#help-23

1 messages · Page 139 of 1

crude star
#

yep

fair dagger
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For n=3 and k=2 you get 6

crude star
#

right

fair dagger
#

Generally you can think about it this way

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When you select the k elements

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for the first element you have n choices

crude star
#

basically it would just b n^k

fair dagger
#

for the second element one is already used, so n-1 choices remain

crude star
#

wait huhhh

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oh because

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oh so for third

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it would be

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n - 2?

fair dagger
#

for the third element you already used 2, so n-2 choices remain

crude star
#

OHHHHHHH

fair dagger
#

^ hence you multiply decending numbers like in the last equation.

crude star
#

wait but

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n - (k - 1)

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still doesnt make sense to me

fair dagger
#

well you just go k down

crude star
#

ok lets say we have n = 3 and k = 2

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it would be 3 - (2 - 1) thats all?

fair dagger
#

3 * 2 * 1

fair dagger
crude star
#

times?

crude star
#

n - (k - 1)

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🤔

fair dagger
#

maybe try constructing some tuples yourself

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We have a, b, c, d, e let's say

crude star
#

A = {a, b, c, d, e}

fair dagger
#

Let's make 3 element tuples

crude star
#

alr so

fair dagger
#

We can start (a, ...), (b, ...), (c, ...), (d, ...), (e, ...).

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That is already 5 or n options.

crude star
#

yep

fair dagger
#

Now let's fill second spot

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What can we do for (a, ...)?

crude star
#

b

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(a, e, ...), (b, a, ...), (c, b, ....), (d, c, ....), (e, d ... ) ?

#

acc how abt this

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jsut shifted one to the right

fair dagger
#

Let's just look at (a, ...) for now maybe

crude star
#

oh al

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alr

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(a, b, ....)

fair dagger
#

What are all the options we have for the first one.

crude star
#

b, c, d, e

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4 options

fair dagger
#

yes

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How many options do we have for (b, ...)?

crude star
#

a, c, d, e

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4

fair dagger
#

yeah

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so always 4

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all 5 but one blocked

crude star
#

oh so its always going to be like that

fair dagger
#

so if we fill 2 spots, we have 5*4 options in total.

crude star
#

right

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and for third spot it would be 3 options right

fair dagger
#

Yes

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5 in total but everytime 2 blocked

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and this is n - (k-1)

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aka 5 - (3-1)=3

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So we mulktiply

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5 * 4 * 3

crude star
#

so k = total - stuff alrdy used

fair dagger
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to find all options

crude star
#

wait i mean k - 1*

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wait no i meant k

fair dagger
fair dagger
crude star
#

but

fair dagger
#

We just multiply all numbers starting at n and stopping at n-(k-1)

crude star
#

oh we started with k = 5 no?

fair dagger
#

This is what they mean.

crude star
#

oh nvm n = 5, k = 3

fair dagger
crude star
#

nvm

fair dagger
crude star
#

so If I had to plug it in the formula what would it be like

fair dagger
#

yes

crude star
#

5(5 - 1) * 5(4 - 1) * 5(3 - 1)

fair dagger
#

no

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nononono 🙂

fair dagger
#

n(n-1)(n-2)

crude star
#

howwwwwwwwwww

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we had n = 5

fair dagger
#

n-2 is n-(k-1), hence stop

crude star
#

n-(k - 1) = 5 -(3 - 1) right

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so thats out stop

fair dagger
#

yes

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3

crude star
#

and we start with

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5(5 - 1)

fair dagger
#

No

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just 5

crude star
#

but it says

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n(n - 1)

fair dagger
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n-1 is just next step

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that is our 4

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tbh it is not perfectly clear.

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I just know from context and experience.

crude star
#

oh so its

5(5 - 1) (4 - 1) ( 3 - 1) ?

fair dagger
#

no

crude star
#

fuck

fair dagger
crude star
#

but ur missing

fair dagger
crude star
#

(n - (k - 1))

fair dagger
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no that is the 3

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n-(k-1)=3

crude star
#

no in n(n - 1)(n - 2)

fair dagger
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(n-2)=(n-(k-1))

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k-1=2

crude star
#

5(5 - 1)(5 - (3 - 1)) ?

fair dagger
#

yes

crude star
#

so why the hell they put ..... in between

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whats the point

fair dagger
#

because it could be more

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of course

crude star
#

oh so lets say

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|A| = 8

fair dagger
#

Let's compute n=20 and k=4

crude star
#

oh a;r

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alr

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so

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20 - (4 - 1) is our stop

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20 - 3 = 17

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so

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20(19)(18)(17) ?

fair dagger
#

yes

crude star
#

oh my god

fair dagger
#

finally 🙂

crude star
#

i finally

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get it

crude star
#

my fault

fair dagger
#

It's okay, I admit the notation is not entirely clear.

