help-36

[icy wind]

yeah it is

[gritty solar]

yup and

if we add k

[icy wind]

ohhh

then it would be 2k

cool

[gritty solar]

yeah

perfect

[icy wind]

how does one write that like fancy tho

for a proof

[gritty solar]

Easiest would be the triangular inequality

[icy wind]

idk that

[gritty solar]

$|z_1 + z_2| \leq |z_1| + |z_2|$

[soft zealot]

NEON

[gritty solar]

this?

[icy wind]

no we haven’t done that

my teacher would b like whattt

[gritty solar]

Hmm

Probably just try z = x + iy

[icy wind]

huh

[gritty solar]

$0 \leq |z + k| \leq 2k \quad |z| \leq k$

[soft zealot]

NEON

[gritty solar]

Substitute z = x + iy and calculate the value of |z + k|

[icy wind]

bro icl i have no idea what that means

[gritty solar]

how do you represent a complex number

[icy wind]

x + iy

[gritty solar]

yeah

so z can be written as x + iy

[icy wind]

yes

[gritty solar]

That is what I ask you to do

write z as x + iy

[icy wind]

ok

[gritty solar]

and find |z + k|

[icy wind]

| x + iy + k|

[gritty solar]

the inequality should be easy to prove since you know |z| =< k

[gritty solar]

yeah

and how do you find the modulus of a complex number

[icy wind]

um

oyhtargoras

pythag

[gritty solar]

$|a + ib| = \sqrt{a^2 + b^2}$

[soft zealot]

NEON

[gritty solar]

yeah

[icy wind]

so (x+k) ^2 + y2

then root

[gritty solar]

yes good

now use (a + b)^2 on the first term

[icy wind]

yup

x^2 + 2x + k^2 + y^2

[gritty solar]

2xk

yes

[icy wind]

ok but now what

[gritty solar]

can you prove that is greater (lesser) than (or equal to) 2k

the square root of that

if you know sqrt(x^2 + y^2) < k

[icy wind]

you genius

wiat

i can get it less than or equal to k but not 2k

[gritty solar]

oh yeah

that

mb

[icy wind]

so how

would i do

[gritty solar]

$x^2 + y^2 + 2kx + k^2$

[soft zealot]

NEON

[gritty solar]

So this is what you have

Remember how I said the maximum real part of z will be k

What would the corresponding imaginary part be?

[icy wind]

um

x + iy

oh sorry it

iy*

[final saddle]

@icy wind Has your question been resolved?

[icy wind]

.close

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[ashen shoal]

what have you done so far

[tranquil pine]

!status

[final saddle]

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

[tranquil pine]

.cloxe

.close

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[warm python]

how would I start this

[warm python]

I mean I could express A in terms of B

by dividign eqn 1 by 2

$\left(\frac{4\sin\left(A\right)}{3\tan\left(A\right)}=\frac{\sec\left(B\right)}{\tan\left(B\right)}\right)$

[soft zealot]

Why am. I here

[warm python]

which is

$\left(\frac{4\sin\left(A\right)}{3\tan\left(A\right)}=\frac{\sec\left(B\right)}{\tan\left(B\right)}\right)$, $\frac{4\cos\left(A\right)}{3}=\frac{1}{\sin\left(B\right)}$

$\frac{4\cos\left(A\right)}{3}=\frac{1}{\sin\left(B\right)}$

[soft zealot]

Why am. I here

[warm python]

now what?

just a hint please

[scarlet sequoia]

oops sorry

[tranquil pine]

sec is cosB

secB= 1/cosB

not 1/sinB

[warm python]

$\frac{4\cos\left(A\right)}{3}=\operatorname{cosec}\left(B\right)$

[soft zealot]

Why am. I here

[warm python]

I guess I have to apply transformations to this?

or would that be wrong

i.e product to sum

though I don't see that helping much

[scarlet sequoia]

sin(A+B) I would compute

[warm python]

ok, so $sin(A+B)=\sin{A}\cos{B}+ \frac{3}{4}$

[soft zealot]

Why am. I here

[scarlet sequoia]

and sinAcosB = sinA/secB

[warm python]

I don't know sec(B) though

oh

nvm

got it

yeah, so sin(A+B)=1

so A+B=90deg

thanks

.close

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[chrome hemlock]

how come my sun rays look so strange

[warm python]

you probably want to change the line equations a bit

and their domains too

[chrome hemlock]

LOL i cant right now

[final tangle]

because they don't aim towards the centre of the sun (or close)

[chrome hemlock]

oh 😭

[final tangle]

that's why it looks strange

[gritty solar]

thats so amazing

[warm python]

this is for a school project ?

[chrome hemlock]

ye i have to investigate how the lines work

and then write a 1000 word report

[craggy plank]

The emerging sun?

