help-36

[storm haven]

1

[tribal ocean]

isnt it /3?

[storm haven]

Wdym

[tribal ocean]

3theta is a multiple of 2pi

[storm haven]

Oh yeah

[gentle zephyr]

w ∈ G3 => w = e^{2ki.pi/3} , k ∈ {0,1,2}

[storm haven]

Now we have it

See this

Then plug w into your equation and set it =0

[tribal ocean]

that was a typo though

[gentle zephyr]

how

[storm haven]

Split into 3 cases first

For each of the 3 possible ws

[gentle zephyr]

case k = 0 is discarded

[storm haven]

Yes

[gentle zephyr]

for k = 1
w = e^{2i.pi/3}

so the argument is 2pi/3 + 2kpi

z1 = cis(2pi/3 + 2kpi)

[storm haven]

You don't have to do that

[storm haven]

Just plug it in this form

[gentle zephyr]

alright if you say so

[tribal ocean]

you can get rid of every w^3

and make it 1

before plugging in

[gentle zephyr]

f(x) = x^6 + x^5 + ax^4 + 2x^3 + ax^2 + x - 1

[tribal ocean]

w^6=1

w^5=w^2

[storm haven]

Yeah

You can also simplify the expression beforehand like what @tribal ocean is doing

Overall you get a pretty neat expresion

[gentle zephyr]

f(w1) = (w1)^6 + (w1)^5 + a(w1)^4 + 2(w1)^3 + a(w1)^2 + w1 - 1

f(w1) = 1 + w1^2 + a.w1 + 2 + a(w1)^2 + w1 - 1

[storm haven]

w1^2 is w2

[gentle zephyr]

what?

[storm haven]

Think about it

[tribal ocean]

ohhh and if you plugged in w2 then w2^2=w1

thats why its symmetric

[storm haven]

What is w1, in Euler form, then square it and you get w2

[storm haven]

Mhmm

[gentle zephyr]

alright

I thought it was a property

[tribal ocean]

its just that 2pi/3 and 4pi/3 add to each other

modulo? 2pi

what is it even called

[gentle zephyr]

dude wait

w ∈ G3
w = e^{2k.i.pi/3}
w1 = e^{2.i.pi/3}
w1^2 = e^{4.i.pi/3}

[olive narwhal]

gng what

we got sweats in math now

[gentle zephyr]

w = e^{2k.i.pi/3}
w2 = e^{4.i.pi/3}

I see

f(w1) = 1 + w1^2 + a.w1 + 2 + a(w1)^2 + w1 - 1
f(w1) = 1 + w2 + a.w1 + 2 + aw2 + w1 - 1

[storm haven]

Now you can find the modulus and arg

[gentle zephyr]

f(w1) = 2 + w2 + a.w1 + a.w2 + w1

[storm haven]

It's no longer useful going with the e^ipi form

But you can set it to 0 and solve for a

[gentle zephyr]

for k = 1
w1 = e^{2i.pi/3}
so the argument is 2pi/3 + 2kpi
w1 = cis(2pi/3 + 2kpi)

[gentle zephyr]

whats the modulus for w1

[storm haven]

1 lol

Cause it's a root of unity

Which is always on the unit circle

[gentle zephyr]

care to explain

[storm haven]

When you multiply complex numbers, their moduluses are multiplied

W^3 = 1

[gentle zephyr]

wait, do I need to get it in z = a + bi form

[storm haven]

Probably

[storm haven]

|w|^3 = 1

|w |= 1

[gentle zephyr]

w1 = cos(2pi/3 + 2kpi) + isin(2pi/3 + 2kpi)

[storm haven]

Why is there that 2kpi hanging around

[gentle zephyr]

don't I need that?

[storm haven]

You don't need it

[gentle zephyr]

no, argument is always in between [0, 2pi)

so its redundant

pointless

[storm haven]

Exactly

[gentle zephyr]

I just find it weird

[storm haven]

Sorry but I gtg

[gentle zephyr]

w1 = cos(2pi/3) + isin(2pi/3)

[storm haven]

You can wait for another helper

[gentle zephyr]

but wait

[lyric obsidian]

noooooooo

I will help for a short period too haha

[gentle zephyr]

w1 = -1/2 + (sqrt{3}/2).i

[lyric obsidian]

hmm

[gentle zephyr]

w1^2 = w2

[lyric obsidian]

Looking at the problem I'd personally start by simplifying the powers once you plug in \omega into the poly

[gentle zephyr]

w2 = -1/2 - (sqrt{3}/2).i

[gentle zephyr]

we already did that

f(w1) = 2 + w2 + a.w1 + a.w2 + w1
0 = 2 + w2 + a.w1 + a.w2 + w1

[lyric obsidian]

ok

So now neat cancellation should happen

[gentle zephyr]

wait

0 = 2 + w2 + a(w1+w2) + w1

w1 = -1/2 + (sqrt{3}/2).i
w2 = -1/2 - (sqrt{3}/2).i

[tribal ocean]

w1+w2 = -1 i guess

[gentle zephyr]

0 = 2 + w2 -a + w1

[tribal ocean]

the a

[tribal ocean]

^

[lyric obsidian]

Can't you just add another w2+w1 term?

[gentle zephyr]

0 = 2 - 1 - a

[lyric obsidian]

w2+a(w1+w2)+w1=(a+1)(w1+w2)

[gentle zephyr]

a = 1

[lyric obsidian]

Ye

My calculations agree

[gentle zephyr]

now what

this is all the possible a ∈ R

[lyric obsidian]

Factorise with real and complex coefficients respectively

[fiery karma]

dont biject r whatever you do

[lyric obsidian]

that's what the R[X] and C[X] parts mean

[gentle zephyr]

is this all the possible values of a or not??

