help-28

[stone mortar]

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[stone mortar]

!chatgpt

<@&286206848099549185>

[spiral vigil]

what else is on that page

[stone mortar]

[spiral vigil]

wow look an explanation of how to do the problem

[stone mortar]

i dont understand how in the 6th line, LHS = RHS

[stone mortar]

[spiral vigil]

it lets us convert the tilted surface integral to an integral on the ground (x and y)

[stone mortar]

is that some theorem?

if so why
therefore, the statement
and not
according to the theorem, the statement

[elder birch]

info

which part do you not understand?

[stone mortar]

how did the above 5 steps implied step 6

[spiral vigil]

you're projecting the vector n onto the k direction

[elder birch]

do you understand up to here?

[stone mortar]

n

o

can you please explain whats going on here

[elder birch]

sure

do you see how gradient of the plane is normal to the plane?

[stone mortar]

yes

[elder birch]

do you understand how surface integral with dot product (flux type integral) works?

[stone mortar]

actually i also have a confusion at this, cuz differentiating a line gives tangent but for a surface, its normal

[elder birch]

do you remember the most general equation of a plane in 3D?

$\vb n\vdot\vb r=0$

[glossy valve]

reverileo

[elder birch]

actually let me ask this first

given a plane with an equation
[ a(x-x_0)+b(y-y_0)+c(z-z_0)=0 ]
do you see why the vector $(a,b,c)$ is normal to the plane?

[glossy valve]

reverileo

[stone mortar]

no

[elder birch]

have you seen this equation?

so are we clear with that?

[stone mortar]

but still can you explain why is this so

[elder birch]

.

[stone mortar]

yes

[elder birch]

do you understand how it's derived?

[stone mortar]

is an equation of a line in 3d

i do for 2d,... so must be similar for 3d

no

[elder birch]

https://youtu.be/E2jbv-cXqXY

could you watch this video and come back

it's really short

[stone mortar]

yes sure

[elder birch]

equation of plane is hard to explain without visualization

[stone mortar]

hm

done

[elder birch]

now do you see this?

[stone mortar]

yes

[elder birch]

in fact, for this question, you don't even need gradient to find normal vector

since the surface is already a plane

do u still want me to explain why gradient is normal to the tangent plane or no?

[stone mortar]

ahh

yes

no

i see it now

[elder birch]

u good?

[stone mortar]

yup

[elder birch]

alright so do u see how we can just take the coefficients of x, y, z of the equation of the plane

and get the normal vector right away

[stone mortar]

it will be left with A+B+C... which is perpendiculatr soo

[stone mortar]

yes

[elder birch]

now you understand up to these steps, right?

[stone mortar]

up to here 👍

I dont understand the next step

this

[elder birch]

because you want to scale down infinitesimal areas

[stone mortar]

ahhhh.........

[elder birch]

i'm trying to think how to explain this well

do you see how dS would be different from dx dy?

because the surface is sorta inclined

[stone mortar]

yes

[elder birch]

that's why you need a scaling factor

[stone mortar]

hmm

k.n

[elder birch]

they skipped steps but do you see why $\dd S=\frac{1}{\cos\theta},\dd A=\frac{1}{\cos\theta}\dd{x}\dd{y}$?

[glossy valve]

reverileo

[elder birch]

where $\theta$ is the angle between $\vu{k}$ and $\vu{n}$

[glossy valve]

reverileo

[stone mortar]

yes

its obvious

[elder birch]

essentially you want to scale based on how slanted (or inclined) the surface is compared to the xy-plane

do you see that?

[stone mortar]

hmm ya

so | k. n | is cos theta?

[elder birch]

$\vu{k}\vdot\vu{n}=\norm{\vu{k}}\norm{\vu{n}}\cos\theta$

[glossy valve]

reverileo

[elder birch]

right?

but k and n are unit vectors

so it's just $\cos\theta=\vu{k}\vdot\vu{n}$

[glossy valve]

reverileo

[stone mortar]

ohhhhhhhhh yes

[elder birch]

that's why you divide by k dot n

[stone mortar]

now i see it

nice

[elder birch]

do you think you can proceed on your own?

or do you need some help?

[stone mortar]

yes I think I can proceed on my own

thanks for the help

[elder birch]

yeah np

[stone mortar]

🥳 🥳🥳🥳

@spiral vigil

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[stuck fiber]

Part (d)

[stuck fiber]

Been a while since you hit me with the otter

I got some weird numbers

First we want the y intercept

so cos(2[0]) + sin(3[0] + pi/3)

cos(0) + sin(pi/3)

So 1 + sqrt(3)/2

Which is already a weird number

[honest hill]

why?

[stuck fiber]

like im guessing im going to have to sub this back into sin or cos at some point

And sin(1 + sqrt(3)/2) isnt going to work out nicely

Same with cos

[honest hill]

no?

[stuck fiber]

Oh

Ok well

We have the x coordinate now

of A

No?

[honest hill]

y

[stuck fiber]

Ah yeye

So (0, 1 + sqrt(3)/2)

We then want to get the slope of the tangent

at this point

So we get the derivative of cos(2x) + sin(3x + pi/3)

Which is -2sin(2x) + 3cos(3x + pi/3)

Then sub in x = 0

-2sin(0) + 3cos(0 + pi/3)

3(1/2)

=3/2

So thats the slope

We now have the line y = 3x/2 + (1 + sqrt(3)/2)

sub in y = 0

0 = 3x/2 + (1 + sqrt(3)/2)

3x/2 = -(1 + sqrt(3)/2)

x = -2(1 + sqrt(3)/2)/3

x = -2/3 - sqrt(3)/3

Yep thats what they wanted!! 😄

Thank you!

