help-49

[runic hamlet]

otherwise you cant claim that you know the degree of q

.close

[midnight plank]

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[runic hamlet]

@twilit field (dunno if I pinged you with the first message)

[twilit field]

I have dnd on

oops

missed these

.reopen

[midnight plank]

✅

[twilit field]

okay, I have to think about this a bit more

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[twilit field]

same as that of p

[midnight plank]

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[kind ledge]

!help

[midnight plank]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[kind ledge]

how on earth do i solve 0,6x + 1,2 = -0,125(x-4)^2 +10

[lyric charm]

!status

[midnight plank]

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

[kind ledge]

2

[lyric charm]

ok show what you got thus far

[kind ledge]

im at -4,8x -8,8 = (x-4)^2

but ima need to restart cause im confised

[lyric charm]

... show your work?

this doesn't look right, but i want to see where you messed up.

[kind ledge]

ngl i wrote fast so its not even clear

i think i messed everything up tho

[lyric charm]

ok then redo it lol

and write it neatly and slowly this time

[kind ledge]

i did

alright

after redoin

i got this now

[lyric charm]

well, so far so good. i would multiply both sides by 5 to get rid of the decimals though.

and in any case you will want to expand this (x-4)^2.

[kind ledge]

alright i expanded

(x-4)(x+4)

[lyric charm]

no

(x-4)^2 is not the same as (x-4)(x+4)...

and that wouldn't be an expansion anyway

[kind ledge]

oh

(x-4)(x-4)?

[lyric charm]

well sure

yes, (x-4)^2 = (x-4)(x-4) but you didn't expand it yet

[kind ledge]

how to expand

[lyric charm]

do you know the distributive law?

or maybe you know the identity for (a-b)^2

[kind ledge]

like x times x

x times -4

-4 times x

-4 times out of-4

[lyric charm]

this is kind of longwinded but it works i suppose

[kind ledge]

alright

[lyric charm]

$(a\pm b)^2 = a^2 \pm 2ab + b^2$ is good to know in general though. to save time and effort.

[grand pond]

Ann

[kind ledge]

i got

x^2 -8x +16

[lyric charm]

ok so that is the right-hand side.

now you have: $$-4.8x + 70.4 = x^2 - 8x + 16$$

[grand pond]

Ann

[lyric charm]

i will repeat my recommendation to multiply both sides by 5 at some point to get rid of the decimals.

[kind ledge]

i would do that but we never learned smth like that, that may be like college lvl

[lyric charm]

????

are you kidding me?

[kind ledge]

no

[lyric charm]

you didn't learn "multiplying both sides by a number"?

[kind ledge]

i didnt

[lyric charm]

but you went and divided both sides by -0.125 in your own work

so what gives

should i frame it as "division by 0.2" instead or what

[kind ledge]

idk where u got 5 from

[lyric charm]

these are decimals with 1 decimal place

it might make sense to multiply by 10 instead

if you want/need something to have "come from" somewhere

[kind ledge]

multiply everything by 10?

[lyric charm]

multiply both sides by 10 yes

i really do not believe you when you say you didn't learn that as an algebraic move.

it is not college-level, i'll tell you that much.

[kind ledge]

wait but

what if the decimals were

70,48

2 numbers

then whaf

[lyric charm]

if you had all the decimals be two decimal places or shorter, then multiplying by 100 would get rid of them all.

2 digits not 2 numbers btw.

[kind ledge]

oh alr

but is it needed

i dont think its important cuz

we can just do

-70,4 both sides

which turns 16 into -54,4

and thrn

+4,8x both sides

[lyric charm]

kind of, but it's not strictly necessary to do it now specifically.

[kind ledge]

turning -8 into -4,8

[kind ledge]

when would it be necessary then

[lyric charm]

-8 + 4.8 isn't -4.8

[kind ledge]

-3,2 mb

[lyric charm]

at whatever point

[kind ledge]

alr

[lyric charm]

you're turning the equation into a quadratic in standard form

it's going to be nice if you can get all coefficients in it as integers

multiplying by 5 will accomplish that

[kind ledge]

true but will it give a different value

for x

?

[lyric charm]

no

[kind ledge]

alright

[lyric charm]

the whole point of algebraic moves is that they preserve all solutions

unless you fuck them up

but then not even god himself can help you.

[kind ledge]

multiply by 10 u mean?

[lyric charm]

if you're allergic to fives then multiply by 10.

what do i care.

[kind ledge]

alr

wait so ill
do it rn

[lyric charm]

show all your work so far

[lyric charm]

where are

-4.8x + 70.4 = x^2 - 8x + 16
and

-70,4 both sides
which turns 16 into -54,4
and thrn
+4,8x both sides

[kind ledge]

no im doing the x 10 thing u said

this

[lyric charm]

ok, good enough.

you can also divide both sides by 2 since all the coefficients are even.

but in any case you now have a quadratic equation in standard form on your hands.

do you know how to solve those? yes or no.

[kind ledge]

yes

i got x = 9,14

i dont need the negative because the coordinate is not supposed to be negative

[lyric charm]

so you decided to just not tell us that this was part of a bigger problem

[kind ledge]

no cuz it was only tbat

that

that i was stuck on

y = 6,69

so i finished

i just had to find

the intercept

of the parabole

and the line

which is

9,14,6,69

[lyric charm]

well that was a horrible way to write it

first you should never drop the brackets when writing down the coordinates of a point

and second, you cannot use the comma both as a decimal point and to separate numbers

you have to write either (9,14; 6,69) or (9.14, 6.69)

[kind ledge]

alr

(9.14;6,69)

ty for helping me tho

.close

[midnight plank]

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[midnight plank]

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[sand ibex]

.close

[midnight plank]

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[rancid vigil]

A test is performed on cattle where 2% are carriers of a disease.

