help-13

[lament vector]

The second braket

[vital jay]

That looks perfect!

For further factoring you'd need a candidate zero

[vital jay]

Although if you're familiar with the quadratic formula you could also use that at this step

Or if you just figure it out by looking at it lmao

Can I ask how you got to that answer?

[lament vector]

The factoring?

[vital jay]

Yeah, how you got x^2-7x+10 = (x-2)(x-5)

[lament vector]

Th3 cross method

What multiples to the last term and adds to the middle

[vital jay]

Like, I figured it out by thinking "I need two values that sum to -7 and multiply to 10... -2 and -5 should do"

Awesome!

[lament vector]

Yea I did that but I put in on a cross to visualize it

[vital jay]

Makes sense!

[lament vector]

So is that it ig

[vital jay]

Yup!

You've shown -4 is a zero, and found the other two zeros

[lament vector]

Yes

[vital jay]

So anything else, or are we done?

[lament vector]

That's it, thx I understand now

[vital jay]

No problem, glad I could help!

[lament vector]

.close

[cedar kiln]

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[clever hazel]

i was attempted

prrof by contradiction

got a bit stuck

found like

subsituting it into one another

[chilly warren]

From dimensional analysis alone, (c) should be the correct answer.

[clever hazel]

but how?

[gentle echo]

dont think dimensional analysis applies here 💔

[chilly warren]

The other equation has mismatched dimension, so the output can change arbitrarily when the rectangle is scaled.

[clever hazel]

and what eve is dimensional analysis

[vestal bluff]

looking at the units

[queen stirrup]

do you know the am-gm inequality

[clever hazel]

no

[queen stirrup]

damn

[chilly warren]

This is probably not an olympiad question.

[clever hazel]

it sorta is

[chilly warren]

huh?

[clever hazel]

its a uni entry test exam

[hexed parrot]

which uni?

[clever hazel]

well

its like the Oxford one

but like im doing tmua but i ran out

so im doing MAT for practice

[hexed parrot]

idnt it a

[clever hazel]

its not b

[hexed parrot]

i mean its A no?

[chilly warren]

For simplicity, let's have a rectangle of size p and q.

[hexed parrot]

lol fair

[clever hazel]

non calculator asw

[clever hazel]

ok

[hexed parrot]

i know

[clever hazel]

its so over man

bro u think if i do fp2 and fp3 for pratice it wld help?

[hexed parrot]

lol

I mean

kinda

I didnt do edexcel so idk the level

have u done further?

[clever hazel]

like

ive done fp1

im doing fp2 and fp3

[hexed parrot]

i mean i dont think itd help too much

mat and tmua are far different imo

[chilly warren]

Then you substitute to the A and P. Also, I changed the parameter to p and q.

[clever hazel]

yh

u get a quadratic right

[chilly warren]

Yes

[clever hazel]

then what?

[chilly warren]

Let's look at the quadratic first.

[hexed parrot]

. dont u have worked sols

[clever hazel]

i do

but like

half the ppl here

[hexed parrot]

didnt make sense?

[clever hazel]

say its illegal to post

[clever hazel]

yh

[hexed parrot]

tf

[clever hazel]

i got the quadratic [art

part

[hexed parrot]

thats cap lol

send the sol

[clever hazel]

i got 2x^2-Px+2A

but they do discriminant

and god knows

why that wld work

cos like

???

[pastel vault]

recall that the discriminant is just the part under the square root, in the quadratic formula

[clever hazel]

yes

[pastel vault]

$x = \frac{-b + \sqrt{\Delta}}{2a}$

[clever hazel]

but why wld we need this

[wraith dagger]

south

[clever hazel]

it makes no sense

[queen stirrup]

the solutions to x are the length and breadth, which are real numbers

[pastel vault]

so if the rectangle has sides y, x instead

the perimeter and the area are the same

[hexed parrot]

if discirminant is negative theres no sols

[pastel vault]

so for any real x, y, you get two possible values of x

[hexed parrot]

Yeh exactly

[pastel vault]

two possible values implies discriminant > 0

[clever hazel]

yh but

like

idk

it jst

isnt working in my head

[queen stirrup]

like you dont understand the solution, or does it feel out of the blue?

[clever hazel]

out of the blue

man it feels like the more maths i do

the worse im getting

[pastel vault]

if you do a ton of these optimisation problems, you'll realise this

[queen stirrup]

yeah this solution is kinda wierd if you havent encountered questions like these befroe

[clever hazel]

bro

its like

idek anymire

its like

like very recently

i was pretty good at maths

now its like

everything i do

i get it wrong

its insane

destroyed my confidence lowk

[pastel vault]

MAT has that effect on people I guess

no one said getting into Ox would be easy

and just imagine how much more in-depth (and quicker) your uni lectures are going to be as an actual student

not to scare you, but people learn through facing challenges

if something doesn't work today, try again tomorrow

if some problem is too abstract, try breaking it down step by step

[clever hazel]

im not trynna get into ox

i jst need

to get like

26-27 marks

in my tmua

[pastel vault]

if some problem uses too many variables, try using numbers to figure it out

[clever hazel]

to save my future

if i fail

its over

[pastel vault]

wdym by 'save your future'?

you always are going to have options

[clever hazel]

gna be stuck at avg wage forever

[pastel vault]

ah

[clever hazel]

work hard for 1 yrr

or work hard forever

[pastel vault]

I noticed that you were using black-and-white thinking just there

what makes you believe that only an 'elite uni' will give you an above average wage?

there are clearly more good jobs than the number of elite uni places

[clever hazel]

no

its the other way round

plus introduction of AI

makes it worse

bro stuck at avg wage

will ruin ur life

idk how ppl do it

but thats cos

[pastel vault]

I know that all of us have a lot of feelings attached to our beliefs and throwing a bunch of statistics at you is only going to make you double down

[clever hazel]

as arrogant as it sounds but im so so so grateful

[pastel vault]

I really do want to understand how you feel

rather than try to prove my own point

[clever hazel]

but due to my father im already

above that

thanks to god

but i cant be reliant on him forever

and if i fail this tmua

my chacnes to get into a top top uni

diminish to 0

and ill go amn avg uni

avg job

avg life

god knows

[pastel vault]

I'm curious: what do you mean by this exactly?