❤️

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[gleaming solstice]

help

[gleaming solstice]

Let $\mathcal{V}$ be the $\mathbb{R}$-vector space of all functions $f : \mathbb{N} \to \mathbb{R}^n$, where $\mathbb{N} = {1,2, \dots}$. Here the addition of functions $f,g \in \mathcal{V}$ is defined as $(f+g)(k) = f(k) + g(k)$ for all $k \in \mathbb{N}$, while scalar multiplication is defined as $(af)(k) = af(k)$ for $a \in \mathbb{R}$ and $f \in \mathcal{V}$. In this exercise we will consider the set

[
\mathcal{S} := { f \in \mathcal{V} \mid f(k+1) = T(f(k)) , \forall k \in \mathbb{N} },
]

where $T : \mathbb{R}^n \to \mathbb{R}^n$ is a linear operator.

\begin{itemize}
\item[(a)] Prove that $\mathcal{S}$ is a subspace of $\mathcal{V}$.
\item[(b)] Let $x \in \mathbb{R}^n$. Show that $f$, defined by $f(k) = T^{k-1}(x)$ for all $k \in \mathbb{N}$, is in $\mathcal{S}$.
\item[(c)] Find a basis for $\mathcal{S}$. Motivate your answer.
\end{itemize}

[glossy valve]

Emmaaaaa

[gleaming solstice]

Here is the problem

(a) $T$ is linear implies that $T$ preserves both homogeneity and additivity, since homogeneity preserves $\mathbf{0}$ element in its image showing ${0}\subset \mathcal{S}$, and since $T$ preserves the additivity and homogeneity, $\mathcal{S}$ preserves these property, hence $\mathcal{S}$ is a subspace of $\mathcal{V}$.

(b) $f(k)=T^{k-1}(x)$ which implies that $T(f(k-1))=T^{k-1}(x)$ this implies $T^k(x)=T[T(f(k-1))]=T[f(k)]=f(k+1)\in \mathcal{S}$

Conversely $x\in \mathbb{R}^n$ we have $x=T^{1-k}f(k)=T^{-k}f(k+1)=f(1)$ this implies $T^k(x)=T^k[f(1)]=f(k+1)\in\mathcal{S}$

But this we have demonstrated that for every $k\in \mathbb{N}_{\geq{1}}$ we have $f(k)\in \mathcal{{S}}$

(c):

Let $\beta_{i,j}=\bigg{f_{i,j}: f(k)=\begin{cases}
e_j,\ \ \text{if } k=i\
0, \ \ \text{if } k\neq i
\end{cases} \ \ \text{for } i,j\in \mathbb{N},\text{and }j\leq{n}\bigg}$ be a infinite dimensional basis for vector space $\mathcal{V}$ then for the subspace of $\mathcal{S}$ of $\mathcal{V}$ then basis of $\mathcal{S}$ can be represented by $\beta_{j}={f_{j}(k):T^{1-k}f(k)=e_j:j=1,2,3,...,n}$

My answer

[glossy valve]

Emmaaaaa

[gleaming solstice]

I know it’s a bit redundant but I am no good with linear algebra so I don’t know if converse is needed to be proven

And I actually don’t know if I can just write answer for a just like that since it’s just a few sentence and bcs of T is linear verification isn’t that necessary I felt but I don’t know

@leaden ermine help me this is very much more elementary compared to analysis

I made a mistake T might not be invertible

@leaden ermine do I have to show converse direction for B, I will use change of basis matrix then

[leaden ermine]

you are probably 99 % right (no guarantees)

[gleaming solstice]

Let $Tx = Ax$ where $A$ is the transformation matrix

Let $\beta$ be a nonstandard basis and a constant matrix $C=[\beta_1,…,\beta_n]$ and clearly $C$ is invertible such that $[x]_\beta=C^{-1}x$

Then $T[x]=AC^{-1}[x]beta$ this follows $[T[x]]\beta=CAC^{-1}[x]_\beta \in S$

[glossy valve]

Emmaaaaa

[gleaming solstice]

So is [T^k[x]]_beta=CA^kC^{-1}[x]_beya

This isn’t necessary right?

[gleaming solstice]

I actually feel these day out of 10 problem I try rarely I got one right though

Unfortunately

[kindred grove]

no you don't have to, not for b) at least

of course it's nice for c) if you know explicitly what's in S

[gleaming solstice]

This simple question cost me a day

[kindred grove]

that's where your converse could be useful

this maneuver will cost us 51 years

[gleaming solstice]

Yes I actually only manage to figure out the basis for S after figuring out what it means for (b)

Is (a) okay

I am thinking maybe it’s too brief

[kindred grove]

also my thinking

like if you're just trying to convince yourself in your head sure

[gleaming solstice]

The point is, whether this argument is considered rigorous. T due to homogeneity does preserve (0)

the left is just homogenous and additive properties

But it’s not a pain to do verification though

[kindred grove]

sounds alright to me

it's really the 2 other that aren't super trivial to see

[gleaming solstice]

Just to be certain, I will use standard method verifying it instead of this..

It’s not hard to just explicitly demonstrate the additive and homogeneous properties

Is the basis correct

I am a bit scared to ask since I didn’t feel like I am entirely sure

The V should be infinite dimensional

Which I eventually figured

And then it’s just some basis in R^n

[kindred grove]

well your c) seems pretty off yes

[gleaming solstice]

I just don’t know if it’s right 😭

But I feel like logically it kinda make sense

[kindred grove]

well question b) is here to instill in you the idea that maybe every f in V is of the form f(k) = T^(k-1)(x) for some x (which you kinda remarked already)

[gleaming solstice]

….