If an animal is sick, the test is positive in 85% of cases
If an animal is healthy, the test is negative in 95% of cases \
M = "the bovine is sick" \
T = "the test is positive" \
An animal is chosen at random
Calculate:\
a) $P(T)$ \
b)$ P_T(M)$

[rancid vigil]

P is the probability

[lyric charm]

\% to get actual percentage signs

[rancid vigil]

ah

[lyric charm]

otherwise tex thinks you're commenting out

[grand pond]

<rajel />

[grim vector]

a tree ?

[opaque briar]

do you need help on both or just b?

[rancid vigil]

a

[grim vector]

formule des probabilités totales

[rancid vigil]

how did you know lol

[grim vector]

that you know fr ?

[rancid vigil]

yep , that the original question is in fr

[grim vector]

M = malade

not sick kekw

[rancid vigil]

oh lmao

nice

[grim vector]

also i'm pretty sure we're the only one using the notation for conditionnal probability like in the b) question

[rancid vigil]

ah i see

so i need to draw i tree

[grim vector]

anyway if you draw a quick tree

[opaque briar]

yeah a tree should do the job

[grim vector]

you can use the formula of total probability for T

which is the sum of the different probability to get to T event

second question using the result of the a)

by the fact that P_A(B) * P(A) = P(A \cap B)

[rancid vigil]

is this a correct tree?

[grim vector]

what does the second leaf represents each?

which one is T

nvm

[rancid vigil]

the positive ones

[grim vector]

yeah its good

[rancid vigil]

the probability of an animal being positive is equals to = P(normal (+)) + P(sick(+))

right ?

[grim vector]

yep

[rancid vigil]

0.05 +0.85 ?

[grim vector]

?

[rancid vigil]

wait

[grim vector]

how do you get P(normal +)

[rancid vigil]

5/98

[grim vector]

whats the formula you using

[rancid vigil]

didnt use nothing tbh lol

[rancid vigil]

oh so we're trying to find P(A^- \cup B) ?

[grim vector]

just an example of the formula, here we have P(normal +) which is P(normal \cap positive)

[rancid vigil]

so P(normal +) = p(normal) * p(+)

[grim vector]

you multiply all the probability till you arrive to positive yeah

which is for normal + ?

[rancid vigil]

ig that what i said

[grim vector]

5/98?

[rancid vigil]

, calc 98%5 + 2%*85%

[grim vector]

Ah the final result ?

[grand pond]

Result:

0.0833

[grim vector]

I was talking about the normal +

But thats not 5/98

[rancid vigil]

yeah , the normal + should be the multiplicaiton of the leaf

[rancid vigil]

yeah i was wrong their

[grim vector]

Yeah thats the way so

[rancid vigil]

ok thx

.close

[midnight plank]

Channel closed

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[midnight plank]

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[light wyvern]

How would u do this question

<@&286206848099549185>

[tawdry laurel]

!15m

[midnight plank]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[wanton geyser]

ive never seen a Q where the interest compounds monthly with an annual rate

[copper notch]

I beleive that's unecesary to consider in order to solve the problem

[wanton geyser]

how would the interest accumulate though

i wonder if you could work backwards?

[copper notch]

@light wyvern

this should solve the problem

[light wyvern]

Indeed but idk what you did in here

[copper notch]

oh no, it's not 30 years

[light wyvern]

Yeah 20 years

[copper notch]

let me correct it

I'll explain it to you

you have an interest of 3.2% per annum

so you after 1 year you will have 103.2% of what you had in the previous year

[light wyvern]

Yh

[copper notch]

100%+3.2%

[copper notch]

in the next year, you will have 103.2% over what you had in the previous year

[light wyvern]

Yh

[copper notch]

so 1.032x1.032

after 10 years you have 1.032^10

now you've gotta add 10k

[light wyvern]

What happens to compounding monthly

I believe if the interest compounds nornthyl then the value of x increases by each month every month in a year

[copper notch]

you have a monthly compound interest that amounts to 3.2% per year

[light wyvern]

Ooh makes sense

[copper notch]

after adding 10k, your interest changes to 2.8% per annum

[light wyvern]

Yh

[copper notch]

that's why I multiplied by that factor

to the 10th power

to account for the next 10 years

[light wyvern]

Yh it does makes sense now

[light wyvern]

Would this be the only method that would be the quickest?

[copper notch]

I think so

[light wyvern]

Appreciate it dude

.close

[midnight plank]

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[midnight plank]

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[twilit field]

I would like my proof for this checked.
\
We first show that if $I$ is in the span of $(T,T^2,\dots, T^{dim(V)})$, $T$ is invertible.
$\sum_{i=1}^{n} a_i T^{i}=I$. Then, factoring $T$ out $T(\sum_{i=0}^{n-1} a_i T^{i})=I$.
\
This proves $T$is invertible.
\
We now prove if $T$ is invertible, $I$ lies in the span of $(T,T^2.\dots, T^{\dim V})$
\
Let $dim(V)=n$.
\
$TT^{-1}=I
\
T^2T^{-1}=T
\
\vdots
\
T^{n+1}T^{-1}=T^{n}$
\
We now consider the set $(I,T,\dots,T^n)$. As this set has length greater than $dim(V)$ it's not Linearly independent.
\
Thus I lies in the span of $(T,T^2,\dots, T^n)$

[grand pond]

wai

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

.close

[midnight plank]

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[quasi field]

yo

[quasi field]

for the two red lines

how do I limit it to only the line on the right?

this is my erquation:

[fallow scarab]

Give the constraint in terms of x instead of y

[quasi field]

I tried that

[quasi field]

it doesnt work for some reason

[fallow scarab]

¯_(ツ)_/¯

[quartz hornet]

{x> -1.42 }

[quartz hornet]

It should work no?