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Which is why other notations exist

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Mathematicians like to write $\prod_{i=n-(k-1)}^n i$

flat frigateBOT
fair dagger
#

Forget about that thouh, you probably won't see that for a while.

crude star
#

nah fuck that. it looks worse

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could u explain this

image.png
fair dagger
crude star
#

I dont understand how they got

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365^-n

fair dagger
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Just 365^n states.

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Maybe think about simpler example first.

crude star
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wait lets say

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we have 2 ppl

fair dagger
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That wouldn't help much

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still 365 dates...

crude star
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nah I mean for 365^n alone

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i was gonna say how would 365^2 make sense

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would it still not be 365

fair dagger
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yeah

crude star
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then why did they say

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365^n

fair dagger
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because 365*365

crude star
#

brah

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i dont get it

fair dagger
#

Imagine as tuples again

crude star
#

alr

fair dagger
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(date1, date2)

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You can now put any of 365 dates in each spots.

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Whatever you put as date1

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there are another 365 options for date2

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now they can be same.

crude star
#

ah

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I see

fair dagger
#

so 365*365

crude star
#

and wb

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365^-n

fair dagger
#

wb?

crude star
#

what about*

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my bad

fair dagger
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That is just 1/365^2

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If you have a dice

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6 sides

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you say probability 1/6

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Here jsut the same

crude star
#

oh trueee

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rightttttt

fair dagger
#

just now 365*365 sides bhappy

crude star
#

and P(A) = 1 - P(A^c)

fair dagger
#

$P(A) = 1 - P(A^c)$

crude star
#

now how in the world

flat frigateBOT
crude star
#

did they know what

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P(A^c)

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was

fair dagger
#

A^c is just the set of date tuples where there are no duplicates

crude star
#

huh

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so permutation again?

fair dagger
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yes

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so we have |A^c| many states

crude star
#

but they said permutation/365^n

fair dagger
#

each of which occur with probsabily 1/365^n

crude star
#

ah

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so I just times the permutation * 1/365^n

fair dagger
#

yes

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Generally if you have equidistirubted stuff

crude star
#

wth does that mean 😭

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equally distributed?

fair dagger
#

$P(A) = \frac{|A|}{|\Omega|}$

flat frigateBOT
crude star
#

what does omega represent here

fair dagger
#

Like dice or birthday.

crude star
fair dagger
#

This may not be true always.

crude star
#

so like

(date1, date2) date1 = date2 ?

fair dagger
#

We assume that all numbers od a dice occur equally often on average.

fair dagger
crude star
fair dagger
#

yes

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and it is 1/6

crude star
#

oh ok

fair dagger
#

because 6 states

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and all equally likely

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If we had an assymetric or just bad dice

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maybe one number would occur way more often

crude star
#

whats a state here

fair dagger
#

then it wouldn't be all 1/6

crude star
#

amount of possible outputs?

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or the space wtvr

fair dagger
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Have you not learn these words?

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A state is a single outcome

crude star
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nope never seen it in my book

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oh u mean

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event

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?

fair dagger
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IIn probability it ids a single outcome.

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No an event is a set of states.

crude star
#

oh

fair dagger
#

This terminology is extremely standard.

crude star
#

oh so like

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lets say I have a coin

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head tails

fair dagger
#

i am surprised your book hasn't mentioned it yet.

crude star
#

state = head or state = tails

fair dagger
#

yes

crude star
#

and event could be something like {H, T, T, H}

fair dagger
crude star
#

and the space would be {H, T}

fair dagger
#

yes but remember a set does not have multientires

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so {H, T, T, H}={H,T}

crude star
#

oh true then what rlly is event

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unless ur referring to the space

fair dagger
#

it is kinda extra boring for dice

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the entire space is an event

crude star
#

oh so u r referring ot the space

fair dagger
#

Let's consider dice aain

crude star
#

alr

fair dagger
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{the number is odd} is an event

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{1,3,5} would be another way to write it

crude star
#

right

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and the number itself would be

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state

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and {1, 2, 3, 4, 5, 6} is the space?

fair dagger
#

yes

crude star
#

ah got it

fair dagger
#

and the later is very very often denoted big omega

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$\Omega$

flat frigateBOT
crude star
#

got it

fair dagger
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it is the set of all states.

crude star
#

noted

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wait nah

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u just

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confused me again

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set of all states = Space no?

fair dagger
#

Again I am surprised why Omega is not used in your book 😕

fair dagger
crude star
#

ohhhhh so omega = space

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we use

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like a weird C

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instead

fair dagger
#

Big C?

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$C$?

flat frigateBOT
crude star
#

nah this C

fair dagger
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oh okay

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that is very non-standard I think.

crude star
#

last one I promis

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can u explain this

fair dagger
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Wikipedia also uses Omega.

fair dagger
#

It is more closely related to the first one.

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^

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Basically you now do this

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Pick k elements of n

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but you do not only disallow duplicates

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but you also consider two tuples the same if they are jsut reordered.

