help-49

[subtle blaze]

suppose $u\in V_C$, and $u = (0_V, 0_V)$ are you happy that $u$ is indeed in $V_C$, with "components" coming from $V$

[grand pond]

Frosst

[twilit field]

yes

[subtle blaze]

now let $v\in V_C$, and $v = (a, b)$, then $v \textcolor{blue}{+} u = (a \textcolor{red}{+} 0_V, b \textcolor{red}{+} 0_V)$

[grand pond]

Frosst

[subtle blaze]

aha! but 0_V is the zero vector in V

and it has the property that a + 0_V = a for all a in V

well, a and b are both in V, so then a + 0_V = a and b + 0_V = b

then $v \textcolor{blue}{+} u = (a, b) = v$

[grand pond]

Frosst

[subtle blaze]

hence the existence of a zero vector $0_{V_C}$ is proven, it is $u$

[grand pond]

Frosst

[subtle blaze]

we immediately get uniqueness as well, because we've already proved from before that the zero vector is unique in any vector space

[twilit field]

this is all so overwhelming to me

I think I need some time to process this

[twilit field]

from where did you get this

[subtle blaze]

by the definition of the zero vector (named 0_V) in V

[twilit field]

got it

[subtle blaze]

and since V is a vector space this is guaranteed to exist

[twilit field]

Got it

thanks

thanks for spending so much time with me on such an easy problem

[subtle blaze]

it's not easy

it's a very very different way of approaching math compared to highschool

[twilit field]

now to prove an inverse element exists

(a,b)+(-a,-b)=(0,0)

I think I'm missing something here

[twilit field]

thanks

<@&286206848099549185>

.close

[midnight plank]

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[twilit field]

ok, so I want to prove the identity element of scalar multiplication now

[twilit field]

by definiton 1(a,b)= (a,b)

is that it

?

[timid spear]

Which channel can I discuss mathematical logic?

[twilit field]

#proofs-and-logic maybe

[prime hornet]

#proofs-and-logic, or #foundations if it's advanced enough

[timid spear]

Get it. Thank you👌

[heavy falcon]

you want to prove that 1v=v for v in V_C?

[twilit field]

yes

[heavy falcon]

ok so use the multiplication definition in this vector space

[twilit field]

I thought I did

[heavy falcon]

with the first element being 1+0i and the second u+iv

[heavy falcon]

where

[twilit field]

here

[heavy falcon]

yes for a,b in V

[twilit field]

so that's it?

[heavy falcon]

you have (1+0i)(a+bi)=1a-0b+i(1b+0a)=a+ib

[heavy falcon]

yes because you are given that V is a vector space so 1,a,b are all in V with 1a=a and 1b=b

[twilit field]

got it

thanks\

now to prove distributivity

k(a,b)=(ka,kb) by definition again

as for distrbutivity,the same argument holds

[heavy falcon]

(a+bi)((u+vi)+(w+zi))=(a+bi)((u+w)+(v+z)i)

now you can use def of multiplication in V_C

[heavy falcon]

then the scalar will distribute over the elements after this because u,v,w and z all are in V which is a vector space over R

so yes its the same way

[twilit field]

thanks!

.close

[midnight plank]

Channel closed

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[subtle blaze]

yes, existence of the inverse element can be constructed from the inverse of the individual components

[twilit field]

thanks

[midnight plank]

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[tidal turret]

find the volume of the solid of revolution obtained by rotating around the $x$ axis the parabola $\y = 3x^2$ with $0 \leq x \leq3$

[grand pond]

calc_and_real_anal

[midnight plank]

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

[tidal turret]

I dont have the y formula but

[dawn dagger]

yea this is washer method

now what

[pearl hull]

That’s the only and the very method you need

then

?

[dawn dagger]

we can sketch the region

,w plot y = 3x^2 between 0 and 3

[grand pond]

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[dawn dagger]

bruh

that's so easy

[tidal turret]

they are asking for volume not arch length

though

[dawn dagger]

i wanted to plot the graph

idc about arc length

[tidal turret]

okay

,, \pi \int_{0}^{3} f^2(x) dx

[grand pond]

calc_and_real_anal

[dawn dagger]

,,\pi \int_0^3 \left ( 3x^2 \right )^2 : \dd x

[grand pond]

now what (𝔸dωn𝓲²s)

[tidal turret]

,, 9\pi \int_{0}^{3} x^4 dx

[grand pond]

calc_and_real_anal

[dawn dagger]

you can cook now on your own i guess...

[prime hornet]

the saga continues...

[tidal turret]

the volume is in pi?

[dawn dagger]

what

[tidal turret]

,,9\pi\left(\frac{3^5}{5} - 0\right)

[grand pond]

calc_and_real_anal

[dawn dagger]

ye

[tidal turret]

,calc 3^7

[grand pond]

Result:

2187

[dawn dagger]

3^7

?

[pearl hull]

…

[dawn dagger]

now what?

[pearl hull]

$3^7$ = ?

[dawn dagger]

,calc 9pi * 3^5/5

[grand pond]

Result:

1374.1326266802

Sufferrrrring rn D:

[burnt flame]

what now?

[dawn dagger]

we taking a break

for now

[tidal turret]

lol sorry

.close

[midnight plank]

Channel closed

Closed by @tidal turret

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[pearl hull]

….

[pearl hull]

Have you understood how it works?

[tidal turret]

washer method?