[chrome hemlock]

investigations are like self learning

the teachers dont teach us anything ab this

idek waht a domain is

[sonic crystal]

woah what is this

[chrome hemlock]

should i do something like this

[warm python]

a domain is the set of inputs a function can take

probably, but change the domain of the function

[chrome hemlock]

bro methods investigation

i have a test right after it

[sonic crystal]

a 1k word report for math 😭

[chrome hemlock]

right ffs

i did so trash in specialist cant be flopping methods too

[warm python]

anyway, you'll want to change the domain of the function to exclude that pafrt of the line inside the sun

[sonic crystal]

such a rip

[chrome hemlock]

im confused

[chrome hemlock]

are u doing chem legit its a bit too hard studied all night for a 76%

[warm python]

about what?

what a domain is?

[chrome hemlock]

yes and what do i change it to?

[warm python]

A domain is the set of inputs a function can accpet

accept

[chrome hemlock]

uhhh

[warm python]

say x=y

[chrome hemlock]

so like when calculating cosine rule angle

[warm python]

what inputs can it accept

[chrome hemlock]

i be like

let lowest degree be 0 degrees and highest be 180?

[warm python]

i.e what values can x take

[chrome hemlock]

in my classpad?

[warm python]

here

[chrome hemlock]

ok so domain is basically just how long the line is

i make

[warm python]

no

[chrome hemlock]

?

oh

[warm python]

it's what inputs a function can accpet

say x=y

what values can x take

[chrome hemlock]

is x=y a line

[warm python]

All reals, right?

x can be any real number

[chrome hemlock]

y=1x

oh

[warm python]

can't it?

[chrome hemlock]

yes

[warm python]

so instead, say x can only take values between -5 and 5

you get a line segment, right?

[chrome hemlock]

then y - -5, -4,-3,-2,-1,0,1,2,3,4,5

[warm python]

you effectively want line segments here

[warm python]

those are just the integral values

[chrome hemlock]

yes

[warm python]

π is a valid input here too

for x=y where x is between -5 and 5

[warm python]

can I see the line equations for these

I'll show you an example

[chrome hemlock]

uh........

i didnt really make equations

i just drew them

wait ima redo the lines

[warm python]

something like this

[chrome hemlock]

so all i have to do is make it point to the sun like dis?

so it dont look funky

and then i add a domain

[warm python]

yeah

[chrome hemlock]

wow

this looks good

right

[warm python]

uh

[final tangle]

depends on where you want the rays to come from
the centre of the sun, or top of the hill

[warm python]

change the domain so that the ray is touching the sun, that will look better

[final tangle]

in pics, they usually put gaps

[warm python]

like this

[warm python]

oh, ok

[final tangle]

looks ugly when touching

[chrome hemlock]

what ive been drawin g it with gaps since i was like 2

[sonic crystal]

make the ray have curvature and shading too

🧌

[chrome hemlock]

OK

now that is A bit too hard

[warm python]

how would you do that in desmos(the shading)

[chrome hemlock]

U FORGET THAT I HAVE TO EXPLAIN THIS IN A 1K WORD REPORT

[warm python]

when is this due?

[chrome hemlock]

3rd of march

LOL

wait no

[sonic crystal]

set up one of those inequalities over a region

[chrome hemlock]

5th of march

[warm python]

yeah. but desmos supports gradients?

[chrome hemlock]

but 4th of march which is monday is a public holiday

and on the 5th of march i have a test too bruh

[sonic crystal]

well u can change the colour

[sonic crystal]

whats this about anyways

[chrome hemlock]

uhhh quadratic lines sin graphs and stuff idk

[warm python]

anyway, create a bunch of rays like this

wish I had projects like this when I was in school

[chrome hemlock]

whats a function

[chrome hemlock]

no u dont

[sonic crystal]

um

[chrome hemlock]

this is so boring

i would much be rather studying for a test

[warm python]

A function is a Map from one set to another

[chrome hemlock]

its a function investigation

i have to design

a drink coaster

[craggy plank]

If I were you, I wouldn’t say that lol. Those "activities" are far more interesting and less stressful than tests

[chrome hemlock]

no but this just feels like a waste of time yk

cus i have to be studying for a test too

[sonic crystal]

y not a rollercoaster

[warm python]

more specifically a relation between 2 sets , wherein each image has only one preimage

[chrome hemlock]

i think it was originally rollercoaster

but they changed it to drink coaster

[sonic crystal]

lmao what a difference

[chrome hemlock]

rollercoaster would be to ez pez

would rather be doing that than this

[warm python]

does this make any sense to you?

[chrome hemlock]

hmm yes i guess

[sonic crystal]

muahhhahahahhahh ext

[chrome hemlock]

is function f(x)

[chrome hemlock]

go to sleep

[warm python]

ok,gtg study chem now, sorry

[sonic crystal]

lmaooo

[chrome hemlock]

aww it ok cya

[sonic crystal]

i should

[chrome hemlock]

thank u for the helpo

[warm python]

np

[sonic crystal]

f(x) = y

[chrome hemlock]

are u doing physics

[sonic crystal]

yes

[chrome hemlock]

i am too lost in physics

im doing nuclear physics

[sonic crystal]

im not doing nuclear yet

[chrome hemlock]

how does bigger nucleus mean more unstable

are u yr 11 or 12

[sonic crystal]

i told u this before lol

12

[chrome hemlock]

i forgot lol

THEN U SHOULD KNOW PLS

[sonic crystal]

im doing electromagnetism we never learnt nuclear

[chrome hemlock]

is electromagnetism the thingyl ike

2 protons

dont touch

[sonic crystal]

no

[chrome hemlock]

oh

or is it like those switches

[sonic crystal]

electromagnetism is the study of electricity stuff

and electric fields

[chrome hemlock]

switches and light bulbs!?!?!

