help-33

[marsh citrus]

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[still temple]

hi guys

[still temple]

can someone?

[blazing rampart]

try and make that one single ratio

7:4 is the same as

14:8

[still temple]

i didi the ratio

[blazing rampart]

so the ratio of houses:flats:bungalows would be 14:8:5

[still temple]

but idk if i have to dived it by 7:4 or mutiply it

[blazing rampart]

wdym

[still temple]

hm

[blazing rampart]

now using that single ratio

to find the houses

you can simply multiply the number of bungalows by 14/5

[blazing rampart]

since the ratio of houses:bungalows is 14:5 ignoring the flats

[still temple]

where you took the 14 from?

[blazing rampart]

uh

the original ratio was 7:4

and I made that 14:8

[still temple]

hm

[blazing rampart]

then I combined the ratios and got rid of the flats

[still temple]

ok i got it

[blazing rampart]

and that way I got a relation between number of bungalows and houses

[still temple]

hm clever guy

but the way i did was totaly messing

i did 7:4=11 and then i timed 11 by 7 and 4, giving me 77 and 44

[polar quail]

Can someone help me with my linear equations I have been sitting here for half a hour stuck on this question

You have to find the intercept

[still temple]

sorry but im nothing good on maths

[polar quail]

This is one I already did if it helps

[still temple]

@blazing rampart just explain one thing to me

[blazing rampart]

yh?

[still temple]

on ratio we devid or times ?

like, 2:3=5, just to me know the amount of 3 and 2 i times them by 5 or divide ?

[blazing rampart]

wait huh

I dont get ur question

[still temple]

its onother example

it like, there is a bag, in the bag have the ratio of 4:5 which is phones and tablets

so to me know the amount of phones what i need to do

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[hollow wharf]

this might just be me being pretty dumb but can someone tell me where the 0.6 came from? i genuinely have no idea

[marsh citrus]

@hollow wharf Has your question been resolved?

[slim spire]

0.6 is not blue

since the problem tells you blue is 0.4 and the rest are some weird ratio

[hollow wharf]

OHH TY BRO omds i was so confused for ages where they pulled 0.6 from

.close

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[inland lichen]

For this question I noticed that I have to multiply 2pir with the density for the mass but I am not sure why. Can anyone explain/

[harsh falcon]

you have kg/m^2
integrating once will give you kg/m, but you want kg
this r comes from (i think) the jacobian

and the 2pi comes from the fact that you also want to integrate from 0 to 2pi

otherwise you would just have a line and not a circle

[inland lichen]

Okay thank you : )

[harsh falcon]

and the jacobian is because we are not in cartesian coordinates

[inland lichen]

okie great thank you!!

[marsh citrus]

@inland lichen Has your question been resolved?

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[silk sparrow]

How do you reduce a value by 80%

Like imagine for a minute the question is just 'reduce 86 by 80%'

[marsh citrus]

@dull hazel Has your question been resolved?

[dull hazel]

I would multiply by 0.8

[silk sparrow]

That would give 80% of your value, rather than reducing by 80%

If I reduce something by 80%, how much do I have left

[marsh citrus]

@dull hazel Has your question been resolved?

[marsh citrus]

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[spring stirrup]

how do i solve this

[spring stirrup]

can someone explain what was done here?

[kind solar]

Solve for the time t such that P(t) = 0.50

i.e. solve 100(1.2)^(-t) = 0.50

(1.2)^(-t) = 0.5/100

ln(1.2^(-t)) = ln(0.5/100)

do you know how to proceed from there?

I mean, there are many different ways to work with logarithms

[spring stirrup]

what exactly is a hlaf life?

[kind solar]

It's the amount of time it takes for the material to decay to 50% of its original mass

[spring stirrup]

ohh

thats why you set the equation to 0.50

because ur trying to solve for the time when its 0.50

i cant remember logs from last semester

howd they get here

OH

NVM

ty

.close

[marsh citrus]

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[marsh citrus]

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[snow nest]

i dont understand III

i mean i do, but the answer sheet says that the graph should have a vertical tangent, how does that make sense??

[upper timber]

Do you mean horizontal?

[snow nest]

no, look

[upper timber]

It’s at 2

[snow nest]

this is the answer sheet, and unless its wrong i have no idea how they got a vertical tangent

[upper timber]

Kinda goofy to make a tangent line there but I guess so

[late geode]

that's answering something different

[snow nest]

so there is no vertical tangent then

[late geode]

there seems to be one at x=-1

[snow nest]

then why did they say x=-1, 3

oh wait

[late geode]

doesn't look like theres a vertical tangent at x=-3

[snow nest]

ok i understamd that the horizontal tangents are at x=-1, 3

[late geode]

no

[snow nest]

oh

what is then

[late geode]

do you know what vertical/ horizontal mean

[snow nest]

vertical is y and horizontal is x

[late geode]

too vague

and unclear

describe it without using x or y

[snow nest]

im so confused what you are asking for, sideways is horizontal and vertical is up and down?

[late geode]

horizontal lines look like _
vertical lines look like |

[snow nest]

yeah

ok so

now what

[late geode]

apply that

draw a horizontal line at the point where x=-2

and it should be clear that's where you have a horizontal tangent

[snow nest]

but if thats true, then the horizontal tangent is x=2 yes?

which makes the vertical tangents x=-1, 3

ohh i switched them around when i first said it mb

[late geode]

to me it doesn't actually look like x=3 is a vertical tangent

[snow nest]

why nbot

[late geode]

the graph doesn't look like its decreasing fast enough

it could be depending on the function, but it's unclear just from looking

[snow nest]

idk what to tell you, my teacher says differently

[snow nest]

fair

thanks for ur help

.close

[marsh citrus]

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[marsh citrus]

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[sour iris]

Need help with number 2

[sour iris]

<@&286206848099549185>

[frigid totem]

whats the question ? solving for x ? if so Id advise you to take the logarithm of both sides @sour iris

[marsh citrus]

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[marsh citrus]

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[sour iris]

So like

[sour iris]

Huh

you can just log base 10 both

What would that do

[frigid totem]

using the properties of the logarithm, that would get rid of the exponent

log(a^b) = b log(a)

[marsh citrus]

@sour iris Has your question been resolved?