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(a,b) and (b,a) we do not count twice

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so for the original example {a,b,c}

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instead of 6

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we only have 3

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(a,b), (b,c), (a,c)

crude star
#

what did u even say there

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wzonly?

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oh only

fair dagger
#

fixed

crude star
#

ah so combination is usually

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permutation/2 ?

fair dagger
#

if you have 2 only

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generally you need to divide by number of reorderings

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if you have 3 elements

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(a,b,c) tuples

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then there are 6=3 * 2 * 1 ways to order differently.

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This is why it is the same formula as before

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but now we also divide by k!

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6=3 * 2 * 1 = k!

crude star
#

wth why k! doe

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I dont get that

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lets say we have

{a, b}

then we got (a, b), (b, a)

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but combination would be

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(a, b)

fair dagger
#

yes

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only 1

crude star
#

and for 3 instead of 6 we have 3

fair dagger
#

yes

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Let's consider n=4 and k=3

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{a,b,c,d}

crude star
#

A = {a, b, c, d}

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(a, ...), (b, ....) , (c, ....), (d, ...) ...

fair dagger
#

First how many permutations like in 1 exist?

crude star
#

in 1?

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wym

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like in (a, ...) alone?

fair dagger
#

^ in your first

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in the first question

crude star
#

u mean like

4(3)(2) ?

fair dagger
#

I just find the names permutation and combination extremely confusing, so I am jsut refering to the first counting situation bhappy

fair dagger
crude star
#

riight so 24

fair dagger
#

yes

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We could write them all out

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but that sounds a bit boring.

crude star
#

fr

fair dagger
#

Here is another way to think about it

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You take 3 elements from 4

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this just means you omit one of them.

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Let's just look at the ones where we specifically omit a.

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I.e. tuples with b, c, d only.

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Which can you find?

crude star
#

ur askin how many permutations?

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or do u want me to write them out

fair dagger
#

yes

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yes

crude star
#

ah shit

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one sec

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(b, c, d), (b, d, c), (c, b, d), (c, d, b), (d, c, b), (d, b, c)

fair dagger
#

yes

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it's 6

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Why 6?

crude star
#

3(2)(1)

fair dagger
#

yeah

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3!

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It's another subpermutation

fair dagger
#

We got 24 before

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but had all those reordered variants

fair dagger
#

where we just reordered

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Now we do not want to have reordered ones count again

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So what we do

crude star
#

then it would just be 3

fair dagger
#

4 bhappy

crude star
#

huh

fair dagger
#

(a,b,c), (a,c,d), (a,b,d), (b,c,d)

crude star
#

why a

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i thought

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ur askn for

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b, c, d

fair dagger
#

oh then there would exist only 1

crude star
#

huhhhhhhhhhhhh

fair dagger
#

didn't you notice?

crude star
#

oh

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ohhhhhhh

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I seee

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what u mean

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my bad

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I was only thinkin about the ones reversed

fair dagger
#

We have 6 dfuplicates

crude star
#

so we remove 5

fair dagger
#

if you omit any of the others you also have 6 duplicates.

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in total we overcounted by a factor of 6 first

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our 24

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had 6 duplicates for each distinct pick out ofr {a,b,c,d}

crude star
#

wait

#

so

fair dagger
#

So 4=24/6

crude star
#

we originally had 4 and then each had 6 duplicates

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permutation ^

fair dagger
#

Well had 24 and then we noticed

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we could group them so that it would be 6 duplicates of 4 things

crude star
#

right

fair dagger
#

here we care about the 4

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So we compute permutation and then divide by duplicates

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The formula is the formula from the beginning

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just now extra division by number of duplicates

crude star
#

so k! = duplicates

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das all

fair dagger
#

yes!

crude star
#

alr got it

#

tysm fam

fair dagger
#

you are welcome

crude star
#

.close

safe radishBOT
#
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fair dagger
#

Tbh your book is kinda confusing.

crude star
#

.reopen

safe radishBOT
#

✅

crude star
#

oh mb u were typin

fair dagger
crude star
#

.close

safe radishBOT
#
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copper stone
#

How can I find all complex numbers z for $z^4 = -7 - 24i$ ?

safe radishBOT
flat frigateBOT
copper stone
#

I tried to expand z^4 and get a system with the real and imaginary part but I don't know how to solve that system either.

safe radishBOT
#

@copper stone Has your question been resolved?

copper stone
#

<@&286206848099549185>

safe radishBOT
#

@copper stone Has your question been resolved?

plucky elk
#

Then take fourth root of both sides

#

Remember demoivre's formula

copper stone
#

uh that looks like trig with complex numbers, we haven't done that yet

copper stone
plucky elk
safe radishBOT
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@copper stone Has your question been resolved?

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safe radishBOT
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@lean otter Has your question been resolved?