[lyric obsidian]

^^

[lyric obsidian]

Yes

[gentle zephyr]

my friend got a = 3

[lyric obsidian]

Uh

Lemme check once more

Nope pretty sure its a=1

[gentle zephyr]

ok then she just sucks at math

[muted prairie]

,w roots of x^6+x^5+3x^4+2x^3+3x^2+x-1

[soft zealot]

Results provided by WolframAlpha

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Requested by a_mog

[gentle zephyr]

no spoilers

[muted prairie]

[lyric obsidian]

spoiled

[lyric obsidian]

don't say that :( maybe they made a silly calculation error

dreyuk you mind taking over

[muted prairie]

well the argument is of the form "i tested this case manually and got this, but someone's friend tested it and got this"

[lyric obsidian]

I gotta finish an assignment due in 3 hrs lmao

[gentle zephyr]

yeah prolly

[muted prairie]

surely its useful to just figure out who made the mistake, if we already have both solutions

[gentle zephyr]

what do you mean

[muted prairie]

well i mean your friend just told you the wrong answer and thats where the confusion is coming from

[gentle zephyr]

I don't get your complicated arguments

[gentle zephyr]

yeah surely

[muted prairie]

the only confusion is that your friends answer doesn't match

[gentle zephyr]

she is not my friend she is just my class mate

[muted prairie]

so if we just have a calculator that can tell us your classmate is wrong

there's not really a reason not to do it

we've already solved the problem

[gentle zephyr]

how do you know she is wrong

[muted prairie]

because i know the third roots of unity, and none of these are it

[gentle zephyr]

can you check with mines now

a = 1

[muted prairie]

well i wouldnt want to spoil

[gentle zephyr]

also, I don't understand why they say f(1) ≠ 0

whats the rationale?

[muted prairie]

ig to force the root to be a primitive 3rd root

w can't be 1

[gentle zephyr]

that

[muted prairie]

it has to be -1/2 +- sqrt(3)i/2

but yeah its a bit weird

[gentle zephyr]

what exactly

[muted prairie]

the f(1) =/= 0 condition

its kinda arbitrary

like they're saying it's not allowed

so you can find an a such that f(1) = 0 but that one isnt' allowed

[gentle zephyr]

a ≠ -2 is not allowed

[unborn ice]

Am I allowed to text here

[gentle zephyr]

I guess, can you help me finish this or

@muted prairie

[muted prairie]

idk what the heck methods ppl tried for this

it looks like people are explicitly writing out the roots

i would just plug in w and simplify

[gentle zephyr]

that's what we did

[muted prairie]

only if you are helping with the problem

otherwise go to #discussion or #chill

[muted prairie]

oh ok

what did we get?

[unborn ice]

Ok

[gentle zephyr]

I think thats what we did

[storm haven]

It is

[muted prairie]

this?

[gentle zephyr]

what? I thought you were leaving

[storm haven]

I came back

The part we need to do now is to factorize it right?

[gentle zephyr]

yes I am stuck

[storm haven]

There should be more values of a than just a=1

Probably tho

[gentle zephyr]

how do I check

[gentle zephyr]

can we start from scratch @muted prairie @storm haven

[muted prairie]

ok

[storm haven]

I looked at the function again and there doesn't seem to be any other value of a

[muted prairie]

lets plug in w

then we will solve for a

[gentle zephyr]

lets do it from scratch @storm haven @muted prairie

alright?

[muted prairie]

ok

[storm haven]

U sure?

But aight 👌

[gentle zephyr]

https://tenor.com/view/nothing-to-do-nothing-but-time-i-have-time-nothing-else-to-do-i-got-nothing-gif-12334945

[muted prairie]

we want f(w)=0

[gentle zephyr]

w ∈ G3 => w = e^{2k.i.pi/3}

[muted prairie]

right

and since we dont want f(1)=0, it means k should be 1 or 2

[gentle zephyr]

f(x) = x^6 + x^5 + ax^4 + 2x^3 + ax^2 + x - 1

[storm haven]

No matter which k we choose, plugging it in gives us the same equation

It's symmetric

[gentle zephyr]

care to elaborate?

[muted prairie]

we want w to be primitive

not equal to 1

[gentle zephyr]

gcd(1,3) = gcd(1,0) = 1
gcd(2,3) = gcd(2,1) = gcd(0,1) = 1
gcd(0,3) = 3

yes k = 1 or k = 2, otherwise its not a primitive root particularly w0, correct?

[muted prairie]

yeah

[gentle zephyr]

the coprimality condition

[muted prairie]

w0 aka 1

[gentle zephyr]

between the k ∈ {0,..., n - 1} and the n in the Gn

we need gcd(n,k) = 1 for it to be a primitive root of unity

[muted prairie]

right

[gentle zephyr]

@muted prairie @storm haven everything good so far?

[muted prairie]

yes

the reason we dont want 1 specifically isnt necessarily because its not primitve but because its given that f(1) =/= 0

then its just the other two roots are primitive

[gentle zephyr]

well w0 = e^0 = 1

[gentle zephyr]

alright 👍 everything good so far, then primality has nothing to do with this

take for example they say w is in G4, then w0 is not possible but w2 is possible, and gcd(2,4) = 2

whatever, we are derailing

lets continue @muted prairie @storm haven

[muted prairie]

ok

so, we know that f(w) needs to be 0

we should write write that out w/ the definition of f

[gentle zephyr]

alright okay

f(x) = x^6 + x^5 + ax^4 + 2x^3 + ax^2 + x - 1

0 = w^6 + w^5 + a.w^4 + 2.w^3 + a.w^2 + w - 1

0 = 1+ w^2 + a.w + 2 + a.w^2 + w - 1

@muted prairie @storm haven

this is what I got after simplifying

is it fine?

[muted prairie]

what happened to w^6 and w^3?

