help-17

[scarlet lichen]

Yo whats up

[desert edge]

Hey hey

[scarlet lichen]

I'd go step by step

if you get this into y=mx+b form

what value of A/B would give you a graph that would intecept, be parallel and be the same line

[desert edge]

So basically the b would be my a in the first equation

[scarlet lichen]

Not exactly

[desert edge]

Like -1/3a

[scarlet lichen]

yep exactly

[desert edge]

Something like that

[scarlet lichen]

yes for the first equation

do the same for the second aswell

[desert edge]

So my slope is the same

[scarlet lichen]

yep exactly

[desert edge]

Okay so I know they’re parallel. So if they’re parallel there’s two options. Either they overlap or they don’t

[scarlet lichen]

for what A/B value would they overlap?

[desert edge]

If they overlap then the constants times the coefficient have to equal the same thing

Basically if -1/3a = -1/6b

They overlap

If those aren’t equal then they’re parallel and never touch

[scarlet lichen]

yep

exactly

A=0.5
B=1 is a example solution

if you want no solution

it would be the opposite

[desert edge]

The opposite or anything else

[scarlet lichen]

and yes, they don't intercept

[scarlet lichen]

no, i mean the opposite of this statement: f -1/3a = -1/6b

if 1/3 A =/= 1/6 B

then they are paralell and have no solution

[desert edge]

I see so there’s no situation where there’s one solution here precisely because the slope is the same

What if the slope wasn’t the same

What logic would we then use to see when there’s one solution etc

[scarlet lichen]

yes

[scarlet lichen]

you would have a solution

but you wouldn't have infinitely many solutions

meaning they won't overlap

[desert edge]

I guess even to see if there’s infinite solutions it wouldn’t be as straightforward as here then because the only reason it was so easy to see here was that I saw they were multiples of each other

[scarlet lichen]

yea exactly

in most cases, if you get down to Y=MX+B

[desert edge]

I guess I just have a hard time writing this out in a way that’s acceptable in an exam setting etc

[scarlet lichen]

it would be pretty obvious if the "M" or the slope are same

[desert edge]

Like mathematically writing it instead of words

[scarlet lichen]

that would be up to your prof

completely depends on them

[desert edge]

So as an algorithm let’s say I should always start by putting in mx+b form

Then if the slopes are same I kind of know what to do as we did here

If they aren’t

Then I’m a bit confused what to do

[scarlet lichen]

if they aren't the same, the lines would intercept

like this for example

[desert edge]

So the moment I see slopes aren’t the same I can just assert that and also know there won’t ever be infinitely many solutions

And also won’t be none

[scarlet lichen]

yes

[scarlet lichen]

also yes

[desert edge]

Ahhh makes sense

Thank you

[scarlet lichen]

one sec here

[scarlet lichen]

thats pretty much it

[desert edge]

Hahah thanks

[scarlet lichen]

a quick tip

would be to think of the graph

it will be helpful imo

[desert edge]

Yeah I guess the visual will help for sure

[scarlet lichen]

try and visualize

Cya

[desert edge]

Ty

[scarlet lichen]

.close

[desert edge]

.close

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[simple carbon]

When should I use nPr and when should I use nCr? (nPr was used for part a and nCr was used in oart b, I just don't understand why)

[simple carbon]

Please @ me when you are available, thank you!

[tawny hollow]

@simple carbon permutations (nPr) cares about order, while combinations (nCr) doesn’t

[simple carbon]

Ok, so with the order or President, ceo etc they have different titles while the committee has equal rank

Thank you

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[narrow tulip]

Hey, so I’m a writer and I’m trying to figure out some area measurements for a map I’m making, I want to figure out the area of the blue-shaded part of the island, and the red-shaded part of the island, the whole island itself is 1,134 km2. The islands and certain recesses in the coast can be disregarded. The measurements also don’t have to be incredibly accurate of course

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@narrow tulip Has your question been resolved?

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@narrow tulip Has your question been resolved?

[blissful sentinel]

since it doesn't need to be incredibly accurate, you could approximate them both as rectangles

something like this

and then all you need to do is approximate what fraction the red one is of the blue one and then multiply that by 1134; if you want to be more precise, you can add more rectangles

another option, since you already have a picture with the things shaded, is to open some sort of image editor that lets you count the number of pixels in the red area and in the blue area and then just do (red)/(red+blue) * 1134

I'm not sure which image editors let you do that though

[narrow tulip]

I think pixel measurement is the way to go, I use procreate so I’ll see if it works

If not I’ll do the other method of course

Thank you for all the help!

[narrow tulip]

Alright so I did the pixel measurements via a website, and I was able to find out that the red area is 484sqkm, and the blue at 700sqkm, which rounds out pretty nicely

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[untold smelt]

So I'm thinking about living in a van for a year and did the basic-ish math that I could have $2k left over after buying everything my friends and i would need to live in a van and make about $240 a month. If we spent $467.30 a month about how long would it take for us not to be able to afford to live in the van anymore. I don't know where to start this math problem since I'm an idiot

[flint idol]

make about $240 a month per person or in total

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[last canyon]

ok so i havd done this question the way i was told to previously, but, i do not get how i can just sub different n's in till i get what i want like keep making n bigger till i get my value

[last canyon]

its the last part of d

this is what i have done

,rotate

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@last canyon Has your question been resolved?

[tawny hollow]

Why not?

[last canyon]

well if i sub a bigger n in i get a bigger value so ln would be bigger than something else

[tawny hollow]

Imagine it like this, the inequality is an estimate. The formula in the first part of d) is exact, nothing* is stopping you from getting a better and better estimate by plugging in larger and larger n

But in this case, you want to show a particular estimate, so by plugging in a reasonable n, you get precisely the bounds of the inequality

You do it two ways, one for 2/3 < ln 2

And one for ln 2 < 17/24

[tawny hollow]

This is reasoned good!

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[merry cairn]

can co terminal angles be negative or positive

[merry cairn]

or just negative

[mild flower]

60 420

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[merry cairn]

wdym?

[merry cairn]

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[fleet thistle]

i did'nt get the second step how 1/4 become 1/2

[fleet thistle]

1/4 is can be taken as 2^-2 so by identity it should become -2 when multiplied by - to become 2 but how 1/2

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@fleet thistle Has your question been resolved?