[pearl hull]

Alr, have a good one

[midnight plank]

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[hazy flint]

@feral sedge sorry to ping but i wanted to ask now that ive found the valus of h and k how can i then find the equation of the ellipse at the origin?

[hazy flint]

sorry i dont understand

this is where i got my h and k from

[pearl hull]

oh

mb

[hazy flint]

and this is what i did

h and k gave me the centre of the ellipse but i want to find the equation of that ellipse when it is centred at the origin

oh wait do i just let h and k = 0 and keep a b c f the same?

[pearl hull]

yep

[hazy flint]

then the x and y term disappear

i tried it and theres an error

[pearl hull]

the xy term wouldn’t disappear

[hazy flint]

yeah i know

[pearl hull]

Since the ellipse has been rotated

[hazy flint]

this is what im left with

because the coefficient of x and y is 0

but the ellipse doesnt exist apparently

[pearl hull]

hm

[hazy flint]

<@&286206848099549185>

[midnight plank]

@hazy flint Has your question been resolved?

[hazy flint]

<@&286206848099549185>

[last slate]

Bro why do you wanna rotate the ellipse, just rotate the axes such that x axis lies on the axis of the ellipse, and shift the origin to the center of ellipse
You would get the standard equation of ellipse

[hazy flint]

yeah ik i dont want to rotate it

the problem is the ellipse is not centred at the origin

i can only rotate the axes if the ellipse is at the origin no?

[last slate]

Then shift the origin

[last slate]

Not really, you can shift origin first and then rotate

[hazy flint]

so like this?

but add a term for the origin translation too

[hazy flint]

could i apply this to the original ellipse (that is not at the origin) then add the h and k terms after?

[last slate]

Yeah

[last slate]

Tbh I have never tried rotation of a 2nd degree equation, can't say about it

[hazy flint]

hm ok

do you know how they dervied cot (2theta) = A-C / B?

[hazy flint]

^

[last slate]

Can you try substituting these new values of x and y into the homogeneous equation?
Might work

[hazy flint]

by homogenous equation do you mean the big equatio of the original ellipse?

[last slate]

Yeah

[hazy flint]

i rotated it but cant seem to put it at the origin

i rotated the axis of the original ellipse and subsituted x+h and y+k to x and y but it doesnt seem to be centred at the origin

oh wait ik why its because the original ellipse wasnt centred at the origin the centre of the rotated ellipse is changed

.close

[midnight plank]

Channel closed

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[last slate]

So you got it ? @hazy flint

[hazy flint]

yeah i think so

ill reopen if theres a problem

[last slate]

Alr

[midnight plank]

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[woven solstice]

(g) I think I miss something, anyone help?

[merry garnet]

yes

you did

[spice citrus]

6x

[merry garnet]

6x

[spice citrus]

du/dx = 6x

[merry garnet]

not 3x

[woven solstice]

oh loll ty

i did too many 3x question

help

√x²-x+1

u= x²-x+1

du/dx = 2x-1

dy/du = 1/2u^-(1/2)

dy/dx = (2x-1)(1/2u^-(1/2))

idk how to multiply the dy/dx lol

[still nacelle]

wdym "to multiply the dy/dx", multiply with what

[woven solstice]

(2x-1)(1/2u^-(1/2))

[still nacelle]

once you have dy/dx, just substitute u back in terms of x

[still nacelle]

yh use latex, writing / is quite ambiguos. Do you mean
$(2x-1)\frac{1}{2}u^{-\frac{1}{2}}$

[grand pond]

lgkoo

[woven solstice]

yea

got it now

[still nacelle]

you got it?

[woven solstice]

yes

[still nacelle]

nice

[woven solstice]

.close

[midnight plank]

Channel closed

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[woven solstice]

ty btw

[midnight plank]

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[young jolt]

How to solve this one?

[last slate]

Hmm so we wanna convert whole thing in terms of tan alpha

Let's see

[young jolt]

Yep i did something but it got messy as you can see

[marsh dove]

Can someone simply question 2.1.2 please

[last slate]

Tan pi/8 = root2 -1

[young jolt]

Yes

[last slate]

Am I correct, can you check I don't remember properly

Okay

[young jolt]

Pi/8 is 22.5

[last slate]

True

[young jolt]

So according to the question the value is already given

But how can I write this in terms of tan pi/8

[last slate]

Alr seo we gotta do smth like use trigonometry circle and simplify cosπ/8,cos3π/8 and so on

Rn not sure about how we gonna get till tan π/8

[young jolt]

Let me try

Bruh

Thanks

[last slate]

Oh yeah

LMAO

nice

I too figured out it now that it is symmetrical

If you got more such questions upload it

[young jolt]

Trig is not leaving me 😌

[last slate]

Oh yeh

[young jolt]

Still can’t figure out a way

[last slate]

Same

[young jolt]

-16cos2x my bad

[last slate]

Oh well

Ended with a 4th degree(in terms of cosx) equation which is of no use

Oh maybe something

Wait a min

[last slate]

What are roots of 9k^2 +12k -5

[young jolt]

,w 9k^2 +12k -5=0

[grand pond]

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[last slate]

I see

[young jolt]

☺️

[last slate]

I used the fact that sec2x = 1+tan2x

[last slate]

Yeah somewhat like that

T^4 =k^2

[young jolt]

Yep the answer is is 1/3

[young jolt]