[sour iris]

Ohh

Wait but then

like what would you do with them after

Cuz u would do that to both sides

[sour iris]

Would you like

Put one to the other side and do the property to make it division

[frigid totem]

yes you would do that for both sides

[sour iris]

but what after that

like to get x

[frigid totem]

simply put all the x on one side

then you can factor by x

[sour iris]

wait

so

after u log

do the logs disappear

[frigid totem]

they dont

but those are constants now

[sour iris]

like does the equation become like x(x-x)= x(x-x)

like you can ignore the logs

i mean

[frigid totem]

after taking the log you have (2x - 3) log (5) = (x - 2) log(9)

right ?

[sour iris]

ye

[frigid totem]

oh yes

you can ignore them

if thats what you mean

they are constants now so they dont pose any issue

[sour iris]

so like you would do (2x - 3) (5) = (x - 2)(9)

[marsh citrus]

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[sour iris]

.reopen

[marsh citrus]

✅

[sour iris]

its this simple?

[frigid totem]

the logs are still there

[sour iris]

so like

what would you do

[frigid totem]

ill send you a picture, easier than typing it out

one sec

[sour iris]

ight

ima brb in like 15

but like ima look after

ty

[frigid totem]

,rotate

[frigid totem]

i write "log" as "ln"

@sour iris

then you can just factor by x

[marsh citrus]

@sour iris Has your question been resolved?

[sour iris]

Uh

[sour iris]

Did the parentheses go away

[frigid totem]

i developped ln(5) and ln(9) yes

so i could get rid of the parenthesis

i re arranged the terms around to only have x on the left side

now you just need to factor by x and finish

[sour iris]

Thats like

We aint ever done that in precalc before

Oh wait can we just ignore the lns cuz theres one on every number

[marsh citrus]

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[night widget]

anyone know where to start with this? im kinda lost, its for a paper due tmr

[night widget]

its for calc 1

[marsh citrus]

@night widget Has your question been resolved?

[night widget]

<@&286206848099549185>

[slim spire]

you start with 60=(circle perimeter)+(rectangle perimeter), and the blue area (circle area)+(triangle area)

all in terms of x's and y's

then move around the 60=... equation to get x= or y= to plug into the blue area, so that a variable goes away

that's when you can finally optimize and set the derivative to 0 and check the endpoints for the maximum

[night widget]

okay thank you im going to give it a shot im not good at math especially calc so i will try my best here is it okay if i ask another question if i get stuck?

[marsh citrus]

@night widget Has your question been resolved?

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[stark trail]

Hello, I was hoping someone could double check if my isomorphism works here :)

[static quarry]

as soon as you show that they are both cyclic groups of order 4 you're done

[quasi patrol]

I love help pages like this because i dont understand half the words in the question

And i can only look in horror

[static quarry]

the more modern term is "doubt channel"

[stark trail]

I don't have any theorem stating this though to use, we're just starting group theory

[static quarry]

show in general that any two finite cyclic groups <a> and <b> of the same order are isomorphic via the map phi(a) = b

[stark trail]

I have a doubt if that is truly necessary for this doubt though

[static quarry]

it's arguably easier to show the general result than it is to get into the weeds of any specific example

but you gotta do you

no matter how you proceed, it's gotta be more interesting than watching people whine about capitalism in discussy haha

[stark trail]

I very very rarely go in discussy

[static quarry]

you are missing very little

[static quarry]

very few doubt clearers to be seen there

doubt spreaders aplenty

[stark trail]

Oh wow even the lusty mods are in there

yeah I'll definitely steer clear

[hasty ruin]

but every once in a while you get another spectacle of someone discovering how to divide by 0

[static quarry]

yea it's like a gathering point for the nefarious

[stark trail]

Sir, I think I am in no need of proceeding, only in having my doubt be cleared

[static quarry]

proceed with clearing your doubt

i proffered one way forward

[quasi patrol]

One time i was in discussion and there was this guy that was doing some philosophising except badly

[static quarry]

that's like 25% of discussy posts

[stark trail]

you always proffered a way into clogging my doubt channel with nonsense

[static quarry]

the other 75% are shitposts

[quasi patrol]

There are people in discussy that seriously like communism

Its a shithole

[static quarry]

you can't make a wilfully blind man see

[stark trail]

U r making my doubt channel into a shit hole

[quasi patrol]

Sorry

Bye

[stark trail]

https://tenor.com/view/modern-family-spray-squirt-annoyed-irritated-gif-11864636400230753109

[static quarry]

it's easy to love if you have never worked or earned anything haha

[stark trail]

I will proffer you a new doubt instead

[static quarry]

by all means

[stark trail]

So

I've done some things with this

but nothing too fruitful

[static quarry]

compute (ab)^(mn)

[stark trail]

If (ab)^k=e I can show that m|k and n|k and hence that mn|k

[stark trail]

=a^(mn)b^(mn)=e

[static quarry]

you don't even need (m,n) = 1 for this

only if you want to compute the actual order of G

[stark trail]

Alright. How shall I proceed?