[ripe jay]

I think you got the formula wrong it should be: log2^-2(x) = -1/2.log2(x)

log a^b (x) = 1/b.log a (x)

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[tawny hollow]

!status

[vocal sleet]

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

[icy jetty]

4

11/9 is what i got

I did change of base

[tawny hollow]

Good start

[icy jetty]

(log3x^3-2)/(log4x^2+3x-6)

then took derivative of both numerator and denomintor

so lhopital

[tawny hollow]

it's the other way around

[icy jetty]

cause plugging in 1 would end up with 0/0

oh

oh yeah i did that, just sent it the other way around on here accidentally

so i did it that way, applied lh rule and got 11/9

[tawny hollow]

it would help if you could show the expression you arrived at after applying that rule

or rather, the actual work, so i can spot any mistakes

[tawny hollow]

it looks correct

[icy jetty]

Accidentally put lim x->1 for the denomintor, but this is just rough work

Ah alright thank you

[tawny hollow]

yeah and 11/9 is the correct answer aswell

[icy jetty]

theres also another question, which im not sure how to start doing

[tawny hollow]

usually the trick here is to notice that u^v = exp(v ln u) for positive u

now since exp is continuous, you can reduce the limit to asking what the limit of v ln u is

and then just plop that value back into exp

to get the original limit

and now if v ln u approaches +infinity, you can show that exp of that also goes to infinty

and if v ln u approaches -infity you can show that exp of that goes to 0

[icy jetty]

This is what I have so far

Not sure what to do with the ln(sinx/x)

[tawny hollow]

uh bad notation here, but ill allow it for now

[tawny hollow]

notice that youll have something of the form 0/0 for the limit

ln is continuous, and you know sinx/x approches 1 as x->0+

so you have the form of ln 1 / 0 i.e 0/0

so you can use Lhoptials rule here or whatever it's called

but uh idk if that helps actually

[icy jetty]

Ill try solving this one as much as I can or move on to another one and come back to it later

Thank you for checking if my answer to the earlier question was correct

[tawny hollow]

i think i see a better way, just putting it here for later

the maclauring expansion of ln(sinx/x) stars with the terms -x^2/6 - x^4/180

so ln(sinx/x) = -x^2/6 - x^4/180 + O(x^5)

this can be useful

[icy jetty]

Thanks, ill come back on here if I need more help

Have a great day

.close

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[neon junco]

I'm not sure what formula to use any help is appreciated

[vocal sleet]

@neon junco Has your question been resolved?

[neon junco]

<@&286206848099549185>

[thick niche]

@neon junco wait im solvinmg it

[neon junco]

okay thank you

[short pilot]

Here for help

[thick niche]

@neon junco ok lets think about total ways

how many ways can A pick his seat

[neon junco]

if he's the last one then 1

[thick niche]

ok wait

im gonna assign each villain

A B C D E F G H I J

a is the most powerufl, j is the weakest

ok?

[neon junco]

okay

so 10

[thick niche]

yes

now what about B

[neon junco]

9

[thick niche]

i dont thin so

[neon junco]

theres 9 seats left after A picks no

[thick niche]

o

no

[neon junco]

i dont get it

[thick niche]

ok so

theres 2 cases

i) A chooses the seat B is in, then B has 9 choices
ii) A chooses a seat that B is not in, then B has 1 choice. But there are 9 seats that A can choose to not sit in B's seat

[neon junco]

so i need to account for both cases

[thick niche]

yuh

[neon junco]

so 1/10

[thick niche]

???

u have 8 ppl left....

[neon junco]

oh

so A is 1/10 and B is 1/8

[thick niche]

nonono we are only counting the total number of possibilites rn, we are gonna do the chances if J sitting in his seat afterwards

i gtg now maybe @short pilot can continue

[short pilot]

Ok

[neon junco]

okay thank you

so C would have 1/6 possibilities?

[neon junco]

.close

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[maiden egret]

I am not sure how to even approach this question

[vagrant gale]

a function is increasing when its derivative is positive, and decreasing when the derivative is negative

[grim lotus]

do you understand the question

basically volume at time t is f(t)

[maiden egret]

is f(t) increasing on (0, sqrt(3)) ?

and decreasing at (-INF, 0), (Sqrt(3), INF)

[magic wasp]

Presumably you only consider t >= 0

[maiden egret]

am I incorrect?

[magic wasp]

You're correct if t >= 0

[maiden egret]

.close

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[midnight basalt]

Hi. So if I have the attached image and I'm simplifying, I got 1+2sinxcosx = ((1+2cosxsinx)/(cosxsinx))/((1)/(cosxsinx))

[drifting scroll]

!status

[vocal sleet]

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

[drifting scroll]

4?

[midnight basalt]

2

If I want to bring the denominator ((1)/(cosxsinx)) to the other side,

[drifting scroll]

do u know latex?

[midnight basalt]

No but there's an online thing which I can use.

Sorry! Let me make it.

[drifting scroll]

dont bother

$1+2\sin x\cos x=\frac{\frac{1+2\cos x\sin x}{\cos x\sin x}}{\frac{1}{\cos x\sin x}}$

this?

[midnight basalt]

Um..yes, I think.

[twin meteor]

π

[drifting scroll]

right?

uh

[midnight basalt]

Oh yes, that.

[drifting scroll]

isnt this just basically an identity?

[midnight basalt]

Wow wth how did you even make that 😭

[desert hornet]

this gives you the answer

[desert hornet]

it is

[midnight basalt]

Yes solving trigonometric identites

[desert hornet]

that's the point

they need to prove the identity

[midnight basalt]

I brought the bottom over

So it became

1+2sinxcosx/cosxsinx

[desert hornet]

$\frac{a}{b}=\frac{\frac{a}{c}}{\frac{b}{c}}$

[twin meteor]

kheerii

[midnight basalt]

So it matched but i feel like that's algebraically wrong

[desert hornet]

why?

[midnight basalt]

Intuition Im not sure

Because if Im bringing the bottom over I'd think it would be multiplying

Not dividing

Multiplying by cosxsinx

Bringing this part over

Would the division sign right above it also come over

Oh, also the original form was

[vocal sleet]

@midnight basalt Has your question been resolved?

[midnight basalt]

How in God's name would I try to prove this?

I tried to factor the left side but it didn't work: (sin^2x+cos^2x)(sin^2x-cos^2x)...

[fluid sorrel]

Is this question solved?

[midnight basalt]

No

I need to prove it

[fluid sorrel]

Ok, just in a bit.

[midnight basalt]

Thank you!

[fluid sorrel]

I hope this helps

[midnight basalt]

Wait, how did the right side turn to that?

[fluid sorrel]

Which step?

[midnight basalt]

Is that equivalent to 1-2cos^2x?

