrow]

right

so that's my next question

😭

[tiny gorge]

should be |f'(c)| <= C

btw, do you need that last line where you divided by |a-b|? what's your definition of lipschitz continuity

[magic sparrow]

I don't

but I wrote it anyways

the line before is the definition

[tiny gorge]

but then you create the annoying "uh except when a=b" case, why do that if you don't have to?

[magic sparrow]

gahhh

you're right

fixed it

[tiny gorge]

natural follow-up question, does there exist a function that is differentiable everywhere and lipschitz continuous, but which does not have a bounded derivative

i.e. is that a necessary and sufficient condition when the function is differentiable everywhere

[magic sparrow]

No

It is an if and only if statement

Lipschitz if and only if bounded derivative

I’ve proven that

[tiny gorge]

yea

[final saddle]

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[tranquil pine]

why is $ln(\prod_{i=0}^N e^{x_i}) = e^{\sum_{i=0}^N x_i}$

[tranquil pine]

.close

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[wanton pine]

Michelle is an experienced target shooter who has scored a perfect bullseye on a target 300m away. If the target was set at her eye level and the bullet left the rifle at a speed of 600ms, how high above the bullseye was she actually aiming her rifle?

[wanton pine]

the hint is to use the trig identity sinθcosθ = 1/2 x sin2θ

obviously the initial horizontal and vertical velocities are 600cosθ and 600sinθ respectively but im not too sure what to do

[cosmic warren]

says eye level

[wanton pine]

yeah the projectile will travel in parabola

[cosmic warren]

Oh wait I'm dumb I read that as "she was aiming at eye level"

[sturdy cypress]

the missing piece is that the rifle is also eye level

like she's not firing from hip

[cosmic warren]

Yeah I get it now, at first I figured she was on a hill or something

[wanton pine]

yeah well that is assumed, i dont really see people shooting rifles at hip level

[sturdy cypress]

you need the angle

[wanton pine]

correct

[sturdy cypress]

then tg or ctg gives you the height

[cosmic warren]

so u got two equations, one for vertical and one for horizontal displacement.

The unknowns of that system will be time and the angle (although you don't actually need the angle itself)

so u can solve it

[wanton pine]

yeah but im doing something wrong

[cosmic warren]

oh mind showing work then

[wanton pine]

aight

i had to iterate that t in, v=u+at was half of the total time so i had to double it, so that's why i crossed out my original answer

sin^-1(9.8/2) is not possible

so idk what ive done wrong

[sturdy cypress]

i don't get what the hint is supposed to do

[wanton pine]

so i can get sin and cos into just sin

[sturdy cypress]

if i have sin and cos, i just do 300(sinθ/cosθ)

[cosmic warren]

1200*600/300 = 1200*2 not (1200/300)*(600/300)

[wanton pine]

ah k, but does that solve everything?

[cosmic warren]

Then 2400 sin(theta)cos(theta) = 9.8 simplifies to 1200sin(2theta) = 9.8, etc.

[wanton pine]

hmm i see ill try it

[cosmic warren]

I don't think you have that though? Cus you have t = 1200 sin(theta)/9.8 and t = 300 sec(theta), so dividing the two gives you sin(theta) cos(theta) rather than tan(theta)

[wanton pine]

i got 0.23 as the angle

surely the angle isnt that small

[cosmic warren]

could be

Next u do 300tan(theta)

and that's yer answer

Looked it up just in case and 1.5 inches for 100 yards is p normal https://tpwd.texas.gov/education/hunter-education/online-course/shooting-skills/sighting-in

[wanton pine]

1.87 is with 300tan

[cosmic warren]

Wait how'd u get 0.23 radians, I got 0.00408. Or did u convert to degrees?

[wanton pine]

degrees

[cosmic warren]

OK bet

But still I get 1.225 for final answer https://www.wolframalpha.com/input?i=300tan(Arcsin(9.8%2F1200)%2F2)

Maybe u rounded off too much?

Actually rounding off would only make it smaller

[wanton pine]

300tan(Arcsin(9.8/1200)/2) where'd u get this?

[cosmic warren]

.

I did see I miss wrote 1200sin(theta) later tho

Wait nvm I didnt

[wanton pine]

hold up, are u saying 1.225 is how high she was aiming?

