help-39

[short orchid]

yes!

[cursive wraith]

(with x >= 9)

ok

so their product

(x-1)(x-9)

is it gonna be greater than (or equal to) 0, or smaller?

[short orchid]

greater than 0!

[cursive wraith]

ok great

so...

[short orchid]

oh ure right 😭

thank you so much 😁😁

[cursive wraith]

just a quick rule of thumb

a(x-x1)(x-x2)

is the sign of "a"

except between the roots

here you have 1(x-1)(x-9)

1 is positive (or non-negative if you care about greater or equal)

so (x-1)(x-9) is greater than equal to 0, except between the roots 1 and 9

(on your graph it would be "+ - +")

[short orchid]

ohhhhh

okayyy thank you so much once again 🙏🏻🙏🏻🙏🏻🙏🏻

[short orchid]

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[onyx helm]

im so confused on mixtures

and ratinoal word problems

i really need someone to teach me it

[pearl pond]

@onyx helm Has your question been resolved?

[plush bramble]

https://m.youtube.com/watch?v=OBVGQt1Eeug

[onyx helm]

thank u

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[surreal sage]

can somebody explain why do get an orthogonal vector in cross products and what does that mean?

[surreal sage]

like dot product dhows how much the two vectors coincide right?

[oak quiver]

if A x B = C
then orthogonal means that C is perpendicular to A and B

[surreal sage]

similarly what does cross product signify?

[surreal sage]

ikt

[versed mica]

it finds a vector perpendicular to the two vectors what are you confused about

[compact ridge]

it gives you a vector that is perpendicular to both your original vectors
only works in 3D

you can Google this stuff

[oak quiver]

i would say that its an operation that is just defined like that

[surreal sage]

and what exactly does that third vector do?

[oak quiver]

ex: in phy torque

[versed mica]

it’s perpendicular to the two vectors

are you asking why it’s useful?

[compact ridge]

we can:

1. find the plane that contains any two vectors in 3D space
2. find the shortest distance from a point to a plane
3. find the intersection of two planes

and a lot of other stuff like this

[surreal sage]

yeah

[versed mica]

there are many applications in physics as well

[surreal sage]

tell me the phy one

physics

that over there got me confused

lol

[versed mica]

also the area of the parallelogram formed by the two original vectors

[oak quiver]

torque

[oak quiver]

.

[versed mica]

is equivalent to the cross product

[surreal sage]

i saw that

anything else?

[oak quiver]

many angular stuff are defined using cross product

[versed mica]

magnetism

[oak quiver]

angular momentum

L = r x p (i think)

[surreal sage]

how does area equate to a vector? like how does a 2d thing become a 1d thing?

[surreal sage]

yeah

[oak quiver]

A x B is the same as A * B * sin theta which is also area

forgot n cap

[surreal sage]

wait! so we have two vectors r and p and cross multiplying them gives another vector L

[compact ridge]

the magnitude of the vector is the area

[surreal sage]

whihc is perpendicular to the previous ones

[oak quiver]

yes

[surreal sage]

how is angular momentum perpendicular to those two

[compact ridge]

you're missing the point, it's the length of the vector

its magnitude

not the vector itself

[oak quiver]

have you learnt rotation and all?

[surreal sage]

could you explain a little

[oak quiver]

if not then i dont think this is the right time

[surreal sage]

not even begun

[oak quiver]

imagine explaining exponents to someone who hasnt done multiplication

[surreal sage]

i was thinking of learning vectors first

then hopping onto phys

yk so that i could better understand the concepts

[versed mica]

it’s defined that way, the moment arm or r vector is perpendicular to the linear momentum vector so that when you take the cross product you get angular momentum which is perpendicular to both

[oak quiver]

i would say just learn it now
learn the applications in phy later

bro if you are his secret twin

then why would you go around and tell everyone?

[surreal sage]

so how does that look like geometrically

like theres a circle

with radius r

[oak quiver]

.

[surreal sage]

is that for me

[oak quiver]

a bit

first learn about linear motion

then motion in 2d

work energy theorem

center of mass

then come to rotation

[surreal sage]

ok

[surreal sage]

but just help me with this one

[versed mica]

sure if you want to think of it as rotating around a circular path

[surreal sage]

whats another way of thinking about it?

[surreal sage]

can you make a diagram?

of the third vector

[oak quiver]

think of it this way
things can rotate only 2 ways clock and anti-clock

now cross-product gives us perpendicular vector

so we just take its magnitude and sign

like one is +10 and one is -5

so net will be +5

we dont take direction into account other than sign

[surreal sage]

yeah i might have to do some physics first

[oak quiver]

lmao

[surreal sage]

but just make a diagram of how that looks like

[oak quiver]

just hold a pencil from its tip

and let it rotate

[surreal sage]

ok

how can i let it rotate on its own?

[oak quiver]

dont hold it too tight

let gravity make it rotate

[surreal sage]

oh

ok

then

lets say it rotated 180 degrees

[oak quiver]

i forgot how to find direction in cross product lol

[surreal sage]

haha

lol man

[oak quiver]

lemme remeber

[surreal sage]

lol

[oak quiver]

if the pencil rotates like this

[surreal sage]

yeah

[oak quiver]

then the torque will be outside of the plane

towards you

[surreal sage]

ooh

i kinda get it

[oak quiver]

no you dont lmao

stick to a good channel maybe like khan academy

[surreal sage]

why would torque be outside?

[oak quiver]

and follow them

and you will understand

BRO

[oak quiver]

.

[surreal sage]

because cross product?