[fluid sorrel]

how?

i can't see, sry

[midnight basalt]

Oh wait

[midnight basalt]

Is this just expanding the left hand side?

[fluid sorrel]

yes

[midnight basalt]

So sin4x-cos4x is equivalent to (sin^2x+cos^2x)(sin^2x-cos^2x)?

[midnight basalt]

How would you do the right-hand side?

This part

[fluid sorrel]

yea

[midnight basalt]

I don't see how you can simplify it into anything. But something is telling me it's to do w/ the pythagorean identity

[fluid sorrel]

yes

[midnight basalt]

I tried multiplying the Pythagorean identity by 2 so i'd get cos^2x but then it turns into 2sin^2x = 2-2cos^2x

[fluid sorrel]

it isnt necessary

[midnight basalt]

Oh! Do you know what I'm suppose to do instead?

[fluid sorrel]

just moment

so

you have

${\sin{x}}^2 + {\cos{x}}^2 = 1$

[midnight basalt]

Yes!

[fluid sorrel]

${\sin^2{x}} + {\cos^2 {x}}= 1$

[twin meteor]

Čeva18

[fluid sorrel]

this

[midnight basalt]

I dont see how you'd manipulate the pythagorean identity to get

1- 2cos^2x

[fluid sorrel]

And one bracket is equal to one

[midnight basalt]

Yes I think so

[fluid sorrel]

so you are left with

$\sin^{2}(x) - \cos^{2}(x)$

[twin meteor]

Čeva18

[fluid sorrel]

And you use again this and get:
$\sin^2(x)=1-\cos^2(x)$

[twin meteor]

Čeva18

[midnight basalt]

Yes

But I can't use it to get 1-2cos^2x

The ratios don't work

[fluid sorrel]

you can

[midnight basalt]

How?

[fluid sorrel]

cuz now this bracket can be rewrited in form:
$(1 - \cos^2{x} - \cos^2{x})$

[twin meteor]

Čeva18

[midnight basalt]

THIS?

[fluid sorrel]

yup

[midnight basalt]

Where did the sin go

[fluid sorrel]

instead $\sin^2(x)$ you write $1-\cos^2(x)$

[twin meteor]

Čeva18

[midnight basalt]

Ohh yea

So you have 1-cos^2x-cos^2x

Which is 1-2cos^2x

[fluid sorrel]

Yeah!

[midnight basalt]

The thing is

How would you manipulate

This to get

[fluid sorrel]

You got this from $\sin^4(x)-\cos^4(x)$

[twin meteor]

Čeva18

[midnight basalt]

Ohh

[midnight basalt]

OH THAT MAKES SENSE NOW

[fluid sorrel]

And used this to get more simple factors.

[midnight basalt]

Hm...but wait

So if I have

(sin^2x+cos^2x)(sin^2x-cos^2x)

That would become

[fluid sorrel]

this is like cheat code, you can you it any time

[midnight basalt]

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

[fluid sorrel]

No.

[midnight basalt]

Oh its used a lot

[fluid sorrel]

Because you forgot + in one bracket.

[midnight basalt]

Oh I meant + and then minus

[fluid sorrel]

$(1-cos^2x+cos^2x)(1-cos^2x-cos^2x)$

[twin meteor]

Čeva18

[midnight basalt]

(1-cos^2x-cos^2x)(1-cos^2x+cos^2x)

[fluid sorrel]

yup

[midnight basalt]

And that would

[fluid sorrel]

and you get same thing

[midnight basalt]

Would I have to expand that out?

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

1-2cos^2x = 1-2cos^2x

WE PROVED IT

Well I should say you proved it

[fluid sorrel]

YEEEEA

[midnight basalt]

Tysm!!

THANK YOU

[fluid sorrel]

no problem

🙂

[midnight basalt]

Omg this cleared up so much things

[fluid sorrel]

i am glad

my english is not so good so i hope you understand me well

[midnight basalt]

It's amazing. I understood everything. I hope my yapping didnt make it annoying

Thank you again though!

/close

[fluid sorrel]

😉

[midnight basalt]

Help- whats the comamnd to close it

.close

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[worldly gazelle]

I need some help with this

[tawny nacelle]

start with the definition of a supremum

[worldly gazelle]

I've looked at the definition but I don't understand it

[tawny nacelle]

okay, do you know the definition of an upper bound?

[worldly gazelle]

An upper bound is greater or equal to every element in the list

[tawny nacelle]

in a set of real numbers, yes

so the supremum is the least upper bound to a set of real numbers

this means that, if A is a bounded set of real numbers, the supremum is:
a) an upper bound of A
b) smaller or equal to than every other upper bound of A

your goal is to prove that the number 8 satisfies a) and b) for the set (3,8)

[worldly gazelle]

I can see that 8 is the upper bound but what the method is for proving it

[tawny nacelle]

use the def of an upper bound

if k is an upper bound for the set A, then k >= x for all x in A

[worldly gazelle]

So would just saying "8 is an upper bound for the set (3,8) as 8>=x for all x in (3,8)" be enough?

[tawny nacelle]

pretty much

by definition (3,8) contains all real numbers x such that 3 < x < 8

so 8 >= x for all x in (3,8)

that's enough to prove a)

[worldly gazelle]

How do you prove b)?

[tawny nacelle]

take an arbitrary upper bound of (3,8), and show that it must be greater than or equal to 8

[worldly gazelle]

No idea how 😢

@tawny nacelle

Can someone help with this <@&286206848099549185>

[blissful sentinel]

What are the upper bounds of (3, 8)?

[worldly gazelle]

8?

[blissful sentinel]

That's one upper bound

There are others

For example 9 is also an upper bound

Because it's greater than every number in (3, 8)

What's the set of all upper bounds?

[worldly gazelle]

Not sure

[tawny nacelle]

ah, my bad lol

[blissful sentinel]

Try some other numbers

[tawny nacelle]

I had to go do smth

[blissful sentinel]

Is 5 an upper bound?

[tawny nacelle]

looks like Eric is here now

I'll let him take over

[blissful sentinel]

no wait

@tawny nacelle come back you do it 😔

[tawny nacelle]

why

you're doing fine

[blissful sentinel]

I'm doing Duolingo

[tawny nacelle]

oh LOL

[blissful sentinel]

@worldly gazelle ^^

[worldly gazelle]

No

I think

[blissful sentinel]

Yeah, why is it not?

[worldly gazelle]

It's less than 8

[blissful sentinel]

yeah, what about 8.5? is that an upper bound?