I got I must fix it

This isn’t as easy I thought

[kindred grove]

just use induction

for x you pick x=f(1)

[gleaming solstice]

yes

[kindred grove]

so if you can show that f(k) = T^(k-1)(f(1)) for all k, you win

base case k=1 there's nothing to show

then for the induction case, you use f(k+1) = T(f(k)) from the def of S

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[mortal kraken]

I want to verify my answer on this assignment my answer is (A) and I did it by u sub of u=1-y/eps

[mortal kraken]

Is the picture too blury

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@mortal kraken Has your question been resolved?

[robust slate]

Yes it’s A

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[plain sand]

could anyone tell me if my answer in red ink will get all 3 marks according to the MS

[robust slate]

Missing this

[plain sand]

what does SAS mean.

[robust slate]

https://thirdspacelearning.com/gcse-maths/geometry-and-measure/congruent-triangles/

[plain sand]

so in words i should just say that SAS has been proven

[robust slate]

Sure ig

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[plain sand]

,close

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[plain sand]

could anyone explain this to me

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[stone mortar]

not exactly mathamatics but stillllllll

[elder birch]

i would say d

it looks like cyclic permutation

[stone mortar]

nice, thanks

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[tall ridge]

can someone help me solve this question I was using the formula F = integral from a to b (density * length * depth) dy

[tall ridge]

and to get the length I drew a line horizontal line inside the triangle

and im trying to get the slope of it however I dont know what points to use

[torn gust]

the points would be width zero 7 meters down and width 7 2 meters down

[tall ridge]

and then my depth would be 7-y?

@torn gust

[torn gust]

uh people do it a bunch of different ways as long as the rest of the integral make sense yea do that

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[upbeat terrace]

hi.. does anyone does do corepetitions?

[unique wagon]

maybe, do you have a particular question to ask about a competition?

[placid oak]

!da2a

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No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

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[vapid parcel]

Can someone confirm my answer for this?

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[vapid parcel]

<@&286206848099549185>

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[quaint rune]

help

[quaint rune]

is this an example of a graph with 2 different Eulerian cycles?

1. v1, v2, v3, v1, v5, v4, v1
2. v1, v5, v4, v1, v2, v3, v1

and that all vertices have even degrees

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[lime valve]

Find the gradient/point at which perfectly splits the area between the curve and line in the lower region and upper region in half
equation of curve = -2x^2 + 4x
Equation of line = y=mx (need to find m)

[royal holly]

equate the areas using integration

[lime valve]

I did

But it didnt work

I found the total area

for the parabola

which was 8/3

meaning area 1 (above the line) = 4/3

I found m = 0.8252

but when I tried calculating the area unedr (lower region) I didn't get a value of 4/3

<@&286206848099549185> please

[fierce quartz]

Bruh

We all sleeping

Also I can’t help you

[lime valve]

y not

[spring glacier]

this should be 18b^2

cuz 3b²/4 becomes 18b²/24 not 12b²/24

[lime valve]

it's a 2b^2 / 4

[spring glacier]

ohh i read it as a 3 mb

[lime valve]

nah all good

[spring glacier]

then thats not the mistake

[lime valve]

does my process look correct

in finding m?

[spring glacier]

yeah the only part i havent checked is the 8/3 and the 4/3

[lime valve]

Because I believe my gradient is right, but idk how to verify it but finding the area of the lower region

[spring glacier]

the process for getting m seems all good tho

[lime valve]

For the lower region of the area do I have to subtract functions from eachother

like I did for the upper region?

Or can I just do integration of mx from x=0 to x=a and then add integration of the curve from x=a to x=2?

sorry if that doesnt make much sense

[spring glacier]

pretty sure you could do that yeah

[lime valve]

Oh

Let me check if that gives me the same 4/3

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[nova wraith]

$\lim_{x \to 0} \frac{\left(\sqrt{1 + x}\right) - 1 - \frac{x}{2}}{x^2}$

[glossy valve]

waffle

[nova wraith]

How would I solve this without l’Hôpital’s rule?

[onyx glen]

expand (1+x)^(1/2) into a power series

using the generalized binomial thm

[nova wraith]

I don't know what's that

I tried rationalizing but it's going too long

[onyx glen]

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

[glossy valve]

Ann

[nova wraith]

$(1+x)^{\frac{1}{2}} = 1+\frac{x}{2}-\frac{x^2}{8} + \frac{x^3}{16}+\dots$

[glossy valve]

waffle

[nova wraith]

<@&268886789983436800> Advert

[onyx glen]

yes

[nova wraith]

Okay I get this

The first two terms go to the left and when divided by x^2 it leaves -1/8 as x goes to 0.

l'Hopital's rule is faster though

[onyx glen]

eh

lhop is going to be hellish if you have any sort of product or nasty power

[nova wraith]

Is it not advisable to use it every time even when the limit is in indeterminate form?

[onyx glen]

its difficulty becomes especially worse if you would have to apply it twice to most anything not nice

if there's trig and roots you'll already drown

so yes it's very inadvisable

last resort

[nova wraith]

I see

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[stark nexus]

Hello

[stark nexus]

I’m a bit confused

What do i do now

[gritty rose]

power rule

u sub if you need to

[stark nexus]

But how

[mighty ridge]

u = x-3 could help you to see why you probably already know this integral

but it's not the only possible choice

[stark nexus]

Ohh

Mb my brain just stopped working

[stark nexus]

What’s the other

[mighty ridge]

well assuming you didn't know any form of integral you could always try stuff like u = (x-3)² and bruteforce from there

wouldn't be very fun tho

[stark nexus]

Yeah I’ll stick to x-3=u

Ty

Just checking, 9 * u^-2

Where do i continue from here

Nvm

Ty

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[white imp]

Can anyone help with this? Would be better if method is drawn, thank you!!