[sonic crystal]

and how electrons move in field

s

[chrome hemlock]

ohms law!?!?!?

[sonic crystal]

and do stuff

[sonic crystal]

thats part of it yea

[chrome hemlock]

apparently that stuffs annoying to learn

lol

[sonic crystal]

thats the easy part

[chrome hemlock]

really what

[sonic crystal]

anwyays

lul

y=f(x) mmmm

function

[chrome hemlock]

did i show u

what i got on my spec test

i did so trash

[sonic crystal]

na u didnt

[chrome hemlock]

i legit studied so much i got the active role and i get this

[sonic crystal]

rip geometry

[chrome hemlock]

does it get better

geometry was too hard

[sonic crystal]

im not sure whats in ur syllabus

[chrome hemlock]

um

we're doign combinations and permutations

rn

[sonic crystal]

ew

[chrome hemlock]

pascals triangle

WYM EW

its easier than proofs

and circles

and then after im doing vectors

[sonic crystal]

yum

wait till u do calculus

[chrome hemlock]

apparently

methods calculus

is piss easy

[sonic crystal]

whats innit

[chrome hemlock]

uhh let me check the textbook rq

is this the calculus stuff

[sonic crystal]

yeah

[chrome hemlock]

idk if this is calculus too

[sonic crystal]

thats just the introduction tho

it gets harder

[craggy plank]

I have some polynomial questions for you to practice if you wants to

[chrome hemlock]

spec 1&2 has no calculus lol

[chrome hemlock]

no i think im fine with polynomials

thanks tho :p

[sonic crystal]

aw

boringg

[chrome hemlock]

sthu u guys are suffering

[sonic crystal]

😂

[chrome hemlock]

nsw is too competitive man

[craggy plank]

yeah, you’re all good. Therefore, it’s time to challenge some difficult tasks to enhance yourself 😂

[sonic crystal]

no.1 state

[chrome hemlock]

the prac test i took was really easy

like really easy

they even took out the circle graph question

i didnt even have to do it

[craggy plank]

Bruh, too easy

How do they distinguish the best and the almighty

If they give tests like this

[chrome hemlock]

uh last year the best girl got straight 100%

[craggy plank]

Pro

[chrome hemlock]

idk

im only assuming

its this easy

because its the start of the year

this looks like a year 9 test

without the radian stuff

[sonic crystal]

come do the hsc

[chrome hemlock]

no

legit in wace i can get like 60% in spec

and its equivalent to like a

90 atar

[sonic crystal]

ext 2 math hsc

[chrome hemlock]

nsw full of try hards

[chrome hemlock]

no.

thats like uni level calculus

[chrome hemlock]

oh wait

[sonic crystal]

too many letters

[chrome hemlock]

is that vectors like this

[sonic crystal]

ti think thats more like ext 1 vectors

[chrome hemlock]

bru do vectors get this hard

no shot

[sonic crystal]

this is ext 2

so yes

[chrome hemlock]

by any chance..

are u chinese...

[sonic crystal]

noh

lol

[chrome hemlock]

indian?

[sonic crystal]

noh

[craggy plank]

I don’t think this is pre-university

[sonic crystal]

yes it is lol

[chrome hemlock]

ext 2 is basically uni

math

[sonic crystal]

this is highschool!!!

[craggy plank]

No, I mean that’s physics

[sonic crystal]

its math

[chrome hemlock]

its vector math

[sonic crystal]

they mix projectile motion with vectors

[chrome hemlock]

we learn vectors in math and physics

[craggy plank]

Well, we don’t calculate air resistance in Math

[sonic crystal]

ur not calculating air resistance

[sonic crystal]

and its a math exam taken by hs students

[chrome hemlock]

poor kids in NSW

[chrome hemlock]

do a lot of schools run math ext 2

or is it just some

[sonic crystal]

i think all of them

[chrome hemlock]

why is not letting me add a slider

[sonic crystal]

hahahhaha im going to sleep

while ur stuck here

desmosing

goodnbyenight

[chrome hemlock]

ok good night

👍

.close

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[native portal]

just do it 2 at a time

or consider a - b - c - d = a-(b+c+d) and add the negative terms first, then it reduces to subtraction of two numbers

[final saddle]

@wind jacinth Has your question been resolved?

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[tranquil pine]

If I have a fair 40-sided dice and a fair 60-sided dice

[tranquil pine]

What’s the probability that the 60 sided die comes up higher

Here’s my answer that was wrong:

P(win) = P(win|roll above a 40) + P(win|roll less than or equal to 40)

= 1/3 * 1 + 2/3 * 1/2

= 2/3

But that’s not the answer

[final saddle]

@tranquil pine Has your question been resolved?