[static quarry]

ah i take that back, you do need it

so ok

show that in fact ab^k is not e if 0 < k < mn

[stark trail]

Okay, I've shown that

if (ab)^k=e

then

mn|k

[static quarry]

yea

that suffices

(ab)^(mn) = e

and if ab^k = e then k >= mn

so the order of ab is mn

[stark trail]

yes indeed

[static quarry]

and ab is an element of G

so <ab> is a subgroup of G

of order mn

so <ab> is..

[stark trail]

ab is an element of G yes

<ab> is a subgroup of G yes

and yes it has order mn

so, it's a subgroup of the same order as the original group

does this imply they're equaal

[static quarry]

sure

even ignoring the group structure

they're finite sets of the same cardinality

and one is contained in the other

so they're equal

[stark trail]

Yeah I'm like shockingly bad with this group stuff and I'm really not sure why

[static quarry]

you'll get used to it

and then this stuff becomes obvious

but then you just move onto harder things

[stark trail]

:p

[static quarry]

like the feit thompson theorem

[stark trail]

let me complete this proof and then I will present you with the cleared doubt

[static quarry]

better hurry, you're missing exciting scintillating discourse such as

[stark trail]

,w scintillating

[elfin berry]

Results provided by WolframAlpha

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

Requested by austinu

[stark trail]

Indeed very scintillating

[stark trail]

Here it is sir. What do you think?

[static quarry]

where do you need abelianness here?

oh wait

[stark trail]

(ab)^k=a^kb^k

[static quarry]

nev mind

all's well

[stark trail]

yes sir

[static quarry]

yea that's legit

[stark trail]

ABCDEFGH

[static quarry]

every fucking goddamn hour

[stark trail]

thanks, I'll probably be in here doubting again later...

[static quarry]

doubt early and often

[stark trail]

.close

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

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[open oriole]

can someone please help me

[wary kite]

close your other channel

[fallow fossil]

so what area is the rectangle

that should be ur 1st step

(the equation)

[wary kite]

what is the formula for the area of the rectangle in the picture

[open oriole]

i am not sure

[quiet anvil]

What is the base and height?

[open oriole]

area = w * l

[quiet anvil]

How wide is the base, if you're considering a point on the graph (x, y)?

[open oriole]

1

[quiet anvil]

Why 1?

[open oriole]

because thats where the x would be

if the nubmrts were labeled?

[quiet anvil]

Why is x = 1?

Wouldn't x just be x?

[open oriole]

oh yes

[quiet anvil]

So the base is what, then?

And the height would be what?

[open oriole]

y

[quiet anvil]

Which makes the area?

[open oriole]

xy

[quiet anvil]

Cool

But you know what y is in terms of x, right?

[open oriole]

dy/dx ?

[quiet anvil]

Hint, (x, y) is on the graph of the function

Defined by what equation?

[open oriole]

the y=8/x^2+1

[quiet anvil]

Yes exactly

[open oriole]

so since we have that ew just need x

right

and then we multiply them

[quiet anvil]

So if the area is xy, then what is it in terms of just x?

What is A(x)? (i.e. the Area function)

[open oriole]

?

[quiet anvil]

Why that?

[open oriole]

i just differrentiated it

[quiet anvil]

You're jumping ahead, and making mistakes as a result.

[open oriole]

do we take that y equation

and then just solve for x

?

[quiet anvil]

We make an equation for the area

A(x) = xy

[open oriole]

we already have y tho right?

[quiet anvil]

= what?

Yes, we already have y, so what is A(x) just in terms of x?

[open oriole]

the Y eaution times X ?

[quiet anvil]

Yes, which is?

[open oriole]

8x/x^2 +1

[quiet anvil]

Exactly

Now, we want to optimize this formula, find the maximum

So we must do what?

[open oriole]

1,4 is the local maxima

so would it be 4?

[quiet anvil]

How did you arrive at that?

[open oriole]

i am not sure

i solved for maxima

but i dont know whether or not

that is what we are lookingfor

[quiet anvil]

Using what method?

[open oriole]

i found where the derivative of the function is equal to 0

[quiet anvil]

What derivative did you come up with?

[quiet anvil]

(yes, x=1 is also the value I arrived at, which gives A(1) = 4)

That is correct, sorry for being a stick in the mud, I just wanted to make sure you didn't skip ahead and make mistakes

[open oriole]

its ok thank you

[quiet anvil]

[marsh citrus]

@open oriole Has your question been resolved?

[marsh citrus]

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[tawny lily]

i just need to know how this happened

[tawny lily]

where did the -1 come from??

[elfin berry]

Info

Info

[placid hound]

*denominator

[tawny lily]

i still dont understand where -1 come from

[placid hound]

Ok, do you agree that :

$$ \dfrac{x^2}{x} = x^{2-1} = x^1$$

[elfin berry]

Info

[tawny lily]

yes

[placid hound]

So when you divide $3^{a}$ by 3, you've got $3^{a-1}$, right?

[tawny lily]

oh shit

[elfin berry]

Info

[tawny lily]

oh i see 😭 ijust didnt realize bc its not a variable

dasasd

[placid hound]

don't worry ^^

now you understand, that's what matters 😋

[tawny lily]

thank u, i get it now

[placid hound]

cool, have a nice day!

[tawny lily]

.close

[marsh citrus]

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[marsh citrus]

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[livid maple]

How do i solve this system?

[marsh citrus]

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

[livid maple]

.status

!status

[marsh citrus]

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

[livid maple]

1

[sinful thistle]

huh

i mean makes sense it looks scary af

tried expanding?

[livid maple]

Yes but it doesn't look good...

[novel juniper]

is this teh OG problem?