[austere cove]

For the first one, it might be easiest to rewrite everything in terms of powers of 2.

For the second one, the question is asking about your understanding of how arcsin works.

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[plucky yoke]

i may be cooked iin gr11 functions

[austere cove]

gr11led as it were?

What's your question?

[plucky yoke]

i need heelp with c

[gritty rose]

you can cancel two terms

[plucky yoke]

sorry how would i do that?

[gritty rose]

$\frac{z-2}{z-2} = 1$

[glossy valve]

riemann

[gritty rose]

and there's another one

[plucky yoke]

oh ok

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[proper zenith]

Why is the answer for the lower limit of -0.618 different when I calculated it to the answer of -0.0758

I’ve calculated it over and over again

But I keep getting 0.233

So I’m not sure what is wrong

[shrewd falcon]

!xy

[full forum]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[proper zenith]

Question g)

[shrewd falcon]

,w solve 2x-x^3=x-x^2

[glossy valve]

Results provided by WolframAlpha

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

Requested by vengeance0548

[shrewd falcon]

yep

[shrewd falcon]

safe to say you're wrong

recalculate it

[proper zenith]

Ok

.close

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[mossy nest]

For a question like this, we are doing first derivatives test because there are multiple x-values correct?

We are searching for minimums because minimum = closest point and maximum = furthest point?

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[tender nacelle]

Kinda. This is an optimization problem. You made a function for the distance (that thing with d = sqrt(something)) and set its derivative equal to zero. The multiple x values you got for this represent local minimum or maximum distances. You do the first derivative to check if these are local minima.

Logically if the distance function is decreasing before hitting a point and then increasing after, it should be a local min, which is what we confirm.

[mossy nest]

So therefore minimums are closest? and maximums would be furthest?

[tender nacelle]

Yeah

I mean here its obvious the points we found are minimums since maximums would be like x = inf but its good to check regardless

[mossy nest]

ahh alright thank you - another question haha sorry

would the first picture be the same as the second?

are the negatives relevant here?

[tender nacelle]

Looks like it. Just be consistent as you solve

[mossy nest]

ahh i understand

[tender nacelle]

If south and east mean (+) that's fine, just dont flip it mid solution

[mossy nest]

thank you

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[wintry edge]

I need help with my research question of How can the Lotka-Volterra equations model the population dynamics of Spotted Hyenas (Crocuta crocuta) and Plains Zebras (Equus quagga) in Kruger National Park, South Africa?
Im trying to understand Lotka-Volterra equations but its difficult and confusing for me...i dont know where exactly to start and what to do i have obtained all my data for using the equations and msking the graphs but i dont know exactly where and what equations to use (if that even makes sense 💀)

[austere cove]

@wintry edge have you taken differential equations before?

About what is your level of math that you are confident with, if not?

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[drifting summit]

I'm trying to get better at proof writing, can someone critque it?

also i'm not sure if we need ot prove for when x = -2

[visual smelt]

.

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@drifting summit Has your question been resolved?

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@drifting summit Has your question been resolved?

[slender onyx]

the final line is right but some intermediate steps are off

$f(I) \ne I$. compute it properly. then $f:I\to f(I)$ being injective (why?) lets us define $f\inv:f(I)\to I$

[glossy valve]

ロケットジャンプ

[slender onyx]

$g(I) \ne I$ as well

[glossy valve]

ロケットジャンプ

[drifting summit]

I think I have a pretty bad understanding of this then

What does it mean when we take a function over an interval

[slender onyx]

we restrict the domain to a subset

[drifting summit]

Could I argue g(I) is a subset of I

[drifting summit]

Hm so like hypothetically we have a domain 0 to infinity, and we want to compute f(x) = x^2

In this case f(I) is equal to I

But I forgot that this isn’t just x^2 for g(I)

So it should actually be [1, inf) for g(I) which then is a subset of I

[slender onyx]

first what is f(I)?

[drifting summit]

oh that part I am wrong and I realize why

I’m just giving a random example like

If I = [0, inf) and f(x) = x^2 then is f(I) = I?

[slender onyx]

when arguing f'(2) exists i wouldnt worry about sets

[drifting summit]

Ah

I was just following the definition so kinda like stupid of me

[slender onyx]

just for this point i dont consider the domain restriction

[drifting summit]

im just using these formal definitions but ig theres not erally a point of doing this

[slender onyx]

i mean very formally yes u need to make sure f(I) is contained by J

btw g should be on J not I

[drifting summit]

Yeah ur right

My idea was just setting I = J

[slender onyx]

ok lets be formal then to be safe 🙂

x^2+1 has domain R and image [1,infty)

its contained in the domain (0,infty) of 1/sqrtx

[drifting summit]

Yep

That part I get now :)

[slender onyx]

ok now im gonna need u to tell me f(I)

[drifting summit]

f(x) we consider has a domain [0, inf) and an image (0, 1]?