[cosmic warren]

Relative to the bullseye

So like

[wanton pine]

yeah i got 1.224 using 4 decimal places

[cosmic warren]

idk maybe I misinterpreted the website, or maybe the problem just used wack values lol

[wanton pine]

i got the same values

[cosmic warren]

Yeah

[wanton pine]

a meter seems justafiable right?

[cosmic warren]

Yeah I think so

I just dunno anything about shooting

but targets are like at least a meter big

So it seems within range

[wanton pine]

300m away is pretty far so 1m above the target when the bullet will take like 0.5s to travel seems fair enough to me

tbh i dont think i will even get a question similar to this in a test

[cosmic warren]

aight

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[wispy condor]

L1: x-y=7, L2: 7x+y=1 the acute angle bisector formula will be?

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[fresh hedge]

How would I solve this? I have no clue where to start.

[thick ledge]

i would start by combining the three ratio statements

you know that the candidate 1 got 4 votes for every vote candidate 2 got

this means that if candidate 1 got 100 votes, candidate 2 got 25 votes

so our ratio of votes is 100 : 25, candidate 1 to candidate 2

we can add in the ratio for candidate three as well, since we know that every 100 votes candidate 1 got, candidate 3 got 13

so the ratio 100 : 25 : 13 would be candidate 1 to candidate 2 to candidate 3

does that help as a starting point? @fresh hedge

[fresh hedge]

Let me try that, one second.

Yeah, it's a good starting point.

What would I do after that? My brain isn't braining right now.

[thick ledge]

using that ratio, you can find the proportion of people that voted for candidate 1

[fresh hedge]

Alright.

It works, thanks.

How do I mark this as solved?

[random forge]

.close

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[deft jewel]

when doing multivariate limits

[deft jewel]

is it vital to identify continuity at a point to evaluate a limit at point BEFORE using the substituion method

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[hoary prism]

let n be a positive integer. Call a nonempty subset S of {1, 2, . . . , n} good if the
arithmetic mean of the elements of S is also an integer. Further let tn denote the number of good
subsets of {1, 2, . . . , n}. Prove that tn and n are both odd or both even.

i came across this question and didnt know how to prove can someone help

[scarlet sequoia]

!status

[final saddle]

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

@hoary prism Has your question been resolved?

[hoary prism]

I don't know where to begin.

[cosmic warren]

Try out a recurrence relation

Like it might help to figure out one

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[turbid owl]

Is this solution correct?

I tried another method but stucked

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[sturdy lichen]

Let S, T, U be sets, with S ⊆ U and T ⊆ U. Prove that S \ T = S ∩ T^c

[sturdy lichen]

I need some help with how to work this problem and turn S\T into something with the compliment of T

.close

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[glass viper]

is the derivative correct

[unique aspen]

looks right

[glass viper]

?

wrong channel

its -23 tho

.close

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[molten roost]

why is my reasoning wrong

[cold gorge]

Are you familiar with the product rule?

[molten roost]

yes, but in this case, why do we not start with chain rule'

[cold gorge]

Well what you've done is hardly chain rule.

You differentiated the two functions separately, that's not chain rule.

Chain rule is typically used in compositions.

Something like this:

[molten roost]

you must differentate together for it to be chain rule?

what about like nested chain rules or so

[cold gorge]

d/dx(f(g(x)) = f'(g(x)) * g'(x)

This is what the chain rule is.

In the question you have f(x) * g(x)

[cold gorge]

Yes, but together for a composite function.

Not for product of two functions.

For that there's product rule.

[molten roost]

?

like

i differentiated the first term and applied the chain rule for the second term,

[cold gorge]

Again, that isn't how chain rule works.

This is a product of two functions.

Chain rule isn't used that way.

If you had something like $\sqrt{x^2+1}$ you'd use chain rule, yes.
Because this is not a product of two functions as you can see.

[soft zealot]

! What the hell am I doing here?

[cold gorge]

This is composition.

[soft zealot]

! What the hell am I doing here?

[cold gorge]

But in your example there's a product involved.

[molten roost]

well could i not treat the term x^4, with other function?

like so

[cold gorge]

Well you only differentiated the second expression here.

Did nothing to x^4

Which isn't constant wither

So you're wrong I'm afraid.

Do you or do you not know what composite functions mean?

I think that's where we should work.

[molten roost]

like i know what you are trying to tell me to do but i dont see why the other wasy would not work. you are insisting I do it like so yes?