[midnight plank]

@quasi field Has your question been resolved?

[midnight plank]

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[midnight plank]

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[molten bay]

I am just checking which singualrity at z=1

how can I use laurents expansion here?

[worthy spruce]

Have you found the laurent series?

[molten bay]

what do you mean?

i tried but do not know how can i do it here

[worthy spruce]

So pretty much just find the laurent series of $\frac{1}{e^{2 \pi i /z} - 1}$

[grand pond]

theaveragejoe6029

[worthy spruce]

and multiply through by z + 2

[worthy spruce]

theres a couple ways of doing this if you want

[molten bay]

Can you tell me list of itM

I only not two ways checking singularity by first thing is like limit and the other one related to lorentz expansion

Is this correct?

[worthy spruce]

I only really know the classification using the laurent series.

[worthy spruce]

so I feel like you're missing a couple steps. but it looks like you got $\frac{1}{\sum^{\infty}_{n=1}\frac{(2 \pi i z^{-1})^n}{n!}}$

??

[molten bay]

Yeah

I got it

[grand pond]

theaveragejoe6029

[worthy spruce]

right, so can you further simplify this to get the summation off the denominator?

[molten bay]

I don't know next

How can I simplyfy?

[worthy spruce]

could you use the geometric series?

[molten bay]

Ohh sure

And n!?

It is not GP@worthy spruce

[worthy spruce]

idk what GP stands for

[molten bay]

Geometric progressions

wait a second

[worthy spruce]

?

you can force it into it

[molten bay]

2πi/z(e^(2πi/z)

[worthy spruce]

Oh I think I see what you were confused on. The "r" is summation. you're going from 1/(1-r) -> sum of stuff

not the other way

[molten bay]

No

This is not geometric series we can't do that

[worthy spruce]

why?

[molten bay]

Remember we have factorial in denominator

It is just expansion of e^z

[worthy spruce]

doesn't matter. just substitute $r = \sum{\frac{(2 \pi i z^{-1})^n}{n!}}$

[molten bay]

,w sun n=1 to infinity (2πi/z-1)/n!

[grand pond]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by andy03738

theaveragejoe6029

[molten bay]

Still not GP

r^2 = ?

[worthy spruce]

$(\sum{\frac{(2 \pi i z^{-1})^n}{n!}})^2$

[molten bay]

But we want each step

[grand pond]

theaveragejoe6029

[molten bay]

I meant this is not 1+r+r^2+r^3 type

[worthy spruce]

what? we dont need it to be in that form. Hold on let me step this through a bit better.

hear me out

[molten bay]

Sure

[worthy spruce]

you agree that we have $\frac{1}{\sum^{\infty}_{n=1}\frac{(2 \pi i z^{-1})^n}{n!}}$?

[grand pond]

theaveragejoe6029

[molten bay]

Yes

So you meant we can substitute the sum?

Upper side and lower side?

[worthy spruce]

then, $r = \sum^{\infty}_{n=0}\frac{(2 \pi i z^{-1})^n}{n!}$

[grand pond]

theaveragejoe6029

[worthy spruce]

so we have $\frac{-1}{1-(r)}$ right?

[molten bay]

Summition of summitions

[worthy spruce]

YES

EXCATLY

🙏

[molten bay]

But where is it written?

[grand pond]

theaveragejoe6029

[molten bay]

It is notwhere written that we need sum n=1 to infinity r

[worthy spruce]

wdym?

[molten bay]

Bro you are wrong here

Check it twice please

[worthy spruce]

arent'y you trying to find the laurent series?

[molten bay]

You are using 1/1-r sum of gp without checking the series properly

[molten bay]

Yes

[worthy spruce]

what is not satisfied?

[molten bay]

in a geometric series we need a term which is common

Here r should be <1 so we can't use

1+1/3+(1/3)^2+(1/3)^3+....

Sum will be 1/(1-1/3)=3/2

So here we need a common thing

Now tell me where is this series looks like geometric

It is clearly not a geometric and you are using it forcefully without

We want sum of it

[molten bay]

We want value of r

.close

[midnight plank]

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[worthy spruce]

fyi, we only need it to converge around the singularity btw. I got that it converges for |z| < pi/20. So this open disc is big enough to check for singularities.

[midnight plank]

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[inland patio]

Consider an $n$th order homogeneous differential equation with constant coefficients and the $n$th degree auxiliary polynomial $g(t)$. So $$g(D)(y)=0,$$where $D$ is the differential operator. Let $V$ be the solution space of this equation. I'm trying to prove $g(t)$ is the minimal polynomial of $D_V$, the restriction of $D$ to $V$. From the above displayed equation, we already know that the minimal polynomial has to divide $g(t)$. And I know $V$ is the null space of $g(t)$, which say, is $n$-dimensional. How do I conclude $g(t)$ is the minimal polynomial?

[grand pond]

psie

[midnight plank]

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[midnight plank]

@inland patio Has your question been resolved?