[slender onyx]

we started at f:R->R but its neither injective nor surjective

we make it injective by restricting the domain to I

surjective by restricting the codomain to the image J=(0,1]

thus $f:I\to J$ has inverse $f\inv:J\to I$

[glossy valve]

ロケットジャンプ

[drifting summit]

ah

Ah

Ur right

J = f(I)

[slender onyx]

wait no im talking out of my ass

should be J=f(I) where f is injective on I

[drifting summit]

Wait so we let J be (0,1] and because we find y0 in J such that y0 = f(x0) we get the rest of the statement

[slender onyx]

yes

heres a slight rewording

we started at f:R->R but its neither injective nor surjective
we restrict the domain to a set I upon which f is injective
we restrict the codomain to J=f(I)=(0,1], making f surjective

then $f:I\to J$ is bijective so $f\inv:J\to I$ exists

[glossy valve]

ロケットジャンプ

[drifting summit]

Yep that makes a lot of sense now

Not gonna lie tho, I’m not sure how necessary putting all of this is for a proof like this

I guess it’s not too difficult just gotta be careful

I realize a lot of the times I do proofs Idon’t know when I need to explain something

[slender onyx]

the domain restriction is very important since being noninjective is the main reason the inverse DNE

much less derivative of the inverse

oh we need I,J open in the inverse rule...

[drifting summit]

oh

[slender onyx]

take I=(0,infty) lol

J=f(I)=(0,1)

[drifting summit]

Are neighborhood considered open by definition btw

[drifting summit]

Ah ok

Also one more thing

Is it necessary to show the case x = -2

[slender onyx]

youll usually want to say open neighborhood to be safe

[drifting summit]

and have our domain restriction (-inf, 0)

[slender onyx]

i think its crazy to have to explore every choice of domain restriction, but ask ur prof

[drifting summit]

Alright thank you!

I guess it would be the same thing

[slender onyx]

no in general u get different f^-1 based on domain restriction

[drifting summit]

oh ok

yeah then I’ll ask her

[slender onyx]

for a simple example sqrtx is the inverse of x^2 restricted to [0,infty)

but if we restrict to (-infty,0] the inverse is -sqrtx

[drifting summit]

Ah I see

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[robust oracle]

anyone just tell me the process

[twin wolf]

hsc?

[robust oracle]

yeah gng

[twin wolf]

nice nice

[robust oracle]

just remind me what i need to do

[twin wolf]

i recognised the font

well have you done the other parts

[robust oracle]

yeah thats what thsc does

yeah

did i need a table for part ii

[twin wolf]

i think u could do that if u wanted to

[robust oracle]

yeah easy so for part 3 i integrate?

[twin wolf]

yeah integrate f' to get f

[robust oracle]

the nsub those lot in

[twin wolf]

yre

[robust oracle]

oh shit i thought of it too much

like we sub in just to find c

then we rearrange

damn

[twin wolf]

ys

r u 3u?

[robust oracle]

man im deeping too much

yeah

[twin wolf]

rip on statistics

stat 2u suck

[robust oracle]

nah stat should be calm im more concerned ab vectors

[twin wolf]

vector 3u should be not too bad

vector 4u is a lot worse

but a lot of 4u kids do 4u to escape 2u stats

[robust oracle]

100

[robust oracle]

idk my schools leninet on it for internals

[twin wolf]

2u hsc gonna be easy this year no sweat

[robust oracle]

say deadass i thought the difficulty spikes according to every odd year

for 3u anyway

[twin wolf]

i am deadass

for 2u

it goes easy hard easy hard

i was cohort 2024

and we got shit on with that 2u exam

[robust oracle]

oh tough

[twin wolf]

it was so cooked

[robust oracle]

yeah everyone shit the bed

[twin wolf]

but that means u guys have an easy year or at least an easier time than last year bcz 2024 2u was def the hardest out of the rest

at least for old syllabus u guys are still doing it i think

[robust oracle]

yeah we are i think that all changes like 27

what school if you dont mind

[twin wolf]

o i graduated

[robust oracle]

yeah which one

[twin wolf]

im not gonna say which one but its public and cracked at adv eng

[robust oracle]

fair

selective?

[twin wolf]

mayhaps

and u?

[robust oracle]

only for like 3u english and all that we pump e4s

[robust oracle]

private

[twin wolf]

ah

[twin wolf]

nice nice

[robust oracle]

ay long question and all

but i js want to know

howd you balance it all

im doing 11 units

n i just need to manage time better

[twin wolf]

good question

[robust oracle]

i need like one more word of affirmation bc ive heard the same response everywhere

[twin wolf]

this doesnt work for everyone but i spent less time on the subjects i knew i was good at and more time on subjects i wasnt so proficient in, not saying you should neglect your good subjects but spending equal time on all ur units isnt alwaays goonna work especially if theres massive imbalances in ur marks

also just know

1 bad assessment wont stop u from getting b6

as long as u do good on the rest

nesa is quite lenient actually

i messed up really bad for 1 of my internals worth like 20% but i still managed to b6

[robust oracle]

according to school ranking though

[twin wolf]

if ur in a good school and good cohort it definitely helps

but you cant really control that

[robust oracle]

bro i actually ate shit for all my first internals last year

and I'm actually going the long way relearning everything and a bit more

[twin wolf]

what subjects do u do?

[robust oracle]

it's not taking too much time

[twin wolf]

just secure rank 1 or try to get at least top 3 ranks for all of ur subjects

if u ccan u will be in a good spot

trials is what separates most people

[robust oracle]

but I just want the most ideal routine for that outcome

[robust oracle]

eng english adv 3u math physics music

[twin wolf]

music tuff

[robust oracle]

wbu

[twin wolf]

i did 2u eng 2u maths ancient his, physics, legal studies i used to do chem but i dropped

[robust oracle]

damn fair

if its calm what atar?