[gentle zephyr]

wait, they are 1

not 0

my bad

[gentle zephyr]

f(x) = x^6 + x^5 + ax^4 + 2x^3 + ax^2 + x - 1
0 = w^6 + w^5 + a.w^4 + 2.w^3 + a.w^2 + w - 1
0 = 1+ w^2 + a.w + 2 + a.w^2 + w - 1

you here ? @muted prairie

[muted prairie]

Yes

Looks good

[gentle zephyr]

now what

[muted prairie]

combine like terms

[gentle zephyr]

what about w^2

[muted prairie]

might as well combine first

that sounds easier

at least combine our constants

[gentle zephyr]

f(x) = x^6 + x^5 + ax^4 + 2x^3 + ax^2 + x - 1
0 = w^6 + w^5 + a.w^4 + 2.w^3 + a.w^2 + w - 1
0 = 1 + w^2 + a.w + 2 + a.w^2 + w - 1
0 = w^2(1+a) + w(a + 1) + 2

[muted prairie]

i think the w coeff is wrong

ok good

yeah now we try to relate w^2 and w

ideally writing w^2 in terms of w

[gentle zephyr]

help

[storm haven]

U don't have to do that

[muted prairie]

no, we can be sneaky instead of principled

[storm haven]

Well in this case we know w^2 + w = -1

Because it's the properties of the third roots of unity

[muted prairie]

yeah sure why would we need some consistent method

just use a hack that'll only work on this exact problem

[storm haven]

But we still get a=1

Like what we got in the first place an hour ago

I think the question might just be a trick question

[gentle zephyr]

please don't fight guys

[storm haven]

Maybe a=1 is the only solution

[muted prairie]

yeah i imagine it is

[storm haven]

Yes I also hope that doesn't happen

[muted prairie]

sorry

[storm haven]

Sorry too man 🤝

[gentle zephyr]

can we go back to finishing this crap?

[storm haven]

Well we can just accept that a=1 is the only solution

And that means w1 and w2 are both roots

[gentle zephyr]

f(x) = x^6 + x^5 + ax^4 + 2x^3 + ax^2 + x - 1
0 = w^6 + w^5 + a.w^4 + 2.w^3 + a.w^2 + w - 1
0 = 1 + w^2 + a.w + 2 + a.w^2 + w - 1
0 = w^2(1+a) + w(a + 1) + 2

[storm haven]

0= (w^2 + w) (a+1) +2

w^2 + w = -1

[gentle zephyr]

care to elaborate

?

[muted prairie]

w^3 = 1

w^3 - 1 = 0

factor left side

[gentle zephyr]

I don't know difference of cubes

[muted prairie]

well

we know w=1 is a root

so we can polynomially divide by w-1

[gentle zephyr]

how?

[muted prairie]

because 1^3 = 1

or 1^3 - 1 = 0 i mean

[gentle zephyr]

f(1)≠0

[muted prairie]

to be clear, a root of w^3 - 1 = 0

not a root of f

[gentle zephyr]

ahh

[muted prairie]

so if yk polynomial division we can do (w^3-1)/(w-1)

[gentle zephyr]

$\polylongdiv{x^3-1}{x-1}$

[soft zealot]

Renato

[muted prairie]

[gentle zephyr]

b^2-4ac = 1-4 = -3 < 0

[muted prairie]

no quadratic formula

[gentle zephyr]

the other roots are imaginary

[muted prairie]

indeed, w and w^2 are imaginary

ok so we have w^3-1 = 0

(w-1)(w^2+w+1) = 0

and since we know w isn't 1, we know w^2+w+1 = 0

[gentle zephyr]

the roots of f are this two roots hiding in this quadratic poly right

cant we divide f by this polynomial?

[muted prairie]

yeah that'll help to factor it

[gentle zephyr]

assuming a = 1

[muted prairie]

but that's after we know a=1

[gentle zephyr]

0= (w^2 + w) (a+1) +2

[gentle zephyr]

w^2 + w = -1

0 = 1 - a + 2

a = 3

[muted prairie]

ah that must be what your friend did

0 = -(a+1) + 2

continue from here

[gentle zephyr]

yeah im bad at math too

[muted prairie]

after setting w^2+w = -1

have to be careful to distribute the negative sign

[muted prairie]

its ok we all are

[gentle zephyr]

0 = - a - 1 + 2

a = 1

is the only possibility, we proved it using math

[muted prairie]

Excellent

[gentle zephyr]

going back at this idea

[muted prairie]

You had the right plan here

It's a little subtle to "prove" we can, but if we do the polynomial division and dont get a remainder, that'll be proof enough

[gentle zephyr]

wdym?

[muted prairie]

well now we know f is x^6+x^5+x^4+2x^3+x^2+x-1

[gentle zephyr]

is hard to explain

[muted prairie]

since f(w)=0, in order to factor f in R[x] we can divide it by the smallest polynomial such that f(w)=0, which is x^2+x+1

[gentle zephyr]

alright okay

this primitive roots of unity thingy really is fascinating

is it normal to be afraid but interested at the same time?

[muted prairie]

idk but i could see it

[gentle zephyr]

$\polylongdiv{x^6+x^5+x^4+2x^3+x^2+x-1}{x^2+x+1}$

[soft zealot]

Renato

[gentle zephyr]

I still get a quartic

[muted prairie]

yay

its RRTing time

[gentle zephyr]

-1 or 1

[muted prairie]

well hopefully not 1

maybe we should check that thouh

idk if we ever did

pro tip: to evaluate f(1), just add up the coefficients

[gentle zephyr]

lmfao

I got lost

f = (x^2+x+1)(x^4+2x-1)

[muted prairie]

ah yeah

[gentle zephyr]

neither 1 or -1 works

[muted prairie]

oh oof

welp

ok well just visualizing the graph

it has a y-intercept of -1 and curves up

so it should have 2 real roots

[gentle zephyr]

you tell me

why RrT doesn't work

[muted prairie]

well if it doesn't work then it must mean there aren't any rational roots

[gentle zephyr]

I just found a glitch in the matrix

[muted prairie]

so all the roots are irrational

which is very possible but just not good

something is wrong lol

[gentle zephyr]

I see

[muted prairie]

first we only get one value of a

[gentle zephyr]

there is an error

in the latex

see

[muted prairie]

it looks like the right factorization though

,w expand (x^2+x+1)(x^4+2x-1)

[soft zealot]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by a_mog

[gentle zephyr]

@muted prairie

its different polys

[muted prairie]

oh

rip

[gentle zephyr]

lol

f(x) = x^6 + x^5 + ax^4 + 2x^3 + ax^2 + x + 1

0 = 1 + w^2 + aw + 2 + aw^2 + w + 1

-4 = w^2 + aw + aw^2 + w

-4 = w^2(1+a) + w(a + 1)

well now things don't simplify easily

we might need to take the long route

[muted prairie]

factor a+1 out of the right

[gentle zephyr]

-4 = (1+a)(w^2+w)