[worldly gazelle]

8.5>=8 so yes

[blissful sentinel]

yeah, so what's the rule for figuring out whether something's an upper bound of (3, 8) or not?

[worldly gazelle]

It needs to be greater than or equal to 8

[blissful sentinel]

yeah, so the set of all upper bounds is what?

[worldly gazelle]

Not sure

[blissful sentinel]

well you just said the rule for figuring out whether something's an upper bound of (3, 8)

so you can use that to figure out the set of all upper bounds of (3, 8)

[worldly gazelle]

Right

{x>=8}?

[blissful sentinel]

yup, another way to write that would be {x ∈ ℝ | x ≥ 8}, or [8,∞)

so now the supremum is the upper bound which is ≤ all other upper bounds

if [8,∞) is the set of all upper bounds of (3, 8), then what is the supremum of (3, 8)

[worldly gazelle]

∞?

[worldly gazelle]

Wait the supremum is less than or equal to all other bounds. So 8??

[blissful sentinel]

yes exactly

8 is in [8,∞) so it's an upper bound, and 8 is ≤ everything in [8,∞)

so it's a supremum

so basically the workflow here is

1. prove that [8,∞) is the set of all upper bounds of (3,8)
2. use this to prove that 8 is the supremum of (3,8)

[worldly gazelle]

Great thanks a lot

.close

[vocal sleet]

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[vocal sleet]

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[worldly gazelle]

Can someone help?

[worn bobcat]

Every x in R is either in (-inf, -3], (-3, -2], (-2, 6] or (6, inf)

and in each of those cases the sign of the first function is quite clear

except on the boundaries where the function is zero

The exact look of the solution probably varies a lot by region but it's a very standard problem

you should probably check the model solution and imitate it

The idea is that a continuous function only changes sign where it's zero

and you can split those functions up into a product of simpler functions

[worldly gazelle]

This is the solution for the 1st one but I'm struggling to understand how the number of negative factors has an effect.

[worn bobcat]

yeah I do think the way I was taught this in high school was a lot more instructive tbh

unfortunately it would require me to draw a diagram

This was the method we used

https://youtu.be/qLOy75XKgoE?si=BcJVGh-FpOPd_es4

oh no the video is awful

arghhh why is everyone not given equal access to high quality math education what a headache

[worldly gazelle]

Yh was about to say that it might just be that my uni is sh1t

[worn bobcat]

this is okay

not quite how we did it but it's okay

the whole pdf can be found here

https://webpages.charlotte.edu/~hbreiter/m1100/testinterval.pdf

So one way to see it is that you divide up the number line into segments based on the zeros of the function, and on each segment you take a test point, and the sign of the function evaluated at that test point is the same as the sign of the function over that whole interval (excluding the boundary where it's zero)

And of course to figure out the sign of the function, you count up how many of the factors give a negative result and if the number is even the negative signs cancel out and otherwise they don't

[tawny hollow]

It’s pretty neat how systematically you can solve this with a bit of logic, for i) you can for example replace each factor with some dummy variables to simplify the process,

so we have the inequality abc >= 0, when is this true for real numbers a,b and c?

[worn bobcat]

(why does that work? inside a given interval, none of the factors become zero, and therefore the value of the function is not zero. and a continuous function cannot change sign without passing through zero)

[worldly gazelle]

I'm a bit confused by the 4th one. How do I find the zeros?

[tawny hollow]

Move the 3 over and simplify

Using basic logic you could actually go ahead and solve it without moving the 3

[worn bobcat]

The whole method relies on the inequality being of the form

something > 0

so if it's

something > 3

you first move the 3 to the left hand side

After that you proceed as normal. You could find the zeros laboriously by solving a quadratic equation, or you could simplify the sum of two fractions

,w simplify x/(x-1) -3

[twin meteor]

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[worn bobcat]

Presumably you know how to simplify to get the first form

and after that, you know the only way for that rational expression to be zero is if the numerator is zero

meaning 3-2x=0

and you should keep in mind that x absolutely cannot be equal to 1

[tawny hollow]

just for the sake of showing what i mean,

we have the inequality x/(x-1) > 3,
multiplying both sides by x-1 we equivalently have

[ x > 3(x-1) and x-1 > 0 ] or [ x < 3(x-1) and x-1 < 0 ]

we can solve these two cases separately and then return to simplify further

[worn bobcat]

you're missing the x=1 case

[tawny hollow]

nope

[worn bobcat]

oh yeah

lol

[worn bobcat]

XD

[tawny hollow]

:P

[worldly gazelle]

I got the first one. Where did the other 2 come from? 😢

[tawny hollow]

this way of solving inequalities really motivates classes you have in logic or set theory imo, i thought it was useless until i started applying it

[worn bobcat]

You get the last form by multiplying by (-1)/(-1)

so is that like expanding by -1? I don't remember the terminology

[worldly gazelle]

What about the 2nd one

[worn bobcat]

They added a clever zero in the numerator to get a nice clean -2

kind of an arbitrary choice imo

sorry

I was wrong

The second form is

(1-2(x-1)) / (x-1)

So clearly what they did

is they turned 3-2x into 1-2(x-1)

I'm not sure what the general technique they're using is

or what the "goal" is

but I personally don't find it too interesting

ohhhh

it might be that they reduced by x

to get rid of the x in the numerator

[tawny hollow]

What are we looking at?

[worn bobcat]

giving

1/(1/x - 1) - 3

and then they just simplified that

idk why I say "they" when it's just a computer program lmao

[worn bobcat]

the second simplification here

[tawny hollow]

Oh

[worn bobcat]

wait no I could be wrong

anyway kind of a stupid problem in my opinion to try to reverse-engineer the behavior of that simplification algorithm based on that

I won't use any more of my precious mental resources on that

[tawny hollow]

lmao

[worldly gazelle]

Found this in the lecture notes

[worn bobcat]

YESSS that's the way I was taught it

[tawny hollow]

whats neat about solving stuff like this is that it nicely covers absolute values of stuff or even equations, its sort of built in

you can solve them systematically without worrying much about a diagram or whatever

[worn bobcat]

I mean our table was slightly different still but that's as close as we're gonna get without me picking up my drawing tablet

one thing you can notice is that it's kind of unnecessary work to compute the sign of the functions at a point where any of the factors is zero

because the value of the function will always be zero in such a case

[tawny hollow]

stuff like this requires you to almost always have everything factored, which sometimes is not possible

[worn bobcat]