[oak quiver]

yes

pls close this now

!done

[pearl pond]

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[surreal sage]

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[uncut timber]

is this possible

a friend of mine told me to solve this

[odd sand]

yes instead of roots think of it as fractions then combine and solve

[sinful nebula]

cotx^3/4 ?

crazy shit

good luck solving this integral @uncut timber

[odd sand]

yea u can use u sub

[calm wing]

not happening lol

[uncut timber]

what the fuck

cotx^3/4 = t

then

hmm

I don't think this is possible

its possible but I don't want to do this much work

[odd sand]

oh wait the u sub doesnt work 😂

[uncut timber]

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[uncut timber]

that works for sqrt(cotx)

[pearl pond]

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[clear osprey]

I was watching a lecture and I'm confused about what happened here

if this is by the assumption of x, that |x-c| < δ, then wouldn't 3|x-2| < δ? why is it 3δ?

[cloud zephyr]

Since we wanna pick a delta, so that |f(x)-L|<epsilon, therefore we can pick delta as epsilon/3 so that finally 3|x-2|<epsilon
not sure if this explains 🤔

[clear osprey]

I'm not sure if I get that 😭

how do we know that delta=epsilon/3 fulfilles 3|x-2|<epsilon

[cloud zephyr]

yay Denascite is here!

[tropic saddle]

c=2. so |x-2| < delta is the assumption

and then the question is, what delta do we have to choose such that 3|x-2| is smaller than epsilon

and if |x-2| < delta, then multiplying both sides by 3 gives that 3|x-2| < 3 delta

so if we choose delta = epsilon/3, then 3 delta = epsilon

which makes 3|x-2| < epsilon

[clear osprey]

but 3|x-2| is just a simplification of f(x)-L, so why also multiply delta?

[tropic saddle]

we have to show that if |x-2| < delta for some delta that we are allowed to choose, then |f(x)-L| < epsilon

that is the definition of continuity which we have to show here

[clear osprey]

Wait I think I got it, is this it?

[tropic saddle]

yes

[clear osprey]

I'm so confused about the order of the work the professor is doing though

that seems like an observation I could only make working backwards, at what point am I supposed to catch this 😭

[tropic saddle]

well solving things is not always the most straightforward thing

[clear osprey]

I guess I'm confused at what he's solving for

I thought we were proving that |f(x)-L|< epsilon, but he ends it with =epsilon so were we solving for what epsilon was??

[tropic saddle]

he is trying to find a delta such that if |x-2| < delta, then |f(x)-L| < epsilon

[clear osprey]

why doesn't it end with =delta then?

[tropic saddle]

if you read the whole inequality from beginning to start it says |f(x)-L| < epsilon

he is "solving" for delta in the sense that in the middle he realizes that if he chooses delta = epsilon/3, then the inequality works out

[clear osprey]

wait I think I'm kind of getting it

thank you for explaining 😭

have a good one

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[sonic rose]

hi guys any ideas how to do this limit where a is some constant

[sonic rose]

i tried l'hopital but it's too complicated

i dont think it works

[cedar perch]

Is a constant?

[pure tiger]

yeah

he said so

[cedar perch]

Mb reading is not a skill I have 💀

[sonic rose]

lol

[pure tiger]

Can you use the formula
$$
\lim _{x \rightarrow 0} \frac{a^x-1}{x}=\ln (a)
$$

[jolly parrot]

Alix

[pure tiger]

Have you tried taylor expansions?

[
e^{2ax} \approx 1 + 2ax + \frac{(2ax)^2}{2} + \dots
]

[jolly parrot]

Alix

[pure tiger]

+O(x^3) but it vanishes as x->0

[pure tiger]

Also forget about this, it's way harder

@sonic rose you there?

[
1 - e^{2ax} \approx 1 - \left(1 + 2ax + \frac{(2ax)^2}{2}\right) = -2ax - 2a^2x^2 + O(x^3)
]

and

[
\left(1 - e^{2ax}\right)^2 \approx \left(-2ax - 2a^2x^2\right)^2 = (2ax)^2 + O(x^3) = 4a^2x^2 + O(x^3)
]

For the denominator

[jolly parrot]

Alix

[pure tiger]

Like you do this again for the numerator and evaluate at zero and you are done

Well I am assuming you have seen taylor expansions

Have you? If you didn't then I guess we'll do something else, there is another method.

[sonic rose]

hi sorry was afk

[pure tiger]

So have you seen taylor expansions?

[sonic rose]

what in the world i wouldnt have thought of that

[sonic rose]

yes but we were never taught to use it in limits

[pure tiger]

did you see the O(x) notation?

[sonic rose]

wouldnt the 2nd degree approximation be not enough though, dont you need the full expasion?

[pure tiger]

No here it works

[sonic rose]

not sure what that is

[pure tiger]

It's a negligeable before x when x goes to 0

Meaning it decreases way faster than f(x)=x when x goes to 0

[sonic rose]

i c

[pure tiger]

Here the rest of the expansion goes to 0 so fast compared to the beginning that they don't matter

[sonic rose]

let me try that then

[pearl pond]

@sonic rose Has your question been resolved?

[sonic rose]

tysm @pure tiger

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[pure tiger]

^^

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[floral knot]

I don’t know the possible values of a

as in can someone help me work out the possible values of a?

[fathom oasis]

What does it mean that the graph of y = f(x+a) passes through the origin?

If you plug in x=0, what should the value of y be?

[floral knot]

it passes through the point 0,0

[floral knot]

the value of y would be

12?

(0+3)(0-2)^2

@fathom oasis

[fathom oasis]

If you plug in x=0 to y = f(x+a), you get 0=f(a)

[floral knot]

I get it now thanks!

i’m confused now on this question…

question 8

is there a similar procedure

[fathom oasis]

Not really

[floral knot]

where should I start?