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[twilit field]

Here can I say that $a_0v+a_1T(v)+ \dots + a_{dim(V)}T^{dim(V)}(v)=0$ for some constants, then $T^{dim(V)}(v)$ lies in the span of the first dim(V) elements

[grand pond]

wai

[twilit field]

similarly for all elemets indexed by m≥dim(V)-1

[runic hamlet]

no

e1,e1,e2 are lin dependent but e2 is not in the span of the first two

[twilit field]

hmm, so what do I do?

[runic hamlet]

one of the elements is in the span of the previous elements

that part is true

you have to use at some point that T is a linear map

[twilit field]

By applying it multiple times to both sides?

[runic hamlet]

yes

[twilit field]

hmm

applying T m- dim(V) times

[twilit field]

How does this help though

[runic hamlet]

maybe start slower and apply T once

the question is, apply it to what

[twilit field]

$\sum_{i=0}^{dim V} a_i T^{i+1} (v)=0$

[grand pond]

wai

[twilit field]

what about T to T^{dim(V)-1}

like applying it multiple times will give me. a subset of the set on the right. but I already knew that

[runic hamlet]

lets do an example

lets assume T^3 v is in the span of v, Tv, T^2 v

what does that mean

[twilit field]

a_1v+a_2Tv+a_3T^2v =T^3v

[runic hamlet]

quantifiers

[twilit field]

$\exists a_1,a_2,a_3 \in \R : a_1v+a_2Tv+a_3T^2v=T^3v$

[grand pond]

wai

[runic hamlet]

and what happens if we apply T to both sides now

[twilit field]

$\exists a_1,a_2,a_3 \in R :a_1Tv+a_2T^2v+a_3T^3=T^4v$

[grand pond]

wai

[runic hamlet]

and what does that mean

[twilit field]

T^4 is in the span of Tv,T^2v,T^4v

[runic hamlet]

typo

and T^3 v is in the span of the first three

so therefore what about T^4 v?

and what about the rest?

then write that down properly

I have to go

[twilit field]

okay, thanks

T^4 is in the span of Tv,T^2 and T^3 v.. BUt T^3 v is in the span of v,T,T^2, so T^4 is in the span of T^2,T,v

[subtle blaze]

yeah but you keep missing the v's

T^4v is in the span of Tv,T^2v and T^3 v.. BUt T^3 v is in the span of v,Tv,T^2v, so T^4v is in the span of T^2v,Tv,v

[twilit field]

so I have to apply a similar idea here

[subtle blaze]

i mean cant you do induction on this

[twilit field]

just for reference

[twilit field]

That did come to mind yea, but won't a direct proof be more isisghtful

[subtle blaze]

i would use induction

the idea is that once you have dimV many of the things

any more will not help increase the span

[twilit field]

dim(V) you mean?

[subtle blaze]

so you show that T^kv can be constructed using T^0v through to T^(dimV-1)v

well i suppose that is exactly what we want to show

[twilit field]

eh, can the dim(V)+1 vector lie outside the span( the first dim (V) vectors can be Linearly dependnent on one of the vectors

[subtle blaze]

is it possible for (T^nv, T^mv) be linearly dependent for some 0 ≤ n,m ≤ dimV-1

what would that mean for the minimal polynomial

[twilit field]

that would mean the minimal polynomial cannot have degree n

[subtle blaze]

hmm i think maybe it doesn't quite work

[subtle blaze]

i dont think this is possible though

since the next one T^(dimV)v can be made using the first dimV many vectors

i think the minimal polynomial can't have more than degree n right?

it's been a while

[twilit field]

no

[subtle blaze]

wait what

[twilit field]

where n is the dim of the vector space right

[subtle blaze]

yeah

[twilit field]

idts, but I'' confirm

[subtle blaze]

doesn't cayley hamilton say there exists a polynomial p of degree dim(im(T)) such that p(T) = 0

[twilit field]

Haven't done that yet

[subtle blaze]

do you know that this isn't the case?

either way you can still proceed with induction

[twilit field]

okay, I guess I will

[twilit field]

so base case will be dim(V)=2, I suppose

[subtle blaze]

i dont see why a direct proof might be more insightful but just because i dont see it doesn't mean there isn't one

why not 1

suppose V is finite dimensional means for any V such that dimV ∈ ℕ₀

[twilit field]

uh, yea, that works too

[subtle blaze]

it would be good housekeeping to also trivially prove it for the dimV = 0 vector space

[twilit field]

We then suppose $span(v,Tv,\dots,T^mv)=span(v,Tv,\dots,T^{dimV-1} v)$

[grand pond]

wai

[subtle blaze]

im pretty sure that's not the first step to induction

[twilit field]

I've verified the base case

[subtle blaze]

okay

[twilit field]

wait, I have to show span(v) = span(v), right

that's it for the base case

we have to induct on both dim(V) and m, don't we

okay, I fix dim (V) and induct on m

That makes more sense

So the base case is trivialy true

[runic hamlet]

induct on m and start the induction at k, where k is the first number so that T^k v depends on the previous vectors

existence of k is guaranteed by one of your theorems

and k<= dim V -1

[twilit field]

k>= dim(V)-1, right

so k = dim(V)

[runic hamlet]

no

[twilit field]

no?

okay

hmm

so do I just assume k to be an arbitrary number

[runic hamlet]

I told you what k is

I clearly defined it

as the smallest number so that T^k v depends on the previous vectors

[crystal plover]

If it's not clear to you why such a k should exist, you can show that too

[twilit field]

I'll do that after I prove this

okay so let $k$ be the smallest number such that $T^kv$ depends on previous vectors. Then $\sum_{i=0}^{k-1}a_i T^{i}v=T^{k}v$ for some scalars, all not zero

[crystal plover]

firstly, you mean Tᵏv
secondly, are you sure about what you wrote?