[clever hazel]

i cant jst

live with my father

forever

like

i cant br like 30 living off his money

[pastel vault]

of course, no one wants you to live with your family forever

[clever hazel]

even if i had an insane job

[pastel vault]

but say if in the 1950s, you could buy a house with 3 times your yearly salary

[clever hazel]

i still wld wish to live with family

their everything

[pastel vault]

and now it's like 10 times, depending on where you live

[pastel vault]

oh

[clever hazel]

bro

obvs i wont

cos

ill have a family of my own

etc etc

but i cant afford

to be avg

i need to get into a top uni

to set myself

to the right path

to succeed

[pastel vault]

that's understandable

what would you say your definition of 'success' is then?

[clever hazel]

to be able to have the life i have now

to be able to have a good job

good wage

and good working hrs

enough money

to support my new family and old

[pastel vault]

okay, and what else do you enjoy about the life you have now?

[clever hazel]

nthn tbh

study wise

idrc

i do it

cos i have to

anything physical

competitive

fighting

sport

gym

thats abt it

[pastel vault]

right

[clever hazel]

thats the only reason i havent popped a vessel from these maths questions

obvs those arent the worse

but ykwim

[pastel vault]

but yep, it sounds like you're pretty miserable right now

[clever hazel]

always am tbh

ppl call me pessimistic or over optimisitc

wtv it is

i feel like

[pastel vault]

or in general, of course you know your own life the best

[clever hazel]

its pretty honest

like

i must succeed

everyone thinks that

ppl have hope though

if i fail

too much debt i owe to my family

i wont be able to cope

with the loss

[pastel vault]

I've noticed that when you talk about success, money comes up a lot

[clever hazel]

thats all there is atp

[pastel vault]

say, think about it this way

[clever hazel]

getting jacked and getting rich abt it

[pastel vault]

if your parents knew they couldn't afford to have you

they probably wouldn't give birth to you anyways

[clever hazel]

yh but theyve invested so much into me

i must pay it back

[pastel vault]

so they've made a choice to raise you

[pastel vault]

I'm also Asian if that helps

I'm southern Chinese

and say, yes, of course it'd be good if they were comfortable when they retire

but there's other ways of showing them that you love them

say, if you're out working and making big bucks all the time

you're not going to have much time to keep them company, right?

[clever hazel]

yh'

thats why i said

i want good working hrs

i dont want

to work some

12 hr shifts

bro 9-5 is so miserable

[pastel vault]

good working hours and a decent wage

[clever hazel]

ive had job exp

realistically

[pastel vault]

I feel you

[clever hazel]

its like a 7-7 if u account transportation etc

if u get 8 hrs of sleep

wake up 6 sleep 10

3 hrs with family

not enough.

[pastel vault]

work is pretty damn miserable these days, right

there's billions of people on earth who think the same thing about work

[clever hazel]

exactly

i wish

[pastel vault]

you're nowhere near alone

[clever hazel]

i was born

centuries before

living in my village

which is a thing

im from a tribal area of pakistan

wldve went out

to battle

probs died at the ripe age of 23

or died at 14 of a common cold

either one

[pastel vault]

but yeah, life is not the most fun these days heh

[clever hazel]

better than slaving away for 50 hrs

yrs*

for someone who knows nthn abt u

if i ever succeed

since im doing maths and econ

hopefully

ill be able to make bank

or even learn how to trade

[pastel vault]

people throughout history have stood up to unfair bosses and unfair working conditions

[clever hazel]

and make a living off that

[pastel vault]

sometimes violently, in fact, but not always

the labour movement is in every single country in the world

if one person resists, that might not do anything

but if a thousand people resist, or ten thousand, or a million people resist

then they can do a lot together

[clever hazel]

If everyone hates work

it wldve happened

for bargaining lower hrs

it wont happen

it cant

[pastel vault]

okay, do people in Pakistan still work Saturdays? I'm guessing if they're not that rich, they do

my point is, a lot of people in the West usually don't work Saturdays
because their ancestors said no, we'll ruin your whole business if you make us work long hours

no workers means no production, and no production means no business

so it really is possible

[clever hazel]

no

im not in pakistan though

i live middle east

[pastel vault]

fair enough, I guess that's not really important to what I wanted to say though

[clever hazel]

yh i get ur point

but theres no way

they can decrease

9-5

to some like

10-3

or sm

😭

[pastel vault]

yeah.........

we were supposed to have that today, according to predictions of the future....

[clever hazel]

predicitions

are always wrong

humans are unpredictable

it will nvr go right

[cedar kiln]

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[crimson sedge]

Eight people, including X and Y, are placed randomly in a row. Calculate the probability that X and Y are next to each other, given that Y is in the 5th position.