[muted prairie]

oml

you're doing good i just figured out the answer is all

[gentle zephyr]

help

[muted prairie]

w^2+w+1=0

[gentle zephyr]

w^3 = 1
w^3 - 1 = 0

[gentle zephyr]

w^2+w = -1

-4 = -(1+a)
4 = 1 + a

3 = a

$\polylongdiv{x^6 + x^5 + 3x^4 + 2x^3 + 3x^2 + x + 1}{x^2+x+1}$

[soft zealot]

Renato

[gentle zephyr]

f = (x^2+x+1)(x^4+2x^2+1)

[muted prairie]

yay

[gentle zephyr]

we are back at the same issue at hand

shit is irrational

[muted prairie]

yea but this time its easier to factor

whenever we have a quartic without odd degree terms

we can factor over x^2 instead of x

[gentle zephyr]

we can depress it

[muted prairie]

subbing like y=x^2 would make it all go through the easiest

[gentle zephyr]

y^2 + 2y + 1 = 0

(y+1)^2

[gentle zephyr]

wait

does this mean I can divide x^4 + 2x^2 +1 by this

[supple ivy]

You've got the polynomial wrong here, it's -1 at the end not +1

[muted prairie]

nah the english latex is wrong

its +1

[supple ivy]

Ohhh ok nws

[gentle zephyr]

I appreciate the help

[muted prairie]

yea we were trying to do the -1 case for a while

[gentle zephyr]

wait but

can we divide by this (y+1)^2 or not

does this quartic leaves remainder 0 when we divide by (x+1)^2

surely, right?

[muted prairie]

nah

remember y=x^2

we gotta sub back

[gentle zephyr]

fuck my life

f = (x^2+x+1)(x^4+2x^2+1)
f = (x^2+x+1)(x^2+1)(x^2+1)

@muted prairie

[muted prairie]

yeah that looks right

thats the valid factorization over the reals

[gentle zephyr]

this is in R[X]

yeah because the discriminant is prolly imaginary for all of them

so in order to factorize in C[X] we need to expand all of this little shits

OK lets go one by one

(-b ± w^2)/2

where w = b^2-4ac

[muted prairie]

i think its sqrt(w)

also reusing w :monke:

[gentle zephyr]

sqrt is not well defined in C I think

like the sqrt function usually goes from R to C

[muted prairie]

well you've redefined w to make it a negative int now

so we're all good i think

[muted prairie]

also, that doesn't just mean you can square instead!

and there is a sqrt from C to C, but thats its own story

[gentle zephyr]

(-b ± w)/2

with w^2 = b^2-4ac

I just made a little tipo

[gentle zephyr]

care to elaborate

[muted prairie]

well w here isn't our root of unity now

that isn't b^2-4ac

so we're reusing the name

[gentle zephyr]

ok

b^2-4ac = -3

sqrt{-3} = 3i

[muted prairie]

ehh

= sqrt{-1} * sqrt{3}

[gentle zephyr]

sqrt{3}.i

(-1 ± sqrt{3}.i)/2

[muted prairie]

yay

its w and w^2

(the 3rd roots of unity)

[gentle zephyr]

-1/2 + (sqrt{3}.i)/2 and -1/2 - (sqrt{3}.i)/2

[gentle zephyr]

you mean if I cube this shit I get 1?

[muted prairie]

yeah

so much uglier than the 4th roots

need * for times

or something like that

instead of .

[gentle zephyr]

https://tenor.com/view/not-quite-my-tempo-whiplash-stop-it-stop-there-gif-16460604

[muted prairie]

we got it by factoring x^2+x+1 after all

which in turn was a factor of x^3-1

[gentle zephyr]

fair nuff

[muted prairie]

anyways the other 2 are luckily easier to factor

[gentle zephyr]

f = (x^2+x+1)(x^4+2x^2+1)
f = (x^2+x+1)(x^2+1)(x^2+1)

its sqrt{-1}

f = (x-(-1/2 + (sqrt{3}.i)/2))(x-(-1/2 - (sqrt{3}.i)/2))(x-i)(x+i)(x-i)(x+i)

@muted prairie

[muted prairie]

looks good

,w expand (x-(-1/2 + (sqrt{3}*i)/2))(x-(-1/2 - (sqrt{3}*i)/2))(x-i)(x+i)(x-i)(x+i)

[soft zealot]

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Requested by a_mog

[gentle zephyr]

alright okay

I appreciate the handhold

[muted prairie]

,w ((-1+sqrt(5))/4 + isqrt(2)sqrt(5+sqrt(5))/4)^5

[soft zealot]

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Upgrade to WolframAlpha Pro!

Requested by a_mog

[muted prairie]

you're welcome

it seems like you could solve a similar one on your own now so it feels productive

[gentle zephyr]

the flip is this

[muted prairie]

mfw 5th roots of unity

[gentle zephyr]

can I?

[muted prairie]

possibly

idk

but when we had to reset you did a lot more of it yourself

[gentle zephyr]

we will see how it goes

I appreciate the help

.solved

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[shy socket]

I do not get this

[gentle zephyr]

well we have this area function that is the area of the rectangle

we took the first derivative of this area function and got that A'(x) changes from positive to negative at x = sqrt{2}

do you happen to know first derivative test?

basically, you are being told that the area of the rectangle is at its maximum at x = sqrt{2}

and in the figure the height of this rectangle is sqrt{4-x^2} and width is 2x

[final saddle]

@shy socket Has your question been resolved?

[shy socket]

Yes

[gentle zephyr]

so, use it

[shy socket]

On what

Isn’t it already used

They already give it to use tho

[gentle zephyr]

the area function has a maximum at x = sqrt(2), yes or no

[shy socket]

Yes

[gentle zephyr]

and Area(x) = 2x.sqrt{4-x^2}

[shy socket]

Yeah

[gentle zephyr]

where 2x is width and sqrt{..} is height

[shy socket]

Yeah

[gentle zephyr]

then use the x = sqrt{2} where the area is at maximum to find maximum length and height

[shy socket]

So we plug it in

To the of equations

[gentle zephyr]

ye

[shy socket]

Oh

Ohhhh ok tysm

[gentle zephyr]

do u get it?

[shy socket]

Yes

,close

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[chrome ermine]

how do i appromixate the 4th root of 624

[tribal ocean]

to what precision?