Tᵏv linearly depends on previous vectors, not Tᵏ

Tᵏ is an operator

Tᵏv is a vector

[grand pond]

wai

[crystal plover]

Okay

[twilit field]

now I apply the operator on both sides m-k times?

so I get $\sum_{i=0}^{k-1} a_iT^{m-k+i}v = T^m v$

[grand pond]

wai

[runic hamlet]

induction

[twilit field]

I'm trying to do induction

I'm just so confused

[runic hamlet]

the claim you are trying to prove is that for all m>=k the vector T^m v depends on v, Tv, ..., T^(k-1) v

base case is done

[twilit field]

this was the base case, right

[runic hamlet]

no

base case is m=k

T^k v = sum a_i T^i v

for some a_i

[runic hamlet]

btw

all not zero
and
not all zero
are not the same thing

and both are wrong here

[subtle blaze]

Wai I swear you type like how I would type if I didn’t proofread before I sent my messages

[runic hamlet]

wai doesnt even read the messages after he sent them

[twilit field]

So I now he hypothesis is , this is true for T^l , where l≥k

[runic hamlet]

again forgot the v

induction hypothesis is that T^l v is in the span of v,Tv,...,T^(k-1)v

induction step: show that T^(l+1) v is in the span of v,Tv,...,T^(k-1)v

[twilit field]

$T^{l} = \sum_{i=0}^{k-1} a_iv^i$
\
Applying $T$ to both sides
\
$T^{l+1} = \sum_{i=0}^{k-1} a_i T^{i+1} (v_i)$
\
but $T^k(v)$ is in the span of $v,T(v)\dots T^{k-1}v$.
thus $T^{l+1}v$ is in span of $v,Tv, T^{k-1}v$

Editing rn, sorry

[grand pond]

wai

[subtle blaze]

but like

do you actually read your own writing

[twilit field]

My brain usually auto corrects the mistakes I make

so I don't notice them

[subtle blaze]

but you should read it

literally read the first one

all 3 of us here have complained that you keep losing the v in Tv

and yet it's missing in the first line

and then sometimes there's a bracket

sometimes there isn't

[twilit field]

$T^{l}v = \sum_{i=0}^{k-1} a_iT^{i}v_i$
\
Applying $T$ to both sides
\
$T^{l+1} v = \sum_{i=0}^{k-1} a_i T^{i+1} v_i$
\
but $T^kv$ is in the span of $v,Tv\dots T^{k-1}v$.
thus $T^{l+1}v$ is in span of $v,Tv, T^{k-1}v$

[subtle blaze]

you always split up your statements so much idk where you're getting your stuff from

i dont even know where the first line comes from

what's l, what's k

[twilit field]

T^l comes from here

[runic hamlet]

btw never use l in latex, use \ell

or just choose a different letter

still lost a v

I genuinely dont know how you manage that all the time

[grand pond]

wai

[runic hamlet]

especially after it was pointed out and you presumably checked for it explicitly

[twilit field]

I did, but as I said, I just miss it

[subtle blaze]

i dont even know why this v has a subscript all of a sudden

how do you manage to put so many mistakes in 1 line and not notice any of them

[twilit field]

which v

[subtle blaze]

the first line

[twilit field]

right

$T^{l}v = \sum_{i=0}^{k-1} a_iT^{i}v$
\
Applying $T$ to both sides
\
$T^{l+1} v = \sum_{i=0}^{k-1} a_i T^{i+1} v_i$
\
but $T^kv$ is in the span of $v,Tv\dots T^{k-1}v$.
thus $T^{l+1}v$ is in span of $v,Tv, T^{k-1}v$

[grand pond]

wai

[subtle blaze]

and then you litearlly just dont look and notice that it's in line 3 as well

like 80% of the time i look at your problems and i can't tell if im the one who doesn't know what's going on or is it you, because i look at what you write and it confuses me

[twilit field]

I think I should focus more on how I present my work at this point

[subtle blaze]

like no offense but of the regulars here you're one of the most un-understandable people here

[twilit field]

I get that, yes

[subtle blaze]

it's actually increidble how you don't confuse yourself with what you write

[runic hamlet]

he does lol

constantly

[subtle blaze]

well i can't tell because im also confused

[twilit field]

okay,I'll close this now

thanks

.close

[midnight plank]

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[fervent burrow]

why am i incorrect

[dreamy lichen]

It would be 60 if AE was parallel to BD

but it's not

[fervent burrow]

Oh

[dreamy lichen]

the triangles ACE and BCD arent similar

[fervent burrow]

Ok thank you

.close

[midnight plank]

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[fervent burrow]

.reopen

[midnight plank]

✅

[fervent burrow]

am i right

[nimble leaf]

Does EDB look like 60?

No, it looks obtuse

[fervent burrow]

idk im not the best at maths

[nimble leaf]

How did you get 60?