[twin wolf]

well i wish i could say there was a manual guide on how to get top ranks for each subject but its kinda different for everything; for physics i rlly sucked at the long response and mcq so i just grinded that and for eng adv it was chill just keep submitting ur essays to ur teachers till its perfect then u can use that scaffold for the hsc

[twin wolf]

90>

[robust oracle]

nah that's tough

[twin wolf]

btw u can get 90> atar with 0 b6s lmao

[robust oracle]

yeah idk for english i just need to flow more

[twin wolf]

by friend did it

my*

if u get high band 5s for everything u can get 90+

[robust oracle]

yeah i know it's js being concistent

yeah my mate got 94 doing fair scaling subjects

[twin wolf]

yes english is unique you need to find your own voice, tips for mod C is always have a discursive prepared

discursive is marked better than imaginative

persuasive is eh i know some people got b6 doing persuasive but its a bit dodgey cuz u dont really do persuasive in year 11-12

[robust oracle]

ight i see it now

[twin wolf]

discursive can also be turned into an imaginative pretty easily

[robust oracle]

yeah theyre usually all internals

[twin wolf]

so if you write discursive ur really preparing for 2/3 possibilties for mod C

and biggest tip i can give u is

trust yourself, u might not think u can write 14 pages or so in english but when the time comes your body will just write without thinking

[robust oracle]

allah 14 pages 😭😭😭

wait is that all accumulatory

like across 3 questions

[twin wolf]

yeah mod a + b + c

even more actually idk remember how much i wrote

but i wrote at least 8 pages for mod b and c

[robust oracle]

yeah that's fair

[twin wolf]

so more like 20 something ish

20+

[robust oracle]

i think i js need to work on fluidity and that's it

[twin wolf]

also physics rlly grind them mcq

i hated mcq for physics

dont study last minute tho 🤣

[robust oracle]

alrighty

did NOT know what mcq meant but it's calm

[twin wolf]

multiple choice q

[robust oracle]

that was my whole outcome for prelims

well actually all y11

[robust oracle]

yeah i saw

[twin wolf]

it only works if ur 10000% cracked at a subject but even then dont risk

i had like maybe 3 hours in total of studying for my ancient history hsc and i was cramming for the night before my ancient exam

not my proudest moment but then i ended up getting 92 external for it

lmfaooooooo

[robust oracle]

woah wtf

[twin wolf]

ya my reaction

[robust oracle]

man i feel like its an aquired thing though you js need intuition

yeah another thing though

wait one second

yeah i js wanted to ask like did you go on social media often or nah

did it ever interfere

[twin wolf]

to be honest

when ur in ur hsc period with all the exams going on

[robust oracle]

actually coudl we go dms i js want to ask a few more

[twin wolf]

u dont think about that stuff

sure sure

[robust oracle]

cheers brody

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[hallow walrus]

If I know $C$ is an $n\times n$ matrix which satisfies $$(C-I)(C-\frac1{4}I)=0$$ does that mean $1$ and $\frac1{4}$ are the only eigenvalues of $C$?

[hallow walrus]

does anything change if C is invertible?

[onyx glen]

$n\times n$

[glossy valve]

Ann

[onyx glen]

[glossy valve]

kheerii

[onyx glen]

C can be proven invertible from this equation alone

so no

[hallow walrus]

oh?

[onyx glen]

yes

you can rearrange it to get a quadratic polynomial equation in C

C^2 - (5/4)C + (1/4)I = 0

4C^2 - 5C = -I

[hallow walrus]

ah right

so does that mean C has only 1 and 1/4 as eigenvalues?

[fast peak]

C could also only have one of those as eigenvalue

[hallow walrus]

right

true

[onyx glen]

try taking an eigenvector v and running it thru (C-I)(C-I/4)

with arbitrary ev λ

[hallow walrus]

if C has only 1 as an eigenvalue that means the eigenspace of 1 is the whole space?

[hallow walrus]

(lambda-1)(lambda-1/4)=0

ok

nice

[hallow walrus]

like I don't know how exactly to interpret (C-I)(C-I/4)=0

does that mean the sum of the two eigenspaces is the whole space?

i.e. $\ker(C-I)\oplus\ker(C-I/4)=\bR^n$

[glossy valve]

kheerii

[fast peak]

take a vector v and write it as a sum v=v1+v2+v3 where v1,v2 are from the eigenspaces and v3 is from the rest

[hallow walrus]

so Cv=v1+v2/4+Cv3

hmm

oh wait I see what you mean

(C-I)(C-I/4)v = (C-I)(C-I/4)v3 = 0

which means it has to be from one of those eigenspaces

[hallow walrus]

this is the basic reasoning for the cayley-hamilton theorem right?

[fast peak]

honestly I forgot how that one is proved. but maybe

[hallow walrus]

well this kinda proof works when A is diagonalisable at least

[sharp flame]

Hayley-Camilton theorem

[hallow walrus]

true

have you started doing linalg yet

[sharp flame]

A vector space is a set and two associated operations; the elements of which obey 10 axioms 🔥

They're like the 10 commandments but they make less sense

Oh fuck I thought this was the lounge lmfao

[hallow walrus]

so true

nah they make sense though

[kindred parcel]

lol i believe wikipedia's most concise definition is "a vector space is a module over a field"

[hallow walrus]

.close

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[regal forge]

not sure how to do this

my idea is that we suppose P(x) is a polynomial such that there is no roots

so the minimum has to be in the region in the boundary of D.

from the minimum modulus theorem

but not sure how to proceed there.