Could use newton's method tho from the sound of it

[final tangle]

you could do linear approximation of a perfect 4th power around 624

[pliant shore]

$(625)^{1/4} (1 - 1/625)^{1/4}$ then there's a Taylor series for $(1 + x)^n$

[soft zealot]

south

[pliant shore]

$5\left(1+\frac{u}{4}-\frac{3u^{2}}{32}+\frac{7u^{3}}{128}\right)$ where $u = -\frac{1}{625}$ is already accurate to $11$ decimal places

[soft zealot]

south

[final saddle]

@chrome ermine Has your question been resolved?

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[shy socket]

How do i solve this

[tribal ocean]

If you think about moving the point (x,y) to the right and rotating the line while keeping the origin fixed

the line will decrease in slope and at some point itll have to touch the graph at exactly one point right?

before going back up and passing through two points again

That means the line's slope is minimized when it is tangent to the graph

[shy socket]

Yes

[tribal ocean]

So the question is asking you to find a point (x,y) for x in [1,8] such that the tangent line through (x,y) on the graph passes through the origin

[shy socket]

So whatever we have rn

[tribal ocean]

what is the tangent line for any (x,y) for your graph?

[shy socket]

the derivative

[tribal ocean]

the slope of it is

what else do you know what the tangent line

[shy socket]

horizontal or vertival

[tribal ocean]

w/ respect to the graph and the point its tangent to

[shy socket]

vertical

[tribal ocean]

basically what im asking is how do you write the equation of the tangent

for a function f

at a point a

1. The tangent line passes through (a, f(a))
2. It has a slope of f'(a)

[shy socket]

like the numerator = 0

the inverse of it or something

im not too sure

[tribal ocean]

Can you use those to come up with the tangent line at a?

[shy socket]

2x-1

x = 1/2

tangent?

[tribal ocean]

Do you know the definition of a tangent line

[shy socket]

Yes

Like a line that intersects a poit of a circle

[tribal ocean]

and do you know how to construct a line if you know a point on it and its slope

[shy socket]

Umm

Like you write out the circle equation

and then u set it = 0 or something?

[tribal ocean]

No need for circles

just talking about lines

[shy socket]

Then i'm nnot too sure

Tangent of a line?

How

[tribal ocean]

Dont worry about tangent

do you know how to write the equation of a line if you know a point on it

[shy socket]

Yes

[tribal ocean]

and its slope m

[shy socket]

point slope

[tribal ocean]

exactly

[fiery bluff]

side note: ||[there's a solution without calculus if you are fine with AM-GM inequality]||

[shy socket]

I have to do it the calculus way because my teacher will not accept my work otherwise

[tribal ocean]

a "tangent" line is a special line for a point x=a on the graph such that

1. The tangent line passes through (a, f(a))
2. It has a slope of f'(a)

[shy socket]

Ok

[tribal ocean]

So what is the equation of that line for x=a?

[tribal ocean]

Given this

[shy socket]

1

[tribal ocean]

We can use point slope form

y - y_0 = m(x - x_0)

[shy socket]

y-y=1(x-x1)

[tribal ocean]

But we want it to be for any a

[shy socket]

what is a?

[tribal ocean]

a specific value of x

for our purpose any value in [1,8]

[shy socket]

Ok

[tribal ocean]

its a variable so we can solve for it later

we dont know which one we need yet

[shy socket]

Ok

So why do we need the tangent line

[tribal ocean]

so whats the line that passes through (a, f(a)) with a slope of f'(a)?

[tribal ocean]

^^^^

[shy socket]

y-f(a) = f'(a)(x-a)

[tribal ocean]

yes, thats exactly right

[shy socket]

Ok

[tribal ocean]

you can go back up to read why we got here if you forgot

or ask more questions

[shy socket]

isn't it 0 when it's tangent

[tribal ocean]

what is

[shy socket]

the slope

[tribal ocean]

Nope, you can be tangent to a curve at any angle

[shy socket]

Ok

So the tangent is the smallest slope

[shy socket]

Now we find the tangent

x value

[tribal ocean]

The tangent line of f will give you the smallest slope in the constraints of this problem

not in general

I tried to give a visual explanation of dragging the line

until it reaches a minimum slope

[shy socket]

Ok

Wait so how do we find the answer

[tribal ocean]

Like if you move (x,y) to the right you can see the line has to become less steep

then at some point it becomes tangent and starts increasing in slope again

[tribal ocean]

I'm sorry if I'm unable to give a satisfying answer as this is more of a visual/geometric reason for why its a tangent line that minimizes slope

y - f(a) = f'(a)(x-a) can be simplified to

[shy socket]

it's ok thank you for helping

HOw do we find where it passes through x tho

[tribal ocean]

y - a^2 + a - 16 = (2a-1)(x-a)

[shy socket]

From the looks of it it looks like inbetween (x,y) and the equation

[tribal ocean]

Out of all tangent lines which one passes through the origin?

[shy socket]

A

[tribal ocean]

I mean these ones

[shy socket]

tbh idk i feel like my teacher didn't even teach this

[tribal ocean]

(0,0) needs to be on the line

Yeah this is more of an applied problem

Problem solving

[shy socket]

Ok

[tribal ocean]

We know the tangent at x=a is
y - a^2 + a - 16 = (2a-1)(x-a)

[shy socket]

Is this the tangent equation

[shy socket]

Ok

[tribal ocean]

yeah after i plugged in f and f'

[shy socket]

Ok

[tribal ocean]

we want one that passes through (0,0)

[shy socket]

Now what do we do

Plug in 0 for x and y into that

[tribal ocean]

yup

[shy socket]

0-a^2+a-16=(2a-1)(0-a)

[tribal ocean]

yes

[shy socket]

Is that the tangent

[tribal ocean]

-a^2+a-16=-2a^2+a

[shy socket]

Yes

[tribal ocean]

keep going

[shy socket]

So we know that is the tangent line

[tribal ocean]

keep going

[shy socket]

We wanna know where it croses the other line

[tribal ocean]

until you get a

[shy socket]

parabola

Oh

A is just any value in the interval

[tribal ocean]

we are solving for which a the tangent line passes through (0,0)

[shy socket]

i got a = +- 4

[tribal ocean]

cool

-4 is out of the range

as we wanted x in [1,8]

[shy socket]

Yes

[tribal ocean]

So your answer is 4

[rocky tusk]

you can also do this by setting
cx = x^2 - x + 16 then solving for the minimum such value of c using the discriminant and of course when the discriminant is zero there is one solution so the line is tangent

[tribal ocean]

thats actually smart ngl

[shy socket]

Idk if my teacher will allow me to solve it like that

[cobalt dawn]

line of min slope is tangential to the curvbe

[shy socket]

Bro wants us to use like some optimization type of solving

[rocky tusk]

i mean this is the way to do it without just intuitively knowing it will occur when it’s tangent

[cobalt dawn]

dont do that

[rocky tusk]

by doing what he said and rotating the lines in your head

[shy socket]

I actually do not know about this lesson bruh

[cobalt dawn]

Your tangent must be such that c=0.. so y=mx...