I should have used cos^2x=t

[last slate]

Cos^2x = 1/3

Hmmm now what

[young jolt]

Cos4x=2cos^2(2x)-1

[last slate]

Oh yeah

[young jolt]

Yep

[last slate]

-7/9 is what I got

Ignore my leg at bottom right corner 🤡

[young jolt]

Me Tooo

[last slate]

Alr imma go do some other shit now

Bye

[young jolt]

Obey bike

[last slate]

Hope someone else would help ya

[young jolt]

Bie

[young jolt]

Thank u

.close

[midnight plank]

Channel closed

Closed by @young jolt

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[last slate]

Noice

[midnight plank]

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[radiant vector]

Stuck on this

i thought abt seperated the 2 from the f(x) but its in the bracket so

[obtuse basin]

You could substitute for x-2

Btw the question is really bad, the answer may not be unique

[radiant vector]

it was in our exam

and I did manage to get it but I dont really know how

apparently u had to like

reverse the domain so

4+2
1+2

and the answers are 6 and 3

but I dont understand the logic and how u would show it

[obtuse basin]

See what happens afterwards

[radiant vector]

in where

no equation

oh well I think I get it ig abt reversing domains

.close

[midnight plank]

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[versed elbow]

uh what?

u = x - 2

[grand pond]

nyxie9151

[versed elbow]

but the bounds c and d are "x" and you have to use your u = x- 2 equation to convert the bounds to be in terms of u too

[midnight plank]

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[wooden knoll]

$|x+k| \leq a \ \ k+a \leq x \leq k-a$

[grand pond]

pixel

[wooden knoll]

correct right?

[chrome vessel]

close but not quite

for example x+k <= a implies x<= a-k

[wooden knoll]

mmmm

$x \leq a-k$

[grand pond]

pixel

[chrome vessel]

what about the other side?

?<=x

[wooden knoll]

how would we get the other side

do we just

flip the signs

ive just started this topic for graphic but i dont really get how it works

[chrome vessel]

here is a hint: |x+k|<=a is the same as saying -a<=x+k<=a

can you figure out why?

[wooden knoll]

$x+k \geq -a \ |x+k| \leq a$

[grand pond]

pixel

[chrome vessel]

🤔

[wooden knoll]

😭 sorry i dont really know

but i wanted to try

[chrome vessel]

think about separating it into two cases: x+k>=0 ot x+k<0

[wooden knoll]

yeah but where does 'a' come in that

x >= -k
x < -k

lol doesnt look right at all

[chrome vessel]

if x+k>=0 then |x+k|=x+k

so you can simplify the equation |x+k|<=a

[wooden knoll]

$x+k \leq a$

[grand pond]

pixel

[chrome vessel]

since we are separating into two cases we are saying "either x >= k or x<-k" and this is a true statement

[chrome vessel]

good! If x+k >=0 we have 0<=x+k<=a

what if x+k<0?

[wooden knoll]

x-k?

[chrome vessel]

can you clarify?

when x+k<0 what does |x+k|<=a turn into

[wooden knoll]

|x+k|<0 = x-k>=-a

god its bothering me how im so clueless

like im saying things but i have no idea what it means

[chrome vessel]

ah, we dont use = signs here, when we talk about implications we use =>, like raining implies i get umbrella is written as "raining=>i get umbrella"

[chrome vessel]

|x+k|<0 is incorrect,

lemme restate what our idea is:

we are given |x+k|<=a, but it is frustrating to deal with the absolute value sign. We have to consider cases, and the most reasonable way to separate into cases is by considering whether x+k is greater than 0 or not

IF x+k>=0 (this is an assumption, not a fact), then we can derive that |x+k|=x+k<=a, and we can combine x+k>=0 and x+k<=a to get 0<=x+k<=a

now what IF x+k<0?

IF x+k<0, what is |x+k|?

[wooden knoll]

well im looking at this cause it looks related to what we're talking about

if x+k<0, |x+k| = -(x+k) ???

or am i understanding it completely wrong

[chrome vessel]

yes!

[wooden knoll]

oh

😭

ok

thats good

we're getting somewhere

[chrome vessel]

-(x+k)<=a, and what does that simplify into?

[wooden knoll]

-x-k<= a

[chrome vessel]

can you remove the negative signs on the left?

multiply it by -1 perhaps?

[wooden knoll]

x+k<= -a

?

[chrome vessel]

well you have to change signs here

[wooden knoll]

wait why though

[chrome vessel]

here is an example: 1<=2 but when we multiply by negative 1 we get -1<=-2

that cant be right, -1 is greater than -2

lets look at it graphically:

the middle point is 0

multiplying by -1 is sort of "flipping" by 0

and you see than b>=a but -b<=-a

is it clearer now?

[wooden knoll]

mm right

[wooden knoll]

okay yeah that looks right

[chrome vessel]

so when we multiply by negative 1 we have to swap signs!

now this is true for all negative numbers

when we multiply by negative numbers, we have to swap signs

if a<b then -2a>-2b, things like that

[wooden knoll]

-(x+k) <= a
x+k >= -a

[chrome vessel]

good!