[cold gorge]

Yes, there's a reason for that which I've been meaning to tell you, if only you told me whether or not you knew what composite functions mean...

[molten roost]

i apologize

[cold gorge]

Well do you know?

[molten roost]

composite functions as such

f(g(x))

or chain rule

f'(g(x)) * g'(x)

[cold gorge]

Good.

Now your original query was why exactly can't you use chain rule here, is that right?

[molten roost]

uhhhhhhhhhhhhhh

?

the x^4 is outter function and (x +1) was inner function. thats how i saw it

[cold gorge]

If the outer function were x^4 and inner were x+1
The composition would be (x+1)^4

This is product, not composition.

Let's look at another example

sin(x) and x^4

Outer and inner respectively

The composition would be sin(x^4)

For this you'd use chain rule.

But!

For something like sin(x) * x^4

You'd use product rule.

They're two different scenarios.

[molten roost]

and what if you had sin^4 * x

[cold gorge]

I'm sorry, do you mean $\sin^4(x)$

[soft zealot]

! What the hell am I doing here?

[molten roost]

yes

[cold gorge]

Chain rule.

It's like this: sin^4(x) = y = z^4, say.
sin(x) = z.
dy/dx = dy/dz * dz/dx

dy/dz would be 4z^3. dz/dx is cos(x). Then you'd sub z back in its place, which would make it look like: 4sin^3(x) cos(x)

Ultimately.

However the point is, you'd need to be able to say when it's a composition and when it's a product.

Do you think you get that yet, or...?

[molten roost]

mb im visual learner so im tryna seee uhhhh

yeah no

i dont see

[molten roost]

these terms are weirddd

[cold gorge]

Well, how about I give you a few examples and you tell me which one you'd use.

[molten roost]

YES

[soft zealot]

! What the hell am I doing here?

[cold gorge]

,texconfigcolor white

[soft zealot]

You have switched to the white colourscheme.

[cold gorge]

Well?

[molten roost]

first one chain rule

second one product rule

coposition = chain rule

[cold gorge]

Composition*

[molten roost]

multipliing 2 terms = product

[soft zealot]

! What the hell am I doing here?

[molten roost]

idk what rule that is

log rules?

[cold gorge]

It's still chain rule. It seems you still don't have an understanding of compositions.

Well I recommend watching a few videos then, maybe?

Chain rule and product rule of course. Their examples and the way you'll see them being solved, will perhaps make you understand better 'visually'.

However for one last time I'll say:
When it's f(g(x)) it's chain rule. When it's f(x)*g(x) it's product rule.

[molten roost]

?

[cold gorge]

Not quite.

You'd still need to differentiate 1 + e^x

[soft zealot]

! What the hell am I doing here?

[molten roost]

would that be quotient rule

[cold gorge]

Well, I wasn't asking which rule you'd use here, it's the answer to the previous one.

But yes for this one it'd be quotient rule.

[molten roost]

i seeeeeeee

[molten roost]

um also, how did we get this, becasue i was thinking just 1/x so we put 1 over the inner function

[cold gorge]

The derivative of ln(x) is 1/x

But we also have an inner function here.

Which is 1+e^x

[molten roost]

ohhhhhhhhhh

[cold gorge]

The derivative of 1+e^x is e^x

[molten roost]

like multiply by g'

[cold gorge]

Exactly.

[molten roost]

ok

um

going back to what we had at the beginning.

im not sure this gives me an answer

[cold gorge]

This is the right answer.

[molten roost]

i dont see it

[cold gorge]

Well of course that's because you'd need to simplify further.

Add the two fractions.

[molten roost]

i

like

multiply the right fraction by the lefts denominator over itself?

i got stuck

[cold gorge]

You've made an error

You added the second term pre and post multiplication.

Not the first and second term.

It's the same term of two different steps twice.

[molten roost]

uhhhhhhhhhhh

im not sure i understand

[final saddle]

@molten roost Has your question been resolved?

[final saddle]

@molten roost Has your question been resolved?

[molten roost]

bro

alright i give i guess

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[raw sundial]

Is c correct

[warm ether]

seems alright

woah, some mad deja vu

[raw sundial]

bro are u sure

I wanted to ask someone else just incase

[warm ether]

is there any reason you should think its wrong

[raw sundial]

idk ill try it

its right

ty

can you help me with another

is this right

[molten roost]

here lemme try

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[rocky flax]

do i use the integral test for for convergance here?