[fathom oasis]

Consider the illustration of the graph of the function. You'd want point R to lie above the x axis and point Q to lie below the x axis. Then the function has three distinct real roots

[floral knot]

yes

I understand that part

what’s next

[fathom oasis]

Let g(x) = f(x)+a.

g(1) <= 0
g(-1) >= 0

[floral knot]

got it

.solved

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[fathom oasis]

L'hospital twice?

[acoustic spindle]

French ?

[feral sedge]

,w L'hopital's rule

[jolly parrot]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by a_mog

[acoustic spindle]

Still don t know how to do it

[feral sedge]

f(x) is the top function and g(x) is the bottom function

[acoustic spindle]

Alr

[pearl pond]

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[hollow pond]

i'm extremely puzzled here (this is geo A work)

[hollow pond]

why is A. incorrect? my teacher released our practice test answer keys to review and study, however A. is marked incorrect. i've recreated the rigid transformation so many times in so many ways and i still dont understand it

im genuinely thinking that it's an error from my teacher

[woven garnet]

It is

[hollow pond]

it's an error from my teacher?

[woven garnet]

Yes

[hollow pond]

thank god

i thought i was going insane

[faint mica]

Wait what

[wet swallow]

What if AB≠BD

[faint mica]

Yh exactly

It is wrong

[signal eagle]

Error from teacher probably

[faint mica]

Imagine d was literally right next to B

[hollow pond]

huh

[wet swallow]

That's what I was thinking

[faint mica]

Now I think I'm going crazy 😭😭

[hollow pond]

bro i understand nothing 😭

[faint mica]

I'm agreeing w David

Like it says what MUST be true

[woven garnet]

Ah right

[faint mica]

Imagine u redrew the diagram so like D was right next to B

[wet swallow]

I if AB≠BD
Then

You can't assume choice A is correct

[opaque arch]

Can someone help me simplify trig expressions

[woven garnet]

New channel

[opaque arch]

I'm in the wrong place arent I

Yeah

[hollow pond]

im a visual learner 😭

[hollow pond]

vertical angles

[wet swallow]

No let me draw with pen and paper

1 sec

[hollow pond]

👍

[wet swallow]

It is possible that the lengths are not equal

The arrow shows the location of the points after rotation by 180⁰

@hollow pond

[hollow pond]

my bad, i was checking something

i understand it now

your drawing was very helpful

[wet swallow]

Thank you

[hollow pond]

i didnt really understand why it was incorrect

[wet swallow]

!done

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[hollow pond]

.close

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[torpid schooner]

What is problem 2 asking me to do?

[torpid schooner]

Is it just telling me to find implicit differentiation

Like this

Sorry I’m kinda slow

[faint mica]

Probably

For part a it's saying do it w implicit differentiation

For part b it's saying re arranged the equation to make y the subject and then differentiate

I think

And part c is make sure these r the same

[torpid schooner]

I have no idea what this means LOL

sorry for late response

Could you clarify please

[faint mica]

So for q 2

[torpid schooner]

🙏

[faint mica]

U wld write 6x to the four + y to the 5 equals 2x as 6x to the 4 - 2x = y to the 5

Then take the fifth root of both sides

And now u have y on one side

And x on the other

And then differentiate

And u shld get the same answer

[torpid schooner]

Do I write it as $6x^4 + y^5 =2x \ as 6x^4-2x=y^5$

Bro what the flip

LOL

[jolly parrot]

David

[torpid schooner]

This

And then use implicit differentiation in the right

Equation

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[torpid schooner]

.reopen

[pearl pond]

✅

[torpid schooner]

Can someone define explicit and implicit functions?

I mean

Implicit and explicit differentiations

Can someone give an example based off of this explanation

[merry carbon]

Explicit functions, those in the form y = f(x), you can differentiate “straight”, to dy/dx = f’(x)

[torpid schooner]

Ah thanks!

.,|+~*|

X,@9/;

Z,@9

?49

Bro

.close

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[plush bramble]

.,|+~*| X,@9/; Z,@9 ?49 Bro

[torpid schooner]

It kept opening up the number section

Instead oof the word

Stupid keyboard phone

Not my fault

It’s keyboard fault

[plush bramble]

bots are out against you

[rough forge]

keyboard killer

[pearl pond]

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[vivid pilot]

is this answer to this 2?

[vivid pilot]

i think so but im not certain

[plush bramble]

use rank nullity theorem

https://en.wikipedia.org/wiki/Rank–nullity_theorem

[vivid pilot]

oh thanks

so it is 2

[feral sedge]

Yes

[vivid pilot]

thank yoiu

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@velvet wolf Has your question been resolved?

[pearl pond]

@velvet wolf Has your question been resolved?

[pearl pond]

@velvet wolf Has your question been resolved?

[pearl pond]

@velvet wolf Has your question been resolved?

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[sterile turtle]

ik how to do this q, i just want to ask what a triangular number is

[sharp vigil]

the triangular numbers are the ones that follow the formula for Sn given there. They're called triangular because you can arrange the sum as a triangle

[sterile turtle]

ahh

ic

thanks

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[sterile turtle]

how can I expain part b

[sterile turtle]

$S_n = \frac{n}{2}(n+1)$

[jolly parrot]

Sho

[sterile turtle]

so the number will end in 4,9,5,0

4,9 for the n+1

5,0 for the n

is there a better way to explain this or nah

[sharp wing]

hmm i think there might be! if it's divisible by 5, either the n/2 term has to be a multiple of 5 or the n+1 term has to be a multiple of 5, so you just need to show what values of n work for either of those conditions

and likewise if it's even, at least n/2 or n+1 has to be a multiple of 2, so you could set up 2 equations to figure out which values of n work!