[livid maple]

Yes

[sinful thistle]

that is what we get

[blazing rampart]

rearrange the first one

[elfin berry]

ren

[novel juniper]

add them and subtract them I guess

then factorise and substitute

[sinful thistle]

which is what i was thinking

equate them? @novel juniper

[novel juniper]

maybe

I was thinking that these reprsent the standard forms of certain conics

,w x^2+10xy+6x+30y+15=0 graph

[elfin berry]

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[novel juniper]

,w 12y^2+12y+4xy+1=0 graph

[leaden monolith]

I’m thinking to make 2y + 1 the subject on the top, sub it into the bottom and solve for y in terms of x

[elfin berry]

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[novel juniper]

graph them and solve perhaps?

[marsh citrus]

@livid maple Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[fossil grail]

the perpendicular line to x+y-1=0 is x-y+C=0, right?

[winged pier]

Is it?

[fossil grail]

IDK, you tell me

that's why I'm asking

[winged pier]

What needs to change in the slope for a line to be perpendicular

[fossil grail]

.close

[marsh citrus]

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[blazing rampart]

lol

[marsh citrus]

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[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[nimble elm]

Is it possible to solve this without using graphs

[blazing rampart]

Uh yes if you're allowed to use lambert W function

[nimble elm]

What is the lambert w function

How to do it using that

[blazing rampart]

Product log

Wait do you want to find the solutions?

[nimble elm]

No

I just wanna know how many solutions have

This was in my yr11 extension class past paper

Is there anyway to know how many solutions there are without using lambert w function and graphs

[blazing rampart]

Uh

[nimble elm]

Like you don’t have to solve for x, just how many solutions there are

[blazing rampart]

Firstly there are two 'obvious' solutions

Or trivial ones

[nimble elm]

Yeah?

[blazing rampart]

But then you have to realise there's a third one too

[nimble elm]

This is the way I did it

[marsh citrus]

@nimble elm Has your question been resolved?

[marsh citrus]

@nimble elm Has your question been resolved?

[slim spire]

I've only seen this done with graphs

[nimble elm]

Ohh okok

All good

Thank u

[marsh citrus]

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[tulip jacinth]

Given a graph of a system of 3 linear equations, how would you decide whether the graph has 0,1, or 2 free variables?

[leaden monolith]

Graph?

A graph of 3 linear equations with 3 variables lives in 3D

[tulip jacinth]

Yeah, like this

[leaden monolith]

💀 ok I guess that is what you’re asking lol

[leaden monolith]

It looks like no point is on all 3 planes at the same time

[tulip jacinth]

Yeah, I’m not sure when the graph has 0,1, or 2 free variables

[leaden monolith]

If a point is on the plane it satisfies the equation of said plane

If a point is to satisfy all 3 equations then it must be on all 3 planes at the same time

[tulip jacinth]

My bad, for graphs like this

[leaden monolith]

Well think of it as every point on that intersection is on all 3 planes aka satisfies all 3 equations

And you got slide around 1 direction (or 1 free variable) and still be on all 3 planes

[tulip jacinth]

What do you mean by slide around 1 direction?

[lilac siren]

Do you recognize the intersections in E as a single line?

[tulip jacinth]

Yeah

[lilac siren]

k how many variables do you require to express a line in 3D?

[tulip jacinth]

3

[lilac siren]

I know you thought of X/Y/Z, but I meant how many variables do you need for the set of points that describe a line

you'd need one variable, because a line is 1-dimensional

if you want to express a plane, you need 2 variables, because a plane is 2-dimensional

for a point you need zero variables to describe its set of points since it's 0-dimensional

An example for a line would be
{(n, 4, 7) | n e R}

[leaden monolith]

If you looked at the shape of the intersection of all 3 planes it’s a line, you can specify how long along this line you are aka 1 free variable

I’d say all 2 planes were on top of each other then the intersection of all 3 planes is just the plane

And you need 2 variables to specify where you are on a plane, unlike on a line where you only needed one

[tulip jacinth]

Ohhh, ok, so a point as an intersection would be 0, a line 1, and a plane 2

[leaden monolith]

Yep

Exactly

[lilac siren]

Yup, have you worked with sets of points describing objects before?

[leaden monolith]

If the planes were parallel you would have no solutions as well

[leaden monolith]

Or like this one

[tulip jacinth]

Nope

[leaden monolith]

You’ve done this without knowing before

The graph of y = x consists of the points (x, y) such that x = y

We write it as {(x, y) ∈ ℝ²| x = y, x and y ∈ ℝ}

If you thought about the x-y plane as a collection of pairs of real numbers

[leaden monolith]

This is how we can describe the graph of y = x

[tulip jacinth]

So an object is a point/line/plane etc

[leaden monolith]

Some collection of points

It just so happens that the things we put after the | make a plane/line/point

[leaden monolith]

The stuff after | conditions the particular points we want

It’s also read as “such that”

[lilac siren]

for instance if you try to express the x-axis as a set of points in 3D, then you may start looking at a few sample points:
(1, 0, 0), (7, 0, 0), (0, 0, 0), (-1, 0, 0), (473872, 0, 0) etc.
and clearly the Y/Z values are 0, but we want to express all of those possible points, meaning for all possible X-values, as a set

and you can do so by introducing a variable, say n, which we declare as a real number

which we can use to construct a set of those points:

{(n, 0, 0) | n e R}

as Frosst mentioned, the "|" lets a condition for the set follow

in this case that n is a real number

[leaden monolith]

That’s a weird way to use set builder notation

[tulip jacinth]

Sorry if I’m not responding, I’m just trying to process everything you guys are saying

[lilac siren]

I feel it may be intuitive at first

[lilac siren]

nw

[lilac siren]

this set includes all points of the form (n, 0, 0) where n is a real number, e.g. since n is allowed to be 3.8, we know the point (3.8, 0, 0) is in the set

Since that's true for all real n, this set would describe the line lying on the x-axis in 3D

[leaden monolith]

This is a bit lazy, (but it’s correct although I wouldn’t use | like that) I would prefer {(x, y, z) ∈ ℝ³| y = 0, z = 0, x ∈ ℝ}

[tulip jacinth]

If n has a condition placed on it, would it only describe a portion of the line?