[potent birch]

did you account for the fact that they can both show the same number?

[final saddle]

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[hard mantle]

Hey y'all I have a question about proofs

[hard mantle]

I want to prove the following question:

Suppose a set D is irreducible, finite, and all its states are recurrent. Does this imply it is closed? Prove or show a counterexample.

So I'm fairly sure it is true, and that I should prove it

And my line of reasoning goes like this:

"If it were not closed, that means that it would not be recurrent"

That would be fairly easy to show

But I'm not sure if proof by contrapositive works like that

[royal gust]

That's exactly what a contrapositivev proof is

[hard mantle]

Or if I have to show that it is not irreducible, not finite, and not reccurent

Like do I negate all the properties or just one?

[main goblet]

I was asked this math questions from A professional And still dont know how to solve
I PROBABLY DONT WANNA FAIL
Consider a right-angled triangle with side lengths that are consecutive positive integers. If the length of the hypotenuse is a prime number, what is the smallest possible value for this prime number?

Take your time, and feel free to utilize any mathematical techniques or principles that you think will help you in solving this problem. Good luck!

[royal gust]

#❓how-to-get-help

[hard mantle]

Dude I'm already in this help channel

Get outta here

[main goblet]

sorry

[hard mantle]

You're all good

See the "Math Help (AVAILABLE)" section

[royal gust]

Your line of reasoning so far would prove:
"If recurrent, then closed"

[hard mantle]

Aha

[royal gust]

Which would be enough for the original question

[hard mantle]

Dope sauce

[royal gust]

That being said, there would be a lot of details that end up being irrelevant.

[hard mantle]

Oh?

[royal gust]

Like, why mention irreducible and finite, if those conditions don't matter?

You could be right, and they don't actually matter. Happens.

[hard mantle]

Hmmm real

Well that's just how the question is posed

Are you a markov-head?

[final saddle]

@hard mantle Has your question been resolved?

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[gaunt helm]

How do I evaluate the integral, from 0 to sqrt(2), of ((x^2)/sqrt(4-x^2))dx using the substitution x=2sin(theta)?

[vital crag]

find x^2, dx and change your bounds

[gaunt helm]

Alright.

I have got the integral, from 0 to pi/4, of (8sin^2θcosθ)/(4-4sin^2θ)
)^(1/2) dθ.

[vital crag]

yea try simplifying

[gaunt helm]

I think I might know what to do.

@vital crag Thank you my good sir.

.close

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[signal gust]

I wanted to ask if my answer for this is correct: I got 54.3

[eternal portal]

no

[static stump]

Nope, It should be 29

[signal gust]

Oh

Could you tell me how to get that?

Alright

[eternal portal]

google inscribed angle theorem

[signal gust]

Gotcha

Thank you!

[eternal portal]

🐑

[signal gust]

I don't have access to it

But thanks for the help

.close

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[quasi plinth]

\text{Demonstrate the following equalities} \text{and interpret them geometrically}

a) $\left | \mathbf{A} - \mathbf{B} \right | = \left | \mathbf{A} + \mathbf{B} \right | \iff \mathbf{A} \cdot \mathbf{B} = 0$

b) $\left | \mathbf{A} + \mathbf{B} \right |^2 = \left | \mathbf{A} \right |^2 + \left | \mathbf{B} \right |^2 \iff \mathbf{A} \cdot \mathbf{B} = 0$ (Pythagorean Theorem)

[soft zealot]

renato

[tranquil pine]

which norm is this?

frobenius?

[quasi plinth]

Norm?

[tiny gorge]

should hold for any norm that is induced by an inner product

[quasi plinth]

Is just norm

[tiny gorge]

what's the context, these are vectors in R^n or what?

[quasi plinth]

,, \left | \mathbf{A}\right | = \sqrt{a_1^2+ a_2^2+ a_3^2 + … + a_n^2}

[soft zealot]

renato

[quasi plinth]

Yes

[tiny gorge]

ah yea

try expressing the given equalities using the dot product

recalling that ||x||^2 = x . x

[tranquil pine]

ok

yeah just expand the norm squared as the dot product

[tribal helm]

hi

[quasi plinth]

Yes

[tranquil pine]

oh it's a vector

[tribal helm]

can someone please help me on this

[tranquil pine]

yeah then this is clean

[tiny gorge]

!occupied

[final saddle]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[tranquil pine]

another problem is being done here

[tribal helm]

oh k

[tranquil pine]

both 1 and 2 can be solved by taking the norm squared and using the defintion of norm squared as dot product

[tribal helm]

im new

so i need help to pen a channel

[quasi plinth]

If A-B = Y then Y.Y =0

(a^2-2ab+b^2)=0

@tranquil pine @tiny gorge

Since norm(x)=x.x=x^2

@here

[tiny gorge]

wait which one are you doing here?