[fervent burrow]

co interior angles add up to 180

dont they

[haughty lion]

what rawr is trying to say is although the diagrams "arent to scale", once you get an answer its very important you check/review your answer and see if it makes sense

this applies to pretty much every subject/course you'll ever take in school or university

[nimble leaf]

Yeah I think I phrased it a little rudely

But that's what I meant

[nimble leaf]

Only for parallel lines

[fervent burrow]

oh

[nimble leaf]

See if it was 60 then you'd have a 120 degree angle BDC in the triangle so BCD would be 0, which is impossible

[haughty lion]

i agree that ABD is 120, because of angles on a straight line

what do you think AED is? @fervent burrow

[fervent burrow]

It wasnt supposed to be used but thank you

got it

.close

[midnight plank]

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[chilly cobalt]

Prove that 1/h_a + 1/h_c + 1/h_b = 1/r_a + 1/r_b + 1/r_c

this probably uses similar angles??

no idea what to do

will be back later

.close

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[mint ravine]

Please help someone

[mint ravine]

I came up with a lot of possible answer choices but each one of them were wrong

i think im overcounting but i don't know where i am overcounting at

or how much im overcounting by

[fallow scarab]

!show

[midnight plank]

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

[mint ravine]

bruh

i did 6 choose 3 times 3 times 3!

6 children to choose from and we need exactly 3 children to choose some flavor

then i multiplied by 3 because there are 3 flavors

and 3! for the rest of the children

please helpp

[midnight plank]

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

[fallow scarab]

,calc 6! / (3! * 3!) * 3 * 3!

[grand pond]

Result:

360

[fallow scarab]

you put 360?

[mint ravine]

yeah that was my first answer choice

it was wrong

[fallow scarab]

did you take into account that the remaining children can't select the initial flavor

because it says "exactly three children"

[mint ravine]

yeah i tried that already

with 3 choose 2

[fallow scarab]

and what number did you get

[mint ravine]

3 choose 2?

oh wait nvm

i got 180

which is 6 choose 3 times 3 times 3 choose 2

<@&286206848099549185>

i dont know what im doing that is wrong

[fallow scarab]

where does 3 choose 2 come from

[mint ravine]

the remaining 3 children choose from the remaining 2 flavors

[fallow scarab]

3 choose 2 = 3. each child chooses independently

[mint ravine]

so 3 choose 2 times 2?

[fallow scarab]

right

but just 3 * 2. 3 for 3 remaining children. and 2 for flavors. 3 choose 2 just happens to equal 3

[mint ravine]

yeah i know that

so

how do you do combinations in latex

[fallow scarab]

#latex-help

[mint ravine]

6 choose 3 times 3 times 3 times 2 then

which is 360

so that's wrong

[fallow scarab]

right

what message shows up when you enter 360

[mint ravine]

im doing this on aops so i don't think there is a mistake in the problem

i think i might js search up the answer atp

[midnight plank]

@mint ravine Has your question been resolved?

[vocal raft]

@mint ravine try 480

[mint ravine]

already tried that lol

[midnight plank]

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[mint ravine]

Can someone please help me with part B

[mint ravine]

(b) Why is Grogg's answer wrong? Should the correct answer be smaller or larger than Grogg's answer (and why)?

I feel like what he did to include both possible cases (adding both expressions) might be incorrect

but idk

because it's technically not mutually exclusive since they can happen both at the same time

please help someone

<@&286206848099549185>

[vast ginkgo]

!15m

[midnight plank]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[vast ginkgo]

but

[mint ravine]

what

[gentle hornet]

It counts the cases where A is present 4 times and B is present 2 times twice

[mint ravine]

yeah it said that in the problem

[foggy sky]

the mistake is in the number 26 instead it should be 25 because we already used the other letter (A or B)

[vast ginkgo]

im no expert but i believe it's to do with the fact that there is one less letter in the alphabet

[mint ravine]

oh yeah

i should have caught that

thanks @foggy sky

and @vast ginkgo

[foggy sky]

also there's another problem we have to take away the intersection

is it clear why?

@mint ravine it's to avoid duble counting the plates that have 4As and 2Bs at the same time

[midnight plank]

@mint ravine Has your question been resolved?

[mint ravine]

so we subtract by 6!/(4!*2!)

sorry for the late response

@foggy sky

[foggy sky]

exactly

[mint ravine]

thanks!

[foggy sky]

don't mention it

[midnight plank]

@mint ravine Has your question been resolved?

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[desert siren]

!status

[midnight plank]

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

@finite wharf Has your question been resolved?

[midnight plank]

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[kind ledge]

!help

[midnight plank]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[kind ledge]

can the first expression even be simplified?

[cedar pawn]

Have you tried factorising the denominator?

[kind ledge]

no because i dont see what can be factorosed

none of them got the same variables

[cedar pawn]

Have you done polynomial division before?

[kind ledge]

yeah

oh i gotta do that?

what do i put then

[cedar pawn]

Do you know what you're trying to do here?

[kind ledge]

yeah find the numerator

of the 6th expression

[cedar pawn]

I meant for the first one

[kind ledge]

yeah reduce

simplified

[cedar pawn]

Right so for the first one, the only way to simplify it would be to see if we can find a factor of y - 3 in the denominator

[kind ledge]

heah

yeah

how do we simplify

[cedar pawn]

So either you can try to factorise out y - 3

Or use polynomial division to take the factor out and see if it works

[kind ledge]

we cant factorise that though

theres nothing that has the same variable

right

[cedar pawn]

We can

But if you can't see how to do it, try polynomial division

[kind ledge]

alr but what am i dividing that with

[cedar pawn]

So we want to remove a factor of y - 3 from the denominator, right?