[rocky flax]

basically just set up integral from the boundaries, solve then do the limit?

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[rocky flax]

<@&286206848099549185>

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[glad ivy]

hello, i'm currently trying to find remainder of 2m^4 - 3m^2 + 1 using synthetic division with -2 as a

[glad ivy]

ill show my work in a bit, my answer was 45 although i don't think this is correct as remainder theorem gave me 21 as remainder

[odd parrot]

i just did it and i got 21 using synthetic division

did you forget to add the 0s

because there is a hidden m^3 term and m term

the coefficients of those are 0

so it should be set up as

[odd parrot]

ok i see the issue

you put 3

as the square term

when it should be -3

[glad ivy]

oh!!

wait, by chance what do you mean?

are you referring to this row?

[odd parrot]

yes

also

if its -2 shouldn't you put

-2

as the divisor

[glad ivy]

oh!!! -3

[odd parrot]

yea sorry it was a typo

[glad ivy]

i cant believe i missed that

thank you so much for explaining to me

[odd parrot]

np

[glad ivy]

i hope youll have a lovely day

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[odd parrot]

u too

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[drowsy kite]

where did v come from

[final saddle]

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

[drowsy kite]

2

[thick ledge]

what have you tried?

[thick ledge]

we would need more context to answer this question

[drowsy kite]

I didnt know what to do so i looked up the answer on symbolab and im going by step by step

[thick ledge]

do you know integration by parts?

[drowsy kite]

I could do it for some problems

integral xsinxdx

integral 2xe^xdx

is what ive done so far

[thick ledge]

$\int\arctan(7x),\dd x=\int1\cdot\arctan(7x),\dd x$

[soft zealot]

light

[drowsy kite]

oh ok that helps

[thick ledge]

cool! let me know if you can figure it out from there

[drowsy kite]

thank you!

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[crude wagon]

Can you help me understand how there are 'k' (1/2) s?

[scarlet sequoia]

the first 1/2 corresponds the the summation of 1/2

the next one corresponds to the summation of 1/4 = 1/2^2

then the same for the summation of 1/8 = 1/2^3

goes all the way until the summation of 1/2^k

so if you count the summations, we did k of them

one for every inverse of a power of 2

[crude wagon]

ohh

makes sense

even if I take s8, i count 1/2s 3 times

[scarlet sequoia]

yes

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[brazen sigil]

what is the degree of F?
x -2 -1 0 1 2 3 4
f(x) 5 -2 -3 -4 -11 -30 - 67

[scarlet sequoia]

you're supposed to find the polynomial f of smallest degree that verifies this?

[final saddle]

@brazen sigil Has your question been resolved?

[brazen sigil]

i think so

its just telling me to find the degree of f from the polynomial function

[opal plinth]

A trick to this is to recursively compute the differences between adjacent outputs

(because all the adjacent x values have the same difference, 1)

So here you have:

• outputs: 5, -2, -3, -4, -11, -30, -67
• 1st differences: 7, 1, 1, 7, 19, 37
• 2nd differences: 6, 0, -6, -12, -18
• 3rd differences: 6, 6, 6, 6
• 4th differences: 0, 0, 0

[brazen sigil]

so is the degree 6?

[opal plinth]

No

Why 6?

[brazen sigil]

i thought maybe because of how often it repeats

[opal plinth]

That has nothing to do with the degree

Take a simpler example: y = 3x+2

Obviously the degree is 1, but let's take some values and try the differences

So at x=0, 1, 2, 3, y = 2, 5, 8, 11

1st differences: 3, 3, 3

2nd differences: 0, 0

Basically if the nth differences are constant, it's a hint that the degree is n

[brazen sigil]

is my degree 3?

[opal plinth]

Possibly

[brazen sigil]

🤔

[opal plinth]

This is just a hint, not a proof

I don't think you can prove that the minimal degree is 3 without just finding the polynomial

[brazen sigil]

how do i know for sure?

[opal plinth]

You have 7 pairs of input and outputs (x and y), so you can make 7 equations based on the expected form of the polynomial

Plenty to find a solution

[brazen sigil]

ok

(x-2)(x+2)(x-3)

i think those are my factors but

i jsut wanna double check

[opal plinth]

I don't understand, is this another question?

[brazen sigil]

yes

i just wanna make sure i got the answer correct

[opal plinth]

Hm yes (x-2)(x+2)(x-3) is a factor of that polynomial

[brazen sigil]

okay ty!