[leaden monolith]

Yes

You can write all sorts of different conditions

[lilac siren]

Absolutely I prefer this representation as well, I think we should introduce it stepwise :]

[leaden monolith]

That’s alright as well, sometimes it’s omitted which set you’re taking elements from, and may be confusing if you don’t see it in its whole entirety

[lilac siren]

If you had a bit of set notation before, definitely experiment with the conditions! You can quickly construct many shapes which may seem complicated at first visually, but might follow simple rules

[tulip jacinth]

What do you mean by a bit of set notation?

[lilac siren]

well the {...} brackets

[tulip jacinth]

Sorry if I sound like a monkey, my first week of linear algebra and I haven’t don’t a lot of math since calc 2 in high school. Are you referring to the image I sent? The problem only had the pictures of graphs

[leaden monolith]

Have you seen this before

[tulip jacinth]

This is a different problem but are these brackets that you’re referring to?

[leaden monolith]

Nope

[tulip jacinth]

I think I saw it introduced today in my lecture slides

[leaden monolith]

It’s very important to be familiar with this notation

It will be used heavily from now on because it is very very convenient

For all areas of math but it will be used extensively in linear algebra as well

[lilac siren]

Depending on what you study you'll likely encounter it very soon in linalg or at the start of analysis, but yeah it's crucial for most fields

[tulip jacinth]

Yeah I just started taking linear algebra last week

[lilac siren]

maybe to give a brief, not definition-wise correct, introduction to them, maybe consider sets like containers which contain elements at first

[leaden monolith]

If you know regex, it’s got the same spirit as regex

[tulip jacinth]

Those elements are lines?

[leaden monolith]

They’re points

But together they (can) make lines and planes

[lilac siren]

Nono the elements of a set can be of arbitrary nature, in linear algebra however you'll most often encounter sets which contain vectors (or interpreted as points)

e.g. the set {1,2,3} contains 3 elements, just natural numbers in this case

[leaden monolith]

The part of the notation that says (x, y, z) ∈ ℝ³ tells you they are points with 3 real numbers describing them

[lilac siren]

and it's elements are 1, 2 and 3

A set cannot contain duplicates, so {1,2,3,3} would be equivalent to the set {1,2,3}

meaning you only regard the individual elements of a set and aren't interested in whether elements would occur multiple times like in a list [1, 2, 3, 3]

(I'm using [...] for list notation for this demonstration)

and another contrast to lists is that order does not matter!

meaning {1,2,3} = {2,1,3}

they contain the same elements

therefore they are the same set.

@tulip jacinth the properties of no duplicates and no order are fairly clear? :)

[tulip jacinth]

Yes 👌

[lilac siren]

Neat, note that we aren't inherently restricted to use numbers or points as elements

it could for example also be words

{"example", "nut"}

which is a little loosely written mathematically, but it's just for demonstration

this set would have two elements

now, I could try to write bigger and bigger lists of natural numbers

eg the range from 1 to 10

{1,2,3,4,5,6,7,8,9,10}

or the range from 1 to 100?

But now you may notice that writing every single element becomes impractical

[tulip jacinth]

Yeah, so that’s why you use the all real numbers sign?

[lilac siren]

ys, I'll introduce variables for them in a sec

[leaden monolith]

We’ve got some standard notation for sets we use very often

[tulip jacinth]

Can I also ask what a span is?

[lilac siren]

just to clarify that a set can contain anything, a set can also contain sets:
{{1}, {2}, {3}}

[leaden monolith]

Cos it’ll be very annoying to have to specify them all the time

[leaden monolith]

This is most easily understood with set notation

Hence why I said you want to be very familiar with set notation

[lilac siren]

It's the set of all possible vectors formed by a linear combination of a given set of vectors

which I could also try describing without sets, but it becomes more tedious

[lilac siren]

But this is another crucial part, since we can let sets contain sets, and you'll be able to write lines as sets, we can have a set that contains lines!

🐛

or a set that contains sets that contain lines!

for whatever reason

Ok but back to how we can write the natural numbers from 1 to 100 as a set

for that we actually need sets that you already are familiar with

(I presume)

N: natural numbers
Z: whole numbers
Q: rational numbers
R: real numbers

these are indeed sets

[tulip jacinth]

Yes, I’m familiar with those 👌

[lilac siren]

N = {1,2,3,4,5,6,...}

ofc we could use "..." Within our set definitions, but that would be a little loose

[tulip jacinth]

Would even numbers be another set

[lilac siren]

eg for the range from 1 to 100 we could write

{1,2,...,99,100}

which seems right, but it's not well written not properly defined

although everyone will understand what you mean

[lilac siren]

Yes!

We can get to how you can construct them in a sec

Ok so how can we introduce a variable for a set?

We'll simply choose some variable name, e.g. n

and we must declare which values n can have

for instance we can write
n e N
meaning "n element N" or "n is a natural number"

But we now have to put that declaration into our set

And to split the part which shows the elements and the part which shows the variables/conditions, we use a |

so { elements go here | variable declarations/conditions go here }

maybe a simple example for that:

{ n | n e N }

now what does that mean? It means that this set contains each elements before the "|" for every possible value of n in N

eg n could be 5, so 5 is in the set

n could be 377, so 377 is in the set

etc

in fact, every natural number is in that set!