[quasi plinth]

A

[tiny gorge]

so ||A-B|| = ||A+B||
first note that this is true if and only if their squares are equal
then expand the squares using the dot product

[quasi plinth]

Ye

[nova iron]

lol

edd answered it

[tranquil pine]

So do this for 1 for instance:

[nova iron]

||A-B||^2 = ||A+B||^2

| | A - B| |^2 = < A - B, A - B>

[tranquil pine]

||A-B||^2 = (A-B)^T (A-B) = ||A||^2 + ||B||^T - 2 A^T B

[tranquil pine]

yeah

that's correct

so A dot B = 0

[terse dagger]

from this you can get both directions

[quasi plinth]

Inner product?

[terse dagger]

you can assume that the first part is true

[nova iron]

yes

dot product

[terse dagger]

and get the result that A . B = 0

and other way around too

[quasi plinth]

Ty

What about b doe?

[nova iron]

try drawing out the vectors as a sum

when a and b are perpendicular, they form two sides of a right triangle

show that the third side is a+b from vector sum

~~ in fact it can even be a - b using a) ~~

then pythag for the sides of a right triangle: a^2 + b^2 = c^2

[soft zealot]

nyxie9151

[final saddle]

@quasi plinth Has your question been resolved?

[quasi plinth]

Sorry i missclicked

I am closing, i understood both

[quasi plinth]

Thx

Thx all. Closing…

.close

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[solid flare]

How do I solve this, I’m blanking atm

[solid flare]

Isn’t ab half of df?

<@&286206848099549185>

[median badge]

Don't know enough about the triangle

[solid flare]

Nvm I got it

.closr

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[primal sierra]

ur right

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[subtle swan]

Just wondering if this is correct or not. And if it is could I have proved it in less steps?

[final saddle]

@subtle swan Has your question been resolved?

[subtle swan]

<@&286206848099549185>

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@subtle swan Has your question been resolved?

[cyan kayak]

@subtle swan Looks good to me.

[subtle swan]

Awesome, thank you!

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[ebon nimbus]

Find g'(x) and g(pi/2). I am supposed to get g'(x)=cosx but i can't figure out how they got this result

[minor nexus]

maybe you can simplify f(x)
id add1 then -1 in the numerator

f(x) = (e^x +1 -1) / (e^x+1)

Then you can simplify into two fractions and integrate

[final saddle]

@ebon nimbus Has your question been resolved?

[minor nexus]

bruh

[ebon nimbus]

it does help but i still didn't get that result

[ebon nimbus]

when i did the derivative of g i got (H(x)-H(-x))'=h(x)-h(-x) where h is f*cos and H its primitive. So it's f(x)*cosx-f(-x)cosx because cos is an even function. The solution said f(x)*cos + f(-x)cosx so i wonder if they mistook cos an an odd function or i lost a - along the way

[lyric summit]

continue my writing and you will get the solution in two lines

[ebon nimbus]

thanks

[lyric summit]

yw

[ebon nimbus]

.close

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[final shadow]

can somebody help me prove that for any constant, positive natural number, n

[final shadow]

,tex $2 > \log_{n-1}n > 1 $

[soft zealot]

j

[minor nexus]

could you take each to be the exponent of n-1?

so (n-1)^2>n>n-1

[final shadow]

Sry im not following tbh

Like

,tex $n - 1 < n < (n-1)^2?$

[soft zealot]

j

[minor nexus]

yup

[final shadow]

What exactly what this do though?

[minor nexus]

well the left side of what you posted is trivially true

[final shadow]

Oh wait sorry so like if you remove the logarithm then your saying you get this?

if thats what ur saying then that makes a lot more sense

[minor nexus]

yeah

[final shadow]

Mmmmm that makes the left side true then

Let me look at the right side quickly

One moment

[final shadow]

Im not really sure how i would prove that n < n^2 - 2n + 1 though

am i overthinking it? bc it seems like not like obvious at a first glace that its true

[minor nexus]

yeah, its not trivial.
how rigourous do they want it?

you could either draw a graph and show it grows much faster

[final shadow]

Its for my math like 'research' project and like my prof prefers algebra > graphs

[minor nexus]

okay, maybe find the solutions to (n-1)^2=n then have the graph and show if greater, then always greater

or if you really wanted, you could try induction

i think thats overkill though

[final shadow]

Yeah induction would be overkill here because this is like a small part in my big project which literally revolves around induction 😭

[final shadow]

Hm okay that makes sense ty for ur help

[minor nexus]

np

[final shadow]

if anybody else also has solutions that would be appreciated too!

[wintry kindle]

your task is insufficiently defined, the statement is only true for n>=3

[final shadow]

it works for 4 too though

[wintry kindle]

ofc since 4 >= 3

[final shadow]

oops im sorry

i misunderstood

Yeah im trying to prove that actually

[wintry kindle]

you've already shown this part with ronush

[final shadow]

Could u walk me through those steps?