[wooden knoll]

okay yes this is making sense

[chrome vessel]

so if x+k<0 we must have -a<=x+k<0

and if x+k>=0 we must have 0<=x+k<=a

so, you can see that no matter what x+k is, whether it be greater than 0 or less than 0, it must be between a and -a

[wooden knoll]

$|x+k| \leq a \ \ $\text{IF } x+k \geq 0 \ |x+k| = x+k \leq a \ \text{Considering that x+k } \geq 0 \text{ and x+k } \leq a \text{ then } 0 \leq x+k \ leq a \ \ \text{IF } x+k<0 \ |x+k| = -(x+k) \ -(x+k) \leq a \ x+k \leq -a \ \ -a \leq x+k \leq a$

[grand pond]

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

[wooden knoll]

does this sum it all up?

[chrome vessel]

PERFECt

[wooden knoll]

hooray

ok ima try |x-k| <= a

its very similar

[chrome vessel]

we are not done yet though

[wooden knoll]

so i should get it right

[wooden knoll]

oh

[wooden knoll]

ill leave this for now then

[wooden knoll]

why not

oh we're finding a domain

so it has to be x

[chrome vessel]

so we have derived that |x+k|<=a must imply that x+k is between a and -a

we haven't garanteed that the implication is the STRONGEST possible condition

For example: if Y is an positive integer, then it implies that Y is a integer.
While true, the implication would be a weaker condition...

[wooden knoll]

honestly i have no idea what that means

but

how do we guarantee the strongest possible condition

for the implication

[chrome vessel]

we have to show that the other way is true as well

that is, x+k being between a and -a implies |x+k|<=a

[wooden knoll]

what do you mean by "the other way"

oh

so work backwards

we start with $-a \leq x+k \leq a$

[grand pond]

pixel

[wooden knoll]

and work towards $|x+k| \leq a$

[grand pond]

pixel

[wooden knoll]

also if thats the case

[chrome vessel]

👍 if the two statements implies each other, then they must be equivalent!

[wooden knoll]

does this look right to you

[chrome vessel]

exactly right, you see proving equivalence requires two steps: we need to prove both |x-k|<=a implies k-a<=x<=k+a and the other way around too

[wooden knoll]

my question is though

how did they like

get 'x' in the middle

if that makes sense

[chrome vessel]

because they simplified the inequalities futher

for example: x+k<=a is equivalent to x<=a-k

you can subtrack k on both side and you will see why

[wooden knoll]

yeah yeah that makes sense

[chrome vessel]

so -a<=x+k<=a would simplify to -a-k<=x<=a-k, can you see why?

[wooden knoll]

yep

[chrome vessel]

great

so now you probably undertand why they got an x in the middle

[wooden knoll]

yep

domain

right?

[chrome vessel]

well yeah, this way you can figure out the domain of x

logical deduction and stuff like this may take some time to click.

[wooden knoll]

yeah but I understand what we've done so far

[chrome vessel]

👍

[wooden knoll]

and I was able to do |x-k| <= a in my book and it worked

[chrome vessel]

great!

[wooden knoll]

lemme show u how we'd solve this

ok so

[chrome vessel]

did you do the reverse direction too? that is: a<=x-k<=a implies |x-k|<=a?

[wooden knoll]

$|2x-4| = 16 \ \ \text{ IF } 2x-4<0 \text{ THEN } -(2x-4) = 16 \ 2x-4 = -16 \ 2x = -12 \ x = -6 \ \ \text{ IF } 2x-4 \geq 0 \text{ THEN } 2x-4= 16 \ 2x = 20 \ x=10$

[grand pond]

pixel

[chrome vessel]

good job, all you need to do is check that they are actually solutions!

[wooden knoll]

oh yeah could you show me how we'd that

from my understanding, I think for us to work backwards and let the domain IMPLY |x-k| <= a, we'd break the domain into two parts

x-k >= -a
x-k <= a

[chrome vessel]

can you try to follow a similar strategy and attempt to prove it? that is: what happens when x-k>=0 and what happens when x-k<0

[chrome vessel]

what do you mean by let the domain imply |x-k| <=a?

[wooden knoll]

well like you see how you said that |x-k| <=a implies the domain, or am i tweaking

and like for us to make this implication stronger

or smthn like that

we have to proive that they're equivalent

[chrome vessel]

oh by "the domain" you probably meant "the domain equals to k-a<=x<=k+a"

[wooden knoll]

oh ya

haha

[chrome vessel]

👍 the proof is very similar i think you are able to cook up something

[weak plover]

can someone explain to me how do i study maths

do i just study the concepts?

[chrome vessel]

post it in math help channels that are not occupied, indicated by the "math help (available)" folder

[chrome vessel]

any progress?

[weak plover]

oh thanks

[midnight plank]

@wooden knoll Has your question been resolved?

[midnight plank]

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[wooden knoll]

Sorry I had to go do something and I kinda forgot what I was doing ngl

[wooden knoll]

what would this be used for

what are we going to do with it

[pearl hull]

Solving questions on your test sheet

[zealous schooner]

Absolute value function has many applications

[zealous schooner]

Yeah this

$\sqrt{x^2}=|x|$

[grand pond]

kheerii

[zealous schooner]

For real numbers

[zenith bough]

Not a chance

[jagged needle]

im stuck

[pearl hull]

Don't

[midnight plank]

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).