Which means: {n | n e N } = N

that set is the same as N, since it contains the same elements :)

Ok, another example with a condition instead, how could we write the range from 1 to 100?

{ n | n e N, 1≤n≤100}

this set contains all natural numbers from 1 to 100

[lilac siren]

Or let's try even natural numbers

{ n | n e N, "n is divisible by 2" }
which is a little loosely written but conceptually right

formally you'd rather write
{ n | n e N, n = 0 mod 2 } for instance

which uses the modulo operator

but now comes a neat trick!

[tulip jacinth]

Oh ok, I was about to ask if you could do n%2=0

[lilac siren]

Exactly that'd be modulo

I assume you have coding background?

given that you use %

so probably Python or C

[tulip jacinth]

Just an intro to python class

[lilac siren]

kk

Sets in Python pretty much work the same way btw :)

they also only have different elements

and the order doesn't matter

[lilac siren]

Ok but the neat trick I wanted to mention is:

{ 2n | n e N }

which also describes all even natural numbers, but much more elegantly

note that you can write before and after the "|" what you want!

we could take all whole numbers divisible by 7 for example:

{ 7x | x e Z }

note that x is just the variable name, the name could be anything

it's the same as

{ 7nut | nut e Z }

they are the same set

Or we could do all odd natural numbers:

{ 2n-1 | n e N }

or all odd natural numbers up to 100:

{ 2n-1 | n e N, 2n-1 ≤ 100 }

the freedom is yours

[tulip jacinth]

Does a set of vectors being linearly dependent or independent rely on whether one of the vectors is in the span?

[lilac siren]

if one of the vectors lies in the span of the others

then that set is linearly dependent

if that quickly makes sense to you visually or set-wise then congrats.

[tulip jacinth]

Thank you so much for the explanation on sets, you made it simple and easy for me to understand 🙏

[lilac siren]

Maybe to give a little visual context say you have the set
{(1, 2, 3)} so a set that contains a single vector

the span of that set would be all vectors as a linear combination of that set

to put it visually, envision the vector from the origin

pointing towards the position x=1 y=2 z=3

then the span of that vector is the line that goes through the origin and that point

meaning the line exactly "lies" on the vector

is it understandable?

[tulip jacinth]

Yes 👍

[lilac siren]

if we then look at another vector, that vector is linearly dependent to (1,2,3) if it lies in the span of (1,2,3)

meaning that vector lies on the mentioned line

for instance (2,4,6) is linearly dependent of (1,2,3)

because (1,2,3) * 2 = (2,4,6)

meaning (2,4,6) is a point on the line (the span)

[tulip jacinth]

Given three sets, my math teacher said that only 2 have to be multiples because you can just “pad it with zeroes to make it dependent”

What does padding it with zeroes mean

[lilac siren]

hm not quite sure what you refer to exactly

[tulip jacinth]

I’ll try to send one of the lecture slides

[lilac siren]

sry I gtg but I hope it was a little enlightening :}

[tulip jacinth]

Np, thank you, I learned a lot from this 🙏

Im supposed to close it now right?

.close

[marsh citrus]

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[marsh citrus]

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[glass pike]

Is there a quicker way to do part e than considering (ab + by + ay)^3

[livid osprey]

yoink

[cyan parcel]

I remember vaguely abt some shorthand to solve these kind of problem

[glass pike]

i think ive figured it out

theres a formula: a^3 + b^3 + c^3 = (a+b+c)^3 -3(a+b+c)(ab+bc+ac) +3abc which is nasty but works

[cyan parcel]

Yeah

[knotty trellis]

its mainly about expaanding the shit out

[cyan parcel]

Seem boring

I hate question that tries to become complicated by forcing u to expanding term

[knotty trellis]

it isnt that bad though

it should be quick

[glass pike]

.close

[marsh citrus]

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[orchid oracle]

Can anyone explain this step to me?

[orchid oracle]

It has something to do with using the z table

but cant figure it out

[cyan parcel]

You can try to ask in #probability-statistics

[knotty trellis]

what distribution is that

gaussian centered at 0?

[orchid oracle]

Yes

[knotty trellis]

So we are given that the blue area is 0.3

if i read that correctly

[orchid oracle]

correct

[knotty trellis]

now the z-tables work with area from far in the left

this area

do you know how to calculate it from the blue area?

blue area is 0.3, what's the green area?

[orchid oracle]

Wait my z table shows this

[knotty trellis]

hmm

can you show me the table you are using?

printscreen it

the table

[orchid oracle]

Its not in english

wait

[jovial star]

Isn't finding the blue area just overcomplicating?
Because we can directly find the green area from the z table

[knotty trellis]

the blue area is given

we are given that the blue area is 0.3

[knotty trellis]

Ah okay

thats cool

now you just need to find where the area is around 0.3

[jovial star]

In the start the area complementing the green area is also given

[knotty trellis]

so where the value is 0.3

[knotty trellis]

well we are working with the steps that were already done i think

[jovial star]

Oh
His table finds values from center
Weird

[knotty trellis]

also the OP has special z score table

[knotty trellis]

it actually just differs by 0.5, so it wouldnt be too hard to do it with standard table anyway

[orchid oracle]

Uh

here?