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[final saddle]

@reef canyon Has your question been resolved?

[reef canyon]

<@&286206848099549185>

[alpine seal]

https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/03%3A_Determinants/3.02%3A_Properties_of_Determinants#thm:switchingrows

[vital crag]

<@&268886789983436800>

[reef canyon]

I still don't get it

we're multiplying the row by 2 and switching rows so why isn't it -2

[mint orbit]

did you find A?

[worn pike]

I can send you a picture of my work solving this problem, if you'd like?

[reef canyon]

yes please

[worn pike]

Does this help?

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@reef canyon Has your question been resolved?

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[glass mica]

how do i do the table in this picture for my question

[glass mica]

,rotate

[glass mica]

<@&286206848099549185>

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[stuck juniper]

Where did I go wrong?

[unreal anvil]

you are dealing with cos(2theta), not sin(2theta)

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[lunar breach]

how do i do this?

[glossy heath]

@tired walrus

pls help

[supple copper]

I would split it into a rectangle and triangle

[glossy heath]

can you go step by step

[lunar breach]

yea

fr

[glossy heath]

nma dougie is acoustic

[lunar breach]

oi mods ban him

[glossy heath]

no don't

we're study buddies

[tired walrus]

!noping

[final saddle]

Please do not ping individual helpers unprompted.

[lunar breach]

w annghost

[glossy heath]

AnnGhost!!!

[lunar breach]

can i get help here

[glossy heath]

W Annghost

[supple copper]

what is the confusion here?

[lunar breach]

shut up icecold

[supple copper]

I gave the first step

[lunar breach]

so annoying

thanks i go it now

.close

[glossy heath]

wait

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[glossy heath]

what next

.reopen

what next witht the numbers

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[vagrant wraith]

is it correct that {(0,1),(1,0)} is a spanning set for {(0,1)}?

[vagrant wraith]

or must spanning set have the same dimension?

[tranquil pine]

what you have listed is a basis for R2

not sure what you mean by {(0, 1)} here

[vagrant wraith]

oops

i meant to ask whether R^2 would be considered a spanning set for span((0,1))

[tranquil pine]

no

[vagrant wraith]

ok ty

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[vale python]

where C is the centralizer subgroup

i am so sorry. ive searched out help on this problem like 5 times today and i feel like I have little to no idea what is going on

I think I have shown gC(a)g^-1 is a subset of C(gag^-1), but i am utterly lost on how to show that C(gag^-1) is a subset of gC(a)g^-1

my prof didnt tell us about the centralizer subgroup, and the textbook has like a single sentence definition with no examples and im just so thoroughly confused

[graceful anvil]

Okay, you know the definition of centralizer, right?

[vale python]

not intimately

there is a single line definition in the textbook and im not sure i fully understand it

unless you mean stabilizer?

[graceful anvil]

Okay, what is the one line definition then?

[vale python]

it says: the stabilizer subgroups of each of the x_i's (the representatives from each of the nontrivial conjugacy classes of G), C(x_i)={g \in G : gx_i=x_ig} are called the centralizer subgroups of the x_i's

[graceful anvil]

Yes, so in other words, can you say that $C(a)={g\in G|\ gag^{-1}=a}$?

[soft zealot]

Tardis

[vale python]

yes

is a just any element in G tho?

[graceful anvil]

Yes

[vale python]

okay

[graceful anvil]

So can you say that $C(gag^{-1})={h\in G|\ hgag^{-1}h^{-1}=gag^{-1}}$?

[soft zealot]

Tardis

[vale python]

sure that makes sense

[graceful anvil]

Okay, so if h is any element of C(gag^{-1}), how do you get a on the RHS of that equality?

[vale python]

wait so is h being an element of C(gag^{-1}) a different h in the line from above where it just says h is an element of G? or are they somehow the same?

[graceful anvil]

They're the same

C(gag^-1) is a subgroup of G

[vale python]

oh of course right, apologies

[vale python]

do you mean how do you get a alone on the RHS of the equality $hgag^{-1}h^{-1}=gag^{-1}$?

[soft zealot]

goobybalooby

[graceful anvil]

Yes

[vale python]

just $g^{-1}hgag^{-1}h^{-1}g=a$ ?

[soft zealot]

goobybalooby

[graceful anvil]

That's correct

So can you conclude from here that something belongs to C(a)?