[tropic oxide]

!status

[cedar kiln]

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

[crimson sedge]

4

i have 2 options of answers

either it is 2x6!/8! or 2x6!/7! and i would like to know which one is right

[smoky cliff]

So

How did you get those?

[tropic oxide]

show your work for either one

[crimson sedge]

one sec

[tropic oxide]

also of note is they can (and imo should) be simplified

[crimson sedge]

i know

but i don't know, if there should be 8! or 7!

[tropic oxide]

,rccw

[tropic oxide]

well first off if you are talking about probabilities then they can never be a bare factorial.

do not confuse "number of ways" with "probability"

[crimson sedge]

okey

[tropic oxide]

anyway,

the total number of arrangements with Y fixed in seat #5 is 7! because you are arranging seven people and not eight

(because Y is already fixed in place)

(she can't move anywhere)

tbh though i would reason in a much more streamlined fashion like so: X has 7 equally likely seats to choose from, and of these 2 will put her next to Y.

[crimson sedge]

shouldn't it be 8! becouse you have 8 arrangements? isn't first part of task P(B) and other part is P(AnB)?

[tropic oxide]

if you really want to do it this way, then 8! will be the denominator of both probabilities

and it'll cancel out

[cedar kiln]

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[prisma warren]

can anyone help me with parts 2 and 3 of this question because i am stuck/

[prisma warren]

anyone available to help?

[cedar kiln]

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[warm plaza]

I'm asked to find the interval of convergence for this power series.
I used the ratio test and found that it approaches |x|/e. Which means if |x|/e < 1 it converges, >1 diverges, and inconclusive if =1.
Meaning it converges for -e < x < e, but I still need to check those endpoints.

Trying x = e, I get the series (n!/(n^n))*e^n which I'm not sure how to find convergence or divergence for. I tried the ratio test for it but got 1 so it's inconclusive

[dire geode]

did you try limit test

[warm plaza]

No, what similar series would I think of

that I would know converges or diverges

[warm plaza]

Do you mean limit comparison test or the test for divergence?
For the latter I’m not sure how I could solve that limit

[dire geode]

root test should also work for x=e

[warm plaza]

how do I advance from here

[dusk finch]

it seems like it will approach 1

,w lim as n -> inf of (e/n) * (n!)^(1/n)

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by methisalwaysright

[dusk finch]

maybe stirling could work

[warm plaza]

we haven’t seen something called stirling

[abstract breach]

this https://en.wikipedia.org/wiki/Stirling's_approximation

I don't know of any other way

[dusk finch]

n! ~ (n/e)^n * sqrt(2pin)

sth like htis

[warm plaza]

but I don’t think we’d do something like this in the class 😭

I’ll just ask my teacher later

[cedar kiln]

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[wise galleon]

the top 2 lines are the problem and the rest are my supposed solution, idk if its correct or not though

[wise galleon]

also, meant to put f_k(x) not f_k(f)

[cedar kiln]

@wise galleon Has your question been resolved?

[civic heron]

Idk if it’s too late but you may have seen that the limit as x->infinity of a fraction of polynomials is the quotient of the coefficient of the highest order term

And both are (basically) polynomials of degree n with leading coefficient 1

[cedar kiln]

@wise galleon Has your question been resolved?

[cedar kiln]

@wise galleon Has your question been resolved?

[dapper raven]

it seems good

i just made a python program to double check and it's good

[warm plaza]

If this means it approaches 1 then it doesn’t help me

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[ripe swallow]

I have no idea what to do

[hot cipher]

Youre basically aproximating the slope of a function at a point

Have you done any work on it?

[ripe swallow]

No

[hot cipher]

Gimme a sec please

The Slope of a function at a point is, logically speaking, the same as the slope of the tangent line to that point.

I hope you know what a tangent is

[ripe swallow]

Yea

Tangent is just one point but I think secant is two

[hot cipher]

Exactly

Youll soonly learn why the slope is important, but for now, we want to approximate it using secants

The way to approximate is by, instead of finding the tangent to a given point (which can take a lot of time or straight up be impossible manually)
We will find the Slope between two points which are really close to each other.

[ripe swallow]

So we would put 2,4 and x,y into slope formula?

I see those two points

[hot cipher]

Something like that, yeah

The closer those two are, the closer the aproximation

Lets say we take the points from f(0) and f(1)
which are
(0, 0^2) and (1, 1^2)

a difference of 1 between the two points x coordinate makes for a horrible approximation, specially at the lower side of this function

If we reduce the difference to 0.2, we get a better one

And if we do a difference of 0.01, we get a really close approximation to the real tangent.

[hot cipher]

Here, they are asking you to find the slope at (2,4) or (2,f(2))

For that, they ask you to find multiple points (3,f(3)), (2.5,f(2.5)) etc...

and calculate the slope for each point with respect to (2,f(2))

[ripe swallow]

Ok i figured it out

Thanks

[hot cipher]

no problem, good luck crunching numbers

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[summer iris]

Hi!!! I just need someone to explain something to me so I could understand.

On this graph, the domain is (-infinity, positive infinity)

And the range is [0, infinity)

Which I don’t understand because I thought the domain is all x’s and clearly there is a point on -6
I’m not sure how the range is 0 either

[hot cipher]

Well, lets go over, do you know what the Domain and Range/Image are?

[summer iris]

Here’s more to the problem.

[summer iris]

Isn’t the domains any number on the x-axis?

[hot cipher]

kinda
its all the values on the x-axis for where the function is defined.