[wintry kindle]

the other part is equivalent to (n-1)² > n as you figured out, which you'd usually do via induction

the induction is quite short

[final shadow]

What exactly does induction mean because i think i have a general sense but like i just wanna confirm

[wintry kindle]

btw is it clear how you got to this step before?

before we do the induction

[final shadow]

I assume by removing like the logarithm

Is it deeper than that

[wintry kindle]

not quite:

you take (n-1)^x for each value in the inequation

but it's important that this only works if the base is > 1

and n-1 > 1 since n >= 3

[final shadow]

Yeah becasse then it would be negative leading to undefined

[final shadow]

Sorry give me a minute to like comprehend these steps really quickly

[wintry kindle]

it could also be 0 <= base < 1

then it wouldn't work either

[wintry kindle]

sure

Line 1 to Line 2 is: x ----> (n-1)^x

Line 2 to Line 3 is just applying logarithm rule for the second value: (n-1)^log_n-1(n) = n

[final shadow]

Ohhhhh

That actually helps a lot

I understand what you mean

[wintry kindle]

k neat

n > n-1 is clear

[final shadow]

Yeah that part is clear to me

[wintry kindle]

and (n-1)² > n via induction

we can do a trick though

which would circumvent induction

we could instead show that the (n-1)² is bigger than n for n=3 AND show that the derivative of (n-1)² is bigger than n.

because then the function (n-1)² will always be bigger than n for n>=3.

is that mostly understandable?

[final shadow]

Yeah the derivative part is understandable and that makes sense to me, and i understand that this works for n >= 3, but why exactly are we picking n = 3 as our value

Is that like just as an example

[wintry kindle]

because your initial task only works for n>=3

your task is: "show that 2 > log_n-1(n) > 1 for n>=3"

otherwise that task is senseless

[final shadow]

Ohh ok alright

Yeah that makes sense

[wintry kindle]

you can attempt for n>=1 (all natural numbers)

but you'll realize it doesn't work

[final shadow]

yeah that doesnt work

[wintry kindle]

so in theory, because they worded it that way

you could just state: "It doesn't work for n=1, therefore it can't be proven"

:D

but we'll do it correctly for n>=3

[final shadow]

that wont work bc this isnt like a question posed by a prof its like one posed by me

i wont get into details tho 😭

i see what ur saying tho lol

[wintry kindle]

oki

[final shadow]

Mhm mhm this makes sense, i can work through the derivatives part and lyk how it goes

Thank you SO MUCH for your help and same to the other person as well

[wintry kindle]

np 🦇

I'll put the solution below hidden:

[final shadow]

Kk also just to confirm, set n=3, then derivative right, or is it the other way around bc im assuming its the latter

[wintry kindle]

||for n=3: (n-1)² = 2² = 4 > 3 = n. And the derivative of (n-1)² = n²-2n-1 is 2n-2. The derivative of n is just 1. So we want to show that 2n-2 > 1: 2n-2 > 1 is the same as n > 3/2 which is true, since we defined the task for n >= 3 and 3 > 3/2.||

[wintry kindle]

first you show the that the term is bigger than the other for n=3

then you show that the derivative is bigger than the other

[final shadow]

Kk got it !!!

Thank you so much

IT WORKED

tysm

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[gray acorn]

how does one do this

integral

[vital surge]

hint: not much to be said here without just giving away the answer

[vital crag]

Yea

Does the integrand have any symmetry on the domain of integration

[gray acorn]

well from what i can see i can just make it into an arctan but theres a random sin

thats blocking it

[vital crag]

No substitution necessary. Just answer the question

Plot it if you need help

[gray acorn]

if i understand well what you mean is that the domain is basically just the same

doesnt it jsut =0?

i might be on drugs

ah its 0

ty!

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[frozen gale]

You have to find cos(cosa) if you know that sin(sin(60+a))=2/3 and a is in between 70 and 100 degrees

[final saddle]

@frozen gale Has your question been resolved?

[frozen gale]

<@&286206848099549185>

sin(60+a)

u mean arcsin?

arcsin(2/3)

did not get the question

arcsin value is in radians

if thats what u asked

i dont know how much arcsin(2/3) lol

um

how do i do it without a calculator

youre not supposed to know the value of arcsin(2/3) tho

what if its on ur midterm

I know how to solve this with a calculator

this is not my question

ok..

did not know people used calcualtors for every math equation

You have to find cos(cosa) if you know that sin(sin(60+a))=2/3 and a is in between 70 and 100 degrees

anyone help?

[frozen gale]

with a pen and a paper

@wintry kindle hey

[wintry kindle]

ys a sec

[wintry kindle]

k I'd probably start by using sin/cos addition theorems

not sure where the task is headed though

either one solves directly for x and is able to insert

or by simplifying the second one already gets the expression cos(cos(x)) within it

sin(60+a) = sin(2pi/6+a) = sin(pi/3+a) = sin(a)*cos(pi/3)+sin(pi/3)*cos(a)

[frozen gale]

then

[wintry kindle]

where sin(pi/3) is I think sqrt(3)/2

and cos(pi/3) 1/2

[frozen gale]

it is

[wintry kindle]

hm thinking whether applying it for the stacked one yields any results sin(sin(a)/2 + sqrt(3)/2*cos(a)) = ...

it at least gives us a cos(cos(...)) term

however we can't directly solve for it

[frozen gale]

i simplified sin(sin60+a))=2/3 to cos(cos(a-30))=sqrt5/3

dont know what to do next tho

how do i get coscosa from coscos(a-30)