[shy socket]

Why is everyone question so different

[rocky tusk]

you get (c + 1)^2 - 64 >= 0 with equality if c = 7 or c = -9 but of course the slope is positive so c =
7 and hence x^2 - 8x + 16 = 0 so (x - 4)^2 = 0

x = 4

[tribal ocean]

They want you to solve different types of problems with your toolset

[tranquil pine]

everyone is using this website

[tribal ocean]

it will require some thinking theres no way around that

[rocky tusk]

it’s ap classroom

[tranquil pine]

oh

[shy socket]

Bruh all this guy told me is write two equation and then replace the 2 variables into 1

Like where do I get the equations from when did we do any of this

[rocky tusk]

@tribal ocean weren't you a helpful in this server before on another account?

your pfp looks familiar

[tribal ocean]

Nope

Its just a very common pfp lol

[rocky tusk]

#help-49 message

oh his was similar

not exactly the same though

[tribal ocean]

This is combining a couple different things you're expected to know plus applying it to something kinda new

If you don't know it or weren't taught it you should do simpler questions from those topics first

[final saddle]

@shy socket Has your question been resolved?

[shy socket]

How do i do this

[topaz raven]

first find the critical points

[shy socket]

I got like

3/4 + - sqrt 150 over 20

something like that

[topaz raven]

did u find the crtical points?

[shy socket]

Yeah

[shy socket]

Those

[topaz raven]

erm... how did u get those

,w critical points of 1/10 (-2x^3 +9x^2 -12x) +7

[topaz raven]

do u know what critical points are?

[soft zealot]

Results provided by WolframAlpha

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

Requested by valentine.94

[tribal ocean]

The critical points are where $f'(m)=0$ and the boundary points $m=0,3$

[soft zealot]

flynger

[tribal ocean]

$f'(m)=-0.6m^2+1.8m-1.2$

[soft zealot]

flynger

[tribal ocean]

$m^2-3m+2=0$

[soft zealot]

flynger

[tribal ocean]

$m=1,2$

[soft zealot]

flynger

[tribal ocean]

Now check all $m=0,1,2,3$

[soft zealot]

flynger

[final saddle]

@shy socket Has your question been resolved?

[final saddle]

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[delicate sinew]

i have gotten the (i)
however i am unsure of the qns (ii)

[green pebble]

pi r^2

r is x in terms of y

then integrate

[green pebble]

,rccw

[delicate sinew]

i dont get which part im trying to find

for part(ii)

[green pebble]

wdym

[lilac moat]

@delicate sinew you don't square the distance between the two curves

[delicate sinew]

this is the part (i)

[lilac moat]

the integral setup should be $\pi\left(\int_{1}^{4}\left(\sqrt[3]{y}^{2}-1^{2}\right)dy+\int_{0}^{1}\left(1^{2}-\sqrt[3]{y}^{2}\right)dy\right)$

[delicate sinew]

for part (ii)

[soft zealot]

Roy

[lilac moat]

so you're subtracting the volumes off, not the distances

[delicate sinew]

yes i understand

i saw that mistake

however what is part(ii) trying to ask me to find

[lilac moat]

It's revolving this shaded area around the y-axis

Pretend it's a 2d region, then just rotate that region around the y-axis to make a 3d solid

[delicate sinew]

so it is this right?

[lilac moat]

well yeah in 3d

but I think you get it

[delicate sinew]

answer will be same as this

bacause even if you rotate

[lilac moat]

no?

[delicate sinew]

the answer you are finding is the same

take a look at this

isn't them producing the same answer

[lilac moat]

these two solids are different (excuse the atrocious drawing...)

[delicate sinew]

,rccw

[lilac moat]

make sure your integral setup for the lower solid of revolution is right - you've missed out $+\pi\int_{0}^{1}\left(1^{2}\right)dy$

[soft zealot]

Roy

[lilac moat]

because we're taking the volume "under" ³√(y) and subtracting it from the volume "under" x = 1

[delicate sinew]

i do not understand this

but im very close to giving this up

[lilac moat]

@delicate sinew is this visual understandable?

[final saddle]

@delicate sinew Has your question been resolved?

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[timber plume]

how does diving by 3! remove the extra factorials?

[candid pulsar]

wdym extra factorials

[abstract bramble]

youre basically dividing by the number of permutations that are considered the same combination

if i want pairs of a, b, c
well, there are 3 * 2 = 6 PERMUTATIONS
a,b
a,c
b,a
b,c
c,a
c,b
but i want pairs, i dont care about the order
there are 2 * 1 = 2 ways to order each pair
so the answer is permutations/orderings = (3 * 2) / (2 * 1) = 3 ways to pair
a-b
b-c
a-c

[tired walrus]

@timber plume do you understand the meaning of factorial in combinatorics

also are you actually still here and in need of help

[split surge]

I didn't understood his problem with that question

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@timber plume Has your question been resolved?

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[runic sparrow]

I would like help understanding what this question asks of me, and how i should tackle such questions. I get stuck pretty much immediately, as i am not sure what they expect as a result.

[versed crater]

Suppose mu and sigma were known, how would you do this?

[runic sparrow]

Question b gives me a sigma and mu, which i (may have, we do not have the answers yet) solved in this manner:

mu = 56.3, sigma = 7.6

the normal distribution is this N(56.3, (7.6)²)

P(X > 68.5) = P(Z > 68.5-56.3 / 7.6) = P(Z< 1.61) = 1-phi(1.61) = 1-0.9463 = 0.0537 which is roughly equal to 5.4%

[versed crater]

Sigma squared!!