If you need to determine the sign of a function that is the product of simpler functions, a chart like that will always be helpful

if one of the factors is a difficult function (one that "cannot be factored" for example) then you may need to do some special analysis

but that's never gonna happen in school problems

because if your function is continuous and you can figure out its zeros then you can just take test points

and include the zeros in the diagram

[worn bobcat]

this diagram is GOOD

holistically puristically worthwhile to learn

[tawny hollow]

well it depends what it's for imo, its a shame because using logic sort of neatly puts this all into a single chorent system and you start grasping it a lot better later on for proofs aswell

[worn bobcat]

The diagram is just a shorter way to write down a bunch of "logic"

[tawny hollow]

and students have a hard time anyways enjoying a logic course

[tawny hollow]

sure, but it does not aid them in the future

[worn bobcat]

You seem to be confusing "logic" with "logic notation"

[tawny hollow]

well not really

i didnt use v for or

[worn bobcat]

Forcing someone to write a simple idea in painful logical symbols is not automatically a good idea in every situation

[tawny hollow]

i explicitly used the word

[worldly gazelle]

Lol

[tawny hollow]

Well that's not what im doing at all, where did i use logic symbols?

Im just pointing out the fact that later on youll realize that you can systematically do all of this by understanding the way logic works in mathematics

[worn bobcat]

Well the only example that you've given of "using logic" is this, and my brain extrapolates that to assume that when you say "use logic" you mean "write out the solution in a painful way"

[tawny hollow]

and it will help a great deal for proofs too

[worn bobcat]

for example I would consider this model solution unnecessarily painful even though it could be much, much worse

[tawny hollow]

While yes painful at times, it's a better option for more complicated looking statements

Im simply offering a method to solve them systematically

you dont have to use them

[worn bobcat]

Because you can take this idea and write a really long and painful proof of the inequality

...or you can draw a beautiful chart that illustrates the same line of reasoning

[tawny hollow]

That's subjective

[worn bobcat]

The chart is a way to solve these problems systematically once you understand what the chart is really saying

[tawny hollow]

Unless of course you dont have it factorized..

[worn bobcat]

In which case you can have a chart with just one row?

it sounds stupid but even in that case I think the chart is a helpful conceptual shorthand

if your method is "find the zeros and appeal to continuity"

and if you can't use that method then I would like to hear an example of such a problem that comes up

[tawny hollow]

Sure i'm not at all trying to argue that, if you find it useful go ahead; i just find it a great way of motivating the upcoming proof courses one might have and the need to understanding the framework used, one way to motivate that is to apply it for solving inequalities or equations

[worn bobcat]

At the risk of sounding discouraging (towards dapper) while I do agree that there's a lot of value in viewing inequalities in all kinds of technical ways, I think that's only helpful after you've mastered the basic conceptual ideas like if you have an upward facing parabola then its sign goes + - +

That way you can solve many kinds of inequalities effectively, if need be

And I think the chart is the best way to conceptualize the central idea that you figure out the sign of the function one segment at a time

[tawny hollow]

basic conceptual ideas like if you have an upward facing parabola then its sign goes + - +
I actually had trouble with this and found it (for me) alot easier to understand it on a more technical and visual-free sense

it's all subjective really

i found it a really useful tool, and thats that really

[tawny hollow]

I was uncomfortable with the idea that if i ever can't graph something or write the right signs for a possibly complicated expression, then i had no other way of solving it

[worldly gazelle]

The chart seems more intuitive to me personally but I see where @tawny hollow is coming from

[tawny hollow]

That's perfect!

[worn bobcat]

The chart doesn't make you draw any graphs

and there's no method that I know of other than the chart to solve inequalities involving polynomials

[tawny hollow]

I wasnt solely talking about the charts above, but just when doing stuff in-general theres an idea to try and picture everything

[worn bobcat]

well I guess you can throw in absolute values and then stuff might get more complicated

[tawny hollow]

well, "my method" would be another one

[worn bobcat]

But when it comes to absolute values, you need to figure out the sign of the expression inside the absolute value

How do you do that? Chart!

And after that? There are no more absolute values

Then? Another chart!

The chart is just a very visual way to solve a sub-problem of a problem that involves absolute values

[tawny hollow]

One big difference here is that i dont even need to care about signs (in the sense you do), what if you have an expression that can be simplified alot better if you just dont toss everything to one side; well then i can safely do that

[tawny hollow]

Ironically i think pictures are probably the most powerful tool

[worn bobcat]

I doubt such a problem exists

[tawny hollow]

but pictures dont need to be phyically drawn pictures

they could be very abstract ideas

[tawny hollow]

i dont have any off hand but it sure has lol

how would you solve inequalities involving more variables? (whatever that means)

I think one pretty neat example is unpacking defintions,

[worn bobcat]

well we know that the solution set to

x^2 + y^2 <= 1

forms a unit disk

And the way we know that is by appealing to the pythagorean theorem and geometric intuition

One can imagine that in general it becomes very difficult to describe the solution set in a "reasonable" or "precise" sense

[tawny hollow]

in say real analysis, say you want to see if an element is in the union then it falls out almost for free sometimes by just following the logic

[tawny hollow]

That's the power of seeing things more logically, you can sometimes avoid the problem of "knowing" what they describe, but once you say know that it's the unit disk for example we're dealing with it's often times easier of course doing it the way you reasoned

[tawny hollow]

and real analysis is a great examples, because the intuition fails us often

[worn bobcat]

What I'm saying is that "solving a multivariable inequality" makes quite little sense for the reason that I outlined (what counts as a "reasonable representation" of the solution set is completely subjective), no amount of logic will help you out

but it's still helpful to be aware of the "unit disk" interpretation

(and needless to say I didn't arrive at the unit disk conclusion by any logical argument)

[tawny hollow]

Well not quite the point i was trying to make. If you need to deduce something and say you dont even know how to think about the object intuitively, then youll have a problem because what else are you supposed to rely on to make any sort of reasonable (mathematical) deduction about the thing youre studying; thats where logic comes in

[worn bobcat]

If you don't know how to think about an object then you can play around with it until you do know how to think about it

and quite often the way one thinks about something is very different from the way one ends up writing it all down in a research paper

[tawny hollow]

Yeah via logic for example

[worn bobcat]

which I see as a problem in mathematics

[tawny hollow]

So how would you go about if you some object in say a really weird space?