[vale python]

its not $hgag^{-1}h^{-1}$ is it?

[soft zealot]

goobybalooby

[graceful anvil]

No

Look at it like this, $(g^{-1}hg)a(g^{-1}h^{-1}g)=a$

[soft zealot]

Tardis

[graceful anvil]

You can easily see that $(g^{-1}hg)^{-1}=(g^{-1}h^{-1}g)$

[soft zealot]

Tardis

[vale python]

yes. so then $g^{-1}hg$ belongs to $C(a)$ ?

[soft zealot]

goobybalooby

[graceful anvil]

That's correct

[vale python]

is that what we were trying to show?

ohh it is, i think i actually sort of see it

kind of

i mean i see it

[graceful anvil]

So $g(g^{-1}hg)g^{-1}\in gC(a)g^{-1}$, which is essentially $h$

[soft zealot]

Tardis

[vale python]

oh okay i didnt see it then lol I was just going to say if $g^{-1}hg \in C(a)$ then $h \in gC(a)g^{-1}$

[soft zealot]

goobybalooby

[vale python]

well i guess those are nearly the same thing

do you think you could look over what i have for the first part? cuz now im worried i did that incorrectly

[graceful anvil]

Alright

[graceful anvil]

That looks correct to me

[vale python]

oh okay. it just seems so different than what we just went over, but if it works it works

thank you so so so so so much

[graceful anvil]

You're welcome

[vale python]

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[grand anchor]

The supermarket offers jars of jam, bottles of mineral water and packets of salt. A gentleman bought 2 jars of jam, 4 bottles of water and 1 packet of salt, and paid €20. Another gentleman bought 1 jar of jam, 2 bottles of water and returned a packet of salt that was in bad condition, and paid €7. A lady bought 3 bottles of water and returned 2 packets of salt, and paid €2. How much was each jar of jam, each bottle of water and each packet of salt?

[grand anchor]

take out this system 2x+4y+z=20, x+2y-z=7, 3y-2z=2

and got has a result x=5, y=2, z=2

can it be possible?

[ocean stratus]

yes I think that's right

[molten raptor]

Plugging in the values work so yes

[grand anchor]

I was hesitating, since Z and Y are unknowns with different values and I don't know if it could make the same difference.

<@&286206848099549185>

[random forge]

bro 2 people already said you were right, why are you pinging helpers

[vital crag]

plug this in to check your answer

[grand anchor]

i was asking why of this question

[vital crag]

two variables being the same doesn't make any difference in solutions of systems of linear equations

[grand anchor]

ok tysm

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[fiery spade]

hello

how can i find a and b?

[vital crag]

,rotate

[fiery spade]

limit for n to infinite

i tried something

in the end the limit is always infinite

[vital crag]

where's the question

[fiery spade]

I just need to find a and b

thats all i was given

[vital crag]

what's $a_n$ ?

[soft zealot]

riemann

[fiery spade]

its a times n

i need to find a and b and n its n

like the substitution for infinite

[tulip coyote]

To clarify, it's
[
\lim_{n\to\infty} \pqty{ \frac{n^2 + n + 1}{2n + 1} - an - b } = 1
]
you have, right?

[soft zealot]

@tulip coyote

[fiery spade]

exactly

[tulip coyote]

You may want to think about polynomial dividing that $\frac{n^2 + n + 1}{2n + 1}$ out

[soft zealot]

@tulip coyote

[fiery spade]

yeah its infinite

do you want to see the previous exercise? its similar to this one

<@&286206848099549185>

helloooo

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[covert hornet]

if f(x) = asin(lx) with l <> 0 then it has a period of T = 2π / l

[covert hornet]

a question here says that the period is T = 2π / abs(l)

which is correct? i always thought it is the first formula but i dont know if period can be negative

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@covert hornet Has your question been resolved?

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@covert hornet Has your question been resolved?

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@covert hornet Has your question been resolved?

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[prisma flower]

i dont understand

[final saddle]

@prisma flower Has your question been resolved?

[final saddle]

@prisma flower Has your question been resolved?

[wicked knoll]

@prisma flower these are three different types of transformations

are you aware of any types of transformations?

[prisma flower]

yes

for them the ansswer i got was translation, reflection and rotation

in the order

[wicked knoll]

thats correct

[prisma flower]

do i have to put any numbers?