[sterile turtle]

basically

hmm

[sterile turtle]

i can use thsi for the second one better, but the first one i just did like this

so

i said $\frac{n}{2}(n+1) = 2m$

[jolly parrot]

Sho

[sterile turtle]

then $n(n+1)=4m$

[jolly parrot]

Sho

[sterile turtle]

therefore n either has to be divisible by 4 or have a remainder of 3

ngl i hate these types of q

[sharp wing]

ah yes that works!

and if you wanted to use it to find which values of n make it divisible by 5, you just use the same logic but with a 5 instead of a 2

[sterile turtle]

alr

thanks

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[abstract halo]

would y_g = c1 (t) + c2 (t^2) + y_p, where y_p can be defined as t^3?

[abstract halo]

we're given t,t^2, t^3 are particular solutions

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[rich jolt]

can anyone show how to arrive at that conclusion

[gilded tide]

What's statement II

[rich jolt]

its not gonna affect what i am asking

[pastel raven]

you need to show the full question bro

affecting or not

[pearl pond]

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[reef kestrel]

How do I calculate this integral? The function is non elementary so I have to use a theorem instead

[bronze coyote]

Try by parts?

I’m not sure either

[reef kestrel]

I dont think its integrable and the hint says to use a theorem instead

The only idea I had was Taylor approximating

[west bluff]

I tried putting this in wolframalpha,first time I've seen it not be able to compute something

Said it needed more time, so yeah

Pretty sure you need a theorem

[reef kestrel]

Yupp but no idea what theorem

The limit of the integrand to infinity is 2/pi, can I use a comparison test with that?

[signal atlas]

Yes I think comparison + squeeze

[reef kestrel]

Does squeeze theorem apply for integrals

[signal atlas]

It applies for limits, which you have

[reef kestrel]

Ah wait maybe I can use mean value theorem for integrals

Yeah I think mean value theorem is easiest. I guess we assume continuity and then we can take the limit of f(c)*(n+2-n) instead of the integral.

Where c is in [n, n+2],

[signal atlas]

Yeah thats a good idea

[reef kestrel]

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[vapid bough]

$\int{\sec x(\sec x\tan x)}$

[jolly parrot]

bush_tree_plant

[vapid bough]

If you let secx = t then we get sec^2x/2

on the other hand, if we let tanx = t, we get tan^2x/2

what is going on here?

are they the same thing?

[glacial sequoia]

yes

(you forgot the + C)

[jolly parrot]

krypton

[pearl pond]

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[lethal estuary]

140cos(a)+60cos(35+a)=160cos(a)

[lethal estuary]

how would i get a here

[cosmic garnet]

is this in degrees or radians

[twilit lion]

Given the 35 + a it's most likely degrees

[feral sedge]

In either case the answer is the same

Apply the cos(a+b) identity to cos(35+a) and then isolate cos(a)

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[reef kestrel]

How do I prove whether this integral converges/diverges?

[midnight haven]

If it has a diverging minorant, it diverges. If it has a converging majorant, it converges. Does this help?

[reef kestrel]

Whats a minorant/majorant

[midnight haven]

Like another function that is in absolute value smaller or larger

[reef kestrel]

Is it like supremum/infimum?

[midnight haven]

No, more like something that is smaller or equal to / greater or equal to, for all values of x.

But to be honest, I don’t know if the integral converges or diverges yet. This is just the way I would think about it

[reef kestrel]

Ah okay so a comparison test right? I'm confused how to deal with it since its generalized

In the solution they first prove that the integral from 0 to pi converges, then they rewrite the integral as a series

[midnight haven]

So you have the solution to the problem already and are just trying to understand it

[reef kestrel]

Yeah and looking for alternate solutions

If I was to solve it myself I wouldnt do it that way so im trying to understand how to think about these questions

I would just directly move to a comparison test, but this seems iffy since the function is not defined in infinitely many points when sin(x)=0 so I'm kind of stuck

[midnight haven]

Yeah I was actually thinking that the first question to attack would be this: Does the integral $\int_{\pi}^{2\pi} \frac{1}{x^2 |\sin x|^{1/2}} dx$ converge?

[jolly parrot]

gautamdb

[midnight haven]

😅

[reef kestrel]

Its generalized at pi and 2pi so we get two integrals to analyse right

Between pi and 3pi/2, and 3pi/2 to 2pi

[midnight haven]

What do you mean by „generalised“? The integrand approaches infinity, yes.

[reef kestrel]

Yup exactly

Thats what its called in my textbook at least

[midnight haven]

But why from pi to 3pi/2, and why from 3pi/2 to 2pi?

I mean, between pi and 2pi, the integrand is nonzero, I don’t get why you are trying to involve the midpoint 3pi/2

[reef kestrel]

In that case we can set a limit as x approaches 2pi and pi

If we separate the integral

[midnight haven]

I think we should start by understanding the solution you have. After looking at the integral from 0 to pi, what i the next step there?

[midnight haven]

Also, is it clear to you how to prove that the integral from 0 to pi converges?

[reef kestrel]

So first we prove that the integral from 0 to pi converges

Yup that proof is just using comparisons

[midnight haven]

So basically the idea is the same as what I said: You find a converging majorant.

But the details were complicated and I didn’t get it immediately

They look at these two integrals: Integral from 0 to pi/2, and from pi/2 to pi.