Maybe this is starting to become a bit unrelated to the topic, but i think its a good discussion to have

[worn bobcat]

Well let's say I was given the definition of the determinant as the sum of terms formed using permutations and their signs

If I was to try to understand why swapping two columns changes the sign, one of the realizations I would probably eventually make is that a transposition is always an odd permutation

And the way I would presumably prove it is by imagining the transposition as a composition of adjacent transpositions

And all of that would involve a lot of geometric thinking (I don't know if that's obvious? it's hard to give an example because I'm communicating via text) that I would eventually write down in more or less rigorous mathematical notation

it's funny speaking in hypotheticals when this is something that I've actually gone through

[tawny hollow]

Interesting! Linear algebra is one of the few areas where i found this to be the opposite for me, and where i was able to almost symbolically go through something or prove it, the geometric viewpoint was still important at times for me and as i mentioned above I believe pictures in some sense is the most important tool, but i think its flawed to think this has to be literal images, there's so many different ways we can abstractly think or reason about it! And one way would often times for me atleast be logically

[worn bobcat]

well when I was younger I had a 3blue1brown phase where I thought math was all about pretty pictures and the symbols were just distraction

but then as you actually study math, you realize that you need to play around with symbols a lot or otherwise you can't really do anything besides wave your hands, and I started getting really frustrated at some of my friends who in my view would skip all the good math in favor of easily digestible handwavy explanations

so something I've realized is that rigorous math is important and reading proofs carefully is important

however, you still have to "bring those symbols to life" in your head somehow, and that's where I distinguish between good kinds of visual aids and handwavy nonsense

Like I know exactly how to translate a sign diagram into a "formal argument" if need be, but in practice I prefer working with sign diagrams because it's just a nicer way to express the underlying ideas

than something like

"Let us consider these sets that partition the real number line. For each x in this set, we get this and this and blabla..."

like that's just sadistic nonsense to be frank

cool the first time you see it

but not really helpful to anyone in practice

[tawny hollow]

In some sense i think math is all about the pictures, but we might have different interpretations of pictures here. To come back to the charts, i think its wonderful you have a way of working with it and I realize now that it seems we both share that comfort but just for different looking methods!

And in this case Dapperjaguar197 (i hope youre not reading all of this lmao) shared that comfort with your method!

[worn bobcat]

I like the word comfort because for me math in the recent years has been all about bringing me comfort from the cruelty of the outside world tbh

I see the sign chart as "moving up in abstraction", you're sacrificing being precise in favor of being understandable

but as long as you know how to convert a more abstract argument into a more precise one if needed, I don't see the issue

Abstraction is at the heart of mathematics

it's the very soul of the art

[vocal sleet]

@worldly gazelle Has your question been resolved?

[tawny hollow]

Tao once named three crude stages of maturing in math, i think I'm probably in the middle stage, where i might be rigorous (eh) and all but I'm still lacking that ability to be comfortable about being understandable as you put it; while knowing how to rigorously convert it later on

So for me atm i really like working with the thing very precisely/directly (or so i belieave lmao)

[worn bobcat]

Wow I haven't read about Tao's three stages before but that makes a lot of sense to me

[tawny hollow]

It surprised me how accurate they were!

[worn bobcat]

My experience has been that you're giving yourself a very serious handicap if you don't allow yourself to be ruthlessly imprecise and schizophrenic at a whiteboard before finally sitting down and trying to convert the crazy ideas into formal arguments and constructions, and then you realize that you overlooked something and go back to the whiteboard

[tawny hollow]

Hopefully I'll learn to do that better, when dealing with courses like real analysis i have a really hard time being imprecise to advance, but I totally see the practically of it; but i often miss something, but it seems you just need alot of experience before you can do this

Thank you btw for sticking out, this is a fun discussion. I'm soon starting 2nd year in undergrad so I'm not really that far into my education; and still have yet to learn

[worn bobcat]

Haha I've been enjoying it too

I just completed my third year and I'm 98% done with my bachelor thesis that I went completely overboard on

[tawny hollow]

Oh congrats!

[worn bobcat]

I feel like I never really understood what math was until I tried to figure stuff out on my own

and I realized that I both love and hate it

[tawny hollow]

Hm, like working with proofs on your own for example?

or even more rudimentary stuff?

[worn bobcat]

Well I was tasked with giving an exposition on a certain result in commutative algebra and me and my advisor realized very quickly that the way the theorem is taught in the few textbooks that there are on the subject is not very good. And once you start really thinking about it, the rabbit hole goes pretty deep. And I ended up generalizing matrix multiplication in a pretty satisfying way. Tbh I don't actually know whether my work is new or not, I mean, probably not, but I discovered it all independently which I'm very proud of

[tawny hollow]

That sounds really cool

[worn bobcat]

Basically I formalize the idea of having let's say functions in matrix A and let's say vectors in matrix B and then defining the matrix product AB as something where the "multiplication" is plugging in the vector into the function

And that allows for a pretty nice proof of the Cayley-Hamilton theorem for example

[vocal sleet]

Channel closed

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[worn bobcat]

rip

[tawny hollow]

Rip!

[worn bobcat]

well it was a good yap

[tawny hollow]

Hm, cool!

[tawny hollow]

Indeed :P

[vocal sleet]

Available help channel!

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[vital wave]

What is R?

[azure spade]

As in R_n?

[vital wave]

no just the R in the picture

yea maybe its that

[azure spade]

It's the area of the rectangles

Yes, you're finding an expression for the area under x^2 * exp(x)

[vital wave]

i dont get how to do it

does n = 4?

so its like A = lim n -> 4 [f(x1)delta x + .... f(xn) delta x?

or do i need to write in the equation instad of just x1

[vocal sleet]

@vital wave Has your question been resolved?

[vital wave]

still there?

[azure spade]

Yes, sorry

Yeah, your f(x) is your function x^2 * exp(x)

$\qty(x_1)^2 \times e^{x_1}\Delta x$

[twin meteor]

@azure spade

[azure spade]

and so on

[vital wave]

as n -> 4?

cause it has bounds

[azure spade]

Yes

[vital wave]

ok so i lterally just write that?

[azure spade]

You can summarise in a summation too

[gaunt sparrow]

n goes to infinity though

[vital wave]

that would fit better in my answebox

[gaunt sparrow]

Not 4

[vital wave]

y inf... its bounded from 0 <= x <= 4

[gaunt sparrow]

x is bounded.

n is the number of subintervals

[azure spade]

^

[vital wave]

oh rly

Take a look at this

How does it look

i added + ... + in between

instead of just + f(x n)

[gaunt sparrow]

Just (x_i)^2

Not f(x_i)^2

f(x) = x^2 e^x

[vital wave]

ooo

[gaunt sparrow]

So you're putting f(x_i) in there.