[wintry kindle]

hm addition theorem again? but it'll be hit or miss

cos(cos(a-30)) = cos(cos(a)*cos(-30)-sin(a)*sin(-30))

[frozen gale]

what do u do about sin(a)

[wintry kindle]

not sure yet

at least I don't see a pattern, I'd just apply all of the trig identities for sin/cos until I see a clearer path here :/

[frozen gale]

crazy how they brought this in 2015 as a 10th grade midterm problem

[wintry kindle]

I might also not remember or know some identity which simplifies it

[frozen gale]

how old are you btw

[wintry kindle]

naw age q's here

[frozen gale]

alright

ima just close the channel pretty sure no one got time to do this lol

thanks tho

.close

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[wintry kindle]

np :c

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[gritty axle]

How to show $Tr(E^n)$ is bounded by 3 ($E \in {0, 1}^{l \times l}$)?

[soft zealot]

[Insert_Name] Bringer of Plague

[gritty axle]

(for $n=1, 2, 3, 4, 5, 6$ it is $0, 0, 3, 0, 0, 3$)

[soft zealot]

[Insert_Name] Bringer of Plague

[gritty axle]

(also the diagonal is all 0's)

(and less then l non-zero elements)

[deft ravine]

Tr is the trace here and E is just a matrix with 0,1 as an entry?

[grim nebula]

,w Tr({{0, 1, 1, 0, 0, 0, 0}, {1, 0, 1, 0, 0, 0, 0}, {1, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}}^n) for n = 69

[soft zealot]

Results provided by WolframAlpha

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

Requested by tearfox

[grim nebula]

yes very bounded

[deft ravine]

Interesting number choice

[grim nebula]

i dont know what you mean

[vital surge]

the placement of 1s and 0s probably

[grim nebula]

oh, thanks for your explanation

[deft ravine]

I think you can find an upper bound for each n that will go to infinity for n goes to infinity.

[final saddle]

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[chrome hemlock]

sxcuse me

[chrome hemlock]

when i complete the square i get 11:4

11/4

but they get 11/2

what am i doing wrong

is it cause i need to take the -11/4 into the bracket with the 2?

so its -11/4/2?

how come in the answer it isnt in the bracket though

then shouldnt there be 2 brackets like this

Where did the 2 go

Wait

Once i divide it by 2 it doesnt have to be in bracket?

[solar glade]

That is wrong

[chrome hemlock]

Which

[solar glade]

This

If u expand the last u never get 2x^2

[chrome hemlock]

Lol

[solar glade]

Unless im drunk

The way to do is

Divide first everything but 2 and multiply everything by 2

2 outside bracket everything inside divided by 2

Then complete the square inside the bracket

[chrome hemlock]

What

Isnt that just reversing what u just did

Dividing and then multiplying it again

[solar glade]

I am just talking about the last thing i saw

[chrome hemlock]

Ohhh

Wait can u help me with my q

[solar glade]

2(x-1/2)^2-1/4-5/2)

[chrome hemlock]

wat

then -1/4 - 10/4

and u get -11/4

[solar glade]

Yes

[chrome hemlock]

but its 11/2

-11/2

[solar glade]

Rly?

Then i am drunk im gonna check again xD

[chrome hemlock]

The answer here

[soft zealot]

Solomaniac

[chrome hemlock]

Where did that clme from

Come

[static stump]

I rectified the mistake in the last step.

[chrome hemlock]

Oh i think i see

Then u take out the -1/4

To subtract from -5?

[static stump]

Add it

[solar glade]

My answer is correct

[static stump]

And it wasn't really a mistake. I just forgot to type a 2.

[solar glade]

But not finished

Yeah

Why that face

Ew

[chrome hemlock]

-1/4 divide 2 is -2/4 -11/2?

[solar glade]

Yes

[chrome hemlock]

-1/2 - 10/2*

[solar glade]

I was getting crazy xd

[chrome hemlock]

=-11/2

Oh ok thank u

Im getting chicken tenders now with my mum

🤤🤤

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[wet helm]

how do i do this?

[stone wagon]

write down the distance from A to P

[magic sparrow]

Write an expression for the distance from P to A, and from P to B

(and then set them equal)

[wet helm]

ohh ok

thx guys

[magic sparrow]

you're welcome

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[tranquil pine]

tips

[tranquil pine]

idk how to start it

i need positive solutions

of that system

if they are positive then d and e are out

but

i also see a is b in some way

and c is f but reversed same as a and b

nvm i found it

you just cube it

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[tranquil pine]

.reopen

[final saddle]

✅

[tranquil pine]

even tho

cubed works

its c apparently but how

[neon slate]

well...the easy way...since it's multiple choice is you could plug in the various answer choices and see which one works 😄

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[tranquil pine]

i need proof

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[tawny ivy]

This is about amortized analysis and I’m a bit bit confused. How can I answer these six options down below?

I know the two answers below I think, but I don’t know how to answer the four above

[final saddle]

@tawny ivy Has your question been resolved?

[final saddle]

@tawny ivy Has your question been resolved?

[final saddle]

@tawny ivy Has your question been resolved?