[zinc spade]

Domain is the x values and range is the y values for a given function.
It is basically the values of x and y the function includes/the function defines

[hot cipher]

x^2 turns to be defined everywhere

[zinc spade]

Like if you had a straight like going from 0 to 10, the domain would be [0,10]

However, if you had a line that starts at 0 and goes infinitely to the right (more +), it would be [0,+infinity)

[summer iris]

Right so how isn’t the domain (-6, infinity)

[hot cipher]

That would imply that the function doesnt exists to the left of -6

[zinc spade]

No.

It would be if the graph stopped at -6.

Well with soft bracket technically would mean stop at but not inclduong -6

Basically, if you can take the function and plug in a value of x and it has a value of y associated with it, then its in the domain.

^ Not like 100% perfect answer but like generally trying to describe it

[hot cipher]

Ill go over an example:

$\frac{1}{x}$ has domain $(-\infty,0) \cup (0, \infty)$
Which means that you can plug in every value except 0

For $x^2$ you can plug in all values.

[summer iris]

Could you show me an example please?

[wraith dagger]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[zinc spade]

Sure

[zinc spade]

Well first of all, do you understand the concept of a function?

As in, do you know the very basic def of a function

If you dont thats okay I can explain

[summer iris]

Yes

[zinc spade]

can you tell me?

[summer iris]

You need to have a single output.

For every input

[zinc spade]

For each input, correct.

So lets look at the parabola you gave us

If we thing of the input, x, if I were to say what y value goes with x=7, could you find what y value is associated with it?

Like generally could you

As in is it possible

[summer iris]

Yes? It should be possible
I’m sorry if the answer is wrong

[zinc spade]

Yes 100% correct.

It is very possible and very solvable as well (although not getting into that rn lol)

Basically, we know the graph goes INFINITELY in the + and - x directions, MEANING that for EVERY x value, there is a y value, and thus the domain is all real numbers. For the range, we see how the vertex is at (-5,0), meaning the MINIMUM value of y is 0 and the MAXIMUM is infinity. So it foes from [-5,+infinity). Does that make sense?

[summer iris]

Kinda???

[zinc spade]

Okay so

for y=x. what is the domain and range?

[summer iris]

D= (-infinity, infinity) R=[5, infinity) ?

[zinc spade]

Close, but not

The domain is correct

The range has an error

Think of the function, less so the image I sent. Go onto desmos and plug in y=x and play around with what you see

then tell me the answer

[summer iris]

Is it [O, infinty )

[zinc spade]

No.

Think about it.

For what Values of Y is there a value x

Or I mean, are there any values of y that are not somwhere on this graph?

LIke if I said go to y=1000000, would there be a point (...,100000)?

[summer iris]

No there wouldn’t be

[zinc spade]

There would be

Im gonna find a youtube video for you to watch

[zinc spade]

Watch these 2 videos:
https://www.youtube.com/watch?v=fyROLkZc75E
More in depth --> https://www.youtube.com/watch?v=KirGQOwjBVI
Optional (Likely to be too complicated): https://www.youtube.com/watch?v=djT6-YamHaA

[summer iris]

Okay thanks

[zinc spade]

I gotta finish a lab, but ping me if you are still stuck. Try the practice problems in the videos before watching solutions

[cedar kiln]

@summer iris Has your question been resolved?

[cedar kiln]

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[spice mountain]

Is my answer to the 2nd part of b wrong?

[hollow trail]

you can write T[0] = 0 so that's false

you can't just arbitrarily specify that it must be nonzero

also "Ax = 0 does not have non-trivial solutions" is irrelevant to whether it's onto

[spice mountain]

But I said x ≠ 0

[spice mountain]

That’s not what I lose points on anyway

Lost*

[hollow trail]

they asked you to find a vector y such that there is no vector x such that T[x] = y

you did not do that

[spice mountain]

Ahhh ok I see

Thanks

[cedar kiln]

@spice mountain Has your question been resolved?

[cedar kiln]

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[brittle tendon]

I just need assistance on part d. i have integrated the numerator and think it is wrong

[brittle tendon]

F(xy) is this

[cedar kiln]

@brittle tendon Has your question been resolved?

[brittle tendon]

.close

[cedar kiln]

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[spice mountain]

What class is this

[cedar kiln]

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[spiral iris]

lim as x app +inf sqrt(x²+1)*arctanx -pi/2x

[spiral iris]

help me nothing works

[long swan]

Multiply by conjugate

Then lhospital

[spiral iris]

not accepted

[long swan]

Still multiply by conjugate

Trust

[spiral iris]

i did

its still inf*0 no matter what i do

[dry root]

could u send the ques again

[spiral iris]

lim as x app +inf of sqrt(x²+1)*arctanx -pi/2x

[dry root]

sry unable to understand

[spiral iris]

clearer?

[dry root]

hmm

[spiral iris]

i tried using arctanx+arctan1/x=pi/2 but it doesnt work

[dry root]

L hopital saves the day

[spiral iris]

not accepted

[dry root]

huh no

let me try another method then

[spiral iris]

our school's sytem is pretty stupid

[sand cradle]

Are you allowed to Taylor expand?

[upper ruin]

What about a factorizing x?

[spiral iris]

didnt study it yet sorry

[spiral iris]

i mean i havent really thought of it but how will that help us?

[upper ruin]

I don't know yet, I'm just suggesting possible ways

[sand cradle]

Do you know some stuff about arctan?