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frosty ridge
#

Any hints or tips for the last part of the question? That is, showing that the decay is not exponential.

image.png
safe radishBOT
frosty ridge
#

The solution to the DEq is $$x=\pm\sqrt{\frac{1}{2t+\frac{1}{x_0^2}}}$$

flat frigateBOT
#

Goose on a Moose

412739317887139840.png
safe radishBOT
#

@frosty ridge Has your question been resolved?

safe radishBOT
#

@frosty ridge Has your question been resolved?

plucky elk
#

Do you know what growth that is?

#

As t gets large, what's the behavior of x roughly?

frosty ridge
#

@plucky elk I suppose I don't know what growth rate it. At least precisely.

plucky elk
#

2t + constant is approximately 2t for very large t

#

Then use exponent rules to simplify

frosty ridge
#

so as t gets large, the growth rate would be 1/sqrt(2t).

#

which is clearly not exponential.

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obtuse harbor
#

I need help with this form of problem

IMG_0133.jpg
safe radishBOT
worthy hemlock
dense sedge
safe radishBOT
#

@obtuse harbor Has your question been resolved?

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safe radishBOT
fair dagger
whole acorn
#

<@&268886789983436800>

#

wth

warped roost
#

.close

gray flume
#

Snipd again

safe radishBOT
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dark marlin
safe radishBOT
dark marlin
#

i'm lost

charred atlas
#

ummm

#

expand

west hedge
#

yes

charred atlas
#

you'll get extra 2

dark marlin
#

im not good with tan's

#

let me try

west hedge
#

tan = 1/cot

dark marlin
#

$$tan^2(x) + 2tancot + cot^2(x)$$

flat frigateBOT
#

puckmyseen

480769458206670848.png
dark marlin
#

?

#

or wil there be a 2 in front of tan and cot as well

west hedge
#

what is tan * cot

dark marlin
#

?

charred atlas
#

1

dark marlin
#

ohhh yea

charred atlas
#

tan = sin/cos

#

cot = cos/sin

dark marlin
#

trig identites

#

ok i got the rest of it

#

thank you very much i forgot about that one

#

.close

safe radishBOT
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bright sable
#

Anyone able to help me with the second part of this question? My reasoning is that I am doing 5C4/16C5, but that doesn't seem to be working. Thanks!

image.png
safe radishBOT
rigid inlet
#

You're missing a piece. You need to account for the remaining kids (that can't not wait to go home)

#

Can wait to go home? Whatever

bright sable
#

ah I see, thanks, I'll try to work that into it

rigid inlet
#

5C4 accounts for the eager kids, but you still have one slot to fill

bright sable
#

would it make sense if I added that to the probability of any random kid out of the pile that's left? so 1/12

rigid inlet
#

You aren't adding anything, no

bright sable
#

oh wow i actually got it

#

it's 55/4368

#

basically what I ended up doing was multiplying what I got by the remaining 11 possible students

#

thanks for the help!

#

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safe radishBOT
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wind abyss
#

.close

safe radishBOT
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wraith prism
#

tan(π/4+x)^(1/x)=t

1/x log tan(π/4+x) =log t

1/tan(π/4+x) × sec^(π/4+x) =logt
Limit tends to infinity e^{1/sin(π/4+x) cos(π/4+x)} =t

IMG-20230914-WA0000.jpg
safe radishBOT
quasi bison
#

where does x approach?

#

infinity??

#

1/x log tan(π/4+x) =log t
1/tan(π/4+x) × sec^(π/4+x) =logt
this smells like bullshit by the way.

wraith prism
#

I uploaded the picture 😮‍💨

quasi bison
#

well, as written the limit just does not exist lol

wraith prism
#

What do you mean?

#

It should be 0?

quasi bison
#

no, i mean the limit doesn't exist.

wraith prism
#

Limit tends to 0

#

Now help me how to solve it

quasi bison
#

do you mean x tends to 0?

wraith prism
#

Yes

quasi bison
#

ok, this is moderately painful.

fleet tendon
#

what's 4

quasi bison
#

you mean what's option 4?

fleet tendon
#

we discussed this earlier

#

we think DNE

#

idk tho

quasi bison
#

as written or with the correction that x -> 0?

#

or do you mean you think option 4 is DNE

fleet tendon
#

(4) looks like another e of sort

quasi bison
#

this feels like not answering my question fully

#

do you think the limit DNE as written, or do you think the limit DNE with arjunn's correction that x should go to 0?

fleet tendon
#

Oh as written I think it might be dne

wraith prism
#

e^2 given answer

cold aurora
quasi bison
#

l'hop is unneccessary imo.

wraith prism
quasi bison
#

but i am using a different kind of lubricant to make this go down

wraith prism
#

And I asked to Wolfram they said something expansion of series at 0 i would love to learn that too

cold aurora
wraith prism
wraith prism
#

At the start of question

cold aurora
#

Well, you didn't differentiate properly.

quasi bison
#

$\log(y) = \frac{\log(1+\tan(x)) - \log(1-\tan(x))}{x}$

flat frigateBOT
quasi bison
#

since numerator -> 0 (as x approaches 0)

cold aurora
#

It's sec^2($\pi$/4 + x).

flat frigateBOT
#

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

1116326425088241714.png
cold aurora
#

Which is 2.

quasi bison
#

BADTEX!!!!!!!!!!!!!!!!!!!!!!

cold aurora
#

Yeah. Yeah. I'm on phone.