[vital wave]

not x_1?

[gaunt sparrow]

Yeah well f(x_1), f(x_2), and so on

[vital wave]

aright yea

so why do they give 0 <= x <= 4

dont i need to put that somewhere>

For example they all have ranges for x

@gaunt sparrow so wouldnt that matter?

[gaunt sparrow]

I mean you wrote it down with $\Delta x$, but this delta x depends on n and the length of the interval

[twin meteor]

Azyrashacorki

[vital wave]

oh im confused

wait so we dont know that info basedon this

so why did they give us the ranges? with no N?

the endpoints are from [0 ,4] ?

but no n? im guessing n = 4

so delta x is 1?

[gaunt sparrow]

No.

[vital wave]

oh

[gaunt sparrow]

Delta x is the length of the subintervals

As you have more subintervals, this length gets small

[vital wave]

yea but it doesnt say

and whyu are they giving us the intervals then?

[gaunt sparrow]

$\Delta x = \frac{b-a}{n}$

[twin meteor]

Azyrashacorki

[vital wave]

i think its cause they want us to use that and start with n = 4

[gaunt sparrow]

No.

[vital wave]

so what do you think n is

[gaunt sparrow]

If 0 <= x <= 4, then it just means that b = 4 and a =0

[vital wave]

but think about it why do they give us that if they dont give us n?

[gaunt sparrow]

They wouldn't give you n

Because you're taking the limit

n is the number of intervals

The point is you're taking thinner and thinner rectangles using the right endpoints

[gaunt sparrow]

AS n grows large, delta x is small

[vital wave]

so why are they giving us 0 <= x < =4

then why do i need that information

[gaunt sparrow]

Because this is a = 0 and b = 4

[vital wave]

like why is tha important for my answer though you kno?

[gaunt sparrow]

So you're dividing the interval [0,4] in n subintervals

Yielding delta x = (4-0)/n

[vital wave]

so the answer stays the same as this?

like what about for 19?

[gaunt sparrow]

It's the exact same thin, but the interval is [1,5] instead.

[vital wave]

yea but do i need the interval in the answer?

[gaunt sparrow]

You'll want to rewrite delta x as (b-a)/n here probably. Otherwise it doesn't really translate the fact that it gets small

[vital wave]

ok

I don’t really get it I’ll just go to tutoring tomorrow

Thanks though

.Close

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[vocal sleet]

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[vocal sleet]

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[waxen hawk]

Isn’t it

[waxen hawk]

I thought this could be solved by squaring the both sides then applying the discriminate

I got a singular result by doing so, and realized that cannot gain me the correct answer.

What’s wrong with my method?

[sharp lily]

you can just x with zero and solve

replace x*]

[waxen hawk]

That’s not my question

[sharp lily]

squaring the root will give you a +-

its just annoying, and im pretty sure you want to only work with variable k so replacing x with zero immediately just makes things easier

[vocal sleet]

@waxen hawk Has your question been resolved?

[waxen hawk]

How it works tho

[mellow void]

did you account for cases where there's only one solution, because the other (in the quadratic you get from squaring both sides) makes the right-hand side sqrt(-junk)?

there's a chance that could happen

[sudden falcon]

did you try 9?

[waxen hawk]

How it works

[mellow void]

@waxen hawk sorry I was slightly wrong. you want x=0 as a solution, so you probably just substituted 0 for x and solved for k. but then in checking that 0 is the only solution to the above for this choice of k, you may find more than just 0 as a solution

but that's only in the quadratic; after checking in the original with the second solution, it doesn't work

hence x=0 is still the only solution

[waxen hawk]

How do I know x=0 would be the only solution by doing so

[mellow void]

uh, by checking the other solution you get from the quadratic, like I said

[waxen hawk]

I see

[waxen hawk]

Which type of animal are you?

[mellow void]

it's a goose

[waxen hawk]

Ohh, tasteful

[mellow void]

it was a cockatiel for a while but it got old

yellow is fine but mid

[sudden falcon]

??

[short pilot]

Uo

Yo

Here for help

3+5x=√(k+6x

Ok

For k=9 right

[sudden falcon]

i mean that's what i got

[vocal sleet]

@waxen hawk Has your question been resolved?

[vocal sleet]

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[vocal sleet]

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[reef isle]

Can anyone help me express this series in sigma notation?

[reef isle]

For the life of me I cannot.

The only thing I can easily pick out is ofc the (-1)^n but The 2/3 throws me off

[neon plover]

Im trying

I cant find it either 💀

[oak magnet]

Ive think about factorial but kinda hard

[neon plover]

i assume it starts at n=2

dawg what is this pattern

this is actually impossible

this one is over boys

[hushed pewter]

I see the pattern

[neon plover]

you found it?

im still triyng

[hushed pewter]

It's not a pretty pattern

But I'm quite sure it's right

[neon plover]

care to tell?

ill try it out

[reef isle]

I think there is a -2^(n+1) ?? in the numerator

[neon plover]

i tried that

but it wouldnt fit

it just doesnt

[reef isle]

I've probably spent like 3 hours on this lol

[neon plover]

thats tough

[reef isle]

I would normally just write a bunch of terms instead of summation notation, but i need that notation for some frobenius method problem

[hushed pewter]

Yeah I found it

[reef isle]

Wow

How?

[neon plover]

wow

bro is keeping it a secret

[reef isle]

Thats what im saying

https://tenor.com/view/frog-king-shrek-gioated-gif-22410043

[neon plover]

https://tenor.com/view/cat-kitten-spilling-milk-milk-spilling-gif-25553835

[reef isle]

I need you SWR

[hushed pewter]

Yes

[neon plover]

god has returned

[reef isle]

Please sir

I got n from 1 to 10 written down in front of me and im just staring at it

[neon plover]

wait

where

[reef isle]

im saying i cant figure it out lol i just got all the expanded terms divided into multiples lol

[neon plover]

what is bro typing

[reef isle]

i can see the pattern but its so hard to express it for me

[hushed pewter]

$y=\sum_{k=0}^{\infty} \frac{(-1)^k x^{k+2}k!(k+2)!}{2^{k+1}}$

[twin meteor]

SWR

[neon plover]

holy

he cooked

[reef isle]

wow

thank you

[neon plover]

holy crpa

its right

how did you even see this??

also i think its upside down

the coefficient terms

[hushed pewter]

You are right.