[pearl hull]

plus, someone is solving your question in your channel

[midnight plank]

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[last slate]

Hey, I have a quick algebra question. In the following equation:
$9^{\sqrt{x}} = \sqrt{27^x}$
Knowing that
$x^{a^{b}} = x^{ab}$
Can we apply it on the left side? Like that?
$9^{\frac{x}{2}}$

[grand pond]

Victor Freitas

[junior flower]

that's like saying sqrt(x) = x/2

so, no

be careful of how the exponents are nested

$(x^a)^b = x^{ab}$ but not $x^{(a^b)} = x^{ab}$

[grand pond]

slayla

[last slate]

Hum, it makes sense. Thanks Slayla!

[junior flower]

no problemo

[last slate]

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[red sparrow]

can somebody please explain how does it become (-1)^2

[frank marsh]

for an n x n matrix, det(kA) = k^n det(A)

[red sparrow]

okay thanks!

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[tranquil turtle]

The position of a particle on a spinning wheel is given by f(t) = 5cos(200t), where t is time in seconds. How many seconds will it take for the wheel to spin 50 revolutions?

[tranquil turtle]

how do I calculate this

[tranquil turtle]

<@&286206848099549185>

[midnight plank]

@tranquil turtle Has your question been resolved?

[tranquil turtle]

no

can somebody please help?

[heady bluff]

is there a diagram or something?

[tranquil turtle]

there isn't

[heady bluff]

assuming co ordinate axis

wait

what would you get if you differentiated wrt t

[tranquil turtle]

this is not calculus

a functions course

[flat spire]

i think its just asking you to find t when the period = 50

[tranquil turtle]

so I set the f(t) to 2pi/50 ?

and solve for t

oh wait

thats wrong

[heady bluff]

oh im not sure then

[tranquil turtle]

okay thanks

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[grizzled garden]

is this the right process

and can i also assume that the line coming down from a

is midway between the origin and (20,0)

[heady bluff]

i mean for the diagram i dont think it really makes a difference

what do you plan on doing next

[grizzled garden]

so i tried

using pythag

as teh hypotenuse is 12?

[heady bluff]

wait

[grizzled garden]

because its radius

[heady bluff]

why

[grizzled garden]

i think that works idk

[heady bluff]

no like what do you get after you find hypotenuse

and hyp for which triangle?

[grizzled garden]

well that depends on whether or not a has the x coord of 10 or not

[heady bluff]

i dont really think you need that

do you know parametric form of tangent of circle?

[grizzled garden]

whatt

whats that

[heady bluff]

(rcostheta, rsintheta)?

you havent learnt that?

[grizzled garden]

oh ye ik that

wouldnt have thought to use that

[heady bluff]

well thats a clue

take 5 mins think about what you could do

if you dont get i t

ping me

[grizzled garden]

ok thanks

[midnight plank]

@grizzled garden Has your question been resolved?

[heady bluff]

@grizzled garden

did you understand what i said?

or you need me to explain?

[grizzled garden]

im usign a diff method now

with discriminant

[heady bluff]

how so

[midnight plank]

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[thin crater]

Can someone explain why the answer is B? I noticed that the average rate of change for f(x) must be positive because the function is increasing so I concluded that f’(C), f’(D), f’(E) could not represent the average rate of change. Considering between f’(A) and f’(B) representing the average rate of change I guessed that it was B because f’(A) looked like it would increase too quickly but I think there is a more precise way to do this and I may have just gotten lucky.

[hoary pulsar]

This feels like a pretty murky question, but the way I'd view it is that if you drew a line from f(-2) to f(4), the slope of that line would be closest to the slope of the tangent line at f(B)

Does that help at all?

I agree that the difference between f'(A) and f'(B) in this question feels pretty subjective

[thin crater]

It solidifes my reasoning so yes it does help

thank you

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[twin ridge]

I am curious how I would even approach this.

[twin ridge]

I'm seeing "y>0" which implies that 0 is the infimum of this function. So it's saying that "For any positive real number, there's a lower number which can be represented by 1/2n"?

Essentially, a proof proving that 1/2n is asymptotic?

[midnight plank]

@twin ridge Has your question been resolved?

[twin ridge]

<@&286206848099549185>

Am I able to say that we can represent y as a fraction 1/y_i, where y_i could be a square root or anything. Because if y_i is a real number then by Archimedes Property, there is always a natural number larger than y_i.

This 1/y_i is basically just doubly inverting it.

So like if y=3, then I could say that 3=1/(1/3) and my y_i is 1/3

I guess I wouldn't need a y_i here then.

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[twin ridge]

.reopen

[midnight plank]

✅

[twin ridge]

Is someone able to assist me with this please. I can't even tell what it is asking.

So example a is saying if you shift S, then your supremum also shifts, and by the same amount.

[surreal moon]

essentially. yea

[twin ridge]

And then I think this question is asking "if you have sup X and add the shift to that number, it is the same supremum as if you shift X by a and then find the supremum

I'm assuming same for inf.

[surreal moon]

$\sup f(X)$, not $\sup X$

[grand pond]

SWR

[twin ridge]

Ah shoot. I hate bijections.

[surreal moon]

Then you don't need to worry. This isn't a bijection.

[twin ridge]

What is it?

I have a function that's mapping X onto R.

[surreal moon]

It's just a function with a bounded range in R

[twin ridge]

Oh ok.

So it's similar, just different notation?

[surreal moon]

similar to what?

[twin ridge]

Similar to if I define $X\subseteq\mathbb{R}$?

[grand pond]

Narutoes

[twin ridge]

This was the previous problem, I just gave up on it cuz nobody jumped on for 2hours.