[spiral iris]

like what

[upper ruin]

arctan(X) + arctan(1/X) = sgn(X)•π/2
Or, if X is positive, arctan(X) + arctan(1/X) = π/2

[sand cradle]

or $\lim _{y\to0}\frac{\arctan (y)}{y}=1$

[wraith dagger]

Kepe

[spiral iris]

whats "sgn"

[sand cradle]

X/|X|

[spiral iris]

yeah ik that

[upper ruin]

The sign function

[spiral iris]

x is positive so its equal to pi/2

yeah ik that i studied it

[sand cradle]

\li \ Ok, let's factor out an $x$ to get [\lim _{x\rightarrow +\infty }x\left(\sqrt{1+\frac{1}{x^{2}}}\arctan x-\frac{\pi }{2}\right),] afterwards we can change this into $y \to 0$ by defining $y = \frac 1x$ and we get [\lim _{y\rightarrow 0^{+}}\frac{\sqrt{1+y^{2}}\arctan \left(\frac{1}{y}\right)-\frac{\pi }{2}}{y}] and by what Alberto gave we get [\lim _{y\rightarrow 0^{+}}\frac{\sqrt{1+y^{2}} \l(\frac{\pi }{2}-\arctan (y)\r)-\frac{\pi }{2}}{y}] and factoring out $\frac \pi 2$ we get [\lim _{y\rightarrow 0^{+}}\frac{\frac \pi 2 \l(\sqrt{1+y^{2}} - 1\r) -\arctan(y) \sqrt{1 + y^2}}{y}][ = \frac \pi 2\lim {y\rightarrow 0^{+}}\frac{\sqrt{1+y^{2}} - 1}{y} - \lim{y \to 0^+}\frac{\arctan(y) \sqrt{1 + y^2}}{y}] and multiplying by the conjugate, we have [ = \frac \pi 2\lim {y\rightarrow 0^{+}}\frac{1+y^{2} - 1}{y \l(\sqrt{1 + y^2} + 1\r)} - \lim{y \to 0^+}\frac{\arctan(y) \sqrt{1 + y^2}}{y}] [= 0 - 1 = \underline{\underline{- 1}}]

Now we could change that arctan(1/y) into arctan(y)

with what Alberto gave

We have y -> 0^+ here so sgn(y) = 1

@spiral iris

[spiral iris]

im thinking

[sand cradle]

And we should basically be done

Because now we can pull this apart

Well ok not yet

[dry root]

[spiral iris]

it still doesnt give anything

[sand cradle]

One sec

Ok now

And now we can break it up into two limits (this is a little abuse of notation, but at the end just read it from right to left and everything is fine):

@spiral iris See what happened there?

Just pulled apart the fraction

[spiral iris]

is it equal to -1?

i think it is

[sand cradle]

we will get to that in a second

[spiral iris]

yes

THE RHS limit is equal to 0 and the LHS is equal to -1

ok but i would never reach this method by myslf 😭

[sand cradle]

Yeah seems so

[spiral iris]

im 90 sure im falling

[sand cradle]

We have a y in the denominator that goes to 0, denom goes to 2

so that's 0

The right summand is 0 because of arctan(y)/y -> 1 and sqrt(1 + y^2) -> 1

And that's a 0 - 1

[spiral iris]

thanks man

ur a life saver

[wraith dagger]

Kepe

[spiral iris]

thanks guys i love you all 💜 @sand cradle @upper ruin @dry root

.close

[cedar kiln]

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[cedar kiln]

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[wet yew]

I'm definitely sleep deprived, but what did they do to get to the 3rd derivative from the 2nd? x-x

[foggy merlin]

product rule

[tropic oxide]

product + chain rules

[wet yew]

oh I'm so dumb, thank you xDD

oki I'ma go for a bit, but might return. Thank you both :D

.close

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[spring stump]

how can you tell whether a graph will fall or rise on either side in non linear functions

[tropic oxide]

wdym by "on either side" exactly

[spring stump]

higher power functions where they go different directions on either side

[ember heath]

could u give an example

[spring stump]

pff um

-3(x^2 -1 )(2x-3)^2

for this one it falls on both sides, but how would you tell that

[crimson delta]

imagine multiplying it out. what is the leading term?

you do not have to fully multiply it out

[spring stump]

so does that only work if both ends are going in the same direction

because what would it look like if one end was rising and one end was falling

[spring stump]

for the leading term

[crimson delta]

the leading term is a x^n. there are now 4 options: a positive/negative and n even/odd
a positive, n even: starts top left, ends top right
a negative, n even: starts bottom left, ends bottom right
a positive, n odd: starts bottom left, ends top right
a negative, n odd: starts top left, ends bottom right

[spring stump]

alright im getting it ya

thank you thank you

.close

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[pastel shore]

Can someone help I don’t really understand mathematical induction for this

I know we let assume n =k

And then h get k(3k-1)/2 as ur assumption

U

But idk where to go from there

[lucid plaza]

just add (3(k+1)-2) to both sides and show that k(3k-1)/2 + (3(k+1)-2) = (k+1)(3(k+1)-1)/2

@pastel shore

[pastel shore]

Ah ok lemme try rq

[pastel shore]

Sorry do u know where I go from here im learning ts for the first time 😭

[lucid plaza]

it's (k+1)(3(k+1)-1)/2 in the first step on the right side

i think i had a typo there or something sorry

[pastel shore]

Ah alr I’ll try again

[lucid plaza]

also 3-2 = 1 not -1 lol

[pastel shore]

U right xD

Now I have this

[lucid plaza]

i can't see the right side

wait i meant right side last time too

[pastel shore]