But you could as well have taken any other number in between other than pi/2

Only pi/2 is convenient because with the substitution t = pi - x you can make the second integral into the first

[reef kestrel]

How does this help in proving the original integral converges? The bounds for integration are different

[midnight haven]

The original integral is split up into parts, integrating over k pi to (k+1) pi, for k from 1 to infinity

So you consider the integral from pi to 2pi, then from 2pi to 3pi, and so on.

Then you realise that for every integral from k pi to (k+1) pi, a majorant is given by

$\int_{k\pi}^{(k+1)\pi} \frac{1}{(k\pi)^2|sin x|^{1/2}} dx$

Oh, mistake

[reef kestrel]

I think I get the rest of the solution

[midnight haven]

I‘ll edit the above integral

[jolly parrot]

gautamdb

[reef kestrel]

It feels unnatural because they started by proving the integral between 0 and pi, whereas when I would do it I would start by converting it to series form

[midnight haven]

Exactly

You are absolutely right

This happens often with proofs

The order one thinks about it is one, the way it is written down is often in reverse

[reef kestrel]

Is it considered incorrect to write it in "thinking" order?

[midnight haven]

No.

I like it more when it is written in thinking order.

Only when the proofs get very long, it is sometimes more convenient to write it in reverse order

Because instead of having to prove each small result you need during the proof, you can just refer to Lemma xyz

[reef kestrel]

Makes sense, but i would argue this wasnt that long that it would be inconvenient to do it chronologically

[midnight haven]

You would argue right

[reef kestrel]

Anyways ill try to redo the question in a while to check ive actually understood

Thank you for the help!

.close

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[hollow halo]

how to find the slope of
E (-2, 1), F(2, 5)

[hollow halo]

I have a formula which is
y2 - y1 / x2 - x1

[unborn abyss]

yep

[carmine schooner]

are E and F points? just plug in the values

where E and F are (x1, y1) and (x2, y2) respectively

[stuck schooner]

bre

is thomas calculus math book hard?

[hollow halo]

so it's like this

based on the formula

[carmine schooner]

yeah, that's right

[hollow halo]

okay thank you

[pearl pond]

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[wild fable]

is there an easy way to do this

[viscid sierra]

what details can you notice in the graph

[wild fable]

uh well its got an inflection point

to the right of the axis

[viscid sierra]

These are all the conclusions:||two distinct roots, one inflection point, constant term positive, even roots, negative leading coefficient,positive local maxima for negative x||
open this only after trying yourself first

[wild fable]

a local max to the left

its got 2 x intercepts

and then

yeah thats about all i can find

[gleaming musk]

idk why but are u australian curriculum, it just looks oddly similar to the typsetting style they use here

sry it is bit of a random question

[wild fable]

yes this is the hsc

[gleaming musk]

3u?

[wild fable]

2u

[gleaming musk]

ah i see

[wild fable]

are u australian?

[gleaming musk]

im prepping for hsc as well rn lol

[wild fable]

lol nice

[gleaming musk]

absolutely swamped by english advanced T=T

[wild fable]

ya

[gleaming musk]

sorry im getting off topic

[wild fable]

u do 4u?

[gleaming musk]

yeah

[viscid sierra]

crazy that you identified an entire curriculum based off one question with nothing supporting the same

[wild fable]

i mean its pretty easy to identify when youve spent hours practicing the same style of questions

hsc got a distinct font

so nsw kiddies would know

[gleaming musk]

its kinda sad but ive done 500 or so trial papers and hsc papers os its become ingrianed into my mind

traumatising to some extent

anyhow i digress

[viscid sierra]

distinct font??😭😭 I can't distinguish it with the discord font

anyways

back to the question

[wild fable]

so its either B or C

ohh

gawt it

multiplciitititiies

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[sand geode]

Cant just cant. Need to study questions like this for my upcoming olympiad

[pearl pond]

@sand geode Has your question been resolved?

[acoustic path]

consider sum of cubes

[sand geode]

How do I even use the sum of cubes in the 1st one

[acoustic path]

(sqrta + sqrtb)^3 etc

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[pearl pond]

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[midnight haven]

Not particularly sure where to start tbh

[random ermine]

expand (X - mu)^2

[midnight haven]

yup did that

I get $E(X^2 + \mu^2 - 2X\mu)$

[gleaming musk]

u need to know some properties

[jolly parrot]

Edmund Cloudsley

[midnight haven]

Oh which one in particular

(sorry im a bit new to stats)

[random ermine]

use linearity of expectation

E(aX + bY) = aE(X) + bE(Y)

[midnight haven]

gotcha

so given this I should get

$$E[X^2] + E[\mu^2] - 2E[X\mu]$$

[jolly parrot]

Edmund Cloudsley

[midnight haven]

How do I get rid of the Mu term and the 2?

[gleaming musk]

do u recognise what can be immediately simplified

[midnight haven]

I don't think so

[gleaming musk]

what is mu?

@midnight haven

[midnight haven]

the mean

[gleaming musk]

ok

[midnight haven]

So the expectation of a constant

should be a constant

[gleaming musk]

yesh

[midnight haven]

therefore this would put mu^2 and the mu outside the expectation

but then how do we remove them after that?

[gleaming musk]

it should turn to this $E[X^{2}] + (E[X])^{2}-2E[X]*E[X]$

[jolly parrot]

Dootud

[midnight haven]

hmm but why so?

[gleaming musk]

we just apply linearity again to the last term

[midnight haven]

ohhhh okay

[gleaming musk]

so this simplifies into what u want

[midnight haven]

I woudl stuill have to look up this concept of linearity

but thanks so much for your assistance

have a good day

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[fathom saffron]

i keep getting -(ax+hy)/(hx+by)

[fathom saffron]

answer is a

my work

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[torpid tiger]

in an arithmetic series, the sum of the 9th and 12th member is 15, what is the sum of the first 20 members of this series

[compact ridge]

do you know the formula for the nth term of an arithmetic sequence?

and the sum up to n terms?