[surreal moon]

It is not necessary that X be a subset of R

X is just any nonempty set

[twin ridge]

So we're in the world of maybes again?

[surreal moon]

maybes?

[twin ridge]

As in an undefined set

[surreal moon]

"undefined" is an ugly word

[twin ridge]

So is this problem lmao

[surreal moon]

It's well-defined. It's just not exactly defined

But that's fine. Let's break it down

[twin ridge]

It is defined as "there are an undefined amount of elements in R"

[surreal moon]

No. Not at all.

X is defined simply as a non-empty set. That's it. That's all you need.

[twin ridge]

X is a nonempty set with elements in R. So it is bounded with a supremum and infimum

[surreal moon]

There's no implication of it being a subset of R, or having any elements of R. It could be, but there's no guarantee. And we don't need these assumptions

[surreal moon]

X is a nonempty set with elements in R
Full stop.

X is not bounded.

[twin ridge]

So the elements are not in R?

[surreal moon]

The elements of the image of f are what are elements of R

[twin ridge]

Ok.

[surreal moon]

And the image of f is what is bounded

Does that make sense, or are you still lost?

[twin ridge]

I understand that.

Do I ever have to refer to just X in this proof, or do I always deal with f(X)?

[surreal moon]

Probably just f(X)

[twin ridge]

Ok cool.

Just gotta make sure I include the f() in everything then

[surreal moon]

Okay, consider this. If $X$ is non-empty, and $f :X\to\bR$, what can you say about $f(X)$?

[grand pond]

SWR

[twin ridge]

If X is nonempty and f is a bijection of X onto R, are we able to assume that X is finite, or could it be infinite?

Nonempty doesn't imply finite

[surreal moon]

no

f is not a bijection

[twin ridge]

f maps X onto R

[surreal moon]

Bijections have nothing to do with this problem

[twin ridge]

So X could be infinite?

[surreal moon]

Where does it say "onto"? What does it say "bijection"?

[twin ridge]

Idk man

I'm taking linear algebra as we speak so idek what onto really means.

[surreal moon]

you gotta be careful when you read this stuff

[twin ridge]

can we just get back to the problem

[surreal moon]

You're making a hundred assumptions aobut these problems

I'm trying

[twin ridge]

Because I can barely read what the problem is.

[surreal moon]

But we seem to keep veering off into bijections

[twin ridge]

I don't know what a lot of these signs means

[surreal moon]

Like which?

Tell me which you're stuck at and I will gladly give you a full explanation of it

[twin ridge]

I always thought -> meant bijection. I thought f(x) and f(X) were different things.

[surreal moon]

f(x) and f(X) are different things. -> does not mean bijection. I'll explain them a bit more

[twin ridge]

The notation in this class doesn't really lend to a good learning environment when they don't explain the notation at all.

[surreal moon]

Quick review: A set is any collection of elements. The elements could be numbers, they could be something else. When you hear "set" don't always think numbers. It'll poison your perception. Sets can be empty, finite, countably infinite, or uncountably infinite. Whenever you read "set", keep this all in mind until the problem says different.

[twin ridge]

I got the set \neq numbers part

[surreal moon]

A function is a way of mapping elements of one set to elements of a another set.

[twin ridge]

So X->R maps elements of X into real numbers?

[surreal moon]

yes. But you need a function name too, like f. So you would say f : X -> R

[twin ridge]

Right.

Does f(x) refer to a specific element while f(X) is the entire image?

[surreal moon]

f : X -> Y is just shorthand for "f is a function that maps elements of X to elements of Y". There is no guarantee here there f is bijective, surjective, or injective. The only guarantee is that f will always map the same input to the same output.

[surreal moon]

Yes. exactly. But it helps to specify that x is an element of X.

[twin ridge]

Right, which is done in the question

[surreal moon]

You can write it as "If $x\in X$, then $f(x)$ blah blah blah..."

[grand pond]

SWR

[surreal moon]

Whereas $f(X)$ is defined as the image (or range) of the function

[grand pond]

SWR

[surreal moon]

$f(X)$ is the set of all $f(x)$ where $x\in X$. You can write this in set notation as $f(X)={f(x) :x\in X}$

[grand pond]

SWR

[twin ridge]

So I have to prove that for f(X), a+sup f(x) = sup (a+f(x))

With the set notation stuff that I don't feel like typing

[surreal moon]

yes. Exactly

[surreal moon]

Yeah I feel ya

[surreal moon]

Err. you have to prove $a+\sup f(X)=\sup(a+f(X))$. You have to be careful with your capitalizing

[grand pond]

SWR

[twin ridge]

Wish me much luck. I got my midterm on this in 6 minutes

[surreal moon]

gg

Well the example kinda just gives you the whole answer

If $X$ is non-empty, what can you say about $f(X)$?

[grand pond]

SWR

[twin ridge]

If $X$ is non-empty, then $f(X)$ is nonempty.

[grand pond]

Narutoes

[surreal moon]

mhm

[twin ridge]

I'm still not entirely sure if that implies if it is finite.

[surreal moon]

And the problem told you that the range of f (which is just f(X)) is bounded.

[twin ridge]

Oh ok

That's fair.

[surreal moon]

It does not. No need to worry about finite or infinite

[twin ridge]

Then there's a supremum and infimum.

[surreal moon]

So if F(X) is bounded and non-empty?

[surreal moon]

yes

[twin ridge]

Teacher's running late apparently.