[kind ledge]

yeah

[brittle grotto]

the bottom may be in the form of (x+a)(y-3) for some number a

[kind ledge]

how did u do it

[brittle grotto]

I didn't do it

[kind ledge]

oh ok nvm

[brittle grotto]

just explaining how even though there are two variables, you may still be able to factor something out

[cedar pawn]

If we want to remove a factor of y - 3 from the denominator, we need to divide the denominator by y - 3

[kind ledge]

so 5xy + 2y -15x -6 / y-3

[brittle grotto]

yes

[kind ledge]

alright

ok this already doesnt work

[brittle grotto]

uhhhh

it should

[brittle grotto]

hmmm

I'll be honest with you

I did this one by guessing that the denominator can be decomposed into (ax+b)(y-3)

that seems a little opaque of a guess

so let me explain why it's reasonable

you have xy, x, y, and a constant term, right?

[kind ledge]

so basically the simplified version is

yeah

[cedar pawn]

I did this one just by seeing how to factorise it, but it's useful to have techniques available when you can't see it immediately

[cedar pawn]

Gonna finish it off?

[kind ledge]

ngl i found the answer without polynomial division

[brittle grotto]

and the xy term is not 1, so it's probably going to yield a since ax*y=5xy

[kind ledge]

its (5x+2)(y-3)

[brittle grotto]

nice

[kind ledge]

but my question is

[brittle grotto]

sometimes, useful observations about problems can help you not need to apply algorithms like polynomial division

[kind ledge]

u see 5xy + 2y -15x -6

?

[brittle grotto]

but it's probably important to understand the general process, and it's undoubtedly important to understand why a guess is a reasonable one

[kind ledge]

am i allowed to reverse 2 and -15 places

?

[brittle grotto]

so make sure it makes sense

[kind ledge]

yh

[brittle grotto]

wdym

[kind ledge]

because i switched placed

the -15x and 2y

changed places

to factorise

and simplify

[cedar pawn]

Addition is commutative so yes

[kind ledge]

alright

[cedar pawn]

a + b = b + a

[brittle grotto]

oh, that's what you mean

[kind ledge]

yeeah

ok anyways

back to this so

we cancel out (y-3)

leaving us with 1/5x+2

wait no

[brittle grotto]

wasn't it 5x+2

[kind ledge]

yeah

[cedar pawn]

I just saw that we want to simplify it, so we need a factor of y - 3, which means it must be (5x + a)(y - 3), and then it was immediately obvious that a = 2

[brittle grotto]

(if such a guess form doesn't line up, then you may need to do some polynomial division)

[cedar pawn]

So yeah you can just see it but that's not particularly helpful if they don't just see it

I was going to explain that part at the end as well but nvm

[kind ledge]

alrighy

dw

im doing the 3rd expression i already did the second

the second is 1/5x+4

[brittle grotto]

I think it's instructive to explain why ax+b is a sensible guess probably, since that's what we both defaulted to doing sort of

oh u turned blue lol

nice

[cedar pawn]

I've been blue before

[brittle grotto]

i remember when i randomly turned blue

[cedar pawn]

But yeah lol

[brittle grotto]

and then last week or something i randomly turned green

[cedar pawn]

I defaulted to 5x + a, but yeah

[kind ledge]

guys

3rd expression is

i cant cancel the x+3

nor the x+8

[cedar pawn]

Try putting them over a common denominator

[kind ledge]

x+3

[cedar pawn]

Is that not a division sign in the 3rd one?

<@&268886789983436800>

[midnight plank]

@kind ledge Has your question been resolved?

[kind ledge]

oh

ah

im not sure

wait

yeah

its a division

[midnight plank]

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[molten bay]

direction ratio of x+y=0

[molten bay]

so here it will be (1,-1,0)?

[dreamy lichen]

so you are looking for the slope?

[molten bay]

no no

I want to understand how direction ratio work

z axis directio ratio will be (0,0,1)

Am i right?

[dreamy lichen]

oh, but this is not a 3d line, is it?

[molten bay]

It is not

But it's a plane equation no?

[dreamy lichen]

yeah

[molten bay]

I know it's line equation but

[dreamy lichen]

doesnt plane have many direction vectors?

[molten bay]

This is the thing where I am confusing

Could you tell me more

I am slightly confusing on my doubt

[dreamy lichen]

I think that a direction ratio here would be (a, b, 0)

for any a, b

[molten bay]

I see

[dreamy lichen]

but like its not standard to look for direction ratio of planes

[molten bay]

(1,-1,0) is called what?

[dreamy lichen]

are you sure you got the question correctly?

[molten bay]

Yeah

[dreamy lichen]

that's just one specific direction vector

[molten bay]

I see

[dreamy lichen]

(1, 1, 0) would work as well

[molten bay]

How?

1+1=0

[dreamy lichen]

oh wait no im stupid

(a, -a, b)

this is the general form

so z can be anything

(2, -2, 1) would work as well

any point that lies in the plane should be valid

so any point (x, y, z) which satisfies x + y = 0

so any point (x, -x, z) for some x, z

[glass lantern]

z has to be 0

[molten bay]

So when I take z=0 it will form a line?

[dreamy lichen]

if you do
x + y = 0, z = 0

then it will be just a single line

3D line can be characterized by 2 equations

[dreamy lichen]

(it would be the intersection of 2 planes, so a line)

[dreamy lichen]

Why?

you could also interpret x + y = 0 as a single 2D line and say just (1, -1)

[glass lantern]

direction ratios are proportional to direction cosines

cos90⁰ = 0

[dreamy lichen]

why 90 tho

[dreamy lichen]

there isnt a unique angle

[glass lantern]

xy plane is at right angle with z axis..?

[dreamy lichen]

with the z-axis, correct

but not with the plane

[glass lantern]

which plane?