Bruh 😭😭

[lucid plaza]

sorry i mess up left and right a lot

[pastel shore]

XD it’s fine

I’ll try again

[lucid plaza]

you did the left side here correct tho

you just gotta do that for k(3k-1)/2 + 3(k+1) - 2

[pastel shore]

Alr

[pastel shore]

I messed up something here prob but this is what I got

Wait I realised

It’s -1

Nit +1

Nit

[lucid plaza]

lmao

[pastel shore]

XD

[pastel shore]

Finally

Tysm for the help

[lucid plaza]

yw nice job

[pastel shore]

Got there eventually

[lucid plaza]

you understand how induction works right?

j wanna make sure

[pastel shore]

Pretty much u prove the base thing by subbing in n =1

Assume n = k

And find the n = k+1 I think

Smthn like that

I’m tryna teach myself it

[lucid plaza]

basically you prove that P(1) holds and, assuming P(k) holds then prove P(k+1). then you let k = 1 since you've already prove P(1), and that proves P(k+1), i.e. P(2), then let k = 2 and that proves P(3), etc etc. if you want i can show you an example

where p is a property or formula or something

[pastel shore]

Sure I’d like to see one

Ye

Bc I get everything until the k + 1 bit

[lucid plaza]

oh wait i copy pasted this from another answer, the example is the one you just did lmao

you can prove n = 1 here, then since you've proven k+1, that means you've proven n = 2

and that means you've proven n = 3

so on and so forth for all natural numbers

[pastel shore]

So to prove k + 1 you just add it to both sides

Is that right?

Bc that’s the part I mess up on

[lucid plaza]

at least for sums/series yes

remember that sigma notation just means a_i + a_(i+1) + ... + a_(n)

[pastel shore]

Ye

So far the only induction problems I’ve done were the tower of Hanoi and these series type ones

[lucid plaza]

adding the case where n = k+1 works there since it's just another way of saying add, for different problems you'll have do other stuff but yea

[pastel shore]

Alr Ty I’m prob done for today

Gonna go to bed

Ty for the help

[lucid plaza]

np don't forget to .close

@pastel shore

[pastel shore]

Almost did Ty

.close

[cedar kiln]

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[idle sky]

is it sufficient to say that e^x = 1 + x/1! + x^2/2! + ... + x^n/n!, therefore, e^1 = LHS, which is less than 3

[smoky cliff]

Yes

Technically

But i suppose they do not want you to do that

As it trivialises the problem

I think this is rather a way to see that e is less or equal to 3

I'd suggest to estimate the terms

[cedar kiln]

@idle sky Has your question been resolved?

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[tiny berry]

how do i solve this integral? apparently i tried simplifying $r_k$, but i ended up getting a complicated expression which seemed unsolvable.

[wraith dagger]

Veds

[tiny berry]

i used integration by parts to simplify the denominator one

[jade blaze]

whats ur approach

working*

[tiny berry]

i seperated $(1-x^7)^(k+1)$ as $(1-x^7)^k * (1-x^7)$

after that i used integration by parts to try to simplify it but i wasnt getting anywhere with that

[wraith dagger]

Veds

[jade blaze]

try substitution

putting t = x^7

[tiny berry]

but for that i will need dt that is 7x^6

[jade blaze]

put x as t1/7 ig

[tiny berry]

nah evn though you substitute, you will have to use by parts but that is complicating it more than simplifying it

i tried doing that right now

[cedar kiln]

@tiny berry Has your question been resolved?

[cedar kiln]

@tiny berry Has your question been resolved?

[cedar kiln]

@tiny berry Has your question been resolved?

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[pale hound]

can I try this question? it says it was closed unsolved

[flint plinth]

you can, but OP might be gone by now

[dire geode]

don't open channels for other people for ^ reason

.close

[cedar kiln]

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[pale hound]

@tiny berry
is the answer 65?

[pale hound]

.close

[cedar kiln]

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[formal copper]

help me with the figure so i can proceed

[pale hound]

okay, lemme tryy

[dire geode]

you can use desmos to plot

[pale hound]

dude, we dont get to use desmos in JEE t~t

[formal copper]

what is it

[dire geode]

this isn't your channel

[dire geode]

https://www.desmos.com/calculator

[formal copper]

i cant decode the language so

[formal copper]

try

[dire geode]

which word confuses you

[pale hound]

I see

[formal copper]

cant write circum circle eqn

and director too

[brazen merlin]

beautiful question 🫠

[formal copper]

thanks did you tried>?

[brazen merlin]

making the figure
🫠

[dire geode]

here's hwo you can get the circle equation from the triangle:
https://math.stackexchange.com/a/213670

[formal copper]

i made the figure want to verify and check my approach

[brazen merlin]

what's your approach ?

[formal copper]

man we dont gotta use this in exam we indians are not used to this kind of stuff

[dire geode]

then you should save everyone time and just show your work

[dire geode]

like it says here

[formal copper]

yaeh will do it

[dire geode]

why not

[pale hound]

dude, is the answer 6?

Just did a bit of scribbling

[dire geode]

which part of this is too advanced

[formal copper]

not allowed

lemme check

myst

[pale hound]

keep ur solutions with u man

[pale hound]

lemme know if its 6

I'll write it properly then

[formal copper]

you got it

pls proceed

[pale hound]

okkay, 5 mins, I'll prepare a detailed solution

[dire geode]

can you not

!nosols

[cedar kiln]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[formal copper]

why cant he?