#

Can't do better.

quasi bison
#

$\sec^2(\pi/4 + x)$ for god's sake!!!! dollars are NOT meant to enclose individual symbols!!!

flat frigateBOT
quasi bison
#

yes you CAN!!!

#

you've typed dollars and a backslash

cold aurora
#

Well, it's understandable and that's all what matters. Lol

quasi bison
flat frigateBOT
quasi bison
#

for g(x) = numerator of that thing

#

noting that g(0)=0 of course.

#

99% chance i'll have to re-explain my thing like 5 times

#

because apparently everybody fucking forgets that defn of derivative is a thing after solving 1000 problems with l'hopital

wraith prism
#

We can do it ann without opening tan(a+b)

quasi bison
#

yes, e^2 is correct.

#

yeah i guess we can.

#

will make differentiation easier a bit.

#

my first thought was taylor, for which my form is somewhat more convenient.

wraith prism
#

So now your turn. I would love to know all other lubricant methods

wraith prism
#

Wolfram did it same

quasi bison
#

$\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$

flat frigateBOT
quasi bison
#

for x close to 0

#

my idea would have been to expand log(1+tan(x)) as tan(x) - tan^2(x)/2 + ... and the same for log(1-tan(x))

#

end up with 2tan(x) + some bullshit

#

disregard the bullshit as negligible

wraith prism
#

2 tanx/x
And limit x tends to 0 so tanx/x =1
logy= 2
y= e^2

#

Bingooooo

safe radishBOT
#

@wraith prism Has your question been resolved?

safe radishBOT
#
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leaden pendant
#

$Draw the Function, showing all important features: y=2x^2+5x-3$

safe radishBOT
flat frigateBOT
#

Ibrahim Adeel

762605816103698442.png
quasi bison
#

badtex!

#

Draw the Function, showing all important features: $y=2x^2+5x-3$

flat frigateBOT
leaden pendant
quasi bison
#

anyway ok

#

!status

safe radishBOT
# 
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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
leaden pendant
#

!

#

2

#

1

#

sorry 1

quasi bison
#

ok

#

do you know how to graph parabolas generally

leaden pendant
#

yep

quasi bison
#

ok then what is troubling you w this one

leaden pendant
#

i only know how to graph in the form of y=a(x+.....)

#

so on

quasi bison
#

do you mean a(x-h)^2 + k?

leaden pendant
#

yeeh y=ax^2+bx+c

quasi bison
#

...

#

but 2x^2 + 5x - 3 is in that form?

leaden pendant
#

how would u draw it by hand

quasi bison
#

i'm confused. you are contradicting yourself.

#

ok so let me just state for the record that my confusion regarding your statements of what you can and cannot do has not been clarified.

leaden pendant
#

i cant graph this

#

sorry im raelly new to this area of maths

quasi bison
#

that is not even a parabola.

#

but to graph your parabola i would:

  • factor it, or find its roots (basically the same thing).
  • mark the x intercepts accordingly.
  • mark the vertex's x coordinate as halfway between those two, and the y coordinate according to the equation
  • mark the y intercept
  • trace the parabola through the 4 points thus marked.
leaden pendant
#

so wt would the qyestion i put down be?

quasi bison
#

i don't understand what you asked just now, please be more precise in your phrasing.

leaden pendant
quasi bison
#

yes of course it would...

#

it's a polynomial function of degree 2

#

its graph is a parabola

leaden pendant
quasi bison
#

the degree of a polynomial is the highest power of x present in it...

leaden pendant
quasi bison
#

just a week or so ago you were asking about logarithmic stuff

leaden pendant
quasi bison
#

yeah so i'm surprised you don't know what a parabola is or couldn't definitively recognize yours as such

leaden pendant
quasi bison
#

ok let me repeat my instructions since they fell on deaf ears earlier

#

to graph your parabola i would:

  • factor it, or find its roots (basically the same thing).
  • mark the x intercepts accordingly.
  • mark the vertex's x coordinate as halfway between those two, and the y coordinate according to the equation
  • mark the y intercept
  • trace the parabola through the 4 points thus marked.
#

read this through 5 times

leaden pendant
quasi bison
#

yes, factor the 2x^2 + 5x - 3. what else could i be POSSIBLY asking you to factor?

leaden pendant
quasi bison
#

ok, then i have affirmed it for you already.