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[regal forge]

<@&286206848099549185>

[regal forge]

<@&286206848099549185>

[regal forge]

<@&286206848099549185>

@fleet drum Could i get some help pls

[quaint prawn]

here are some hints

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[regal forge]

is there better hints?

@quaint prawn im sorry, i'm confused

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[alpine hull]

How do I solve for d?

[round gust]

well

the direction changes when the sign of the velocity changes

if velocity > 0 then it's going right

if its <0 its going left

[upper valley]

$\frac{dx}{dt} = V(t)$

[glossy valve]

SELVATOR

[round gust]

so you need to find the point where the velocity changes signs

[upper valley]

$\frac{dv}{dt} = a$

[glossy valve]

SELVATOR

[upper valley]

Maybe it changes direction on V = 0

U need to solve delta

And chose the positive result

Negative result is unacceptable

[round gust]

there won't be any negative answer

[upper valley]

6t - 36 = 0
t = 36/6
t = 6

We have $6t^2 - 36t$ $\to$ $ t(6t - 36) = 0$

[glossy valve]

SELVATOR

[upper valley]

$t = 0$ unacceptable and $6t - 36 = 0$

[glossy valve]

SELVATOR

[upper valley]

@alpine hull

[alpine hull]

I just plugged in the numbers that were on the sign chart I drew

[round gust]

yes it's correct

[upper valley]

Yes, but i doubt 0

Because it is the point of movement

[round gust]

yeah i guess you can disregard t=0

and assume that the particle starts moving at t=0 to the left

and only changes direction at t=6 to the right

[alpine hull]

from my understanding it’s moving from -1 -> 0 positively then at 0 it goes to rest then changes direction from the intervals 0 -> 6 and goes negatively

or are y’all saying -1 doesn’t matter it starts at 0 bc thats when its at rest and it’s pointing down while going to 6 and when we go to 6 it’s pointing up

[round gust]

yeah but because this is physics usually time starts at 0

so you would only consider the functions for t>=0

cuz negative time isnt really a thing

[alpine hull]

so is drawing that little line helpful or should i not

[alpine hull]

this is for my calc 1 class unless calc 1 uses physics

[round gust]

it's correct and helps you visualise the problem

[round gust]

well

[alpine hull]

so i should interpret it as this?

[round gust]

if this were a purely theorerical question

[upper valley]

-1 is point before record start

[round gust]

it would be correct

[upper valley]

We don't know if t = -1s exist

[round gust]

but because they've based the problem on a real life scenario, you have to start the time at t=0

[alpine hull]

okay i guess that makes sense so i don’t to plug in and check for numbers lower than 0?

[round gust]

yeah

you can view t=0 as the moment when the movement of the particle starts

clearly you cant talk about t<0 since there was no movement

[alpine hull]

and its as simple as from 0-6 arrow is pointing down and after 6 its pointing up so thats the change in direction

[round gust]

yeah

[alpine hull]

alright thanks 🙏🏻

[round gust]

you're welcome

[alpine hull]

also when they ask this question i plug it into velocity equation?

jus to make sure

@round gust

[upper valley]

It depends

[round gust]

well yeah

as we said, the direction changes when the sign of the velocity changes

so you plug in t= whatever and see if v is positive or negative

[alpine hull]

will it ever be for acceleration?

[round gust]

a=0 could mean that the object is moving at a constant speed

or it could mean that the object is standing still

or it could mean a point where the acceleration changes from positive to negative, which would indicate a maximum/minimum of velocity

so if the question were: find the maximum/minimum velocity of the particle

you would set a(t)=0

and then whatever t you get, you plug it into v(t) to get the maximum/minimum velocity

[alpine hull]

Is this valid?

for when the particle changes direction

[upper valley]

I think so

[round gust]

so when does it change direction

[alpine hull]

at 1?

[round gust]

at 1 and 3 or what

[alpine hull]

1 and 3 yes

[upper valley]

Yes

V(1) = 0, V(3) = 0

[round gust]

yeah

[upper valley]

You can calculate delta if you want.

And check

[round gust]

i guess...

but there's no need

you forgot a t there

it should be -4t

but the factoring is correct

[alpine hull]

whoops

[upper valley]

$x_1 = \frac{4 - 2}{2}$

[glossy valve]

SELVATOR

[upper valley]

$x_2 = \frac{4 + 2}{2}$

[glossy valve]

SELVATOR

[upper valley]

$x_1 = 1$ and $x_2 = 3$

[glossy valve]

SELVATOR

[upper valley]

Ok good

Any other questions?

[alpine hull]

not right now thanks idk how to close this

[round gust]

.close

[alpine hull]

.close

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[torpid perch]

do you know how to do it or no

recall the eigenvectors equations

A . x = λ . x
(A - λI) . x = 0

so find the nullspace for (A-λI)

but first plug λ = 1

then do the same for λ = 5

[final roost]

find the eigenvectors of each eigenvalue->create a 2x2 matrix where each column is an eigenvector->answer

is this correct? (asking for myself. im studying this irl as well)

[regal forge]

hey, could you please help me understand this

[weak pelican]

Why do you find the null space?

[torpid perch]

is so you find all the possible vectors x that satisfy (A - λI) . x = 0

you get a span

[torpid perch]

which other way is to do it?