N(mu, sigma²)

[runic sparrow]

Oh yeah i did write that down, didn't type it out properly

[versed crater]

Okay

So just replace every instance of 56.3 with mu and 7.6 with sigma

[runic sparrow]

That makes sense.

Is there any more to it?

.close

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[versed crater]

No that’s literally it haha

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[sharp wraith]

i think ive shown that a has to be = 2, but not sure about the n is prime part

[plucky rover]

Are you familiar with the factorization of x^n - a^n

[sharp wraith]

x = a seems to be a root so (x-a) is a factor? then i could somewhat generalise the rest with polynomial division or smth

[plucky rover]

Wasn't a question for you lol

Was a question for the OP

[lilac moat]

Isn't he the op

[plucky rover]

Oh they changed username

Oof

Thanks

[plucky rover]

Good

Now what happens if n is not prime

[sharp wraith]

im assuming smth happens but i dont rly see it

is it to do with the second factor, i.e. not the (x-a) part

[plucky rover]

Well let's put it this way

If you have x^n - a^n

And n = pq

You can write it as x^pq - a^pq

Do you see what you can do now

(think substitution)

Also tag me when replying so I can get back to you faster

[sharp wraith]

ig u can maybe made a sub like X = x^p and A = a^p, giving us X^q - A^q

which gives us another thing in the form x^n -a ^n @plucky rover

[plucky rover]

Yes

And then this is divisible by x^p - a^p

Now do you see how to use this

[sharp wraith]

is it that if we let a^n = 1 from the start, and repeating the substitution process, we will certainly get (x^p-1)( ... ) which means that there is a non one factor, implying that it will never be prime, hence, n has to be prime

@plucky rover

like assume, for contradiction, n is not prime, then 2^pq - 1 (since we've established a = 2), then letting u = 2^p, we get u^q - 1 which has a factor of (u-1), which means 2^pq _ 1 isn't prime, therefore, n has to be prime

[plucky rover]

Yup

[sharp wraith]

tysm you have been super helpful

.close

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[cedar obsidian]

What fundamental concept i should read from text to solve this question correctly?

[cedar obsidian]

i got that vm

nvm

.close

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[cedar obsidian]

.close

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[severe verge]

\begin{flushleft}
Given the following functions $f(x) = x^3 (x^2-1)^4 (x-2)^5$ and $g(x) = ln |f(x)|$. Determine whether these statements are true or false: \
A. The equation $2025f(x) = f'(x)$ has 1 real solution \
B. The equation $2025f(x) = f'(x)$ has 2 real solutions \
C. The equation $2025f(x) = f'(x)$ has 3 real solutions \
D. The equation $2025f(x) = f'(x)$ has 4 real solutions \
E. $g'(x) = \frac{f'(x)}{f(x)}$ $\forall x \notin {0; -1; 1; 2}$ \
F. $g''(x) < 0$ $\forall x \notin {0; -1; 1; 2}$ \
\end{flushleft}

[soft zealot]

Carbonara

[severe verge]

this appeared in my mid-term exam and it is the only problem i think i got wrong

for A, B, C and D i got (f(x)*e^2025)' = 0 and got stuck

E and F seems very dodgy

alr E is correct

[halcyon ether]

x=0,1,-1 and 2 is triples, double and 5th roots of f(x) which should mean they're also roots of f'(x) right

[severe verge]

hmm

oh wait yeah

god that's an awful blunder

[halcyon ether]

It should imply f(x)=f'(x) have 4 roots i think

[severe verge]

there might be more

at least 4 roots

[halcyon ether]

Yeah at least 4 lol

I'm thinking 2025 = f'(x)/f(x) and integrate both side to get 2025x+c=g(x) but I'm not sure if it leads to anywhere

[severe verge]

hmm

wait a sec

I can draw a rough graph of g(x)

and then let 2025x + c intersect it

does that help?

[halcyon ether]

I'm not sure tho

[severe verge]

got another idea

e^(2025x+c) = |f(x)|

now e^(2025x+c) tends to 0 when x goes to negative infinity and equals to some positive value when x = 0

and |f(x)| goes from +inf to 0 when x goes from -inf to -1

so we have at least 1 more root between -inf and -1

so there are at least 5 roots

[severe verge]

wait what this is wrong

nvm i was right

so A, B, C and D are all wrong

[halcyon ether]

Lmao I was questioning myself in desmos

[severe verge]

yeah the growth of e^2025x is way too slow

[halcyon ether]

I was wonder why it's wrong

Yeah

[severe verge]

all my friends told me there are 4 roots

clearly that isnt the case

so only F left

[final saddle]

@severe verge Has your question been resolved?

[severe verge]

<@&286206848099549185>

[final saddle]

@severe verge Has your question been resolved?

[final saddle]

@severe verge Has your question been resolved?

[loud sundial]

can't you compute g'' directly?

[final saddle]

@severe verge Has your question been resolved?

[severe verge]

cam you elaborate on that?

like, this is not exactly beautiful

[loud sundial]

$g(x)=3\ln|x|+2\ln|x-1|+2\ln|x+1|+5\ln|x-2|$

[soft zealot]

Civil Service Pigeon

[loud sundial]

$g'(x)=\frac{3}{x}+\frac{2}{x-1}+\frac{2}{x+1}+\frac{5}{x-2}$

[soft zealot]

Civil Service Pigeon

[loud sundial]

$g'(x)=-\frac{3}{x^2}-\frac{2}{(x-1)^2}-\frac{2}{(x+1)^2}-\frac{5}{(x-2)^2}$

[soft zealot]

Civil Service Pigeon

[loud sundial]

now consider the sign of each term

oh I made a goof and put 2 instead of 4

the same idea still holds

[severe verge]

i am stupid

[loud sundial]

❌

[severe verge]

i wrote multiply instead of plus

great

my brain at midnight never works

oh well thanks for clearing up my awful blunders

.close

[final saddle]

Channel closed

Closed by @severe verge

Use .reopen if this was a mistake.

[bright mica]

[final saddle]

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[craggy plank]

What’s the question?