#

your factorization looks correct to me.

leaden pendant
quasi bison
#

the what or the why?

leaden pendant
quasi bison
#

y = a(x-r)(x-s) has its vertex at x = (r+s)/2

leaden pendant
quasi bison
#

there are various ways of explaining why this is the case.

#

if you've memorized the -b/(2a) thing, you can expand a(x-r)(x-s), get that you have -a(r+s) for b, and thus -b/(2a) will be (r+s)/2

#

or you could appeal to the symmetry of the parabola, or something.

leaden pendant
quasi bison
#

sure.

#

the vertex lies on the AoS.

leaden pendant
quasi bison
#

expand a(x-r)(x-s)

#

also don't swallow the damn minus sign

leaden pendant
#

i got somehting different when expanding a(x-e)(x-s)

#

hello

#

.close

safe radishBOT
#
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safe radishBOT
#
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thin halo
#

hello, given a 3x3 matrix A with 3 eigenvalues, 2, -1, 0, with corresponding eigenvectors of v1, v2, v3,
lets say i were to do A^2, does the sign of my (-1) eigenvalue become positive in my diagonal matrix?

safe radishBOT
idle widget
#

yeah i believe so because you can use diagonalization to turn it into A^n=PD^nP^{-1}, but you can distribute powers into diagonal matrices so, yes, -1 would square to 1

thin halo
#

so it would be correct to say that for every power of n that is even, i will get a change of sign in its corresponding eigenvalue in the diagonal matrix?

idle widget
#

yeah i guess so, its like the space flips along the v_2 axis under the transformation A

#

each time you successsively apply A is oscilates back and forth

#

along the axis created by v_2 through the origin

thin halo
#

okay! uhm i have another question regarding that

idle widget
#

ok

thin halo
#

my P doesnt change no matter how many A^n i do right?

idle widget
#

no it wouldnt change

thin halo
#

and the vectors are not unique too, so if i were to lets say A^3, and ill just bomb some scalar vectors eg, 2_v1, -_v2, v3 they still do diagonalise A right?

#

^ in this case i scalar-multiplied the eigenvalues towards the eigenvectors

idle widget
#

yeah any element of the eigenspace could be put into P i believe

#

and its always the same

thin halo
#

okay!

idle widget
#

because theres nothing special about any given eigenvector

#

it is derived from the equation Av_i=lambda_i v_i

#

and any multiple of v_i satisfes it

thin halo
#

yeah

idle widget
#

so its kinda arbitrary

#

its nice to choose one of unit length tho

#

usually

thin halo
#

one unit length means the scalar of the original lamda?

idle widget
#

just so you can make an orthonormal basis

#

i mean choos v_i so it have norm = 1

#

but you dont have to

#

its just the standard

thin halo
#

yeah its good to know tbh im trying to get the foundation up

idle widget
#

smart with linear algebra

#

lots of people complain about this subject at first

#

but its pretty neat

#

and every so important

thin halo
#

it surely gets easier but yeah the terminology is kinda hard to grasp at the start with all the subspaces rowspace etc

idle widget
#

its nice to be able to look those things up

thin halo
#

surely

#

also question

#

if a 3x3 matrix has 3 eigenvalues it means its distinct right?

#

so it goes along with 3linearly independent eigenvectors, so P definitely has an inverse and A is definitely diagonalisable

#

oh wait, if i have an eigenvalue of 0, A is not diagonalisable anymore? and P doesnt exist?

idle widget
#

think about an eigenvalue of zero for a second

#

Av_0=0

#

it would mean that the eigenvector gets mapped to the zero vector

#

and so the matrix in the case of a zero eigenvector shrinks the dimension of the space

thin halo
#

yes

idle widget
#

i think it actually implies that the eigenvector is zero as well

#

so you have no linearly independent basis of eigenvectors and so it cannot be diagonalizable

thin halo
#

so as long as i have an eigenvalue of 0, no matter how many n eigenvalues i have that equals to nxn matrix, that matrix is not diagonalisable?

thin halo
idle widget
#

well eigenvector are non-zero by definition

thin halo
#

oh because we are interested in finding its non-trival solutions?

idle widget
#

yeah

#

it basically just pointless to consider the zero vector

#

since it doesnt given any new information

thin halo
#

is there a case whereby
i'll use back my same example, eigenvalues of 0, -1, 2
and matrix A is diagonalisable where,
E_-1 gives 2 basis of eigenvectors
and E_2 gives 1 basis of eigenvector

#

oh wait nvm then the diagonal would be -1 -1 2?