[dire geode]

part of the rules

read #rules

[formal copper]

which rule can you specify i have lack of time

i just wanted to solve this ques so did i joined this channel

[pale hound]

I hv not shown all the calcns here, u can work them out

[chrome elk]

!nosols again

[cedar kiln]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[chrome elk]

Also your name is in the picture

[pale hound]

its the watermark

but he needed the solution man

questions like these cant be helped with just thru chat

Ur rules must have exceptions

[worldly chasm]

No

[solar pecan]

Well I doubt it

[worldly chasm]

He needed help

you gave him the solution

these are different things

[chrome elk]

We should not be setting a precedent where we hand out solutions and people copy them

[pale hound]

its explained properly there tho

but im new here, so ill just listen to u guys

[solar pecan]

You have to guide him tho, smth helper spend like 1-2 hours to walk through a problem with helpee

[calm sierra]

our server works by helping people through conversation (think socratic method), not by being chegg

[pale hound]

im sorry chat

[tropic oxide]

im really tempted to say skill issue, cause yes they can.

[calm sierra]

1-2hr is a bit extreme, no need to push yourself like that

[pale hound]

@formal copper you can just dm me if u dont understand any line

[calm sierra]

engage in moderation 🙂

[solar pecan]

I meant ||lnrd||

[pale hound]

i see, but i must let u know im recognised by indian maths Olympiad association, i doubt the word "skill issue"

Im sorry

just dont make it an argument

[calm sierra]

pls dont take the skill issue thing to heart

[chrome elk]

You're the one making it an argument, with all due respect

[tropic oxide]

with all due respect: good at oly math ≠ good at teaching math.

[calm sierra]

but now u know how we expect helpers to conduct themselves here

[pale hound]

i said, im sorry

[formal copper]

thanks i got it

and sorry chat

[pale hound]

again, my grading rates state different words which includes, the ability to teach fellow students

[formal copper]

it was my fault not his

[pale hound]

jesus, just stop it

[chrome elk]

You're the one continuing it my guy

You can't say things and expect people to not respond

Stop talking

[calm sierra]

we mods already got our msg across, no need for anyone else to pile on

[brazen merlin]

@pale hound i support that thing that question like these take a lot of efforts to explain through chat
and you are right at this thing

[tropic oxide]

!done @formal copper

[cedar kiln]

If you are done with this channel, please mark your problem as solved by typing .close

[chrome elk]

If it's too much effort for you, don't do it

[formal copper]

yeah he's right actually

[pale hound]

these people wont understand, theyll link wolfram and desmos and say they are helping in understanding "visually"

[brazen merlin]

shut up

[brazen merlin]

right

[pale hound]

whats ur problem man

u couldnt even solve it

[brazen merlin]

[chrome elk]

I wasn't even involved what do you mean

[pale hound]

get me kicked out if u can

[pale hound]

try and explain me

lets see how u can

[chrome elk]

I have better things to do, unsurprisingly

[pale hound]

with all ur "skills"

[tropic oxide]

myst mate you're definitely dragging it out

[pale hound]

like poking in other peoples' stuff

[pale hound]

id like to stop it here

[dire geode]

cool everyone's done

.close

[cedar kiln]

Channel closed

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[chrome elk]

No, like helping people who actually need the help

[brazen merlin]

😏

[tiny berry]

yesss thank you so much

[cedar kiln]

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[formal copper]

.close

[native heath]

skilled in math \neq skilled in teaching

[native heath]

oh nvm ann beat me to it

and its an available channel

oh well

.close

[cedar kiln]

Channel closed

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[pale hound]

See dude, i already said im sorry, besides ur statement has also been answered at a later reply 🙂

.close

[civic heron]

Why not

[warm plaza]

If the result of the root test is 1 then it’s inconclusive and I’d need another method to test convergence

[chrome elk]

Gonna ask you guys to claim another channel since this one's technically occupied and might close randomly

[cedar kiln]

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[ember heath]

If a,b,c and a',b',c' are reciprocal system of vectors then
a.b' +b.c' + c.a' = 3. I need to tell if this is true or false

[dire geode]

what's the definition of reciprocal system of vectors

[formal copper]

its a teacher association

[ember heath]

i googled it and i got
a.a' = 1
b.b'=1
c.c'=1

[flint plinth]

so for example if a,b,c are unit vectors and you take a' = a, b' = b, c' = c, then this qualifies?

[ember heath]

yeah

[flint plinth]

does the result necessarily hold in that special case?

[ember heath]

yes it should right cause cos theta will always be 1?

[flint plinth]

well is a.b always equal to 1?

[ember heath]

no..cause a is not always parallel to b

[flint plinth]

in fact the definition seems to require that it is not:

(assuming this is the right definition)

[ember heath]

what is not

[flint plinth]

a.b' cannot be 1, it is required to be 0

according to the above

however, we need to be sure that your definition of "reciprocal system of vectors" is the same as the above

is it defined in your book or notes?

[ember heath]

no ive never even been taught it formally it just came in one of the questions

[flint plinth]

they just introduced the term without defining it?

or was there an earlier question that defined it?

[ember heath]

yea

wait ill send the original one

[modern carbon]

Guys, in multi-variable calculus while testing our 3d graph in a specific point according to the discriminant formula test, how can I check it if the discriminant =0? (if it was a mixima, minima, saddle, or neither )

[flint plinth]

pls see #❓how-to-get-help , you should open your own channel as this one is in use

[modern carbon]

mb

[modern carbon]

sry

[flint plinth]

nw

[flint plinth]

and they've never used the term before this? not even in any of the previous questions?