[knotty trellis]

not quite what i meant

[orchid oracle]

or am i using this table wrong

[knotty trellis]

the values inside the table give the area

[jovial star]

Ik
But ig that makes sense why he found the blue area

[knotty trellis]

all those values are giving the area

now you are given that your area is 0.3

[knotty trellis]

try finding something thats close to 0.3 inside of that table

as close as possible ideally

[orchid oracle]

wait

[knotty trellis]

not bad

but there is one closer

[orchid oracle]

one to the left

[knotty trellis]

mhm

.2996

[orchid oracle]

yes

[knotty trellis]

now check which input it corresponds to

which row and column

and what number is it

[orchid oracle]

0.84

[knotty trellis]

and thats it

[orchid oracle]

But .2996 =/= .3

[knotty trellis]

its the closest we can get with this precision of table

[orchid oracle]

or is it negligible

oh ok

[knotty trellis]

note that if you did it with .3023, you'd get 0.85

so we know that the number must lie somewhere between 0.84 and 0.85

that's good enough precision for our purposes ig

[orchid oracle]

I see

Thank you!

[knotty trellis]

you are welcom

e

[orchid oracle]

.close

[marsh citrus]

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[marsh citrus]

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[untold jungle]

in what scenarios does a function only have increasing values

[untold jungle]

please help, thank you

[devout mauve]

what do you mean with "scenarios"

[untold jungle]

cases

situations

questions

when does a function only have increasing values

[low ruin]

You can use the definition of strictly increasing function (available on wikipedia). You can also use differential calculus to get dy/dx

dy/dx is the gradient

So if the gradient is positive the function is increasing

[untold jungle]

yep i know how to figure out increasing, decreasing and constant values

but this question specifically asks me why this specific function only has increasing x values

[low ruin]

Use dy/dx then

[devout mauve]

!original

[marsh citrus]

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.

[untold jungle]

yep , done that

but it asks me to explain why though

[low ruin]

Since dy/dx is positive for all x we can say it is increasing for the given domain

[untold jungle]

ahh thank you

.close

[marsh citrus]

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[still temple]

How can I prove that for dy/dx + ay = f(x) all solutions can be written as y = y_h + y_p? (y_h is the homogenous solution Ce^(-ax) and y_p is the particular solution)

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[urban bobcat]

"Let f(x, y, z) = ...
(a) Specify the tangent plane to the level surface f(x, y, z) = 19 at point A.
(b) At what rate of change f′v (A) does f increase or decrease in the direction (−6, 2, −3) away from the point A?
(c) In what limits does the directional derivative f′v (A) depending on the departure direction v away from A?"

[urban bobcat]

i have done a) and i got w = 4ex + 8ey + z -12e + 23

[marsh citrus]

@urban bobcat Has your question been resolved?

[urban bobcat]

<@&286206848099549185>

[hallow crown]

Howd you get to thus

This

[urban bobcat]

i got help with it a few days ago

w = fx(A)(x-xa) + fy(A)(y-ya) + fz(A)(z-za) + f(A)

this is the formula we used

[hallow crown]

Yea but what do i see e in here

[urban bobcat]

and then we found out that fx(A) = 4e and fy(A) = 8e and fz(A) = 1

becuse the original equation has an e and when we darive it does not dissepear?

[hallow crown]

Youd have a term of e^(x(z + 4))

So you plug in z = -4 it disappears

you got fx(x, y, z) = x^2 * y^8 * e^(x(z + 4)) * (3 + x(z + 4)) right

And fy(x, y, z) = 8x^3 * y^7 * e^(x(z + 4))

And fz(x, y, z) = x^4 * y^8 * e^(x(z + 4))

[urban bobcat]

i dont know i have written down fx(a) = 4e

[hallow crown]

Should be that and when you plug in A you get fx(A) = 3, fy(A) = 1, fz(A) = 1

[urban bobcat]

why is it that?

[hallow crown]

A = (1, 1, -4) right and you have z + 4 over there so it cancels

[urban bobcat]

oky so my answer is compleatly wrong?

[hallow crown]

Well if you dont mind reverify the first order partial derivatives please

[urban bobcat]

well i dont know if that is correct so i cant verify it

[hallow crown]

Should be correct normally

So youd then write the linear approximation at point A for f(x,y,z)

w = f(A) + fx(A) (x - x0) + fy(A) (y - y0) + fz(A) (z - z0)

[urban bobcat]

yes

[hallow crown]

Which would give you w = 19 + 3(x - 1) + (y - 1) + (z + 4)

[urban bobcat]

how?

[hallow crown]

I plugged in the corresponding values here

f(A) is 19, fx(A) is 3, fy(A) and fz(A) are both 1

[urban bobcat]

okay i guess

[hallow crown]

But do reverify afterwards tho please just to avoid errors

[urban bobcat]

how do i verify?

[hallow crown]

Calculate the partial derivatives fx fy and fz

[urban bobcat]

okay and then what?

[hallow crown]

Whatd you get

[urban bobcat]

fx = 3x^2y^8e^x(z+4) + x^3y^8(z+4)e^x(z+4)

[hallow crown]

Yea its correct thats what i got as well

Which would get you 3 if you plug in the coordinates of A

[urban bobcat]

yes okay

[urban bobcat]

so is this the plane tangent?

[hallow crown]

Yea

[urban bobcat]

okay how do i do b?

[hallow crown]

I guess at first calculate the direction derivative of f for v

And v would be (-6, 2, -3)

Then youd plug in As coordinates

[urban bobcat]

okay is it this?

[hallow crown]

Yep

[urban bobcat]

is it (4,8,1)?

[hallow crown]

Lemme check

Should be -19/7 is the rate of change

[urban bobcat]

what does that mean?