[surreal moon]

So if f(X) has a supremum, then just use your example 2.4 whatever, and congrats, you're done

[twin ridge]

This class was like a pre-summer class. Actual summer semester started today, so the parking lot was packed.

Ok cool. Basically just copy paste?

The problem before this one is with a scalar rather than a shift. Is it a similar proof, or very different since they stipulate that the scalar is negative?

[surreal moon]

Don't even need to copy-paste. Just show how f(X) works as the set S in example 2.4.

[twin ridge]

Gotcha.

[surreal moon]

?

[twin ridge]

This was the one before.

[surreal moon]

Pretty similar, yeah

[twin ridge]

Ok cool. He just got here.

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[distant moat]

it is correct to use newton's first law for an object laying on a surface, right?

[distant moat]

like this

[hearty rune]

the way under first law is in fact just newtons second law in action

a=0 so N-G=0 so N=G, magnitude wise

idk whats going on under second law, doesnt make much sense

[distant moat]

Yeah, I think I missunderstood what I was supposed to do there

and sett a = g

[hearty rune]

but yeah, the way under first law, is newtons second law F=ma

[distant moat]

what?

[hearty rune]

the writing under 'first law'
is just the correct way to use newtons second law

[distant moat]

are you sure?

[hearty rune]

yes

if everything is at rest

which it should be

[distant moat]

but Newton's first law says that an object remains still, or moves with a constant velocity only if the sum of the forces equal 0

[hearty rune]

it does, sure, i dont see your point really you could apply any of the three laws to this, its not like theyre independent

the logic would be the same

[distant moat]

yeah, but the stuff I wrote under Newton 1 is supposed to be under Newton 1, not 2

[hearty rune]

saying the sum of f=0 therefore v is constant
it exactly the same as saying the sum of f is 0, therefore a is 0

[distant moat]

yeah I guess

how do I use Newton 3?

is it just everything flipped?

N = 98,1 N
N* = -98,1 N
G = 98,1 N
G* = -98,1 N

like this?

[west iron]

are we on some hypothetical planet where the force of gravity is 10× stronger

or you mean kg m/s²

[distant moat]

whaat?

[distant moat]

I wrote that

[west iron]

was the kg invisible?

[distant moat]

oh here

no that was just a mistake

[distant moat]

you can see that I wrote the correct term here

[distant moat]

there

fixed

[west iron]

N and G should have opposite signs then probably

[distant moat]

wdym

they do

[west iron]

i'm not sure where they do but it isn't here

[distant moat]

then wdym by oppoisate signs?

[west iron]

N = 98,1 and G = -98,1 or something

[distant moat]

what

why

[west iron]

because they act in opposite directions

[distant moat]

but they aren't action reaction pairs

it would make sense to do so if the object was moving up

if they had opposite signs then the sum of the forces wouldn't be 0

You know what I mean?

thanks y'all

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[normal plover]

@sharp coral sorry its not letting me reopn

do you know why they introduce matrices in this question?

whats the point of using matrices

[sharp coral]

their intersection is the solution to the system of linear equations, and matrices are a convenient way of solving systems of linear equations

[normal plover]

ohhh

so theyre finidng the intersection

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[stable owl]

Can someone please explain why Q x Z does not contain any isolated points? Thanks!

[stable owl]

I know that Q is dense in R, but Z is not.

[floral apex]

i dont think there is anything special here

Q doesnt contain any isolated points

why should Q x Z

[stable owl]

Would you mind elaborating why we don't need to consider Z?

[floral apex]

Well Z isnt going to cause the problem here

maybe you think of IDK lets plot this on a cartesian real plane

lets say Z is horizontally

Q is vertically

so you get the visual of like

every unit step in any horizontal direction from the original

were going to land on a line of points

fair?

it goes vertically up and down

[stable owl]

Ohhh

So we are going from -1 to 0 to 1...

[floral apex]

that line is dense, vertically

[stable owl]

On the domain of Z

[floral apex]

and your neighborhood will have some nonzero radius

[floral apex]

yea

[stable owl]

And every epsilon ball would contain some points in Q since it is dense

[floral apex]

so, i mean, Z is not contributing any issues so its maybe important to visualize but its not the problem

[floral apex]

yea, unless you can propose a way to find a neighborhood which doesnt

[misty gorge]

also, Q x Z is homogeneous, so if (0,0) isn't an isolated point, then there is no isolated point

[floral apex]

which would be impossible

unless it has no vertical ... width

[stable owl]

wait would you mind elaborating this???

[floral apex]

assuming youre comfortable with Q being dense

[stable owl]

Sounds cool

[misty gorge]

geometrically (very vaguely), things are sort of the same everywhere. There is a simply transitive action of Q x Z on itself by translations, and these translations are all homeomorphisms

[stable owl]

oh!!!

Okiii

Thanks so much @floral apex @misty gorge

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[twilit field]

what assumptions am I allowed to make here

[twilit field]

like which properties can I assume hold true

[prime hornet]

at least the vector space axioms, surely

the field axioms too

it’s probably best to just write out x, y, and z in coordinate form and apply field axioms from there

[twilit field]

uh

let me check if axler has covered them prior to this

the only thing he has done is defined addition on $F^n$

[grand pond]

ƒ(Why am. I here)=I don't Know

[prime hornet]

how did he define them

it’s almost certainly enough

[prime hornet]

.