[ember heath]

nope

the questions are all independant

not like a comprehension

[flint plinth]

strange, ok let's assume the definition i gave is right, i've seen it in a few places so far

and i haven't seen any conflicting definitions

so ok, what happens if a,b,c are an orthonormal set of vectors and you just take a' = a, b' = b, c' = c

that satisfies all the conditions, right?

[ember heath]

orthonormal means mutually perpendicular?

[flint plinth]

yep, and also they all have length 1

[ember heath]

oh ok yes

[flint plinth]

so does that satisfy a.a' + b.b' + c.c' = 3?

oops sorry i meant:

a.b' + b.c' + c.a' = 3

[ember heath]

no

it will be = to 0 right?

[flint plinth]

yep all three terms on the left side are zero so they sum to zero

[ember heath]

ohk so by taking an example we can say that it is wrong

got it

thank you

[flint plinth]

right

one counterexample is enough since they're asking if it's true for an arbitrary such system

[ember heath]

got it

ill close this now?

[flint plinth]

sure

[ember heath]

kk

.close

[cedar kiln]

Channel closed

Closed by @ember heath

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[ember heath]

thanks again

[flint plinth]

yw

[cedar kiln]

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[halcyon flax]

Can someone explain me 16:21 from https://www.youtube.com/watch?v=d-o3eB9sfls ? A video from 3 Blue 1 Brown about the Basel Problem. I understand why it is x1/4, but I do not understand why it is x3/4. He said, so that it is complete like 3/4+1/4=1 I think, but I do not understand why it needs to be complete and even if it is right if it is complete.

[dusk finch]

sum of all terms = even terms + odd terms
ill write that as S = E + O
you said you understand E = S/4
that means S = S/4 + O, subtract S/4 and you get O = 3/4 S

it needs to add up to 1 because even terms and odd terms add up to the full sum basically

[cedar kiln]

@halcyon flax Has your question been resolved?

[halcyon flax]

@dusk finch I see, thanks for the help! You explained it quite precise 😅

.close

[cedar kiln]

Channel closed

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[cedar kiln]

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[viral fox]

Hello great math fellows, I'd have a quick help regarding linear algebra.
How may I prove that [x, iy] = -[ix, y] where [,] is a real inner product.

[hybrid torrent]

hey

[pale hound]

consider conjugate linearity in the RHS term [ix,y] and linearity in the LHS term, [x,iy]

[pale hound]

hi

@viral fox I hope its in sesquilinear form, coz when done with linearity in both sides, we dont get the negation

[viral fox]

May I just use the fact as [x,x]=[ix,ix], then conclude that for any two vectors a, and b , if we rotate both vectors, then their inner products remain the same, hence we can find i as a rotation by pi/2 op?

[pale hound]

yeah it works that too ig

[viral fox]

Actually <a,b>=[a,b]+i[x,iy] , and [a,b]=[b,a] as [,] is real, but <a,b> satisfies all axioms in the complex field.

[pale hound]

yeahh

[viral fox]

Much appreciated. thank you for your time and consideration. Really appreciate it. Math bless us all. 🤘🤓🍻

.close

[cedar kiln]

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[tulip crow]

y=6(x-4)^3-2?

[flint plinth]

your expression will do the "stretch by 6" before shifting down 2

also are you sure that "stretch" means "vertical stretch"? it's ambiguous as stated

[tulip crow]

yes

[cedar kiln]

@tulip crow Has your question been resolved?

[cedar kiln]

Channel closed

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[sturdy star]

i need help understanding truth tables i understand basic ones such as p -> q but if it adds in more such as [p ->q] ~ [q ^ r] i don't understand

[dusk goblet]

have you made a truth table yet without filling this in?

[sturdy star]

yes

[dusk goblet]

can you send it?

[sturdy star]

this isn’t that exact question but it’s similar to what i’m having issues with

[dusk goblet]

are you required to add columns for (pvq) and notp?

its generally advisable

[sturdy star]

they are not graded but i can add them in

[dusk goblet]

so what should the column for pvq look like

assuming the first column is p and the second is q

[sturdy star]

t t t f

[dusk goblet]

yep

now notp should be simple

i'll assume you have no issues there

when can an implication ever be false?

[sturdy star]

if both are false

[dusk goblet]

not quite

anything follows from a false statement

if today is monday the moon is made of cheese

true

an implication p -> q is only false when p is true but q is not

an easy way to see this is to notice that p -> q is equivalent to notp or q

so negating it gives p and not q

[sturdy star]

would the answer for it would be F F T T or is that just for notp

[dusk goblet]

for the (pvq) --> ~p?

[sturdy star]

yes

[dusk goblet]

well we only need to look at the first 3 rows when pvq is true and look to see if its ever the case that ~p is false, i.e., p is true and in those cases we will have that the implication is false so you are right

oh wait yes no you're right

you said F F T T

[sturdy star]

thank you so much

[dusk goblet]

you're welcome

[sturdy star]

.close

[cedar kiln]

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[cedar kiln]

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[tight widget]

Question and answer.
I don't get how there can be 3 square having vertices (1,0) and (2,0). I can only visualize two, one in 1st quadrant and other in 4th quadrant.

[tight widget]

OMG, didn't thought about that given points can be diagonals. Thank you!

[neat dune]

no problem!

[tight widget]

.close

[cedar kiln]

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[crimson solstice]

hm I need help with some theory

suppose you have 2 vector spaces