[torpid tiger]

yes

but the first member and the change are absent from the data

sorry

difference

[compact ridge]

you can write an equation for them

it is actually possible using only this information

[torpid tiger]

so

the general arithmetic formula being: a1 + (n-1)d

[compact ridge]

yes

[torpid tiger]

we have 2a1 + 11d + 8d = 15

[compact ridge]

great, that's correct

[torpid tiger]

2a1 +19d =15

[compact ridge]

yep and now can you use the formula for the sum?

[torpid tiger]

yes

n/2[2a1 + (n-1)d]

for the first 20 sentences that would be

10[2a1 + 19d]

oh wait

thats the same thing

10 time 15

thank you man

[compact ridge]

no worries!

[torpid tiger]

🙏

[compact ridge]

yeah because the 1st term + 20th term is constant
so is the 2nd term + 19th term the same

so it's the (k)th term and (21 - k)th term in general

here we just have k = 9

that's how Gauss added up 1 + 2 + 3 + .... + 100 actually

[torpid tiger]

indeed (; we have learned the proof for the sum of the arithmetic series formula, it is practically a re-written form of gauss's formula for integers

thank you once more you have been extremely helpful

[compact ridge]

no problemo again

.close

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[naive orbit]

"Express the side length of a square as a function of the length d of the square's diagonal"

what does it mean? does it want me to create a function like:
d is length of the square's diagonal and f(d) is side length of the square?

[frank goblet]

Lets call the side length a.

It would be like
d(a)=8a+3/a
(Not the solution ofc)

To find it, you can use pythagoras

And then solve for d

[naive orbit]

and d(a) is diagonal length?

[frank goblet]

d is the length of the diagonal

the (a) means it depends on the value of a

We say d is a function of a

[naive orbit]

okay thanks

.close

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[warm osprey]

Is this a valid proof for proving that sup{A+k} = sup{A} + k?

[pearl pond]

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[stoic imp]

\textbf{3.} Let the subspaces $\ \mathcal{W} = { \mathbf{x} \in \mathbb{R}^4 \mid x_1 - x_2 + x_3 = 0, ; x_1 + 2x_3 = 0, ; x_2 + x_3 + x_4 = 0 }$ and $\ \mathcal{H} = { \mathbf{x} \in \mathbb{R}^4 \mid 2x_3 - x_4 = 0 }$.

Find, if possible, subspaces $\mathcal{S}$ and $\mathcal{T}$ of $\mathbb{R}^4$ such that:
[
\dim(\mathcal{S}) = \dim(\mathcal{T}), \quad \mathcal{S} \cap \mathcal{T} = \mathcal{H}^{\perp}, \quad \mathcal{W} \subset \mathcal{S}, \quad \text{and} \quad \mathbb{R}^4 = (\mathcal{S} + \mathcal{T}) \oplus \langle (1, 2, 2, 1) \rangle.
]

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[pearl pond]

@stoic imp Has your question been resolved?

[stoic imp]

please help

<@&286206848099549185>

[pearl pond]

@stoic imp Has your question been resolved?

[finite cliff]

Help

Help

[stoic imp]

<@&286206848099549185>

[stoic imp]

<@&286206848099549185>

[deep temple]

what

[stoic imp]

exactly my thoughts

[rugged night]

is this correct or wrong

[stoic imp]

H^1?

fuck my life bro

[rugged night]

idk 😭

[pearl pond]

@stoic imp Has your question been resolved?

[frozen lynx]

What a predicament

May god send you someone to get you out of this

Although i want to help, i only understand half of the question

Anyways, gn

[stoic imp]

thank u for the kind words bachira

[frozen lynx]

Anytime megumin

[pearl pond]

@stoic imp Has your question been resolved?

[stoic imp]

https://media.discordapp.net/attachments/268882317391429632/1223598233763319819/1192815796490080286.gif

[frozen lynx]

Unreal

[pearl pond]

@stoic imp Has your question been resolved?

[frozen lynx]

I suggest pinging helpers

[coarse harbor]

0. Both $\mathcal{S}$ and $\mathcal{T}$ must be proper subspaces, otherwise the sum $(\mathcal{S} + \mathcal{T}) \oplus \langle (1, 2, 2, 1) \rangle$ wouldn't be direct. Even more, $\dim(\mathcal{S} + \mathcal{T}) = 3$.

1. $\mathcal{S}$ contains both $\mathcal{H}^{\perp}$ and $\mathcal{W}$. It's easy to calculate that $\mathcal{W} = \langle (2, 1, -1, 0) \rangle$, and $\mathcal{H}^{\perp} = \langle (0, 0, 2, -1) \rangle$ (check that yourself!), hence $\mathcal{S} = \langle (2, 1, -1, 0), \ (0, 0, 2, -1), \ v \rangle$ for some vector $v$.

2. Since $\mathcal{T}$ contains $\mathcal{H}^{\perp}$, we see that $\mathcal{T} = \langle (0, 0, 2, -1), \ u_1, \ u_2 \rangle$.

3. From (1) and (2) we know that $\mathcal{S} + \mathcal{T} = \langle (2, 1, -1, 0), \ (0, 0, 2, -1), \ v, \ u_1, \ u_2 \rangle$. From (0) we conclude that $\dim(\langle v, \ u_1, \ u_2 \rangle) = 1$, hence $v=k_0 w$, $u_1 = k_1 w$, $u_2 = k_2 w$ for some $w \in \mathbb{R}^4 \setminus 0$.