#

man im fried tbh

idle widget
#

i think i mightve misspoke, you can have a diagonalizable matrix with an eigenvalue zero of course because you did it in this example, but a matrix with eigenvalue 0 is not invertible

#

thats probably what i meant to say

thin halo
#

oh!

idle widget
#

we do a little trolling

#

lol

thin halo
#

okay ive cleared up my stuffs i guess

#

thanks for helping out today @idle widget

#

🫶

#

,close

#

.close

safe radishBOT
#
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untold topaz
#

Why does Horizontal Asymptote not exist?

wolf.png
safe radishBOT
icy lance
#

why do you think there is one

untold topaz
icy lance
#

3x^2/3 tends to +inf, 2/x tends to 0, so overall it tends to +inf

untold topaz
icy lance
#

they arent multiplied, theyre added?

untold topaz
#

Added?

icy lance
#

the limit would be +inf+0

#

so just +inf

untold topaz
icy lance
#

lim 2/x = 0

untold topaz
#

Yes

#

But how did u get 'add'

icy lance
#

lim[3x^2/3 +2/x] = lim[3x^2/3] + lim[2/x]

untold topaz
#

Or is that a rule? 🥲

icy lance
#

where did you even get [3x^2/3]/2x from?

#

its easier to just separate the limit

untold topaz
icy lance
#

3x^2/3 +2/x and [3x^2/3]/2x arent equivalent

#

so idk where it came from in your limit

untold topaz
#

Ohhh

#

...

#

NOw I see.

#

I need sleep

#

Thanks so much pal

#

.close

safe radishBOT
#
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icy lance
#

if you wanted to specify 3 then it would rather be 8.00*10^3

fleet condor
#

to show they are significant

safe radishBOT
#
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sterile dew
safe radishBOT
sterile dew
#

does anyone see how you can find angle ACD?

#

given angle E is 40, and the three big arcs are the same length

#

there's so little information that I think it's straight up impossible. i mean you don't even know how far E is from the circle or how long AD is

#

but the book says it's 15 degs

safe radishBOT
#

@sterile dew Has your question been resolved?

sterile dew
#

<@&286206848099549185>

full tiger
#

i think 15 degs

sterile dew
#

bro how???

#

@full tiger

full tiger
#

wait for me

#

i think so

z4694209546545_6a1d8d166d4a77694209c9bd421ef8d1.jpg
sterile dew
#

@full tiger omg thank you

#

literally all i had to do was realise that triangle BEC is isosceles

#

.close

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crude star
safe radishBOT
crude star
#

How would I do this

safe radishBOT
#

@crude star Has your question been resolved?

crude star
#

<@&286206848099549185>

safe radishBOT
#

@crude star Has your question been resolved?

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brazen helm
#

$\text{lcm}{a,b}\cdot \gcd{a,b}=ab$

safe radishBOT
brazen helm
#

How do I prove this?

sacred verge
flat frigateBOT
#

bigpufik

510744681043460102.png
karmic hedge
#

Venn diagram

brazen helm
#

.close

safe radishBOT
#
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sacred verge
#

.repon

#

.reopen

#

venn diagram is for sets , how does it prove this ?

karmic hedge
#

The set is the prime factors of a number

#

Or rather a multi set

sacred verge
#

I see. But venn is just a digram, not a proof.

#

you can't prove something by drawing a diagram

karmic hedge
#

Yeah it’s a hint

sacred verge
#

yeah

#

this guy closed the question a bit too fast lol

#

because I'm sure their teacher won't be happy with only a diagram as proof

karmic hedge
#

Lol

brazen helm
#

no teacher

#

but i think i know its ok dw lmaoo

sacred verge
#

well venn diagram is only a hint, not the full proof

brazen helm
#

I know:)

sacred verge
#

then prove it lol

karmic hedge
#

The result becomes quite immediate with the Venn diagram, I think that’s why he was fast

sacred verge
#

but yeah, whatever

karmic hedge
#

Using the diagram

#

We aren’t just drawing a diagram and say we are done

sacred verge
#

this is not easy to prove, he needs a theorem called integer factorization. not sure this is what class

safe radishBOT
#
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shy flame
safe radishBOT
shy flame
#

please can someone help me with the same direction part

#

what should i do if they are in the same direction

thin swan
#

like

#

vector 1 + vector 2 + vector 3 points at the same direction as the vector with the force of 6.8 N (to the right)

#

or at least I think thats what it means

safe radishBOT
#

@shy flame Has your question been resolved?

shy flame
#

<@&286206848099549185> please i am completely stuck on this question

lean otter
#

Wat question

safe radishBOT
#

@shy flame Has your question been resolved?

safe radishBOT
#

@shy flame Has your question been resolved?

safe radishBOT
#
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loud oasis
#

ok

safe radishBOT
loud oasis
#

@scarlet temple

#

there

#

ok

scarlet temple
#

here we are

loud oasis
#

whats the first thing to do

scarlet temple
#

want me to rewrite the system?