[tawny ivy]

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[tranquil pine]

im looking to understand the general procedure in converting between coordinate system unit vectors? I of course cant (and will not) memorise the above table

but, I think i should understand the procedure at least

[vital crag]

Jacobian all of them

[tranquil pine]

I know how to convert magnitudes i suppose

[tranquil pine]

Jacobian?

do u have some article on this i can

read

sounds very interesting

[vital crag]

https://math.libretexts.org/Bookshelves/Calculus/Supplemental_Modules_(Calculus)/Vector_Calculus/3%3A_Multiple_Integrals/3.8%3A_Jacobians
Example 3.8.3

And the formula above it

Oh there's more to it than just integration hang on

[tranquil pine]

hmmm but what exactly is rho^2 sin(theta) signifying?

[vital crag]

Oh nevermind it's just the picture that derives them all
https://math.libretexts.org/Courses/Mount_Royal_University/MATH_2200%3A_Calculus_for_Scientists_II/7%3A_Vector_Spaces/5.7%3A_Cylindrical_and_Spherical_Coordinates#:~:text=To convert a point from,r2%2Bz2).

[vital crag]

That's the volume correction when converting from Cartesian integral to spherical

Oh you wanted the unit vectors goddamnit

[tranquil pine]

yeah

[vital crag]

Okay you gotta convert the Jacobian of the unit vectors then

https://math.stackexchange.com/a/158830

That's cylindrical even though OP says spherical

[tranquil pine]

oh what the top answerer said is very interesting tho

like using the gradient

thats so great because i just derived the gradients for the 3 coordinate systems

[tranquil pine]

hmm ok so as an attempt at understanding this lets try and convert the unit vectors in cylinderical here to spherical using the method they used

so ig first we note down the relation between cylinderical and spherical

uh

[vital crag]

Yea this is the prerequisite for those derivatives

[vital crag]

Yea that top answer lays the blueprint for Cartesian to spherical as well

[tranquil pine]

ok i think i got the idea now thanks riemann a bunch i have been kind of losing it with figuring this out LOL

i have to go now but ill expand this later and then if i get stuck open another help channel

.close

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[proper flicker]

Hello, I need to prove that
I thought about using Abel's test but I couldnt prove that this series is monotonic, though I believe it is for all n >=3.

[final saddle]

@proper flicker Has your question been resolved?

[final saddle]

@proper flicker Has your question been resolved?

[proper flicker]

<@&286206848099549185>

.close

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[rustic trench]

how do i determine if it's convergent or divergent? [\sum_{n=30}^{\infty} \frac{1}{n \ln (\ln (n))}]

[soft zealot]

dgh

[scarlet sequoia]

have you tried integral test?

[rustic trench]

no how do we do it

[scarlet sequoia]

sum(f(n)) when f is decreasing has same convergence type as int(f(x)dx)

[rustic trench]

is there a mathematical reason for that

[scarlet sequoia]

there is proof

[scarlet sequoia]

i can show you the proof

[scarlet sequoia]

it's just when you graph f(x)

int(f(x)dx) between n and n+1 is bounded between f(n) and f(n+1)

$\forall x\in [n,n+1], f(n+1)\leq f(x) \leq f(n)$

[soft zealot]

rafilou2003

[scarlet sequoia]

then you integrate yes

[signal vector]

Oh alr, you were proving it

[scarlet sequoia]

its based on the cauchy test for convergence

thus $\int_n^{n+1}f(n+1)dx \leq \int_n^{n+1}f(x)dx \leq \int_n^{n+1}f(n)dx$

[soft zealot]

rafilou2003

[scarlet sequoia]

thus $f(n+1) \leq \int_n^{n+1}f(x)dx \leq f(n)$

[soft zealot]

rafilou2003

[scarlet sequoia]

Then we sum

[rustic trench]

why is f(n+1)\leq f(x) \leq f(n)?

[signal vector]

Its a decreasing function

[scarlet sequoia]

because by our assumption f is decreasing

[rustic trench]

okay

[scarlet sequoia]

then we sum

[rustic trench]

so we need to prove f(x) is decreasing first?

then we'd know f(n) is greater?

[scarlet sequoia]

I mean both x and ln(ln(x)) are increasing and positive starting from 30

so their product is increasing

so the inverse is decreasing

[rustic trench]

i see

yes

so which step do i need to do first

integrate?

that looks like a complicated function

[scarlet sequoia]

find the integral of 1/(xlnlnx)

[rustic trench]

should i just plug it into the integral calculator? bc no way i can do it myself

[scarlet sequoia]

I mean we could do something else

multiply by 1/ln(x)

could be fun

since 1/ln(x) < 1

if integral of 1/(xln(x)lnln(x)) diverges

then surely our initial integral diverges

[rustic trench]

so is the idea that we take a part of our function and see if its value is greater or smaller than the whole function for all x?

[scarlet sequoia]

and now the trick for finding $\int_{30}^\infty \frac{dx}{x\ln(x)\ln(\ln(x))}$

[soft zealot]

rafilou2003

[scarlet sequoia]

not sure if I understood you there