[hallow crown]

Directional derivative is the scalar product of the gradient and a unit vector

[urban bobcat]

i got ∇f(A)=(4,8,1)

[hallow crown]

How

[urban bobcat]

by putting in a into the gradiant

[hallow crown]

It should be (3, 1, 1)

Howd you get (4, 8, 1)

[urban bobcat]

oh wait i did it wrong

i get 3 for the first one now

[hallow crown]

Others should be 1 1

[urban bobcat]

and 1 for the last

[hallow crown]

Yep

[urban bobcat]

but how did you get 1 on the second one?

i get 8

[hallow crown]

Whatd you get as the partial derivative fy

[urban bobcat]

8x^3y^7e^x(z+4)

[hallow crown]

Ow my bad

Sorry

[urban bobcat]

its okay

[hallow crown]

This would be w = 19 + 3(x - 1) + 8(y - 1) + (z + 4) my bad

Youre right so its gonna be (3, 8, 1)

Now just take the scalar product of the gradient at point A and the normalized form of the vector v

[urban bobcat]

do i take (3,8,1) * v first?

then i get -5

i get |v| = 7

[hallow crown]

Yep so itd be -5/7

[urban bobcat]

yes i got that too

so now c?

[hallow crown]

If i understood the question correctly my guess would be to take an arbitrary unit vector u this time and get the directional derivative in terms of that unit vector u

[urban bobcat]

this?

[hallow crown]

Yea just this time for some random unit vector u

[urban bobcat]

cant we use 3,8,1?

no wait

thats wrong

becuse f(A) is that

so i can just use 3 random numbers?

[hallow crown]

Youd want to write u in some general form

Like u = (x, y, z) for x^2 + y^2 + z^2 = 1 so it can be a unit vector

Or use some other coordinate system like spherical

That way youd get u = (rsin(theta)cos(phi), rsin(theta)sin(phi), rcos(theta)) with r = 1, theta between 0 and pi, phi between 0 and 2pi

Then youd calculate the directional derivative and get it in terms of whatever parameters you chose

[urban bobcat]

okay so we can use this one?

[hallow crown]

Yea

[urban bobcat]

okay so we take that times (3,8,1)

[hallow crown]

Yep

[urban bobcat]

wait how tho we cant time a vector with a line?

[hallow crown]

Wdym?

[urban bobcat]

how do we multiply them?

what is x, y and z?

[hallow crown]

Yea

u is a standard vector

So you can just multiply it

By the gradient at point A

[urban bobcat]

okay so 3x^2+ 8y^2 + z^2 = 1?

[hallow crown]

x^2 + y^2 + z^2 = 1 would be a condition so u can be a unit vector so it should be left untouched

[hallow crown]

The left hand side of this equation is your direction derivative

So f'_u(x, y, z) is equal to that

[urban bobcat]

i dont know what that means

what am i supposed to multiply?

[hallow crown]

This

[urban bobcat]

okay and f(a) is (3,8,1) and u is x^2+y^2z^2=1?

[urban bobcat]

then i just get this

[hallow crown]

u is just some (x0, y0, z0)

Where the x0, y0 and z0 are part of a sphere of radius 1

So they verify the equation of a sphere

Your directional derivative would be the scalar product of u and the gradient and it would depend on the coordinates of u

[urban bobcat]

okay so what do i write as the answer?

just that the asnwer depends on u?

[hallow crown]

Calculate the directional derivative, otherwise can you retranslate the question

[urban bobcat]

In what limits does the directional derivative vary f′v(A) depending on the departure direction v away from A?

[hallow crown]

Whatd they mean in what limits

[urban bobcat]

that is what i dont understand

he writes it so weird

i gues its the limits where Fv´(A) is at its max and minimum?

[hallow crown]

That could be it yea

[urban bobcat]

but i dont really know how to find the limits

[hallow crown]

Ive used google translate it could also mean the range

[urban bobcat]

dont they depend on the angle or somthing?

ohh okay

[hallow crown]

I guess using spherical coordinates would be better because in that case youd only get two variables instead of 3 that you can control, which are two angles

Actually my bad even if youre using standard coordinates itd depend on two variables

Because its a unit vector

So the directional derivative you got was 3x + 8y + z, you could substitute z by x and y using that sphere equation

But i think itd still be a bit complicated than spherical

[urban bobcat]

yes okay

i might just give up on this question

"Show that γ constitutes the intersection curve between the surfaces S1 and S2 if...
(a) Calculate ∇F1(r(t))◦r′(t) and ∇F2(r(t))◦r′(t) for all t ∈ R. What is the geometric mean?
(b) State the tangent line T to the curve γ at the point A = ( 5/2,3/2,15/4)."

could you just take a look at this i just want to know if i got the right answer on a i gotis to be 0 but i dont understand what he means by what is the geometric mean?

[hallow crown]

I dont know tbh lemme check what it could be

Wait are you sure its geometric mean and not geometric interpretation or something

[urban bobcat]

ohh sorry its meaning or signification could also be it

[hallow crown]

Oh if thats the case then those dot products compare how aligned the tangent of the curve Y and the vectors perpendicular to each surfaces are (gradients)

At each point t

Of the curve Y

[urban bobcat]

hmm okay

[hallow crown]

So in your case it was 0 right

0 for both of the products?

[urban bobcat]

yes

[hallow crown]

If thats the case then the gradients were perpendicular to the tangent of the curve which would mean that curve itself is tangent to the surfaces at all points t

[urban bobcat]

ohh okay

is that related to the geometric meaning?

[hallow crown]

I guess its a geometric interpretation of the results you got

Since both products are zero for all points of the curve

So its a conclusion

[urban bobcat]

ohh okay!

you dont happen to know how to solve b?

[hallow crown]

Let me take a look

I guess just a linear approximation of the curve at point A

L(t) = r(t0) + t * r'(t0) sometjing like this

With t0 where A is at

In the curve

[urban bobcat]

hmm okay