[twilit field]

I'll do that

thanks

so I have $[(x_1,x_2,x_3....,x_n)+(y_1,y_2,y_3......y_n)]+(z_1,z_2.....z_n)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

from this it follows trivially

is that it?

so this gives $(x_1+y_1,x_2+y_2......x_n+y_n)+(z_1,z_2,z_3......z_n$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

or $(x_1+y_1+z_1,x_2+y_2+z_2....x_n+y_n+z_N$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

I can similarly expand the RHS

.close

[midnight plank]

Channel closed

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[midnight plank]

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[bleak pier]

Hints

[midnight plank]

@bleak pier Has your question been resolved?

[midnight plank]

@bleak pier Has your question been resolved?

[midnight plank]

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[twilit field]

show that $1x=x ;\forall x \in F^n$

[grand pond]

ƒ(Why am. I here)=I don't Know

[half ingot]

Huh?

[twilit field]

I have to prove this

[half ingot]

I just need help with this page 😭

I chose the “I need help one”

[twilit field]

#❓how-to-get-help

[twilit field]

this channel is occupied, please open your own channel

try #help-31

[half ingot]

Thanks, I’ll try to figure out how to navigate!

[twilit field]

so $1(x_1,x_2....x_n)$ is what I have

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

which is (1x_1, 1x_2......1x_n)$

which is $(x_1,x_2.....x_n)$

is that it?

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

.close

[midnight plank]

Channel closed

Closed by @twilit field

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[prime hornet]

that's how Axler defined scalar multiplication on F^n, yeah?

if so, then you're just done lol

[twilit field]

wat

you're joking, right

why would that be a question then?

😭

[midnight plank]

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[sage helm]

Having trouble with (b)

I'm pretty sure no such collection should exist

[west iron]

Any intuition as to why?

[sage helm]

Because like

I should be able to assign a natural to every set

Which means its countable

I can't think of a collection for which I can't do that

[west iron]

ok well yes

[sage helm]

I know it's a little handwavy

Since that's the fucking definition

[west iron]

yeah lol fair enough

uh

how about assigning a natural to each set

we find a different countable set to assign to each interval

[sage helm]

Huh?

[west iron]

use something other than the naturals

[sage helm]

Oh

[west iron]

can we associate every interval with a unique member of some countable set

[sage helm]

Well the next logical choice would be rationals I guess

[visual tiger]

So you're not required to GIVE such a collection, but you can definitely show that one exists

[sage helm]

It exists?

[west iron]

sure, is there a way of associating open intervals with rationals that stands out?

[sage helm]

Wait I'm confused, does it exist or not, because I don't think it does

[west iron]

❌

[visual tiger]

Ah wait there's a flaw in my logic

[west iron]

it does not exist

[visual tiger]

ok nvm

[sage helm]

Stands out Let me think

I remember the proof of showing that the rationals are countable

Using finite length sets

[visual tiger]

I even forgot I proved that as well

[sage helm]

Wait actually

If the rationals are dense, for any (a, b) I can assign some arbitrary rational in that interval to it

So we have a 1-1 correspondence now

[west iron]

yeah exactly

1-1 correspondence to a subset of rationals anyways

[sage helm]

Yeahh

But since that would be a subst of Q, it is also countable

Right so that concludes it I suppose

Thanks for your help

.close

[midnight plank]

Channel closed

Closed by @sage helm

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[west iron]

np

having lag problems

so didnt see

lol

[midnight plank]

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[lament fox]

The binary operation $a-b$ is associative right for \bf{Z}?

[grand pond]

The د

[lament fox]

since $a-b+c=a-(b-c)$

[grand pond]

The د

[lament fox]

.close

[midnight plank]

Channel closed

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[midnight plank]

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[solid island]

can i just ask

[solid island]

for this question

if they asked a question saying

how many arrangements can be made where the vowels must be included

is it 5! * 5c3

[fleet moss]

send the whole problem

oh i see

[fleet moss]

yes i think this is right

[solid island]

ohhh

okok thank u

and then

can i ask for this question

[fleet moss]

go ahead

[solid island]

could someone just try it

[fleet moss]

okay so

[solid island]

cos

i got an answer

[fleet moss]

mhm

[solid island]

but its not matching up w the actual answers

but then

i dont think

the actual answers

r right

cos they arent like verified 😭😭

[last slate]

what is it ?

[solid island]

well

i got 360

[fleet moss]

im getting 30 * 5c3

[solid island]

it says 480 appar

[fleet moss]

,w 30 * (5c3)

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by esclown

[fleet moss]

wow thank you

[last slate]

lol

[fleet moss]

,w 30 * (5 choose 3)

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by esclown

[solid island]

i got 360

[fleet moss]

ok i was off as well it seems

[solid island]

for that pne

cos ok

so it says

sepearted by

at least two consonants

right

so i just took the cases where

the two vowels r together

and then the two vowels r separated by 1 consonant

[fleet moss]

well in this case its easier to do directly

[solid island]

oh i guess so

its just because from a previous questin

i alr did

two vowels together

wait lemme take a photo

[fleet moss]

if you have the two vowels separated by two consonants, youre left with only one consonant. you have 5 choices of where to put that last consonant

[solid island]

oh wait u rrihgt

its much quicker

to do it direclty

ok lemme try it that way

oh

shi

i still got 360

[fleet moss]

lowkey

im also getting 360