4. If $k_0 \neq 0$ and $k_1 \neq 0$, then from (3) we obtain that $\mathcal{S} \cap \mathcal{T} = \langle (0, 0, 2, -1), \ w \rangle$, which obviously has dimension 2. But we want dim=1 so $k_0 = 0$ or $k_1=k_2=0$.

5. If $k_0=0$, we have $\mathcal{S}=\langle (2, 1, -1, 0), \ (0, 0, 2, -1) \rangle$, and $\mathcal{T}=\langle (0, 0, 2, -1), w \rangle$. We want the sum from (0) to be direct, so $w$ must be a vector linearly independent with $(2, 1, -1, 0), \ (0, 0, 2, -1), \ (1, 2, 2, 1)$. For example, $w=(0, 0, 0, 1)$.

6. If $k_1=k_2=0$, then $\mathcal{S}=\langle (2, 1, -1, 0), \ (0, 0, 2, -1), \ w \rangle$, and $\mathcal{T}=\langle (0, 0, 2, -1) \rangle$. Again, $w=(0, 0, 0, 1)$ works.

[jolly parrot]

EQUENOS

[frozen lynx]

@stoic imp the wait seems to be over?

[stoic imp]

I wanted to ask, how did you figure that dim(S) has exactly 3 vectors inside

still reading point 2, but was wondering about that in point 1, everything else is very clear

oh I see

dim(S+T)=dim(S)+dim(T)-dim(SnT

dim(S+T)=dim(S)+dim(T)-dim(SnT)
3 = 2dim(S)-dim(Hperp)
3 +1 = 2 dim(S)
dim(S)=2??

[coarse harbor]

Oh, right, I forgot that they want dim(S)=dim(T)

Yes, in this case dim(S)=2

[stoic imp]

The whole argument changes if dim(T)=dim(S)=2

Look at (6) you said T = <(0,0,2,-1)>

[stoic imp]

Also I noticed there is a typo in the set builder

[coarse harbor]

Yes, point 6 would've been removed if I remembered about dim(T)=dim(S)

Anyways, the scheme of proof still works

[stoic imp]

but I still need to find T

and is dim 2 somehow

also you say (0,0,0,1) works but if we are going to extend a basis, is better if we rref the column space and append the canonical vectors of R^4

[stoic imp]

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[supple citrus]

why is the red one wrong?

[supple citrus]

the 2 is for accounting CV and VC both

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[supple citrus]

😥

[supple citrus]

ples 😞

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[buoyant sigil]

can someone draw the new figure and explain to me 😢 Im on the verge of tears

[pearl pond]

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[buoyant sigil]

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[pearl pond]

@buoyant sigil Has your question been resolved?

[buoyant sigil]

NO 😡 OKAY LISTEN 😭 LIFE IS DIFFICULT ALRIGHT MAN? THANKS TO DISCORD I HAD TO GET MY DAD TO HELP 😰 HE HELPED AND I FINALLY FINISHED

THIS SERVER IS SO FAKE! 😡 When I was a little boy...my mama told me 🤫 Son, we gonna have to buy jewels for lit

😢

[plush bramble]

[buoyant sigil]

HRY

YOU

I NEED HELP

😭

[tall flint]

you just said your dad helped

[versed mica]

cry more

[plush bramble]

dads are the best

[buoyant sigil]

yeah i need help on another question 😁

my dad is a tired man I want him to get rest

[buoyant sigil]

knief you better meet me at room 312 I am waiting for a boxxing session

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[elfin jewel]

yo

[elfin jewel]

anyone know how to solve this ?

ik the first one is correct doing matrix multiplication

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[dense badger]

how many similar triangles can be formed using n integer numbers starting from 1? (the numbers represent side length)

[gleaming musk]

are repeated numbers allowed?

like can u have 111, 222, 333, 444...nnn

[dense badger]

yes

The main obstacle for me to solve this problem is prime numbers.

[tardy reef]

Primes should be the easiest sides in the problem. If a side is prime, then the ratio of similarity is restricted to integers

For composite numbers as sides, the ratio of similarity extends to rational numbers.

You should count the triangles such that gcd(a,b) = gcd(b,c) = gcd(a,c) = 1, as they would encompass all the possible triangles with non similarity, and extend that set to include the similarity

Of course, that set isnt exhaustive of all possible triangles since it wont count triangles like 8,9,12 and so on

[dense badger]

triangles with such conditions would have the most number of similar triangles

[tardy reef]

yes, and most easy to count similar triangles for

[dense badger]

I tried to sort all prime number triangles first then triangles with one side having two factors and then triangles with sides having 3 factors and so on..

[dense badger]

Not all three sides of the triangle have to have the same number of factor though

Is my method on spot?

But my worry is how can I detemine how many primes are there

@tardy reef do you have any idea?

[tardy reef]

By my logic, the number of triangles becomes a sum of totient functions, so I dont think thats a viable solution

number of primes in [1,n] and values of totient(k) in [1,n]; knowing either of them is pretty much impossible, without exhaustive calculations

ngl, only similar problem I have done is where the chosen set of integers was for the values of the three angles, and that is a pretty small number, easy to calculate

this problem on the other hand seems too beyond me to solve casually in small time.

[pearl pond]

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[dense badger]

what does that totient function mean?