[granite tapir]

The question is to show <pr_a,pr_b> is the identity arrow of a×b, I think

[plucky rover]

Are you familiar with the universal property of products

[granite tapir]

Not yet

[final saddle]

@granite tapir Has your question been resolved?

[plucky rover]

Well what exactly are the tools available to you

[final saddle]

@granite tapir Has your question been resolved?

[granite tapir]

For example I have nice textboks, I foudn the property you mentioned but I don't yet understand it

I am also curious about constructive logic, so I am reading about it at the moment, and non-classical logics, and trying to figure out an explanation of inductive proof

[granite tapir]

from Conceptual Mathematics - A first introduction to categories, by Lawvere / Schanuel

[granite tapir]

maybe it goes like this:
for any arrow
a⟶d,
b⟶d,
there is a unique arrow c⟶d,

so for
a⟶a×b
b⟶b×a
there is
c⟶a×b unique
a×b ⟶ a×b is identity arrow,
so since c⟶a×b is unique and c=a×b,
it is the identity arrow

What I'm confused about is an arrow a to a is not only the identity arrow?
like why prove a × b ⟶ a × b is the identity if it's an arrow that goes from A ⟶ A?

Ah! I know one example, in ℕ succ : ℕ → ℕ, n ↦ n+1, yet it's not the ientity!

[granite tapir]

Merci @plucky rover 🌸

[plucky rover]

Okay I'm gonna keep it real

Why are you doing category theory at this stage

Because it looks like you have very little understanding of algebra, topology or even set theory in the first place

[final saddle]

@granite tapir Has your question been resolved?

[granite tapir]

because categories show structure and motivation and I enjoy learning category theory

I often think "why is this not shown first?", when learning category patterns, since it's underlying the entire thing of mathematics

[granite tapir]

like you learn something in one context and you learn it for every other contexts

[plucky rover]

That structure and motivation means little when you do not have the underlying foundations to appreciate it

Like I agree that category theory is fun

[granite tapir]

categories is the underlying foundations

[plucky rover]

But I'd suggest doing some set theory and abstract algebra first

[granite tapir]

and logic

it's just maybe not very accessible yet, because most texts are technical

[craggy cedar]

do you know some discrete math?

[plucky rover]

Accessible in general? Or to you specifically?

Because you have avenues to fix the latter

[granite tapir]

I'm learning set theory and abstract algebra also, another thing is I percieve mathematics like interconected graph, I don't know whats the name, like ever node connected to every node, so start anywhere you like is good approach

[granite tapir]

yes

[plucky rover]

While I agree that maths is an interconnected graph, you can't exactly jump in to things without properly doing the prerequisites

[granite tapir]

I can

[plucky rover]

Well knock yourself out then

[granite tapir]

I have been studying like this for a few years already, and there are also books about this sort of approach, it's a valid way of learning, it works for me and I am having a lot of fun and learning a lot

as you see I could complete the proof from before, so

and learnt a lot while doing so, and the small categories insight I gained applies to all of mathematics contexts

.close

[final saddle]

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[warm python]

We first examine S

$\langle SS^{-1}v,v \rangle = \langle Sv, S^*S^{-1}v. \rangle = \langle Sv, v \rangle$

[soft zealot]

wai

[warm python]

so $\langle v,v \rangle =\langle Sv,v \rangle$?

[soft zealot]

wai

[warm python]

which would mean S=I which makes no sense

[desert mantle]

I have no clue how you are doing that first step

[warm python]

okay,I see the issue now

[warm python]

yup, messed up the defn of adjoint

$\langle SS^{-1}v,v \rangle = \langle S^{-1}v, S^*S^{-1}v. \rangle = \langle S^{-1}v, v \rangle$

[soft zealot]

wai

[warm python]

this is better I think

[desert mantle]

no

[warm python]

what's wrong

[scarlet sequoia]

what operator are you applying "adjoint" definition on?

[warm python]

S

[scarlet sequoia]

so

why do we end up with S*S^{-1}

on the right part of the scalar product

[warm python]

ah

got it

[desert mantle]

you really need to learn to be more precise and notice these kinds of errors yourself

[warm python]

tq

.close

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[dense garnet]

I know that if $U=2B-1$ where $B$ is Bernoulli, then $U|X|$ has the same distribution as $X$ if $X$ is symmetric. Now the claim is that $UX$ has a normal distribution $N(0,1)$ if $X$ has one. How is this in agreement with the fact that I first mentioned?

[soft zealot]

psie

[tired walrus]

U is a coin flip with values 1 and -1

[dense garnet]

Yep

[tired walrus]

and also meant to be indep of X, yes?

[dense garnet]

Ah, yes.

Thanks.

[tired walrus]

where exactly do you see the disagreement btw

[dense garnet]

Well, X is not nonnegative, so U|X| != UX, right?

[tired walrus]

do you wanna go bureaucratic on this

Assumption 1: U is a coin flip (returns values ±1 with equal probability).
Assumption 2: U is independent of X.

Claim 1: If X is symmetric, then U|X| has the same distribution as X.
Claim 2: If X ~ N(0,1), then UX (without modulus) ~ N(0,1) also.

i don't see any contradiction or disagreement between claims 1 and 2

[dense garnet]

Well, how can you reach the conclusion that UX ~ N(0,1) also if X has no modulus. This doesn't seem in agreement with claim 1, doesn't it?

[tired walrus]

it neither agrees nor disagrees with claim 1

it's just logically independent of it

[dense garnet]

Aha, ok. So a totally different claim then. 🥲

.close

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[native sluice]

The first question’s answer is c/2, the second question’s answer is a+b-c, I am stuck at the second part, I have used tangent properties TQ = radius of circle 2, but I still can’t get the ideal result

[final saddle]

@native sluice Has your question been resolved?

[final saddle]

@native sluice Has your question been resolved?

[native sluice]

<@&286206848099549185> Please help

[shell condor]

I'm here now, give me some time

[native sluice]

Okk, thank you very much

[shell condor]

Okay so let's see what we have, we know that BTA is 90 degrees, because it's in the semicircle

TPO2 and TQO2 are 90 degrees, because radius from point of tangent to centre of circle is a right angle

Also PO2 and QO2 are equal (radius), so PTQO2 is a square

[native sluice]

It’s a square but in a weird position