[weak pelican]

Idk how to do it

My teacher wants us to use RREF to solve

[torpid perch]

yes

lets go step by step

[weak pelican]

So 5-1 and 1-1

[torpid perch]

plug A - λI to λ = 1

[weak pelican]

Yes

So 5-1 and 1-1

[torpid perch]

yes the diagonal elements are only affected

[weak pelican]

So the final matrix is just 1

In top left corner

[torpid perch]

,, A - I = \begin{pmatrix} 4 & 0 \ 2 & 0 \end{pmatrix}

[weak pelican]

Yes then reduce it to only having 1 in top left corner

Ok so you have free variable x = 2

[glossy valve]

938c2cc0dcc05f2b68c4287040cfcf71

[weak pelican]

Just reduce whole thing

10
00

[torpid perch]

sure

but you are skipping some steps

[weak pelican]

Null space is
0
1

You got the right answer

[torpid perch]

,w nullspace {{4,0},{2,0}}

[glossy valve]

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[torpid perch]

true

nullspace is

<(0,1)>

[weak pelican]

Are all Eigen vectors finding null spaces ?

[final roost]

is null space the same as the eigenvector?

[weak pelican]

My tutor at school was telling me that too

Like (A-I)v = 0

He said v is the null space

[torpid perch]

all the vectors, v that satisfy that equation are multiples of the vectors of the nullspace I think

like <(0,1)> meaning (0,x) where x is free, x ∈ R

[weak pelican]

Hmm

Thanks for your help

I’ll be on a little bit later to finish the assignment

If you’re here.. I’ll ping you

Unless you want me to show you another one right now?

[torpid perch]

if you want

[weak pelican]

If you want to find out the eigen value Is part of the matrix and you plug it in and get the identity matrix that means it is not part of it correct?

Like you do (A-lambda(I))v and do RREF

And get identity matrix

Meaning it is not part of the matrix

?

Let me know if that makes sense

[torpid perch]

I think that would mean that lambda is not an eigenvalue

[weak pelican]

Ok great

[torpid perch]

and thus, the nullspace is trivial , only the zero vector

[weak pelican]

What do you mean by only the zero vector

[torpid perch]

,w eigenvalues {{5,0},{2,1}}

[glossy valve]

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[weak pelican]

That is not part of the null space

[torpid perch]

the nullspace is a vector subspace, zero vector is always inside it

take for example a lambda that is not 1 nor 5

solve for v, in : (A-λI).v = 0

you will get that the nullspace is trivial (only zero vector inside)

and if you row reduce the matrix you get a matrix with full rank, as you said you get identity matrix

[weak pelican]

Oh true

So if the nullity is 0 then there is still a zero vector lol

Yeah we have some basic questions in this but I’m trying to figure out how to do the different versions of them

Like when you have the eigen value and find a vector

Or find if the vector is eigen vector

And also find if the eigen Value is part of the matrix

@torpid perch so we are basically trying to find these three things I belive

It seems all we have learned so far was RREF

And spaces

And it has been 3 months

Basis for spaces

Expanding the dimension

Finding if a vector is part of the matrix by augmenting the matrix

[torpid perch]

lmao too many questions

[weak pelican]

Sorry mein

[torpid perch]

is better if you close this and open another one for another question

you also give the opportunity of other people to help u

you can always ping me if no one replies

but I need to go eat something

[weak pelican]

Me too lol

I’m waiting for my school cafeteria to start serving dinner

I’ll ttyl

.close

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[weak pelican]

i applied the eigen value to (A-3I)

then did RREF

and got the identity matrix

ok so it is no then

it is not an eigenvalue

ok lmc

?

[fast peak]

check your work

[weak pelican]

ok

1-3 = -2 , -2-3 = -5, 1-3 = -2

yes that is correct i checked the reduction with a calculator

ok

i was confuseed bc my professor in calss used an example of a 3x3 matrix with a left triangle of 0's

[fast peak]

check again

[weak pelican]

?

ooh

oops idk how that happene

[weak pelican]

i mistook a 1 for a 0

in my reduction

i did RREF wrong though

yes

it is a vector in the matrix

when you use an eigen value

it is a null space of that matrix

what

[weak pelican]

i got vector [3, -2, 1]

ok

i disagree

[fast peak]

check again

[fast peak]

this whole definition is just a mess

what does "using an eigenvalue" mean

an eigenvector is a nonzero vector v so that Av=lambda v for some number lambda

thats the definition you should remember

for one its a bit circular cause lambda being an eigenvalue is defined by having an eigenvector

[weak pelican]

3, 2 ,1

[fast peak]

yes

[weak pelican]

ok

[fast peak]

also really the point of eigenvectors is that matrix*vector is the same as number* vector

we could, sure

[weak pelican]

so there are like 3 things to solve from it?

[fast peak]

there are obvious issues with the fact that it doesnt generalize to infinite dim but we'll ignore those

the question is, why would we care about randomly det(A-lambda I)=0

[weak pelican]

find if the eigen value is part of the matrix

find if the vector is an eigenvector

like you can reverse these things

[fast peak]

in functional analysis you encounter lots of infinite dim spaces

[weak pelican]

oh you were right my bad the eigenspace is the null space

when you apply the eigen value

[fast peak]

please stop saying "applying the eigenvalue"

[weak pelican]

lol what do you call it then

[fast peak]

I dont even know what you mean

do you mean subtracting lambda I ?

[weak pelican]

like you do (A-lamdaI)

yea

[fast peak]

call it subtracting lambda I for all I care. but really the fact that you need the phrase in the first place is worrying

[weak pelican]

well how do you describe what you're fdoig then

subtracting lambda

[fast peak]

computing the eigenspace

[weak pelican]

oh

so it's a sequence

subtract lambda, RREF, find null space = computing eigenspace

[fast peak]

yes

[weak pelican]

cool

[fast peak]

not every step always needs to have a name

[weak pelican]

thats true