Oops

[neon plover]

can you explain how you saw this?

or is it just experience

cause damn man

[mellow void]

damn I wish I could've made my lagrange polynomial joke answer sooner lol

[hushed pewter]

I took ratio of successive terms, then I liked by how much the ratio changed

[neon plover]

k!(k+2)! is crazy to see

intuitively

i could never honestly

[hushed pewter]

Nah like I looked at the numbers of the ratio

Then compared their differences

And the differences kept increasing by 1

[mellow void]

sad, I also did that but failed to recognize anything useful

[neon plover]

which difference?

[reef isle]

Wow

[neon plover]

increased by 1

[hushed pewter]

So then you need the 1+2+...+n formula

[neon plover]

i tried that too

[reef isle]

This was a good 2-3 hours for me lol

[hushed pewter]

Then that's how much the ratio increased by, which means I need to do product of that

Which led to factorial

Another approach I was considering was taking second derivative and see if that led anywhere

Not sure if it would have, but it was another considering

Maybe even third derivative

[neon plover]

derivative of what 😭

[mellow void]

of the power series itself

[neon plover]

we dont got the power series though

oh wait

nvm

im dumb

[hushed pewter]

Actually yeah. Derivative would have been the better play

[reef isle]

Really?

[mellow void]

I tried it but I failed to see the pattern still

[hushed pewter]

Maybe

[mellow void]

2, -4, 2, -4/9, 1/18, -1/225?

[reef isle]

Well I can finally get my second frobenius solution. Thank you so much @hushed pewter

Cute cat pfp too

[hushed pewter]

Normally I don't give answers like this, but I gotta rush to bed

[hushed pewter]

Multiply by (k!)^2 where k starts at 0

[mellow void]

I should be more hopeful when digging for gold, I think

good night

[vocal sleet]

@reef isle Has your question been resolved?

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[vernal flame]

how do i prove this??

[rugged orchid]

what theorems have you come across that you may be able to use?

[vernal flame]

a sequence diverges if there exists some epsilon >0 such that for every N there exists some n>N such that |An-a|>=Epsilon

[brisk moss]

you need to do more than just show the sequence diverges

[vernal flame]

it's also always greater than 0?

[brisk moss]

you need to show it “diverges to infinity” if you will

[brisk moss]

no

it doesn’t need to be

[vernal flame]

oh yeah...

[brisk moss]

a_n, b_n, and a_n + b_n could all have negative entries and this still make sense

[vernal flame]

yeah

ok how about
for every AN there exists n>=N st. An>=AN

[brisk moss]

what

that means A has a subsequence that’s increasing or something

i don’t see any relevance

[vernal flame]

i guess i just think if it has a subsequence that's monotone increasing then it's not bounded above

wait no

yeah i'm wrong

[brisk moss]

what are you trying to do or say right now?

[vernal flame]

definition for diverges to infinity...

[brisk moss]

oh

that means for every real number M, there is a positive integer N such that for every n >= N, a_n > M

for every “bound” M, the sequence is eventually (namely after N) above the bound

[vernal flame]

oh that makes sense

wait bruh how did i not think of that

death

ok how do i close a channel....

.close

[vocal sleet]

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[untold arrow]

Consider an ( n \times n ) matrix ( A ) with ( n \geqslant 2 ):
[ A = \begin{pmatrix}
a & 1 & 1 & \cdots & 1 \
1 & a & 1 & \cdots & 1 \
1 & 1 & a & \cdots & 1 \
\vdots & \vdots & \vdots & & \vdots \
1 & 1 & 1 & \cdots & a
\end{pmatrix}. ]

(I) Find an invertible matrix ( P ) and a diagonal matrix ( D ) such that ( P^{-1}AP = D ).

(II) Find ( r(A^*) ).

[twin meteor]

nino

[untold arrow]

I dont know how to handle this matrix

[mild flower]

how would you do it for 2x2?

[untold arrow]

just find its eigenvectors

[mild flower]

sure

why can't you do that here?

[untold arrow]

[mild flower]

do it for 2x2 and 3x3 and pay attention to their similarities

the similarities in the process, not necessarily the result

[mild flower]

you know, I did say to do that yourself; I didn't say to make a machine do it

at any rate

what happens with 3x3?

[untold arrow]

it's a bit complex to do the char poly of 3×3

[twin meteor]

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

[untold arrow]

$(\lambda-1)(\lambda+1)(\lambda-2)=0$

[twin meteor]

nino

[vocal sleet]

@untold arrow Has your question been resolved?

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[safe spindle]

this is a short question

when i calculated the eigenvectors, i found (1 0 0) and (0 1 0)

they have (1 0 0) and (1 1 0)

i cant tell if im wrong or they're wrong

or if its technically the same?

yea if anyone can clarify that would be great

[vocal sleet]

@safe spindle Has your question been resolved?

[mellow void]

A(u+v) = Au + Av = mu + mv = m(u+v); your eigenvector justifies theirs with u = (1 0 0) and v = (0 1 0)

but indeed (0 1 0) is in the span of {(1 0 0), (1 1 0)}

so it's all the same

[safe spindle]

i see

thanks!

.close

[vocal sleet]

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[safe spindle]

in this question they gave a counter example

but i dont know how to come up with a counter example in the exam

does anyone have any tips

or is there a way to disprove this without a counter example

i am at a loss cuz i cant think counter examples up on the spot

[finite flax]

what's eigenvalue

do they mean determinant?

[left sparrow]

do you know the relationship between eigenvalues of T and eigenvalues of T^k

[safe spindle]

isnt it that if T has an eigen value lambda, then t^2 has an eigenvalue lambda^2

[left sparrow]

exactly

so what we want is a matrix with eigenvalue l, such that l^2 = 1

I can think of two such numbers: namely, 1 and -1

[safe spindle]

AH

yea thats pretty simple

i confused myself

lmao thanks for walking me through it

[left sparrow]

👍

[safe spindle]

.close

[vocal sleet]

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[vast shale]

Um hello I need help on questions number 4 and 5

[vocal sleet]

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

[vast shale]

Tbh I don’t know where to begin

1

[drifting jackal]

For question 4, it's asking about point P, where is it located?

[vast shale]

Yup

[drifting jackal]

Where is it located

[vast shale]

Negative 3?

[drifting jackal]

No I mean it's on an axis, correct?

Which one?

[vast shale]

Y axis

[drifting jackal]

So knowing that it's on the y axis, that means one of the coordinate values is 0, which one? The x or y value?

[vast shale]

X axis.

[drifting jackal]

X value is 0, right
So that means you know it's going to be (0, ___)