loud oasis
#

when solving a systems of euqations problem

loud oasis
#

sure

scarlet temple
#

so let me show you the whole thing

loud oasis
#

ok

#

give me thed problem

#

and ill break it down

#

so sorry about them and the moderators

#

anywho

#

you can send

#

the example

loud oasis
#

cant leave this ticket alove

#

alone

loud oasis
scarlet temple
#

the original problem is: 2x+y-3z=1, x+y+4z=4,x+2y=41
Now I divided it into two simpler systems, with the first and second part of it, and the first and third part of it. Solving the first part by subtracting gave me x-7=-3 but idk how to solve the second part which is: 2x+y-3z=1,x+2y=41

scarlet temple
scarlet temple
loud oasis
#

ok im writing it down

scarlet temple
#

💯

#

thank you

#

take your time

devout shale
#

$$2x+y-3z=1$$ $$x+y+4z=4$$ $$x+2y=41$$

flat frigateBOT
#

Austin

779131727338930196.png
loud oasis
#

and there was no explaination

scarlet temple
#

the one I'm trying to solve doesn't include the middle part tho

devout shale
#

just making it more clear what the system is

loud oasis
#

can you go back

loud oasis
scarlet temple
devout shale
loud oasis
#

<@&268886789983436800>

scarlet temple
#

I'm just dumb

loud oasis
devout shale
#

no worries, I'd start by manipulating the bottom equation in order to solve for x or y by itself

loud oasis
#

for his wrongdoing

#

thank you

devout shale
#

and then substitute that into the top two equations

scarlet temple
#

or am I wrong

devout shale
#

are you for some reason not allowed to use substitution here?

scarlet temple
#

cause they want us to solve it with elimination

devout shale
#

ah okay that is also fine

scarlet temple
#

nice

devout shale
#

so it looks like

#

try subtracting equation 3 from equation 2

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then write down what that gives you

scarlet temple
#

actually I started off by subtracting equation 1 from 2

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cuz they both had that y and I eliminated it

#

is it wrong?

#

it gave me x-7z=-3

devout shale
#

it isn't wrong, you could also do that

scarlet temple
#

nice

devout shale
#

Could you show me your work for whatever you tried?

scarlet temple
#

sure I just gotta go on my phone

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gimme a sec

#

it's pretty awful but I was clueless when I started off today

#

it's sending the picture

devout shale
#

so that works, you've now eliminated y from equation 1, now just go ahead and eliminate y from another equation (either equation 2 or 3) and then set up your new system. Since you eliminated y and now have two equations, this will be a 2 equation system with only x's and z's

#

from there you can choose to eliminate either x or z, and then once you have solved for that one variable you can just plug it into the other equations to get the rest

scarlet temple
devout shale
#

you have not yet made the second system

#

atleast I don't see it on your paper

scarlet temple
#

my bad I meant this

devout shale
#

x-7z=-3 is half of it

scarlet temple
#

yeah exactly

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the one after it

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idk how to solve it

#

how to eliminate y in this case

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in the first case it was simply subtracting

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but now what?

devout shale
#

which equation do you want to eliminate y from now

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either equation 2 or 3

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it doesn't matter

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but pick one

scarlet temple
#

2

devout shale
#

okay

#

so

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you can take equation 3 and divide it by 2

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this gives

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$\frac{x}{2}+y=\frac{41}{2}$

scarlet temple
#

actually no wait a sec

flat frigateBOT
#

Austin

779131727338930196.png
scarlet temple
#

I just realized the way I separated the system I'm supposed to find 3 right?

devout shale
#

no it doesn't matter. you just have to eliminate y from two of the 3 equations and make a new smaller system

scarlet temple
#

oh alright

scarlet temple
devout shale
# flat frigate **Austin**

so after doing this, you can subtract this from equation 2. Then you will have eliminated y from it also. Write down that new equation with your x-7z=-3 and show me what you got

scarlet temple
#

oh ok

safe radishBOT
#

@loud oasis Has your question been resolved?

scarlet temple
devout shale
scarlet temple
#

damn

#

my bad

#

it's my first time so I was pretty much clueless

devout shale
#

yes that looks good

scarlet temple
#

nice

devout shale
#

now create a system of 2 equations with 2 variables (x and z)

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and do the same thing

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eliminate either x or z

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doesn't matter which

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at the point that you have only one variable left, just solve for it

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and then plug back in

scarlet temple
#

alright man I'll be back

safe radishBOT
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safe radishBOT
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devout raven
#

how can u get unit normal vector given unit tangent vector?

safe radishBOT
devout raven
#

cuz ik derivative of tangent vector would be normal

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but how would u do it ONLY given the UNIT tangent vector

safe radishBOT
#

@devout raven Has your question been resolved?

devout raven
#

<@&286206848099549185>

plucky elk
#

Normal vector and tangent vectors are orthogonal

devout raven
#

but

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what if i want to get the unit normal

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cuz if i j took derivative of unit tangent

#

and divided by mag of that

#

thatd be wrong

plucky elk
#

Show some work

safe radishBOT
#

@devout raven Has your question been resolved?

devout raven
#

u divide by r' mag

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then divide by r" mag

#

isnt that diff then j r"/r" mag

safe radishBOT
#
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lean otter