[wicked knoll]

yes

[prisma flower]

how do i do it

[wicked knoll]

i'll go through them one by one

for the first one you need the translation vector

which is basically a way of writing how much the shape has moved

[prisma flower]

yea'

[wicked knoll]

so for that just pick any point and see how far it moved

[prisma flower]

so +1?

[wicked knoll]

you need to write how much it has changed in each axis

so if it moves 2 right, 3 up, it would be

(2, 3)

[prisma flower]

oh

[wicked knoll]

so what do you think this one will be

[prisma flower]

for the second one>?

[wicked knoll]

for the first one

[prisma flower]

let mse

for all the points they go by +1

[wicked knoll]

thats technically correct but not fully

think about it like this

how many points does it move right

and how many points does it move up

[prisma flower]

for this one theres no up

[wicked knoll]

exactly

[prisma flower]

so just 0?

[wicked knoll]

yup

so to fully complete the question you would write

"Translation by the vector (1, 0)"

[prisma flower]

how do i know what number to put first

like when i get 2 numvbers

[wicked knoll]

x-axis comes first, y-axis comes second

like a co-ordinate

[prisma flower]

oh i see

also
for the second the letters have different

[wicked knoll]

for future reference if you get one that moves left or down, those would be minus numbers

[prisma flower]

moving to the right

[wicked knoll]

yeah so the second one is a reflection

[prisma flower]

so then theres no numbers

[wicked knoll]

yes

instead you need to find the line in which it is reflected in

[prisma flower]

oh

okay

[wicked knoll]

do you know which line that is?

[prisma flower]

no, is it supposed to be referred to as a letter

[wicked knoll]

somewhat

okay so to find the line

try and find the line where the shape on either side is the exact same

its hard to explain in writing lemme get a diagram

[prisma flower]

ok

[wicked knoll]

you see in this image how both shapes are equal in distance from the line

[prisma flower]

yeah

[wicked knoll]

try and find the line that does the same on your diagram

[prisma flower]

ok

is it just in the middle?

[wicked knoll]

yup

[prisma flower]

alright

[wicked knoll]

for some reason your questions don't have a scale on their axis so im just going to assume that each box goes up by one

[prisma flower]

yeah im pretty sure

[wicked knoll]

so now we need a way to mathematically describe that line

[prisma flower]

yeah

[wicked knoll]

im going to go back to the image i sent you as an example

[prisma flower]

okay

[wicked knoll]

every point on the line has an x co-ordinate of -2

this means that the line can be described as x = -2

can you try describing the line in your question?

[prisma flower]

im a little confused

[wicked knoll]

whats confusing you?

[prisma flower]

the -2

oh wait nevermind

why is it

negative?

[wicked knoll]

i'd just like to clarify we are both talking about this image

[prisma flower]

ah

[wicked knoll]

in your question, the line is not x = -2

[prisma flower]

yeah my fault i got confused

[wicked knoll]

thats alright

do you know what the actual answer is?

[prisma flower]

im not sure how to describe the line

[wicked knoll]

okay so

lemme draw something up real quick

[prisma flower]

alright

[wicked knoll]

so we agree the red line is the line of reflection

[prisma flower]

yup

[wicked knoll]

whenever we have a vertical line in a graph

it will always be in the form x = a

[prisma flower]

oh ok

[wicked knoll]

where a is the x co-ordinate of the line

[prisma flower]

ok i see

[wicked knoll]

for a horizontal line it will be y = a

where a is the y co-ordinate

its basically saying that in this case, x will equal the same no matter what

[prisma flower]

yeah

[wicked knoll]

so for this case, since the x co-ordinate of the line is 3

we can say the line is x = 3

[prisma flower]

alright that makes sense

[wicked knoll]

so to complete the question you would say "Reflection in the line x = 3"

[prisma flower]

yeah

[wicked knoll]

if you get confused on these lines again i recommend using

https://www.desmos.com/calculator

you can plug in some values and get a graph back

[prisma flower]

woah ok thanks

[wicked knoll]

trust me this is very useful in maths

i use it decently often

anyways final question

[prisma flower]

alrgiht ill use it

[wicked knoll]

do u think u could paste it back in theres been a lot of messages lol

[prisma flower]

ok

[wicked knoll]

cool okay so the last one is rotation

[prisma flower]

yea

[wicked knoll]

for this one we dont need a number, or a line

do you know what we do need

[prisma flower]

degrees?