[dreamy lichen]

x + y = 0

[glass lantern]

and why are we counting the angle with that...

direction cosines consider angles made with each of the axes

[dreamy lichen]

How do you define angle of a plane with an axis?

the plane x + y = 0 passes through the z-axis

each point on the plane gives a slighlty different angle

and by slightly different i mean varying from 0 to 90

It's weird to talk about direction ratio of a plane tbh

it's certainly not unique

[glass lantern]

it is but

[glass lantern]

I suggest you double check this please

[dreamy lichen]

.

[glass lantern]

none of the lines you drew

exist on the xy plane

[dreamy lichen]

and why do we need to talk about x-y plane

[glass lantern]

it's by definition that any vector on xy plane must be perpendicular to the z axis

[dreamy lichen]

okay but we arent on the xy plane here

[glass lantern]

we're looking for the direction ratio of xy plane..?

[dreamy lichen]

the plane x + y = 0 isnt the xy plane

[dreamy lichen]

its this plane

[glass lantern]

oh

true

[dreamy lichen]

its what you get when you plot x + y = 0 in 2D and then extend it upwards and downwards to 3D

[glass lantern]

my bad

[glass lantern]

yea

idk what I was thinking

sorry man

[dreamy lichen]

np

[dreamy lichen]

uh btw i was searching online

and i didnt find anyone talking about direction ratio of a plane

so perhaps you just mean the 2D line x + y = 0

or you are talking about the direction ratio of a normal to the plane x + y = 0

[midnight plank]

@molten bay Has your question been resolved?

[molten bay]

xy=k

[dreamy lichen]

what is this supposed to be

thats a hyperbola in 2D plane

or hyperbola extended upwards & downwards in 3D space

What are you actually trying to do?

Those certainly arent examples from a textbook...

[molten bay]

Yes they are not examples of test books actually I am just thinking about how direction ratios work and how we write symmetric form and the specific values of 11 and 0

@dreamy lichen

I mean I am just thinking when we move to the to 3D how we select component Z

so when xy=a it will form a plane?

Tell me one more thing x+y=0 so here normal vector will be?@dreamy lichen

[dreamy lichen]

nope

[dreamy lichen]

e.g. (1, 1, 0)

this is how xy = 1 looks like

it's just hyperbola extended above and below the xy plane

by taking product of x and y, its no longer linear basically

so its not plane, line or anything like that

[molten bay]

Got it

[dreamy lichen]

that table is extremely weird

[molten bay]

xy=0

[dreamy lichen]

this would be 2 striaght lines

x = 0 and y = 0

[molten bay]

It is 2 degree so yes it can't form a line it has to be 0

xy=0

[dreamy lichen]

why do they divide by 0

[molten bay]

It is fine to me untill 4

[dreamy lichen]

x/1 = y/0 = z/0

are they dividing by 0

or am i just reading it wrong

[molten bay]

Yeah

See what they are writing under the table

[sudden yacht]

(ii), (iii) and (iv) are completely illegal lol

[dreamy lichen]

indeed

they shouldnt write it like that

the forms below work only if the denominators are non-zero

if they are zero, then they should just put a comma and write, z = z1 as another equaton

but dividing by 0 is just wrong

[molten bay]

Tell me about normal vector

[dreamy lichen]

seems like it wasnt covered yet in your book

[molten bay]

Suppose I have x axis and (x,0,0) will be direction ratio

[dreamy lichen]

but its basically a vector thats perpendicular a given plane

[molten bay]

Yes

and it has cofficients of x,y

[sudden yacht]

And z

[molten bay]

x axis normal vector will be?

normal vector exist in 3d only?

[dreamy lichen]

they are a 3D thing

but they are associated to planes

not lines

and x axis is a line

[molten bay]

Hmm i see

[sudden yacht]

In whatever dimension

[molten bay]

So x axis normal vector would be?

[dreamy lichen]

But yeah, you can have a normal vector to
(n-1)D subspace of nD space in general

[sudden yacht]

Any vector perpendicular to it

Such as (0, 1, 0) or (0, 0, 1)

[dreamy lichen]

if we're inside the 2D plane, it would be (0, t) for any t

so (0, 1), (0,-1) are both fine

[molten bay]

Hmm right i am understanding now

[dreamy lichen]

if we're talking about 3D, it can be anything of form (0, a, b)

[sudden yacht]

If you're given a plane in the form ax + by + cz + d = 0, then a normal vector to it is (a, b, c)

[molten bay]

This is confusing?

[sudden yacht]

Why?

[molten bay]

Will it be surely 90°

If we take y,z together

[dreamy lichen]

yes, do you know dot products?

[molten bay]

Yes

[dreamy lichen]

2 vectors are perpendicular if their dot product is 0

so we can take the direction vector of x-axis

(1, 0, 0)

[sudden yacht]

Well, if you wanted it to be perpendicular (to the z-axis), of course the angle is 90°

[dreamy lichen]

and dot product it with (0, a, b)

(1, 0, 0) . (0, a, b) = 1*0 + 0*a + 0*b = 0

so it's perpendicular

all the points (0, a, b) lie in the blue plane

red line is the x-axis

you can see that if we draw a vector from origin to any point on the plane, it will be perpendicular

[molten bay]

You guys explain brilliantly which I never understood by text books

Thank you very much both of you lots of love

[dreamy lichen]

np

[sudden yacht]

Yw

[dreamy lichen]

https://www.desmos.com/3d

[molten bay]

(0,a,b) but if I take (0,2,0)

Will it be on the blue plane?

[dreamy lichen]

btw you can try using this to graph stuff

[dreamy lichen]

Yep