[tardy reef]

Euler's totient function counts the positive integers up to a given integer n that are relatively prime to n

[pearl pond]

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[dense badger]

I see so this might be a problem that can be only solved using a computer

[dense badger]

btw this problem just pop up on my mind while I was doing a similar type of problem

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[midnight haven]

trying to prove this

i went like

*by induction

(n+1 + 1) ^2 = (x + 1)^2 + 3

(x + 1)^2 + 3 < 2n^2 + 3

idk where to go after this

im thinking

(x + 1)^2 + 3 < 2n^2 + 3(x + 1)^2

3 < 3(x + 1)^2

(x + 1)^2 + 3 < 2n^2 + 6n^2

3(x+1)^2 < 6n^2

(x+1)^2 + 3 < 4*2n^2

ughh but thats wrong i needed n to be n+1

im stuck

[worthy lance]

Lets go back

First off, you gotta test the base case

Then assume the premise

[midnight haven]

yes

true

(n+1)^2 < 2n^2

lets say n = 3

4^2 < 2*(3)^2

16 < 18

✅

[worthy lance]

16 not 8

[midnight haven]

yeah

nvm that

[worthy lance]

Now we assume the premise

[midnight haven]

yes

[worthy lance]

So (k+1)^2 < 2k^2

[midnight haven]

yep

[worthy lance]

Now you have to prove for k+1

Ok let’s do it

(k+2)^2 < 2(k+1)^2

Right?

[midnight haven]

yep

right

[worthy lance]

Ok expand both sides

[midnight haven]

k^2 + 4 + 4k < 2k^2 + 2 + 4k

[worthy lance]

The right sise is wrong

Ok

4k and 4k are equal so we can eliminate them from the inequality

[midnight haven]

2 < k^2 where k>=3

that works

i think going from left side and try to make right kinda makes things little tricky

straight up putting values on both sides

nice

@worthy lance can u help with this approach!

(n+1 + 1) ^2 = (x + 1)^2 + 3

(x + 1)^2 + 3 < 2n^2 + 3

is it possible to get the same proof

via this

.close

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[barren sorrel]

Can someone explain this to me?

[barren sorrel]

How do you get -4 out of this

This is vectors btw

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[copper lily]

Is my height for this is 0m??

[cloud zephyr]

yea, i think it is 🙂

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[river oak]

I have been experimenting around with continued fractions and ended up with this one. Using code I managed to find out, that this series seems to diverge. How would I proof that x_n+1 > x_n for all n?

[midnight haven]

Have you considered an inductive proof perhaps?

[river oak]

I have, but it doesn't exactly feel very elegant. I feel like there should be a way to proof this using algebra

[midnight haven]

hmm I see.

btw did the inductive proof work?

$$\frac{n}{1 + \frac{n - 1}{2 + \frac{n - 2}{3 + ...}}}$$

[jolly parrot]

Edmund Cloudsley

[midnight haven]

$$\frac{n + 1}{1 + \frac{n}{2 + \frac{n - 1}{3 + ...}}}$$

[jolly parrot]

Edmund Cloudsley

[midnight haven]

well one argument could be

(and I am not entirely sure if I'm correct)

is that the numerator would be larger than the denominator in every single case

or iteration rather

therefore the X_n+1 must be larger than X_n

bit of an intuitive way to put it

[river oak]

Yeah that's what I got, but I don't know how I would proof that the numerator outgrows the denominator

Or since it seems to grow infinitely, what the growth would look like (in terms of a function)

To maybe explain how I got there: I stumbled upon this very interesting result

I then saw this in a stackexchange forum page:

So I wondered, since he came up with a way to turn series into a continued fraction, maybe the other way would be true too - turning a continued fraction into a series.

And maybe, using this series, it is possible to determine that it would diverge and therefore proove it?

Using some terrible python code:

[midnight haven]

hmm that is an interesting thought

or

you could also perhaps one of the iterations in terms of another iteration

like we do in integration sometimes

$$x_n = \frac{n}{1 + \frac{n - 1}{2 + \frac{n - 2}{3 + \cdots}}}$$
$$
x_{n-1} = \frac{n - 1}{2 + \frac{n - 2}{3 + \cdots}}$$

[jolly parrot]

Edmund Cloudsley

[pearl pond]

@river oak Has your question been resolved?

[river oak]

This also seems like a dead end :/

Also it doesn't seem to be correct, for instance:

$$x{n-1} = \frac{n - 1}{1 + \frac{n - 2}{2 + \cdots}}$$

[jolly parrot]

NinTobias

[river oak]

^ supposed to be x_n-1

[pearl pond]

@river oak Has your question been resolved?

[river oak]

I have made... a bit of progress. Consider the function

expanding it yields

where, if you take the limit as n->infinity, you get 20.

but adding more terms yields

So the more terms get added, the bigger the limit - and I think, x_n where n-> infinity diverges.

not at all rigorous - I'd be pleased and enlightened to know how to proof this rigorously (or even if it is possible to find a closed form for this).

[pearl pond]

@river oak Has your question been resolved?

[pearl pond]

@river oak Has your question been resolved?

[coarse harbor]

I think I have a rigorous proof but it's not very elegant

[jolly parrot]

EQUENOS

EQUENOS

EQUENOS

EQUENOS

EQUENOS

EQUENOS

EQUENOS

[coarse harbor]

Here I wanted to include the calculation of the senior coefficients of Q_{2k}(x) and R_{2k}(x) but to be fair that's not sufficient...

(Btw they're equal to k(k+1) and 1 respectively)

But I forgot that other coefficients might also rapidly grow with n... Nonetheless, I think it's very possible to show that other coefficients don't make much impact

[river oak]

Wow, this is quite dense. Gotta make a pot of coffee and work through it.

[coarse harbor]

Btw, the bot might accidentally close the channel soon: #help-39 message

I suggest pressing the x while the coffee is brewing

[river oak]

thanks!

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[pearl pond]

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[sterile turtle]

uhh wait a sec imma send my working out