[wicked knoll]

thats right

[prisma flower]

is it just 360

[wicked knoll]

close

[prisma flower]

wait no

[wicked knoll]

a 360 degree turn will return a shape to the same place it originally was in

[prisma flower]

i forgot yeah

so its just 180?

[wicked knoll]

yep

have you ever used trace paper?

[prisma flower]

no

[wicked knoll]

okay

well basically the trick i use for rotation is

i imagine the shape on a piece of paper

(or just actually draw it)

and for every 90 degrees i turn the paper right once

may work for you as well idk

[prisma flower]

wait i think i remember doing that in middle school yeah

ok

[wicked knoll]

so yeah the answer will be

"Rotation by 180 degrees clockwise"

[prisma flower]

alright

ok ill include the clockwise

[wicked knoll]

i probably went into more detail than is necessary for these questions but its better to understand it

[prisma flower]

its ok it helped me understand

[wicked knoll]

im glad 🙂

[wicked knoll]

one final note

for 180 degrees both clockwise and anticlockwise are the same so it doesn't really matter

[prisma flower]

true

[wicked knoll]

but for any other angle of rotation you gotta make sure you put one or the other

[prisma flower]

alright, thanks so much ❤️

[wicked knoll]

its all good

think u need to close it not me lol

[prisma flower]

alr

.close

[final saddle]

Channel closed

Closed by @prisma flower

Use .reopen if this was a mistake.

[final saddle]

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

what does that E like that mean

[crude wagon]

in

z is member of C

[full halo]

yeah but what does it mean z in 0,inf

[crude wagon]

(a-ib)-(a+ib)

[full halo]

-2bi

[crude wagon]

what does tali che means?

[full halo]

such as

[rotund citrus]

such that*

[crude wagon]

z- z conjugate is a complex number, idk what means to lie in (0, infy)

you can say, if b<0, the imaginary part of this complex number is positive

[full halo]

why

[crude wagon]

^

[full halo]

oh

[crude wagon]

-2bi, where (-2b) is the imaginary component

but for that to be in (0, infy), take b<0

[full halo]

but it doesnt ask of just imaginary

[crude wagon]

post the full problem ?

[final saddle]

@full halo Has your question been resolved?

[full halo]

<@&286206848099549185>

[final saddle]

@full halo Has your question been resolved?

[final saddle]

@full halo Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

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[final saddle]

Available help channel!

Send your question here to claim the channel.

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

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

[old iris]

how to integrate this

[old iris]

Help pls

[final saddle]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[thick ledge]

what is a?

[final saddle]

@old iris Has your question been resolved?

[final saddle]

Channel closed

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[worthy shale]

multiplying top and bottom by sec^4x seems to work
then u sub

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[crimson cobalt]

my first pic is me plugging in my points into point-slope form after i found my slope of the tan line, 2nd pic is the question and 3rd is me double checking that hte derivative i got was correct

[final saddle]

@crimson cobalt Has your question been resolved?

[crimson cobalt]

.close

[final saddle]

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

Can someone please explain to me how we solved this

[cobalt lodge]

which equality do you not understand

because it’s literally just plug it in then expand and expand and simplify and then lhopital

[spice field]

How can you compare something with factorial

How did they get to 1/n^3

[cobalt lodge]

it’s a comparison test, you compare to whatever you want to compare against

question asks you to compare against 1/n^3 so you just slap 1/n^3 in there

[spice field]

No the question doesn't mention it

[cobalt lodge]

oh, to get there you just simplify the fraction

do you know the expression for a factorial

[spice field]

5! = 5x4x3x2x1?

[cobalt lodge]

yes

so simplify the fraction in the question

[spice field]

so n , n+1 , n+2

for the numerator?

[cobalt lodge]

you mean denominator?

[tired walrus]

black band : image ratio ≈ 3 : 1

[spice field]

Sorry?

[tired walrus]

the image you sent had black bands on the top and bottom

[cobalt lodge]

the complaint is that you took a vertical screenshot of a landscape image so you have like 85% blank space

[tired walrus]

which were bigger than the image itself

[spice field]

Oh Okay sorry

[spice field]

Can you help me simplify ?

(n-1)!

[cobalt lodge]

expand both factorials

[spice field]

so I just plug numbers of the series?

[cobalt lodge]

it has nothing to do with the series

just the fraction

[spice field]

I don't know how to expand it